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|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
1706.06978 | 50 | [11] F. Maxwell Harper and Joseph A. Konstan. 2015. The MovieLens Datasets: History and Context. ACM Transactions on Interactive Intelligent Systems 5, 4 (2015). [12] Kaiming He, Xiangyu Zhang, Shaoqing Ren, and Jian Sun. 2015. Delving deep into rectifiers: Surpassing human-level performance on imagenet classification. In Proceedings of the IEEE International Conference on Computer Vision. 1026â1034. [13] Ruining He and Julian McAuley. 2016. Ups and Downs: Modeling the Vi- sual Evolution of Fashion Trends with One-Class Collaborative Filtering. In Proceedings of the 25th International Conference on World Wide Web. 507â517. https://doi.org/10.1145/2872427.2883037
[14] Gao Huang, Zhuang Liu, Laurens van der Maaten, and Kilian Q. Weinberger. Densely connected convolutional networks. | 1706.06978#50 | Deep Interest Network for Click-Through Rate Prediction | Click-through rate prediction is an essential task in industrial
applications, such as online advertising. Recently deep learning based models
have been proposed, which follow a similar Embedding\&MLP paradigm. In these
methods large scale sparse input features are first mapped into low dimensional
embedding vectors, and then transformed into fixed-length vectors in a
group-wise manner, finally concatenated together to fed into a multilayer
perceptron (MLP) to learn the nonlinear relations among features. In this way,
user features are compressed into a fixed-length representation vector, in
regardless of what candidate ads are. The use of fixed-length vector will be a
bottleneck, which brings difficulty for Embedding\&MLP methods to capture
user's diverse interests effectively from rich historical behaviors. In this
paper, we propose a novel model: Deep Interest Network (DIN) which tackles this
challenge by designing a local activation unit to adaptively learn the
representation of user interests from historical behaviors with respect to a
certain ad. This representation vector varies over different ads, improving the
expressive ability of model greatly. Besides, we develop two techniques:
mini-batch aware regularization and data adaptive activation function which can
help training industrial deep networks with hundreds of millions of parameters.
Experiments on two public datasets as well as an Alibaba real production
dataset with over 2 billion samples demonstrate the effectiveness of proposed
approaches, which achieve superior performance compared with state-of-the-art
methods. DIN now has been successfully deployed in the online display
advertising system in Alibaba, serving the main traffic. | http://arxiv.org/pdf/1706.06978 | Guorui Zhou, Chengru Song, Xiaoqiang Zhu, Ying Fan, Han Zhu, Xiao Ma, Yanghui Yan, Junqi Jin, Han Li, Kun Gai | stat.ML, cs.LG, I.2.6; H.3.2 | Accepted by KDD 2018 | null | stat.ML | 20170621 | 20180913 | [
{
"id": "1704.05194"
}
] |
1706.06905 | 51 | [49] S. Hershey, S. Chaudhuri, D. P. W. Ellis, J. F. Gemmeke, A. Jansen, C. Moore, M. Plakal, D. Platt, R. A. Saurous, B. Seybold, M. Slaney, R. Weiss, and K. Wilson, âCNN architectures for large-scale audio classiï¬cation,â in International Conference on Acoustics, Speech and Signal Processing (ICASSP), 2017.
[50] D. P. Kingma and J. Ba, âAdam: A method for stochastic optimization,â in ICLR, 2015.
[51] S. Ioffe and C. Szegedy, âBatch normalization: Accelerating deep network training by reducing internal covariate,â arXiv preprint arXiv:1502.03167, 2015.
[52] Y. Gao, O. Beijbom, N. Zhang, and T. Darrell, âCompact bilinear pooling,â in CVPR, 2016.
[53] H.-D. Wang, T. Zhang, and J. Wu, âThe monkeytyping solution to the youtube-8m video understanding challenge,â arXiv preprint arXiv:1706.05150, 2017. | 1706.06905#51 | Learnable pooling with Context Gating for video classification | Current methods for video analysis often extract frame-level features using
pre-trained convolutional neural networks (CNNs). Such features are then
aggregated over time e.g., by simple temporal averaging or more sophisticated
recurrent neural networks such as long short-term memory (LSTM) or gated
recurrent units (GRU). In this work we revise existing video representations
and study alternative methods for temporal aggregation. We first explore
clustering-based aggregation layers and propose a two-stream architecture
aggregating audio and visual features. We then introduce a learnable non-linear
unit, named Context Gating, aiming to model interdependencies among network
activations. Our experimental results show the advantage of both improvements
for the task of video classification. In particular, we evaluate our method on
the large-scale multi-modal Youtube-8M v2 dataset and outperform all other
methods in the Youtube 8M Large-Scale Video Understanding challenge. | http://arxiv.org/pdf/1706.06905 | Antoine Miech, Ivan Laptev, Josef Sivic | cs.CV | Presented at Youtube 8M CVPR17 Workshop. Kaggle Winning model. Under
review for TPAMI | null | cs.CV | 20170621 | 20180305 | [
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1706.06927 | 51 | 2 4 6 8 2 4 6 8 2 4 6 8 #c 6 2 2 2 4 4 6 8 8 8 10 12 4 2 8 10 4 14 10 14 L 54 101 121 150 65 89 130 141 46 80 120 158 MO 74 123 161 52 114 178 220 E 1.3k 3.9k 3.9k 4.5k 6.2k 9.6k 3.1k 5.9k 1.1k 2.3 3.5k 3.4k - 1.6k 2.6k 3.5k 1.3k 55.5k 166k 201k Prep 8.1 8.1 7.18 8.26 19.19 17.9 17.66 18.42 23.74 24.44 27.04 23.74 - 30.37 30.09 32.22 45.64 45.65 47 46.46 Search 0.23 0.8 0.6 0.91 1.32 2.29 0.73 1.26 0.23 0.54 0.91 0.69 - 0.4 0.64 0.86 0.33 13.12 41.36 55.57 Total 8.33 8.9 7.78 9.17 20.51 20.19 18.39 19.68 23.97 24.98 27.95 24.43 - 30.77 30.73 33.08 45.97 58.77 | 1706.06927#51 | Combined Task and Motion Planning as Classical AI Planning | Planning in robotics is often split into task and motion planning. The
high-level, symbolic task planner decides what needs to be done, while the
motion planner checks feasibility and fills up geometric detail. It is known
however that such a decomposition is not effective in general as the symbolic
and geometrical components are not independent. In this work, we show that it
is possible to compile task and motion planning problems into classical AI
planning problems; i.e., planning problems over finite and discrete state
spaces with a known initial state, deterministic actions, and goal states to be
reached. The compilation is sound, meaning that classical plans are valid robot
plans, and probabilistically complete, meaning that valid robot plans are
classical plans when a sufficient number of configurations is sampled. In this
approach, motion planners and collision checkers are used for the compilation,
but not at planning time. The key elements that make the approach effective are
1) expressive classical AI planning languages for representing the compiled
problems in compact form, that unlike PDDL make use of functions and state
constraints, and 2) general width-based search algorithms capable of finding
plans over huge combinatorial spaces using weak heuristics only. Empirical
results are presented for a PR2 robot manipulating tens of objects, for which
long plans are required. | http://arxiv.org/pdf/1706.06927 | Jonathan Ferrer-Mestres, Guillem Francès, Hector Geffner | cs.RO, cs.AI | 10 pages, 2 figures | null | cs.RO | 20170621 | 20170621 | [] |
1706.06978 | 51 | [14] Gao Huang, Zhuang Liu, Laurens van der Maaten, and Kilian Q. Weinberger. Densely connected convolutional networks.
[15] Diederik Kingma and Jimmy Ba. 2015. Adam: A Method for Stochastic Optimiza- tion. In Proceedings of the 3rd International Conference on Learning Representations. [16] Mu Li, Ziqi Liu, Alexander J Smola, and Yu-Xiang Wang. 2016. DiFacto: Dis- tributed factorization machines. In Proceedings of the 9th ACM International Conference on Web Search and Data Mining. 377â386.
[17] Laurens van der Maaten and Geoffrey Hinton. 2008. Visualizing data using t-SNE. Journal of Machine Learning Research 9, Nov (2008), 2579â2605.
[18] Julian Mcauley, Christopher Targett, Qinfeng Shi, and Van Den Hengel Anton. Image-Based Recommendations on Styles and Substitutes. In Proceedings of the 38th International ACM SIGIR Conference on Research and Development in Infor- mation Retrieval. 43â52.
[19] H. Brendan Mcmahan, H. Brendan Holt, et al. 2014. Ad Click Prediction: a View from the Trenches. In Proceedings of the 19th ACM SIGKDD International Conference on Knowledge Discovery and Data Mining. 1222â1230. | 1706.06978#51 | Deep Interest Network for Click-Through Rate Prediction | Click-through rate prediction is an essential task in industrial
applications, such as online advertising. Recently deep learning based models
have been proposed, which follow a similar Embedding\&MLP paradigm. In these
methods large scale sparse input features are first mapped into low dimensional
embedding vectors, and then transformed into fixed-length vectors in a
group-wise manner, finally concatenated together to fed into a multilayer
perceptron (MLP) to learn the nonlinear relations among features. In this way,
user features are compressed into a fixed-length representation vector, in
regardless of what candidate ads are. The use of fixed-length vector will be a
bottleneck, which brings difficulty for Embedding\&MLP methods to capture
user's diverse interests effectively from rich historical behaviors. In this
paper, we propose a novel model: Deep Interest Network (DIN) which tackles this
challenge by designing a local activation unit to adaptively learn the
representation of user interests from historical behaviors with respect to a
certain ad. This representation vector varies over different ads, improving the
expressive ability of model greatly. Besides, we develop two techniques:
mini-batch aware regularization and data adaptive activation function which can
help training industrial deep networks with hundreds of millions of parameters.
Experiments on two public datasets as well as an Alibaba real production
dataset with over 2 billion samples demonstrate the effectiveness of proposed
approaches, which achieve superior performance compared with state-of-the-art
methods. DIN now has been successfully deployed in the online display
advertising system in Alibaba, serving the main traffic. | http://arxiv.org/pdf/1706.06978 | Guorui Zhou, Chengru Song, Xiaoqiang Zhu, Ying Fan, Han Zhu, Xiao Ma, Yanghui Yan, Junqi Jin, Han Li, Kun Gai | stat.ML, cs.LG, I.2.6; H.3.2 | Accepted by KDD 2018 | null | stat.ML | 20170621 | 20180913 | [
{
"id": "1704.05194"
}
] |
1706.06708 | 52 | For each i in {1,...,n}\ O, the number of index-i row moves is odd by the parity constraint. Furthermore, by the definition of O, this number is not one. Thus each i in {1,...,n}\O contributes at least three moves. Therefore the number of type-2 moves is at least 3(|{1,...,n}\O]) = 3(nâ|O]). Consider the moves of rows i with i ⬠O. Since the [;s are all distinct, there must be at least one column move between every consecutive pair of such moves. Thus the total number of type-3 moves (column moves) is at least |O| â 1. Furthermore, the number of type-3 moves is |O| â 1 if and only if the consecutive pairs of row 1 ⬠O moves have exactly one column move between them. Such a pair of is has exactly one column move between the two row-i moves only if the corresponding pair of J;s is at Hamming distance one. Therefore, if we consider the l;s for i ⬠O in the order in which row-i moves occur in m1,..., mMâ, then the number of type-3 moves is exactly |O| â 1 if and only if those l;s in that order have each |; at Hamming distance exactly one from the next (and more otherwise). | 1706.06708#52 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06905 | 52 | [54] F. Li, C. Gan, X. Liu, Y. Bian, X. Long, Y. Li, Z. Li, J. Zhou, and S. Wen, âTemporal modeling approaches for large-scale youtube-8m video understanding,â arXiv preprint arXiv:1707.04555, 2017.
[55] S. Chen, X. Wang, Y. Tang, X. Chen, Z. Wu, and Y.-G. Jiang, âAg- gregating frame-level features for large-scale video classiï¬cation,â arXiv preprint arXiv:1707.00803, 2017.
[56] M. Skalic, M. Pekalski, and X. E. Pan, âDeep learning methods for efï¬cient large scale video labeling,â arXiv preprint arXiv:1706.04572, 2017.
[57] A. Miech, âLOUPE tensorï¬ow toolbox for learnable pooling module,â https://github.com/antoine77340/LOUPE, 2017.
8 | 1706.06905#52 | Learnable pooling with Context Gating for video classification | Current methods for video analysis often extract frame-level features using
pre-trained convolutional neural networks (CNNs). Such features are then
aggregated over time e.g., by simple temporal averaging or more sophisticated
recurrent neural networks such as long short-term memory (LSTM) or gated
recurrent units (GRU). In this work we revise existing video representations
and study alternative methods for temporal aggregation. We first explore
clustering-based aggregation layers and propose a two-stream architecture
aggregating audio and visual features. We then introduce a learnable non-linear
unit, named Context Gating, aiming to model interdependencies among network
activations. Our experimental results show the advantage of both improvements
for the task of video classification. In particular, we evaluate our method on
the large-scale multi-modal Youtube-8M v2 dataset and outperform all other
methods in the Youtube 8M Large-Scale Video Understanding challenge. | http://arxiv.org/pdf/1706.06905 | Antoine Miech, Ivan Laptev, Josef Sivic | cs.CV | Presented at Youtube 8M CVPR17 Workshop. Kaggle Winning model. Under
review for TPAMI | null | cs.CV | 20170621 | 20180305 | [
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"id": "1602.07261"
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"id": "1706.05150"
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1706.06978 | 52 | [20] Steffen Rendle. 2010. Factorization machines. In Proceedings of the 10th Interna- tional Conference on Data Mining. IEEE, 995â1000.
[21] Ying Shan, T Ryan Hoens, Jian Jiao, Haijing Wang, Dong Yu, and JC Mao. Deep Crossing: Web-scale modeling without manually crafted combinatorial features. [22] Nitish Srivastava, Geoffrey E Hinton, Alex Krizhevsky, Ilya Sutskever, and Ruslan Salakhutdinov. 2014. Dropout: a simple way to prevent neural networks from overfitting. Journal of Machine Learning Research 15, 1 (2014), 1929â1958. [23] Andreas Veit, Balazs Kovacs, et al. 2015. Learning Visual Clothing Style With Heterogeneous Dyadic Co-Occurrences. In Proceedings of the IEEE International Conference on Computer Vision.
[24] Ronald J Williams and David Zipser. 1989. A learning algorithm for continually running fully recurrent neural networks. Neural computation (1989), 270â280.
[25] Ling Yan, Wu-jun Li, Gui-Rong Xue, and Dingyi Han. 2014. Coupled group lasso for web-scale ctr prediction in display advertising. In Proceedings of the 31th International Conference on Machine Learning. 802â810. | 1706.06978#52 | Deep Interest Network for Click-Through Rate Prediction | Click-through rate prediction is an essential task in industrial
applications, such as online advertising. Recently deep learning based models
have been proposed, which follow a similar Embedding\&MLP paradigm. In these
methods large scale sparse input features are first mapped into low dimensional
embedding vectors, and then transformed into fixed-length vectors in a
group-wise manner, finally concatenated together to fed into a multilayer
perceptron (MLP) to learn the nonlinear relations among features. In this way,
user features are compressed into a fixed-length representation vector, in
regardless of what candidate ads are. The use of fixed-length vector will be a
bottleneck, which brings difficulty for Embedding\&MLP methods to capture
user's diverse interests effectively from rich historical behaviors. In this
paper, we propose a novel model: Deep Interest Network (DIN) which tackles this
challenge by designing a local activation unit to adaptively learn the
representation of user interests from historical behaviors with respect to a
certain ad. This representation vector varies over different ads, improving the
expressive ability of model greatly. Besides, we develop two techniques:
mini-batch aware regularization and data adaptive activation function which can
help training industrial deep networks with hundreds of millions of parameters.
Experiments on two public datasets as well as an Alibaba real production
dataset with over 2 billion samples demonstrate the effectiveness of proposed
approaches, which achieve superior performance compared with state-of-the-art
methods. DIN now has been successfully deployed in the online display
advertising system in Alibaba, serving the main traffic. | http://arxiv.org/pdf/1706.06978 | Guorui Zhou, Chengru Song, Xiaoqiang Zhu, Ying Fan, Han Zhu, Xiao Ma, Yanghui Yan, Junqi Jin, Han Li, Kun Gai | stat.ML, cs.LG, I.2.6; H.3.2 | Accepted by KDD 2018 | null | stat.ML | 20170621 | 20180913 | [
{
"id": "1704.05194"
}
] |
1706.06708 | 53 | The number of type-4 moves is at least 0. Adding these bounds up, we see that there are at least (|O|) + 3(n â |O|) + (JO| -â 1) +0 = 3nâ1-â|O| = k+(nâ|O]) moves. Since nâ|O| > 0 and the number of moves is at most k, we can conclude that (1) |O| = n and (2) the number of moves of each type is exactly the minimum possible computed above. Since |O| = n we know that O = {1,...,n}. But then looking at the condition for obtaining the minimum possible number of type-3 moves, we see that the l;s for i ⬠O = {1,...,n} in the order in which row-i flips occur in mj,...,m: are each at Hamming distance exactly one from the next. Thus, there is a reordering of 1),...,/, in which each 1; is Hamming distance one from the next; in other words, the cubical graph specified by bitstrings l1,...,/n has a Hamiltonian path and l,...,ln is a âyesâ instance to the Promise Cubical Hamiltonian Path problem. | 1706.06708#53 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06927 | 53 | (a) Manipulating objects, one single table.
(b) Manipulating objects, three tables.
TABLE II: Per-instance results for one and three tables. Each row shows results for one instance. Three leftmost columns report instance characteristics. #o denotes number of objects on the table, #g number of different goals, #c is a proxy for the number of objects that initially obstruct the achievement of the goal, as described in the text. Remaining columns report length of the computed plan (L), number of nodes expanded during the search (E), and, in seconds, preprocessing, search and total time. TO and MO denote time- and memory-outs.
up with relative ease with the number of different speciï¬ed goals. Finally, preprocessing times scale up linearly with the number of objects, regardless of the number of goals, thanks to the low-polynomial cost of the IW(2) pass on which the preprocessing is based, as detailed above.
resolve procedural calls into efï¬cient table lookups. We have shown that the compilation process is efï¬cient and independent of the number of objects, that the compiled problems are compact, and that the planning algorithm can generate long plans effectively.
# VII. DISCUSSION | 1706.06927#53 | Combined Task and Motion Planning as Classical AI Planning | Planning in robotics is often split into task and motion planning. The
high-level, symbolic task planner decides what needs to be done, while the
motion planner checks feasibility and fills up geometric detail. It is known
however that such a decomposition is not effective in general as the symbolic
and geometrical components are not independent. In this work, we show that it
is possible to compile task and motion planning problems into classical AI
planning problems; i.e., planning problems over finite and discrete state
spaces with a known initial state, deterministic actions, and goal states to be
reached. The compilation is sound, meaning that classical plans are valid robot
plans, and probabilistically complete, meaning that valid robot plans are
classical plans when a sufficient number of configurations is sampled. In this
approach, motion planners and collision checkers are used for the compilation,
but not at planning time. The key elements that make the approach effective are
1) expressive classical AI planning languages for representing the compiled
problems in compact form, that unlike PDDL make use of functions and state
constraints, and 2) general width-based search algorithms capable of finding
plans over huge combinatorial spaces using weak heuristics only. Empirical
results are presented for a PR2 robot manipulating tens of objects, for which
long plans are required. | http://arxiv.org/pdf/1706.06927 | Jonathan Ferrer-Mestres, Guillem Francès, Hector Geffner | cs.RO, cs.AI | 10 pages, 2 figures | null | cs.RO | 20170621 | 20170621 | [] |
1706.06978 | 53 | [26] Shuangfei Zhai, Keng-hao Chang, Ruofei Zhang, and Zhongfei Mark Zhang. 2016. Deepintent: Learning attentions for online advertising with recurrent neural networks. In Proceedings of the 22nd ACM SIGKDD International Conference on Knowledge Discovery and Data Mining. ACM, 1295â1304.
[27] Song J et al. Zhou C, Bai J. 2018. ATRank: An Attention-Based User Behavior Mod- eling Framework for Recommendation. In Proceedings of 32th AAAI Conference on Artificial Intelligence. | 1706.06978#53 | Deep Interest Network for Click-Through Rate Prediction | Click-through rate prediction is an essential task in industrial
applications, such as online advertising. Recently deep learning based models
have been proposed, which follow a similar Embedding\&MLP paradigm. In these
methods large scale sparse input features are first mapped into low dimensional
embedding vectors, and then transformed into fixed-length vectors in a
group-wise manner, finally concatenated together to fed into a multilayer
perceptron (MLP) to learn the nonlinear relations among features. In this way,
user features are compressed into a fixed-length representation vector, in
regardless of what candidate ads are. The use of fixed-length vector will be a
bottleneck, which brings difficulty for Embedding\&MLP methods to capture
user's diverse interests effectively from rich historical behaviors. In this
paper, we propose a novel model: Deep Interest Network (DIN) which tackles this
challenge by designing a local activation unit to adaptively learn the
representation of user interests from historical behaviors with respect to a
certain ad. This representation vector varies over different ads, improving the
expressive ability of model greatly. Besides, we develop two techniques:
mini-batch aware regularization and data adaptive activation function which can
help training industrial deep networks with hundreds of millions of parameters.
Experiments on two public datasets as well as an Alibaba real production
dataset with over 2 billion samples demonstrate the effectiveness of proposed
approaches, which achieve superior performance compared with state-of-the-art
methods. DIN now has been successfully deployed in the online display
advertising system in Alibaba, serving the main traffic. | http://arxiv.org/pdf/1706.06978 | Guorui Zhou, Chengru Song, Xiaoqiang Zhu, Ying Fan, Han Zhu, Xiao Ma, Yanghui Yan, Junqi Jin, Han Li, Kun Gai | stat.ML, cs.LG, I.2.6; H.3.2 | Accepted by KDD 2018 | null | stat.ML | 20170621 | 20180913 | [
{
"id": "1704.05194"
}
] |
1706.06708 | 54 | All thatâs left is to complete the ï¬rst two steps of the proof. We prove these two steps in the lemmas below:
Lemma 4.8. Move sequence m,...,my must flip row i an odd number of times if i ⬠{1,...,n}, and an even number of times otherwise.
Proof: Consider the transformation
# Mp OrsomMz,ot=Mpo-+-0mM,0a1,0b,0b20+-+0bdn.
This transformation, while not necessarily the identity transformation, must transform Cp into another solved Rubikâs Square configuration Câ.
Consider the 2n = k + 1 indices max(m,n) + 1,...,max(m,n) + 2n. At least one such index i must exist for which no move in m,,...,my is an index-i move. Let u be such an index. | 1706.06708#54 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06927 | 54 | # VII. DISCUSSION
The presented work is closest to [6, 31]. What distinguishes our approach is that combined task and motion planning problems are fully mapped into classical AI planning problems encoded in an expressive planning language. Motion planners and collision checkers are used at compile time but not at planning time. The approach is sound (classical plans map into valid executable robot plans) and probabilistically complete (with a sufï¬cient number of conï¬gurations sampled, robot plans have a corresponding classical plan). For the approach three elements are essential. First, an ex- to be effective, pressive planning language that supports functions and state constraints. Second, a width-based planning algorithm that can plan effectively for models expressed in such a language without requiring the use of accurate but expensive heuristic estimators. Third, a preprocessing stage that computes the ï¬nite graphs of robot bases and arm conï¬gurations, the pos- sible object conï¬gurations, and the tables that allow us to | 1706.06927#54 | Combined Task and Motion Planning as Classical AI Planning | Planning in robotics is often split into task and motion planning. The
high-level, symbolic task planner decides what needs to be done, while the
motion planner checks feasibility and fills up geometric detail. It is known
however that such a decomposition is not effective in general as the symbolic
and geometrical components are not independent. In this work, we show that it
is possible to compile task and motion planning problems into classical AI
planning problems; i.e., planning problems over finite and discrete state
spaces with a known initial state, deterministic actions, and goal states to be
reached. The compilation is sound, meaning that classical plans are valid robot
plans, and probabilistically complete, meaning that valid robot plans are
classical plans when a sufficient number of configurations is sampled. In this
approach, motion planners and collision checkers are used for the compilation,
but not at planning time. The key elements that make the approach effective are
1) expressive classical AI planning languages for representing the compiled
problems in compact form, that unlike PDDL make use of functions and state
constraints, and 2) general width-based search algorithms capable of finding
plans over huge combinatorial spaces using weak heuristics only. Empirical
results are presented for a PR2 robot manipulating tens of objects, for which
long plans are required. | http://arxiv.org/pdf/1706.06927 | Jonathan Ferrer-Mestres, Guillem Francès, Hector Geffner | cs.RO, cs.AI | 10 pages, 2 figures | null | cs.RO | 20170621 | 20170621 | [] |
1706.06708 | 55 | Consider the effect of transformation my 0---0m 10a ,0b,0b20---0b, on the cubie in position u,u). If we write t = aj 0b, 0b0---obp as a sequence of x;s and ys (using the definitions of a; and b;), then every move in ¢ flips rows and columns with indices of absolute value at most max(m, 1). Thus no term in the transformation (my 0 +++ 0m, 0 a1 0 by 0 bg 0-++ 0 by) flips row or column uw. We conclude that the cubie in position (u,w) is unmoved by this transformation. Applying this ransformation to Co yields Câ. So since this cubie starts with top sticker red in configuration Co, he final configuration Câ also has this cubieâs top sticker red. Since Câ is a solved configuration, he entire top face in Câ must be red.
Next consider the cubie in position (u,r) for any r. Since no row or column with index +u is ever flipped in transformation my 0-+-0m 041 0b, 0b20-+-obn, this cubie is only ever affected by flips of row r. Furthermore, every flip of row r flips this cubie and therefore switches the color of
15 | 1706.06708#55 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06927 | 55 | For the experiments, we have considered the type of pick and place problems that have been used in recent work [6, 31]. For these problems, it is sufï¬cient to sample robot base conï¬gurations that are close to the physical tables, and arm trajectories that can pick up and place objects in the local space of the robot at a height that corresponds to the height of the tables. This part of the problem is not modeled explicitly in the Functional STRIPS planning encodings, which implicitly assume a ï¬nite graph of robot bases and one of robot arm conï¬gurations computed at preprocessing. In the future, we want to represent this information explicitly in the planning encoding so that the preprocessing stage can be fully general and automatic. This requires a general representation language for CTMP problems so that the compilation will be a mapping between one formal language and another. Unfortunately, there are no widely accepted and shared formal models and languages for CTMP, which makes it difï¬cult to compare approaches empirically or to organize â CTMP competitionsâ, that in the case of AI planning or SAT solving | 1706.06927#55 | Combined Task and Motion Planning as Classical AI Planning | Planning in robotics is often split into task and motion planning. The
high-level, symbolic task planner decides what needs to be done, while the
motion planner checks feasibility and fills up geometric detail. It is known
however that such a decomposition is not effective in general as the symbolic
and geometrical components are not independent. In this work, we show that it
is possible to compile task and motion planning problems into classical AI
planning problems; i.e., planning problems over finite and discrete state
spaces with a known initial state, deterministic actions, and goal states to be
reached. The compilation is sound, meaning that classical plans are valid robot
plans, and probabilistically complete, meaning that valid robot plans are
classical plans when a sufficient number of configurations is sampled. In this
approach, motion planners and collision checkers are used for the compilation,
but not at planning time. The key elements that make the approach effective are
1) expressive classical AI planning languages for representing the compiled
problems in compact form, that unlike PDDL make use of functions and state
constraints, and 2) general width-based search algorithms capable of finding
plans over huge combinatorial spaces using weak heuristics only. Empirical
results are presented for a PR2 robot manipulating tens of objects, for which
long plans are required. | http://arxiv.org/pdf/1706.06927 | Jonathan Ferrer-Mestres, Guillem Francès, Hector Geffner | cs.RO, cs.AI | 10 pages, 2 figures | null | cs.RO | 20170621 | 20170621 | [] |
1706.06708 | 56 | 15
its top face. Since the transformation in question converts configuration Cp into configuration Câ, both of which have every cubieâs top face red, the row in question must be flipped an even number of times.
For i ⬠{1,...,n}, the transformation a 0b, 0b20---obn, when written out fully in terms of y;s and js, includes exactly one flip of row y;. Thus move sequence m1,...,7â¢, must flip each of these rows an odd number of times. Similarly, for 7 ¢ {1,...,n}, the transformation aj 0 bj 0b) 0--+-obn, when written out fully in terms of ys and x;s, does not include any flips of row y; at all. Thus move sequence m 1,..., mx must flip each of these rows an even number of times.
Lemma 4.9. If i;,i2 ⬠O (with i, 4 iz), then the number of column moves x; between the unique Yi, and Yig Moves in sequence m1,..., My is at least the Hamming distance between l;, and li,. | 1706.06708#56 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06927 | 56 | have been an essential ingredient for progress. We believe that Functional STRIPS can actually serve both roles: as integrated representation language the basis for a general, for CTMP problems and as a convenient language of the compilation representations. This work is a ï¬rst step towards this goal where we have shown that the compilation is indeed feasible and effective both representationally and computationally.
# REFERENCES
[1] S. Cambon, F. Gravot, and R. Alami. aSyMov: Towards more realistic robot plans. In Proc. ICAPS, 2004. [2] Stephane Cambon, Rachid Alami, and Fabien Gravot. A hybrid approach to intricate motion, manipulation and task planning. The International Journal of Robotics Research, 28(1):104â126, 2009.
[3] N. Dantam, Z. Kingston, S. Chaudhuri, and L. Kavraki. Incremental task and motion planning: a constraint-based In Proc. of Robotics: Science and Systems, approach. 2016.
[4] C. Dornhege, P. Eyerich, T. Keller, S. Tr¨ug, M. Brenner, Semantic Attachments for Domain- and B. Nebel. Independent Planning Systems. In Proc. ICAPS, pages 114â121, 2009. | 1706.06927#56 | Combined Task and Motion Planning as Classical AI Planning | Planning in robotics is often split into task and motion planning. The
high-level, symbolic task planner decides what needs to be done, while the
motion planner checks feasibility and fills up geometric detail. It is known
however that such a decomposition is not effective in general as the symbolic
and geometrical components are not independent. In this work, we show that it
is possible to compile task and motion planning problems into classical AI
planning problems; i.e., planning problems over finite and discrete state
spaces with a known initial state, deterministic actions, and goal states to be
reached. The compilation is sound, meaning that classical plans are valid robot
plans, and probabilistically complete, meaning that valid robot plans are
classical plans when a sufficient number of configurations is sampled. In this
approach, motion planners and collision checkers are used for the compilation,
but not at planning time. The key elements that make the approach effective are
1) expressive classical AI planning languages for representing the compiled
problems in compact form, that unlike PDDL make use of functions and state
constraints, and 2) general width-based search algorithms capable of finding
plans over huge combinatorial spaces using weak heuristics only. Empirical
results are presented for a PR2 robot manipulating tens of objects, for which
long plans are required. | http://arxiv.org/pdf/1706.06927 | Jonathan Ferrer-Mestres, Guillem Francès, Hector Geffner | cs.RO, cs.AI | 10 pages, 2 figures | null | cs.RO | 20170621 | 20170621 | [] |
1706.06927 | 57 | [5] G. Franc`es and H. Geffner. Modeling and computation in planning: Better heuristics from more expressive lan- guages. In Proc. ICAPS, 2015.
[6] C. Garrett, T. Lozano-P´erez, and L. Kaelbling. FFRob: An efï¬cient heuristic for task and motion planning. In Proc. Int. WAFR, 2014.
[7] C. Garrett, T. Lozano-P´erez, and L. Kaelbling. FFRob: An efï¬cient heuristic for task and motion planning. In Algorithmic Foundations of Robotics XI, pages 179â195. Springer, 2015.
[8] H. Geffner. Functional STRIPS: A more ï¬exible lan- guage for planning and problem solving. In J. Minker, editor, Logic-Based Artiï¬cial Intelligence, pages 187â 205. Kluwer, 2000.
[9] H. Geffner and B. Bonet. A Concise Introduction to Models and Methods for Automated Planning. Morgan & Claypool Publishers, 2013.
[10] T. Geffner and H. Geffner. Width-based Planning for In Proc. AIIDE-2015, General Video-Game Playing. 2015. | 1706.06927#57 | Combined Task and Motion Planning as Classical AI Planning | Planning in robotics is often split into task and motion planning. The
high-level, symbolic task planner decides what needs to be done, while the
motion planner checks feasibility and fills up geometric detail. It is known
however that such a decomposition is not effective in general as the symbolic
and geometrical components are not independent. In this work, we show that it
is possible to compile task and motion planning problems into classical AI
planning problems; i.e., planning problems over finite and discrete state
spaces with a known initial state, deterministic actions, and goal states to be
reached. The compilation is sound, meaning that classical plans are valid robot
plans, and probabilistically complete, meaning that valid robot plans are
classical plans when a sufficient number of configurations is sampled. In this
approach, motion planners and collision checkers are used for the compilation,
but not at planning time. The key elements that make the approach effective are
1) expressive classical AI planning languages for representing the compiled
problems in compact form, that unlike PDDL make use of functions and state
constraints, and 2) general width-based search algorithms capable of finding
plans over huge combinatorial spaces using weak heuristics only. Empirical
results are presented for a PR2 robot manipulating tens of objects, for which
long plans are required. | http://arxiv.org/pdf/1706.06927 | Jonathan Ferrer-Mestres, Guillem Francès, Hector Geffner | cs.RO, cs.AI | 10 pages, 2 figures | null | cs.RO | 20170621 | 20170621 | [] |
1706.06708 | 58 | We know from Theorem that, if i ⬠{1,2,...,n} and j ⬠{1,2,...,m}, then the top color of the cubie in location (j,i) of configuration C; is red if and only if (I;) j = 1. Thus, if J;, and |, differ in bit j, then in configuration C; one of the two cubies in positions (j,7,) and (j, i2) will have top face red and the other will have top face blue. Applying the above useful fact, we see that at least one index-j column move will occur in sequence m ,...,m,- between the unique y;, and yi, moves. Since this column move has index +j, every difference in J;, and 1;, will contribute at least one distinct column move between the unique y;, and y;, moves. Assuming the useful fact, we can conclude that the number of column moves between the unique y;, and yj, moves is at least the Hamming distance between |;, and l;,, as desired. | 1706.06708#58 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06927 | 58 | [10] T. Geffner and H. Geffner. Width-based Planning for In Proc. AIIDE-2015, General Video-Game Playing. 2015.
[11] F. Gravot, S. Cambon, and R. Alami. aSyMov: a planner that deals with intricate symbolic and geometric prob- lems. In Robotics Research. The Eleventh International Symposium, pages 100â110. Springer, 2005.
[12] K. Hauser. The minimum constraint removal problem with three robotics applications. The International Jour- nal of Robotics Research, 2013.
[13] J. Hoffmann and B. Nebel. The FF Planning System: Fast Plan Generation Through Heuristic Search. JAIR, 14:253â302, 2001.
[14] L. Kaelbling and T. Lozano-P´erez. Hierarchical task and
motion planning in the now. In Proc. ICRA, pages 1470â 1477. IEEE, 2011.
[15] Nathan Koenig and Andrew Howard. Design and use paradigms for gazebo, an open-source multi-robot sim- In Intelligent Robots and Systems, 2004.(IROS ulator. 2004). Proceedings. 2004 IEEE/RSJ International Con- ference on, volume 3, pages 2149â2154. IEEE, 2004. | 1706.06927#58 | Combined Task and Motion Planning as Classical AI Planning | Planning in robotics is often split into task and motion planning. The
high-level, symbolic task planner decides what needs to be done, while the
motion planner checks feasibility and fills up geometric detail. It is known
however that such a decomposition is not effective in general as the symbolic
and geometrical components are not independent. In this work, we show that it
is possible to compile task and motion planning problems into classical AI
planning problems; i.e., planning problems over finite and discrete state
spaces with a known initial state, deterministic actions, and goal states to be
reached. The compilation is sound, meaning that classical plans are valid robot
plans, and probabilistically complete, meaning that valid robot plans are
classical plans when a sufficient number of configurations is sampled. In this
approach, motion planners and collision checkers are used for the compilation,
but not at planning time. The key elements that make the approach effective are
1) expressive classical AI planning languages for representing the compiled
problems in compact form, that unlike PDDL make use of functions and state
constraints, and 2) general width-based search algorithms capable of finding
plans over huge combinatorial spaces using weak heuristics only. Empirical
results are presented for a PR2 robot manipulating tens of objects, for which
long plans are required. | http://arxiv.org/pdf/1706.06927 | Jonathan Ferrer-Mestres, Guillem Francès, Hector Geffner | cs.RO, cs.AI | 10 pages, 2 figures | null | cs.RO | 20170621 | 20170621 | [] |
1706.06708 | 59 | We now prove the useful fact by contradiction. Assume that the useful fact is false, i.e., that 1ere exists some i1,%2 ⬠O and j ⬠{1,2,...,m} such that the top colors of the cubies in locations j,i1) and (j,i2) are different in C, and such that no index-j column move is made between the unique y;, and yj, moves in sequence m1,..., Mx".
Consider these two cubies. Starting in configuration Cy, we can reach configuration Câ by applying transformation mp 0 -+-0 m0 a, = mpro-+-o myo (a1)! 0 (ag))2 0 +++ 0 (x3), Note that this transformation consists of some (but not necessarily all) of the moves 21, %2,...,0%m âollowed by the move sequence â¢m1,...,â¢m,/. We will consider the effect of this transformation on 1e two cubies. | 1706.06708#59 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06927 | 59 | [16] F. Lagriffoul, D. Dimitrov, A. Safï¬otti, and L. Karlsson. Constraint propagation on interval bounds for dealing with geometric backtracking. In Proc. IROS, pages 957â 964. IEEE, 2012.
[17] Steven M LaValle. Planning algorithms. Cambridge, 2006.
[18] F. Lin and R. Reiter. State constraints revisited. Journal of logic and computation, 4(5):655â677, 1994.
[19] N. Lipovetzky and H. Geffner. Width and Serialization In Proc. ECAI, pages of Classical Planning Problems. 540â545, 2012.
[20] N. Lipovetzky and H. Geffner. Best-ï¬rst Width Search: Exploration and Exploitation in Classical Planning. In Proc. AAAI-2017, 2017.
[21] N. Lipovetzky and H. Geffner. A polynomial planning In Proc. ICAPS- algorithm that beats LAMA and FF. 2017 (to appear), 2017.
[22] N. Lipovetzky, M. Ramirez, and H. Geffner. Classical Planning with Simulators: Results on the Atari Video Games. In Proc. IJCAI-2015, 2015. | 1706.06927#59 | Combined Task and Motion Planning as Classical AI Planning | Planning in robotics is often split into task and motion planning. The
high-level, symbolic task planner decides what needs to be done, while the
motion planner checks feasibility and fills up geometric detail. It is known
however that such a decomposition is not effective in general as the symbolic
and geometrical components are not independent. In this work, we show that it
is possible to compile task and motion planning problems into classical AI
planning problems; i.e., planning problems over finite and discrete state
spaces with a known initial state, deterministic actions, and goal states to be
reached. The compilation is sound, meaning that classical plans are valid robot
plans, and probabilistically complete, meaning that valid robot plans are
classical plans when a sufficient number of configurations is sampled. In this
approach, motion planners and collision checkers are used for the compilation,
but not at planning time. The key elements that make the approach effective are
1) expressive classical AI planning languages for representing the compiled
problems in compact form, that unlike PDDL make use of functions and state
constraints, and 2) general width-based search algorithms capable of finding
plans over huge combinatorial spaces using weak heuristics only. Empirical
results are presented for a PR2 robot manipulating tens of objects, for which
long plans are required. | http://arxiv.org/pdf/1706.06927 | Jonathan Ferrer-Mestres, Guillem Francès, Hector Geffner | cs.RO, cs.AI | 10 pages, 2 figures | null | cs.RO | 20170621 | 20170621 | [] |
1706.06708 | 60 | Since the two cubies start in locations (j, i1) and (j, i2), the only moves that could ever aï¬ect these cubies are of the forms xj, xâj, yi1, yâi1, yi2, and yâi2. Furthermore, by the deï¬nition of O, no moves of the form yâi1 or yâi2 occur and the moves yi1 and yi2 each occur exactly once. Finally, we have by assumption that no moves of the form xj or xâj (index-j column moves) occur between moves yi1 and yi2.
Putting these facts together, we see that the effect of transformation my 0---om 0a, on these two cubies is exactly the same as the effect of some transformation of the following type: (1) some number of moves of the form 2; or z_j;, followed by (2) the two moves y;, and yj. in some order, followed by (3) some number of moves of the form a; or x_;. | 1706.06708#60 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06927 | 60 | [23] T. Lozano-P´erez and L. Kaelbling. A constraint-based method for solving sequential manipulation planning problems. In Proc. IROS, pages 3684â3691. IEEE, 2014. [24] D. McDermott. The 1998 AI Planning Systems Com- petition. Artiï¬cial Intelligence Magazine, 21(2):35â56, 2000.
[25] B. Nebel, C. Dornhege, and A. Hertle. How much does a household robot need to know in order to tidy up? In Proc. AAAI Workshop on Intelligent Robotic Systems, Bellevue, WA, 2013.
[26] S. Nedunuri, S. Prabhu, M. Moll, S. Chaudhuri, and L. Kavraki. SMT-based synthesis of integrated task and motion plans from plan outlines. In IEEE Int. Conf on Robotics and Automation (ICRA), pages 655â662, 2014. [27] M. Quigley, K. Conley, B. Gerkey, J. Faust, T. Foote, J. Leibs, R. Wheeler, and A. Y Ng. ROS: an open- source Robot Operating System. In ICRA workshop on open source software, volume 3, page 5. Kobe, 2009. | 1706.06927#60 | Combined Task and Motion Planning as Classical AI Planning | Planning in robotics is often split into task and motion planning. The
high-level, symbolic task planner decides what needs to be done, while the
motion planner checks feasibility and fills up geometric detail. It is known
however that such a decomposition is not effective in general as the symbolic
and geometrical components are not independent. In this work, we show that it
is possible to compile task and motion planning problems into classical AI
planning problems; i.e., planning problems over finite and discrete state
spaces with a known initial state, deterministic actions, and goal states to be
reached. The compilation is sound, meaning that classical plans are valid robot
plans, and probabilistically complete, meaning that valid robot plans are
classical plans when a sufficient number of configurations is sampled. In this
approach, motion planners and collision checkers are used for the compilation,
but not at planning time. The key elements that make the approach effective are
1) expressive classical AI planning languages for representing the compiled
problems in compact form, that unlike PDDL make use of functions and state
constraints, and 2) general width-based search algorithms capable of finding
plans over huge combinatorial spaces using weak heuristics only. Empirical
results are presented for a PR2 robot manipulating tens of objects, for which
long plans are required. | http://arxiv.org/pdf/1706.06927 | Jonathan Ferrer-Mestres, Guillem Francès, Hector Geffner | cs.RO, cs.AI | 10 pages, 2 figures | null | cs.RO | 20170621 | 20170621 | [] |
1706.06708 | 61 | Consider the eï¬ect of any such transformation on the two cubies. In step (1), each move of the form xj or xâj either ï¬ips both cubies (since they both start in column j) or ï¬ips neither, so the two cubies are each ï¬ipped an equal number of times. Furthermore, the row index of the two cubies is either positive for both or negative for both at all times throughout step (1). In step (2), either each of the two cubies is ï¬ipped exactly once (if their row indices at the start of step (2) are
16
both positive) or neither of the two cubies is ï¬ipped at all (if their row indices at the start of step (2) are negative); again, the number of ï¬ips is the same. Finally, in step (3), both cubies are in the same column (column j if they were not ï¬ipped in step (2) and column âj if they were), so each move of the form xj or xâj either ï¬ips both cubies or ï¬ips neither; the two cubies are ï¬ipped an equal number of times. Thus we see that the two cubies are ï¬ipped an equal number of times by such a transformation. | 1706.06708#61 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06927 | 61 | The LAMA planner: Guiding cost-based anytime planning with landmarks. JAIR, 39(1):127â177, 2010.
[29] A. Shleyfman, A. Tuisov, and C. Domshlak. Blind Search for Atari-like Online Planning Revisited. In Proc. IJCAI- 2016, 2016.
[30] T. C. Son, P. Huy Tu, M. Gelfond, and A. Morales. Conformant Planning for Domains with Constraints: A In Proc. AAAI-05, pages 1211â1216, New Approach. 2005.
[31] S. Srivastava, E. Fang, L. Riano, R. Chitnis, S. Russell, and P. Abbeel. Combined task and motion plan- ning through an extensible planner-independent interface layer. In Proc. ICRA, pages 639â646, 2014.
[32] I. Sucan and S Chitta. MoveIt! At http://moveit.ros.org. [33] J. Wolfe, B. Marthi, and S. Russell. Combined Task and Motion Planning for Mobile Manipulation. In Proc. ICAPS, pages 254â258, 2010. | 1706.06927#61 | Combined Task and Motion Planning as Classical AI Planning | Planning in robotics is often split into task and motion planning. The
high-level, symbolic task planner decides what needs to be done, while the
motion planner checks feasibility and fills up geometric detail. It is known
however that such a decomposition is not effective in general as the symbolic
and geometrical components are not independent. In this work, we show that it
is possible to compile task and motion planning problems into classical AI
planning problems; i.e., planning problems over finite and discrete state
spaces with a known initial state, deterministic actions, and goal states to be
reached. The compilation is sound, meaning that classical plans are valid robot
plans, and probabilistically complete, meaning that valid robot plans are
classical plans when a sufficient number of configurations is sampled. In this
approach, motion planners and collision checkers are used for the compilation,
but not at planning time. The key elements that make the approach effective are
1) expressive classical AI planning languages for representing the compiled
problems in compact form, that unlike PDDL make use of functions and state
constraints, and 2) general width-based search algorithms capable of finding
plans over huge combinatorial spaces using weak heuristics only. Empirical
results are presented for a PR2 robot manipulating tens of objects, for which
long plans are required. | http://arxiv.org/pdf/1706.06927 | Jonathan Ferrer-Mestres, Guillem Francès, Hector Geffner | cs.RO, cs.AI | 10 pages, 2 figures | null | cs.RO | 20170621 | 20170621 | [] |
1706.06708 | 62 | We can conclude that the two cubies are flipped an equal number of times by transformation Mp O+++omy,0 a4. In configuration Cy, the two cubies have different colors on their top faces, so after transformation mz o--+ 0m, 0 a, flips each of the two cubies an equal number of times, the resulting configuration still has different colors on the top faces of the two cubies. But the resulting configuration is Câ, which has red as the top face color of every cubie. Thus we have our desired contradiction. Therefore the useful fact is true and the desired result holds.
# 4.6 Conclusion
Theorems 4.2 and 4.6 and Corollaries 4.3 and 4.7 show that the polynomial-time reductions given are answer preserving. As a result, we conclude that
Theorem 4.10. The Rubikâs Square and Group Rubikâs Square problems are NP-complete.
# (Group) STM/SQTM Rubikâs Cube is NP-complete
# 5.1 Reductions | 1706.06708#62 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 63 | # (Group) STM/SQTM Rubikâs Cube is NP-complete
# 5.1 Reductions
Below, we introduce the reductions used for the Rubikâs Cube case. These reductions very closely mirror the Rubikâs Square case, and the intuition remains exactly the same: the bi terms commute, and so if the input Promise Cubical Hamiltonian Path instance is a âyesâ instance then the bis can be reordered so that all but k moves in the deï¬nition of t will cancel; therefore in that case t can be both enacted and reversed in k moves. | 1706.06708#63 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 64 | There are, however, several notable diï¬erences from the Rubikâs Square case. The ï¬rst diï¬erence is that in a Rubikâs Cube, the moves xi, yi, and zi are all quarter turn rotations rather than self- inverting row or column ï¬ips. One consequence is that unlike in the Rubikâs Square case, the term ai does not have the property that (ai)â1 = ai. A second diï¬erence is that in a Rubikâs Square, the rows never become columns or visa versa. In a Rubikâs Cube on the other hand, rotation of the faces can put rows of stickers that were once aligned parallel to one axis into alignment with another axis. To avoid allowing a solution of the puzzle due to this fact in the absence of a solution to the input Promise Cubical Hamiltonian Path instance, the slices in this construction which take the role of rows 1 through n in the Rubikâs Square case and the slices which take the role of columns 1 through m in the Rubikâs Square case will be assigned entirely distinct indices. | 1706.06708#64 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 65 | To prove that the STM/SQTM Rubikâs Cube and Group STM/SQTM Rubikâs Cube problems are NP-complete, we reduce from the Promise Cubical Hamiltonian Path problem of Section 3.2 as described below.
Suppose we are given an instance of the Promise Cubical Hamiltonian Path problem consisting of n biststrings l1, . . . , ln of length m (with ln = 00 . . . 0). To construct a Group STM/SQTM Rubikâs Square instance we need to compute the value k indicating the allowed number of moves and construct the transformation t in RCs.
The value k can be computed directly as k = 2n â 1.
17 | 1706.06708#65 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 66 | The value k can be computed directly as k = 2n â 1.
17
The transformation t will be an element of group RCs where s = 6n + 2m. Deï¬ne ai for 1 ⤠i ⤠n to be (x1)(li)1 ⦠(x2)(li)2 ⦠· · · ⦠(xm)(li)m where (li)1, (li)2, . . . , (li)m are the bits of li. Also deï¬ne bi = (ai)â1 ⦠zm+i ⦠ai for 1 ⤠i ⤠n. Then we deï¬ne t to be a1 ⦠b1 ⦠b2 ⦠· · · ⦠bn.
Outputting (t, k) completes the reduction from the Promise Cubical Hamiltonian Path problem to the Group STM/SQTM Rubikâs Cube problem. To reduce from the Promise Cubical Hamilto- nian Path problem to the STM/SQTM Rubikâs Cube problem we simply output (Ct, k) = (t(C0), k). As with the Rubikâs Square case, these reductions are clearly polynomial-time reductions. | 1706.06708#66 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 67 | # 5.2 Promise Cubical Hamiltonian Path solution â (Group) STM/SQTM Ru- bikâs Cube solution
In this section, we prove one direction of the answer preserving property of the reductions. This proof is not substantively diï¬erent from the proof of the ï¬rst direction for the Rubikâs Square problems (in Section 4.3). The diï¬erences in these proofs are all minor details that are only present to account for the diï¬erences (listed above) between the Rubikâs Square and Rubikâs Cube reductions.
Lemma 5.1. The transformations bi all commute.
Proof: Consider any such transformation bi. The transformation bi can be rewritten as (ai)â1 ⦠zm+i ⦠ai. For any cubie not moved by the zm+i middle term, the eï¬ect of this transformation is the same as the eï¬ect of transformation (ai)â1 ⦠ai = 1. In other words, bi only aï¬ects cubies that are moved by the zm+i term. | 1706.06708#67 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 68 | A cubie aï¬ected by this term was either moved into the z slice with index (m + i) by ai or was already there. ai consists of some number of clockwise x turns. Thus, in order to be moved into a position with z = (m + i), a cubie would have to start in a position with y = â(m + i) on the +z face or in a position with y = (m + i) on the âz face.
Thus, the cubies affected by b; must either have y coordinate +(m +7) and lie on one of the tz faces or have z coordinate (m+) and lie on one of the other four faces. This is enough to see hat the cubies affected by b; are disjoint from those affected by b; (for 7 #7). In other words, the ransformations b; all commute.
Theorem 5.2. If l1, . . . , ln is a âyesâ instance to the Promise Cubical Hamiltonian Path problem, then (t, k) is a âyesâ instance to the Group SQTM Rubikâs Cube problem. | 1706.06708#68 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 69 | Proof: Suppose l1, . . . , ln is a âyesâ instance to the Promise Cubical Hamiltonian Path problem. Let m be the length of li and note that ln = 00 . . . 0 by the promise of the Promise Cubical Hamiltonian Path problem. Furthermore, since l1, . . . , ln is a âyesâ instance to the Promise Cubical Hamiltonian Path problem, there exists an ordering of these bitstrings li1, li2, . . . , lin such that each consecutive pair of bitstrings is at Hamming distance one, i1 = 1, and in = n (with the ï¬nal two conditions coming from the promise).
By Lemma 5.1, we know that t = a1 ⦠b1 ⦠b2 ⦠· · · ⦠bn can be rewritten as
t = a1 ⦠bi1 ⦠bi2 ⦠· · · ⦠bin.
Using the deï¬nition of bi, we can further rewrite this as
t = a1 ⦠((ai1)â1 ⦠zm+i1 ⦠ai1) ⦠((ai2)â1 ⦠zm+i2 ⦠ai2) ⦠· · · ⦠((ain)â1 ⦠zm+in ⦠ain),
18
# or as | 1706.06708#69 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 70 | 18
# or as
t = (a1 ⦠(ai1)â1) ⦠zm+i1 ⦠(ai1 ⦠(ai2)â1) ⦠zm+i2 ⦠(ai2 ⦠(ai3)â1) ⦠· · · ⦠(ainâ1 ⦠(ain)â1) ⦠zm+in ⦠(ain).
We know that i1 = 1, and therefore that a1 ⦠(ai1)â1 = a1 ⦠(a1)â1 = 1 is the identity element. Similarly, we know that in = n and therefore that ain = an = (x1)(ln)1 ⦠(x2)(ln)2 ⦠· · · ⦠(xm)(ln)m = (x1)0 ⦠(x2)0 ⦠· · · ⦠(xm)0 = 1 is also the identity.
Thus we see that
t = zm+i1 ⦠(ai1 ⦠(ai2)â1) ⦠zm+i2 ⦠(ai2 ⦠(ai3)â1) ⦠· · · ⦠(ainâ1 ⦠(ain)â1) ⦠zm+in. | 1706.06708#70 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 71 | Consider the transformation aip ⦠(aip+1)â1. This transformation can be written as
aip ⦠(aip+1)â1 = (x1)(lip )1 ⦠(x2)(lip )2 ⦠· · · ⦠(xm)(lip )m ⦠(x1)â(lip+1 )1 ⦠(x2)â(lip+1 )2 ⦠· · · ⦠(xm)â(lip+1 )m.
Because xu always commutes with xv, we can rewrite this as
aip ⦠(aip+1)â1 = (x1)(lip )1â(lip+1 )1 ⦠(x2)(lip )2â(lip+1 )2 ⦠· · · ⦠(xm)(lip )mâ(lip+1 )m.
Since lip diï¬ers from lip+1 in only one position, call it jp, we see that (lip)j â (lip+1)j is zero unless j = jp, and is ±1 in that ï¬nal case. This is suï¬cient to show that aip ⦠(aip+1)â1 = (xjp)sp where sp = ±1.
Thus we see that | 1706.06708#71 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 72 | Thus we see that
t = zm+i1 ⦠(xj1)s1 ⦠zm+i2 ⦠(xj2)s2 ⦠· · · ⦠(xjnâ1)snâ1 ⦠zm+in,
or (by left multiplying) that
(zm+in)â1 ⦠(xjnâ1)âsnâ1 ⦠· · · ⦠(xj2)âs2 ⦠(zm+i2)â1 ⦠(xj1)âs1 ⦠(zm+i1)â1 ⦠t = 1.
We see that t can be reversed by k = 2n â 1 terms of the form (z;)~!, xj, and (a;)71, which are all SQTM moves. In other words, (t,k) is a âyesâ instance to the Group SQTM Rubikâs Cube problem. | 1706.06708#72 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 73 | Corollary 5.3. If l1, . . . , ln is a âyesâ instance to the Promise Cubical Hamiltonian Path problem, then (Ct, k) is a âyesâ instance to the STM/SQTM Rubikâs Cube problem and (t, k) is a âyesâ instance to the Group STM/SQTM Rubikâs Cube problem.
Proof: This follows immediately from Theorem 5.2 and Lemmas 2.2 and 2.3.
# 5.3 Coloring of Ct
As in the Rubikâs Square case, it will be helpful for the second direction of the proof to know the coloring of the Cubeâs conï¬guration. As before, we deï¬ne b = b1 ⦠· · · ⦠bn (so that t = a1 ⦠b) and determine the colors of the stickers in conï¬guration Cb = b(C0).
Consider the example instance of Promise Cubical Hamiltonian Path with n = 5 and m = 3 introduced in the Rubikâs Square section and reproduced below:
19
l1 = 011 l2 = 110 l3 = 111 l4 = 100 l5 = 000 | 1706.06708#73 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 75 | Figure 6: The faces of Cb for the example input l1, . . . , ln. In this ï¬gure, the top and bottom faces are the +z and âz faces, while the faces in the vertical center of the ï¬gure are the +x, +y, âx, and ây faces from left to right.
In this section, we prove the following useful theorem, which formalizes the pattern of colors from the example (Figure 6):
Theorem 5.4. In Cb, the stickers have the following coloring:
20
+z: The stickers on the +z face with (x, y) coordinates (j, â(m + i)) where i â {1, . . . , n} and the jth bit of li is one are all red. All other stickers are white.
âz: The stickers on the âz face with (x, y) coordinates (j, â(m + i)) where i â {1, . . . , n} and the jth bit of li is one are all orange. All other stickers are blue. | 1706.06708#75 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 76 | +y: The stickers on the +y face with (x, z) coordinates (j, (m + i)) where i â {1, . . . , n} and either li doesnât have a jth bit (i.e. j < 0 or j > m) or the jth bit of li is zero are all red. All other stickers are green.
ây: The stickers on the ây face with (x, z) coordinates (j, (m + i)) where i â {1, . . . , n} and either li doesnât have a jth bit (i.e. j < 0 or j > m) or the jth bit of li is zero are all orange. All other stickers are yellow.
+x: The stickers on the +x face with (y, z) coordinates (âj, (m + i)) where i â {1, . . . , n} and the jth bit of li is one are all white. All other stickers with z coordinate in {1, . . . , n} are green. All other stickers are orange. | 1706.06708#76 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 77 | âx: The stickers on the âx face with (y, z) coordinates (âj, (m + i)) where i â {1, . . . , n} and the jth bit of li is one are all blue. All other stickers with z coordinate in {1, . . . , n} are yellow. All other stickers are red.
The proof of this theorem is involved and uninsightful. In addition, no other result from this section will be used in the rest of this paper. As a result, the reader should feel free to skip the remainder of this section.
To formally derive the coloring of conï¬guration Cb, we need to have a formal description of the eï¬ect of transformation bi. For example, Figure 7 shows the +x, +y, and +z faces of a Rubikâs Cube in conï¬gurations C0, a2(C0), (zm+2 ⦠a2)(C0), and b2(C0) = ((a2)â1 ⦠zm+2 ⦠a2)(C0) where a2 and zm+2 = z5 are deï¬ned in terms of l2 = 110 as in the reduction.
The exact behavior of a Rubikâs Cube due to bi is described by Lemmas 5.5 through 5.7: | 1706.06708#77 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 78 | The exact behavior of a Rubikâs Cube due to bi is described by Lemmas 5.5 through 5.7:
Lemma 5.5. Suppose i â {1, . . . , n}. Then the eï¬ect of bi on the stickers from the ±z faces of a Rubikâs Cube can be described as follows:
⢠If the jth bit of li is one, then the sticker starting on the +z face with (x, y) coordinates (j, â(m + i)) ends up on the +x face with (y, z) coordinates (âj, (m + i)).
⢠If the jth bit of li is one, then the sticker starting on the âz face with (x, y) coordinates (j, â(m + i)) ends up on the âx face with (y, z) coordinates (âj, (m + i)).
⢠All other stickers on the ±z faces stay in place.
Proof: As noted in the proof of Lemma 5.1, a sticker is aï¬ected by bi = (ai)â1 ⦠zm+i ⦠ai if and only if it is moved by the zm+i term. | 1706.06708#78 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 79 | Consider the stickers originally on the +z face. bi starts with ai, which rotates the x slices with x coordinates j such that bit j of li is one. Therefore, the stickers on the +z face with x coordinates of this form are rotated to the +y face, and all the other stickers are left in place. After that, the only stickers from the +z face which are moved by the zm+i term of bi are the stickers which were on the +y face with z coordinate (m + i) and x coordinate j such that bit j of li is one. In other words, the only stickers from the +z face moved by the zm+i term, are those starting at (x, y) coordinates (j, â(m + i)) where bit j of li is one.
21
(a)
(b)
(c)
(d)
Figure 7: Applying b2 to C0 step by step.
All other stickers starting on the +z face are not aï¬ected by the zm+i term, and are therefore not moved by bi. On the other hand, consider any sticker of this form: a sticker starting on the +z face at (x, y) coordinates (j, â(m+i)) where bit j of li is one. Such a sticker is moved by ai to (x, z)
22 | 1706.06708#79 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 80 | 22
coordinates (j, (m + i)) of face +y. It is then moved by zm+i to (y, z) coordinates (âj, (m + i)) of face +x. Finally, (ai)â1 does not aï¬ect the sticker since it is on the +x face at the time and (ai)â1 consists of rotations of x slices.
Thus, if the jth bit of li is one, then the sticker starting on the +z face with (x, y) coordinates (j, â(m+i)) ends up on the +x face with (y, z) coordinates (âj, (m+i)). All other stickers starting on the +z face remain in place.
The exact same logic applies to the stickers originally on the âz face, allowing us to conclude that the lemma statement holds, as desired.
Lemma 5.6. Suppose i â {1, . . . , n}. Then the eï¬ect of bi on the stickers from the ±y faces of a Rubikâs Cube can be described as follows: | 1706.06708#80 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 81 | ⢠If the jth bit of li does not exist (i.e. j < 0 or j > m) or if the jth bit of li is zero, then the sticker starting on the +y face with (x, z) coordinates (j, (m + i)) ends up on the +x face with (y, z) coordinates (âj, (m + i)).
⢠If the jth bit of li does not exist (i.e. j < 0 or j > m) or if the jth bit of li is zero, then the sticker starting on the ây face with (x, z) coordinates (j, (m + i)) ends up on the âx face with (y, z) coordinates (âj, (m + i)).
⢠All other stickers on the ±y faces stay in place.
Proof: As noted in the proof of Lemma 5.1, a sticker is aï¬ected by bi = (ai)â1 ⦠zm+i ⦠ai if and only if it is moved by the zm+i term. | 1706.06708#81 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 82 | Consider the stickers originally on the +y face. bi starts with ai, which rotates the x slices with x coordinates j such that bit j of li is one. Therefore, the stickers on the +y face with x coordinates of this form are rotated to the âz face, and all the other stickers are left in place. After that, the only stickers from the +y face which are moved by the zm+i term of bi are the stickers with z coordinate (m + i) which were not moved from the +y face. In other words, the only stickers from the +z face moved by the zm+i term, are those starting at (x, y) coordinates (j, â(m + i)) where bit j of li either does not exist (i.e. j < 0 or j > m) or is zero. | 1706.06708#82 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 83 | All other stickers starting on the +y face are not aï¬ected by the zm+i term, and are therefore not moved by bi. On the other hand, consider any sticker of this form: a sticker starting on the +y face at (x, z) coordinates (j, (m + i)) where bit j of li either does not exist or is zero. Such a sticker is not moved by ai. It is then moved by zm+i to (y, z) coordinates (âj, (m + i)) of face +x. Finally, (ai)â1 does not aï¬ect the sticker since it is on the +x face at the time and (ai)â1 consists of rotations of x slices.
Thus, if the jth bit of li does not exist (i.e. j < 0 or j > m) or if the jth bit of li is zero, then the sticker starting on the +y face with (x, z) coordinates (j, (m + i)) ends up on the +x face with (y, z) coordinates (âj, (m + i)). All other stickers starting on the +y face remain in place.
The exact same logic applies to the stickers originally on the ây face, allowing us to conclude that the lemma statement holds, as desired. | 1706.06708#83 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 84 | The exact same logic applies to the stickers originally on the ây face, allowing us to conclude that the lemma statement holds, as desired.
Lemma 5.7. Suppose i â {1, . . . , n}. Then the eï¬ect of bi on the stickers from the ±x faces of a Rubikâs Cube can be described as follows:
⢠If the jth bit of li is one, then the sticker starting on the +x face with (y, z) coordinates (j, (m + i)) ends up on the âz face with (x, y) coordinates (j, â(m + i)).
23
⢠If the jth bit of li does not exist (i.e. j < 0 or j > m) or if the jth bit of li is zero, then the sticker starting on the +x face with (y, z) coordinates (j, (m + i)) ends up on the ây face with (x, z) coordinates (j, (m + i)).
⢠If the jth bit of li is one, then the sticker starting on the âx face with (y, z) coordinates (j, (m + i)) ends up on the +z face with (x, y) coordinates (j, â(m + i)). | 1706.06708#84 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 85 | ⢠If the jth bit of li does not exist (i.e. j < 0 or j > m) or if the jth bit of li is zero, then the sticker starting on the âx face with (y, z) coordinates (j, (m + i)) ends up on the +y face with (x, z) coordinates (j, (m + i)).
⢠All other stickers on the ±x faces stay in place.
Proof: As noted in the proof of Lemma 5.1, a sticker is aï¬ected by bi = (ai)â1 ⦠zm+i ⦠ai if and only if it is moved by the zm+i term.
Consider the stickers originally on the +x face. bi starts with ai, which aï¬ects none of the stickers on the +x face. After that, the zm+i term moves exactly those stickers from the +x face that had z coordinate (m + i). As a result, these stickers are all aï¬ected by bi, and all others are not. | 1706.06708#85 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 86 | Consider a sticker starting on the +x face with (y, z) coordinates (j, (m + i)). This sticker is unaï¬ected by ai and then moved to the ây face by zm+i. In particular, it is moved to (x, z) coordinates (j, (m + i)). After that, there are two cases:
Case 1: If bit j of li does not exist (i.e. j < 0 or j > m) or if the jth bit of li is zero, then the sticker is unaï¬ected by (ai)â1. This shows that if the jth bit of li does not exist or if the jth bit of li is zero, then the sticker starting on the +x face with (y, z) coordinates (j, (m + i)) ends up on the ây face with (x, z) coordinates (j, (m + i)). | 1706.06708#86 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 87 | Case 2: If bit j of li is one, then after being moved to the ây face by zm+i, the sticker in question is moved to the âz face by (ai)â1. In particular, the sticker ends up at (x, y) coordinates (j, â(m + i)). This shows that if the jth bit of li is one, then the sticker starting on the +x face with (y, z) coordinates (j, (m + i)) ends up on the âz face with (x, y) coordinates (j, â(m + i)).
As previously mentioned, all stickers starting on the +x face other than those addressed by the above cases stay in place due to bi. Together with the statements shown in the two cases, this is exactly what we wished to show.
The same logic applies to the stickers originally on the âx face, allowing us to conclude that the lemma statement holds, as desired.
We can apply the above lemmas to ï¬gure out the eï¬ect of transformation b1 ⦠b2 ⦠· · · ⦠bn on
conï¬guration C0. In particular, this allows us to learn the coloring of conï¬guration Cb. At this point, we can prove Theorem 5.4, which is restated below for convenience: | 1706.06708#87 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 88 | Theorem 5.4. In Cb, the stickers have the following coloring:
+z: The stickers on the +z face with (x, y) coordinates (j, â(m + i)) where i â {1, . . . , n} and the jth bit of li is one are all red. All other stickers are white.
âz: The stickers on the âz face with (x, y) coordinates (j, â(m + i)) where i â {1, . . . , n} and the jth bit of li is one are all orange. All other stickers are blue.
+y: The stickers on the +y face with (x, z) coordinates (j, (m + i)) where i â {1, . . . , n} and either li doesnât have a jth bit (i.e. j < 0 or j > m) or the jth bit of li is zero are all red. All other stickers are green.
24
ây: The stickers on the ây face with (x, z) coordinates (j, (m + i)) where i â {1, . . . , n} and either li doesnât have a jth bit (i.e. j < 0 or j > m) or the jth bit of li is zero are all orange. All other stickers are yellow. | 1706.06708#88 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 89 | +x: The stickers on the +x face with (y, z) coordinates (âj, (m + i)) where i â {1, . . . , n} and the jth bit of li is one are all white. All other stickers with z coordinate in {1, . . . , n} are green. All other stickers are orange.
âx: The stickers on the âx face with (y, z) coordinates (âj, (m + i)) where i â {1, . . . , n} and the jth bit of li is one are all blue. All other stickers with z coordinate in {1, . . . , n} are yellow. All other stickers are red.
Proof: Cb is obtained from C0 by applying transformation b1 ⦠b2 ⦠· · · ⦠bn. Each bi aï¬ects a disjoint set of stickers. Using this fact together with the description of the eï¬ect of one bi, we can obtain the description of the coloring of Cb given in the above theorem statement. | 1706.06708#89 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 90 | For example, consider the stickers that end up on the +z face. According to Lemma 5.7, if the jth bit of li is one, then bi moves the sticker starting on the âx face with (y, z) coordinates (âj, (m + i)) to the +z face with (x, y) coordinates (j, â(m + i)). Since the bis each aï¬ect disjoint sets of stickers, the stickers on the +z face with (x, y) coordinates (j, â(m + i)) where i â {1, . . . , n} and the jth bit of li is one are all stickers that started on the âx face. Since the âx face is red in C0, these stickers are all red. We know from Lemmas 5.5 through 5.7 that no stickers other than the ones that started there and those described above are moved to the +z face by bi. Therefore all other stickers on the +z face started there. Since the +z face is white in C0, these stickers are all white. Putting this together, we obtain exactly the ï¬rst bullet point of the theorem statement: The stickers on the +z face with (x, y) coordinates (j, â(m + i)) where i â {1, . . . , n} and the | 1706.06708#90 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 91 | jth bit of li is one are all red. All the other stickers are white.
The logic for the other ï¬ve faces is exactly analogous, and is omitted here for brevity.
This concludes the description of Cb in terms of colors. The coloring of conï¬guration Ctâthe conï¬guration that is actually obtained by applying the reduction to l1, . . . , lnâcan be obtained from the coloring of conï¬guration Cb by applying transformation a1. This is shown for the previously given example in Figure 8.
5.4 (Group) STM/SQTM Rubikâs Cube solution â Promise Cubical Hamilto- nian Path solution: proof outline
We wish to prove the following:
Theorem 5.8. If (Ct, k) is a âyesâ instance to the STM Rubikâs Cube problem, then l1, . . . , ln is a âyesâ instance to the Promise Cubical Hamiltonian Path problem.
By Lemmas 2.2 and 2.3, this will immediately also imply the following corollary: | 1706.06708#91 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 92 | By Lemmas 2.2 and 2.3, this will immediately also imply the following corollary:
Corollary 5.9. If (t, k) is a âyesâ instance to the Group STM/SQTM Rubikâs Cube problem or (Ct, k) is a âyesâ instance to the STM/SQTM Rubikâs Cube problem, then l1, . . . , ln is a âyesâ instance to the Promise Cubical Hamiltonian Path problem.
The intuition behind the proof of this theorem is similar to that used in the Rubikâs Square case, but there is added complexity due to the extra options available in a Rubikâs Cube. Most of the added complexity is due to the possibility of face moves (allowing rows of stickers to align in several directions over the course of a solution).
25 | 1706.06708#92 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 93 | Figure 8: The +x, +y, and +z faces of Ct for the example input l1, . . . , ln.
Below, we describe an outline of the proof, including several high-level steps, each of which is described in more detail in an additional subsection.
To prove the theorem, we consider a hypothetical solution to the (C;,k) instance of the STM Rubikâs Cube problem. A solution consists of a sequence of STM Rubikâs Cube moves m1,..., mx with kâ < k such that Câ = (mp o---0m1)(C;) is a solved configuration of the Rubikâs Cube.
One very helpful idea that is used several times throughout the proof is the idea of an index u such that no move mi is an index-u move.
Definition 5.10. Define u⬠{m+n+1,m+n+2,...,m+n-+ (2n)} to be an index such that M4,-.-,Mg contains no index-u move.
Notice that a value for u satisfying this definition must exist because variable u has 2n = k+1 > k > kâ possible values and each of the kâ moves m; disqualifies at most one possible value from being assigned to variable wu. | 1706.06708#93 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 94 | Step 1 of the proof is a preliminary characterization of the possible index-(m +7) moves among m,...,my for i ⬠{1,...,n}. Consider the following definition:
Deï¬nition 5.11. Partition the set {1, . . . , n} into four sets of indices Z, O, T , and M (where Z, O, T , and M are named after âzeroâ, âoneâ, âtwoâ, and âmoreâ) as follows:
© ie Z if and only if my,...,my contains exactly zero index-(m + 1) moves © iâ¬O if and only if m.,.. .,;mMy contains exactly one index-(m +i) move © iâ¬T if and only if m,...,mx contains exactly two index-(m + +) moves e©iâ¬M if and only if m,...,my contains at least three index-(m +1) moves
In Step 1, we prove the following list of results, thereby restricting the set of possible index- (m +2) moves (for i ⬠{1,..., n}) among mj,..., Mp:
26
⢠Z is empty.
⢠If i â O, then the sole index-(m + i) move must be a counterclockwise z turn. | 1706.06708#94 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 95 | 26
⢠Z is empty.
⢠If i â O, then the sole index-(m + i) move must be a counterclockwise z turn.
⢠If i â T , then the two index-(m + i) moves must be a clockwise z turn and a z ï¬ip in some order.
⢠If i â O ⪠T , then any move of z slice (m + i) must occur at a time when faces +x, +y, âx, and ây all have zero rotation and any move of z slice â(m + i) must occur at a time when these faces all have rotation 180â¦.
Step 2 of the proof concerns the concept of paired stickers:
Deï¬nition 5.12. Suppose p1, p2, and q are all distinct positive non-face slice indices. Then we say that two stickers are (p1, p2, q)-paired if the two stickers are on the same index-j slice, the two stickers are on the same quadrant of a face, one of the stickers has coordinates ±q and ±p1 within that face, and the second sticker has coordinates ±q and ±p2 within the face.
In particular, we prove the following useful properties of paired stickers: | 1706.06708#95 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 96 | In particular, we prove the following useful properties of paired stickers:
⢠If two stickers are (p1, p2, q)-paired, then they remain (p1, p2, q)-paired after one move unless the move is an index-p1 move or an index-p2 move which moves one of the stickers.
e Suppose i1,%2 ⬠O and j ⬠{1,2,...,m}. Then consider any pair of stickers that are (m+ i1,m + ig, 7)-paired in Cy. If there are no face moves of faces +2, +y, âx, and ây and no index-j moves that affect either of the stickers between the index-(m + i;) O-move and the index-(m + i2) O-move, then the two stickers remain (m+ i1,m + ig, j)-paired in Câ.
Step 3 of the proof uses a counting argument to significantly restrict the possible moves in my,..-., Mx. In particular, consider the following classification of moves into disjoint types:
⢠âO-movesâ: index-(m + i) moves with i â O
⢠âT -movesâ: index-(m + i) moves with i â T | 1706.06708#96 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 97 | ⢠âT -movesâ: index-(m + i) moves with i â T
⢠âM -movesâ: index-(m + i) moves with i â M
⢠âJ-movesâ: index-j moves with j â J = {1, . . . , m}
⢠âvertical face movesâ: face moves of faces +x, +y, âx, or ây
⢠âother movesâ: all other moves
We show using the results from Steps 1 and 2 that there must be a J-move or two vertical face moves between each pair of O-moves in m1,...,my. As a result, we can count the number of moves of each type as follows:
Let cO, cT , cM , cvertical, cJ , and cother be the number of moves of each type. We derive the following constraints:
⢠cO = |O|
⢠cT = 2|T |
⢠cM ⥠3|M |
27
⢠cJ + 1 2 cvertical ⥠|O| â 1
cother ⥠0
Adding these together, we ï¬nd that | 1706.06708#97 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 98 | ⢠cM ⥠3|M |
27
⢠cJ + 1 2 cvertical ⥠|O| â 1
cother ⥠0
Adding these together, we ï¬nd that
1 1 Ki vertical = Co + er + eM + 5 Cvertical + CJ + Cother 2 |O| + 2|T| + 3)M|+ ((O| -â1)=k+|M|.
The above shows that kâ > k, but we also know that kâ < k. Thus, equality must hold at each step. Working out the details, we find that co |O|, er = 2|T|, cs |O| â 1, and CM = Cvertical = Cother = 0. Thus, the counting argument in this step shows that the only moves in m,,...,m other than O-moves and T-moves are the |O| â 1 quantity of J-moves which are between O-moves.
In Step 4, we further restrict the possibilities. In particular, we show the following: | 1706.06708#98 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 99 | In Step 4, we further restrict the possibilities. In particular, we show the following:
⢠Since there are no face moves, the index-(m + i) O-move for i â O can only be a counterclock- wise z turn of slice (m + i). Similarly the index-(m + i) T -moves for i â T are a clockwise z turn and a z ï¬ip of slice (m + i).
⢠Consider the elements i â O in the order in which their O-moves occur. We show that if i1 is immediately before i2 in this order, then it must be the case that li1 diï¬ers from li2 in exactly one bit.
⢠Furthermore, the one J-move between two consecutive O-moves of slices m + i1 and m + i2 must rotate the x slice whose index is the unique index j at which strings li1 and li2 diï¬er. | 1706.06708#99 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 100 | At this point, we are almost done. Consider the elements i â O in the order in which their O-moves occur. The corresponding bitstring li in the same order have the property that each li is at Hamming distance one from the next. In Step 5, we use the ideas of paired stickers to show that T is empty, and as a result conclude that O = {1, . . . , n} and therefore that the above ordering of the lis is an ordering of all the lis in which each li has Hamming distance one from the next. In other words, we show our desired result: that l1, . . . , ln is a âyesâ instance to the Promise Cubical Hamiltonian Path problem.
# 5.5 Step 1: restricting the set of possible index-(m + i) moves
As stated in the proof outline, we will prove the following list of results in this section
⢠Z is empty.
⢠If i â O, then the sole index-(m + i) move must be a counterclockwise z turn.
⢠If i â T , then the two index-(m + i) moves must be a clockwise z turn and a z ï¬ip in some order. | 1706.06708#100 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 101 | ⢠If i â T , then the two index-(m + i) moves must be a clockwise z turn and a z ï¬ip in some order.
⢠If i â O ⪠T , then any move of z slice (m + i) must occur at a time when faces +x, +y, âx, and ây all have zero rotation and any move of z slice â(m + i) must occur at a time when these faces all have rotation 180â¦.
We begin with a preliminary result concerning the coloring of the solved configuration Câ =
We begin with a preliminary result concerning the coloring of the solved configuration Câ = (my o+++0m1)(C}).
28
Lemma 5.13. The solved Rubikâs Cube configuration Câ has the same face colors as Co. | 1706.06708#101 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 102 | 28
Lemma 5.13. The solved Rubikâs Cube configuration Câ has the same face colors as Co.
Proof: Consider the sticker with both coordinates u on any face of Co. No index-u moves occur within my: 0---omy, by definition of u. No index-u moves occur within t = a, 0b, 0---ob, because t is defined entirely using moves of slices whose indices have absolute values at most m+n and u>m-+n. Asa result, the sticker in question is never moved off of the face it starts on by the transformation my o-+-omy,ot. Applying transformation mz o+--0om,0t to Co yields Câ, so the sticker is on the same face in Câ as it is in Co. Since both Co and Câ are solved configurations, we conclude that configuration Câ has the same face colors as Cp.
Using this, we can show the ï¬rst desired result:
# Lemma 5.14. Z is empty.
Proof: Suppose for the sake of contradiction that m1,...,m, contains no index-(m +7) move for some i ⬠{1,...,n}. | 1706.06708#102 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 103 | Proof: Suppose for the sake of contradiction that m1,...,m, contains no index-(m +7) move for some i ⬠{1,...,n}.
Then consider the sticker with coordinates (x, z) = (u,m+7) on the +y face of Cy. Configuration Câ can be obtained from configuration C; by transformation mj, o-+-0m,0a,. We know, however, that moves m1,...,m include no index-(m +7) or index-w moves. Similarly, since a, = (a1) ° ie) (am), we see that a; consists of no index-j moves with 7 > m. Since both m +i and u are greater than m, we can conclude that transformation myo -++o my, 0 a, can be built without any index-(m +i) or index-u moves. As a result, this transformation does not move the sticker in question to a different face.
We then see that the sticker with coordinates (x, z) = (u,m-+i) on the +y face of Cj is also on the +y face of Câ. By Theorem [5.4] we see that the color of this sticker in Cy is red. However, the +y face of Câ is supposed to be the same color as the +y face of Co: green. | 1706.06708#103 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 104 | By contradiction, we see as desired that mj,...,â¢m,/ must contain some index-(m +7) move for alli ⬠{1,...,n}
The rest of what we wish to show concerns index-(m + i) moves where i â O ⪠T . For any i, we can restrict our attention to a speciï¬c set of stickers as in the following deï¬nition:
Deï¬nition 5.15. Deï¬ne the special stickers to be the 48 stickers in Cb with coordinates ±u and ±(m + i) (eight special stickers per face).
Notice that by Theorem 5.4, all but 8 special stickers have the same color as the color of their starting face in C0. This motivates the following further deï¬nition:
Deï¬nition 5.16. Deï¬ne the correctly placed stickers to be the 40 special stickers which have the same color as their starting face has in C0. Deï¬ne the misplaced stickers to be the other 8 special stickers. | 1706.06708#104 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 105 | Of the 8 misplaced stickers, the two on the +y face have the color of the âx face in C0, the two on the âx face have the color of the ây face in C0, the two on the ây face have the color of the +x face in C0, and the two on the +x face have the color of the +y face in C0. In short, starting at Cb, the 8 misplaced stickers must each move one face counterclockwise around the z axis in order to end up on the face whose color in C0 matches the color of the sticker.
Next, consider the effect that move sequence m,,...,7â¢m,/ must have on the special stickers
Lemma 5.17. When starting in configuration Ch, move sequence m,,...,mx must move the mis- placed stickers one face counterclockwise around the z axis and must return each of the correctly placed stickers to the face that sticker started on.
29 | 1706.06708#105 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 106 | 29
Proof: Configuration Câ, which has the same coloring scheme as configuration Co, can be reached rom configuration Cy by applying transformation my o-+-0omy,0a,. Therefore, the 8 misplaced ickers must be moved counterclockwise one face around the z axis and the 40 correctly placed ickers must stay on the same face due to this transformation. Notice that the only moves which ransfer special stickers between faces are index-u and index-(m +7) moves. The only other moves 1at even affect special stickers are face moves. As previously argued, a, contains no index-u moves. In fact, a1 does not contain face moves or index-(m + 7%) moves either and so a; does not move any of the special stickers at all. nH
In other words, the effect of this transformation (my 0 ---0 mj, 0 a 1) on the special stickers is he same as the effect of just the transformation m1,...,m,. Thus, m1,...,m must move the misplaced stickers one face counterclockwise around the z axis and must return each of the correctly placed stickers to the face that sticker started on.
This allows us to directly prove the next two parts of our desired result:
Lemma 5.18. If i â O, then the sole index-(m + i) move must be a counterclockwise z turn. | 1706.06708#106 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 107 | Lemma 5.18. If i â O, then the sole index-(m + i) move must be a counterclockwise z turn.
Proof: Consider the result of move sequence m1,...,â¢z- when starting in configuration Cy. We showed above that the 8 misplaced stickers must each move one face counterclockwise around the z axis and the correctly placed stickers must stay on the same face. Furthermore, the only moves which cause special stickers to change faces are index-u or index-(m +i) moves. Since m1,..., Mx includes no index-u moves and includes exactly one index-(m +i) move (for i ⬠O), we see that the special stickers only change faces during the sole index-(m +7) move in m4,...,â¢gv.
Every slice with index ±(m+i) contains exactly 8 special stickers; therefore the sole index-(m+i) move must cause exactly 8 of the special stickers to change faces. | 1706.06708#107 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 108 | Every slice with index ±(m+i) contains exactly 8 special stickers; therefore the sole index-(m+i) move must cause exactly 8 of the special stickers to change faces.
In order for the 8 misplaced stickers to change faces and for the correctly placed stickers not to, it must be the case that the single index-(m + 7%) move relocates exactly the 8 misplaced stickers. These stickers are on the +x and +y faces. Since the single index-(m + 7) move affects 8 stickers on the +z and +y faces and moves each of these stickers exactly one face counterclockwise around the z axis, it must be the case that this move is a counterclockwise z slice turn. As desired, the sole index-(m + 7%) move is a counterclockwise z turn.
Lemma 5.19. If i â T , then the two index-(m + i) moves must be a clockwise z turn and a z ï¬ip in some order. | 1706.06708#108 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 109 | Proof: As before, consider the result of move sequence m1,..., mg: when starting in configuration Cy. The 8 misplaced stickers must each move one face counterclockwise around the z axis and the correctly placed stickers must stay on the same face. Since the only moves which cause special stickers to change faces are index-u or index-(m + 7%) moves, the only moves among mj,...,â¢x which move special stickers between faces are the two index-(m +7) moves (for i ⬠Tâ).
Notice that every slice with index ±(m + i) contains exactly 8 special stickers, so each of the
two index-(m + i) moves must cause exactly 8 of the special stickers to change faces. We proceed by casework:
⢠If exactly one of the two index-(m + i) moves is an x or y move, then at least one of the correctly placed stickers from the +z face is moved from that face and never returned there. Note that correctly placed stickers are supposed to end up on their starting faces.
⢠If both index-(m + i) moves are x moves, then the misplaced stickers from the +y face never leave that face. Note that misplaced stickers are supposed to move from their starting faces.
30 | 1706.06708#109 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 110 | 30
⢠If both index index-(m + i) moves are y moves, then the misplaced stickers from the +x face never leave that face. Note that misplaced stickers are supposed to move from their starting faces.
⢠If the ï¬rst index-(m + i) move is an x move and the second is a y move, then each of the misplaced stickers from the +x face end up on the ±x or ±z faces. Note that misplaced stickers from the +x face are supposed to move to the +y face.
⢠If the ï¬rst index-(m + i) move is a y move and the second is an x move, then each of the misplaced stickers from the +y face end up on the ±y or ±z faces. Note that misplaced stickers from the +y face are supposed to move to the âx face.
Since all these cases lead to contradiction, we can conclude that the only remaining case holds: both index-(m + i) moves must be z moves. | 1706.06708#110 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 111 | Since all these cases lead to contradiction, we can conclude that the only remaining case holds: both index-(m + i) moves must be z moves.
Next suppose for the sake of contradiction that the 8 special stickers which are moved by one index-(m + i) move are not the same as the special stickers moved by the other index-(m + i) move. Any special sticker moved by exactly one of these moves will change faces and must therefore be a misplaced sticker. That sticker must move one face counterclockwise around the z axis. Since each of the two index-(m + i) moves includes at least one sticker that is not moved by the other index-(m + i) move we can conclude that the two index-(m + i) moves are both counterclockwise z turns. Then any sticker moved by both index-(m + i) moves is moved two faces counterclockwise around the z axis. This is not the desired behavior for any of the special stickers so none of the stickers can be moved by both index-(m + i) moves. Thus there are a total of 16 diï¬erent special stickers, each of which is moved by exactly one of the two index-(m + i) moves. All 16 of these stickers end up on a diï¬erent face from the one they started at. This is a contradiction since there are only 8 misplaced stickers. | 1706.06708#111 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 112 | We conclude that the two moves affect the same 8 stickers. The only way to rotate a tota! of one quarter rotation counterclockwise with two moves is using one clockwise turn and one flip. Thus, as desired, the two index-(m + 7) moves for i ⬠T must be a clockwise z turn and a z flip in some order.
Finally, we have only one thing left to prove in this section:
Lemma 5.20. If i â O ⪠T , then any move of z slice (m + i) must occur at a time when faces +x, +y, âx, and ây all have zero rotation and any move of z slice â(m + i) must occur at a time when faces +x, +y, âx, and ây all have rotation 180â¦. | 1706.06708#112 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 113 | Proof: As before, consider the result of move sequence m1,..., mg: when starting in configuration Cy. The 8 misplaced stickers must each move one face counterclockwise around the z axis and the correctly placed stickers must stay on the same face. The only moves which cause special stickers to change faces are index-u or index-(m + 7%) moves, though face moves also move special stickers. The only moves among m,...,7â¢, which move special stickers between faces are the one or two index-(m-+7) moves (for i ⬠OUT). Furthermore, as shown in the proofs of Lemmas the special stickers which are affected by these moves are exactly the misplaced stickers. In other words, throughout the entire move sequence mj ,..., mxâ, the only moves which affect the correctly placed stickers are the face moves.
Let mj be any index-(m + i) move. Note that according to Lemmas 5.18 and 5.19, mj rotates a z slice.
Consider the six correctly placed stickers on one of the ±x or ±y faces. Since these stickers are only ever aï¬ected by face moves, their coordinates within the face are completely determined by
31 | 1706.06708#113 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 114 | 31
the total rotation of the face so far. If the total rotation so far is 0, then the six correctly placed stickers are in the positions with coordinates tu and +(m +i) and with z 4 (m+ 1). If the total rotation so far is 90°, then the six correctly placed stickers are in the positions with coordinates tu and +(m +i) and with x 4 â(m +i) for the +y faces or y 4 (m+ i) for the +2 faces. If the total rotation so far is 180°, then the six correctly placed stickers are in the positions with coordinates u and +(m +i) and with z 4 â(m-++7). If the total rotation so far is 270°, then the six correctly placed stickers are in the positions with coordinates tu and +(m+i) and with « 4 (m+) for the +y faces or y #4 â(m +7) for the +a faces. | 1706.06708#114 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 115 | The only way for move m, to avoid affecting these stickers if m,; rotates z slice (m+ 4) is for the stickers to be in the positions with z 4 (m +i). In other words, the total rotation of the face mus be 0. The only way for move m,; to avoid affecting these stickers if m; rotates z slice â(m + 7%) is for the stickers to be in the positions with z 4 â(m +i). In other words, the total rotation of the face must be 180°. Note that this logic applies to each of the +a and +y faces. In other words, if mz, is some move with index (m+), then each of the +a and ty faces must have rotation 0 and if mj, is some move with index â(m +â), then each of the +a and +y faces must have rotation 180°.
# 5.6 Step 2: exploring properties of paired stickers
As stated in the proof outline, this step of the proof explores the properties of paired stickers.
Lemma 5.21. If two stickers are (p1, p2, q)-paired, then they remain (p1, p2, q)-paired after one move unless the move is an index-p1 move or an index-p2 move which moves one of the stickers. | 1706.06708#115 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 116 | Proof: Consider the eï¬ect of any move on the two stickers.
If the move doesnât aï¬ect either sticker, then the two stickers maintain their coordinates (and therefore also stay on the same face quadrant and slice). Thus the two stickers remain (p1, p2, q)- paired.
If the move moves both stickers, then they both rotate the same amount. In other words, as far as those two stickers are concerned, the eï¬ect of the move is the same as the eï¬ect of rotating the entire Rubikâs Cube. When rotating the Rubikâs Cube, two stickers sharing a slice continue to share a slice, two stickers sharing a face quadrant continue to share a face quadrant, and each sticker maintains the same set of coordinate absolute values as it had before. Thus, the two stickers remain (p1, p2, q)-paired. | 1706.06708#116 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 117 | Clearly, the only way for the stickers to no longer be (p1, p2, q)-paired is for the move to affec exactly one of the stickers. The possible moves affecting the stickers are face moves, index-q moves, index-p; moves, and index-pg moves. Among these, face moves and index-q moves necessarily affect either both stickers in the pair or neither. Thus, the only way for the stickers to stop being (pi, p2, 7)-paired is via an index-p; move or an index-py move which moves one of the stickers.
Using this, we prove the following lemma:
Lemma 5.22. Suppose i1,i2 ⬠O and j ⬠{1,2,...,m}. Then consider any pair of stickers that are (m + i1,m + ig, j)-paired in Cy. If there are no face moves of faces +x, +y, âx, and ây and no index-j moves that affect either of the stickers between the index-(m + %i,) O-move and the index-(m + iz) O-move, then the two stickers remain (m+ i1,m +4 i2,j)-paired in Câ. | 1706.06708#117 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 118 | Proof: Consider two (m + i1, m + i2, j)-paired stickers in Cb. Suppose that there exists neither an index-j move aï¬ecting one of the stickers nor a face move of face +x, +y, âx, or ây between the
32
index-(m + i1) and index-(m + i2) O-moves. Let mα be the index-(m + i1) O-move and let mβ be the index-(m + i2) O-move. Without loss of generality, suppose mα occurs before mβ.
Since there are no +x, +y, âx, or ây face moves between mα and mβ, we know that the rotations of these faces remain the same at the times of both moves. Applying the results from Step 1, either mα and mβ are both counterclockwise turns of z slices (m + i1) and (m + i2) or mα and mβ are counterclockwise turns of z slices â(m + i1) and â(m + i2). | 1706.06708#118 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 119 | Configuration Câ can be obtained from configuration C, by applying transformation my, 0 ---o m 20m ,0a,. Since a; consists of some number of x-slice turns, we can represent this transformation as a sequence of moves. We know that since the stickers are (m+ i,m + ig, j)-paired in Cy, they must remain (m + i1,m + ig, j)-paired until immediately before the first index-(m + i1) or index- (m+ ig) move: mq. We will show below that because of our assumption, the stickers will also end up (m+ i1,m + ig, j)-paired immediately after mg in all cases. | 1706.06708#119 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 120 | The first case is that the stickers are on face +z or face âz immediately before mq. In that case, move â¢Mq, which is a z move, will not affect either sticker. As a result, the two stickers will remain (m+i1,m-+i2, j)-paired after ma. With the exception of mq and mg, the only moves in m1, ..., Mx which move these two stickers between faces are index-7 moves. But by assumption, there are no index-j moves occurring between mg and mg which affect the stickers. Thus, immediately before mg, the two stickers will still be (m+i1,m-+ ig, j)-paired and will still be on face +z or face âz. As a result, mg will also not affect the stickers. Therefore, they will remain (m + i;,m + ig, j)-paired immediately after mg. | 1706.06708#120 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 121 | The second case is that the stickers are on face +x, +y, âx, or ây immediately before ma. Between mg and mg, the only moves in m,...,â¢m,- which move these two stickers are face moves and index-j moves. No matter how mq affects the two stickers, they will both remain on the four faces +aâ, +y, âx, and ây. By assumption, neither sticker will be moved by an index-j move until mg. Since the stickers are on faces +a, +y, âx, or ây, the assumption tells us that neither sticker will be moved by a face move until mg either. Thus, the next move after mq which affects either sticker is mg. | 1706.06708#121 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 122 | Note that immediately before mα, the ï¬rst sticker has z coordinate (m + i1) if and only if the second sticker has z coordinate (m + i2). Similarly, the ï¬rst sticker has z coordinate â(m + i1) if and only if the second sticker has z coordinate â(m + i2). This is simply a consequence of the deï¬nition of (m + i1, m + i2, j)-paired stickers. We know that mα and mβ together either rotate z slices (m + i1) and (m + i2) counterclockwise one turn or rotate z slices â(m + i1) and â(m + i2) counterclockwise one turn. Thus in any case we see that over the course of the moves from mα to mβ, either both stickers are rotated counterclockwise one turn around the z axis or neither is. As far as the two stickers are concerned, that is equivalent to a rotation of the entire Rubikâs Cube. That means that in this case as well, the two stickers remain (m + i1, m + i2, j)-paired immediately after mβ. | 1706.06708#122 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 123 | We see that in both cases the two stickers remain (m+ i1,m + ig, j)-paired immediately after mg. Since there are no index-(m + i) or index-(m + i2) moves after mg, we know that the two stickers will continue to be (m+ i1,m + i2, j)-paired until Câ.
# 5.7 Step 3: classifying possible moves with a counting argument
As stated in the proof outline, this step uses a counting argument to restrict the possible moves in M4, .++, Mgr.
To begin, we show the following:
33
Lemma 5.23. There must be a J-move or two vertical face moves between each pair of O-moves in M1,..., MR.
Proof: Consider any pair of O-moves mα and mβ which occur in that order. Suppose mα is an index-(m + i1) move and mβ is an index-(m + i2) move. Let j be an index such that (li1)j diï¬ers from (li2)j. | 1706.06708#123 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 124 | Notice that the (m+i1,m+i2, j)-paired stickers on face +y with (x, z) coordinates (j,m+i,) and (j,m+i2) have different colors in Cy (see Theorem . Therefore they cannot be (m+i1,m+i2, j)- paired in Câ. By the contraposative of Lemma|5 we see that at least one index-j move affecting one of these stickers or at least one face move of faces +xâ, +y, â2, or ây must occur between ma and mg.
know
We know from the results of Step 1 that at the times of mα and mβ, the four faces ±x and ±y must either each have total rotation of 0 or total rotation of 180â¦. Thus between the two moves, either the rotations of all four faces must change, or the rotation of any face that changes must also change back. Therefore it is impossible for exactly one face move of faces +x, +y, âx, and ây to occur between these two moves.
In other words, we have shown that at least one J-move or at least two vertical face moves must occur between mg and mg.
# Corollary 5.24. If | 1706.06708#124 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 125 | In other words, we have shown that at least one J-move or at least two vertical face moves must occur between mg and mg.
# Corollary 5.24. If
⢠mα and mβ are index-(m + i1) and index-(m + i2) O-moves,
⢠li1 and li2 diï¬er in bit j, and
⢠there are no vertical face moves between mα and mβ,
then there must be an index-j J-move between mα and mβ.
Proof: This follows directly from one of the cases in the previous proof.
With that done, we can count the number of moves of each type as follows: There is exactly one O-move for each i â O (the sole index-(m + i) move), so therefore cO = |O|. There are exactly two T -move for each i â T (the two index-(m + i) moves), so therefore
There is exactly one O-move for each i ⬠O (the sole index-(m+i) move), so therefore co = |O}. There are exactly two T-move for each i ⬠T (the two index-(m + i) moves), so therefore cr = 2\TI.
cT = 2|T |. | 1706.06708#125 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 126 | cT = 2|T |.
There are at least three M -moves for each i â M (the index-(m + i) moves), so therefore cM ⥠3|M |.
Consider the O-moves in order. Between the cO = |O| diï¬erent O-moves there are |O| â 1 gaps. As shown above, each such gap must contain either at least one J-move or at least two vertical face moves. Therefore the number of J-moves plus half the number of vertical face moves upper-bounds the number of gaps: cJ + 1 Finally, cother ⥠0.
34
Putting this together, we see the following:
J khâ =co +er +m + Cvertical + CJ + Cother 1 1 Cor Cr + CM («. t 9 vertical + 9 vertical + Cother 1 > |O| + 2|T| + 3|M| + (jO| â1)+ 3 (vertical 1 2|0 t 2|T| t 3|M| 1+ 9 vertical 1 =2n-1+ |M| + 9 vertical 1 =k+ |M| + 9 vertical >k | 1706.06708#126 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 127 | The above shows that kâ > k, but we also know that kâ < k. Thus, equality must hold at each step. In particular, cag must equal 3|M|, (ey + 4d Cvertical) must equal |O| â 1, Cother must equal 0, and |M|+ $Cvertical must equal 0.
2 cvertical = 0, we can conclude that both |M | and cvertical are equal to 0. Thus cM = 3|M | = 0 also holds. All together, this shows that cO = |O|, cT = 2|T |, cJ = |O| â 1, and cM = cvertical = cother = 0.
# 5.8 Step 4: further restricting possible move types
As stated in the proof outline, we will prove the following list of results in this section:
⢠Since there are no face moves, the index-(m + i) O-move for i â O can only be a counterclock- wise z turn of slice (m + i). Similarly the index-(m + i) T -moves for i â T are a clockwise z turn and a z ï¬ip of slice (m + i). | 1706.06708#127 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 128 | ⢠Consider the elements i â O in the order in which their O-moves occur. We show that if i1 is immediately before i2 in this order, then it must be the case that li1 diï¬ers from li2 in exactly one bit.
⢠Furthermore, the one J-move between two consecutive O-moves of slices m + i1 and m + i2 must rotate the x slice whose index is the unique index j at which strings li1 and li2 diï¬er.
Lemma 5.25. [fi ⬠O, the single index-(m +i) move in m1,...,m is a counterclockwise z turn of slice (m +7).
Proof: We have already seen that the move in question must be either a counterclockwise z turn of slice (m +7) or a counterclockwise z turn of slice â(m +7). Furthermore, the slice being rotated is slice (m +7) if at the time of the move each vertical face (Ax and +y) has the total rotation 0. We have already seen, however, that none of the moves in mj,...,â¢m, are face moves. Thus the total rotation of each face is always 0, and as desired, the move in question is a counterclockwise z turn of slice (m+ i). | 1706.06708#128 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 129 | Lemma 5.26. [fi ⬠T, the two inderz-(m+i) moves in m,...,mg are a clockwise z turn of slice (m+ 1) and a z flip of slice (m +i).
35
Proof: This proof follows analagously to the previous.
Lemma 5.27. Suppose that mα and mβ are two O-moves of slices (m + i1) and (m + i2) with no other O-moves between them. It must be the case that li1 diï¬ers from li2 in exactly one bit.
Proof: We have already seen that there must be at least one J-move between mα and mβ. In fact, there has to be exactly one J-move in each of the |O| â 1 âgapsâ between O-moves, so there can only be one J-move between mα and mβ.
We saw in Corollary however, that if Jj, and 1;, differ in bit 7, then there must be an index-j J-move between mq and mg. As desired, we conclude that J;, and 1;,, must differ in at most one bit j. Since the bitstrings are all distinct, this is exactly what we were trying to show. | 1706.06708#129 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 130 | Lemma 5.28. Suppose that mα and mβ are two O-moves of slices (m + i1) and (m + i2) with no other O-moves between them. If li1 diï¬ers from li2 in bit j, then it must be the case that the one J-move between mα and mβ must rotate the x slice with index j.
Proof: We know that the J-move in question must rotate a slice with index ±j. We want to show that the move rotates the x slice with index j in particular.
Consider the pair of stickers in Cb at (x, z) coordinates (j, m + i1) and (j, m + i2) on the +y face and also the pair of stickers in Cb at (x, y) coordinates (j, â(m + i1)) and (j, â(m + i2)) on the +z face. These two pairs of stickers are both (m + i1, m + i2, j)-paired. Furthermore, each of these two pairs contain stickers of two diï¬erent colors (see Theorem 5.4). | 1706.06708#130 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 131 | To transition from C;, to Câ, we apply transformation m,0---o0m 0a}. In other words, we apply a sequence of moves starting with some number of x turns (making up a1) and then proceeding hrough move sequence m1,...,7â¢mx. Because the solution contains no face moves, the only moves in this list before m, which affect the four stickers in question are rotations of the x slice with index j. No matter how much or how little this slice rotates, one of the two pairs of stickers will ye on face +z or âz. | 1706.06708#131 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 132 | Consider that pair. Move mq will be a counterclockwise z turn and therefore will not affect either sticker in the pair. That pair of stickers cannot be (m+%1,m + ig, j)-paired in Câ since they 1ave different colors. Since mg is the only other index-(m + 71) or index-(m + ig) move, we can conclude from Lemma [5.21] thaâ one of the two stickers must be affected by mg. In order for that o be the case, however, the sole J-move between mq and mg must move the stickers in this pair off of the +z face. Notice that the J-move between mg and mg must rotate a slice with index +). Since there are no face moves in the solution, the only option which meets the requirements is to aave the J-move rotate the x slice with index j.
# 5.9 Step 5: showing T is empty
As stated in the proof outline, the purpose of this step is to show that T is empty. That on its own is suï¬cient to complete the proof. | 1706.06708#132 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 133 | As stated in the proof outline, the purpose of this step is to show that T is empty. That on its own is suï¬cient to complete the proof.
Lemma 5.29. When applying the move sequence a1, m1, . . . , mk to Cb, the stickers with z = i and 1 ⤠x ⤠n of face +y for i â O immediately after the O-move of slice (m + i) are the ones which started in the corresponding positions z = i and 1 ⤠ây ⤠n of the face +x in Cb.
Proof: Let mα be the O-move of slice (m + i).
36
Consider the stickers in positions z = i and 1 ⤠x ⤠n of face +y for i â O immediately after the move mα. These stickers were moved there by mα from positions z = i and 1 ⤠ây ⤠n of the face +x. | 1706.06708#133 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 134 | All O-moves and T -moves prior to mα aï¬ect z slices whose indices are not i. All J-moves and all moves comprising a1 aï¬ect non-face x slices and therefore donât aï¬ect the +x face. As a result, no move in a1, m1, . . . , mk before mα aï¬ects the stickers with z = i and ân ⤠y ⤠â1 on the +x face. Thus, the stickers in positions z = i and 1 ⤠ây ⤠n of the face +x immediately before mα are the same as the stickers in those positions in conï¬guration Cb.
As desired, the stickers with z = i and 1 < x < n of face +y for i ⬠O immediately after the move mq are the ones which started in the corresponding positions z = i and 1 < ây < n of the ace +x in Cp. | 1706.06708#134 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 135 | Lemma 5.30. When applying the move sequence a1, m1, . . . , mk to Cb, the stickers with z = i and 1 ⤠x ⤠n of face +y for i â T after the second T -move rotating a slice with index (m + i) are the ones which started in the corresponding positions z = i and 1 ⤠ây ⤠n of the face +x in Cb.
Proof: Let mα and mβ be the two T -moves of slice (m + i).
Consider the stickers in positions z = i and 1 ⤠x ⤠n of face +y immediately after mβ. These stickers were moved there by mβ either from positions z = i and 1 ⤠y ⤠n of the face âx or from positions z = i and 1 ⤠âx ⤠n of face ây (depending on whether the second T -move is the turn or the ï¬ip). | 1706.06708#135 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 136 | In either case, none of the moves between mα and mβ could have aï¬ected any of these stickers (since the moves in that interval are all either O- or T - moves moving z slices of other indices or J-moves moving x slices with indices 1 through n). Therefore immediately before mα, these stickers were in positions z = i and 1 ⤠ây ⤠n of the face +x. Once again, no moves before that could aï¬ect these stickers, so these stickers must have started in that position in Cb.
As desired, the stickers with z = 1 and 1 < x < n of face +y immediately after the move mg are the ones which started in the corresponding positions z = 7 and 1 < ây < n of the face +a in Cp.
# Theorem 5.31. T is empty. | 1706.06708#136 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
1706.06708 | 137 | # Theorem 5.31. T is empty.
Proof: Note that O cannot be empty since then the number of J-moves would be |O| â 1 = â1. Suppose for the sake of contradiction that i1 â T . Consider the second T -move of the z slice with index (m + i1) in move sequence a1, m1, . . . , mk. Call this move mα. The move mα cannot be seperated from every O-move by J-moves because if that were the case, there would be two J-moves without an O-move between them (or in other words there would be two O-moves with at least two J-moves between them). Thus there must be some O-move mβ of slice (m + i2) that is not seperated from mα by any J-move. | 1706.06708#137 | Solving the Rubik's Cube Optimally is NP-complete | In this paper, we prove that optimally solving an $n \times n \times n$
Rubik's Cube is NP-complete by reducing from the Hamiltonian Cycle problem in
square grid graphs. This improves the previous result that optimally solving an
$n \times n \times n$ Rubik's Cube with missing stickers is NP-complete. We
prove this result first for the simpler case of the Rubik's Square---an $n
\times n \times 1$ generalization of the Rubik's Cube---and then proceed with a
similar but more complicated proof for the Rubik's Cube case. | http://arxiv.org/pdf/1706.06708 | Erik D. Demaine, Sarah Eisenstat, Mikhail Rudoy | cs.CC, cs.CG, math.CO, F.1.3 | 35 pages, 8 figures | null | cs.CC | 20170621 | 20180427 | [] |
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