BackgroundA binary decision tree $T$ is a rooted tree where each internal node (and root) is labeled by an index $j \in \{1,..., n\}$ such that no path from root to leaf repeats an index, the leafs are labeled by outputs in $\{A,B\}$, and each edge is labeled by $0$ for the left child and $1$ for the right child. To apply a tree to an input $x$: Start at the rootif you are at leaf, you output the leaf label $A$ or $B$ and terminateRead the label $j$ of your current node, if $x_j = 0$ then move to the left child and if $x_j = 1$ then move to the right child.jump to step (2)The tree is used as a way to evaluate a functions, in particular we say a tree $T$ represents a total function $f$ if for each $x \in \{0,1\}^n$ we have $T(x) = f(x)$. The query complexity of a tree is its depth, and the query complexity of a function is the depth of the smallest tree that represents it.ProblemGiven a binary decision tree T output a binary decision tree T' of minimal depth such that T and T' represent the same function.QuestionWhat is the best known algorithm for this? Are any lower bounds known? What if we know that the $\text{depth}(T') = O(\log \text{depth}(T))$? What about if we only require $T'$ to be of approximately minimal depth?Naive approachThe naive approach is given $d = \text{depth}(T)$ to recursively enumerate all binary decision trees of depth $d - 1$ while testing if they evaluate to the same thing as $T$. This seems to require $O(\frac{d 2^n n!}{(n - d)!})$ steps (assuming that it takes $d$ steps to check what $T(x)$ evaluates to for an arbitrary $x$). Is there a better approach?MotivationThis question is motivated by a previous question on the trade off between query complexity and time complexity. In particular, the goal is to bound the time separation for total functions. We can make a tree $T$ from a time optimal algorithm with runtime $t$, and then we would like to convert it to a tree $T'$ for a query optimal algorithm. Unfortunately, if $t \in O(n!/(n - d)!)$ (and often $d \in \Theta(n)$) the bottleneck is the conversion. It would be nice if we could replace $n!/(n - d)!$ by something like $2^d$.
Algorithm for optimizing decision trees
ds.algorithms;query complexity;decision trees
I have 3 answers, all giving somewhat different hardness results.Let $f: \{0,1\}^n \rightarrow \{0,1\}$ be some function.Answer 1Given a decision tree $T$ computing $f$ and a number, it is NP-hard to tell if there exists a decision tree $T'$ computing $f$ of size at most that number. (Zantema and Bodlaender '00)Answer 2Given a decision tree $T$ computing $f$, it is NP hard to approximate the smallest decision tree computing $f$ to any constant factor. (Sieling '08)Answer 3Let $s$ be the size of the smallest decision tree computing $f$.Given a decision tree $T$ computing $f$, assuming $NP \subsetneq DTIME(2^{n^\epsilon})$ for some $\epsilon < 1$, one cannot find an equivalent decision tree $T'$ of size $s^k$ for any $k \ge 0$.I think that this stronger answer (relying on a weaker assumption) can be made from known results in the learning theory of Occam algorithms for decision trees, via the following argument:  Is it possible to find a decision tree on $n$ variables in time$n^{\log s}$, where $s$ is the smallest decision tree consistent with examples coming from a distribution (PAC model). (Blum '92)  Assuming $NP \subsetneq DTIME(2^{n^\epsilon})$ for some $\epsilon < 1$, we cannot PAC learn size $s$ decision trees by size $s^k$ decision trees for any $k \ge 0$. (Alekhnovich et al. '07)These two results seem to imply a hardness result for your problem. On the one hand (1), we can find a large decision tree; on the other hand (2), we shouldn't be able to minimize it to get an equivalent small one, of size $s^k$, even when one exists of size $s$.