



"""simple docstring"""







# This Python program implements an optimal binary search tree (abbreviated BST)
# building dynamic programming algorithm that delivers O(n^2) performance.
#
# The goal of the optimal BST problem is to build a low-cost BST for a
# given set of nodes, each with its own key and frequency. The frequency
# of the node is defined as how many time the node is being searched.
# The search cost of binary search tree is given by this formula:
#
# cost(1, n) = sum{i = 1 to n}((depth(node_i) + 1) * node_i_freq)
#
# where n is number of nodes in the BST. The characteristic of low-cost
# BSTs is having a faster overall search time than other implementations.
# The reason for their fast search time is that the nodes with high
# frequencies will be placed near the root of the tree while the nodes
# with low frequencies will be placed near the leaves of the tree thus
# reducing search time in the most frequent instances.
import sys
from random import randint





class    UpperCamelCase     :




	"""simple docstring"""






	def __init__(			self							,UpperCAmelCase_							,UpperCAmelCase_					):
			_lowercase							:      Any			=					key
			_lowercase							:      int			=					freq






	def __str__(			self					):
			return f"""Node(key={self.key}, freq={self.freq})"""







def       __SCREAMING_SNAKE_CASE      (   __UpperCAmelCase					,							__UpperCAmelCase					,							__UpperCAmelCase					,							__UpperCAmelCase					,							__UpperCAmelCase					,							__UpperCAmelCase	):
		if i > j or i < 0 or j > len(__UpperCAmelCase	) - 1:
				return

		_lowercase							:      List[Any]			=					root[i][j]
		if parent == -1:  # root does not have a parent
				print(F"""{key[node]} is the root of the binary search tree."""	)
		elif is_left:
				print(F"""{key[node]} is the left child of key {parent}."""	)
		else:
				print(F"""{key[node]} is the right child of key {parent}."""	)

		print_binary_search_tree(__UpperCAmelCase					,							__UpperCAmelCase					,							__UpperCAmelCase					,							node - 1					,							key[node]					,							__UpperCAmelCase	)
		print_binary_search_tree(__UpperCAmelCase					,							__UpperCAmelCase					,							node + 1					,							__UpperCAmelCase					,							key[node]					,							__UpperCAmelCase	)


def       __SCREAMING_SNAKE_CASE      (   __UpperCAmelCase	):
		nodes.sort(key=lambda __UpperCAmelCase	: node.key	)

		_lowercase							:      int			=					len(__UpperCAmelCase	)

		_lowercase							:      List[str]			=					[nodes[i].key for i in range(__UpperCAmelCase	)]
		_lowercase							:      List[str]			=					[nodes[i].freq for i in range(__UpperCAmelCase	)]

		# This 2D array stores the overall tree cost (which's as minimized as possible);
		# for a single key, cost is equal to frequency of the key.
		_lowercase							:      str			=					[[freqs[i] if i == j else 0 for j in range(__UpperCAmelCase	)] for i in range(__UpperCAmelCase	)]
		# sum[i][j] stores the sum of key frequencies between i and j inclusive in nodes
		# array
		_lowercase							:      str			=					[[freqs[i] if i == j else 0 for j in range(__UpperCAmelCase	)] for i in range(__UpperCAmelCase	)]
		# stores tree roots that will be used later for constructing binary search tree
		_lowercase							:      List[str]			=					[[i if i == j else 0 for j in range(__UpperCAmelCase	)] for i in range(__UpperCAmelCase	)]

		for interval_length in range(2					,							n + 1	):
				for i in range(n - interval_length + 1	):
						_lowercase							:      List[Any]			=					i + interval_length - 1

						_lowercase							:      int			=					sys.maxsize  # set the value to "infinity"
						_lowercase							:      Optional[Any]			=					total[i][j - 1] + freqs[j]

						# Apply Knuth's optimization
						# Loop without optimization: for r in range(i, j + 1):
						for r in range(root[i][j - 1]					,							root[i + 1][j] + 1	):  # r is a temporal root
								_lowercase							:      List[str]			=					dp[i][r - 1] if r != i else 0  # optimal cost for left subtree
								_lowercase							:      int			=					dp[r + 1][j] if r != j else 0  # optimal cost for right subtree
								_lowercase							:      Tuple			=					left + total[i][j] + right

								if dp[i][j] > cost:
										_lowercase							:      Union[str, Any]			=					cost
										_lowercase							:      Tuple			=					r

		print("""Binary search tree nodes:"""	)
		for node in nodes:
				print(__UpperCAmelCase	)

		print(F"""\nThe cost of optimal BST for given tree nodes is {dp[0][n - 1]}."""	)
		print_binary_search_tree(__UpperCAmelCase					,							__UpperCAmelCase					,							0					,							n - 1					,							-1					,							__UpperCAmelCase	)


def       __SCREAMING_SNAKE_CASE      (   ):
		# A sample binary search tree
		_lowercase							:      int			=					[Node(__UpperCAmelCase					,							randint(1					,							50	)	) for i in range(10					,							0					,							-1	)]
		find_optimal_binary_search_tree(__UpperCAmelCase	)


if __name__ == "__main__":
		main()








from .chunk_utils import chunk_layer
from .data_transforms import make_atomaa_masks
from .feats import atomaa_to_atomaa, frames_and_literature_positions_to_atomaa_pos, torsion_angles_to_frames
from .loss import compute_predicted_aligned_error, compute_tm
from .protein import Protein as OFProtein
from .protein import to_pdb
from .rigid_utils import Rigid, Rotation
from .tensor_utils import dict_multimap, flatten_final_dims, permute_final_dims





