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Ignore `A003` Co-authored-by: Christian Clauss <cclauss@me.com> Co-authored-by: pre-commit-ci[bot] <66853113+pre-commit-ci[bot]@users.noreply.github.com> Co-authored-by: Dhruv Manilawala <dhruvmanila@gmail.com>
145 lines
5.3 KiB
Python
145 lines
5.3 KiB
Python
#!/usr/bin/env python3
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# This Python program implements an optimal binary search tree (abbreviated BST)
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# building dynamic programming algorithm that delivers O(n^2) performance.
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#
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# The goal of the optimal BST problem is to build a low-cost BST for a
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# given set of nodes, each with its own key and frequency. The frequency
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# of the node is defined as how many time the node is being searched.
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# The search cost of binary search tree is given by this formula:
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#
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# cost(1, n) = sum{i = 1 to n}((depth(node_i) + 1) * node_i_freq)
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#
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# where n is number of nodes in the BST. The characteristic of low-cost
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# BSTs is having a faster overall search time than other implementations.
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# The reason for their fast search time is that the nodes with high
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# frequencies will be placed near the root of the tree while the nodes
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# with low frequencies will be placed near the leaves of the tree thus
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# reducing search time in the most frequent instances.
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import sys
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from random import randint
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class Node:
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"""Binary Search Tree Node"""
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def __init__(self, key, freq):
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self.key = key
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self.freq = freq
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def __str__(self):
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"""
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>>> str(Node(1, 2))
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'Node(key=1, freq=2)'
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"""
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return f"Node(key={self.key}, freq={self.freq})"
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def print_binary_search_tree(root, key, i, j, parent, is_left):
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"""
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Recursive function to print a BST from a root table.
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>>> key = [3, 8, 9, 10, 17, 21]
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>>> root = [[0, 1, 1, 1, 1, 1], [0, 1, 1, 1, 1, 3], [0, 0, 2, 3, 3, 3], \
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[0, 0, 0, 3, 3, 3], [0, 0, 0, 0, 4, 5], [0, 0, 0, 0, 0, 5]]
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>>> print_binary_search_tree(root, key, 0, 5, -1, False)
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8 is the root of the binary search tree.
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3 is the left child of key 8.
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10 is the right child of key 8.
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9 is the left child of key 10.
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21 is the right child of key 10.
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17 is the left child of key 21.
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"""
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if i > j or i < 0 or j > len(root) - 1:
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return
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node = root[i][j]
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if parent == -1: # root does not have a parent
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print(f"{key[node]} is the root of the binary search tree.")
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elif is_left:
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print(f"{key[node]} is the left child of key {parent}.")
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else:
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print(f"{key[node]} is the right child of key {parent}.")
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print_binary_search_tree(root, key, i, node - 1, key[node], True)
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print_binary_search_tree(root, key, node + 1, j, key[node], False)
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def find_optimal_binary_search_tree(nodes):
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"""
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This function calculates and prints the optimal binary search tree.
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The dynamic programming algorithm below runs in O(n^2) time.
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Implemented from CLRS (Introduction to Algorithms) book.
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https://en.wikipedia.org/wiki/Introduction_to_Algorithms
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>>> find_optimal_binary_search_tree([Node(12, 8), Node(10, 34), Node(20, 50), \
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Node(42, 3), Node(25, 40), Node(37, 30)])
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Binary search tree nodes:
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Node(key=10, freq=34)
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Node(key=12, freq=8)
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Node(key=20, freq=50)
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Node(key=25, freq=40)
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Node(key=37, freq=30)
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Node(key=42, freq=3)
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<BLANKLINE>
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The cost of optimal BST for given tree nodes is 324.
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20 is the root of the binary search tree.
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10 is the left child of key 20.
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12 is the right child of key 10.
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25 is the right child of key 20.
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37 is the right child of key 25.
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42 is the right child of key 37.
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"""
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# Tree nodes must be sorted first, the code below sorts the keys in
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# increasing order and rearrange its frequencies accordingly.
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nodes.sort(key=lambda node: node.key)
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n = len(nodes)
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keys = [nodes[i].key for i in range(n)]
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freqs = [nodes[i].freq for i in range(n)]
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# This 2D array stores the overall tree cost (which's as minimized as possible);
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# for a single key, cost is equal to frequency of the key.
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dp = [[freqs[i] if i == j else 0 for j in range(n)] for i in range(n)]
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# sum[i][j] stores the sum of key frequencies between i and j inclusive in nodes
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# array
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total = [[freqs[i] if i == j else 0 for j in range(n)] for i in range(n)]
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# stores tree roots that will be used later for constructing binary search tree
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root = [[i if i == j else 0 for j in range(n)] for i in range(n)]
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for interval_length in range(2, n + 1):
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for i in range(n - interval_length + 1):
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j = i + interval_length - 1
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dp[i][j] = sys.maxsize # set the value to "infinity"
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total[i][j] = total[i][j - 1] + freqs[j]
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# Apply Knuth's optimization
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# Loop without optimization: for r in range(i, j + 1):
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for r in range(root[i][j - 1], root[i + 1][j] + 1): # r is a temporal root
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left = dp[i][r - 1] if r != i else 0 # optimal cost for left subtree
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right = dp[r + 1][j] if r != j else 0 # optimal cost for right subtree
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cost = left + total[i][j] + right
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if dp[i][j] > cost:
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dp[i][j] = cost
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root[i][j] = r
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print("Binary search tree nodes:")
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for node in nodes:
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print(node)
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print(f"\nThe cost of optimal BST for given tree nodes is {dp[0][n - 1]}.")
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print_binary_search_tree(root, keys, 0, n - 1, -1, False)
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def main():
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# A sample binary search tree
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nodes = [Node(i, randint(1, 50)) for i in range(10, 0, -1)]
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find_optimal_binary_search_tree(nodes)
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if __name__ == "__main__":
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main()
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