Update

divide_and_conquer/convex_hull.py

@@ -7,7 +7,7 @@
 
 Two algorithms have been implemented for the convex hull problem here.
 1. A brute-force algorithm which runs in O(n^3)
-2. A divide-and-conquer algorithm which runs in O(n^3)
+2. A divide-and-conquer algorithm which runs in O(n log(n))
 
 There are other several other algorithms for the convex hull problem
 which have not been implemented here, yet.

dynamic_programming/longest_common_subsequence.py

@@ -50,7 +50,7 @@ def longest_common_subsequence(x: str, y: str):
 
     seq = ""
     i, j = m, n
-    while i > 0 and i > 0:
+    while i > 0 and j > 0:
         if x[i - 1] == y[j - 1]:
             match = 1
         else:

dynamic_programming/rod_cutting.py

@@ -1,57 +1,193 @@
-from typing import List
+"""
+This module provides two implementations for the rod-cutting problem:
+1. A naive recursive implementation which has an exponential runtime
+2. Two dynamic programming implementations which have quadratic runtime
 
-def rod_cutting(prices: List[int],length: int) -> int:
+The rod-cutting problem is the problem of finding the maximum possible revenue
+obtainable from a rod of length ``n`` given a list of prices for each integral piece
+of the rod. The maximum revenue can thus be obtained by cutting the rod and selling the
+pieces separately or not cutting it at all if the price of it is the maximum obtainable.
+
+"""
+
+
+def naive_cut_rod_recursive(n: int, prices: list):
+	"""
+	Solves the rod-cutting problem via naively without using the benefit of dynamic programming.
+	The results is the same sub-problems are solved several times leading to an exponential runtime
+
+	Runtime: O(2^n)
+
+	Arguments
+	-------
+	n: int, the length of the rod
+	prices: list, the prices for each piece of rod. ``p[i-i]`` is the
+	price for a rod of length ``i``
+
+	Returns
+	-------
+	The maximum revenue obtainable for a rod of length n given the list of prices for each piece.
+
+	Examples
+	--------
+	>>> naive_cut_rod_recursive(4, [1, 5, 8, 9])
+	10
+	>>> naive_cut_rod_recursive(10, [1, 5, 8, 9, 10, 17, 17, 20, 24, 30])
+	30
+	"""
+
+	_enforce_args(n, prices)
+	if n == 0:
+		return 0
+	max_revue = float("-inf")
+	for i in range(1, n + 1):
+		max_revue = max(max_revue, prices[i - 1] + naive_cut_rod_recursive(n - i, prices))
+
+	return max_revue
+
+
+def top_down_cut_rod(n: int, prices: list):
 	"""
-	Given a rod of length n and array of prices that indicate price at each length.
-	Determine the maximum value obtainable by cutting up the rod and selling the pieces
-
-	>>> rod_cutting([1,5,8,9],4) 
+	Constructs a top-down dynamic programming solution for the rod-cutting problem
+	via memoization. This function serves as a wrapper for _top_down_cut_rod_recursive
+
+	Runtime: O(n^2)
+
+	Arguments
+	--------
+	n: int, the length of the rod
+	prices: list, the prices for each piece of rod. ``p[i-i]`` is the
+	price for a rod of length ``i``
+
+	Note
+	----
+	For convenience and because Python's lists using 0-indexing, length(max_rev) = n + 1,
+	to accommodate for the revenue obtainable from a rod of length 0.
+
+	Returns
+	-------
+	The maximum revenue obtainable for a rod of length n given the list of prices for each piece.
+
+	Examples
+	-------
+	>>> top_down_cut_rod(4, [1, 5, 8, 9])
 	10
-	>>> rod_cutting([1,1,1],3) 
-	3
-	>>> rod_cutting([1,2,3], -1)
-	Traceback (most recent call last):
-   	ValueError: Given integer must be greater than 1, not -1
-   	>>> rod_cutting([1,2,3], 3.2)
-	Traceback (most recent call last):
-   	TypeError: Must be int, not float
-   	>>> rod_cutting([], 3)
-	Traceback (most recent call last):
-   	AssertionError: prices list is shorted than length: 3
-
-
-
-	Args:
-		prices: list indicating price at each length, where prices[0] = 0 indicating rod of zero length has no value
-	    length: length of rod
-
-	Returns:
-	    Maximum revenue attainable by cutting up the rod in any way. 
-	"""
-
-	prices.insert(0, 0)
-	if not isinstance(length, int):
-	    raise TypeError('Must be int, not {0}'.format(type(length).__name__))
-	if length < 0: 
-	    raise ValueError('Given integer must be greater than 1, not {0}'.format(length))
-	assert len(prices) - 1 >= length, "prices list is shorted than length: {0}".format(length) 
-
-	return rod_cutting_recursive(prices, length)
-
-def rod_cutting_recursive(prices: List[int],length: int) -> int:
-	#base case 
-	if length == 0:
+	>>> top_down_cut_rod(10, [1, 5, 8, 9, 10, 17, 17, 20, 24, 30])
+	30
+	"""
+	_enforce_args(n, prices)
+	max_rev = [float("-inf") for _ in range(n + 1)]
+	return _top_down_cut_rod_recursive(n, prices, max_rev)
+
+
+def _top_down_cut_rod_recursive(n: int, prices: list, max_rev: list):
+	"""
+	Constructs a top-down dynamic programming solution for the rod-cutting problem
+	via memoization.
+
+	Runtime: O(n^2)
+
+	Arguments
+	--------
+	n: int, the length of the rod
+	prices: list, the prices for each piece of rod. ``p[i-i]`` is the
+	price for a rod of length ``i``
+	max_rev: list, the computed maximum revenue for a piece of rod.
+	``max_rev[i]`` is the maximum revenue obtainable for a rod of length ``i``
+
+	Returns
+	-------
+	The maximum revenue obtainable for a rod of length n given the list of prices for each piece.
+	"""
+	if max_rev[n] >= 0:
+		return max_rev[n]
+	elif n == 0:
 		return 0
-	value = float('-inf')
-	for firstCutLocation in range(1,length+1):
-		value = max(value, prices[firstCutLocation]+rod_cutting_recursive(prices,length - firstCutLocation))
-	return value
+	else:
+		max_revenue = float("-inf")
+		for i in range(1, n + 1):
+			max_revenue = max(max_revenue, prices[i - 1] + _top_down_cut_rod_recursive(n - i, prices, max_rev))
+
+		max_rev[n] = max_revenue
+
+	return max_rev[n]
+
+
+def bottom_up_cut_rod(n: int, prices: list):
+	"""
+	Constructs a bottom-up dynamic programming solution for the rod-cutting problem
+
+	Runtime: O(n^2)
+
+	Arguments
+	----------
+	n: int, the maximum length of the rod.
+	prices: list, the prices for each piece of rod. ``p[i-i]`` is the
+	price for a rod of length ``i``
+
+	Returns
+	-------
+	The maximum revenue obtainable from cutting a rod of length n given
+	the prices for each piece of rod p.
+
+	Examples
+	-------
+	>>> bottom_up_cut_rod(4, [1, 5, 8, 9])
+	10
+	>>> bottom_up_cut_rod(10, [1, 5, 8, 9, 10, 17, 17, 20, 24, 30])
+	30
+	"""
+	_enforce_args(n, prices)
+
+	# length(max_rev) = n + 1, to accommodate for the revenue obtainable from a rod of length 0.
+	max_rev = [float("-inf") for _ in range(n + 1)]
+	max_rev[0] = 0
+
+	for i in range(1, n + 1):
+		max_revenue_i = max_rev[i]
+		for j in range(1, i + 1):
+			max_revenue_i = max(max_revenue_i, prices[j - 1] + max_rev[i - j])
+
+		max_rev[i] = max_revenue_i
+
+	return max_rev[n]
+
+
+def _enforce_args(n: int, prices: list):
+	"""
+	Basic checks on the arguments to the rod-cutting algorithms
+
+	n: int, the length of the rod
+	prices: list, the price list for each piece of rod.
+
+	Throws ValueError:
+
+	if n is negative or there are fewer items in the price list than the length of the rod
+	"""
+	if n < 0:
+		raise ValueError(f"n must be greater than or equal to 0. Got n = {n}")
+
+	if n > len(prices):
+		raise ValueError(f"Each integral piece of rod must have a corresponding "
+						 f"price. Got n = {n} but length of prices = {len(prices)}")
 
 
 def main():
-	assert rod_cutting([1,5,8,9,10,17,17,20,24,30],10) == 30
-	# print(rod_cutting([],0))
+	prices = [6, 10, 12, 15, 20, 23]
+	n = len(prices)
+
+	# the best revenue comes from cutting the rod into 6 pieces, each
+	# of length 1 resulting in a revenue of 6 * 6 = 36.
+	expected_max_revenue = 36
+
+	max_rev_top_down = top_down_cut_rod(n, prices)
+	max_rev_bottom_up = bottom_up_cut_rod(n, prices)
+	max_rev_naive = naive_cut_rod_recursive(n, prices)
+
+	assert expected_max_revenue == max_rev_top_down
+	assert max_rev_top_down == max_rev_bottom_up
+	assert max_rev_bottom_up == max_rev_naive
+
 
 if __name__ == '__main__':
 	main()
-