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April 22, 2017 00:29
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Dynamic programming code from CLRS Intro to Algorithms pseudocode..
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#!/usr/bin/env python | |
# -*- coding: utf-8 -*- | |
""" | |
This file contains Python implementations of dynamic programming | |
from Intro to Algorithms (Cormen et al.). | |
The aim here is not efficient Python implementations | |
but to duplicate the pseudo-code in the book as closely as possible. | |
Also, since the goal is to help students to see how the algorithm | |
works, there are print statements placed at key points in the code. | |
The performance of each function is stated in the docstring, and | |
loop invariants are expressed as assert statements when they | |
are not too complex. | |
This file contains: | |
cut_rod() | |
memo_cut_rod() | |
memo_cut_rod_aux() | |
bottom_up_cut_rod(): | |
matrix_chain_order() | |
""" | |
import sys | |
NEG_REV = -1 | |
calls = 0 | |
def cut_rod(prices, n): | |
""" | |
Args: | |
prices: an array of prices 0..n for rods of various lengths | |
n: length of rod | |
Returns: | |
max_rev: the maximum revenue possible from cutting a rod of length n. | |
Note: This is not a good value to return in a real world | |
application: what we would really like returned is what | |
cuts to make to *achieve* that maximum revenue! | |
Our textbook handles this in the extended version of bottom up rod | |
cutting. | |
Note2: CLRS calls variable q. | |
""" | |
if n > len(prices): | |
print("Not enough prices for rod of len " + str(n)) | |
return None | |
global calls | |
calls += 1 | |
if n == 0: | |
return 0 | |
max_rev = NEG_REV # CLRS uses negative infinity, but that hardly seems | |
# necessary: if rods do not sell at a positive price, | |
# best to just go out of business immediately! | |
for i in range(0, n): | |
max_rev = max(max_rev, | |
prices[i] + cut_rod(prices[:n - i - 1], n - i - 1)) | |
print("Handling rod of len " + str(n) + "; max revenue currently is " + | |
str(max_rev)) | |
print("Calls = " + str(calls)) | |
return max_rev | |
def memo_cut_rod(prices, n): | |
""" | |
Args: | |
prices: an array of prices 0..n for rods of various lengths | |
n: length of rod | |
Returns: | |
max_rev: maximum revenue possible | |
""" | |
if n == 0: | |
return 0 | |
max_rev = NEG_REV | |
revs = [NEG_REV for i in range(n)] | |
return memo_cut_rod_aux(prices, n, revs) | |
def memo_cut_rod_aux(prices, n, revs): | |
""" | |
Args: | |
prices: an array of prices 0..n for rods of various lengths | |
revs: revenues previously calculated | |
n: length of rod | |
Returns: | |
max_rev: maximum revenue | |
""" | |
global calls | |
max_rev = NEG_REV | |
if revs[n - 1] >= 0: | |
print("Using memo value with n = " + str(n) | |
+ " and revs[n - 1] = " | |
+ str(revs[n - 1])) | |
return revs[n - 1] | |
if n == 0: | |
max_rev = 0 | |
else: | |
calls += 1 | |
for i in range(0, n): | |
max_rev = max(max_rev, | |
prices[i] | |
+ memo_cut_rod_aux(prices[:n - i - 1], | |
n - i - 1, | |
revs)) | |
print("Handling rod of len " + str(n) | |
+ "; max revenue currently is " + | |
str(max_rev)) | |
revs[n - 1] = max_rev | |
print("Calls = " + str(calls)) | |
return max_rev | |
def bottom_up_cut_rod(prices, n): | |
""" | |
Args: | |
prices: an array of prices for different rod lengths | |
n: length of rod | |
Returns: | |
max_rev: maximum revenue possible | |
""" | |
if n > len(prices): | |
print("Not enough prices for rod of len " + str(n)) | |
return None | |
max_rev = NEG_REV | |
revs = [0 for i in range(n + 1)] # CLRS only initializes first elem, | |
# but no harm doing all | |
for j in range(1, n + 1): # 0th element holds a 0 | |
max_rev = NEG_REV | |
for i in range(j): | |
print("Checking max_rev between " + str(max_rev) | |
+ " and j = " + str(j) + "; i = " + str(i) | |
+ "; j - i - 1 = " + str(j - i - 1)) | |
print("prices[i] = " + str(prices[i]) + | |
"; revs[j - i - 1] = " + str(revs[j - i - 1])) | |
max_rev = max(max_rev, prices[i] + revs[j - i - 1]) | |
print("Maximum revenue calculated at step " + str(j) | |
+ " = " + str(max_rev)) | |
revs[j] = max_rev | |
return revs[n] | |
def ext_bottom_up_cut_rod(prices, n): | |
""" | |
This calculates the maximum revenue, but also returns a list of what | |
cuts generate it. | |
Args: | |
prices: an array of prices for different rod lengths | |
n: length of rod | |
Returns: | |
max_rev: maximum revenue possible | |
""" | |
if n > len(prices): | |
print("Not enough prices for rod of len " + str(n)) | |
return None | |
revs = [0 for i in range(n + 1)] # CLRS only initializes first elem, | |
# but no harm doing all | |
cuts = [0 for i in range(n + 1)] | |
for j in range(1, n + 1): # 0th element holds a 0 | |
print("\n********\n Working on foot: " + str(j)) | |
max_rev = NEG_REV | |
for i in range(j): | |
prev_revs = revs[j - i - 1] | |
print("Revs = " + str(revs)) | |
print("Comparing max_rev of " + str(max_rev) | |
+ " with " + str(prices[i]) | |
+ " plus prev_rev of " + str(prev_revs)) | |
if max_rev < prices[i] + prev_revs: | |
max_rev = prices[i] + prev_revs | |
cuts[j] = i + 1 | |
revs[j] = max_rev | |
return (revs, cuts, max_rev) | |
def print_cut_rod(prices, n): | |
""" | |
Args: | |
prices: price array | |
n: length of rod | |
Returns: | |
None | |
Prints optimal cuts | |
""" | |
(revs, cuts, max_rev) = ext_bottom_up_cut_rod(prices, n) | |
while n > 0: | |
print("Make a cut of: " + str(cuts[n])) | |
n = n - cuts[n] | |
print("And we achieve revenue of " + str(max_rev)) | |
# some price arrays for testing rod cutting: | |
p1 = [3] | |
p2 = [3, 7] | |
p3 = [3, 7, 9] | |
p4 = [3, 7, 9, 12] | |
p7 = [3, 5, 10, 11, 14, 17, 20] | |
p10 = [1, 5, 8, 9, 10, 17, 17, 20, 24, 30] | |
""" | |
On to determining optimal matrix multiplication order. | |
""" | |
BIG_NUM = sys.maxsize | |
def matrix_chain_order(p): | |
""" | |
Args: | |
p: a list of dimensions so that | |
matrix i's dimensions are found at | |
p[i - 1] and p[i]. | |
Returns: | |
list of m and list of s for parenthisation | |
""" | |
n = len(p) - 1 | |
m = [[BIG_NUM for x in range(n)] for x in range(n)] | |
for i in range(n): | |
m[i][i] = 0 | |
s = [[-1 for x in range(n)] for x in range(n)] | |
for l in range(2, n + 1): | |
print("Working on chain length: " + str(l)) | |
for i in range(0, n - l + 1): | |
j = i + l - 1 | |
print("i = " + str(i) + "; j = " + str(j)) | |
m[i][j] = BIG_NUM | |
for k in range(i, j): | |
print("k = " + str(k)) | |
print("Cost = m[i][k] (" + str(m[i][k]) + | |
") + m[k+1][j] (" + str(m[k + 1][j]) + | |
") + p[i] (" + str(p[i]) + | |
") * p[k+1] (" + str(p[k + 1]) + | |
") * p[j+1] (" + str(p[j + 1]) + ")") | |
q = (m[i][k] + m[k + 1][j] | |
+ (p[i] * p[k+1] * p[j+1])) | |
print("Comparing q = " + str(q) | |
+ " with m[i][j] = " | |
+ str(m[i][j]) + " with i = " | |
+ str(i) + " and j = " + str(j) | |
+ " and k = " + str(k)) | |
if q < m[i][j]: | |
m[i][j] = q | |
s[i][j] = k | |
return (m, s) | |
def print_optimal_parens(s, i, j): | |
""" | |
Args: | |
s: the list returned by matrix_chain_order() | |
i: start index | |
j: end index | |
Returns: None; prints results. | |
""" | |
if i == j: | |
print(" A", end="") | |
else: | |
print("(", end="") | |
print_optimal_parens(s, i, s[i][j]) | |
print_optimal_parens(s, s[i][j] + 1, j) | |
print(")", end="") | |
# an M of size n holds the dimensions for n - 1 matrices! | |
# dims 2 ... n - 1 are the 2nd dim of one M and the 1st dim of the | |
# next one. | |
M2 = [10, 100, 5] | |
M3 = [10, 100, 5, 50] | |
M4 = [10, 100, 5, 50, 80] | |
clrs_test = [30, 35, 15, 5, 10, 20, 25] # this is the example from page 376 | |
def optimal_bst(p, q, n): | |
""" | |
Args: | |
p: probabilities of actual items | |
q: probabilities of failed searches (1 bigger than p!) | |
n: number of items in p | |
Returns: | |
e: expected search costs (a 2D list) | |
root: of the constructed BST | |
""" | |
# three 2D arrays: e is expected search costs, w is probabilities, | |
# and root is ? | |
e = [[BIG_NUM for x in range(n + 1)] for x in range(n + 1)] | |
w = [[BIG_NUM for x in range(n + 1)] for x in range(n + 1)] | |
root = [[BIG_NUM for x in range(n)] for x in range(n)] | |
for i in range(0, n + 1): | |
e[i][i] = q[i] | |
print("i = " + str(i) + "; e[i][i] = " + str(e[i][i])) | |
w[i][i] = q[i] | |
print("i = " + str(i) + "; w[i][i] = " + str(w[i][i])) | |
for l in range(0, n): | |
for i in range(0, n - l + 1): | |
j = i + l - 1 | |
w[i][j] = w[i][j - 1] + p[j] + q[j] | |
for r in (i, j + 1): | |
t = e[i][r] + e[r][j] + w[i][j] | |
if t < e[i][j]: | |
e[i][j] = t | |
root[i][j] = r | |
return (e, root) | |
# probabilities from CLRS: | |
p = [.15, .10, .05, .10, .20] | |
q = [.05, .10, .05, .05, .05, .10] | |
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