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dynamic_programming.py
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# -*- coding: utf-8 -*-
"""
Created on Sun Oct 6 17:09:55 2019
@author: Sai
"""
import time
# Rod cutting problem - Chap 15.1 of CLRS
rod_length = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
price_array = [0, 1, 5, 8, 9, 10, 17, 17, 20, 24, 30]
# Recursive solution Pg 363
def cut_rod_recursive(price, length):
if length == 0:
return 0
q = -1
for idx in range(1,length+1):
q = max(q, price[idx]+cut_rod_recursive(price, length-idx))
return q
# Test cases from Pg 362
assert cut_rod_recursive(price_array, 7) == 18
assert cut_rod_recursive(price_array, 9) == 25
assert cut_rod_recursive(price_array, 10) == 30
# Print some runtimes for arbitrary price array
t = time.time()
cut_rod_recursive([2]*51, 10)
print('Recursive version run time for length = 10: '+str(time.time()-t))
t = time.time()
cut_rod_recursive([2]*51, 15)
print('Recursive version run time for length = 15: '+str(time.time()-t))
t = time.time()
cut_rod_recursive([2]*51, 20)
print('Recursive version run time for length = 20: '+str(time.time()-t))
#t = time.time()
#cut_rod_recursive([2]*51, 23)
#print('Recursive version run time for length = 23: '+str(time.time()-t))
# Top-down memoized solution Pg 365
def cut_rod_memoized(price, length):
optimal_price = [-1]*(length+1)
optimal_price[0] = 0
return cut_rod_memoized_aux(price, length, optimal_price)
def cut_rod_memoized_aux(price, length, optimal_price):
if optimal_price[length] == -1:
# Calculate recursively
q = -1
for idx in range(1, length+1):
q = max(q, price[idx]+cut_rod_memoized_aux(price, length-idx, optimal_price))
optimal_price[length] = q
return q
else:
# Lookup in the array
return optimal_price[length]
# Test cases from Pg 362
assert cut_rod_memoized(price_array, 7) == 18
assert cut_rod_memoized(price_array, 9) == 25
assert cut_rod_memoized(price_array, 10) == 30
# Print some runtimes for arbitrary price array
t = time.time()
cut_rod_memoized([2]*51, 10)
print('Memoized version run time for length = 10: '+str(time.time()-t))
t = time.time()
cut_rod_memoized([2]*51, 15)
print('Memoized version run time for length = 15: '+str(time.time()-t))
t = time.time()
cut_rod_memoized([2]*51, 20)
print('Memoized version run time for length = 20: '+str(time.time()-t))
t = time.time()
cut_rod_memoized([2]*101, 100)
print('Memoized version run time for length = 100: '+str(time.time()-t))
# Bottom-up version Pg 366
def cut_rod_bottom_up(price, length):
optimal_price = [-1]*(length+1)
optimal_price[0] = 0
for i in range(1, length+1):
q = -1
for j in range(1, i+1):
q = max(q, price[j]+optimal_price[i-j])
optimal_price[i] = q
return optimal_price[length]
# Test cases from Pg 362
assert cut_rod_bottom_up(price_array, 7) == 18
assert cut_rod_bottom_up(price_array, 9) == 25
assert cut_rod_bottom_up(price_array, 10) == 30
# Print some runtimes for arbitrary price array
t = time.time()
cut_rod_bottom_up([2]*51, 10)
print('Bottom up version run time for length = 10: '+str(time.time()-t))
t = time.time()
cut_rod_bottom_up([2]*51, 15)
print('Bottom up version run time for length = 15: '+str(time.time()-t))
t = time.time()
cut_rod_bottom_up([2]*51, 20)
print('Bottom up version run time for length = 20: '+str(time.time()-t))
t = time.time()
cut_rod_bottom_up([2]*101, 100)
print('Bottom up version run time for length = 100: '+str(time.time()-t))
# Bottom up cut-rod with printing of the cuts
def cut_rod_bottom_up_print(price, length):
optimal_price = [-1]*(length+1)
optimal_price[0] = 0
cuts = [0]*(length+1)
for i in range(1, length+1):
q = -1
for j in range(1, i+1):
if price[j]+optimal_price[i-j] > q:
q = price[j]+optimal_price[i-j]
cuts[i] = j
optimal_price[i] = q
cut_length = length
while cut_length > 0:
print(cuts[cut_length])
cut_length = cut_length - cuts[cut_length]
return optimal_price[length]
# Test cases from Pg 362
assert cut_rod_bottom_up_print(price_array, 7) == 18
assert cut_rod_bottom_up_print(price_array, 9) == 25
assert cut_rod_bottom_up_print(price_array, 10) == 30
# Longest common subsequence - Pg 394-395
def lcs(seq1, seq2):
# Array to store the length of the LCS of the substrings seen so far
length = [[0]*(len(seq2)+1) for i in range(len(seq1)+1)]
# Array to store the steps taken. Used to reconstruct the LCS. The values mean the following:
# 1 - drop the last character in seq1
# 2 - drop the last character in seq2
# 3 - last characters are the same, so the character is added to the LCS
step = [[0]*(len(seq2)+1) for i in range(len(seq1)+1)]
for i1, c1 in enumerate(seq1, 1):
for i2, c2 in enumerate(seq2, 1):
# print(i1, i2, c1, c2)
if c1 == c2:
length[i1][i2] = length[i1-1][i2-1] + 1
step[i1][i2] = 3
# print(3)
elif length[i1-1][i2] > length[i1][i2-1]:
length[i1][i2] = length[i1-1][i2]
step[i1][i2] = 1
# print(1)
else:
length[i1][i2] = length[i1][i2-1]
step[i1][i2] = 2
# print(2)
# print(length)
# print(length)
# print(step)
p1 = len(seq1)
p2 = len(seq2)
LCS = ''
while p1 > 0 and p2 > 0:
if step[p1][p2] == 1:
p1 -= 1
elif step[p1][p2] == 2:
p2 -= 1
elif step[p1][p2] == 3:
# print(p1, p2, seq1[p1-1], seq2[p2-1])
LCS += seq1[p1-1]
p1 -= 1
p2 -= 1
return LCS[::-1]
seq1 = 'TCG'
seq2 = 'TAC'
#print(lcs(seq1, seq2))
assert lcs(seq1, seq2) == 'TC'
seq1 = 'TGACTGGGT'
seq2 = 'GGGG'
#print(lcs(seq1, seq2))
assert lcs(seq1, seq2) == 'GGGG'
# Test case from Pg 391 of CLRS
seq1 = 'ACCGGTCGAGTGCGCGGAAGCCGGCCGAA'
seq2 = 'GTCGTTCGGAATGCCGTTGCTCTGTAAA'
#print(lcs(seq1, seq2))
assert lcs(seq1, seq2) == 'GTCGTCGGAAGCCGGCCGAA'