Modify your bottom-up function DPchange to return not only the best number of coins but also the corresponding combination of denominations as a vector. Do this by creating a new function DPChange_bt that is largely the same as DPChange but declares an array coinsUsed that keeps track of the coin on the best path to a given entry M (between 1 and ?+1). DPChange_bt should return both bestNumCoins and coinsUsed.
Modify your bottom-up function DPchange to return not only the best number of coins but also the corresponding combination of denominations as a vector. Do this by creating a new function DPChange_bt that is largely the same as DPChange but declares an array coinsUsed that keeps track of the coin on the best path to a given entry M (between 1 and ?+1). DPChange_bt should return both bestNumCoins and coinsUsed.
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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Modify your bottom-up function DPchange to return not only the best number of coins but also the corresponding combination of denominations as a
![# Bottom up DPchange
def DPChange (M, c, d):
bestnumcoins - [0] + [np.inf for in range (M)]
for m in range (1, M+1):
print ("\n")
for i in range (d):
if m >= c[i]:
print (f"M = {m}, c = {c[i]}")
if bestnumCoins[m-c[i]] + 1 < bestnumCoins[m]:
bestnumCoins [ m]
bestnumCoins[m-c[i]] + 1
print (bestnumCoins)
return bestnumCoins [M]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F12c18db6-f6e8-4129-b933-2f1b3972f7de%2F5dcc0e55-d3c4-484e-90c5-304e388314b1%2Fpwn5hog_processed.png&w=3840&q=75)
Transcribed Image Text:# Bottom up DPchange
def DPChange (M, c, d):
bestnumcoins - [0] + [np.inf for in range (M)]
for m in range (1, M+1):
print ("\n")
for i in range (d):
if m >= c[i]:
print (f"M = {m}, c = {c[i]}")
if bestnumCoins[m-c[i]] + 1 < bestnumCoins[m]:
bestnumCoins [ m]
bestnumCoins[m-c[i]] + 1
print (bestnumCoins)
return bestnumCoins [M]
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