Smax { Fli – 1, jl, vị + F[i – 1, j- wil } if j-w¿ > 0 if j-wi <0 F[i – 1, j] d we define the initial conditions as follows: F[0,j] = 0 for j>0 and F[i,0] = 0 for i 20 Solve the knapsack instance given below using the recurrence given above. ply the dynamic programming to the following instance of the knapsack problem where the capacity = 4. Fill out the cells in the table using the recurrence given. You will not receive any credit if you do t use the recurrence relation. Write your results to the table below. сараcity j item weight value i 1 2 3 4 $25 $22 1 1 25 25 25 25 $26 $27 3 1 25 25 47 47 4 3 26 51 4 26 = 3 j= 3 F[3,3] = = 3 j = 4 F[3,4] = = 4 j = 2 F[4,2] = = 4 j = 3 F[4,3] = = 4 j = 4 F[4, 4] = 2.

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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Consider the knapsack problem: given n items of known weights w,., w, and values v,., v, and
a knapsack of capacity W, find the most valuable subset of the items that fit int the knapcak . In order to
design a dynamic programming algorithm, we have driven the recurrence relation below that expresses a
solution to an instance of the knapsack problem in terms of solutions to its smaller subinstances. For 1<isn
and 1< j<W the value of F[i, j] is as follows:
F[i, j] =
smax { F[i – 1, j], vị + F[i – 1, j – w;] } if j-w; > 0
if j-w < 0
F[i – 1, j]
and we define the initial conditions as follows: F[0,j] = 0 for j>0 and F[i,0] = 0 for i > 0
2.1)
Solve the knapsack instance given below using the recurrence given above.
Apply the dynamic programming to the following instance of the knapsack problem where the capacity
W = 4. Fill out the cells in the table using the recurrence given. You will not receive any credit if you do
not use the recurrence relation. Write your results to the table below.
сараcity j
item
weight
i
1
2
3
value
$25
$22
1
1
2
2
25
25
25
25
$26
$27
3
1
25
25
47
47
4
26
51
4
26
i = 3 j= 3
F[3,3] =
i = 3 j= 4
F[3,4] =
i = 4 j = 2
F[4, 2] =
i = 4 j= 3
F[4, 3] =
i = 4 j= 4
F[4, 4] =
1.
2.
3.
Transcribed Image Text:Consider the knapsack problem: given n items of known weights w,., w, and values v,., v, and a knapsack of capacity W, find the most valuable subset of the items that fit int the knapcak . In order to design a dynamic programming algorithm, we have driven the recurrence relation below that expresses a solution to an instance of the knapsack problem in terms of solutions to its smaller subinstances. For 1<isn and 1< j<W the value of F[i, j] is as follows: F[i, j] = smax { F[i – 1, j], vị + F[i – 1, j – w;] } if j-w; > 0 if j-w < 0 F[i – 1, j] and we define the initial conditions as follows: F[0,j] = 0 for j>0 and F[i,0] = 0 for i > 0 2.1) Solve the knapsack instance given below using the recurrence given above. Apply the dynamic programming to the following instance of the knapsack problem where the capacity W = 4. Fill out the cells in the table using the recurrence given. You will not receive any credit if you do not use the recurrence relation. Write your results to the table below. сараcity j item weight i 1 2 3 value $25 $22 1 1 2 2 25 25 25 25 $26 $27 3 1 25 25 47 47 4 26 51 4 26 i = 3 j= 3 F[3,3] = i = 3 j= 4 F[3,4] = i = 4 j = 2 F[4, 2] = i = 4 j= 3 F[4, 3] = i = 4 j= 4 F[4, 4] = 1. 2. 3.
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