Dynamic programming vs Greedy algo cont KNAPSACK KNAPSACK - - PowerPoint PPT Presentation

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Jan 19, 2023 180 likes •267 views

Dynamic programming vs Greedy algo cont KNAPSACK KNAPSACK KNAPSACK KNAPSACK Input: a number W and a set of n items, the i-th item has a weight w i and a cost c i a subset of items with total weight W Output: Objective: maximize

SLIDE 1

Dynamic programming vs Greedy algo – con’t

Input: Output: Objective: a number W and a set of n items, the i-th item has a weight wi and a cost ci a subset of items with total weight · W maximize cost Version 1: Items are divisible. KNAPSACK KNAPSACK KNAPSACK KNAPSACK

SLIDE 2

KNAPSACK – divisible: a greedy solution

KNAPSACK-DIVISIBLE(n,c,w,W)

1. sort items in decreasing order of ci/wi
2. i = 1
3. currentW = 0
4. while (currentW + wi < W) {
5. take item of weight wi and cost ci
6. currentW += wi
7. i++
8. }
9. take W-currentW portion of item i

Correctness: Running time:

SLIDE 3

KNAPSACK – indivisible

Version 2: Items are indivisible. Does previous algorithm work for this version of KNAPSACK?

SLIDE 4

KNAPSACK – indivisible: a dyn-prog solution

The heart of the algorithm: S[k][v] =

SLIDE 5

KNAPSACK – indivisible: a dyn-prog solution

maximum cost of a subset of the first k items, where the weight of the subset is at most v The heart of the algorithm: S[k][v] =

SLIDE 6

KNAPSACK – indivisible: a dyn-prog solution

maximum cost of a subset of the first k items, where the weight of the subset is at most v The heart of the algorithm: S[k][v] = KNAPSACK-INDIVISIBLE(n,c,w,W)

1. init S[0][v]=0 for every v=0,…,W
2. init S[k][0]=0 for every k=0,…,n
3. for v=1 to W do
4. for k=1 to n do
5. S[k][v] = S[k-1][v]
6. if (wk · v) and

(S[k-1][v-wk]+ck > S[k][v]) then

7. S[k][v] = S[k-1][v-wk]+ck
8. RETURN S[n][W]

SLIDE 7

KNAPSACK – indivisible: a dyn-prog solution

maximum cost of a subset of the first k items, where the weight of the subset is at most v The heart of the algorithm: S[k][v] = Running time: KNAPSACK-INDIVISIBLE(n,c,w,W)

1. init S[0][v]=0 for every v=0,…,W
2. init S[k][0]=0 for every k=0,…,n
3. for v=1 to W do
4. for k=1 to n do
5. S[k][v] = S[k-1][v]
6. if (wk · v) and

(S[k-1][v-wk]+ck > S[k][v]) then

7. S[k][v] = S[k-1][v-wk]+ck
8. RETURN S[n][W]

SLIDE 8

KNAPSACK – indivisible: a dyn-prog solution

maximum cost of a subset of the first k items, where the weight of the subset is at most v The heart of the algorithm: S[k][v] = How to

utput a

solution ? KNAPSACK-INDIVISIBLE(n,c,w,W)

1. init S[0][v]=0 for every v=0,…,W
2. init S[k][0]=0 for every k=0,…,n
3. for v=1 to W do
4. for k=1 to n do
5. S[k][v] = S[k-1][v]
6. if (wk · v) and

(S[k-1][v-wk]+ck > S[k][v]) then

7. S[k][v] = S[k-1][v-wk]+ck
8. RETURN S[n][W]