Question

proof of the 0/1 knapsack problem has a optimal sub-structure

Answer 1

What is 0/1 knapsack problem?

Let there is a knapsack (or rucksack ) of capacity W.

n items are placed on this knapsack. {Weights and values of n items (given as array weights[ ] and values[ ])

where weights[0..n-1] are the weights items

values[0…n-1] are the values of items

}

The 0/1 knapsack problem here is to sort out highest value subset of values[ ], where the total of weights of the subset <=W, capacity of knapsack.

Here 0 denotes don’t take anything from knapsack

or

take full items in knapsack

What is a solution for this problem?

0/1 Knapsack problem possess 2 properties of a dynamic programming problem:

- overlapping sub-problems property

- Optimal Substructure property

While discussing second property we can see that:

By considering every subset of items (then 2 scenarios are there:

1. item is in optimal subset
2. not in optimal set

) and finding the weight sum and value sum of every subset this problem can be solved. Then take only subsets whenever weight sum < W. Then from those, take the subset with maximum value.

Proof:

Let I be the set of items,

val means valuesv

wt means weight.

Let some solutions X′⊆I−{k}is good solution than X−{k}.

Then X′≠X, the value is high, val(X′)>val(X−{k}),

and k∉X′

and wt(X′)≤W−wt(k)

Now add k to X′ to obtain X′′=X′∪{k}

wt(X′′)=wt(X′)+wt(k)≤wt(X′)≤W

so X″ weighs less than W the capacity of knapsack, so X″ is a solution where knapsack weighs < max weight.

val(X′′)=val(X′)+val(k)

By assumption, val(X′)>val(X−{k})

so, adding val(x) to 2 sides of this final eq, val(X′)+val(k)>val(X−{k})+val(k)=val(X)

Hence val(X′′)>val(X)

Contradiction!!

We assumed X was better solution but X’’ is the best solution where knapsack weighs < max weight.