Mathematics of knapsack problems

Defined

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$I=1,2,\dots ,N$ is the set of goods.

When

the weight of each item $iinI$ is $w_i$ value and the upper limit of the total weight of $v_i$



item is $W$, the following is called the knapsack problem.



$$
max ∑{i in I}v_iw_i s.t. ∑{i in I}v_iw_ile Wxin mathbb{N}(forall
in I)
$$ where $x_i$


represents the number of items to be placed in the knapsack.

solution

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The problem is that if you perform a total search, you will try two options of "select or not select items" for the number of items, and the computational complexity is $O(2^{|I|})Become$. However, an efficient solution method by the greedy method is known, and the solution method is shown here. The recurrence formula in this problem is

$$V(i,w)=begincases0quad if,i=0,or,w=0V(i-1,w)quadif,w_i>wmax(V(i-1),V(i-1,w-w_i)+v_i)quadotherwiseendcases$$

Become. where $V(i,w)$ represents the maximum value of the total value that can be realized using goods with a subscript of $i$ or less if the sum of their weights is less than or equal to $w$. This formula is

• If "no items can be selected" or "the maximum weight is $displaystyle0displaystyle0$", the sum of the values of the selected items is $displaystyle0displaystyle0$ because there are no items to pack.
• If the weight of the item $displaystylei$ exceeds $displaystylew$, the item $displaystylei$ cannot be added, so the total value is the maximum value of the item up to one previous subscript limit.
• If the weight of item $displaystylei$ does not exceed $displaystylew$, the item $displaystylei$ should be the least less, either the maximum value of adding the item $displaystylei$ or the maximum value of the addition of the item.

It shows that. The pseudocode is as follows. The maximum value of the sum is V(| I|, W). In addition, enumerating the selected items requires the addition of code.