If we wanted to see if any disjointed subset of a set $X = [w_1, ..., w_n]$ exists such as the sum of its elements equal exactly a given value $M$ (0-1 Knapsack problem) we could employ a DP solution or a simple greedy algorithm.

A simple implementation of which could be the following. Sort the set $X$ in non-decreasing order and try taking taking the values of $X_{sorted}$ one by one only if said number is smaller than the difference of $M$ with the sum of the currently selected numbers. By the end of it we should be able to see if it is possible to form the number $M$ using a combination of that set. This will of course not work all the time.

My question is, how things would change if we knew that the numbers $w_1, w_2$ and so on are in fact guaranteed to be multiples of 2. Would then be another way to check if any given integer $M$ can be formed or not with a combination of those numbers?


1 Answer 1


Obviously, even numbers cannot add up to an odd $M$. Other than that, we could just set $v_i = w_i /2$ and $N = M/2$. Then finding a subset of $w_1, \dots w_n$ which adds up to $M$ would be the same as finding a subset of $v_1, \dots, v_n$ which adds up to $N$, where the latter is the unconstrained knapsack problem.

  • $\begingroup$ What advantage would dividing by 2 give me exactly? $\endgroup$
    – dearn44
    Commented Nov 11, 2016 at 20:10
  • $\begingroup$ Dividing by 2 transforms your problem in to the usual Knapsack, showing that your problem is not easier than Knapsack. There is no practical advantage. $\endgroup$ Commented Nov 12, 2016 at 12:10
  • $\begingroup$ This is the wrong reduction direction. $\endgroup$
    – Parham
    Commented Sep 21, 2017 at 12:52

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