November 14th Subset Sums

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November 14th Subset Sums

Post  Eddie on Fri Nov 14, 2008 11:16 pm

Subset Sums
JRM

For many sets of consecutive integers from 1 through N (1 <= N <= 39), one can partition the set into two sets whose sums are identical.

For example, if N=3, one can partition the set {1, 2, 3} in one way so that the sums of both subsets are identical:

* {3} and {1,2}

This counts as a single partitioning (i.e., reversing the order counts as the same partitioning and thus does not increase the count of partitions).

If N=7, there are four ways to partition the set {1, 2, 3, ... 7} so that each partition has the same sum:

* {1,6,7} and {2,3,4,5}
* {2,5,7} and {1,3,4,6}
* {3,4,7} and {1,2,5,6}
* {1,2,4,7} and {3,5,6}

Given N, your program should print the number of ways a set containing the integers from 1 through N can be partitioned into two sets whose sums are identical. Print 0 if there are no such ways.

Your program must calculate the answer, not look it up from a table.
PROGRAM NAME: subset
INPUT FORMAT
The input file contains a single line with a single integer representing N, as above.
SAMPLE INPUT (file subset.in)

7

OUTPUT FORMAT

The output file contains a single line with a single integer that tells how many same-sum partitions can be made from the set {1, 2, ..., N}. The output file should contain 0 if there are no ways to make a same-sum partition.
SAMPLE OUTPUT (file subset.out)

4

Spoiler:
The Solution
This is a dynamic programming problem, very similar to the knapsack problem.
First of all, if nís sum is an odd number, there are 0 subsets. Next, in an array dp[i][j], we compute the maximum amount of sets we can form using using 0..i (i<=n) that sums up to j (j<=sum/2). Our base case would be dp[0][0]=1, since there is 1 way to form nothing with nothing. Our final result would be in dp[n][sum/2]/2 (thatís how many sets that sums up to sum/2).
And so we get this relationship:
// since we are getting the max for 0..i, anything thatís i is at least i-1
dp[i][j] = dp[i-1][j];
// if the current number, i, can be used, use it
if (j>=i) dp[i][j] += dp[i-1][j-i];

Thatís pretty much the knapsack problem. I think Tyson got it the first time, or something close. Very Happy
If you feel ambitious, compact this down to a 1D array.

Eddie
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Re: November 14th Subset Sums

Post  a13x on Mon Nov 24, 2008 10:01 pm

still don't get it completely, but since when does anyone have to understand how something works to use it?

a13x
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