November 14th Subset Sums
Page 1 of 1 • Share •
November 14th Subset Sums
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 samesum partitions can be made from the set {1, 2, ..., N}. The output file should contain 0 if there are no ways to make a samesum partition.
SAMPLE OUTPUT (file subset.out)
4
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 samesum partitions can be made from the set {1, 2, ..., N}. The output file should contain 0 if there are no ways to make a samesum 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 n’s 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 (that’s 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 that’s i is at least i1
dp[i][j] = dp[i1][j];
// if the current number, i, can be used, use it
if (j>=i) dp[i][j] += dp[i1][ji];
That’s pretty much the knapsack problem. I think Tyson got it the first time, or something close.
If you feel ambitious, compact this down to a 1D array.
Eddie Admin
 Number of posts : 65
Location : Canada
Grade : 12
L337ness :
Registration date : 20080827
Re: November 14th Subset Sums
still don't get it completely, but since when does anyone have to understand how something works to use it?
a13x Alumni
 Number of posts : 299
Grade : >12
L337ness :
Registration date : 20080911
Page 1 of 1
Permissions in this forum:
You cannot reply to topics in this forum

