问题
What I want to do
I want to find a subset of an array that sums to a target T. I also want to use to a dynamic programming approach (and a bottom-up solution at that) to do this.
What I currently have
Currently I only found a way to see if amongst all subsets of size N, whether or not there is at least one subset that has the desired sum. See code below.
public boolean solve(int[] numbers, int target) {
//Safeguard against invalid parameters
if ((target < 0) || (sum(numbers) < target)){
return false;
}
boolean [][] table = new boolean [target + 1] [numbers.length + 1] ;
for (int i = 0; i <= numbers.length; ++i) {
table[0][i] = true;
}
/* Base cases have been covered.
* Now look set subsets [1..n][target] to be true or false.
* n represents the number of elements from the start that have a subset
* that sums to target
*/
for (int i = 1; i <= target; ++i){
for (int j = 1; j <= numbers.length; ++j){
/* Mark index j as one of the numbers in the array
* which is part of the solution with the given subtarget */
table [i][j] = table[i][j-1];
if (i >= numbers[j-1])
table[i][j] = table[i][j] || table[i - numbers[j-1]] [j-1];
}
}
return table[target][numbers.length];
}
Where I am stuck
Right now, I know if there is a solution, but I can't think of a way to actually output a solution.
I am not looking for anyone to provide me specific code, but pseudocode is welcome as are hints to how a solution may be saved.
回答1:
The algorithm you provided can stay the same, you don't need to store anything else besides the DP-table table[][]. You just need an additional post-processing phase in which you step "backwards" through table[][] to get the solution set.
Just to recall:
You've computed the table table[i][j], which stores for every value 0<=i<=t(:=target) and every 0<=j<=n(:=numbers.length) whether there is a subset of numbers in numbers[0..j-1] that sum to i.
Consider the subset S corresponding to table[i][j] (, which is true). Note that:
-
The subset S contains the number
numbers[j]only iftable[ i-numbers[j] ][j-1]is true.(Proof: recursively take the solution subset S' for
table[ i-numbers[j] ][j-1], and addnumbers[j]) -
On the other hand, this subset S does not contain the number
numbers[j]only iftable[ i-numbers[j] ][j-1]is false.(Proof: assume S contains
numbers[j], trownumbers[j]out of S, this impliestable[ i-numbers[j] ][j-1], contradiction)
numbers[n-1] is in the subset summing to t.
-
If so, recursively compute whether
numbers[n-2]is in the subset summing to t-numbers[n-1], -
else recursively compute whether
numbers[n-2], is in the subset summing to t
回答2:
Here are the two Java solutions for the subset sum problem.
First using Recursive Approach.
Second using Dynamic Programming Approach.
/*
Question: Given a set of non-negative integers, and a value sum, determine if there is a subset of the given set
with sum equal to given sum.
Examples: set[] = {3, 34, 4, 12, 5, 2}, sum = 9
Output: True //There is a subset (4, 5) with sum 9.
Let isSubSetSum(int set[], int n, int sum) be the function to find whether there is a subset of set[] with
sum equal to sum. n is the number of elements in set[].
*/
package SubsetSumProblem;
import java.util.Scanner;
public class UsingResursiveAndDPApproach {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
try{
System.out.println("Enter the number of elements in the array");
int n =in.nextInt();
System.out.println("Enter the elements of the array");
int[] a=new int[n];
for(int i=0;i<n;i++)
a[i]=in.nextInt();
System.out.println("Enter the sum, which you need to find");
int sum = in.nextInt();
System.out.println("Using recursion, the result is: "+usingRecursion(a,a.length,sum));
System.out.println("Using Dynamic Programming, the result is: "+usingDP(a,sum));
}
finally{
in.close();
}
}
private static boolean usingRecursion(int[] a,int length, int sum) {
// 1. Base Cases
if(sum==0)
return true;
if(length==0 && sum!=0)
return false;
// 2. To avoid unnecessary steps, we will optimize the recursion method by avoiding
// recursive calls to areas where we are definite that we can SAFELY ignore the case since
// the SOLUTION does not exist there.
// If last element is greater than sum, then ignore it
if(a[a.length-1]>sum)
return usingRecursion(a,length-1,sum);
// 3. This is the recursion step where we will call the method again and again
/* else, check if sum can be obtained by any of the following
(a) including the last element
(b) excluding the last element */
return (usingRecursion(a, length-1, sum-a[length-1])|| usingRecursion(a, length-1, sum));
}
/*
Analysis:
Time Complexity = O(2^n)
Space Complexity = // Don't know
*/
private static boolean usingDP(int[] a, int sum) {
// using boolean matrix for DP
boolean dp[][] = new boolean[a.length+1][sum+1]; // +1 in row and column
// if the length of the array is variable (and sum is 0) then fill TRUE, since the SUM=0
for(int row=0;row<dp.length;row++){
dp[row][0] = true; // NOTE: dp[length=VARIABLE][sum=0], thus we satisfy the condition where length is VARIABLE
// and the SUM=0
}
// if the SUM is variable and length is 0 then FALSE, since (sum=variable && length=0)
for(int column=1;column<dp[0].length;column++){
dp[0][column] = false; // NOTE: dp[length=0][sum=VARIABLE], thus we satisfy the condition where
// (length=0 && sum=variable)
}
for(int i=1;i<dp.length;i++){
for(int j=1;j<dp[0].length;j++){
/* Check if sum can be obtained by any of the following
(a) including the last element
(b) excluding the last element */
// VERY VERY IMP: This is same as "excluding the last element" which is represented in DP
dp[i][j] = dp[i-1][j]; // the current position[i][j] would be same as previous position.
// the previous position means that SUM is ACHIEVED OR NOT-ACHIEVED
// int the previous position then it will ofcourse be ACHIEVED or NOT-ACHIEVED
// in the current position.
// VERY VERY IMP: This is same as "including the last element" which is represented in DP
// if the column[ sum is represented in column of the matrix i.e this sum exist] > = sum-a[last_index]
// then decrease the sum
if(j>=a[i-1]) // i.e sum >= array[last index element]. If it is true then include this last element by
// deducting it from the total sum
dp[i][j] = dp[i][j] || dp[i-1][j-a[i-1]]; // VERY VERY IMP NOTE: Here dp[i][j] on R.H.S represent
// dp[i-1][j] which we have assigned in the previous step
}
}
return dp[a.length][sum];
}
/*
Analysis:
Time Complexity = O(a.length*sum)
Space Complexity = O(a.length*sum)
*/
}
回答3:
Here is my solution is an iterative dp, but with only one dimension: Hope it can help you.
#include <iostream>
#include <cstring>
using namespace std;
const int maxN=1000;
int memo[maxN];
int pi[maxN];
int main(){
int a[]={7,8,5,1,4};
memset(memo,-1,sizeof memo);
memset(pi,-1,sizeof pi);
int n;
cin>>n;
memo[0]=0;
pi[0]=0;
for(int i=0;i<(int)sizeof(a)/4;i++){
for(int num=n;num>=0;num--){
if(num-a[i]>=0 and memo[num-a[i]]!=-1 and (memo[num]==-1 or memo[num]>1+memo[num-a[i]])){
memo[num]=1+memo[num-a[i]];
pi[num]=num-a[i];
}
}
}
int N=n;
while(N!=0){
cout<<N-pi[N]<<" ";
N=pi[N];
}
cout<<endl;
cout<<memo[n]<<endl;
return 0;
}
来源：https://stackoverflow.com/questions/18818406/find-a-solution-to-subset-sum-using-dynamic-programming