Maximum sum of values with constant sum of indices - algorithm

Question

I am looking for an efficient algorithm for the following problem:
There is an array with values, i.e. (note that index 0 is omitted on purpose)
Index 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
Value 17, 12, 5, 22, 3, 12, 6, 13, 7, 0, 2, 15
What I need to find is a subset of indices under these constraints:
The number of indices is constant (i.e. 3)
The sum of indices is constant (i.e. 20)
Each index may only appear once (so [2, 9, 9] is not a valid solution)
The sum of values is maximum.
For example if the subset length is 3 and the sum is 20, all possible solutions would be
Indices: [1, 7, 12] Sum of values: 17 + 6 + 15 = 38
Indices: [1, 8, 11] Sum of values: 17 + 13 + 2 = 32
Indices: [1, 9, 10] Sum of values: 17 + 7 + 0 = 24
Indices: [2, 6, 12] Sum of values: 12 + 12 + 15 = 39
Indices: [2, 7, 11] Sum of values: 12 + 6 + 2 = 20
Indices: [2, 8, 10] Sum of values: 12 + 13 + 0 = 25
Indices: [3, 5, 12] Sum of values: 5 + 3 + 15 = 23
Indices: [3, 6, 11] Sum of values: 5 + 12 + 2 = 19
Indices: [3, 7, 10] Sum of values: 5 + 6 + 0 = 11
Indices: [3, 8, 9] Sum of values: 5 + 13 + 7 = 25
Indices: [4, 5, 11] Sum of values: 22 + 3 + 2 = 27
Indices: [4, 6, 10] Sum of values: 22 + 12 + 0 = 34
Indices: [4, 7, 9] Sum of values: 22 + 6 + 7 = 35
Indices: [5, 6, 9] Sum of values: 3 + 12 + 7 = 22
Indices: [5, 7, 8] Sum of values: 3 + 6 + 13 = 22
of which [2, 6, 12] is the optimal solution because it has the maximum sum of values.
At the moment I run through all possible combinations using a slightly modified partition algorithm which grows exponentially as the sum of indices grows, so I wonder if there is any better way?

Answer 1

Solution O(I.S.K)
Let's do some naming first:
I is the greatest index (12 in your example)
S is the sum of values whose indices are selected (20 in your example)
K is the number of selected indices
V[] the array of values linked to the indices
maxsum(s, i, k) the maximal sum reachable by using k indices, all differents, whose value is less than or equal to i and whose sum is s.
Then you want to find maxsum(S, I, K)
Your problem exhibits some good properties:
optimal sub-structure
redundant sub-problems
For instance, when trying to compute maxsum(s, i, k) I can either not use index i, in which case the value is maxsum(s, i-1, k). Or I could use index i. In this case, I want to solve the sub-problem: what is the maximum sum reachable by indices less than or equal to i-1 and whose sum is s-i using k-1 such indices. This is the value: V[i] + maxsum(s-i, i-1, k-1).
As we want to reach the maximal sum we end up having: (Edit: corrected maxsum(s-i, i-1, k) to maxsum(s-i, i-1, k-1))
maxsum(s, i, k) = max{ maxsum(s, i-1, k) ; V[i] + maxsum(s-i, i-1, k-1) }
This is typical of a problem solvable by dynamic programming.
Here is an example C++ program solving the problem in O(I.S.K) (space and time).
We can improve the space complexity to O(I.S) at the price of a bigger time complexity: O(I.S.K²).
How to use the program
g++ -std=c++14 -g -Wall -O0 dp.cpp -o dp
./dp input.txt
Where input.txt is a file with the following format:
first line contains three integers: I S K
second line contains I integers, the values of the indices
Example run
---- K=1 ----
17 12 5 22 3 12 6 13 7 0 2 15
[ 1][ 2][ 3][ 4][ 5][ 6][ 7][ 8][ 9][10][11][12]
[ 1] 17 17 17 17 17 17 17 17 17 17 17 17
[ 2] 12 12 12 12 12 12 12 12 12 12 12
[ 3] 5 5 5 5 5 5 5 5 5 5
[ 4] 22 22 22 22 22 22 22 22 22
[ 5] 3 3 3 3 3 3 3 3
[ 6] 12 12 12 12 12 12 12
[ 7] 6 6 6 6 6 6
[ 8] 13 13 13 13 13
[ 9] 7 7 7 7
[10] 0 0 0
[11] 2 2
[12] 15
[13]
[14]
[15]
[16]
[17]
[18]
[19]
[20]
---- K=2 ----
17 12 5 22 3 12 6 13 7 0 2 15
[ 1][ 2][ 3][ 4][ 5][ 6][ 7][ 8][ 9][10][11][12]
[ 1]
[ 2] 12 12 12 12 12 12 12 12 12 12 12
[ 3] 29 29 29 29 29 29 29 29 29 29 29
[ 4] 22 22 22 22 22 22 22 22 22 22
[ 5] 17 39 39 39 39 39 39 39 39 39
[ 6] 34 34 34 34 34 34 34 34 34
[ 7] 27 27 29 29 29 29 29 29 29
[ 8] 8 24 24 24 24 24 24 24
[ 9] 25 25 25 30 30 30 30 30
[10] 34 34 34 34 34 34 34
[11] 15 28 28 28 28 28 28
[12] 9 35 35 35 35 35
[13] 18 18 29 29 29 32
[14] 25 25 25 25 27
[15] 19 19 19 24 24
[16] 13 13 13 37
[17] 20 20 20 20
[18] 13 13 27
[19] 7 15 21
[20] 9 28
---- K=3 ----
17 12 5 22 3 12 6 13 7 0 2 15
[ 1][ 2][ 3][ 4][ 5][ 6][ 7][ 8][ 9][10][11][12]
[ 1]
[ 2]
[ 3]
[ 4]
[ 5] 17 17 17 17 17 17 17 17 17 17
[ 6] 34 34 34 34 34 34 34 34 34 34
[ 7] 51 51 51 51 51 51 51 51 51
[ 8] 44 44 44 44 44 44 44 44 44
[ 9] 39 39 41 41 41 41 41 41 41
[10] 42 42 42 42 42 42 42 42
[11] 37 51 51 51 51 51 51 51
[12] 30 46 46 46 46 46 46 46
[13] 39 40 52 52 52 52 52
[14] 20 35 47 47 47 47 47
[15] 37 37 42 42 42 42 44
[16] 31 37 37 37 41 41
[17] 40 40 40 40 40 54
[18] 21 47 47 47 47 49
[19] 41 41 41 41 44
[20] 22 35 35 35 39
index: 12 sum: 20
index: 6 sum: 8
index: 2 sum: 2
max sum: 39
The source code
#include <cstdio>
#include <iomanip>
#include <iostream>
#include <limits>
#include <valarray>
#include <vector>
using namespace std;
auto const INF = numeric_limits<double>::infinity();
struct matrix {
matrix(size_t rows, size_t cols, double value)
: cells(value, rows*cols)
, rows(rows)
, cols(cols)
, value(value)
{}
double& operator() (int r, int c)
{
if(r < 0 || c < 0)
return value;
return cells[r*cols+c];
}
valarray<double> cells;
size_t rows;
size_t cols;
double value;
};
int main(int argc, char* argv[]) {
if(argc > 1)
freopen(argv[1], "r", stdin);
// I: max index
// S: sum of indices
// K: number of indices in the sum S
int I, S, K;
cin >> I >> S >> K;
// load values
vector<double> V(I+1, 0);
for(int i=1; i<=I; ++i)
cin >> V[i];
// dynamic programming:
// --------------------
// maxsum(i, s, k) is the maximal sum reachable using 'k' indices, less
// than or equal to 'i', all differents, and having a sum of 's'
//
// maxsum(i, s, k) =
// -oo if i > s
//
// -oo if i < s && k == 1
//
// V[s] if i >= s && s <= I && k == 1
// -oo if (i < s || s > I) && k == 1
//
// max { V[i] + maxsum(i-1, S-i, k-1), maxsum(i-1, S, k) }
vector<matrix> maxsum(K+1, matrix(S+1, I+1, -INF));
// initialize K=1
for(int s=0; s<=I && s<=S; ++s) {
for(int i=s; i<=I; ++i) {
maxsum[1](s, i) = V[s];
}
}
// K > 1
for(int k=2; k<=K; ++k) {
for(int s=2; s<=S; ++s) {
for(int i=1; i<=I; ++i) {
auto l = V[i] + maxsum[k-1](s-i, i-1);
auto r = maxsum[k](s, i-1);
maxsum[k](s, i) = max(l, r);
}
}
}
// display the whole dynamic programming tables (optional)
for(int k=1; k<=K; ++k) {
cout << "---- K=" << k << " ----\n";
cout << " ";
for(int i=1; i<=I; ++i) {
cout << setw(3) << V[i] << ' ';
}
cout << '\n';
cout << " ";
for(int i=1; i<=I; ++i) {
cout << '[' << setw(2) << i << ']';
}
cout << '\n';
for(int s=1; s<=S; ++s) {
cout << '[' << setw(2) << s << "] ";
for(int i=1; i<=I; ++i) {
if(maxsum[k](s, i) == -INF) {
cout << " ";
} else {
cout << setw(3) << maxsum[k](s, i) << ' ';
}
}
cout << '\n';
}
}
// output the indices belonging to the solution by working backward in the
// dynamic programming tables
int t_S = S;
int t_I = I;
for(int k=K; k>=1; --k) {
if(t_I <= 0 || t_S <= 0) {
cout << "error...\n";
break;
}
auto m = maxsum[k](t_S, t_I);
int i;
for(i=t_I; i>=1; --i) {
if(maxsum[k](t_S, i) != m)
break;
}
cout << "index: " << setw(3) << (i+1) << ' ';
cout << "sum: " << setw(3) << t_S << '\n';
t_I = i;
t_S = t_S - i - 1;
}
cout << "max sum: " << maxsum[K](S, I) << '\n';
}

Answer 2

Take the arrays, and sort them by value instead of by index (keeping the index-value pairs preserved). Now, starting at the end of the array, take the last k numbers in the indices array, where k is the number of indices you have to have, and sum them up. If it equals the desired sum, great- you are done. If not, take note of the difference (desired sum - actual sum), and add that to the (n - k)th index. Find that index in the index array (ordered by value, mind you), now find your new sum of values (you can optimize this by subtracting out the old index's value and add the new one, instead of recomputing the sum of all k values).
You now have one valid solution, and a lower bound. You know the indices of the rest of the valid solution that can even possibly beat this score must come after the smallest index's value in the value-sorted array. That is:
Both sorted by value-
indices: | bunch of indices | index we found | more | k-1 'random' indices |
values: | bunch of values | value for ^ | more | k-1 largest values |
So we only have to search 'more' and the k-1 largest values for valid indices that satisfy the criteria and also have values that form a larger sum. To do this, we rinse and repeat, moving the smallest of the (n-k-1) elements backwards one, so we effectively try all combination of these elements, but in the order of decreasing subset-sums of our set of k elements. This allows us to continually narrow the space we search as we find larger sums, because know for certain that any sum that contains a smaller value than that of best solution will have a smaller sum (because the rest of the set is already as large as possible).
Pseudo Code:
pair_array = input() // each pair consists of index and value
sort_by_value(pair_array)
best_sum = 0
markers = [n - (k-1) .. n] // mark the k-1 indices being summed
while True:
sum_of_indices = sum_indices(pair_array[markers])
value_sum = sum_value(pair_array[markers])
if pair_array.contains(desired_sum - sum_of_indices): // this lets us effectively reduce our search by a factor of N, given contains uses a hashtable
value_sum += pair_array(pair_array.index(desired_sum - sum_of_indices)).value
if value_sum > best_sum:
best_sum = value_sum
pair_array.remove(0 .. n - (k-1)) // this greatly reduces the combinations checked
if has_next_combination(markers, pair_array):
next_greatest_combination(markers, pair_array) // pick new markers, in a reverse-binary counting fashion (most significant bit first way)
else:
print(best_sum)
break

Answer 3

One small trick I can think if you try to find lets say 3 indexes, is instead of iterating for the 3 indexes you can calculate the 3rd index when you know the first two indexes. For example when you know that
p1 = 1, p2 = 7 => p3 = 20 - (p1 + p2) = 12
This can be generalized when having N indexes the last one can always be inferred from the N-1 previous indexes.
I tried this in Python:
Index = [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 ]
Value = [17, 12, 5, 22, 3, 12, 6, 13, 7, 0, 2, 15 ]
maxSum = 0 SumInd = 20
for p1 in range(1,len(Index)+1):
for p2 in range(p1,len(Index)+1):
if (p1 + p2 < SumInd) and (p1 != p2):
p3 = SumInd - (p1 + p2) #If you know p1 and p2 calculate p3 index
if (p3 < len(Index)+1) and (p2 != p3) and (p1 != p3):
fooSum = Value[p1-1]+Value[p2-1] + Value[p3-1]
print(p1,p2,p3 , "Sum is ",fooSum)
if maxSum < fooSum:
maxSum = fooSum
print("Max Sum is ", maxSum)
Of course you need to save the indexes when you find maxSum. Also this implementation calculate many similar pairs (e.g. [1,9,10] and [9,10,1]). Perhaps a better solution could eliminate this pairs.
EDIT: Big Improvement I found a way to eliminate most unnecessary checks. Let's say you need 3 indexes. The first one checks the whole range of possible values. Let's say it is index1.
The other two indexes must sum at 20 - ind1 let's call this rest. The list of indexes is always ordered so you can have a back index showing the smallest value(first item in the list bigger than index1) and a front index that show the biggest values (last item in the list). So index2 = backIndex , index3 = frontIndex.
If the rest is smaller than the sum of index2,3 you could increase the back index (get the next bigger value) or if it is larger you decrease the front index until the two indexes meet each other where you break and increase index1. This eliminates checking (1,7,12) and (1,12,7) two times.
The code is here in Python:
maxSum = 0
SumInd = 20
for index_1 in range(1,len(Index)):
rest = SumInd - index_1
backIndex = index_1+1
frontIndex = len(Index)
while backIndex < frontIndex:
if rest > (backIndex + frontIndex):
backIndex = backIndex + 1
elif rest < (backIndex + frontIndex):
frontIndex = frontIndex - 1
else:
fooSum = Value[index_1-1]+Value[backIndex-1] + Value[frontIndex-1]
print("Checking for ",index_1,backIndex,frontIndex,' Sum of values:',fooSum)
if maxSum < fooSum:
indList = [index_1-1,backIndex,frontIndex]
maxSum = fooSum
backIndex = backIndex + 1 #To avoid Inf loop
print("Max Sum is ", maxSum,"at",indList)
and gives these results:
Checking for 1 7 12 Sum of values: 38
Checking for 1 8 11 Sum of values: 32
Checking for 1 9 10 Sum of values: 24
Checking for 2 6 12 Sum of values: 39
Checking for 2 7 11 Sum of values: 20
Checking for 2 8 10 Sum of values: 25
Checking for 3 5 12 Sum of values: 23
Checking for 3 6 11 Sum of values: 19
Checking for 3 7 10 Sum of values: 11
Checking for 3 8 9 Sum of values: 25
Checking for 4 5 11 Sum of values: 27
Checking for 4 6 10 Sum of values: 34
Checking for 4 7 9 Sum of values: 35
Checking for 5 6 9 Sum of values: 22
Checking for 5 7 8 Sum of values: 22
Max Sum is 39 at [1, 6, 12]
This can always be generalized for N indexes. The first N-2 indexes can search the whole range of the list (like index 1 in the case above, also it should be noted that all these indexes start checking from previous index value plus one until the end of the list to eliminate many duplicate checks).
The last two indexes can be calculated like I showed in my code and avoid many duplicate checks.