I have a simple algorithmic question:
If I have certain elements that integer values like:
1 1 1 1 1 1 1 1 1 1 1 1 10 12 2
and I have to make the sum 12, the minimum number of elements needed would 1, I would just use 12.
Thus, my question is how would you: find the minimum number of elements to make some sum, and if you can't output -1.
Please suggest an algorithm I can look into so I can solve this efficiently. I've already tried brute force but it is much to slow for my needs.
The problem is np-complete and can be reduced to subset sum or knapsack problem. There is pseudo polynomial time algorithm that can solve it using dynamic programming. Following is a solution similar to knapsack analogy:-
1. Knapsack capacity = Sum
2. Items have same weight and value
3. Maximize profit
4. if max_profit == Sum then there is a solution
5. else Sum cannot be made from the items given.
6. Evaluate the minimum items needed using matrix alongside the DP.
7. Can also reconstruct all solutions and get the minimum one.
Time Complexity : - O(Sum*Items)
Java Implementation :-
public class SubSetSum {
static int[][] costs;
static int[][] minItems;
public static void calSets(int target,int[] arr) {
costs = new int[arr.length][target+1];
minItems = new int[arr.length][target+1];
for(int j=0;j<=target;j++) {
if(arr[0]<=j) {
costs[0][j] = arr[0];
minItems[0][j] = 1;
}
}
for(int i=1;i<arr.length;i++) {
for(int j=0;j<=target;j++) {
costs[i][j] = costs[i-1][j];
minItems[i][j] = minItems[i-1][j];
if(arr[i]<=j) {
costs[i][j] = Math.max(costs[i][j],costs[i-1][j-arr[i]]+arr[i]);
if(costs[i-1][j]==costs[i-1][j-arr[i]]+arr[i]) {
minItems[i][j] = Math.min(minItems[i][j],minItems[i-1][j-arr[i]]+1);
}
else if(costs[i-1][j]<costs[i-1][j-arr[i]]+arr[i]) {
minItems[i][j] = minItems[i-1][j-arr[i]]+1;
}
}
}
}
// System.out.println(costs[arr.length-1][target]);
if(costs[arr.length-1][target]==target) {
System.out.println("Minimum items need : "+minItems[arr.length-1][target]);
}
else System.out.println("No such Set found");
}
public static void main(String[] args) {
int[] arr = {1,1,1,1, 1 ,1 ,1, 1 ,1, 1 ,1 ,1, 10 ,12, 2};
calSets(12, arr);
}
}
here is a recursive approach that should be rather fast:
1) if your input vector is of length 1, either return 1 if the value is equal the target, or return -1 if it doesn't. similarly, if your target is less than any of your items in your input vector, return -1. 2) otherwise, loop on (unique) values in your input vector (in descending order, for performance): 2a) remove the value for your vector, and substract it from your target. 2b) recursively call this function on the new vector and the new target note: you can pass down the algorithm a max.step parameter, so that if you have already found a solution with length K, you would stop the recursive calls at that depth, but not beyond. remember to decrease your max.step value in each recursive call. 3) collect all the values from the recursive calls, take the minimum (which is not -1) and add 1 to it and return, or, if all values in the loop are -1, return -1.
Disclaimer: This is an advertisement for nice but relatively simple mathematics which leads to very clever and fast counting formulas and algorithms. I'm aware that you can find a much simpler and efficient solution using usual programming. I just like the fact that using properly a Computer Algebra System you can do it in a one liner: Lets get 19 with this list:
sage: l = [1,1,1,2,5,2,1,3,12,1,3]; goal = 19
sage: prod((1+t*x^i) for i in l).expand().collect(x).coefficient(x,goal).low_degree(t)
3
What about 25:
sage: goal=25
sage: prod((1+t*x^i) for i in l).expand().collect(x).coefficient(x,goal).low_degree(t)
5
36 is not feasible:
sage: goal=36
sage: prod((1+t*x^i) for i in l).expand().collect(x).coefficient(x,goal).low_degree(t)
0
Here are some details: Just expand the product
(1+t*x^l[0]) (1+t*x^l[1]) ... (1+t*x^l[n])
Where your list is l. Then to find the minimum number of element required to get the sum S, collect the coefficients of x^S and return the minimum degree of a term in t.
Here is how it could be done in sage:
sage: var("x t")
(x, t)
sage: l = [1,1,1,2,5,2,1,3,12,1,3]
sage: s = prod((1+t*x^i) for i in l)
sage: s = expand(s).collect(x)
Now
sage: print(s)
t^11*x^32 + 5*t^10*x^31 + 2*(t^10 + 5*t^9)*x^30 + 2*(t^10 + 5*t^9 + 5*t^8)*x^29 + (11*t^9 + 20*t^8 + 5*t^7)*x^28 + (t^10 + 4*t^9 + 25*t^8 + 20*t^7 + t^6)*x^27 + 2*(3*t^9 + 10*t^8 + 15*t^7 + 5*t^6)*x^26 + (2*t^9 + 17*t^8 + 40*t^7 + 20*t^6 + 2*t^5)*x^25 + (2*t^9 + 12*t^8 + 30*t^7 + 40*t^6 + 7*t^5)*x^24 + (11*t^8 + 30*t^7 + 35*t^6 + 20*t^5 + t^4)*x^23 + 2*(2*t^8 + 13*t^7 + 20*t^6 + 13*t^5 + 2*t^4)*x^22 + (t^8 + 20*t^7 + 35*t^6 + 30*t^5 + 11*t^4)*x^21 + (t^10 + 7*t^7 + 40*t^6 + 30*t^5 + 12*t^4 + 2*t^3)*x^20 + (5*t^9 + 2*t^7 + 20*t^6 + 40*t^5 + 17*t^4 + 2*t^3)*x^19 + 2*(t^9 + 5*t^8 + 5*t^6 + 15*t^5 + 10*t^4 + 3*t^3)*x^18 + (2*t^9 + 10*t^8 + 10*t^7 + t^6 + 20*t^5 + 25*t^4 + 4*t^3 + t^2)*x^17 + (11*t^8 + 20*t^7 + 5*t^6 + 5*t^5 + 20*t^4 + 11*t^3)*x^16 + (t^9 + 4*t^8 + 25*t^7 + 20*t^6 + t^5 + 10*t^4 + 10*t^3 + 2*t^2)*x^15 + 2*(3*t^8 + 10*t^7 + 15*t^6 + 5*t^5 + 5*t^3 + t^2)*x^14 + (2*t^8 + 17*t^7 + 40*t^6 + 20*t^5 + 2*t^4 + 5*t^2)*x^13 + (2*t^8 + 12*t^7 + 30*t^6 + 40*t^5 + 7*t^4 + t)*x^12 + (11*t^7 + 30*t^6 + 35*t^5 + 20*t^4 + t^3)*x^11 + 2*(2*t^7 + 13*t^6 + 20*t^5 + 13*t^4 + 2*t^3)*x^10 + (t^7 + 20*t^6 + 35*t^5 + 30*t^4 + 11*t^3)*x^9 + (7*t^6 + 40*t^5 + 30*t^4 + 12*t^3 + 2*t^2)*x^8 + (2*t^6 + 20*t^5 + 40*t^4 + 17*t^3 + 2*t^2)*x^7 + 2*(5*t^5 + 15*t^4 + 10*t^3 + 3*t^2)*x^6 + (t^5 + 20*t^4 + 25*t^3 + 4*t^2 + t)*x^5 + (5*t^4 + 20*t^3 + 11*t^2)*x^4 + 2*(5*t^3 + 5*t^2 + t)*x^3 + 2*(5*t^2 + t)*x^2 + 5*t*x + 1
Ok this is a huge expression. The nice feature here is that If I take the coefficient say of x^17 I get:
sage: s.coefficient(x, 17)
2*t^9 + 10*t^8 + 10*t^7 + t^6 + 20*t^5 + 25*t^4 + 4*t^3 + t^2
which says the following: the term 10*t^7 tells me that there are 10 different way to obtains the sum 17 using 7 number. Another example, there are 25 way to get 17 using 4 number (25*t^4).
Also since this expression ends with t^2 I learn that I only need two number to get 17. Unfortunately this doesn't tells which numbers.
If you want to understand the trick, look at Wikipedia article on generating functions and This Page.
Note 1: this is not the most efficient since I compute much more than what you need. The huge expression actually described and somehow computed all possible choices (that is 2^the length of the list). But it's a one liner:
sage: prod((1+t*x^i) for i in l).expand().collect(x).coefficient(x,17).low_degree(t)
2
And still relatively efficient:
sage: %timeit prod((1+t*x^i) for i in l).expand().collect(x).coefficient(x,17).low_degree(t)
10 loops, best of 3: 42.6 ms per loop
Note 2: After thinking carefully about it I also realized the following: Generating series is just a compact encoding of what you would have written if you tried to implement a dynamic programming solution.
I don't think this solution is optimal, but it's very easy to understand and use, you sort the elements in decreasing order, then you take each element and try to fit it in your number. If you have the sequence [5,6,2,7] and you need to make the 15 number, you'll reorder the sequence [7,6,5,2] and take 7, then you need to extract 8 so you'll take 6, then you'll need 2 more, check 5 but it's too big and you'll skip it and check the last number, 2, which it's perfect and finishes your number. So you'd print out 3. This is the worst case of the algorithm which is O(n). But in your example with 12, it'll be O(1), because you'll pick 12 from the first checkup of the ordered sequence. (running time applies only for the program of choosing items, not sorting)
resolve_sum(ordered_items[], number) {
count = 0;
aux = number;
i = 0;
while (aux - ordered_items[i] <= 0) {
count = count + 1;
aux = aux - ordered_items[i];
i = i + 1;
}
if (aux == 0) return count;
else return -1;
}
I haven't included an algorithm for sorting, you can choose one that you know best or try to learn a new efiecient one. Link with sorting algorithms and their running time. This is just a sample code you can use in C/C++ or Java or what you need. I hope it isn't way too much brute force.
