Typically, you can quantify how well an algorithm will scale by using big-O notation to analyze its growth rate. When you say that your algorithm works by “brute force,” it’s unclear to what extent it will scale. If your “brute force” solution works by listing all possible subsets or combinations of a set of data, then it almost certainly will not scale (it will have asymptotic complexity O(2n) or O(n!), respectively). If your brute force solution works by finding all pairs of elements and checking each, it may scale reasonably well (O(n2)). Without more information about how your algorithm works, though, it’s difficult to say.
You may want to look at this excellent post about big-O as a starting point for how to reason about the long-term scalablility of your program. Typically speaking, anything that has growth rate O(n log n), O(n), O(log n), or O(1) scale extremely well, anything with growth rate O(n2) or O(n3) will scale up to a point, and anything with growth rate O(2n) or higher will not scale at all.
Another option would be to look up the problem you’re trying to solve to see how well-studied it is. Some problems are known to have great solutions, and if yours is one of them it might be worth seeing what others have come up with. Perhaps there is a very clean, non-brute-force solution that scales really well! Some other problems are conjectured to have no scalable algorithms at all (the so-called NP-hard problems). If that’s the case, then you should be pretty confident that there’s no way to get a scalable approach.
And finally, you can always ask a new question here at Stack Overflow describing what you’re trying to do and asking for input. Maybe the community can help you solve your problem more efficiently than you initially expected!
EDIT: Given the description of the problem that you are trying to solve, right now you are doing one for loop per variable from 0 up to the number you’re trying to target. The complexity of this algorithm is O(Uk), where k is the number of variables and U is the sum. This approach will not scale very well at all. Introducing each new variable in the above case will make the algori2thm run 100 times slower, which definitely will not scale very well if you want 100 variables!
However, I think that there is a fairly good algorithm whose runtime is O(U2k) that uses O(Uk) memory to solve the problem. The intuition is as follows: Suppose that we want to sum up 1, 2, and 4 to get 10. There are many ways to do this:
2 * 4 + 1 * 2 + 0 * 1
2 * 4 + 0 * 2 + 2 * 1
1 * 4 + 3 * 2 + 0 * 1
1 * 4 + 2 * 2 + 2 * 1
1 * 4 + 1 * 2 + 4 * 1
1 * 4 + 0 * 2 + 6 * 1
0 * 4 + 5 * 2 + 0 * 1
0 * 4 + 4 * 2 + 2 * 1
0 * 4 + 3 * 2 + 4 * 1
0 * 4 + 2 * 2 + 6 * 1
0 * 4 + 1 * 2 + 8 * 1
0 * 4 + 0 * 2 + 10 * 1
The key observation is that we can write all of these out as sums, but more importantly, as sums where each term in the sum is no greater than the previous term:
2 * 4 + 1 * 2 + 0 * 1 = 4 + 4 + 2
2 * 4 + 0 * 2 + 2 * 1 = 4 + 4 + 1 + 1
1 * 4 + 3 * 2 + 0 * 1 = 4 + 2 + 2 + 2
1 * 4 + 2 * 2 + 2 * 1 = 4 + 2 + 2 + 1 + 1
1 * 4 + 1 * 2 + 4 * 1 = 4 + 2 + 1 + 1 + 1 + 1
1 * 4 + 0 * 2 + 6 * 1 = 4 + 1 + 1 + 1 + 1 + 1 + 1
0 * 4 + 5 * 2 + 0 * 1 = 2 + 2 + 2 + 2 + 2
0 * 4 + 4 * 2 + 2 * 1 = 2 + 2 + 2 + 2 + 1 + 1
0 * 4 + 3 * 2 + 4 * 1 = 2 + 2 + 2 + 1 + 1 + 1 + 1
0 * 4 + 2 * 2 + 6 * 1 = 2 + 2 + 1 + 1 + 1 + 1 + 1 + 1
0 * 4 + 1 * 2 + 8 * 1 = 2 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1
0 * 4 + 0 * 2 + 10 * 1 = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1
So this gives an interesting idea about how to generate all possible ways to sum up to the target. The idea is to fix the first coefficient, then generate all possible ways to make the rest of the sum work out. In other words, we can think about the problem recursively. If we list the variables in order as x1, x2, …, xn, then we can try fixing some particular coefficient for x1, then solving the problem of summing up sum - c_1 x_1 using just x2, …, xn.
So far this doesn’t seem all that fancy – in fact, it’s precisely what you’re doing above – but there is one trick we can use. As long as we’re going to be thinking about this problem recursively, let’s think about the problem in the opposite manner. Rather than starting with sum and trying to break it down, what if instead we started with 0 and tried to build up everything that we could?
Here’s the idea. Suppose that we already know in advance all the numbers we can make using just sums of x1. Then for every number k between 0 and sum, inclusive, we can make k out of x2 and x1 out of any combination where k – c2 x2 is something that can be made out of combinations of x1. But since we’ve precomputed this, we can just iterate up over all possible legal values of c2, compute k – c2 x2, and see if we know how to make it. Assuming we store a giant U x (k + 1) table of boolean values such that table entry [x, y] stores “can we sum up the first y values, inclusive, in a way that sums up to precisely U?,” we can fill in the table efficiently. This is called dynamic programming and is a powerful algorithmic tool.
More concretely, here’s how this might work. Given k variables, create a U x (k + 1) table T of values. Then, set T[0][0] = true and T[x][0] = false for all x > 0. The rationale here is that T[0][0] means “can we get the sum zero using a linear combination of the first zero variables?” and the answer is definitely yes (the empty sum is zero!), but for any other sum made of no a linear combination of no variables we definitely cannot make it.
Now, for i = 1 .. k, we’ll try to fill in the values of T[x][i]. Remember that T[x][i] means “can we make x as a linear combination of the first i variables?” Well, we know that we can do this if there is some coefficient c such that k – c xi can be made using a linear combination of x1, x2, …, xi – 1. But for any c, that’s just whether T[x – c xi][i – 1] is true. Thus we can say
for i = 1 to k
for z = 0 to sum:
for c = 1 to z / x_i:
if T[z - c * x_i][i - 1] is true:
set T[z][i] to true
Inspecting the loops, we see that the outer loop runs k times, the inner loop runs sum times per iteration, and the innermost loop runs also at most sum times per iteration. Their product is (using our notation from above) O(U2 k), which is way better than the O(Uk) algorithm that you had originally.
But how do you use this information to list off all of the possible ways to sum up to the target? The trick here is to realize that you can use the table to avoid wasting a huge amount of effort searching over every possible combination when many of them aren’t going to work.
Let’s see an example. Suppose that we have this table completely computed and want to list off all solutions. One idea is to think about listing all solutions where the coefficient of the last variable is zero, then when the last variable is one, etc. The issue with the approach you had before is that for some coefficients there might not be any solutions at all. But with the table we have constructed above, we can prune out those branches. For example, suppose that we want to see if there are any solutions that start with xk having coefficient 0. This means that we’re asking if there are any ways to sum up a linear combination of the first k – 1 variables so that the sum of those values is sum. This is possible if and only if T[sum][k – 1] is true. If it is true, then we can recursively try assigning coefficients to the rest of the values in a way that sums up to sum. If not, then we skip this coefficient and go on to the next.
Recursively, this looks something like this:
function RecursivelyListAllThatWork(k, sum) // Using last k variables, make sum
/* Base case: If we've assigned all the variables correctly, list this
* solution.
*/
if k == 0:
print what we have so far
return
/* Recursive step: Try all coefficients, but only if they work. */
for c = 0 to sum / x_k:
if T[sum - c * x_k][k - 1] is true:
mark the coefficient of x_k to be c
call RecursivelyListAllThatWork(k - 1, sum - c * x_k)
unmark the coefficient of x_k
This recursively will list all the solutions that work, using the values in the table we just constructed to skip a huge amount of wasted effort. Once you’ve built this table, you could divvy this work up by farming out the task to multiple computers, having them each list a subset of the total solutions, and processing them all in parallel.
Hope this helps!