Partitioning an Integer into 9 non-negative integers with limits

Question

There's an equation: 1a + 2b + 3c + 4d ... + 9i = 9

Constraints: 1 <= a + b + c + ... + i <= 10⁴

where a,b,..,i are non-negative integers and each of the integers have a particular range.

For example: 1 <= a <= 5, 2 <= b <= 3, and so on.

I need to find out the number of different sets of values of those variables, or simply, the number of ways of solving that equation.

There's a recursive method of solving this, but that's very slow. I'm unable to think of a way to solve this efficiently under the given constraint.

Your first constraint seems to be useless. Since all numbers are non-negative, and `1a + 2b + 3c + 4d ... + 9i = 9`, then `a + b + c + ... + i` would definitely be at least `1` and at most `9`. — Shahbaz, Aug 07 '12 at 10:37
@HighPerformanceMark: `(0,0,0,0,...,1)` satisfies it. So does `(9,0,...,0)`. It is not a set, but I *think* that is what the op is after. — amit, Aug 07 '12 at 10:42

score 2 · Answer 1 · edited May 23 '17 at 10:34

2

If you think about it, the solutions are actually very limited.

Since all numbers are non-negative, and you have:

1a + 2b + 3c + 4d ... + 9i = 9

This implies that:

0 <= a <= 9
0 <= b <= 4
0 <= c <= 3
0 <= d <= 2
0 <= e <= 1
0 <= f <= 1
0 <= g <= 1
0 <= h <= 1
0 <= i <= 1

That is, there are only 10*5*4*3*2*2*2*2*2 = 19200 cases to consider.

You can iterate through these cases and find out which ones satisfy your other constraints and which ones have

1a + 2b + 3c + 4d ... + 9i = 9

Hint: start by giving values from i to a. This way, a high value of i or h for example, immediately reduces the possible range of values of the smaller numbers.

Make sure you apply MSalters' method before computing. Even though in this case it is not necessary, since the problem is too easy, but in general it helps a lot.

edited May 23 '17 at 10:34

Community

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answered Aug 07 '12 at 10:42

Shahbaz

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to be exact, you need to check 5*4*3*2*2*2*2*2=1920 possibilities, and if there is still feasible solution possible - the value of `a` is already defined. – amit Aug 07 '12 at 10:46
@amit, that's exactly what you'll get if you start assigning values to the ones with the highest coefficient first. But thanks for pointing that out. – Shahbaz Aug 07 '12 at 10:48
@Shabaz: Yeap, I missed the line you mentioning it. Anyway, this should do for a problem with this dimensionality and constraints.+1 for simplicity. – amit Aug 07 '12 at 10:50

score 1 · Answer 2 · answered Aug 07 '12 at 10:43

You'd basically want to substitute the variables for which the range is restricted, so a' = a+1, 0 <= a' <= 4 and b' = b+2, 0 <= b' <= 1. Starting at zero makes the math easier. It also allows you to rewrite the equation as 1a' + 2b' + ... + 9i = 4. Since all terms are non-negative, this greatly restricts the search space. For instance, it means that e up to i must all be 0. This reduces the equation to just `1a' + 2b' + 3c + 4d = 4.

score 1 · Answer 3 · answered Aug 07 '12 at 10:56

Another solution would be to solve 1a + 2b + 3c + 4d ... + 9i = 1 (trivial) and the find all solutions for 1a + 2b + 3c + 4d ... + 9i = N+1 given the solution for 1a + 2b + 3c + 4d ... + 9i = N. That's basically a => a+1 or a=>a-1, b=>b+1 or b=>b-1, c=>c+1 etc.

This is nicely recursive, but only takes 8 iterations to get to N=9 and in each iteration you're just incrementing or decrementing 9 variables.

Partitioning an Integer into 9 non-negative integers with limits

3 Answers3