Question

The idea is, given an n number of spaces, empty fields, or what have you, I can place in either a number from 0 to m. So if I have two spaces and just 01 , the outcome would be:
(0 1)
(1 0)
(0 0)
(1 1)

if i had two spaces and three numbers (0 1 2) the outcome would be

(0 1)
(1 1)
(0 2)
(2 0)
(2 2)
(2 1)

and so on until I got all 9 (3^2) possible outcomes.

So i’m trying to write a program that will give me all possible outcomes if I have n spaces and can place in any number from 0 to m in any one of those spaces.

Originally I thought to use for loops but that was quickly shotdown when I realzed I’d have to make one for every number up through n, and that it wouldn’t work for cases where n is bigger.

I had the idea to use a random number generator and generate a number from 0 to m but that won’t guarantee I’ll actually get all the possible outcomes.

I am stuck 🙁

Ideas?

Any help is much appreciated 🙂

Answer 1

Basically what you will need is a starting point, ending point, and a way to convert from each state to the next state. For example, a recursive function that is able to add one number to the smallest pace value that you need, and when it is larger than the maximum, to increment the next larger number and set the current one back to zero.

Take this for example:

#include <iostream>
#include <vector>

using namespace std;

// This is just a function to print out a vector.
template<typename T>
inline ostream &operator<< (ostream &os, const vector<T> &v) {
    bool first = true;
    os << "(";
    for (int i = 0; i < v.size (); i++) {
        if (first) first = false;
        else os << " ";
        os << v[i];
    }
    return os << ")";
}

bool addOne (vector<int> &nums, int pos, int maxNum) {
    // If our position has moved off of bounds, so we're done
    if (pos < 0)
        return false;

    // If we have reached the maximum number in one column, we will
    // set it back to the base number and increment the next smallest number.
    if (nums[pos] == maxNum) {
        nums[pos] = 0;
        return addOne (nums, pos-1, maxNum);
    }

    // Otherwise we simply increment this numbers.
    else {
        nums[pos]++;
        return true;
    }
}

int main () {
    vector<int> nums;
    int spaces = 3;
    int numbers = 3;

    // populate all spaces with 0
    nums.resize (spaces, 0);

    // Continue looping until the recursive addOne() function returns false (which means we
    // have reached the end up all of the numbers)
    do {
        cout << nums << endl;    
    } while (addOne (nums, nums.size()-1, numbers));

    return 0;
}