Question

Is there an efficient way to run nested for loops, avoiding any of the counters having the same value. Obviously, I could run some if statements, as below, but is there a more efficient way?

for i = 1 to 20:
    for j = 1 to 20:
        if (i == j):
            continue
        else:            
            for k = 1 to 20:
                if (i == k) or (j == k):
                    continue
                else:
                    do something useful with these different numbers

EDIT: The variables are not interchangeable, so [2, 1, 0] is different to [0, 1, 2]. The “do something useful” will be about 6 numerical checks on the numbers, involving adding, squaring, and square rooting them.

Thanks, and sorry about the possibly unusual pseudocode (and the constant editing).

Answer 1

That looks like the most efficient way you can do it. What’s wrong with how you have it?

I’m just going to ignore the part where I can’t think of any reason to ever need to do this…. so if you have a specific case please share? Unless your trying to compare an item to all other items in a list that isn’t itself I guess?

which could be done as

list = {1,2,3,4,1,2,3,4} \\where list[0] will return 1, and list.size() will return 8
for(int i = 0; i < list.size()-1; i++){
    for(int j = i + 1; j < list.size(); j++){
        System.out.println(list[i] + "," + list[j]);
    }
}

That way you dont repeat comparisons between things you have already compared.

For three nests

list = {1,2,3,4,1,2,3,4} \\where list[0] will return 1, and list.size() will return 8
for(int i = 0; i < list.size()-2; i++){
    for(int j = i + 1; j < list.size()-1; j++){
        for(int k = j + 1; j < list.size(); k++){
            System.out.println(list[i] + "," + list[j] + "," + list[k]);
    }
}