Howto create combinations of several vectors without hardcoding loops in C++?

Question

I have several data that looks like this:

Vector1_elements = T,C,A
Vector2_elements = C,G,A
Vector3_elements = C,G,T
..... up to ...
VectorK_elements = ...

#Note also that the member of each vector is always 3.

What I want to do is to create all combination of elements in Vector1 through out VectorK. Hence in the end we hope to get this output (using Vector1,2,3):

TCC
TCG
TCT
TGC
TGG
TGT
TAC
TAG
TAT
CCC
CCG
CCT
CGC
CGG
CGT
CAC
CAG
CAT
ACC
ACG
ACT
AGC
AGG
AGT
AAC
AAG
AAT

The problem I am having now is that the following code of mine does that by hardcoding the loops. Since number of Vectors can be varied, we need a flexible way to get the same result. Is there any?

This code of mine can only handle up to 3 Vectors (hardcoded):

#include <iostream>
#include <vector>
#include <fstream>
#include <sstream>
using namespace std;


int main  ( int arg_count, char *arg_vec[] ) {

    vector <string> Vec1;
          Vec1.push_back("T");
          Vec1.push_back("C");
          Vec1.push_back("A");

    vector <string> Vec2;
          Vec2.push_back("C");
          Vec2.push_back("G");
          Vec2.push_back("A");

    vector <string> Vec3;
          Vec3.push_back("C");
          Vec3.push_back("G");
          Vec3.push_back("T");



     for (int i=0; i<Vec1.size(); i++) {
        for (int j=0; j<Vec2.size(); j++) {
            for (int k=0; k<Vec1.size(); k++) {
                cout << Vec1[i] << Vec2[i] << Vec3[k] << endl;
            }
        }
     }



    return 0;
}

Answer 1

You can implement this like an odometer, which leads to the following (works for different-sized vectors):

Say you have K vectors in an array v: v[0], v[1], ... v[K-1]

Keep an array of iterators it (size K) into your vectors, starting with it[i] = v[i].begin(). Keep incrementing it[K-1] in a loop. When any iterator hits the end() of the corresponding vector, you wrap it around to begin() and increment the previous iterator also (so when it[K-1] wraps around, you increment it[K-2]). These increments may "cascade" so you should do them in a loop backwards. When it[0] wraps around, you're done (so your loop condition could be something like while (it[0] != v[0].end())

Putting all that together, the loop that does the work (after setting up the iterators) should be something like:

while (it[0] != v[0].end()) {
  // process the pointed-to elements

  // the following increments the "odometer" by 1
  ++it[K-1];
  for (int i = K-1; (i > 0) && (it[i] == v[i].end()); --i) {
    it[i] = v[i].begin();
    ++it[i-1];
    }
  }

If you're interested in complexity, the number of iterator increments that get performed is easy to calculate. For simplicity here I'll assume each vector is the same length N. The total number of combinations is N^K. The last iterator gets incremented each time, so that is N^K, and moving back through the iterators this count gets divided by N each time, so we have N^K + N^K-1 + ... N¹; this sum equals N(N^K - 1)/(N-1) = O(N^K). This also means that the amortized cost per-combination is O(1).

Anyway, in short, treat it like an odometer spinning its digit wheels.