Simplifying C++ For Loop - c++

Question

at the moment I have the following code:
for(int i = 0; i < 4; i++){
cout << rowNo[i] << endl;
}
for(int i = 0; i < 4; i++){
for(int j = 0; j < 4; j++){
cout << rowNo[i] << '.';
cout << rowNo[j] << endl;
}
}
for(int i = 0; i < 4; i++){
for(int j = 0; j < 4; j++){
for(int k = 0; k < 4; k++){
cout << rowNo[i] << '.';
cout << rowNo[j] << '.';
cout << rowNo[k] << endl;
}
}
}
for(int i = 0; i < 4; i++){
for(int j = 0; j < 4; j++){
for(int k = 0; k < 4; k++){
for(int l = 0; l < 4; l++){
cout << rowNo[i] << '.';
cout << rowNo[j] << '.';
cout << rowNo[k] << '.';
cout << rowNo[l] << endl;
}
}
}
}
Where rowNo[] is an array {1,2,3,4}
And I was wondering two things:
Can this be simplified, so maybe put into some sort of recursive loop?
Following that, can this then be made for an array of size N?

Answer 1

Your looking for Cartesian_product
With
bool increment(std::vector<std::size_t>& v, std::size_t maxSize)
{
for (auto it = v.rbegin(); it != v.rend(); ++it) {
++*it;
if (*it != maxSize) {
return true;
}
*it = 0;
}
return false;
}
then you can do:
void print_cartesian_product(const std::vector<int>&v, int n)
{
std::vector<std::size_t> indexes(n);
do {
print(v, indexes);
} while (increment(indexes, v.size()));
}
Demo

Answer 2

You are actually trying to print a number encoded in base4 with digit {1, 2, 3, 4}. To achieve it, You only need to define a function to increment by one. I propose a generic solution in the term of amount of number to print and base.
Like others, I use a number to mean "empty digit", and I use zero which is quite convenient.
Complete source code :
#include <iostream>
#include <vector>
bool increment_basep(std::vector<int>& number, int p)
{
int i = 0;
while(i < number.size() && number[i] == p)
{
number[i] = 1;
++i;
}
if(i >= number.size())
return false;
++number[i];
return true;
}
void print_vect(std::vector<int>& number)
{
for(int i = number.size() -1 ; i >= 0; --i)
{
if(number[i] != 0)
std::cout << number[i];
}
std::cout << std::endl;
}
int main() {
int n = 4;
int p = 4;
std::vector<int> num4(n);
std::fill(num4.begin(), num4.end(), 0);
while(increment_basep(num4, p))
{
print_vect(num4);
}
return 0;
}
The increment return whether or not the computation has overflown. When we overflow we know we need to stop.

Answer 3

First solution it comes my mind is that on every loop to put in a buffer and finally to print all the buffers.
I think there are some other ingenious methods
for(int i = 0; i < 4; i++){
put in buffer1 rowNo[i]
for(int j = 0; j < 4; j++){
put in buffer2 rowNo[i],rowNo[j]
for(int k = 0; k < 4; k++){
put in buffer3 rowNo[i],rowNo[j],rowNo[k]
for(int l = 0; l < 4; l++){
put in buffer4 rowNo[i],rowNo[j],rowNo[k],rowNo[l],endl.
}
}
}
}
print(buffer1);
print(buffer2);
print(buffer3);
print(buffer4);

Answer 4

The following is the simplest code I came up with. There has to be a more direct way of doing this though...
It basically introduces a "ghost" index -1, corresponding to an empty place in a number. The ternary operators in the loops conditions are there to avoid duplicates.
int main()
{
int N = 4;
int rowNo[4] = {1, 2, 3, 4};
for (int i = -1; i < N; i++)
for (int j = (i > -1 ? 0 : -1); j < N; j++)
for (int k = (j > -1 ? 0 : -1); k < N; k++)
for (int l = (k > -1 ? 0 : -1); l < N; l++)
{
if (i > -1) std::cout << rowNo[i] << '.';
if (j > -1) std::cout << rowNo[j] << '.';
if (k > -1) std::cout << rowNo[k] << '.';
if (l > -1) std::cout << rowNo[l];
std::cout << std::endl;
}
}
It can of course be generalized to an array of arbitraty size, possibly with some code generation script.