How to print all permutation using backtracking? - c++

Question

This code prints all the permutation of N-1 items. But I could not understand one thing:
when n=N, it is returning where it is called and make flag[n-1] = false. Thus, i = N-1 and breaks the loop. But how is the rest of the permutation printing or returning when n=N-2 to 0?
void perm(int n) {
if (n == N) {
for (int i = 0; i < N ; i++) {
cout<<a[i]<<" ";
}
cout<<endl;
return;
}
for (int i = 0; i < N; i++) {
if (flag[i]) continue;
a[n] = i;
flag[i] = true;
cout<<i<<endl;
perm(n + 1);
cout<<i<<endl;
flag[i] = false;
}
}

Answer 1

You need to consider that function calls ended up nested.
Each indentation below shows a nested call:
main()
entering perm(0)
entering perm(1)
i = 0
entering perm(2)
let's say N is 2, this will print and return
now perm(1) continues
i becomes 1
entering perm(2)
...
Since return only returns from the current function call, not all function calls, the permutations continue printing.
In order to gain familiarity with this, try this:
void perm(int n) {
std::cout << "Entering perm " << n << std::endl;
if (n == N) {
for (int i = 0; i < N ; i++) {
cout<<a[i]<<" ";
}
cout<<endl;
std::cout << "Exiting perm " << n << std::endl;
return;
}
for (int i = 0; i < N; i++) {
if (flag[i]) continue;
a[n] = i;
flag[i] = true;
cout<<i<<endl;
std::cout << "about to call perm" << std::endl;
perm(n + 1);
std::cout << "finished call to perm" << std::endl;
cout<<i<<endl;
flag[i] = false;
}
std::cout << "Exiting perm " << n << " (2)"<< std::endl;
}