Determining a valid DNA sequence using recursion - c++

Question

I've been stuck on this problem that I need to solve for a school assignment a quite a while now, over a week and I still don't have any idea of what to do.
DNA forms the building blocks of our genetic sequence. DNA is made up of two chains of nucleotides. Each nucleotide is one of four possibilities:
A - Adenine
T - Thymine
C - Cytosine
G - Guanine
When combined in the DNA Sequence, A and T can bind and C and G can bind. Other combinations are not valid.
For this assignment, you are going to ask the user for two sequences, then return the number of valid pairs by comparing corresponding positions in the sequence. In other words, you will look at the first character in each. If these are an A & T or a C & G, it s a valid pair. You will then move on to the second pair. You can ignore any invalid characters, but you should not assume that the length of the sequences are equal.
I'm not sure how to go about making a recursion statement that would help this assignment. I'm not sure how I would go about finding the base case for this problem and making a function using recursion that gets back to the base case.
#include "util.h"
using namespace std;
int valid(string first, string second)
{
int valid_pairs;
return valid_pairs;
}
int main()
{
string first;
string second;
cout << "Please enter the first sequence: ";
cin >> first;
cout << "Please enter the second sequence: ";
cin >> second;
cout << "Valid pairs: " << valid(first, second) << endl;
return 0;
}

Answer 1

Recursion is mostly a matter of writing down the possible situations and doing The Right Thing for each one in turn.
In your case, those possibilities are
At least one of the strings is empty (in which case there are no pairs at all)
first[0] and second[0] make a valid pair
first[0] and second[0] don't make a valid pair
Since you know that valid(first.substr(1), second.substr(1)) is going to be the number of valid pairs without the two initial characters, you can now write down the solution.
Doing that left as an exercise.
(First writing the function bool is_valid(char c1, char c2)that determines whether two characters make a valid pair will simplify this function.)