These notes were written by Stuart Reges for his recursion lecture

We want to write a method called stutter that will take an integer value like 348 and return the integer 334488 (the value you get by replacing each digit with two of that digit). How do we write this? We again started with simple cases. What are easy numbers to stutter? Someone suggested 0, although that's actually a hard number to stutter. We can't really return 00, although we'll consider it acceptable to return 0. So what are some easy numbers to stutter? 1, 2, any one digit number.

So how do we stutter one-digit numbers? To turn a digit n into nn, we'd need to do something like this:

        n0  10 of n
         n  plus 1 of n
        --
        nn
So we need a total of 11 times n. Some people saw this without thinking about individual digits and it's pretty obvious when you think about it that 1 times 11 is 11, 2 times 11 is 22, 3 times 11 is 33, and so on. I said to assume for now a precondition that n is greater than or equal to 0. So we have our base case:

        //pre : n >= 0
        public int stutter(int n) {
            if (n < 10)
                return 11 * n;
            else
                return ...;
        }
So what do we do with numbers like 348? We need some way to break it up into smaller chunks, a way to split off one or more digits. Someone suggested that division by 10 would help:

        n = 348

             34 | 8
         -------+-------
         n / 10 | n % 10
What next? Here you have to think recursively. The original number was 348. We've split it up into 34 and 8. What if we stuttered those values? You have to make the leap of faith and believe that the method actually works. If it does, then it turns 34 into 3344 and 8 into 88. How do we combine them to get 334488? We can't do simple addition:

        3344
          88
        ----
        wrong
We need to shift the 88 over to the right relative to the 3344. We do this by multiplying 3344 by 100:

        334400
            88
        ------
        334488
But that means we've solved the problem:

        //pre : n >= 0
        public int stutter(int n) {
            if (n < 10)
                return 11 * n;
            else
                return 100 * stutter(n / 10) + stutter(n % 10);
        }
I mentioned that the final version of the method in Stutter.java includes an extra case for negative numbers. So we remove the precondition and add a case for negative n. For a number like -348, we return -334488 by stuttering the positive number (stutter 348) and then negating the result.