11.8. Making A Function Call Itself¶

11.8.1. A Visual Representation¶

To help visualize what happens during recursion, let's start with the base case ['L']:

Nothing complicated here. combineEntries sees only one item in the array, so it returns 'L'.

Now consider an array with two elements, ['L', 'C']:

In this case, combineEntries executes the else statement. We have no code for this yet, but we can still consider the logic:

1. combineEntries returns 'L' and calls itself again using what is left inside the array (['C']).
2. When passed ['C'], which is the base case, combineEntries returns 'C'.
3. The strings 'L' and 'C' get combined and returned as the final result.

Next, consider an array with three elements ['L', 'C', '1']:

As before, combineEntries executes the else statement, and we can follow the logic:

1. combineEntries returns 'L' and calls itself again using what is left inside the array (['C', '1']).
2. When passed ['C', '1'], combineEntries returns 'C' and calls itself again using what is left inside the array (['1']).
3. When passed ['1'], which is the base case, combineEntries returns '1'.
4. The strings 'C' and '1' get combined and returned.
5. The strings 'L' and 'C1' get combined and returned as the final result.

As we make the array longer, combineEntries calls itself more times. Each call evaluates a smaller and smaller section of the array until reaching the base case. This sets up a series of return events - each one selecting a single entry from the array. Rather than building 'LC101' from left to right, recursion constructs the string starting with the base case and adding new characters to the front:

Recursive processes all follow this approach. Each call to the function reduces a problem into a slighly smaller piece. The reduction continues until reaching the simplest possible form---the base case. The base case is then solved, and this creates a starting point for completing all of the previous steps.

11.8.2. A Function Calls Itself¶

So how do we code the else statement in combineEntries? Recall what needs to happen each time the statement runs:

1. Select the first element in the array,
2. Call combineEntries again with a smaller array.

Bracket notation takes care of part a: arrayName[0].

For part b, remember that the slice method returns selected entries from an array. To return everything BUT the first entry in arr = ['L', 'C', '1', '0', '1'], use arr.slice(1).

Let's add the bracket notation and the slice method to our function:

1function combineEntries(arrayName){
2   if (arrayName.length <= 1){
3      return arrayName[0];
4   } else {
5      return arrayName[0]+combineEntries(arrayName.slice(1));
6   }
7}

Each time the else statement runs, it extracts the first element in the array with arrayName[0], then it calls itself with the remaining array elements (arrayName.slice(1)).

For combineEntries(['L', 'C', '1', '0', '1']);, the sequence would be:

To get the final result, proceed up the chain:

See this recursion in action.

11.8.3. Check Your Understanding¶

1function decreasingSum(integer) {
2   if (integer === 1){
3      return integer;
4   } else {
5      //call decreasingSum function again
6   }
7}