Recursion exercises solutions

• Define multiplication over positive integers, using only increment by 1, decrement by 1, add (from the previous page) and recursion.

  int mult(int x, int y) { if (x == 0 || y == 0) { // The base case is the simplest possible multiplication return 0; } else { // The purpose of the inductive case is to shrink x-1 down // to the smallest nonnegative integer. It uses the subproblem // mult(x-1, y) to solve the larger problem mult(x,y). return add(x, mult(x-1, y)); } }

• Write a recursive definition of indexOf from the first page, for an unsorted Vector. You will need an extra parameter:

  // Don't be intimidated by the size of this solution, it's mostly // comments. // First, note that the indexOf method will first be called with 0, // which means we will start looking with the first element of // the array: Vector v; Item i; // ... some stuff, then: v.indexOf(i, 0); // Now, we can define a recursive search that "shrinks" the size // of the vector left to search one item at a time. int Vector::indexOf(Item item, int startIdx) { // In this function, we have two base cases, corresponding to // the two possible outcomes of our search if ( startIdx >= length() ) { // This is to catch ourselves if we run off the end // of the vector return INVALID_INDEX; } else if ( item.equals( get(startIdx) ) ) { // This is if we've found the item return startIdx; } else { // Otherwise, we need to search the rest of the vector. return indexOf(item, startIdx + 1); } }

  Why is this version less desirable than the iterative version?

  Answer: because each recursive call will require a new activation record on the stack. Searching down a 10,000 element vector, for example, will require 10,000 activation records. This takes a lot of memory.