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The general solution to the Towers of Hanoi game is naturally recursive: Part I: Move the smaller n 1 disks from peg A to peg B Part II: Move the remaining disk from peg A to peg C Part III: Move the smaller n 1 disks from peg B to peg C The first and third steps are recursive: Apply the complete solution to n 1 disks The basis to this recursive solution is the case where n = 0 In that case, do nothing The solution for the case of n = 1 disk is: 1 Move the disk from peg A to peg C The solution for the case of n = 2 disks is: 1 Move the top disk from peg A to peg B 2 Move the second disk from peg A to peg C 3 Move the top disk from peg B to peg C.

The solution for the case of n = 3 disks is: 1 Move the top disk from peg A to peg C 2 Move the second disk from peg A to peg B 3 Move the top disk from peg C to peg B 4 Move the remaining disk from peg A to peg C 5 Move the top disk from peg B to peg A 6 Move the second disk from peg B to peg C 7 Move the top disk from peg A to peg C Here, steps 1 3 constitute Part I of the general solution, step 4 constitutes Part II, and steps 5 7 constitute Part III Since the general recursive solution requires the substitution of different peg labels, it is better to use variables.

Then, naming this three-step algorithm hanoi(n, x, y, z), it becomes: Part I: Move the smaller n 1 disks from peg x to peg z Part II: Move the remaining disk from peg x to peg y Part III: Move the smaller n 1 disks from peg z to peg y The general solution is implemented in Example 915 EXAMPLE 915 The Towers of Hanoi.

Figure 7-1

This program prints the solution to the Towers of Hanoi problem of moving three disks from peg A to peg C via peg B:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

public class TestHanoiTowers { public static void main(String[] args) { HanoiTowers(3, 'A', 'B', 'C'); } public static void HanoiTowers(int n, char x, char y, char z) { if (n==1) { // basis System.out.printf("Move top disk from peg %c to peg %c.%n", x, z); } else { HanoiTowers(n-1, x, z, y); // recursion HanoiTowers(1, x, y, z); // recursion HanoiTowers(n-1, y, x, z); // recursion } } }

way the grid behaves and displays items. The following illustration shows the Properties dialog box for the Analysis Grid. Three specific tabs are available in the Analysis Grid Properties dialog box:

CHAP. 9]

The output is:

The downward pressure on short-term interest rates due to an increase in the money supply is also evident when we consider the effect that an increase in bank reserves has upon bank lending. In purchasing government securities and supplying more reserves to the banking system, the Fed increases the supply of excess reserves. Banks can encourage more borrowers to apply for bank loans by lowering their lending rates. Consumer spending on large-ticketed items such as houses and cars is interest-sensitive since individuals are likely to take out loans to pay for major purchases. Business investment purchases of new buildings and equipment is also interest sensitive. Thus, as depicted below, a Fed increase in the money supply should lower the rate of interest, increase interest-sensitive spending, and result in a higher level of spending and gross domestic output. M i total spending gross domestic product

Move Move Move Move Move Move Move top top top top top top top disk disk disk disk disk disk disk from from from from from from from peg peg peg peg peg peg peg A A C A B B A

To solve the problem for three disks, the call at line 3 passes 3 to n, 'A' to x, 'B' to y, and 'C' to z. Since n > 1, line 10 executes next, passing 2 to n, 'A' to x, 'B' to z, and 'C' to y. Again, since n > 1, line 10 executes next, passing 1 to n, 'A' to x, 'B' to y, and 'C' to z. In that call, n = 1, so line 8 executes, printing the first line of output:

That call returns to where the previous call left off at line 10, proceeding to line 11, where n = 2, x = 'A', y = 'C', and z = 'B'. That prints the second line of output:

Then line 12 executes, this time passing 1 to n, 'C' to x, 'A' to y, and 'B' to z. In that call, n = 1 again, so line 8 executes, printing the third line of output: