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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 9.15. EXAMPLE 9.15 The Towers of Hanoi
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This program prints the solution to the Towers of Hanoi problem of moving three disks from peg A to peg C via peg B:
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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 } } }
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The output is:
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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
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to to to to to to to
peg peg peg peg peg peg peg
C. B. B. C. A. C. C.
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:
Move top disk from peg A to peg C.
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:
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