Tower of Hanoi

1. Tower of Hanoi AE1APS Algorithmic Problem Solving John Drake

2. Outline of the course • Invariants – Chapter 2 • River Crossing – Chapter 3 • Logic Puzzles – Chapter 5 • Matchstick Games - Chapter 4 • Sum Games – Chapter 4 • Induction – Chapter 6 • Tower of Hanoi – Chapter 8

3. A recap on induction • Breaking larger problems down into sub-problems • Using small instances of a problem to help us solve larger instances • Find a trivial base case where size = 0 • Show for an arbitrary number n how to solve n + 1 (induction step)

4. Tower of Hanoi - Introduction • This problem is discussed in many maths texts, and is used in computer science and AI as an illustration of recursion and problem solving • You probably know the problem, but we will look at it slightly differently • Induction gives a systematic way of finding a solution • However this solution is undesirable • A better solution is obtained by observing the invariants of the inductive solution

5. Begin with the solution • We will start at the end • We begin with the solution (as you probably already know it) and you can see where we are going • Another reason is to understand why a correct solution has been found, if no information about the solution method is provided

6.  Problem specification • There is a temple in Bramah, where there are 3 giant poles fixed in the ground. • On the first pole, God placed 64 discs, each of different size in decreasing order of size • The Brahmin monks were given the task of moving the discs, one per day, from one pole to another • However, no disc must be above a smaller disc, when on the same pole

7. Tower of Hanoi

8. Triangle representation • Almost every book (and webpage) I have seen draws the 3 poles in a line • We can call the poles A, B, C • However there is a simple symmetry which is being ignored here

9. Triangle representation • We draw them in different positions • A better way is to draw and equilateral triangle, and the symmetry is obvious. • Now the moves of the discs can easily be described as clockwise or anti-clockwise.

10. Iterative solution (assumptions) • There is a very easy solution to the Towers of Hanoi problem • It is easy to remember and execute • We assume the problem is to move the discs from one pole to the next in the clockwise direction • The days are numbered from 0, 1, 2… • On day 0, the discs are placed in their initial position, and the monks begin moving the discs on day 1.

11. Iterative solution 1 • On every alternate day, the smallest disk is moved • The smallest disc should cycle around the poles. • The direction of rotation depends on the number of discs • It the total number is odd, it cycles clockwise, otherwise if the total is even, it cycles anticlockwise

12. Iterative solution 2 • On every other day, another disc is moved (other than the smallest disc) • As no disc can be on a smaller disc, there is only one possible move for the disc • The algorithm terminates when no further moves are possible • That is, on a even numbered day when all the discs are on the same pole

13. Iterative solution – 4 discs • Day 0

14. Iterative solution – 4 discs • Day 1

15. Iterative solution – 4 discs • Day 2

16. Iterative solution – 4 discs • Day 3

17. Iterative solution – 4 discs • Day 4

18. Iterative solution – 4 discs • Day 5

19. Iterative solution – 4 discs • Day 6

20. Iterative solution – 4 discs • Day 7

21. Iterative solution – 4 discs • Day 8

22. Iterative solution – 4 discs • Day 9

23. Iterative solution – 4 discs • Day 10

24. Iterative solution – 4 discs • Day 11

25. Iterative solution – 4 discs • Day 12

26. Iterative solution – 4 discs • Day 13

27. Iterative solution – 4 discs • Day 14

28. Iterative solution – 4 discs • Day 15

29. Why? • Presenting the solution like this, provides us with no help in understanding how the solution was constructed • How would we give a formal mathematical verification of the correctness of the algorithm • By observing a number of invariants, we show how to derive the algorithm from the inductive solution.

30. Inductive solution • Suppose the task is to move M discs from one pole to another specific pole. • Base case. When there are 0 discs, no steps are needed to complete the task • For the inductive step, we assume we can move n discs from A to B, and the problem is to show how to move n+1 discs from A to B • We get stuck with this approach

31. Stuck • First move n top discs from A to B • After doing this, all the discs are on B • We have no hypothesis about moving discs from this pole • i.e. The information that we can move n discs from A to B does not help us in trying to move n+1 from A to B

32. Example with n = 3 B C A

33. Example with n = 3 B C A

34. Still stuck • Alternatively, we can move the smallest from pole A to pole C • We then move n discs from A to B • Once again we have exhausted all possibilities of using the inductive hypothesis • Because the n discs are on pole B and we have no hypothesis about moving discs from this pole.

35. Too Specific • We have been too specific about the inductive hypothesis • The way out of this is to introduce some parameters which model the start and final positions of the discs

36. Rotational symmetry • We make a crucial decision • Rather than name the poles A B C, we observe that the problem exhibits a rotational symmetry • This is obvious when the poles are arranged in a triangle, but obscured when the poles are placed in a line.

37. An additional parameter • One additional parameter needs to be introduced, the direction of movement • We only need to say if a particular disc needs to be moved in a clockwise or anti- clockwise direction • The generalisation of the problem becomes, how to move n discs from one pole to the next in the direction d • (where d is either clockwise or anti-clockwise)

38. Inductive hypothesis • The inductive hypothesis we use is that it is possible to move the n smallest discs from one pole to another in direction d beginning from any valid starting point • That is, a starting position in which the discs are distributed arbitrarily over the poles, but no disc is on top of a disc smaller than itself • This is a far more general definition

39. Base and inductive case • For n = 0, the sequence of moves is empty. • In the case of n+1 discs, we assume we have a method of moving the n smallest discs from one pole to one of the other two poles. • We must show how to move n+1 discs from one pole to another pole in direction d, where d is either clockwise or anticlockwise. • Assume the discs are numbered 1 upwards (with the smallest disc numbered 1)

40. Using the inductive hypothesis • There is little choice in exploiting the inductive hypothesiswe can begin by moving the n smallest discs in the direction d, or direction ¬d. • The solution is to move the n discs in direction ¬d, then the n+1thdisc can be moved in direction d • We can then use the inductive hypothesis again to move the n smallest discs in the direction ¬d. • This places the n discs above the n+1th disc, and all of the n+1 smallest discs have now been moved from their original position to a new pole in direction d

41. Example with n = 3

42. Example with n = 3

43. Example with n = 3

44. Example with n = 3

45. For n+1 (i.e. 3+1 = 4) • We can use the previous solution to get a solution for n+1 using the inductive step

46. Example with n = 4

47. Example with n = 4

48. Example with n = 4

49. Example with n = 4

50. Psuedocode • The following code summarises the inductive solution to the problem • Hn(d) is the sequence of pairs <k,d’> where n in the number of discs, k is a disc number and d and d’ are directions • Directions are Boolean values, true representing clockwise, and false representing anti-clockwise • The pair <k, d’> means, move disc number k from its current position in the direction d’ • The pair is therefore an action. • We use the ; to separate movements.