Towers of Hanoi

1. Towers of Hanoi

2. Introduction • Introduction • This problem is discussed in many maths texts, • And in computer science an 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.

3. Begin with the solution • Begin with the solution • 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.

4.  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.

5. Triangle representation. • Almost every book (and webpage) I have see 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. • There is no difference between B and C (assuming all the discs start on pole A) • But 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.

6. 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. • We assume 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.

7. 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, anticlockwise.

8. 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. • Try executing this algorithm yourself on 4 discs.

9. 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.

10. Inductive solution • Obviously, induction on the number of discs. • Let us call the poles A B C • 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.

11. Stuck 1 • 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. • (illustrate this on the board)

12. Stuck 2 • 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.

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

14. Rotational Symmetry • We make a crucial decision. • Rather than name the poles A B C, • We observer 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.

15. Additional Parameter • One additional parameter needs to be introduced. • This is the direction of movement. • i.e. we only need to say if a particular disc needs to be moved in a clockwise or counter clockwise direction. • Also, the generalization 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 anticlockwise)

16. 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 arbitrarly over the poles, • But no disc is on top of a disc smaller than itself).

17. 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. • Lets assume the discs are numbered 1 upwards • (the smallest disc being numbered 1)

18. Using the inductive hypothesis • There is little choice in exploiting the inductive hypothesis. • We begin by moving the n smallest discs in the direction d, or direction (not)d. • The solution is to move the n discs in direction (not)d, • Then the n+1 th disc can be moved in direction d. • Finally we can use the inductive hypothesis again for a second time, • To move the n smallest discs in the direction (not) d. • This places the n discs above the n+1 th disc, • And all of the n+1 smallest discs have now been moved from there original position to a new pole in direction d. (draw a picture)

19. Code • 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 counter 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.

20. Hn.d • Hn.dmeans the sequence to move the n smallest discs in direction d. • Taking the pairs in order from left to right, the complete sequence Hn.d, • Prescribes how to move the n smallest discs, one by one, from one pole to another in direction d • Write out the rules

21. Hn.d • H0.d = [] • Hn+1.d = Hn.(not)d ; [<n+1, d>] ; Hn.(not)d • Notice that the procedure name, H, occurs on both sides of the equation for Hn+1.d • For this reason we call this a recursive solution (i.e. the procedure calls itself, think of Java).

22. How to use • This inductive procedure gives us a way to generate the solution to the problem, • for any n. • We simply use the rules as rewrite rules, • until all occurrences of H have been eliminated. • i.e. we are left with just a sequence of actions • (e.g. move disc 3 clockwise)

23. H2,cw • Let us use cw and aw to mean clockwise and anticlockwise, • Rather than true false, to improve readability. • Show working on board p112 • Exercise – you do H3.aw • Can you see a pattern in the number of moves required.

24. Recursion vs Iteration • The recursive solution is not the solution the monks were using. • There solution was iteratively (repeatedly) executing a simple procedure depenent only on the current state. • Conversely, The implementation of the inductive solution involves maintaining a stack of the sequence of moves yet to be executed. • In java this can lead to “out of memory” runtime errors. • There are interesting memory and speed issues here, • Which you can address in a future course on Complexity.

25. The iterative solution • Remember that the iterative solution had two main elements. • First, that the smallest disc cycles around the poles • (i.e. keeps going around in one direction) • Second, that the disc to be moved alternates between the smallest disc and some other disc. • We now show how these two properties can be derived from the inductive solution.

26. Cyclic moves • In this section we show that the smallest disc cycles around the poles. • In fact we can show that all discs cycle around the poles !!! • And we can show which direction it is

27. The key - invariant • For all pairs <k, d’> in the sequence Hn+1.d, • The Boolean value even.k=d’ is invariant. • Applying the formula, Hn+1.d, • The parameter “n+1” is replaced by “n” • And “d” is replaced by “not d” • (n:=n-1; and d:= not d; ) • Since even.(n+1) = not.even.n • The value of even(n+1) = d remains constant.

28. Initial conditions • Initial conditions • Even.k = d’ is true or false, • And this depends on the initial values of n and d. • Suppose these are N and D, then • Even.k = d = even.n = D

29. In words • In words • If we require to move an even number of discs in a clockwise direction, • All even numbered discs should cycle in a clockwise direction, • And all odd numbers discs cycle in an anticlockwise direction. • The converse is true. • In particular, the smallest disc (number 1), • Cycles in a direction opposite to D, • If N is even, • And in the same direction as D if N is odd.

30. Alternate discs • The second fact was that the disc to be moved alternates between • The smallest disc and some other disc. • This is not difficult to see this. • Two consecutive moves of the smallest disc can be done in one move. • Two consecutive moves of any other disc do not change the state of the puzzle. • We want to give a formal proof that the sequence Hn.d has this property.

31. Alternating • Let us call a sequence of numbers alternating if it has two properties. • First, consecutive elements alternate between 1 and a value greater than one. • Second, when the sequence is not empty, it begins and ends with the value 1. • We write alt.ks if the sequence ks has these properties.

32. Formally • discn.k is the sequence of discs moved on successive days. • Obtained by taking the first component of pairs in Hn.d, • And ignoring the second directional component. • By definition of H we get, • disc0.d = [] • discn+1.d = discn.not.d ; [n+1] ; discn.not.d