Download Towers of Hanoi - Computing Science and Mathematics

Towers of Hanoi
• 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.
• 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.
• 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.
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• 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.
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• 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.
• 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.
• 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.
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• 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.
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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.
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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)
• 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.
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.
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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.
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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)
• 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).
• 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)
• 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)
• 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.
• Hn.d
• Hn.d means 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
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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).
• 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)
• 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.
• 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.
• 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.
• 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
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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”
Since even.(n+1) = not.even.n
The value of even(n+1) = d remains constant.
• 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
• 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.
• 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.
• 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.
• 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