Transcript Towers of Hanoi - School of Computer Science

AE1APS Algorithmic Problem Solving
John Drake
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Invariants – Chapter 2
River Crossing – Chapter 3
Logic Puzzles – Chapter 5
Matchstick Games - Chapter 4
◦ Sum Games – Chapter 4
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Induction – Chapter 6
◦ Tower of Hanoi – Chapter 8
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Breaking larger problems down into subproblems
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)
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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
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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
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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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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
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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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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.
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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
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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
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Day 0
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Day 1
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Day 2
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Day 3
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Day 4
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Day 5
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Day 6
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Day 7
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Day 8
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Day 9
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Day 10
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Day 11
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Day 12
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Day 13
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Day 14
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Day 15
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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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Suppose the task is to move M discs from one
pole to another specific pole.
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Base case. When there are 0 discs, no steps
are needed to complete the task
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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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First move n top discs from A to B
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After doing this, all the discs are on B
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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
A
B
C
A
B
C
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Alternatively, we can move the smallest from
pole A to pole C
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We then move n discs from A to B
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Once again we have exhausted all
possibilities of using the inductive hypothesis
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Because the n discs are on pole B and we
have no hypothesis about moving discs from
this pole.
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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
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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.
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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)
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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
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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)
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There is little choice in exploiting the inductive
hypothesis we 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+1th disc 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
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We can use the previous solution to get a
solution for n+1 using the inductive step
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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.
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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
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Hn(d)
H0(d) = []
– base case (i.e. 0 moves needed)
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Hn+1(d) = Hn(¬d) ; [<n+1, d>] ; Hn(¬d)
– induction step (move n discs in a given
direction, move the n+1th disc in the opposite
direction then move n discs again in the same
direction)
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Notice that the procedure name, H, occurs on
both sides of the equation for Hn+1(d)
◦ Hn+1(d) = Hn(¬d) ; [<n+1, d>] ; Hn(¬d)
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For this reason we call this a recursive
solution (i.e. the procedure calls itself, think
of Java)
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This inductive procedure gives us a way to
generate the solution to the problem for any
size 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)
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Let us use cw and aw to mean clockwise and
anti-clockwise, rather than true and false to
improve readability
Calculate H2(cw) i.e. how to move 2 discs, one
pole in a clockwise direction
Equation (1)
◦ H0(d) = []
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Equation (2)
◦ Hn+1(d) = Hn(¬d) ; [<n+1, d>] ; Hn(¬d)
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<1,aw>
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<2,cw>
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<1,aw>
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Exercise – Can you do H3(aw)?
Can you see a pattern in the number of
moves required for n discs? i.e. H0, H1, H2, H3…
Show this as a formula in terms of n?
If this is easy, try exercise 8.3 a) (without
cheating and looking at the answer first) 