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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 Induction – Chapter 6 ◦ Tower of Hanoi – Chapter 8

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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)

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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. 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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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

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A B C

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A B C

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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.

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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+1 th 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+1 th 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 H n (d) is the sequence of pairs 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 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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H n (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 H n (d) prescribes how to move the n smallest discs, one by one, from one pole to another in direction d

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H n (d) H 0 (d) = [] – base case (i.e. 0 moves needed) H n+1 (d) = H n (¬d) ; [ ] ; H n (¬d) – induction step (move n discs in a given direction, move the n+1 th 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 H n+1 (d) ◦ H n+1 (d) = H n (¬d) ; [ ] ; H n (¬d) 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 H 2 (cw) i.e. how to move 2 discs, one pole in a clockwise direction Equation (1) ◦ H 0 (d) = [] Equation (2) ◦ H n+1 (d) = H n (¬d) ; [ ] ; H n (¬d)

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Exercise – Can you do H 3 (aw)? Can you see a pattern in the number of moves required for n discs? i.e. H 0, H 1, H 2, H 3… 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)

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