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AE1APS Algorithmic Problem Solving John Drake.  Coursework 4 deadline – ◦ Thursday 29 th November at 1pm  Same submission guidelines as before, hard-

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Presentation on theme: "AE1APS Algorithmic Problem Solving John Drake.  Coursework 4 deadline – ◦ Thursday 29 th November at 1pm  Same submission guidelines as before, hard-"— Presentation transcript:

1 AE1APS Algorithmic Problem Solving John Drake

2  Coursework 4 deadline – ◦ Thursday 29 th November at 1pm  Same submission guidelines as before, hard- copy during lectures or soft-copy via 

3  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

4  Induction is the name given to a problem solving technique based on using the solution to small instances of a problem to solve larger instances of a problem.  In fact we can solve problems of arbitrary size

5  We need to measure the "size" of an instance of a problem  E.g. with matchstick problems, we might call the size the number of matches

6  It is not always obvious what the size should be for some types of problems  The size must be non–negative i.e. whole numbers 0, 1, 2, …  We call these numbers "natural numbers", as they are naturally used for counting

7  Note that some mathematicians do not include 0 in the natural numbers  There are good reasons why we should include 0. Zero has important significance in maths (and the history of maths)  One reason in when counting the size of sets, the null set is defined to have size of 0  We will see examples of including 0 later

8  We solve the problem in two steps  First we solve the problem instance of size 0  This is usually very easy (often called trivial)  Then, we show how to solve, a problem of size n+1 (where n is an arbitrary natural number), given we have a solution to a problem instance of size n.

9  The first step is called the base case of the basis of induction.  The second step is the induction step

10  By this process, we can solve problems of size 0  We also know how to solve instances of size 1 (as it is one larger than 0)  We can apply the induction step to solve problems of size 2 and so on.

11  The induction step allows us to solve problems of size one greater than the current size problem  By repeating this process we can solve problems of any size

12  Straight lines are drawn across a sheet of paper, dividing the paper up into regions. Show that it is possible to colour each region black or white, so that no two adjacent regions have the same colour

13  P120

14  The number of lines is the size of a problem  We have to show how to solve the problem when there are zero lines  We have to show how to solve a problem when there are n+1 lines, assuming we can solve the problem with n lines  Let us call the situation when no two adjacent regions have the same colouring "a satisfactory colouring"

15  This is trivial.  With 0 lines, there is one region  This can be coloured black or white and will be a satisfactory colouring.

16  Assume a number of lines (n) are drawn on the paper, and the regions have a satisfactory colouring

17  Now we add an additional line (i.e. n + 1)  This will divide some existing regions into two new regions, which will have the same colour, therefore the colouring is not satisfactory.

18  The task is to show how to modify the colouring so that it does become satisfactory.

19  Inverting any satisfactory colouring will also give a satisfactory colouring. ◦ i.e. changing black regions for white, and vice versa will also be satisfactory  Let us call the two sides of the new line left and right. Inverting colours in either region will still give a satisfactory colouring in that half (left or right)

20  Hence, inverting the colours in one half will guarantee a satisfactory colouring for the whole piece of paper

21  The algorithm for colouring is non-deterministic in several ways ◦ The initial colouring is not specified (could be black or white) ◦ The order in which the lines are added is unspecified ◦ The naming of the left and right regions is not specified (symmetry)  This means the final colouring can be arrived at in different ways, but it is guaranteed to be satisfactory

22  A square piece of paper is divided into a grid of size 2 n by 2 n, where n is a natural number  Individual squares are called grid squares and a single grid square is covered  E.g. n = 3

23  A triomino is an "L" shape made of 3 grid squares. Show that it is possible to cover the remaining grid squares using triominoes

24  The obvious size to use is n  The base case is when n=0, and the grid has size 1 by 1 (i.e. 2 0 by 2 0 )  This square will of course be covered  The base case is solved

25  We want to solve a grid of 2 n+1 by 2 n+1  We know how to cover a 2 n by 2 n grid, how do we use this to help us cover a grid size larger by one  We make the inductive hypothesis that this is possible

26  A grid of size 2 n+1 by 2 n+1 can be divided into 4 grids of size 2 n by 2 n by drawing horizontal and vertical lines  These 4 regions can be called bottom-left, bottom-right, top-left and top-right  One square is already covered, we will assume this is in the bottom left square, why?

27  The bottom left grid is of size 2 n by 2 n, of which one square has already been covered. By the induction hypothesis, the remaining squares in the bottom-left grid can be covered

28

29  This leaves us having to cover the remaining 3 grids  None of these have any squares covered  We can apply the inductive hypothesis if just one of the squares in each of the 3 remaining grids is covered

30  This can be done by placing a triomino at the junction of the 3 grids

31  What if we had chosen n = 1 to be our base case?  Our proof would have left out the 0 case, so we would still need to prove that separately  However including 0 provides a slightly deeper insight  "Humans" start counting at 1, computer scientists and mathematicians start counting at 0


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