Download presentation

Presentation is loading. Please wait.

Published byScarlett Perry Modified over 2 years ago

2
1Copyright © Gyora Benedek, 2003

3
2 Solving Lights Out with BFS by Dr. Gyora Benedek

4
3Copyright © Gyora Benedek, 2003 The Lights Out game Lights Out™ is a hand-held puzzle game by Tiger Electronics.

5
4Copyright © Gyora Benedek, 2003 Basic definitions Lights Out consists of a 5 by 5 grid of buttons which also have lights in them. By pressing a button, its light and those of the (non-diagonally) adjacent buttons will change. Given a pattern of lights, you have to switch them all off by pressing the correct buttons. Try to do it in as few moves as possible.Demo

6
5Copyright © Gyora Benedek, 2003 Example 1a BeforeAfter

7
6Copyright © Gyora Benedek, 2003 Example 1b AfterBefore

8
7Copyright © Gyora Benedek, 2003 Example 2a AfterBefore

9
8Copyright © Gyora Benedek, 2003 Example 2b

10
9Copyright © Gyora Benedek, 2003 Example 2c

11
10Copyright © Gyora Benedek, 2003 Example 2d

12
11Copyright © Gyora Benedek, 2003 Example 2e

13
12Copyright © Gyora Benedek, 2003 Example 2f

14
13Copyright © Gyora Benedek, 2003 Example 2g - end

15
14Copyright © Gyora Benedek, 2003 Notes Notation: Let lit=1; unlit=0; There are 2^25 positions – only 2^23 have solutions. Repeating same move = undo. Moves order does not matter. Can be solved with Linear Algebra (beyond our scope).

16
15Copyright © Gyora Benedek, 2003 Many Variants Different board sizes Different neighborhood definitions 3 or more colors Restrictions on moves Links to Solutions, Tricks, Research papers, Algorithms, etc http://www.geocities.com/jaapsch/puzzles/lights.htm#desc http://qcunix1.qc.edu/~lteitelm/algs/lo_links.html

17
16Copyright © Gyora Benedek, 2003 Lit Only variant You may only press lit buttons.demo

18
17Copyright © Gyora Benedek, 2003 Example 3a

19
18Copyright © Gyora Benedek, 2003 Example 3b

20
19Copyright © Gyora Benedek, 2003 Example 3c

21
20Copyright © Gyora Benedek, 2003 Example 3d - end

22
21Copyright © Gyora Benedek, 2003 Example 4a

23
22Copyright © Gyora Benedek, 2003 Example 4b

24
23Copyright © Gyora Benedek, 2003 Example 4c

25
24Copyright © Gyora Benedek, 2003 Example 4c

26
25Copyright © Gyora Benedek, 2003 Example 4d

27
26Copyright © Gyora Benedek, 2003 Example 4e

28
27Copyright © Gyora Benedek, 2003 Example 4f

29
28Copyright © Gyora Benedek, 2003 Example 4g

30
29Copyright © Gyora Benedek, 2003 Example 4h

31
30Copyright © Gyora Benedek, 2003 Example 4i

32
31Copyright © Gyora Benedek, 2003 Example 4j

33
32Copyright © Gyora Benedek, 2003 Example 4k

34
33Copyright © Gyora Benedek, 2003 Example 4l - end

35
34Copyright © Gyora Benedek, 2003 Lit Only variant You may only press lit buttons. Moves’ order does matter. The set of “Lit Only solvable puzzles” is obviously a subset of “solvable puzzles”. Can be proved that all solvable puzzles are Lit Only solvable. Lit Only solution may be longer.

36
35Copyright © Gyora Benedek, 2003 How to solve Lit Only? We want to be able to compute an optimal solution to a given puzzle efficiently. We can make a lengthy preparation step and use a lot of memory. Preparation puts all positions in a dictionary: key= position (25 bits) value= solution length (8 bits) GetSolLen(pos) returns solution length for pos.

37
36Copyright © Gyora Benedek, 2003 Background The Lit Only game corresponds to a directed graph G where each vertex is a position and the moves are the edges. We set one vertex (all 0 position) as the target T and are interested in the distance of each vertex from T. Vertices (positions) from which there is no path to T are unsolvable.

38
37Copyright © Gyora Benedek, 2003 Background (2) Usually a graph is given explicitly. Here the edges are implicit, but given a vertex we can find all the out edges and also all the in edges.

39
38Copyright © Gyora Benedek, 2003 How to solve a position void Solve(tPos Pos){ int NewL, L = GetSolLen(Pos); tPos NewPos; if (L==NoSolution) {print “No Solution”; return;} while (L>0){ for all valid moves let NewPos be result of move if ((NewL=GetSolLen(NewPos))

40
39Copyright © Gyora Benedek, 2003 Notes on Solve Moves are printed from left to right. Assuming that GetSolLen( ) returns correct values, L will decrease exactly by 1 in each iteration of the while loop. Time complexity O(nMoves · SolutionLen) here nMoves=25 so we get O(SolutionLen)

41
40Copyright © Gyora Benedek, 2003 Preparation To fill the dictionary we start from the target T (all 0) and go backwards. We use BFS (Breadth First Search) on G with its edges reversed.

42
41Copyright © Gyora Benedek, 2003 BFS Queue Open; Dictionary Closed; int L; tPos P1, P2; Open.Enqueue({T,0});// If many sources Closed.Insert(T, 0);// loop to enqueue all While (not Open.IsEmpty( )){ {P1,L} = Open.Dequeue( ); For each edge (P1,P2) if (not Closed.Find(P2)){ Open.Enqueue({P2,L+1}); Closed.Insert(P2,L+1);}

43
42Copyright © Gyora Benedek, 2003 BFS demo 0 Open T AB CD E F T,0 @While P1=? L=?

44
43Copyright © Gyora Benedek, 2003 BFS demo 0 Open T AB CD E F empty @For P1=T L=0

45
44Copyright © Gyora Benedek, 2003 BFS demo 0 11 Open T AB CD E F A,1 B,1 @While P1=T L=0

46
45Copyright © Gyora Benedek, 2003 BFS demo 0 11 Open T AB CD E F B,1 @For P1=A L=1

47
46Copyright © Gyora Benedek, 2003 BFS demo 0 11 22 Open T AB CD E F B,1 C,2 D,2 @While P1=A L=1

48
47Copyright © Gyora Benedek, 2003 BFS demo 0 11 22 Open T AB CD E F C,2 D,2 @For P1=B L=1

49
48Copyright © Gyora Benedek, 2003 BFS demo 0 11 222 Open T AB CD E F D,2 E,2 @For P1=C L=2 Note that T and D did not change; they were closed already

50
49Copyright © Gyora Benedek, 2003 BFS demo 0 11 222 3 Open T AB CD E F E,2 F,3 @For P1=D L=2

51
50Copyright © Gyora Benedek, 2003 BFS demo 0 11 222 3 Open T AB CD E F F,3 @For P1=E L=2 No edges to D.

52
51Copyright © Gyora Benedek, 2003 BFS demo 0 11 222 3 Open T AB CD E F empty @For P1=F L=3 T and F already closed.

53
52Copyright © Gyora Benedek, 2003 BFS correctness When a vertex {v, L} is added to Closed, L is the shortest distance from one of the sources to v. There is no need to modify them anymore. At any time in the Open queue there are vertexes with L and L+1 only, so that those with L are before those with L+1.

54
53Copyright © Gyora Benedek, 2003 BFS Complexity Each node enters Closed at most once. Each edge is traveled at most once. Time complexity = O(m) Where m is the number of edges in the relevant connected component; assuming O(1) for each operation on the dictionary.

55
54Copyright © Gyora Benedek, 2003 BFS for Lit Only Given a position P1 we need to find all edges (P2,P1) in G. To get a P2 press an unlit button in P1. To find all positions from which P1 can be reached, press all unlit buttons in P1 (starting from P1 each time).Demo

56
55Copyright © Gyora Benedek, 2003 Example 5: step back from T

57
56Copyright © Gyora Benedek, 2003 Example 5 Now we are 1 step away from T

58
57Copyright © Gyora Benedek, 2003 Example 5 This is also 1 step away from T…

59
58Copyright © Gyora Benedek, 2003 Example 5 There are 25 different positions 1 step away from T.

60
59Copyright © Gyora Benedek, 2003 Example 6: step back… The only possibility

61
60Copyright © Gyora Benedek, 2003 Example 6: step back… The only position from which we can reach the above position.

62
61Copyright © Gyora Benedek, 2003 Implementing Closed A Hash table? Actually there are 2^25 =32M positions. An array of 2^25 Bytes with Pos as index will do. Init: For all i do Closed[i]=255; Max distance must be < 255.

63
62Copyright © Gyora Benedek, 2003 Implementing Open Option1: a queue containing pos (25 bits) and L (8 bits). This may be very large (~ 4*2^23 ). Option2: no queue. Run on Closed and look for each pos with L. When done L++. The order in which positions are handled may be different, the result the same. Takes long time when few positions per L.

64
63Copyright © Gyora Benedek, 2003 Implementing Open (2) Option3: Combine the two options. Implement a queue in a relatively short array. Use this queue as in Option1. If this array is full revert to Option2. Limited memory overhead, saves time when queue is not full.

65
64Copyright © Gyora Benedek, 2003 Time Complexity Option1: O(nPositions + nMoves·nSolvablePositions) = O(n+m) =c · (2^25+25·2^23) Option2: O(nPositions · MaxSolutionLen + nMoves · nSolvablePositions)

66
65Copyright © Gyora Benedek, 2003 Optimizations It so happens that all solvable positions are uniquely defined by the 1 st 23 bits. Closed can be 2^23 = 8MBytes ‘only’. By using board symmetry this can be reduced to ~ 1/8; slightly more than 1MBytes.

67
66Copyright © Gyora Benedek, 2003 Problems High Memory & Time complexity! Not practical for larger boards (6x6). A few years ago even 5x5 was not practical.

68
67Copyright © Gyora Benedek, 2003

Similar presentations

OK

CSE332: Data Abstractions Lecture 16: Topological Sort / Graph Traversals Tyler Robison Summer 2010 1.

CSE332: Data Abstractions Lecture 16: Topological Sort / Graph Traversals Tyler Robison Summer 2010 1.

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on introduction to object-oriented programming python Ppt on water cycle for class 7 Ppt on indian musical instruments Ppt on national parks and sanctuaries Ppt on surface area and volume of cylinder and cone Ppt on earthquake engineering Ppt on water scarcity in the world Ppt on carbon and its compounds for 10 class Ppt on principles of object-oriented programming encapsulation Ppt on front office department