CS 312: Algorithm Analysis

1. This work is licensed under a Creative Commons Attribution-Share Alike 3.0 Unported License. CS 312: Algorithm Analysis Lecture #32: Intro. to State-Space Search Slides by: Eric Ringger, with contributions from Mike Jones, Eric Mercer, and Sean Warnick

2. Announcements • HW #22.5 due now • Project #6 • Early: today • Due: Monday • Project #7: TSP – last project! • We will have a bake-off competition! • Help session: Tuesday

3. A little perspective “NP-completeness is not a death certificate – it is only the beginning of a fascinating adventure.” -- Dasgupta et al. (our textbook authors)

4. Objectives • Formulate algorithmic solutions to (intractable) problems as state-space search • Use back-tracking search in a state-space • Gain insight into carefully describing states for more efficiency • Recognize the need for feasibility functions

5. Graph Coloring Problem • Assign colors to the vertices of a graph • So that no adjacent vertices share the same color • Vertices i and j are adjacent if there is an edge from vertex i to vertex j. • Find all m-colorings of a graph • i.e., find all ways to color a graph with at most m colors.

6. States • State is another name for a partial solution.

7. State-Space Search Note the resemblance to the enumeration of computation paths pursued by a Non-deterministic TM.

8. State-Space Search • What to store in each state? • What to store in the initial state? • How to expand a state into successor states? • How to search? • Properties of the state space: • Is it a graph with shared sub-structure? • Is it simply a tree?

9. Example: Graph Coloring m = 3 colors Initial State: State Expansion: State:

10. State-Space Search • Brute-force state-space search • No pruning • Pure enumeration • Backtracking state-space search • Pruning • i.e., Test partial states and eliminate those that are not feasible

11. Brute-force Algorithmfor State-space Exploration procedure explore(state s) Input: current state s Output: none; prints solutions for each s’  extensions_of(s) ifcriterion(s’) then output “Solution:” s’ else // possible solution not yet fully specified explore(s’) Outer-most call: explore(init_state())

12. Example: Graph Coloring m=3 Red Yellow Blue

13. Example: Graph Coloring m=3 Red Yellow Blue

14. Example: Graph Coloring m=3 Red Yellow Blue

15. Example: Graph Coloring m=3 Red Yellow Blue

16. Example: Graph Coloring m=3 Red Yellow Blue

17. Example: Graph Coloring m=3 Red Yellow Blue

18. Example: Graph Coloring m=3 Red Yellow Blue

19. Aside: Coloring a Map • Assign colors to countries so that no two bordering regions are the same color. • Map coloring is reducible to planar graph coloring. Idaho Wyoming Nevada Utah Colorado Arizona New Mexico

20. Color Theorem • How many colors do you need for a planar map? • First proposed in 1852 • Answer: Four • Proved by Haken and Appel in 1976 • Used a computer program • Checked 1,476 graphs, i.e., cases in the proof • Required 1,200 hours of run-time

21. State-space size • Consider how large the set of leaf states in the search space could be (without pruning) if we don’t approach the formulation of a problem carefully.

22. n-Queens Problem • Given: n-queens and an n n chess board • Find: A way to place all n queens on the board s.t. no queens are attacking another queen. • Can you formulate an algorithmic solution to this problem as state-space search? • How would you represent a state?

23. First State-Space Idea • Consider all possible placements of the queens on a chess board. • Represent a state as a chess board. • How many placements are there?

24. First State-Space Idea • Consider all possible placements of the queens on a chess board. • Represent a state as a chess board. • How many placements are there? Recall:

25. Second State-Space Idea • Don’t place 2 queens in the same row.

26. 1 2 3 4 Queen in Row #1 (Q1) 5 6 7 8 1 2 3 4 5 6 7 8 (1,1,1,1,1,1,1,1)

27. 1 2 3 4 5 6 Queen in Row #5 (Q5) 7 8 1 2 3 4 5 6 7 8 (1,1,1,1,1,1,1,1)

28. 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 (1,1,1,1,1,1,1,2)

29. 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 (1,1,1,1,1,1,1,3)

30. Second State-Space Idea • Don’t place 2 queens in the same row. • Represent a state as a list of columns. • Now how many positions must be checked? Represent a positioning as an ordered tuple (x1, …, x8) Where each element is an integer 1, …, 8.

31. Third State-Space Idea • Don’t place 2 queens in the same row or in the same column. • Represent a state as a permutation of (1,2…8) • Now how many positions must be checked?

32. 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 (1,2,3,4,5,6,7,8)

33. 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 (1,2,3,4,5,6,8,7)

34. 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 (1,2,3,4,5,8,6,7)

35. 1 2 3 4 5 6 7 8 1 2 3 4 5 6 7 8 (1,2,3,4,5,8,7,6)

36. Third State-Space Idea • Don’t place 2 queens in the same row or in the same column. • Represent a state as a permutation of (1,2…8) • Now how many positions must be checked? • It is easy to enumerate all permutations • We went from C(n2, n) to nn to n! • And we’re happy about it! • We applied explicit constraints in the enumeration of successor states to shrink our state space up front.

37. Lessons • So far, efficiency of state-space search depends on: • The number of successor states satisfying the explicit constraints • The time to generate the successor states

38. Back-tracking • Aside from intelligently describing states, • We check a solution after placing all 8 queens: • criterion function (i.e., a solution test): yes or no • We also check after each placement of a queen. • partial-criterion function (i.e., a feasibility test): yes or no • Idea: Determine we are on a dead-end road without exploring the entire road

39. Four Queens Problem 1 2 3 4 Easier to manage in a lecture. Let’s illustrate in more detail how back-tracking works.

40. Four Queens Problem x1 = 1 x2 = 2 x3 = 3 x4 = 4 1 2 3 4 Each branch of the tree represents a decision to place a queen. The criterion function can only be applied to leaf states.

41. Four Queens Problem x1 = 1 x2 = 2 x3 = 3 x4 = 4 1 2 3 4 Is this leaf state a solution?

42. Four Queens Problem x1 = 1 x2 = 2 x3 = 3 x3 = 4 x4 = 4 x4 = 3 1 2 3 4 Is this leaf state a solution?

43. Four Queens Problem Using only a criterion function (at the leaves), we would build out the entire state space.

44. Four Queens Problem x1 = 1 x2 = 2 x3 = 3 x3 = 4 x4 = 4 x4 = 3 1 2 3 4 A better question: Is any child of this state ever going to be a solution? (before placing queens #3 and #4)

45. Four Queens Problem x1 = 1 x2 = 2 x3 = 3 x3 = 4 x4 = 4 x4 = 3 1 2 3 4 A better question: Is any child of this state ever going to be a solution? No. 1 is attacking 2. Adding queens won’t fix the problem.

46. Four Queens Problem x1 = 1 x2 = 2 1 2 3 4 The partial criterion or feasibility function checks whether a state may eventually have a solution. Will this state have a solution? no. So we prune and backtrack.

47. Four Queens Problem x1 = 1 x2 = 2 x2 = 3 1 2 3 4 Will this state have a solution? Maybe.

48. Four Queens Problem x1 = 1 x2 = 2 x2 = 3 x3 = 2 1 2 3 4 Will this state have a solution? No. Etc.

49. Four Queens Problem x1 = 1 x2 = 2 x2 = 3 x3 = 2 x3 = 4 1 2 3 4 Will this state have a solution? No. Etc.

50. Four Queens Problem x1 = 1 x2 = 2 x2 = 4 x2 = 3 x3 = 2 x3 = 4 1 2 3 4 And so on …