A DYNAMIC PROGRAMMING APPROACH. Guy Hawerstock Dan Ilan

Transcription

1 THE n-queens PROBLEM A DYNAMIC PROGRAMMING APPROACH Guy Hawerstock Dan Ilan

2 A Short Chess Review A queen in chess can move anywhere along her row, column or diagonals. A queen can strike any unit by moving to its square.

3 The 8-queen Problem How many ways are there to place 8 queens on a chessboard so that no 2 queens will be on the same row, column or diagonal?

4 A Legal placement of 8-queens Notice that the 8-queen problem is a counting problem. The problem entertained the thoughts of mathematicians since the middle of the 19 th century. Used today as a benchmark problem.

5 From 8-queens to n-queens A generalization of the chess problem. How many ways are there to place n-queens on a n n chessboard so that no two queens will be placed on the same row, column or diagonal? Is there a feasible answer to the problem?

6 A Naive Approach Brute Force The simplest naive implementation: Given an n n Chessboard, try every possible placement of n-queens on the board. Sum up the placements that were legal. The Run-Time is clearly non-feasible (even when n is relatively small): The number of different placements: n 2 n

7 A Hard Problem Reductions from Other NP-hard problems exist. We ll show an algorithm that solves the problem in O(f(n) 8 n ) f(n) is a low-order polynomial.

8 Basic Terminology - Lines Line a maximal co-linear set of squares on the chessboard (row/column/diagonal) Closed Line if a queen is placed in one of the line s squares, the line is considered closed.

9 Lines How Many Lines are there in a n n chessboards? n rows. n columns. 2n 1diagonals from left-to-right 2n 1 diagonals from right-to-left A total of 6n 2 lines. 4 distinct lines pass through each square. 1 row, 1 column, 2 diagonals

10 Terminology - Candidates Candidate - a placement of at most n-queens on a n n chessboard. Feasible Candidate a candidate that doesn t place 2 queens that threaten each other. Feasible Candidate Candidate

11 Terminology - Completion A completion of a candidate is a placement of the remaining queens that result in a solution. It s possible that a candidate won t have a valid completion. It s also possible a candidate will have more than one possible completion.

12 Completion Example Candidate Possible Solution Candidate s Possible Completion

13 Lemma I Claim: Two Feasible candidates with the same number of closed lines contain the same number of queens. Proof: 4 lines pass in each square each queen closes 4 lines. n queens can t close more than 4n lines. If n queens close less than 4n lines, at least 2 queens share a closed line clash and contradiction to feasibility. A feasible candidate with K closed lines has K / 4 placed queens. 2 feasible candidates with K closed lines each have the same number of placed queens (K / 4).

14 Two candidates with the same completion Red and Green are 2 distinct feasible candidates with same set of closed lines. Blue is a completion fitting both.

15 Lemma II Claim: C 1, C 2 are feasible candidate with same set of closed lines a completion of C 1 is also a completion of C 2. Preliminaries: S is the shared set of closed lines. k is the number of placed queens in C 1. C is a completion of C 1.

16 Lemma II, proof C contains n k queens, since C C 1 is a solution. Due to Lemma I, C 2 also contains k queens. C C 2 contains n placed queens. Since C did not cause a clash when added to C 1 it will not cause a clash when added to C 2. (Since S is the same for both). Therefore C is a completion for C 2 as well.

17 Equivalency of Candidates Assume C 1, C 2 are feasible candidates with the same set of closed lines. Every completion for one is a completion for the other (Lemma II). Let an Equivalence Class be defined as sets with equivalent sets of closed lines. A given equivalence class will have the same completion for all members of the class. Counting Problem + Equivalence Overlapping Sub Problems

18 The Data Structure The Dynamic Programming Algorithm we ll show uses a set containing tuples of the form S, i. S - a set of closed lines. Defines an Equivalence Class. Used as Key (unique in set). i - an integer. Describes the number of feasible candidates reached by the algorithm so far that belong to the equivalence class S.

19 Algorithm Dynamic Programming Approach 1. [Initialization] Q φ, 1 2. [Square Selection] Choose an unexamined square. Let T be the set of four lines that pass through the square. 3. [Iteration] For every S, i Q such that S T = φ do: 1. [Compaction] if S T, j Q, replace j with i + j. 2. [Creation] Otherwise, add S T, i to Q. 4. [Termination] if an unexamined square remains, go to Square Selection. Otherwise, terminate

20 Example: Q Pre: * φ, Q Post: * φ, 1, T 1, 1 + T 1 : QSquare 1 * is T chosen. 1, 1 } φ, 1 is taken from Q. T 1 φ = φ. T(1) is a feasible addition to φ. T 1 φ, j Q

21 Example: Q Pre: * φ, 1, T(1), Q Post: * φ, 1, T 1, 1, T 2, 1 + T 2 : Square 2 is chosen. φ, 1 is taken from Q. TQ 2 Q φ = * Tφ. 2 T(2), 1 } is a feasible addition to φ. T 2 φ, j Q

22 Example: Q Pre: * φ, 1, T(1), Q Post: * φ, 1, T 1, 1, T 2, 1 + T 2 : Square 2 is chosen. T 1, 1 is taken from Q. T 2 T 1 φ. T(2) is not a feasible addition to T(1).

23 Example: Q Pre: * φ, 1, T 1, 1, T 2, Q Post: * φ, 1, T 1, 1, T 2, 1, T 3, T 3 : Square 3 is chosen. φ, 1 is taken from Q. φq TQ3 = * Tφ. 3 T(3), 1 } is a feasible addition to φ. T 3 φ, j Q

24 Example: Q Pre: * φ, 1, T 1, 1, T 2, Q Post: * φ, 1, T 1, 1, T 2, 1, T 3, 1 +, T 3 T(1), T 3 : Square 3 is chosen. T 1, 1 is taken from Q. QT(1) Q T* T3 3= φ. T(1), T(3) 1 is } a feasible addition to T(1). T 3 T(1), j Q

25 Example: Q Pre: * φ, 1, T 1, 1, T 2, Q Post: * φ, 1, T 1, 1, T 2, 1, T 3, 1, T 3 T(1), 1 +, T 2 T(3), 1 + T 3 : Square 3 is chosen. T 2, 1 is taken from Q. QT(2) Q T* T3 2= φ. T(3), 1 is } a feasible addition to T(2). T 2 T(3), j Q

26 Example: (Compaction) Q Pre: * , T(3) T(4) T(14) T(17), i, , T 2 T 6 T 12, j, Q Post: * , T(3) T(4) T(14) T(17), i +, j, , T 2 T 6 T 12, j, Square 18 is chosen. T 2 T 6 TQ 12 Q, j is * taken T(3) from T(4) Q. T(14) (T(2) T(17), T(6) i + T(12)) j + T 18 T(18) is a feasible addition tot(2) T(6) T(12). T(2) T(6) T(12) T 18 = T(3) T(4) T(14) T(17) = φ.

27 Determining number of solutions A feasible candidate is a solution if and only if it s set of closed lines contain all the rows and columns of the chessboard. Let M be the set of rows and columns. i, M S Define σ S, i = 0, oterwise In order to find the number of solutions: Run the Algorithm. Calculate σ S, i S,i Q

28 Space Complexity Analysis depends upon maximal size of Q and size of tuple. Number of Tuples: Let the maximum number of different lines on the chessboard be α. Maximum number of equivalence classes is 2 α. Size of Tuple: Number of possible candidates is O 2 n2. Therefore the size of an element in Q is O(n 2 ) Overall space complexity is O(2 α n 2 ). α is bound by 6n 2, space complexity is O(64 n n 2 ).

29 Run Time Complexity Square Selection Step is being called n 2 times. Each execution of the step takes O g n 2 α 2 α since Iteration requires passing over all of Q. g n is a low order polynomial that depends upon exact implementation. Overall run-time complexity is O f n 64 n f(n) = g n n 2, still a low-order polynomial. Can we reduce α?

30 Line Exhaustion A line l consists from at most n squares. There are at most n queen placements that close l. When the algorithm passes through all of l s squares, l is considered Exhausted. What happens then?

31 Line Exhaustion rows & columns Let S, i be a tuple in Q when l (which is a row or column) is exhausted. If l S everything is fine. If l S, the Algorithm will fail to find a completion to S (no further queen can close l) In this case, we can Simply discard S, i.

32 Line Exhaustion diagonal Let C 1, C 2 be two candidates. We say they differ in l if l is closed in one and only one of them. Claim: if C 1, C 2 exist in Q at the same time during the algorithm, and differ only in exhausted diagonals, then any completion of one is a completion of the other as well.

33 Diagonal Line Exhaustion Lemma Proof Mark differing exhausted diagonals as D. Let C be a completion of C 1. Proof: C 1, C 2 have the same number of queens. (Equal Rows by Lemma I) C won t place queens on any of the lines in D as they are Exhausted Lines. Both C 1 and C 2 are the same except for D therefore they will have the same completion C.

34 Improving the Algorithm Add the following lines after Iteration: 4. [Exhaustion Loop] For every line l that became exhausted, Do the following for every S, i Q. 1. [Deletion] if l is a row or column, and l S, remove S, i from Q. 2. [Reduction & Conjunction] if l is diagonal and l S, replace S, i with S *l+, i. If there s another S *l+, j Q, merge them (replace them both with S l, i + j )

35 Semi-Exhausted Lines By Improving the Algorithm, it s guaranteed that 2 equivalence classes in Q differ only in Semi- Exhausted Lines. Semi-Exhausted lines lines which are closed on some candidates in Q, but aren t completely exhausted yet. Size of Q is bound by the power-set of maximal semi-exhausted lines. Can we Bound the number to less than 6n 2?

36 Row-Major Bound Claim: When processing the chessboard in row-major order, the number of semi-exhausted lines is bound by 3n + c, for constant c

37 Row-Major Bound The 1 st line: First adds row, column, diagonal. (3) Non-first, non-last add column and 2 diagonals. (3) Last adds column, diagonal and exhausts row. (1) After Line ends 3n 2 semi-exhausted lines exist. On non 1 st, non last lines: First Adds row. Non first, non-last adds nothing. Last adds diagonal, but exhausts row and other diagonal. After the line ends, no change is made to number of semiexhausted lines. Last Line only reduces number of semi-exhausted lines.

38 Exhaustion and complexity The addition of the exhaustion mechanism to the algorithm reduces α. The upper boundary of α is now 3n, instead of 6n. Thus, the new complexities are: Space Complexity O 8 n n 2 Run-Time Complexity O f(n)8 n

39 Summary The n-queen problem was shown and discussed. The problem is considered hard, but we ve seen a dynamic programming approach to it that gave the first non-trivial upper bound to the problem. Questions