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Backtracking algorithms

Backtracking algorithms
ECE 250 Algorithms and Data Structures Backtracking algorithms Douglas Wilhelm Harder, M.Math. LEL Department of Electrical and Computer Engineering University of Waterloo Waterloo, Ontario, Canada ece.uwaterloo.ca dwharderalumni.uwaterloo.ca © 2006-2013 by Douglas Wilhelm Harder. Some rights reserved.Backtracking algorithms 2 Outline In this topic, we will cover: – Traversals of trees and graphs – Backtracking Backtracking algorithms 3 Backtracking Suppose a solution can be made as a result of a series of choices – Each choice forms a partial solution – These choices may form either a tree or DAG • Separate branches may recombine or divergeBacktracking algorithms 4 Backtracking With Dijkstra’s algorithm, we keep track of all current best paths – There are at most V – 1 paths we could extend at any one time – These can be tracked with a relatively small table Suppose we cannot evaluate the relative fitness of solutions – There may just be too many to record efficiently – Are we left with a brute-force search? Suppose there are just too many partial solutions at any one time to keep track of – At any point in time in a game of chess or Go (围棋 or碁), there are a plethora of moves, each valid, but the usefulness of each will varyBacktracking algorithms 5 Sudoku In the first case, consider the game Sudoku: 53 – The search space is 9 5 4 6 6 3 1 9 2 5 4 6 8 1 8 4 2 6 1 9 5 8 1 1 2 3 8 2 7 http://www.sudokuoftheday.com/Backtracking algorithms 6 Sudoku At least for the first entry in the first square, only 1, 3, 7 fit 5 4 6 6 3 1 9 2 5 4 6 8 1 8 4 2 6 1 9 5 8 1 1 2 3 8 2 7 5 1 4 6 5 3 4 6 5 7 4 6 6 3 1 6 3 1 6 3 1 9 2 5 4 9 2 5 4 9 2 5 4 6 8 1 6 8 1 6 8 1 8 4 8 4 8 4 2 6 1 2 6 1 2 6 1 9 5 8 1 9 5 8 1 9 5 8 1 1 2 3 1 2 3 1 2 3 8 2 7 8 2 7 8 2 7Backtracking algorithms 7 Sudoku nd If the first entry has a 1, the 2 entry in that square could be 7 or 8 5 1 4 6 6 3 1 9 2 5 4 6 8 1 8 4 2 6 1 9 5 8 1 1 2 3 8 2 7 5 1 4 6 5 1 4 6 7 6 3 1 8 6 3 1 9 2 5 4 9 2 5 4 6 8 1 6 8 1 8 4 8 4 2 6 1 2 6 1 9 5 8 1 9 5 8 1 1 2 3 1 2 3 8 2 7 8 2 7Backtracking algorithms 8 Sudoku nd If the first entry has a 3, the 2 entry in that square could be 7 or 8 5 3 4 6 6 3 1 9 2 5 4 6 8 1 8 4 2 6 1 9 5 8 1 1 2 3 8 2 7 5 3 4 6 5 3 4 6 7 6 3 1 8 6 3 1 9 2 5 4 9 2 5 4 6 8 1 6 8 1 8 4 8 4 2 6 1 2 6 1 9 5 8 1 9 5 8 1 1 2 3 1 2 3 8 2 7 8 2 7Backtracking algorithms 9 Sudoku nd If the first entry has a 7, the 2 entry in that square could be 8 5 7 4 6 6 3 1 9 2 5 4 6 8 1 8 4 2 6 1 9 5 8 1 1 2 3 8 2 7 5 7 4 6 8 6 3 1 9 2 5 4 6 8 1 8 4 2 6 1 9 5 8 1 1 2 3 8 2 7Backtracking algorithms 10 Sudoku 5 4 6 6 3 1 9 2 5 4 6 8 1 8 4 2 6 1 9 5 8 1 In the next child, there are no 1 2 3 8 2 7 options available for the 5 1 4 6 5 3 4 6 5 7 4 6 next entry 6 3 1 6 3 1 6 3 1 9 2 5 4 9 2 5 4 9 2 5 4 6 8 1 6 8 1 6 8 1 8 4 8 4 8 4 2 6 1 2 6 1 2 6 1 9 5 8 1 9 5 8 1 9 5 8 1 1 2 3 1 2 3 1 2 3 8 2 7 8 2 7 8 2 7 5 1 4 6 5 1 4 6 5 3 4 6 5 3 4 6 5 7 4 6 5 7 4 6 7 6 3 1 8 6 3 1 7 6 3 1 8 6 3 1 1 6 3 1 8 6 3 1 9 2 5 4 9 2 5 4 9 2 5 4 9 2 5 4 9 2 5 4 9 2 5 4 6 8 1 6 8 1 6 8 1 6 8 1 6 8 1 6 8 1 8 4 8 4 8 4 8 4 8 4 8 4 2 6 1 2 6 1 2 6 1 2 6 1 2 6 1 2 6 1 9 5 8 1 9 5 8 1 9 5 8 1 9 5 8 1 9 5 8 1 9 5 8 1 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 8 2 7 8 2 7 8 2 7 8 2 7 8 2 7 8 2 7 5 1 4 6 7 6× 3 1 Any candidate solution built from this 9 2 5 4 partial is infeasible 6 8 1 8 4 – We can ignore this branch 2 6 1 9 5 8 1 1 2 3 8 2 7Backtracking algorithms 11 Sudoku 5 4 6 6 3 1 9 2 5 4 6 8 1 8 4 2 6 1 9 5 8 1 Three other branches lead to 1 2 3 8 2 7 similar dead ends 5 1 4 6 5 3 4 6 5 7 4 6 6 3 1 6 3 1 6 3 1 9 2 5 4 9 2 5 4 9 2 5 4 6 8 1 6 8 1 6 8 1 8 4 8 4 8 4 2 6 1 2 6 1 2 6 1 9 5 8 1 9 5 8 1 9 5 8 1 1 2 3 1 2 3 1 2 3 8 2 7 8 2 7 8 2 7 5 1 4 6 5 1 4 6 5 3 4 6 5 3 4 6 5 7 4 6 7 6 3 1 8 6 3 1 7 6 3 1 8 6 3 1 8 6 3 1 9 2 5 4 9 2 5 4 9 2 5 4 9 2 5 4 9 2 5 4 6 8 1 6 8 1 6 8 1 6 8 1 6 8 1 8 4 8 4 8 4 8 4 8 4 2 6 1 2 6 1 2 6 1 2 6 1 2 6 1 9 5 8 1 9 5 8 1 9 5 8 1 9 5 8 1 9 5 8 1 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 8 2 7 8 2 7 8 2 7 8 2 7 8 2 7 5 1 4 6 5 3 4 6 5 7 4 6 7 6× 3 1 7 6× 3 1 8 6× 3 1 9 2 5 4 9 2 5 4 9 2 5 4 6 8 1 6 8 1 6 8 1 8 4 8 4 8 4 2 6 1 2 6 1 2 6 1 9 5 8 1 9 5 8 1 9 5 8 1 1 2 3 1 2 3 1 2 3 8 2 7 8 2 7 8 2 7Backtracking algorithms 12 Sudoku 5 4 6 6 3 1 9 2 5 4 6 8 1 8 4 2 6 1 9 5 8 1 With the other two, there is only 1 2 3 8 2 7 one candidate for each of the 5 1 4 6 5 3 4 6 5 7 4 6 last two 6 3 1 6 3 1 6 3 1 9 2 5 4 9 2 5 4 9 2 5 4 6 8 1 6 8 1 6 8 1 entries 8 4 8 4 8 4 2 6 1 2 6 1 2 6 1 9 5 8 1 9 5 8 1 9 5 8 1 1 2 3 1 2 3 1 2 3 8 2 7 8 2 7 8 2 7 5 1 4 6 5 1 4 6 5 3 4 6 5 3 4 6 5 7 4 6 7 6 3 1 8 6 3 1 7 6 3 1 8 6 3 1 8 6 3 1 9 2 5 4 9 2 5 4 9 2 5 4 9 2 5 4 9 2 5 4 6 8 1 6 8 1 6 8 1 6 8 1 6 8 1 8 4 8 4 8 4 8 4 8 4 2 6 1 2 6 1 2 6 1 2 6 1 2 6 1 9 5 8 1 9 5 8 1 9 5 8 1 9 5 8 1 9 5 8 1 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 8 2 7 8 2 7 8 2 7 8 2 7 8 2 7 5 1 4 6 5 3 4 6 8 6 7 3 1 8 6 7 3 1 3 9 2 5 4 1 9 2 5 4 6 8 1 6 8 1 8 4 8 4 2 6 1 2 6 1 9 5 8 1 9 5 8 1 1 2 3 1 2 3 8 2 7 8 2 7Backtracking algorithms 13 Sudoku It may seem that this is a reasonably straight-forward method; however, the decision tree continues to branch quick once we start filling the second squareBacktracking algorithms 14 Sudoku 54 A binary tree of this height would have around 2– 1 nodes – Fortunately, as we get deeper into the tree, more get cutBacktracking algorithms 15 Implementation Our straight-forward implementation takes a 9 × 9 matrix – Default entries are values from 1 to 9, empty cells are 0 – Two helper functions: bool next_location( int99, int &i, int &j ) • Finds the next location empty location returning false if none is found bool is_valid( int99, int i, int j, int value ) • Checks if there are any conflicts created if matrixij is assigned value – The backtracing function: • Finds the next unoccupied cell • For each value from 1 to 9, it checks if it is valid to insert that it there – If so, backtracking is called recursively on the matrix with that entry set The main function creates the initial matrix and calls backtrackBacktracking algorithms 16 Implementation // Find the next empty location in 'matrix' // If one is found, assign 'i' and 'j' the indexes of that entry // Otherwise, return false // - In this case, the values of 'i' and 'j' are undefined bool next_location( int matrix99, int &i, int &j ) for ( int i1 = 0; i1 3; ++i1 ) // If 'value' already appears in for ( int j1 = 0; j1 3; ++j1 ) // - the row 'm' for ( int i2 = 0; i2 3; ++i2 ) // - the column 'n' for ( int j2 = 0; j2 3; ++j2 ) // - the 3x3 square of entries it (m, n) appears in i = 3i1 + i2; // return true, otherwise return false j = 3j1 + j2; bool is_valid( int matrix99, int m, int n, int value ) // Check if 'value' already appears in either a row or column // return 'true' if we find an for ( int i = 0; i 9; ++i ) // unoccupied entry if ( (matrixmi == value) ( matrixin == value ) ) if ( matrixij == 0 ) return false; return true; // Check if 'value' already appears in either a row or column int ioff = 3(m/3); int joff = 3(n/3); return false; // all the entries are occupied for ( int i = 0; i 3; ++i ) for ( int j = 0; j 3; ++j ) if ( matrixioff + ijoff + j == value ) return false; // 'value' already in the 3x3 square return true; // 'value' could be added Backtracking algorithms 17 Implementation bool backtrack( int matrix99 ) int i, j; // If the matrix is full, we are done if ( next_location( matrix, i, j ) ) return true; for ( int value = 1; value = 9; ++value ) if ( is_valid( matrix, i, j, value ) ) // Assume this entry is part of the // solutionrecursively call backtrack matrixij = value; // If we found a solution, return // otherwise, reset the entry to 0 if ( backtrack( matrix ) ) return true; else matrixij = 0; // No solution foundreset the entry to 0 return false; Backtracking algorithms 18 Implementation int main() int matrix99 = 5, 0, 4, 0, 0, 0, 0, 6, 0, 0, 6, 0, 3, 0, 0, 1, 0, 0, 0, 9, 2, 5, 4, 0, 0, 0, 0, 6, 0, 8, 0, 0, 1, 0, 0, 0, 0, 0, 0, 8, 0, 4, 0, 0, 0, 0, 0, 0, 2, 0, 0, 6, 0, 1, 0, 0, 0, 0, 9, 5, 8, 1, 0, 0, 0, 1, 0, 0, 2, 0, 3, 0, 0, 8, 0, 0, 0, 0, 2, 0, 7 ; // If found, print out the resulting matrix if ( backtrack( matrix ) ) for ( int i = 0; i 9; ++i ) for ( int j = 0; j 9; ++j ) std::cout matrixij " "; std::cout std::endl; return 0; Backtracking algorithms 19 Implementation In this case, the traversal: – Recursively calls backtrack 874 times • The last one determines that there are no unoccupied entries – Checks if a placement is valid 7658 times 5 3 4 1 7 8 9 6 2 8 6 7 3 2 9 1 4 5 1 9 2 5 4 6 3 7 8 6 7 8 9 3 1 5 2 4 2 1 5 8 6 4 7 9 3 3 4 9 2 5 7 6 8 1 4 2 3 7 9 5 8 1 6 7 5 1 6 8 2 4 3 9 9 8 6 4 1 3 2 5 7Backtracking algorithms 20 Backtracking This should give us an idea, however: – Perform a traversal – Do not continue traversing if a current node indicates all descendants are infeasible solutionsBacktracking algorithms 21 Classical applications Classic applications of this algorithm technique include: – Eight queens puzzle – Knight’s tour – Logic programming languages – Crossword puzzlesBacktracking algorithms 22 Eight queens puzzle Arrange eight queens on a chess board so that no queen can take another ♕Backtracking algorithms 23 Knight’s tour Have a knight visit all the squares of a chess board either as a path or a cycle ♘Backtracking algorithms 24 Logic programming languages Consider the Prolog programming language where we state facts: parent(beatrix, willem). female(juliana). parent(willem, catharina). parent(claus, willem). male(bernhard). parent(maxima, catharina). parent(beatrix, friso). female(beatrix). parent(willem, alexia). parent(claus, friso). male(claus). parent(maxima, alexia). male(firso). parent(willem, ariane). parent(beatrix, constantijn). parent(claus, constantijn). female(mabel). parent(maxima, ariane). male(constantijn). parent(juliana, beatrix). female(laurentien). parent(firso, luana). parent(bernhard, beatrix). parent(mabel, luana). female(catharina). parent(firso, zaria). parent(juliana, irene). parent(bernhard, irene). female(alexia). parent(mabel, zaria). parent(juliana, margriet). female(ariane). parent(bernhard, margriet). parent(constantijn, eloise). parent(juliana, christina). female(luana). parent(laurentien, eloise). parent(bernhard, christina). female(zaria). parent(constantijn, claus_ii). parent(laurentien, claus_ii). spouses(willem, maxima). female(eloise). parent(constantijn, leonore). spouses(firso, mabel). male(claus_ii). parent(laurentien, leonore). spouses(constantinjn, laurentien). female(leonore). spouses(beatrix, claus). spouses(juliana, bernhard).Backtracking algorithms 25 Logic programming languages You can now define relationships between individuals % Relationships mother(M, X) :- parent(M, X), female(M). father(F, X) :- parent(F, X), male(F). sister(S, X) :- sibling(S, X), female(S), \+ (S = X). brother(B, X) :- sibling(B, X), male(B), \+ (S = X). grandparent(G, X) :- parent(G, P), parent(P, X). % Symmetric spouses(X, Y) :- spouses(Y, X); sibling(X, Y) :- parent(P, X), parent(P, Y), \+ (X = Y). cousin(X, Y) :- parent(A, X), parent(B, Y), sibling(A, B). % Antisymmetric uncle(U, X) :- male(U), sibling(U, Y), parent(Y, X). uncle(U, X) :- male(U), spouse(U, Z), sibling(Z, Y), parent(Y, X). aunt(A, X) :- female(A), parent(Y, X), sibling(A, Y). aunt(A, X) :- female(A), spouse(A, Z), sibling(Z, Y), parent(Y, X).Backtracking algorithms 26 Logic programming languages Given these relationships, you can now make queries: cousin(zaria, alexia). uncle(constantijn, alexia). aunt(laurentien, alexia). Backtracking can be used to determine whether the above relationships hold given the stated factsBacktracking algorithms 27 Parsing Question: how do we define a programming language? – Why are any of the following never valid? a + b c3) d?e; 54f """; g = "Hello world"; – Programming languages are defined by grammars – The C++ programming language grammar is available here: http://www.nongnu.org/hcb/Backtracking algorithms 28 Parsing and grammars Consider just the conditional statements from the pre-processor – Square brackets is used to indicate something is optional group ::= group-part group group-part group-part ::= if-section control-line if-section ::= if-group elif-groups else-group endif-line if-group ::= if constant-expression new-line group ifdef identifier new-line group ifndef identifier new-line group elif-groups ::= elif-group elif-groups elif-group elif-group ::= elif constant-expression new-line group else-group ::= else new-line group endif-line ::= endif new-lineBacktracking algorithms 29 Parsing and grammars We cannot work with a full grammar for C++ – Instead, we will consider some vastly oversimplified versions digit ::= 0 1 2 3 4 5 6 7 8 9 pos ::= digit digitpos int ::= pos +pos -pos std ::= int.pos .pos -.pos +.pos sci ::= inteint intEint stdeint stdEint float ::= std sci nondigit ::= A B … Z a b … z _ identifier ::= nondigit identifiernondigit identifierdigit declaration ::= int identifier = identifier; int identifier = int; double identifier = int double identifier = floatBacktracking algorithms 30 Parsing and grammars As you can see, each of these defines a tree – Some of these trees are recursively defined digit ::= 0 1 2 3 4 5 6 7 8 9 pos ::= digit digitpos int ::= pos +pos -pos std ::= int.pos .pos -.pos +.pos sci ::= inteint intEint stdeint stdEint float ::= std sci nondigit ::= A B … Z a b … z _ identifier ::= nondigit identifiernondigit identifierdigit declaration ::= int identifier = identifier; int identifier = int; double identifier = int double identifier = floatBacktracking algorithms 31 Parsing and grammars Suppose we are trying to parse the string int var0 = 3532700; double var1 = 3.5e-27; double var2 = -44.203; What if we’re parsing garbage? double var0 = 3.5g-27; int 1var = 44203; double var2 = 0.0 double var3 = 1.0; Note that in C/C++, 052 defines the octal number equal to 42 10Backtracking algorithms 32 Backjumping In some cases, the following may occur: – Determining that one leaf does not constitute a solution may simultaneously determine that the corresponding sub-tree does not contain a solution, either – In this case, return to the closest ancestor such that it has not yet been determined that all descendants have been ruled out – This is described as backjumpingBacktracking algorithms 33 Searching a maze In trying to find your way through a maze, one simple rule works quite nicely: – The right-hand rule: Touch a wall with your right hand, and continue forward always keeping your right hand touching a wall until you get out.Backtracking algorithms 34 Searching a maze This works well in finding your way through a maze It doesn’t work if you’re trying to get into the maze or out of a mazeBacktracking algorithms 35 Searching a maze Consider the following algorithm: – If the goal is reached, we are done – If there is only one move into a previously unoccupied cell, move to it and flag it as occupied – If there is more than one move into a previously unoccupied cell, push that position onto a stack, and take the right-most available path – If there are no more moves, check the stack: • If the stack is empty, there is no path to the goal • If the stack is not empty, pop to top position and continue the algorithm from that pointBacktracking algorithms 36 Searching a maze In each example, the solution is always found – In the normal maze, less work is required due to backjumpingBacktracking algorithms 37 References Wikipedia, http://en.wikipedia.org/wiki/Backtracking 1 Cormen, Leiserson, and Rivest, Introduction to Algorithms, McGraw Hill, 1990, §11.1, p.200. rd 2 Weiss, Data Structures and Algorithm Analysis in C++, 3 Ed., Addison Wesley, §9.2, p.342-5. These slides are provided for the ECE 250 Algorithms and Data Structures course. The material in it reflects Douglas W. Harder’s best judgment in light of the information available to him at the time of preparation. Any reliance on these course slides by any party for any other purpose are the responsibility of such parties. Douglas W. Harder accepts no responsibility for damages, if any, suffered by any party as a result of decisions made or actions based on these course slides for any other purpose than that for which it was intended.
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