21 recursion and backtracking

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3/5/09

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Intermediate / Advanced

Programming

Recursion and Backtracking

Lynn Robert Carter 2009-03-03

© Copyright 2009, Lynn Robert Carter

Intermediate/Advanced ProgrammingRecursion and Backtracking

Agenda

N Queens

+ Finding solutions

+ Backtracking

+ The N Queens solution

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The N Queens Puzzle

+ Consider a board, like a chess board, but

one where you can specify the board size

+ How many chess queens can be placed

on this “board” so that no queen can “take”

another?+ How many different placements of queens

can be made for a given sized board?

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A board of size 5

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What does “take” mean?

+ In chess a piece “takes” another piece

when it can move to the same square as

the piece it is going to take

+ The various pieces on a chess board

move in very different ways+ Some pieces, such are rooks (castles) can

move all the way across the board, but

only up, down, left, or right

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How does a Queen move?

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♕



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The moves of a Queen

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♕


Safe places for a second Queen

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♕

♕ ♕

♕

♕

♕♕

♕

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Agenda

N Queens

Finding solutions

+ Backtracking


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Finding solutions

+ Start in the first row, we

have a choice of five

possible moves

+ Any of these five choices

could lead to a solution

+ To move forward, we

have to choose one and

then see if that can lead

to a solution

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The first move

+ To be systematic, pick the

first one; place a queen in

the first square

+ We know than no other

queen can be in that row

+ So start with the second

row and see which

squares are “safe”

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♕


Possible safe places

+ The first column is not

safe

+ The second square in the

second row is not safe

+ All of the others in the row

are safe

+ Let’s use the first one

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♕



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Examine row three

+ Which of the squares in

row three are “safe”?

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♕

♕


Examine row three

+ The only “safe” square in

this situation is the right

most square

+ So place a queen there

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♕

♕



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Examine row four

+ Where are the safe

squares in row four?

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♕

♕

♕


Examine row four

+ There is just one possible

square for the fourth row

+ Once again, place the

next queen there!

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♕

♕

♕



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Examine the last row

+ Once again, there is only

one choice

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♕

♕

♕

♕


A solution

+ Using this process, we

have found a solution

+ How many more solutions

are possible?

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♕

♕

♕

♕

♕



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Agenda

N Queens

Finding solutions

Backtracking


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4 Queens

+ Consider the 4 Queens case

+ There are four possible first

moves

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4 Queens row 2 choice

+ Start with a Queen in the

first row and first column

+ There are just two possible

choices for the second row

+ Let’s pick the first one

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♕


4 Queens row 2

+ Given this sized board and

these two queens, where

can a queen go for row 3?

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♕

♕



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4 Queens row 3

+ There is no “safe” move in

this situation, so be have to

back up and try a different

choice

+ This “backing up” is known

as “backtracking”

+ So, let’s back up to the most

recent point where we had a

choice, the choice in row 2

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♕

♕


Back up to row two

+ We know that the first of

these two choices does not

lead to a solution

+ Let’s try the second of the

two choices

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♕



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A second try at row three

+ With queens positioned as

shown here, where are the

possible “safe” moves in row

three?

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♕

♕


Possible row three moves

+ With queens positioned as

shown here, there is just

one possible “safe” move,

so let’s take it

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♕

♕



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Possible row four moves

+ Where are the “safe” moves

in row four?

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♕

♕

♕


Another failure

+ There are no safe moves

here, so… we have to back

track again

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♕

♕

♕



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Once again at row two

+ Now we know that neither of

the two choices for row 2

will lead us to a solution

+ Therefore, we must back up

to row 1 and try a different

choice there

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♕


Backup all the way to row 1

+ We now know that it is not

possible for 4 Queens to

have a solution with a queen

in the first row and first

column

+ Let’s try the queen in the

first row and second column

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+ With a queen in this position

in the first row, there is only

one choice for row 2

+ So, let’s make that move

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♕



+ Queens in these positions

for the first two rows means

that there is again just one

choice for row three


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♕

♕



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+ Queens in these positions

for the first three rows

means once again just one

choice is possible


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♕

♕

♕


4 Queens solution

+ A queen in the first row and

second column leads to a

solution

+ How many more solutions

are possible?

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♕

♕

♕

♕



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Number of possible solutions

+ We know about two of the

choices in the first row

+ What is your intuition about

the other two positions?

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A tree of possible solutions

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♕ ♕ ♕ ♕

♕

♕

♕

♕

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♕

♕

♕

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♕

♕

♕

♕

♕

♕

♕

♕

♕

♕

♕

♕

♕

♕

♕

♕

♕

♕

♕

♕

♕

♕

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♕ ♕ ♕ ♕

12 more

not shown

60 more

not shown

252 more

not shown



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A solution strategy

+ Build a tree of all possible valid queen

placements

+ Traverse the whole tree

+ Whenever there is a board where the

number of queens is equal the number of rows in the board, add that board to a

collection of solutions

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Possible valid placements?

+ Assume you are given a valid board and a

row number where you are to try to place

as many queens as possible

+ How can you do this?

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Possible valid placements

+ Perform a loop that iterates once for each

possible placement in the new row using

the index col

+ Make a temporary copy of the board

+ Try to place the queen in the columnsspecified by col

+ If that placement is valid, use it to try to place

queens in the rest of the rows (recursively)

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Possible placement example

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♕ ♕ ♕ ♕

♕

♕

♕

♕

♕

♕

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addQueen

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/*** Method to add a Queen to the board * @param b The Board * @param rowNum The starting row number to try to add the queen* * @return The board after placing the Queen, if possible else null*/public Board addQueen(Board b, int rowNum) {// base case: number of Queens on the board is equal to the board sizeif(b.numQueens() == b.getSize()) return b;

for(int x =0; x < b.getSize(); x++){ Board tempBoard = new Board(b); // try to place the queen in the board at column x within row, rowNum if(tempBoard.place(new Board.Position(rowNum, x))){ // If it was added successfully // recursively try adding a queen into the next row Board nextBoard = addQueen(tempBoard, rowNum+1); // If it comes back null, no solution is possible if(nextBoard != null) // The board is not null, so this is a solution return nextBoard; }}return null;

}


getAllSolutions

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/** * Method to get all possible solutions to the Queens problem* * @param b The board to place the Queens at* @param c All possible boards* @param rowNum The initial row number to start placing queens*/public void getAllSolutions(Board b, Collection<Board> c, int rowNum){

if(rowNum < b.getSize()){ for(int x =0; x< b.getSize(); x++){ Board tempBoard = new Board(b); if(tempBoard.place(new Board.Position(rowNum, x))){ // a queen is placed at that row // recursively call the method on the next row getAllSolutions(tempBoard, c, rowNum+1); } } // getting here means that we have run past the end of the row // so we return without adding any more boards to the collection return;}// add the solved board to the Collection of boards// this only happens when rowNum >= b.getSize()c.add(b);

}



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The Class Design of NQueens

+ The program consists of

+ A class that defines the board (Board)

+ A class that solve the problem (NQueens)

+ A class that implements a Swing GUI window

(NQueensSwing)

+ A mainline class for the Swing version of the

puzzle solver (NQueensGUI)

+ A mainline class for a text version (Driver)

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The NQueen Board-1

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public class Board {protected enum BoardState implements Cloneable { UNTHREATENED , THREATENED , QUEEN };protected BoardState[][] board;private int n;private int numQueens;public static class Position { int x = 0, y = 0; public Position(int i, int j) { x = i; y = j; } public boolean equals(Position p) { return x == p.x && y == p.y; }}public int getSize(){ return this.n;}



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The NQueen Board-2

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/** * Constructs an empty board sizeXsize * * Complains if you attempt to construct a negative board Boards of size 0 * are allowed */public Board(int size) { if (size < 0) { System.out .println("Negative boards are not allowed."); return; }

n = size; numQueens = 0; board = new BoardState[n][n]; for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) board[i][j] = BoardState.UNTHREATENED ;}public Board(Board b) { board = new BoardState[b.n][b.n]; for (int i = 0; i < b.n; i++) for (int j = 0; j < b.n; j++) board[i][j]=b.board[i][j]; n = b.n; numQueens = b.numQueens;}


isThreatened and more

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/*** Determines if a position can be reached in one or zero moves by any of* the queens currently on the board* * @param p* the position to determine the state of* @return true if Position p is being threatened*/public boolean isThreatened(Position p) {return board[p.x][p.y] != BoardState.UNTHREATENED ;}/*** accessor for the number of queens currently placed on the board*/public int numQueens() {return numQueens;}/*** returns true if the maximum number of queens have been placed.*/public boolean done() {if (n < 1 || n == 2 || n == 3) return (numQueens == 0);return (numQueens == n);}



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place

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/*** Places a queen on the board at Position P if that is an empty* unthreatened place* * @param p* the position to place the queen in* * @return true if placement is sucsessful*/public boolean place(Position p) {if (board[p.x][p.y] != BoardState.UNTHREATENED )return false;

// mark the row and col as threatenedfor (int i = 0; i < n; i++) { board[p.x][i] = BoardState.THREATENED ; board[i][p.y] = BoardState.THREATENED ;}// mark the diagonals as threatenedfor (int x = p.x, y = p.y; x < n && y < n; x++, y++) board[x][y] = BoardState.THREATENED ;for (int x = p.x, y = p.y; x >= 0 && y >= 0; x--, y--) board[x][y] = BoardState.THREATENED ;for (int x = p.x, y = p.y; x >= 0 && y < n; x--, y++) board[x][y] = BoardState.THREATENED ;for (int x = p.x, y = p.y; x < n && y >= 0; x++, y--) board[x][y] = BoardState.THREATENED ;board[p.x][p.y] = BoardState.QUEEN ;numQueens++;return true;

}


Agenda

N Queens

Finding solutions

Backtracking

The N Queens solution

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A sample solution

+ Download the NQueens archive from the

blackboard and get it to work

+ A solution written by Hatem Alismail

+ Study the solution and be able to answer

some questions about this code

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Intermediate / Advanced

Programming

Recursion and Backtracking

Lynn Robert Carter 2009-03-03

© Copyright 2009, Lynn Robert Carter

21 recursion and backtracking

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