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POHow to find all available maze paths?
 

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  1. POHow to find all available maze paths?
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    <p>I am trying to write a program that is given a maze and tries to find the way out. M is the entrance, E is the exit, 1s are walls, and 0s are pathways. It is supposed to find each path and put P in the path. It is supposed to find all available paths. Right now it finds part of a path.</p> <p>Here is the code:</p> <pre><code>public class Maze { private int size; private String[][] board; private int total; //# of boards private int eX; private int eY; private int mX; private int mY; public Maze( int size, String[][] board ) { this.size = size; this.board = board; total = 0; } private void find( String c ) { int x=0, y=0; for( int i = 0; i &lt; size; i++ ) { for( int j = 0; j &lt; size; j++ ) { if( board[i][j].equals(c) ) { x = i; y = j; } } } if( c.equals("M") ) { mX = x; mY = y; } else if( c.equals("E") ) { eX = x; eY = y; } } public void findPath( ) { find( "M" ); find( "E" ); findNext( mX, mY ); } public void findNext( int x, int y ) { String last = board[x][y]; if( board[x][y].equals("P") ) board[x][y] = "1"; board[x][y] = "P"; if( rightAvailability(x,y) ) { findNext(x+1, y); } else if( leftAvailability(x,y) ) { findNext(x-1, y); } else if( aboveAvailability(x,y) ) { findNext(x, y+1); } else if( belowAvailability(x,y) ) { findNext(x, y-1); } else { total++; printBoard(); } board[x][y]= last; } public boolean rightAvailability( int x, int y ) { if( x+1 &gt;= size ) return false; else if( board[x+1][y].equals("1") ) return false; else if( board[x+1][y].equals("P") ) return false; else return true; } public boolean leftAvailability( int x, int y ) { if( x-1 &lt; 0) return false; else if( board[x-1][y].equals("1") ) return false; else if( board[x-1][y].equals("P") ) return false; else return true; } public boolean aboveAvailability( int x, int y ) { if( y+1 &gt;= size ) return false; else if( board[x][y+1].equals("1") ) return false; else if( board[x][y+1].equals("P") ) return false; else return true; } public boolean belowAvailability( int x, int y ) { if( y-1 &lt; 0) return false; else if( board[x][y-1].equals("1") ) return false; else if( board[x][y-1].equals("P") ) return false; else return true; } public void printBoard() { System.out.println( "The board number " +total+ " is: "); for(int i=0; i&lt; size; i++ ) { for(int j=0; j&lt; size; j++ ) { if( (i==mX) &amp;&amp; (j==mY) ) { System.out.print("M"); } else if( (i==eX) &amp;&amp; (j==eY) ) { System.out.print("E"); } else if( board[i][j].equals("1") ) { System.out.print("1"); } else if( board[i][j].equals("0") ) { System.out.print("0"); } else { System.out.print("P"); } } System.out.println(); } } } </code></pre> <p>Here is the tester:</p> <pre><code>public class MazeTester { public static void main(String[] args) { int size = 11; String[][] board = new String[][] { {"1","1","1","1","1","1","1","1","1","1","1"}, {"1","0","0","0","0","0","1","0","0","0","1"}, {"1","0","1","0","0","0","1","0","1","0","1"}, {"E","0","1","0","0","0","0","0","1","0","1"}, {"1","0","1","1","1","1","1","0","1","0","1"}, {"1","0","1","0","1","0","0","0","1","0","1"}, {"1","0","0","0","1","0","1","0","0","0","1"}, {"1","1","1","1","1","0","1","0","0","0","1"}, {"1","0","1","M","1","0","1","0","0","0","1"}, {"1","0","0","0","0","0","1","0","0","0","1"}, {"1","1","1","1","1","1","1","1","1","1","1"}, }; Maze m = new Maze( size, board ); m.findPath(); } } </code></pre> <p>Here is the current output:</p> <pre><code>The board number 1 is: 11111111111 1PPPPP1PPP1 1P1PPP1P1P1 EP1PPPPP1P1 101111101P1 10101PPP1P1 10001P1PPP1 11111P1PP01 101M1P1PP01 100PPP1PP01 11111111111 The board number 2 is: 11111111111 1PPPPP1PPP1 1P1PPP1P1P1 EP100PPP1P1 101111101P1 10101PPP1P1 10001P1PPP1 11111P1PP01 101M1P1PP01 100PPP1PP01 11111111111 The board number 348 is: 11111111111 1PPPPP10001 1P1PPP10101 EP1PPPPP101 1011111P101 10101PPP101 10001P10001 11111P10001 101M1P10001 100PPP10001 11111111111 </code></pre> <p>EDIT: I added if( board[x][y].equals("P") ) board[x][y] = "1"; at the beginning of findIndex. I also fixed the x &lt;= 0 problem. I updated the output to what I am getting now (it is actually print 348 similar boards).</p>
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    1. USJosh Curren
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      1. POHow to find all available maze paths?
      1. USDaniel Schmitz
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    1. COWhen you say it needs to find "all possible paths", doesn't this imply the maze will contain cycles? A strict interpretation of "all possible paths" in the presence of cycles may require an infinite number of paths - do you mean all paths that don't visit a node twice?
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      1. POHow to find all available maze paths?
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    2. COCorrect- it doenst visit a node twice.
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      1. USJosh Curren
    3. COIn leftAvailability, you're still using y-1 instead of x-1.
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      1. POHow to find all available maze paths?
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