Magnet Puzzle | Backtracking-9 - GeeksforGeeks

26 Apr, 2020
The puzzle game Magnets involves placing a set of domino-shaped magnets (or electrets or other polarized objects) in a subset of slots on a board so as to satisfy a set of constraints. For example, the puzzle on the left has the solution shown on the right:Each slot contains either a blank entry (indicated by ‘x’s), or a “magnet” with a positive and negative end. The numbers along the left and top sides show the numbers of ‘+’ squares in particular rows or columns. Those along the right and bottom show the number of ‘-’ signs in particular rows or columns. Rows and columns without a number at one or both ends are unconstrained as to the number of ‘+’ or ‘-’ signs, depending on which number is not present. In addition to fulfilling these numerical constraints, a puzzle solution must also satisfy the constraint that no two orthogonally touching squares may have the same sign (diagonally joined squares are not constrained).
You are given top[], bottom[], left[], right[] arrays indicates the count of + or – along the top(+), bottom(-), left(+) and right(-) edges respectively. Values of -1 indicate any number of + and – signs. Also given matrix rules[][] contain any one T, B, L or R characters. For a vertical slot in the board, T indicates its top end and B indicates the bottom end. For a horizontal slot in the board, L indicates left end and R indicates the right end.
Examples:
Input : M = 5, N = 6
        top[] = { 1, -1, -1, 2, 1, -1 }
        bottom[] = { 2, -1, -1, 2, -1, 3 }
        left[] = { 2, 3, -1, -1, -1 }
        right[] = { -1, -1, -1, 1, -1 }
        rules[][] = { { L, R, L, R, T, T },
                      { L, R, L, R, B, B },
                      { T, T, T, T, L, R },
                      { B, B, B, B, T, T },
                      { L, R, L, R, B, B }};
Output : + - + - X - 
         - + - + X + 
         X X + - + - 
         X X - + X + 
         - + X X X - 

Input : M = 4, N = 3
        top[] = { 2, -1, -1 }
        bottom[] = { -1, -1, 2 }
        left[] = { -1, -1, 2, -1 }
        right[] = { 0, -1, -1, -1 }
        rules[][] = { { T, T, T },
                      { B, B, B },
                      { T, L, R },
                      { B, L, R } };
Output : + X +
         – X –
        + – +
        – + –

We can solve this problem using Backtracking.
# Write Python3 code hereM = 5N = 6top = [ 1, -1, -1, 2, 1, -1 ]bottom = [ 2, -1, -1, 2, -1, 3 ]left = [ 2, 3, -1, -1, -1 ]right = [ -1, -1, -1, 1, -1 ]  rules = [["L","R","L","R","T","T" ],                      [ "L","R","L","R","B","B" ],                      [ "T","T","T","T","L","R" ],                      [ "B","B","B","B","T","T" ],                      [ "L","R","L","R","B","B" ]];               def canPutPatternHorizontally(rules,i,j,pat):          if j-1>=0 and rules[i][j-1] == pat[0]:        return False    elif i-1>=0 and rules[i-1][j] == pat[0]:        return False    elif i-1>=0 and rules[i-1][j+1] == pat[1]:        return False    elif j+2 < len(rules[0]) and rules[i][j+2] == pat[1]:        return False          return True        def canPutPatternVertically(rules,i,j,pat):          if j-1>=0 and rules[i][j-1] == pat[0]:        return False    elif i-1>=0 and rules[i-1][j] == pat[0]:        return False    elif j+1 < len(rules[0]) and rules[i][j+1] == pat[0]:        return False          return True      def doTheStuff(rules,i,j):          if rules[i][j] == "L" or rules[i][j] == "R":                      #        option 1 +-        if canPutPatternHorizontally(rules,i,j,"+-"):            rules[i][j] = "+"            rules[i][j+1] = "-"                          solveMagnets(rules,i,j)        #        option 2 -+          #        option 3 xx              def checkConstraints(rules):          pCountH = [0 for i in range(len(rules))]    nCountH = [0 for i in range(len(rules))]    for row in range(len(rules)):        for col in range(len(rules[0])):            ch = rules[row][col]            if ch == "+":                pCountH[row] += 1            elif ch == "-":                nCountH[row] += 1                pCountV = [0 for i in range(len(rules[0]))]    nCountV = [0 for i in range(len(rules[0]))]    for col in range(len(rules[0])):        for row in range(len(rules)):            ch = rules[row][col]            if ch == "+":                pCountV[col] += 1            elif ch == "-":                nCountV[col] += 1                            for row in range(len(rules)):        if left[row] != -1:            if pCountH[row] != left[row]:                return False        if right[row] != -1:            if nCountH[row] != right[row]:                return False                                      for col in range(len(rules[0])):        if top[col] != -1:            if pCountV[col] != top[col]:                return False        if bottom[col] != -1:            if nCountV[col] != bottom[col]:                return False        #                    #  if (top[col] != -1 and pCountH[col] != top[col]) or (bottom[col] != -1 and nCountH[col] != bottom[col]) :        #      return False          return True                                                      def solveMagnets(rules,i,j):          if i == len(rules) and j == 0:          # check the constraint before printing        if checkConstraints(rules):            print(rules)    elif j >= len(rules[0]):                   solveMagnets(rules,i+1,0)      # normal cases    else:                   if rules[i][j] == "L":                          #  option 1 +-            if canPutPatternHorizontally(rules,i,j,"+-"):                rules[i][j] = "+"                rules[i][j+1] = "-"                                  solveMagnets(rules,i,j+2)                                  rules[i][j] = "L"                rules[i][j+1] = "R"                          # option 2 -+            if canPutPatternHorizontally(rules,i,j,"-+"):                rules[i][j] = "-"                rules[i][j+1] = "+"                                  solveMagnets(rules,i,j+2)                                  rules[i][j] = "L"                rules[i][j+1] = "R"              # option 3 xx            if True or canPutPatternHorizontally(rules,i,j,"xx"):                rules[i][j] = "x"                rules[i][j+1] = "x"                                  solveMagnets(rules,i,j+2)                                  rules[i][j] = "L"                rules[i][j+1] = "R"           #        vertical check        elif rules[i][j] == "T":                          #        option 1 +-            if canPutPatternVertically(rules,i,j,"+-"):                rules[i][j] = "+"                rules[i+1][j] = "-"                                  solveMagnets(rules,i,j+1)                                  rules[i][j] = "T"                rules[i+1][j] = "B"              #        option 2 -+            if canPutPatternVertically(rules,i,j,"-+"):                rules[i][j] = "-"                rules[i+1][j] = "+"                                  solveMagnets(rules,i,j+1)                                  rules[i][j] = "T"                rules[i+1][j] = "B"              #        option 3 xx                              if True or canPutPatternVertically(rules,i,j,"xx"):                rules[i][j] = "x"                rules[i+1][j] = "x"                                  solveMagnets(rules,i,j+1)                                  rules[i][j] = "T"                rules[i+1][j] = "B"                          else:            solveMagnets(rules,i,j+1)    # Driver code         solveMagnets(rules,0,0)
Source :https://people.eecs.berkeley.edu/~hilfingr/programming-contest/f2012-contest.pdf
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