예제 #1
0
파일: Sudoku_main.py 프로젝트: Szubie/Misc
def main():
    complete=False
    count=0
    itercount=0
    while complete==False and count<=10:
        Sudoku_validCheck.validChecker(Sudoku_board.board.tileList())
        if Sudoku_validCheck.validSetter(Sudoku_board.board.tileList())==True:
            count-=1
        for item in Sudoku_board.board.tileList():
            if Sudoku_eliminationTest.elimination(item) !=-1:
#                print Sudoku_eliminationTest.elimination(item)
#                count-=1  BUG?
                item.setValue(Sudoku_eliminationTest.elimination(item))
#                Sudoku_validCheck.validCheck(Sudoku_eliminationTest.getAllNeighbourTiles(item)) #To update the valid and invalid tiles after a change has been made.
                Sudoku_validCheck.validCheck(Sudoku_board.board.tileList())
                #If things are happening, continue this thread.
#        Sudoku_view.view()
#        print 
        if isComplete()==True:
            Sudoku_view.view()
            complete=True
            return True
#        if continues()==False:
 #           return
        count+=1
        itercount+=1
#    Sudoku_view.view()
#    print
    
    return "Couldn't find a solution in "+str(itercount)+" iterations."
예제 #2
0
def solveSudoku(tileList):
    """ Takes a partially filled-in grid and attempts to assign values to
    all unassigned locations in such a way to meet the requirements
    for Sudoku solution (non-duplication across rows, columns, and boxes) """

 #   index; #Index unused, inaccurate?
    index=findUnassignedLocation(tileList)
#    print index.name()
 
    # If there is no unassigned location, we are done
    if (findUnassignedLocation(tileList)==False):
       Sudoku_view.view()
       return True # success!
 
    # consider digits 1 to 9
    for num in index.validValues():
        if (isValid(index,num)):
            index.setValue(num)
            Sudoku_validCheck.validCheck(Sudoku_board.board.tileList())
            if Sudoku_main.main()==True: #TEST#
                Sudoku_view.view()
                return True
            else:
                Sudoku_main.reset()
                Sudoku_main.createScenario3()
                index.setValue(num)
                if solveSudoku(tileList):
                    return True
                index.setValue(-1)
    return False # this triggers backtracking

#solveSudoku(Sudoku_board.board.tileList())
#elapsed = timeit.default_timer() - start_time

#print elapsed
예제 #3
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def backtrack(strippedList):
    Sudoku_validCheck.validCheck(Sudoku_board.board.tileList())
    for tile in sortedList:
        for validValue in tile.validValues():
            tile.setValue(validValue)
            Sudoku_validCheck.validCheck((Sudoku_board.board.tileList())) #Keep things up to date here...
            Sudoku_main.main()
            if Sudoku_main.isComplete()==True:
                return 
            else:
                Sudoku_main.reset()
                Sudoku_main.createScenario3() #Ah, this solves the mystery
                Sudoku_view.view()
예제 #4
0
def attempt3():
    Sudoku_validCheck.validCheck(Sudoku_board.board.tileList())
    Sudoku_eliminationTest.elimination(Sudoku_board.board.tileList()[0])
        
    for item in Sudoku_board.board.tileList():
        col=Sudoku_board.board.columnList()[item.col()-1]
        row=Sudoku_board.board.rowList()[item.row()-1]
        square=Sudoku_board.board.squareList()[item.square()-1]
        for num in item.validValues():
            if not (col.valueExists(num) or row.valueExists(num) or square.valueExists(num)):
                item.setValue(num)
                Sudoku_validCheck.validCheck(Sudoku_board.board.tileList())
                Sudoku_eliminationTest.elimination(Sudoku_board.board.tileList()[item])
예제 #5
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def recursive(n=0):
    validNums=[1,2,3,4,5,6,7,8,9]
    for j in range(n,len(Sudoku_board.board.tileList())):
        for num in validNums:
            if num in Sudoku_board.board.tileList()[j].validValues():
                Sudoku_board.board.tileList()[j].setValue(num)
                Sudoku_validCheck.validChecker(Sudoku_board.board.tileList())
                recursive(n=n+1)
                print n   #How to backtrack?
        print n
        Sudoku_view.view()
    
    if n==len(Sudoku_board.board.tileList())-1 and num in Sudoku_board.board.tileList()[n].validValues():
        Sudoku_board.board.tileList()[n].setValue(num)
        Sudoku_view.view()
        return True
        
    if n==len(Sudoku_board.board.tileList())-1 and num not in Sudoku_board.board.tileList()[n].validValues():
        return False
예제 #6
0
def main():
    Continue=True
    while Continue==True:
        reset()
        scen=Sudoku_scenario.createScenario()
        if scen==3:
            start_time = timeit.default_timer()
            Sudoku_validCheck.validChecker(Sudoku_board.board.tileList())
            Sudoku_solver.solveSudoku(Sudoku_board.board.tileList())
            elapsed = timeit.default_timer() - start_time
            print elapsed
        else:
            start_time = timeit.default_timer()
            Sudoku_recursive2.solveSudoku(Sudoku_board.board.tileList())
            elapsed = timeit.default_timer() - start_time
            print elapsed
        Continue=continues()
        
    print "Thanks for playing!"
    return 
예제 #7
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def attempt():
    validNums=[1,2,3,4,5,6,7,8,9]
    Sudoku_validCheck.validCheck(Sudoku_board.board.tileList())
    Sudoku_eliminationTest.elimination(Sudoku_board.board.tileList()[0])
    for item in Sudoku_board.board.tileList():
        if item.value()==-1:
            print item.name()
            col=Sudoku_board.board.columnList()[item.col()-1]
            row=Sudoku_board.board.rowList()[item.row()-1]
            square=Sudoku_board.board.squareList()[item.square()-1]
            for num in validNums:
                if not (col.valueExists(num) or row.valueExists(num) or square.valueExists(num)): #But how to backtrack here...?
                    item.setValue(num)
                    Sudoku_validCheck.validCheck(Sudoku_board.board.tileList())
                    Sudoku_eliminationTest.elimination(item)
                else:
                    continue
            

    return "Done"
예제 #8
0
파일: SolverClass.py 프로젝트: Szubie/Misc
    def solveSudoku(self,tileList):
        """ Takes a partially filled-in grid and attempts to assign values to
        all unassigned locations in such a way to meet the requirements
        for Sudoku solution (non-duplication across rows, columns, and boxes) """
        Sudoku_validCheck.validChecker(Sudoku_board.board.tileList())
    
        index=self.findUnassignedLocation(tileList)
    
        # If there is no unassigned location, we are done
        if (self.findUnassignedLocation(tileList)==False):
            Sudoku_view.view()
            return True # success!
 
        # consider valid digits
        for num in index.validValues():
            if (self.isValid(index,num)):
                index.setValue(num)
                if self.solveSudoku(tileList):
                    return True
                index.setValue(-1)
        return False # this triggers backtracking
예제 #9
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def attempt2(iteration, validNumbers=[1,2,3,4,5,6,7,8,9], keyMax=9):
    Sudoku_validCheck.validCheck(Sudoku_board.board.tileList())
    Sudoku_eliminationTest.elimination(Sudoku_board.board.tileList()[0])
    
    keyMax
    key=0
    iteration=0
    
    while iteration<len(Sudoku_board.board.tileList()):
        col=Sudoku_board.board.columnList()[Sudoku_board.board.tileList()[iteration].col()-1]
        row=Sudoku_board.board.rowList()[Sudoku_board.board.tileList()[iteration].row()-1]
        square=Sudoku_board.board.squareList()[Sudoku_board.board.tileList()[iteration].square()-1]
        while key<keyMax:
            if key==keyMax:
                if not (col.valueExists(validNumbers[key-1]) or row.valueExists(validNumbers[key-1]) or square.valueExists(validNumbers[key-1])):
                    Sudoku_board.board.tileList()[iteration].setValue(validNumbers[key])
                    Sudoku_validCheck.validCheck(Sudoku_board.board.tileList())
                    Sudoku_eliminationTest.elimination(Sudoku_board.board.tileList()[iteration])
                    return attempt2(iteration+1)
                else:
                    return attempt2(iteration-1,validNumbers[1:0],keyMax=keyMax-1)
            else:
                if not (col.valueExists(validNumbers[key-1]) or row.valueExists(validNumbers[key-1]) or square.valueExists(validNumbers[key-1])):
                    Sudoku_board.board.tileList()[iteration].setValue(validNumbers[key])
                    Sudoku_validCheck.validCheck(Sudoku_board.board.tileList())
                    Sudoku_eliminationTest.elimination(Sudoku_board.board.tileList()[iteration])
                    return attempt2(iteration+1)
                key+=1
        iteration+=1
예제 #10
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def main():
    Continue=True
    while Continue==True:
        reset()
        Sudoku_scenario.createScenario()
        Sudoku_view.view()
        print 
        start_time = timeit.default_timer()
        unassignedTiles=Sudoku_validCheck.validChecker(Sudoku_board.board.tileList())
        Sudoku_solver.solveSudoku(unassignedTiles)
        elapsed = timeit.default_timer() - start_time
        print elapsed

        Continue=continues()
        
    print "Thanks for playing!"
    return 
예제 #11
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"""I can think of a brute force back-tracking algorithm that will check all valid numbers for every square...Horrifically inefficient, but possible."""
#How exactly to implement, though? 81 nested for loops sounds ridiculous...
#Recursion?
import operator

import Sudoku_board
import Sudoku_main
import Sudoku_eliminationTest
import Sudoku_view
import Sudoku_validCheck

Sudoku_main.createScenario3()
Sudoku_validCheck.validCheck(Sudoku_board.board.tileList())

tileList = Sudoku_board.board.tileList()
    
def merge(left, right, compare):
    result=[]
    i , j = 0 , 0    
    while i<len(left) and j<len(right):
        if compare(len(left[i].validValues()), len(right[j].validValues())):
            result.append(left[i])
            i+=1
        else:
            result.append(right[j])
            j+=1
    while (i<len(left)):
        result.append(left[i])
        i+=1
    while (j<len(right)):
        result.append(right[j])