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euler96.py
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euler96.py
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"""
Euler 96: Su Doku
Grid 01
003 020 600
900 305 001
001 806 400
008 102 900
700 000 008
006 708 200
002 609 500
800 203 009
005 010 300
"""
import qa
v=qa.v
srt=qa.srt()
class sudoku:
# initialize puzzle from text
def __init__(self,puzzle):
self.grid=[int(cell) for cell in puzzle]
self.unsolved=[i for i,cell in enumerate(self.grid) if cell==0]
self.couldBe=map(lambda x: [1,2,3,4,5,6,7,8,9] if x == 0 else None,self.grid)
_=self.updatePass()
if False:
print "Starting Grid:"
self.printGrid()
print "{0!s} unsolved cells".format(len(self.unsolved))
print ""
# return puzzle grid as a string
def getPuzzle(self):
return ''.join([str(cell) for cell in self.grid])
# set puzzle grid from a string
def setPuzzle(self,puzzle_string):
for old,new in zip(self.grid,[int(cell) for cell in puzzle_string]):
if old!=new and old!=0:
print "New puzzle setting contradicts original puzzle!"
self.grid=[int(cell) for cell in puzzle_string]
# return indexes of row r as a list
def getRow(self,r):
if r>8: return None
return range(r*9,(r+1)*9)
# return indexes of column c as a list
def getCol(self,c):
if c>8: return None
return [i for i in range(81) if i%9==c]
# return indexes of box (r,c), r,c in (0,2)
def getBox(self,r,c):
if r>2 or c>2:return None
return [i for i in range(81) if (i/3)%3==c and (i/27)==r]
# prints the sudoku grid all pretty-like
def printGrid(self):
row=''
for i, cell in enumerate(self.grid):
row+=str(cell)
if i%9==8:
print row
row=''
if (i/9)%3==2 and i<80:
print '---|---|---'
elif i%3==2:
row+="|"
print row
# prints a list of unsolved squares and what they could be
def printUnsolved(self):
print "Unsolved:"
for i in self.unsolved:
print "({0!s},{1!s}) - {2!s}".format(i/9,i%9,self.couldBe[i])
# winnow couldbe list for unsolved squares
def updatePass(self):
updated=0
for i in self.unsolved:
thisRow=set([self.grid[x] for x in self.getRow(i/9)])
thisCol=set([self.grid[x] for x in self.getCol(i%9)])
thisBox=set([self.grid[x] for x in self.getBox(i/27,(i%9)/3)])
cb_was=len(self.couldBe[i] or [])
self.couldBe[i]=list(set(self.couldBe[i]).difference(thisRow).difference(thisCol).difference(thisBox))
if len(self.couldBe[i])<cb_was:
updated+=1
return updated
# solve squares with only one couldBe entry
# or whose couldBe contains a value that doesn't exist elsewhere in the row/col/box
def solvePass(self):
if len(self.unsolved)==0:
return None
solved=0
for i in self.unsolved:
solution=None
thisRowCouldBe=set([c for x in self.getRow(i/9) if x!=i and self.couldBe[x] for c in self.couldBe[x]])
thisColCouldBe=set([c for x in self.getCol(i%9) if x!=i and self.couldBe[x] for c in self.couldBe[x]])
thisBoxCouldBe=set([c for x in self.getBox(i/27,(i%9)/3) if x!=i and self.couldBe[x] for c in self.couldBe[x]])
if len(self.couldBe[i])==1:
solution=self.couldBe[i][0]
elif set(self.couldBe[i]).difference(thisRowCouldBe):
solution=set(self.couldBe[i]).difference(thisRowCouldBe).pop()
elif set(self.couldBe[i]).difference(thisColCouldBe):
solution=set(self.couldBe[i]).difference(thisColCouldBe).pop()
elif set(self.couldBe[i]).difference(thisBoxCouldBe):
solution=set(self.couldBe[i]).difference(thisBoxCouldBe).pop()
if solution:
self.grid[i]=solution
self.couldBe[i]=None
self.unsolved.remove(i)
_=self.updatePass()
solved+=1
return solved
def probSolve(self):
for unsolved_cell in self.unsolved:
for solution_option in self.couldBe[unsolved_cell]:
print "trying solution with ({0!s},{1!s}) = {2!s}".format(unsolved_cell/9,unsolved_cell%9,solution_option)
old_puzzle=self.getPuzzle()
tryPuzzle=sudoku(old_puzzle[:unsolved_cell]+str(solution_option)+old_puzzle[unsolved_cell+1:])
trySolution=tryPuzzle.solve()
if trySolution:
self.setPuzzle(trySolution)
return self.getPuzzle()
return None
def solve(self):
if v:tick=qa.tick()
counter=0
solves=1
while solves>0:
solves=self.solvePass()
counter+=1
if solves==None:
if v:tick=qa.tock(tick,"Solved in {0!s} passes".format(counter))
return self.getPuzzle()
if v:tick=qa.tock(tick,"Deadlocked in {0!s} passes".format(counter))
return None
"""
New solving methods:
Swaps: identify locations where n cells in the same unit each contain the same n numbers; even if
these cells contain other numbers that aren't explicitly forced out of them, they can't really take
those values because they are the only place where those values could appear.
2-group push: if the only possible cells in a box for some number are in the same row/col, we can
eliminate other instances of that number in the couldBe list in that row/col.
Probabilistic: try something and see if it solves, if not go back to where you were. effectively
recursion.
well, probabilistic solving should take care of the rest.
"""
sudokutxt=open("sudoku.txt","r")
def getPuzzles(f):
puzzles=[]
i=-1
for row in f.readlines():
if row.__contains__('Grid'):
puzzles.append('')
i+=1
else:
puzzles[i]+=row.replace('\r\n','')
return puzzles
puzzles=getPuzzles(sudokutxt)
solves=[]
i=1
for puzzle in puzzles:
print "Puzzle {0!s}".format(i)
p=sudoku(puzzle)
this_solve=p.solve()
if this_solve:
solves.append(int(this_solve[0:3]))
else:
this_solve=p.probSolve()
if this_solve:
solves.append(int(this_solve[0:3]))
i+=1
print solves
print sum(solves)
""""""
"""
[483, 245, 462, 137, 523, 176, 143, 487, 814, 761, 976, 962, 397, 639, 697, 361, 359, 786, 743, 782, 428, 425, 348, 124, 361, 581, 387, 345, 235, 298, 761, 132, 698, 852, 453, 516, 945, 365, 134, 193, 814, 384, 469, 316, 586, 954, 159, 861, 294, 351]
24702
"""