Following this code, I'm trying to write a piece to find out if a sudoku has one or more than one solutions (we can safely assume that there is at least one solution for sure). This is the code:
grid = []
while True:
row = list(input('Row: '))
ints = []
for n in row:
ints.append(int(n))
grid.append(ints)
if len(grid) == 9:
break
def possible(x, y, n):
for i in range(0, 9):
if grid[i][x] == n and i != y: # Checks for number (n) in X columns
return False
for i in range(0, 9):
if grid[y][i] == n and i != x: # Checks for number (n) in X columns
return False
x0 = (x // 3) * 3
y0 = (y // 3) * 3
for X in range(x0, x0 + 3):
for Y in range(y0, y0 + 3): # Checks for numbers in box(no matter the position, it finds the corner)
if grid[Y][X] == n:
return False
return True
def solve():
global grid
for y in range(9):
for x in range(9):
if grid[y][x] == 0:
for n in range(1, 10):
if possible(x, y, n):
grid[y][x] = n
solve()
grid[y][x] = 0
return
print(grid)
solve()
The problem is, this code will print out all of the solutions for a given sudoku, even if the puzzle has, say, 100 solutions. I am not interested in that. I just want to know if there is more than one answer or not. So basically, I want the solve function to stop immediately once more than one solution is found, and inform the user about that. How can I do that? How can I stop the code to go all the way down to the very last solution?
A recursive solve function could be rewritten as a generator:
# print(solution) --> yield solution
# solve(...) --> yield from solve(...)
Then to collect all solutions:
for s in solve(...):
print(s)
and finally to abort after N (e.g. 2) solutions:
for i,s in enumerate(solve(...), start=1):
print(s)
if i >= N:
has_N_solutions = True
break
else:
# else = no break
has_N_solutions = False