Recursively Generate Every Tic-Tac-Toe game in Python
I am working on a project where I generate every possible tic-tac-toe array. As a proof of concept, I am working on code to fill an array with 9 subarrays. Each subarray will have two values, the first one being 0 or 1 (for x and o respectively), and the second one being from 1 to 9 (representing when it was placed).
An example of an array I would like to get out would look like:
[[0, 0], [1, 1], [0, 2], [1, 3], [0, 4], [1, 5], [0, 6], [1, 7], [0, 8]]
I have already written code, using 9 for loops, each one nested in the one above, which gives me the desired results (every possible array, and each one unique). But I am trying to write code, using recursion, and avoiding writing tons of nested loops.
When I run the code below, it is only able to generate the array above, and cannot create other combinations. My code is below:
print("running...")
allGames =
checkCurrentGame = [5, 5, 5, 5, 5, 5, 5, 5, 5]
stepsDown = 0
def cleanGame(move, currentGame):
for j in range(9):
if (currentGame[j][1] >= move):
currentGame[j] = [5, 0]
def completeMove(moveNumber, currentGame):
global stepsDown
stepsDown = stepsDown + 1
for i in range(9):
cleanGame(moveNumber, currentGame)
if (currentGame[i][0] == 5):
currentGame[i][0] = i % 2
currentGame[i][1] = moveNumber
allGames.append(currentGame)
break
if (stepsDown < 9):
generateGame(currentGame)
def generateGame(currentGame):
for i in range(9):
completeMove(i, currentGame)
generateGame([[5, 0], [5, 0], [5, 0], [5, 0], [5, 0], [5, 0], [5, 0], [5, 0], [5, 0]])
for x in range(len(allGames)):
print(allGames[x])
python arrays recursion
add a comment |
I am working on a project where I generate every possible tic-tac-toe array. As a proof of concept, I am working on code to fill an array with 9 subarrays. Each subarray will have two values, the first one being 0 or 1 (for x and o respectively), and the second one being from 1 to 9 (representing when it was placed).
An example of an array I would like to get out would look like:
[[0, 0], [1, 1], [0, 2], [1, 3], [0, 4], [1, 5], [0, 6], [1, 7], [0, 8]]
I have already written code, using 9 for loops, each one nested in the one above, which gives me the desired results (every possible array, and each one unique). But I am trying to write code, using recursion, and avoiding writing tons of nested loops.
When I run the code below, it is only able to generate the array above, and cannot create other combinations. My code is below:
print("running...")
allGames =
checkCurrentGame = [5, 5, 5, 5, 5, 5, 5, 5, 5]
stepsDown = 0
def cleanGame(move, currentGame):
for j in range(9):
if (currentGame[j][1] >= move):
currentGame[j] = [5, 0]
def completeMove(moveNumber, currentGame):
global stepsDown
stepsDown = stepsDown + 1
for i in range(9):
cleanGame(moveNumber, currentGame)
if (currentGame[i][0] == 5):
currentGame[i][0] = i % 2
currentGame[i][1] = moveNumber
allGames.append(currentGame)
break
if (stepsDown < 9):
generateGame(currentGame)
def generateGame(currentGame):
for i in range(9):
completeMove(i, currentGame)
generateGame([[5, 0], [5, 0], [5, 0], [5, 0], [5, 0], [5, 0], [5, 0], [5, 0], [5, 0]])
for x in range(len(allGames)):
print(allGames[x])
python arrays recursion
add a comment |
I am working on a project where I generate every possible tic-tac-toe array. As a proof of concept, I am working on code to fill an array with 9 subarrays. Each subarray will have two values, the first one being 0 or 1 (for x and o respectively), and the second one being from 1 to 9 (representing when it was placed).
An example of an array I would like to get out would look like:
[[0, 0], [1, 1], [0, 2], [1, 3], [0, 4], [1, 5], [0, 6], [1, 7], [0, 8]]
I have already written code, using 9 for loops, each one nested in the one above, which gives me the desired results (every possible array, and each one unique). But I am trying to write code, using recursion, and avoiding writing tons of nested loops.
When I run the code below, it is only able to generate the array above, and cannot create other combinations. My code is below:
print("running...")
allGames =
checkCurrentGame = [5, 5, 5, 5, 5, 5, 5, 5, 5]
stepsDown = 0
def cleanGame(move, currentGame):
for j in range(9):
if (currentGame[j][1] >= move):
currentGame[j] = [5, 0]
def completeMove(moveNumber, currentGame):
global stepsDown
stepsDown = stepsDown + 1
for i in range(9):
cleanGame(moveNumber, currentGame)
if (currentGame[i][0] == 5):
currentGame[i][0] = i % 2
currentGame[i][1] = moveNumber
allGames.append(currentGame)
break
if (stepsDown < 9):
generateGame(currentGame)
def generateGame(currentGame):
for i in range(9):
completeMove(i, currentGame)
generateGame([[5, 0], [5, 0], [5, 0], [5, 0], [5, 0], [5, 0], [5, 0], [5, 0], [5, 0]])
for x in range(len(allGames)):
print(allGames[x])
python arrays recursion
I am working on a project where I generate every possible tic-tac-toe array. As a proof of concept, I am working on code to fill an array with 9 subarrays. Each subarray will have two values, the first one being 0 or 1 (for x and o respectively), and the second one being from 1 to 9 (representing when it was placed).
An example of an array I would like to get out would look like:
[[0, 0], [1, 1], [0, 2], [1, 3], [0, 4], [1, 5], [0, 6], [1, 7], [0, 8]]
I have already written code, using 9 for loops, each one nested in the one above, which gives me the desired results (every possible array, and each one unique). But I am trying to write code, using recursion, and avoiding writing tons of nested loops.
When I run the code below, it is only able to generate the array above, and cannot create other combinations. My code is below:
print("running...")
allGames =
checkCurrentGame = [5, 5, 5, 5, 5, 5, 5, 5, 5]
stepsDown = 0
def cleanGame(move, currentGame):
for j in range(9):
if (currentGame[j][1] >= move):
currentGame[j] = [5, 0]
def completeMove(moveNumber, currentGame):
global stepsDown
stepsDown = stepsDown + 1
for i in range(9):
cleanGame(moveNumber, currentGame)
if (currentGame[i][0] == 5):
currentGame[i][0] = i % 2
currentGame[i][1] = moveNumber
allGames.append(currentGame)
break
if (stepsDown < 9):
generateGame(currentGame)
def generateGame(currentGame):
for i in range(9):
completeMove(i, currentGame)
generateGame([[5, 0], [5, 0], [5, 0], [5, 0], [5, 0], [5, 0], [5, 0], [5, 0], [5, 0]])
for x in range(len(allGames)):
print(allGames[x])
python arrays recursion
python arrays recursion
edited Nov 24 '18 at 15:53
גלעד ברקן
13.2k21543
13.2k21543
asked Nov 22 '18 at 20:23
Luke SchultzLuke Schultz
464
464
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
If I understand your question correctly this should do, however this is not recursion -
import itertools
[zip(p, range(0, 9)) for p in itertools.product([0, 1], repeat=9)]
The code first generates a board (9 0's or 1's) -
itertools.product([0, 1], repeat=9)
and then add the index data to it.
I recommend having a look in itertools
This is not good, since it contains many invalid entries. For example the very first entry is[(0, 0), (0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (0, 6), (0, 7), (0, 8)]
.
– handras
Nov 23 '18 at 6:46
Also, it only includes games in wich the cells are filled sequentially. So it misses a lot of valid games in which e.g. the first player start in the middle.
– handras
Nov 23 '18 at 7:10
It contains all possible entries - len([zip(p, range(0, 9)) for p in itertools.product([0, 1], repeat=9)]) = 512=2^9.
– Tom Ron
Nov 23 '18 at 10:53
But 512 is not all the possible entries. You need to take into account which cell was filled first, second, ... last. That means a factor of 9!. Here is a representation of a valid game:[(0,8), (1,7), (0,6), (1,5), (0,4), (1,3), (0,2), (1,0), (0,0)]
(note that 0 and 1 alternating in filling cells). Can you find it in your list of 512? It's not there... Furthermore you did not address the fact that your code will yield incorrect entries.
– handras
Nov 23 '18 at 11:29
1
It looks like you may have misunderstood the question. 2^9 will give us all combinations of 9 zeros and ones. Tic-tac-toe games are a significantly smaller subset of that and the OP's game representation also includes a number in each cell's data for "when it [the move] was placed."
– גלעד ברקן
Nov 24 '18 at 16:02
add a comment |
This is fun. Here's a method that plays the game, branching off each possible move (or at any rate, that's the intention).
Although my preference is usually to concat and return a pure result, to allow for an early exit and inspection, this recursion accumulates in a global variable.
import copy
import json
import time
s = set()
rs =
def f():
"""
111
111000
111000000
1001001
10010010
100100100
1010100
100010001
"""
wins = [0007,0070,0700,0111,0222,0444,0124,0421]
def winner(board):
# check xs or os individual board for a win
for i in wins:
if not (i ^ (board & i)):
return True
return False
# xs board, os board, move-number, game
def g(xs, os, move, result):
# allow for early exit
global rs
if (len(rs) > 20000):
return
# base case, win or draw
if winner(xs) or winner(os) or move == 9:
#print "{0:b}".format(xs)
#print "{0:b}".format(os)
#print (result, xs, os)
#print
# confirm we're not duplicating results
enc = json.dumps(result)
if enc in s:
print "Duplicate!"
print result
s.add(enc)
# accumulate results
rs.append((result, xs, os))
return
board = xs | os
for i in xrange(9):
# candidate move
m = 1 << i
# valid move
if not (m & board):
_result = copy.deepcopy(result)
# 'O' plays on odd numbered moves
if move & 1:
_result[i] = [1, move]
g(xs, os | m, move + 1, _result)
# 'X' plays on even numbered moves
else:
_result[i] = [0, move]
g(xs | m, os, move + 1, _result)
# start the recursion
g(0, 0, 0, [[-1, -1]] * 9)
start_time = time.time()
f()
print("--- %s seconds ---" % (time.time() - start_time))
Output:
"""
rs[2002]
=> ([[0, 0], [1, 1], [-1, -1],
[-1, -1], [1, 5], [0, 2],
[0, 4], [1, 3], [-1, -1]], 97, 146)
x--
---
---
xo-
---
---
xo-
--x
---
xo-
--x
-o-
xo-
--x
xo-
xo-
-ox
xo-
"""
add a comment |
I recommend you watch this video on youtube. The professor teaches really nicely about recursion. I think it will benefit for you more, then getting a working code. He uses sudoku as an example, but both games are just 2D arrays.
If you watched the video, you will know how to modify this example to better suit your needs.
A pseudo code:
def place(who, when, where, table):
if (None, None) not in table: # the table is full
print(table)
return
if table[where] not (None, None): # this cell is already marked
return
table[where] (who, when) # can place our mark on the cell
# make the recursive calls
for i in range(9):
place(who.opponent, when+1, i, table)
for cell in range(9):
empty = [(None, None) for _ in range(9)]
place(firstplayer, 0, cell, empty)
add a comment |
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3 Answers
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active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
If I understand your question correctly this should do, however this is not recursion -
import itertools
[zip(p, range(0, 9)) for p in itertools.product([0, 1], repeat=9)]
The code first generates a board (9 0's or 1's) -
itertools.product([0, 1], repeat=9)
and then add the index data to it.
I recommend having a look in itertools
This is not good, since it contains many invalid entries. For example the very first entry is[(0, 0), (0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (0, 6), (0, 7), (0, 8)]
.
– handras
Nov 23 '18 at 6:46
Also, it only includes games in wich the cells are filled sequentially. So it misses a lot of valid games in which e.g. the first player start in the middle.
– handras
Nov 23 '18 at 7:10
It contains all possible entries - len([zip(p, range(0, 9)) for p in itertools.product([0, 1], repeat=9)]) = 512=2^9.
– Tom Ron
Nov 23 '18 at 10:53
But 512 is not all the possible entries. You need to take into account which cell was filled first, second, ... last. That means a factor of 9!. Here is a representation of a valid game:[(0,8), (1,7), (0,6), (1,5), (0,4), (1,3), (0,2), (1,0), (0,0)]
(note that 0 and 1 alternating in filling cells). Can you find it in your list of 512? It's not there... Furthermore you did not address the fact that your code will yield incorrect entries.
– handras
Nov 23 '18 at 11:29
1
It looks like you may have misunderstood the question. 2^9 will give us all combinations of 9 zeros and ones. Tic-tac-toe games are a significantly smaller subset of that and the OP's game representation also includes a number in each cell's data for "when it [the move] was placed."
– גלעד ברקן
Nov 24 '18 at 16:02
add a comment |
If I understand your question correctly this should do, however this is not recursion -
import itertools
[zip(p, range(0, 9)) for p in itertools.product([0, 1], repeat=9)]
The code first generates a board (9 0's or 1's) -
itertools.product([0, 1], repeat=9)
and then add the index data to it.
I recommend having a look in itertools
This is not good, since it contains many invalid entries. For example the very first entry is[(0, 0), (0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (0, 6), (0, 7), (0, 8)]
.
– handras
Nov 23 '18 at 6:46
Also, it only includes games in wich the cells are filled sequentially. So it misses a lot of valid games in which e.g. the first player start in the middle.
– handras
Nov 23 '18 at 7:10
It contains all possible entries - len([zip(p, range(0, 9)) for p in itertools.product([0, 1], repeat=9)]) = 512=2^9.
– Tom Ron
Nov 23 '18 at 10:53
But 512 is not all the possible entries. You need to take into account which cell was filled first, second, ... last. That means a factor of 9!. Here is a representation of a valid game:[(0,8), (1,7), (0,6), (1,5), (0,4), (1,3), (0,2), (1,0), (0,0)]
(note that 0 and 1 alternating in filling cells). Can you find it in your list of 512? It's not there... Furthermore you did not address the fact that your code will yield incorrect entries.
– handras
Nov 23 '18 at 11:29
1
It looks like you may have misunderstood the question. 2^9 will give us all combinations of 9 zeros and ones. Tic-tac-toe games are a significantly smaller subset of that and the OP's game representation also includes a number in each cell's data for "when it [the move] was placed."
– גלעד ברקן
Nov 24 '18 at 16:02
add a comment |
If I understand your question correctly this should do, however this is not recursion -
import itertools
[zip(p, range(0, 9)) for p in itertools.product([0, 1], repeat=9)]
The code first generates a board (9 0's or 1's) -
itertools.product([0, 1], repeat=9)
and then add the index data to it.
I recommend having a look in itertools
If I understand your question correctly this should do, however this is not recursion -
import itertools
[zip(p, range(0, 9)) for p in itertools.product([0, 1], repeat=9)]
The code first generates a board (9 0's or 1's) -
itertools.product([0, 1], repeat=9)
and then add the index data to it.
I recommend having a look in itertools
answered Nov 22 '18 at 20:37
Tom RonTom Ron
1,305819
1,305819
This is not good, since it contains many invalid entries. For example the very first entry is[(0, 0), (0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (0, 6), (0, 7), (0, 8)]
.
– handras
Nov 23 '18 at 6:46
Also, it only includes games in wich the cells are filled sequentially. So it misses a lot of valid games in which e.g. the first player start in the middle.
– handras
Nov 23 '18 at 7:10
It contains all possible entries - len([zip(p, range(0, 9)) for p in itertools.product([0, 1], repeat=9)]) = 512=2^9.
– Tom Ron
Nov 23 '18 at 10:53
But 512 is not all the possible entries. You need to take into account which cell was filled first, second, ... last. That means a factor of 9!. Here is a representation of a valid game:[(0,8), (1,7), (0,6), (1,5), (0,4), (1,3), (0,2), (1,0), (0,0)]
(note that 0 and 1 alternating in filling cells). Can you find it in your list of 512? It's not there... Furthermore you did not address the fact that your code will yield incorrect entries.
– handras
Nov 23 '18 at 11:29
1
It looks like you may have misunderstood the question. 2^9 will give us all combinations of 9 zeros and ones. Tic-tac-toe games are a significantly smaller subset of that and the OP's game representation also includes a number in each cell's data for "when it [the move] was placed."
– גלעד ברקן
Nov 24 '18 at 16:02
add a comment |
This is not good, since it contains many invalid entries. For example the very first entry is[(0, 0), (0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (0, 6), (0, 7), (0, 8)]
.
– handras
Nov 23 '18 at 6:46
Also, it only includes games in wich the cells are filled sequentially. So it misses a lot of valid games in which e.g. the first player start in the middle.
– handras
Nov 23 '18 at 7:10
It contains all possible entries - len([zip(p, range(0, 9)) for p in itertools.product([0, 1], repeat=9)]) = 512=2^9.
– Tom Ron
Nov 23 '18 at 10:53
But 512 is not all the possible entries. You need to take into account which cell was filled first, second, ... last. That means a factor of 9!. Here is a representation of a valid game:[(0,8), (1,7), (0,6), (1,5), (0,4), (1,3), (0,2), (1,0), (0,0)]
(note that 0 and 1 alternating in filling cells). Can you find it in your list of 512? It's not there... Furthermore you did not address the fact that your code will yield incorrect entries.
– handras
Nov 23 '18 at 11:29
1
It looks like you may have misunderstood the question. 2^9 will give us all combinations of 9 zeros and ones. Tic-tac-toe games are a significantly smaller subset of that and the OP's game representation also includes a number in each cell's data for "when it [the move] was placed."
– גלעד ברקן
Nov 24 '18 at 16:02
This is not good, since it contains many invalid entries. For example the very first entry is
[(0, 0), (0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (0, 6), (0, 7), (0, 8)]
.– handras
Nov 23 '18 at 6:46
This is not good, since it contains many invalid entries. For example the very first entry is
[(0, 0), (0, 1), (0, 2), (0, 3), (0, 4), (0, 5), (0, 6), (0, 7), (0, 8)]
.– handras
Nov 23 '18 at 6:46
Also, it only includes games in wich the cells are filled sequentially. So it misses a lot of valid games in which e.g. the first player start in the middle.
– handras
Nov 23 '18 at 7:10
Also, it only includes games in wich the cells are filled sequentially. So it misses a lot of valid games in which e.g. the first player start in the middle.
– handras
Nov 23 '18 at 7:10
It contains all possible entries - len([zip(p, range(0, 9)) for p in itertools.product([0, 1], repeat=9)]) = 512=2^9.
– Tom Ron
Nov 23 '18 at 10:53
It contains all possible entries - len([zip(p, range(0, 9)) for p in itertools.product([0, 1], repeat=9)]) = 512=2^9.
– Tom Ron
Nov 23 '18 at 10:53
But 512 is not all the possible entries. You need to take into account which cell was filled first, second, ... last. That means a factor of 9!. Here is a representation of a valid game:
[(0,8), (1,7), (0,6), (1,5), (0,4), (1,3), (0,2), (1,0), (0,0)]
(note that 0 and 1 alternating in filling cells). Can you find it in your list of 512? It's not there... Furthermore you did not address the fact that your code will yield incorrect entries.– handras
Nov 23 '18 at 11:29
But 512 is not all the possible entries. You need to take into account which cell was filled first, second, ... last. That means a factor of 9!. Here is a representation of a valid game:
[(0,8), (1,7), (0,6), (1,5), (0,4), (1,3), (0,2), (1,0), (0,0)]
(note that 0 and 1 alternating in filling cells). Can you find it in your list of 512? It's not there... Furthermore you did not address the fact that your code will yield incorrect entries.– handras
Nov 23 '18 at 11:29
1
1
It looks like you may have misunderstood the question. 2^9 will give us all combinations of 9 zeros and ones. Tic-tac-toe games are a significantly smaller subset of that and the OP's game representation also includes a number in each cell's data for "when it [the move] was placed."
– גלעד ברקן
Nov 24 '18 at 16:02
It looks like you may have misunderstood the question. 2^9 will give us all combinations of 9 zeros and ones. Tic-tac-toe games are a significantly smaller subset of that and the OP's game representation also includes a number in each cell's data for "when it [the move] was placed."
– גלעד ברקן
Nov 24 '18 at 16:02
add a comment |
This is fun. Here's a method that plays the game, branching off each possible move (or at any rate, that's the intention).
Although my preference is usually to concat and return a pure result, to allow for an early exit and inspection, this recursion accumulates in a global variable.
import copy
import json
import time
s = set()
rs =
def f():
"""
111
111000
111000000
1001001
10010010
100100100
1010100
100010001
"""
wins = [0007,0070,0700,0111,0222,0444,0124,0421]
def winner(board):
# check xs or os individual board for a win
for i in wins:
if not (i ^ (board & i)):
return True
return False
# xs board, os board, move-number, game
def g(xs, os, move, result):
# allow for early exit
global rs
if (len(rs) > 20000):
return
# base case, win or draw
if winner(xs) or winner(os) or move == 9:
#print "{0:b}".format(xs)
#print "{0:b}".format(os)
#print (result, xs, os)
#print
# confirm we're not duplicating results
enc = json.dumps(result)
if enc in s:
print "Duplicate!"
print result
s.add(enc)
# accumulate results
rs.append((result, xs, os))
return
board = xs | os
for i in xrange(9):
# candidate move
m = 1 << i
# valid move
if not (m & board):
_result = copy.deepcopy(result)
# 'O' plays on odd numbered moves
if move & 1:
_result[i] = [1, move]
g(xs, os | m, move + 1, _result)
# 'X' plays on even numbered moves
else:
_result[i] = [0, move]
g(xs | m, os, move + 1, _result)
# start the recursion
g(0, 0, 0, [[-1, -1]] * 9)
start_time = time.time()
f()
print("--- %s seconds ---" % (time.time() - start_time))
Output:
"""
rs[2002]
=> ([[0, 0], [1, 1], [-1, -1],
[-1, -1], [1, 5], [0, 2],
[0, 4], [1, 3], [-1, -1]], 97, 146)
x--
---
---
xo-
---
---
xo-
--x
---
xo-
--x
-o-
xo-
--x
xo-
xo-
-ox
xo-
"""
add a comment |
This is fun. Here's a method that plays the game, branching off each possible move (or at any rate, that's the intention).
Although my preference is usually to concat and return a pure result, to allow for an early exit and inspection, this recursion accumulates in a global variable.
import copy
import json
import time
s = set()
rs =
def f():
"""
111
111000
111000000
1001001
10010010
100100100
1010100
100010001
"""
wins = [0007,0070,0700,0111,0222,0444,0124,0421]
def winner(board):
# check xs or os individual board for a win
for i in wins:
if not (i ^ (board & i)):
return True
return False
# xs board, os board, move-number, game
def g(xs, os, move, result):
# allow for early exit
global rs
if (len(rs) > 20000):
return
# base case, win or draw
if winner(xs) or winner(os) or move == 9:
#print "{0:b}".format(xs)
#print "{0:b}".format(os)
#print (result, xs, os)
#print
# confirm we're not duplicating results
enc = json.dumps(result)
if enc in s:
print "Duplicate!"
print result
s.add(enc)
# accumulate results
rs.append((result, xs, os))
return
board = xs | os
for i in xrange(9):
# candidate move
m = 1 << i
# valid move
if not (m & board):
_result = copy.deepcopy(result)
# 'O' plays on odd numbered moves
if move & 1:
_result[i] = [1, move]
g(xs, os | m, move + 1, _result)
# 'X' plays on even numbered moves
else:
_result[i] = [0, move]
g(xs | m, os, move + 1, _result)
# start the recursion
g(0, 0, 0, [[-1, -1]] * 9)
start_time = time.time()
f()
print("--- %s seconds ---" % (time.time() - start_time))
Output:
"""
rs[2002]
=> ([[0, 0], [1, 1], [-1, -1],
[-1, -1], [1, 5], [0, 2],
[0, 4], [1, 3], [-1, -1]], 97, 146)
x--
---
---
xo-
---
---
xo-
--x
---
xo-
--x
-o-
xo-
--x
xo-
xo-
-ox
xo-
"""
add a comment |
This is fun. Here's a method that plays the game, branching off each possible move (or at any rate, that's the intention).
Although my preference is usually to concat and return a pure result, to allow for an early exit and inspection, this recursion accumulates in a global variable.
import copy
import json
import time
s = set()
rs =
def f():
"""
111
111000
111000000
1001001
10010010
100100100
1010100
100010001
"""
wins = [0007,0070,0700,0111,0222,0444,0124,0421]
def winner(board):
# check xs or os individual board for a win
for i in wins:
if not (i ^ (board & i)):
return True
return False
# xs board, os board, move-number, game
def g(xs, os, move, result):
# allow for early exit
global rs
if (len(rs) > 20000):
return
# base case, win or draw
if winner(xs) or winner(os) or move == 9:
#print "{0:b}".format(xs)
#print "{0:b}".format(os)
#print (result, xs, os)
#print
# confirm we're not duplicating results
enc = json.dumps(result)
if enc in s:
print "Duplicate!"
print result
s.add(enc)
# accumulate results
rs.append((result, xs, os))
return
board = xs | os
for i in xrange(9):
# candidate move
m = 1 << i
# valid move
if not (m & board):
_result = copy.deepcopy(result)
# 'O' plays on odd numbered moves
if move & 1:
_result[i] = [1, move]
g(xs, os | m, move + 1, _result)
# 'X' plays on even numbered moves
else:
_result[i] = [0, move]
g(xs | m, os, move + 1, _result)
# start the recursion
g(0, 0, 0, [[-1, -1]] * 9)
start_time = time.time()
f()
print("--- %s seconds ---" % (time.time() - start_time))
Output:
"""
rs[2002]
=> ([[0, 0], [1, 1], [-1, -1],
[-1, -1], [1, 5], [0, 2],
[0, 4], [1, 3], [-1, -1]], 97, 146)
x--
---
---
xo-
---
---
xo-
--x
---
xo-
--x
-o-
xo-
--x
xo-
xo-
-ox
xo-
"""
This is fun. Here's a method that plays the game, branching off each possible move (or at any rate, that's the intention).
Although my preference is usually to concat and return a pure result, to allow for an early exit and inspection, this recursion accumulates in a global variable.
import copy
import json
import time
s = set()
rs =
def f():
"""
111
111000
111000000
1001001
10010010
100100100
1010100
100010001
"""
wins = [0007,0070,0700,0111,0222,0444,0124,0421]
def winner(board):
# check xs or os individual board for a win
for i in wins:
if not (i ^ (board & i)):
return True
return False
# xs board, os board, move-number, game
def g(xs, os, move, result):
# allow for early exit
global rs
if (len(rs) > 20000):
return
# base case, win or draw
if winner(xs) or winner(os) or move == 9:
#print "{0:b}".format(xs)
#print "{0:b}".format(os)
#print (result, xs, os)
#print
# confirm we're not duplicating results
enc = json.dumps(result)
if enc in s:
print "Duplicate!"
print result
s.add(enc)
# accumulate results
rs.append((result, xs, os))
return
board = xs | os
for i in xrange(9):
# candidate move
m = 1 << i
# valid move
if not (m & board):
_result = copy.deepcopy(result)
# 'O' plays on odd numbered moves
if move & 1:
_result[i] = [1, move]
g(xs, os | m, move + 1, _result)
# 'X' plays on even numbered moves
else:
_result[i] = [0, move]
g(xs | m, os, move + 1, _result)
# start the recursion
g(0, 0, 0, [[-1, -1]] * 9)
start_time = time.time()
f()
print("--- %s seconds ---" % (time.time() - start_time))
Output:
"""
rs[2002]
=> ([[0, 0], [1, 1], [-1, -1],
[-1, -1], [1, 5], [0, 2],
[0, 4], [1, 3], [-1, -1]], 97, 146)
x--
---
---
xo-
---
---
xo-
--x
---
xo-
--x
-o-
xo-
--x
xo-
xo-
-ox
xo-
"""
answered Nov 24 '18 at 12:55
גלעד ברקןגלעד ברקן
13.2k21543
13.2k21543
add a comment |
add a comment |
I recommend you watch this video on youtube. The professor teaches really nicely about recursion. I think it will benefit for you more, then getting a working code. He uses sudoku as an example, but both games are just 2D arrays.
If you watched the video, you will know how to modify this example to better suit your needs.
A pseudo code:
def place(who, when, where, table):
if (None, None) not in table: # the table is full
print(table)
return
if table[where] not (None, None): # this cell is already marked
return
table[where] (who, when) # can place our mark on the cell
# make the recursive calls
for i in range(9):
place(who.opponent, when+1, i, table)
for cell in range(9):
empty = [(None, None) for _ in range(9)]
place(firstplayer, 0, cell, empty)
add a comment |
I recommend you watch this video on youtube. The professor teaches really nicely about recursion. I think it will benefit for you more, then getting a working code. He uses sudoku as an example, but both games are just 2D arrays.
If you watched the video, you will know how to modify this example to better suit your needs.
A pseudo code:
def place(who, when, where, table):
if (None, None) not in table: # the table is full
print(table)
return
if table[where] not (None, None): # this cell is already marked
return
table[where] (who, when) # can place our mark on the cell
# make the recursive calls
for i in range(9):
place(who.opponent, when+1, i, table)
for cell in range(9):
empty = [(None, None) for _ in range(9)]
place(firstplayer, 0, cell, empty)
add a comment |
I recommend you watch this video on youtube. The professor teaches really nicely about recursion. I think it will benefit for you more, then getting a working code. He uses sudoku as an example, but both games are just 2D arrays.
If you watched the video, you will know how to modify this example to better suit your needs.
A pseudo code:
def place(who, when, where, table):
if (None, None) not in table: # the table is full
print(table)
return
if table[where] not (None, None): # this cell is already marked
return
table[where] (who, when) # can place our mark on the cell
# make the recursive calls
for i in range(9):
place(who.opponent, when+1, i, table)
for cell in range(9):
empty = [(None, None) for _ in range(9)]
place(firstplayer, 0, cell, empty)
I recommend you watch this video on youtube. The professor teaches really nicely about recursion. I think it will benefit for you more, then getting a working code. He uses sudoku as an example, but both games are just 2D arrays.
If you watched the video, you will know how to modify this example to better suit your needs.
A pseudo code:
def place(who, when, where, table):
if (None, None) not in table: # the table is full
print(table)
return
if table[where] not (None, None): # this cell is already marked
return
table[where] (who, when) # can place our mark on the cell
# make the recursive calls
for i in range(9):
place(who.opponent, when+1, i, table)
for cell in range(9):
empty = [(None, None) for _ in range(9)]
place(firstplayer, 0, cell, empty)
edited Nov 26 '18 at 21:32
answered Nov 22 '18 at 20:32
handrashandras
601118
601118
add a comment |
add a comment |
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