Creating a randomized set of numbers in an array without the same number repeating three times in a row
Hi so i have to basically create a game like candy crush saga and i created a 2d array to start and i was able to create the game board with the corresponding pictures based on the indices of my array. However i have to make sure that to start off i cant have 3 of the same pictures or candy or whatever in a row. To create the array in the first place i used:
board = [[random.randint(1,6) for i in range(7)] for j in range(9)]
to create a 9x7 board and basically each number from 1-6 corresponds to a specific picture but i cant figure out how to iterate through and check if there are three of the same numbers in a row. Any ideas? Thanks
python-3.x
edited Nov 25 '18 at 4:24
Mateen Ulhaq
11.5k114794
asked Nov 25 '18 at 4:14
ssalssal
153
add a comment |
Hi so i have to basically create a game like candy crush saga and i created a 2d array to start and i was able to create the game board with the corresponding pictures based on the indices of my array. However i have to make sure that to start off i cant have 3 of the same pictures or candy or whatever in a row. To create the array in the first place i used:
board = [[random.randint(1,6) for i in range(7)] for j in range(9)]
to create a 9x7 board and basically each number from 1-6 corresponds to a specific picture but i cant figure out how to iterate through and check if there are three of the same numbers in a row. Any ideas? Thanks
python-3.x
edited Nov 25 '18 at 4:24
Mateen Ulhaq
11.5k114794
asked Nov 25 '18 at 4:14
ssalssal
153
When you say "in a row" do you literally mean in a row? Or do you mean rows/columns/possibly diagonals?
– ggorlen
Nov 25 '18 at 4:20
sorry for not clarifying, i mean like in a row either vertically or horizontally no diagonals
– ssal
Nov 25 '18 at 4:21
row_counts = [collections.Counter(row) for row in board]will give you number of occurences for each item.
– Mateen Ulhaq
Nov 25 '18 at 4:27
This will tell you which ones have three or more:[[k for k, v in row.items() if v >= 3] for row in row_counts]
– Mateen Ulhaq
Nov 25 '18 at 4:28
3 in a row?? like 3 beside each other? or 3 in the actual row cause i need to change if there are 3 beside each other (in a row) either horizontally or vertically
– ssal
Nov 25 '18 at 4:30
add a comment |
Hi so i have to basically create a game like candy crush saga and i created a 2d array to start and i was able to create the game board with the corresponding pictures based on the indices of my array. However i have to make sure that to start off i cant have 3 of the same pictures or candy or whatever in a row. To create the array in the first place i used:
board = [[random.randint(1,6) for i in range(7)] for j in range(9)]
to create a 9x7 board and basically each number from 1-6 corresponds to a specific picture but i cant figure out how to iterate through and check if there are three of the same numbers in a row. Any ideas? Thanks
python-3.x
edited Nov 25 '18 at 4:24
Mateen Ulhaq
11.5k114794
asked Nov 25 '18 at 4:14
ssalssal
153
Hi so i have to basically create a game like candy crush saga and i created a 2d array to start and i was able to create the game board with the corresponding pictures based on the indices of my array. However i have to make sure that to start off i cant have 3 of the same pictures or candy or whatever in a row. To create the array in the first place i used:
board = [[random.randint(1,6) for i in range(7)] for j in range(9)]
to create a 9x7 board and basically each number from 1-6 corresponds to a specific picture but i cant figure out how to iterate through and check if there are three of the same numbers in a row. Any ideas? Thanks
python-3.x
python-3.x
edited Nov 25 '18 at 4:24
Mateen Ulhaq
11.5k114794
asked Nov 25 '18 at 4:14
ssalssal
153
edited Nov 25 '18 at 4:24
Mateen Ulhaq
11.5k114794
asked Nov 25 '18 at 4:14
ssalssal
153
edited Nov 25 '18 at 4:24
Mateen Ulhaq
11.5k114794
edited Nov 25 '18 at 4:24
Mateen Ulhaq
11.5k114794
edited Nov 25 '18 at 4:24
Mateen Ulhaq
11.5k114794
11.5k114794
asked Nov 25 '18 at 4:14
ssalssal
153
asked Nov 25 '18 at 4:14
ssalssal
153
asked Nov 25 '18 at 4:14
ssalssal
153
153
When you say "in a row" do you literally mean in a row? Or do you mean rows/columns/possibly diagonals?
– ggorlen
Nov 25 '18 at 4:20
sorry for not clarifying, i mean like in a row either vertically or horizontally no diagonals
– ssal
Nov 25 '18 at 4:21
row_counts = [collections.Counter(row) for row in board]will give you number of occurences for each item.
– Mateen Ulhaq
Nov 25 '18 at 4:27
This will tell you which ones have three or more:[[k for k, v in row.items() if v >= 3] for row in row_counts]
– Mateen Ulhaq
Nov 25 '18 at 4:28
3 in a row?? like 3 beside each other? or 3 in the actual row cause i need to change if there are 3 beside each other (in a row) either horizontally or vertically
– ssal
Nov 25 '18 at 4:30
add a comment |
When you say "in a row" do you literally mean in a row? Or do you mean rows/columns/possibly diagonals?
– ggorlen
Nov 25 '18 at 4:20
sorry for not clarifying, i mean like in a row either vertically or horizontally no diagonals
– ssal
Nov 25 '18 at 4:21
row_counts = [collections.Counter(row) for row in board]will give you number of occurences for each item.
– Mateen Ulhaq
Nov 25 '18 at 4:27
This will tell you which ones have three or more:[[k for k, v in row.items() if v >= 3] for row in row_counts]
– Mateen Ulhaq
Nov 25 '18 at 4:28
3 in a row?? like 3 beside each other? or 3 in the actual row cause i need to change if there are 3 beside each other (in a row) either horizontally or vertically
– ssal
Nov 25 '18 at 4:30
When you say "in a row" do you literally mean in a row? Or do you mean rows/columns/possibly diagonals?
– ggorlen
Nov 25 '18 at 4:20
When you say "in a row" do you literally mean in a row? Or do you mean rows/columns/possibly diagonals?
– ggorlen
Nov 25 '18 at 4:20
sorry for not clarifying, i mean like in a row either vertically or horizontally no diagonals
– ssal
Nov 25 '18 at 4:21
sorry for not clarifying, i mean like in a row either vertically or horizontally no diagonals
– ssal
Nov 25 '18 at 4:21
row_counts = [collections.Counter(row) for row in board] will give you number of occurences for each item.– Mateen Ulhaq
Nov 25 '18 at 4:27
row_counts = [collections.Counter(row) for row in board] will give you number of occurences for each item.– Mateen Ulhaq
Nov 25 '18 at 4:27
This will tell you which ones have three or more:
[[k for k, v in row.items() if v >= 3] for row in row_counts]– Mateen Ulhaq
Nov 25 '18 at 4:28
This will tell you which ones have three or more:
[[k for k, v in row.items() if v >= 3] for row in row_counts]– Mateen Ulhaq
Nov 25 '18 at 4:28
3 in a row?? like 3 beside each other? or 3 in the actual row cause i need to change if there are 3 beside each other (in a row) either horizontally or vertically
– ssal
Nov 25 '18 at 4:30
3 in a row?? like 3 beside each other? or 3 in the actual row cause i need to change if there are 3 beside each other (in a row) either horizontally or vertically
– ssal
Nov 25 '18 at 4:30
add a comment |
1 Answer
1
active
oldest
votes
This solution will not win a beauty contest, but it generates the board in one pass, which should offer significant performance advantages over generating entire boards and running counts on them until you manage to get a lucky board (there's about a 10% chance a random board will be valid).
def generate_board(rows=9, cols=7):
board =
for i in range(rows):
board.append([0] * cols)
for j in range(cols):
while not board[i][j]:
c = random.randint(1, 6)
if (i < 2 or board[i-2][j] != c or board[i-1][j] != c) and
(j < 2 or board[i][j-2] != c or board[i][j-1] != c):
board[i][j] = c
return board
A few sample output boards:
[6, 1, 3, 3, 2, 5, 5] [3, 5, 6, 5, 2, 1, 1] [2, 2, 1, 5, 4, 6, 6] [4, 1, 1, 5, 2, 6, 2]
[4, 2, 5, 5, 3, 5, 3] [4, 2, 2, 1, 4, 3, 1] [5, 3, 6, 2, 3, 4, 5] [1, 3, 6, 1, 2, 4, 2]
[5, 2, 6, 5, 1, 4, 6] [2, 4, 1, 4, 3, 3, 6] [4, 1, 4, 1, 4, 1, 3] [4, 3, 4, 3, 1, 6, 3]
[5, 6, 2, 6, 3, 6, 6] [4, 5, 5, 3, 4, 5, 5] [5, 5, 3, 3, 1, 6, 3] [2, 4, 1, 5, 5, 1, 5]
[2, 1, 4, 2, 4, 3, 5] [4, 6, 5, 1, 5, 3, 4] [1, 3, 2, 3, 6, 2, 4] [4, 6, 4, 3, 6, 4, 2]
[1, 3, 4, 1, 6, 3, 1] [6, 2, 1, 5, 1, 2, 1] [6, 3, 3, 2, 4, 6, 6] [2, 3, 3, 6, 2, 1, 5]
[2, 1, 1, 3, 2, 5, 4] [6, 5, 5, 4, 5, 3, 3] [2, 4, 2, 2, 3, 6, 3] [2, 1, 3, 6, 1, 2, 6]
[3, 1, 2, 3, 3, 5, 1] [5, 3, 6, 1, 5, 4, 1] [4, 2, 1, 6, 5, 3, 6] [3, 6, 5, 2, 1, 5, 5]
[2, 6, 3, 1, 5, 6, 4] [6, 6, 4, 2, 3, 4, 2] [5, 4, 6, 6, 4, 2, 6] [5, 3, 1, 6, 3, 6, 2]
Try it!
answered Nov 25 '18 at 4:41
ggorlenggorlen
7,1883826
add a comment |
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1 Answer
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1 Answer
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oldest
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oldest
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oldest
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This solution will not win a beauty contest, but it generates the board in one pass, which should offer significant performance advantages over generating entire boards and running counts on them until you manage to get a lucky board (there's about a 10% chance a random board will be valid).
def generate_board(rows=9, cols=7):
board =
for i in range(rows):
board.append([0] * cols)
for j in range(cols):
while not board[i][j]:
c = random.randint(1, 6)
if (i < 2 or board[i-2][j] != c or board[i-1][j] != c) and
(j < 2 or board[i][j-2] != c or board[i][j-1] != c):
board[i][j] = c
return board
A few sample output boards:
[6, 1, 3, 3, 2, 5, 5] [3, 5, 6, 5, 2, 1, 1] [2, 2, 1, 5, 4, 6, 6] [4, 1, 1, 5, 2, 6, 2]
[4, 2, 5, 5, 3, 5, 3] [4, 2, 2, 1, 4, 3, 1] [5, 3, 6, 2, 3, 4, 5] [1, 3, 6, 1, 2, 4, 2]
[5, 2, 6, 5, 1, 4, 6] [2, 4, 1, 4, 3, 3, 6] [4, 1, 4, 1, 4, 1, 3] [4, 3, 4, 3, 1, 6, 3]
[5, 6, 2, 6, 3, 6, 6] [4, 5, 5, 3, 4, 5, 5] [5, 5, 3, 3, 1, 6, 3] [2, 4, 1, 5, 5, 1, 5]
[2, 1, 4, 2, 4, 3, 5] [4, 6, 5, 1, 5, 3, 4] [1, 3, 2, 3, 6, 2, 4] [4, 6, 4, 3, 6, 4, 2]
[1, 3, 4, 1, 6, 3, 1] [6, 2, 1, 5, 1, 2, 1] [6, 3, 3, 2, 4, 6, 6] [2, 3, 3, 6, 2, 1, 5]
[2, 1, 1, 3, 2, 5, 4] [6, 5, 5, 4, 5, 3, 3] [2, 4, 2, 2, 3, 6, 3] [2, 1, 3, 6, 1, 2, 6]
[3, 1, 2, 3, 3, 5, 1] [5, 3, 6, 1, 5, 4, 1] [4, 2, 1, 6, 5, 3, 6] [3, 6, 5, 2, 1, 5, 5]
[2, 6, 3, 1, 5, 6, 4] [6, 6, 4, 2, 3, 4, 2] [5, 4, 6, 6, 4, 2, 6] [5, 3, 1, 6, 3, 6, 2]
Try it!
answered Nov 25 '18 at 4:41
ggorlenggorlen
7,1883826
add a comment |
This solution will not win a beauty contest, but it generates the board in one pass, which should offer significant performance advantages over generating entire boards and running counts on them until you manage to get a lucky board (there's about a 10% chance a random board will be valid).
def generate_board(rows=9, cols=7):
board =
for i in range(rows):
board.append([0] * cols)
for j in range(cols):
while not board[i][j]:
c = random.randint(1, 6)
if (i < 2 or board[i-2][j] != c or board[i-1][j] != c) and
(j < 2 or board[i][j-2] != c or board[i][j-1] != c):
board[i][j] = c
return board
A few sample output boards:
[6, 1, 3, 3, 2, 5, 5] [3, 5, 6, 5, 2, 1, 1] [2, 2, 1, 5, 4, 6, 6] [4, 1, 1, 5, 2, 6, 2]
[4, 2, 5, 5, 3, 5, 3] [4, 2, 2, 1, 4, 3, 1] [5, 3, 6, 2, 3, 4, 5] [1, 3, 6, 1, 2, 4, 2]
[5, 2, 6, 5, 1, 4, 6] [2, 4, 1, 4, 3, 3, 6] [4, 1, 4, 1, 4, 1, 3] [4, 3, 4, 3, 1, 6, 3]
[5, 6, 2, 6, 3, 6, 6] [4, 5, 5, 3, 4, 5, 5] [5, 5, 3, 3, 1, 6, 3] [2, 4, 1, 5, 5, 1, 5]
[2, 1, 4, 2, 4, 3, 5] [4, 6, 5, 1, 5, 3, 4] [1, 3, 2, 3, 6, 2, 4] [4, 6, 4, 3, 6, 4, 2]
[1, 3, 4, 1, 6, 3, 1] [6, 2, 1, 5, 1, 2, 1] [6, 3, 3, 2, 4, 6, 6] [2, 3, 3, 6, 2, 1, 5]
[2, 1, 1, 3, 2, 5, 4] [6, 5, 5, 4, 5, 3, 3] [2, 4, 2, 2, 3, 6, 3] [2, 1, 3, 6, 1, 2, 6]
[3, 1, 2, 3, 3, 5, 1] [5, 3, 6, 1, 5, 4, 1] [4, 2, 1, 6, 5, 3, 6] [3, 6, 5, 2, 1, 5, 5]
[2, 6, 3, 1, 5, 6, 4] [6, 6, 4, 2, 3, 4, 2] [5, 4, 6, 6, 4, 2, 6] [5, 3, 1, 6, 3, 6, 2]
Try it!
answered Nov 25 '18 at 4:41
ggorlenggorlen
7,1883826
add a comment |
This solution will not win a beauty contest, but it generates the board in one pass, which should offer significant performance advantages over generating entire boards and running counts on them until you manage to get a lucky board (there's about a 10% chance a random board will be valid).
def generate_board(rows=9, cols=7):
board =
for i in range(rows):
board.append([0] * cols)
for j in range(cols):
while not board[i][j]:
c = random.randint(1, 6)
if (i < 2 or board[i-2][j] != c or board[i-1][j] != c) and
(j < 2 or board[i][j-2] != c or board[i][j-1] != c):
board[i][j] = c
return board
A few sample output boards:
[6, 1, 3, 3, 2, 5, 5] [3, 5, 6, 5, 2, 1, 1] [2, 2, 1, 5, 4, 6, 6] [4, 1, 1, 5, 2, 6, 2]
[4, 2, 5, 5, 3, 5, 3] [4, 2, 2, 1, 4, 3, 1] [5, 3, 6, 2, 3, 4, 5] [1, 3, 6, 1, 2, 4, 2]
[5, 2, 6, 5, 1, 4, 6] [2, 4, 1, 4, 3, 3, 6] [4, 1, 4, 1, 4, 1, 3] [4, 3, 4, 3, 1, 6, 3]
[5, 6, 2, 6, 3, 6, 6] [4, 5, 5, 3, 4, 5, 5] [5, 5, 3, 3, 1, 6, 3] [2, 4, 1, 5, 5, 1, 5]
[2, 1, 4, 2, 4, 3, 5] [4, 6, 5, 1, 5, 3, 4] [1, 3, 2, 3, 6, 2, 4] [4, 6, 4, 3, 6, 4, 2]
[1, 3, 4, 1, 6, 3, 1] [6, 2, 1, 5, 1, 2, 1] [6, 3, 3, 2, 4, 6, 6] [2, 3, 3, 6, 2, 1, 5]
[2, 1, 1, 3, 2, 5, 4] [6, 5, 5, 4, 5, 3, 3] [2, 4, 2, 2, 3, 6, 3] [2, 1, 3, 6, 1, 2, 6]
[3, 1, 2, 3, 3, 5, 1] [5, 3, 6, 1, 5, 4, 1] [4, 2, 1, 6, 5, 3, 6] [3, 6, 5, 2, 1, 5, 5]
[2, 6, 3, 1, 5, 6, 4] [6, 6, 4, 2, 3, 4, 2] [5, 4, 6, 6, 4, 2, 6] [5, 3, 1, 6, 3, 6, 2]
Try it!
answered Nov 25 '18 at 4:41
ggorlenggorlen
7,1883826
This solution will not win a beauty contest, but it generates the board in one pass, which should offer significant performance advantages over generating entire boards and running counts on them until you manage to get a lucky board (there's about a 10% chance a random board will be valid).
def generate_board(rows=9, cols=7):
board =
for i in range(rows):
board.append([0] * cols)
for j in range(cols):
while not board[i][j]:
c = random.randint(1, 6)
if (i < 2 or board[i-2][j] != c or board[i-1][j] != c) and
(j < 2 or board[i][j-2] != c or board[i][j-1] != c):
board[i][j] = c
return board
A few sample output boards:
[6, 1, 3, 3, 2, 5, 5] [3, 5, 6, 5, 2, 1, 1] [2, 2, 1, 5, 4, 6, 6] [4, 1, 1, 5, 2, 6, 2]
[4, 2, 5, 5, 3, 5, 3] [4, 2, 2, 1, 4, 3, 1] [5, 3, 6, 2, 3, 4, 5] [1, 3, 6, 1, 2, 4, 2]
[5, 2, 6, 5, 1, 4, 6] [2, 4, 1, 4, 3, 3, 6] [4, 1, 4, 1, 4, 1, 3] [4, 3, 4, 3, 1, 6, 3]
[5, 6, 2, 6, 3, 6, 6] [4, 5, 5, 3, 4, 5, 5] [5, 5, 3, 3, 1, 6, 3] [2, 4, 1, 5, 5, 1, 5]
[2, 1, 4, 2, 4, 3, 5] [4, 6, 5, 1, 5, 3, 4] [1, 3, 2, 3, 6, 2, 4] [4, 6, 4, 3, 6, 4, 2]
[1, 3, 4, 1, 6, 3, 1] [6, 2, 1, 5, 1, 2, 1] [6, 3, 3, 2, 4, 6, 6] [2, 3, 3, 6, 2, 1, 5]
[2, 1, 1, 3, 2, 5, 4] [6, 5, 5, 4, 5, 3, 3] [2, 4, 2, 2, 3, 6, 3] [2, 1, 3, 6, 1, 2, 6]
[3, 1, 2, 3, 3, 5, 1] [5, 3, 6, 1, 5, 4, 1] [4, 2, 1, 6, 5, 3, 6] [3, 6, 5, 2, 1, 5, 5]
[2, 6, 3, 1, 5, 6, 4] [6, 6, 4, 2, 3, 4, 2] [5, 4, 6, 6, 4, 2, 6] [5, 3, 1, 6, 3, 6, 2]
Try it!
answered Nov 25 '18 at 4:41
ggorlenggorlen
7,1883826
edited Nov 25 '18 at 5:34
answered Nov 25 '18 at 4:41
ggorlenggorlen
7,1883826
answered Nov 25 '18 at 4:41
ggorlenggorlen
7,1883826
answered Nov 25 '18 at 4:41
ggorlenggorlen
7,1883826
7,1883826
add a comment |
add a comment |
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When you say "in a row" do you literally mean in a row? Or do you mean rows/columns/possibly diagonals?
– ggorlen
Nov 25 '18 at 4:20
sorry for not clarifying, i mean like in a row either vertically or horizontally no diagonals
– ssal
Nov 25 '18 at 4:21
row_counts = [collections.Counter(row) for row in board]will give you number of occurences for each item.– Mateen Ulhaq
Nov 25 '18 at 4:27
This will tell you which ones have three or more:
[[k for k, v in row.items() if v >= 3] for row in row_counts]– Mateen Ulhaq
Nov 25 '18 at 4:28
3 in a row?? like 3 beside each other? or 3 in the actual row cause i need to change if there are 3 beside each other (in a row) either horizontally or vertically
– ssal
Nov 25 '18 at 4:30