Mixing
Cost Function-Using the counter function for nonets of a Sudoku grid
I am currently trying to build a cost function for Sudoku. I have the following code for my random 9x9 grid assignment:
m = np.zeros((9,9)) #9x9 grid of all zeros
vals = np.arange(1,10) #set of values from 1 to 9 aranged randomly
for i in range(0,9):
m[i,:] = np.random.choice(vals,size=9,replace=False) #randomly choses a set of 9 values for the row
print(m.astype(int)) #prints as integers
Subsequently I have created a score for the number of duplications in each column for this grid:
m = m
i = 1
for i in range(0,9): #produces the sum of the columns score values collectively
print(sum([r - 1 for r in Counter((m[:,i])).values()]))
I am wondering whether there is a way to produce the scores for each nonet? (each 3x3 grouping of the 9x9 Sudoku grid)
I feel like I will have to take a subsection for each 3x3 grid and then produce the duplications using the Counter function in the same way as above, but I am unsure as to how to do this.
python sudoku
asked Nov 22 '18 at 10:56
Hew.rHew.r
84
add a comment |
I am currently trying to build a cost function for Sudoku. I have the following code for my random 9x9 grid assignment:
m = np.zeros((9,9)) #9x9 grid of all zeros
vals = np.arange(1,10) #set of values from 1 to 9 aranged randomly
for i in range(0,9):
m[i,:] = np.random.choice(vals,size=9,replace=False) #randomly choses a set of 9 values for the row
print(m.astype(int)) #prints as integers
Subsequently I have created a score for the number of duplications in each column for this grid:
m = m
i = 1
for i in range(0,9): #produces the sum of the columns score values collectively
print(sum([r - 1 for r in Counter((m[:,i])).values()]))
I am wondering whether there is a way to produce the scores for each nonet? (each 3x3 grouping of the 9x9 Sudoku grid)
I feel like I will have to take a subsection for each 3x3 grid and then produce the duplications using the Counter function in the same way as above, but I am unsure as to how to do this.
python sudoku
asked Nov 22 '18 at 10:56
Hew.rHew.r
84
add a comment |
I am currently trying to build a cost function for Sudoku. I have the following code for my random 9x9 grid assignment:
m = np.zeros((9,9)) #9x9 grid of all zeros
vals = np.arange(1,10) #set of values from 1 to 9 aranged randomly
for i in range(0,9):
m[i,:] = np.random.choice(vals,size=9,replace=False) #randomly choses a set of 9 values for the row
print(m.astype(int)) #prints as integers
Subsequently I have created a score for the number of duplications in each column for this grid:
m = m
i = 1
for i in range(0,9): #produces the sum of the columns score values collectively
print(sum([r - 1 for r in Counter((m[:,i])).values()]))
I am wondering whether there is a way to produce the scores for each nonet? (each 3x3 grouping of the 9x9 Sudoku grid)
I feel like I will have to take a subsection for each 3x3 grid and then produce the duplications using the Counter function in the same way as above, but I am unsure as to how to do this.
python sudoku
asked Nov 22 '18 at 10:56
Hew.rHew.r
84
I am currently trying to build a cost function for Sudoku. I have the following code for my random 9x9 grid assignment:
m = np.zeros((9,9)) #9x9 grid of all zeros
vals = np.arange(1,10) #set of values from 1 to 9 aranged randomly
for i in range(0,9):
m[i,:] = np.random.choice(vals,size=9,replace=False) #randomly choses a set of 9 values for the row
print(m.astype(int)) #prints as integers
Subsequently I have created a score for the number of duplications in each column for this grid:
m = m
i = 1
for i in range(0,9): #produces the sum of the columns score values collectively
print(sum([r - 1 for r in Counter((m[:,i])).values()]))
I am wondering whether there is a way to produce the scores for each nonet? (each 3x3 grouping of the 9x9 Sudoku grid)
I feel like I will have to take a subsection for each 3x3 grid and then produce the duplications using the Counter function in the same way as above, but I am unsure as to how to do this.
python sudoku
python sudoku
asked Nov 22 '18 at 10:56
Hew.rHew.r
84
asked Nov 22 '18 at 10:56
Hew.rHew.r
84
asked Nov 22 '18 at 10:56
Hew.rHew.r
84
asked Nov 22 '18 at 10:56
Hew.rHew.r
84
asked Nov 22 '18 at 10:56
Hew.rHew.r
84
84
add a comment |
add a comment |
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