Qlstudio - 2 months ago 8x

Python Question

I tried to solve the following problem with Python:

But I got stuck at generating a single valid table. I was expecting the program to display a valid matrix, but in order for the program to continue and not print

`None`

`7`

My code so far:

`from pprint import pprint`

import sys

import random

sys.setrecursionlimit(10000)

def getPossiblesForSquare(sqx,sqy,matrix):

'''Gets the possible entries of matrix[sqy][sqx].

Assumes it equals 0.'''

assert matrix[sqy][sqx]==0

# get the row that it is on

rowon=matrix[sqy]

# columns are a little trickier

colon=[matrix[row][sqx] for row in range(5)]

# find the possibilities!

possibles=list(range(1,7))

for item in list(set(rowon+colon)): # remove duplicates

if not (item == 0) and (item in possibles):

del possibles[possibles.index(item)]

random.shuffle(possibles)

return possibles

def getPossiblesForMatrix(matrix):

'''Gets all the possible squares for a matrix.'''

possiblesdict={}

for y in range(6):

for x in range(6):

if matrix[y][x]==0:

possiblesdict[(y,x)]=getPossiblesForSquare(x,y,MATRIX)

return possiblesdict

def flattenList(matrix):

result=[]

for i in matrix:

if not isinstance(i,list):

result+=[i]

else:

result+=flattenList(i)

return result

def getPossibleMatrix(startMatrix, iteration=0, yon=1, prevZeroInd=None):

if 0 not in flattenList(startMatrix):

print('RESULT:\n\n')

return startMatrix

else:

# find&fill in the first blank one

ind=flattenList(startMatrix).index(0)

y=ind//6

x=ind%6

if (x,y)==prevZeroInd:

startMatrix[y][x]=7

else:

possibles=getPossiblesForSquare(x,y,startMatrix)

if len(possibles)==0:

startMatrix[y][x]=7

else:

startMatrix[y][x]=possibles[0]

if iteration <= 6:

return getPossibleMatrix(startMatrix, iteration+1, yon, (x,y)) # <<BUG

else:

if yon!=4:

return getPossibleMatrix(startMatrix, 0, yon+1, (x,y))

MATRIX=[[1,2,3,4,5,6],

[2,0,0,0,0,5],

[3,0,0,0,0,4],

[4,0,0,0,0,3],

[5,0,0,0,0,2],

[6,5,4,3,2,1]]

result=getPossibleMatrix(MATRIX)

pprint(result)

Answer

Essentially your script encounters problems here:

```
for item in list(set(rowon + colon)): # remove duplicates
if not (item == 0) and (item in possibles):
del possibles[possibles.index(item)]
```

- At the third iteration, for the third cell your if condition is evaluated as true for all possible values
`[1 to 6]`

(if you output the matrix you will see that the`set()`

you are creating contains all elements), so you always return zero, re-check the values, return zero.*ad infinitum*

If you're looking to brute-force a solution out of this, you might want to update the `sqx`

and `sqy`

to go to a different cell when possibles is empty.

Another additional small mistake I located was:

```
# you are checking values 1-5 and not 1-6!
possibles = list(range(1, 6)) # should be range(7) [exclusive range]
```

- Don't forget that range is exclusive, it doesn't include (excludes) the upper limit.

There exist of course, different ways to tackle this problem.

Read this for the general, alternate view of how to solve this. If you do not want to see a possible solution, skip the 'code' part.

The solution matrix (one of the possible ones) has this form (unless I am making a horrible mistake):

```
MATRIX = [[1, 2, 3, 4, 5, 6],
[2, 3, 6, 1, 4, 5],
[3, 1, 5, 2, 6, 4],
[4, 6, 2, 5, 1, 3],
[5, 4, 1, 6, 3, 2],
[6, 5, 4, 3, 2, 1]]
```

The Logic is as follows:

You must observe the symmetry present in the matrix. Specifically, every row and every column displays a 'flip and reverse' symmetry. For example, the first and last rows are connected by this equation :

```
row[0] = REV(flip(row[n]))
```

Similarly, all additional rows (or columns) have their corresponding counterpart:

```
row[1] = REV(flip(row[n-1]))
```

and so on.

So, for `n=6`

this essentially boils down to finding the `(n / 2) -1`

(because we already know the first and last row!) and afterwards flipping them (not the finger), reversing them and assigning them to their corresponding rows.

In order to find these values we can observe the matrix as a combination of smaller matrices:

These make the first two (unknown) rows of the matrix:

```
sub_matrix = [[1, 2, 3, 4, 5, 6],
[2, 0, 0, 0, 0, 5],
[3, 0, 0, 0, 0, 4],
[6, 5, 4, 3, 2, 1]]
```

and the other two rows can be made by finding the correct values for these two.

Observe the restrictions in hand:

In column `[1][1]`

and `[1][m-1]`

we cannot:

- place a
`2`

or a`5`

In columns `[1][2]`

and `[1][m-2]`

we cannot:

- place the previous values (
`[2, 5]`

) along with (`[3, 4]`

) so, we cannot have a value`in`

`[2,3,4,5]`

For the inner columns we're left with the set `set(1-6) - set(2-5) = [1, 6]`

and since we get a normal row and a ** single** inverted and flipped row for this, we can arbitrarily select a value and add it as a column value.

By using another list we can keep track of the values used and fill out the rest of the cells.

**Note**: I did not use `numpy`

for this. You **can** and **should** though. (Also, `Python 2.7`

)

Also, I did not use recursion for this (you can try to, by finding the same matrix for bigger values of `n`

[I believe it's a good fit for a recursive function].

First, in order to not type this all the time, you can create a `n x n`

matrix as follows: (This isn't much of a spoiler.)

```
# our matrix must be n x n with n even.
n = 6
# Create n x n matrix.
head = [i for i in xrange(1, n + 1)] # contains values from 1 to n.
zeros = [0 for i in xrange(1, n-1)] # zeros
tail = [i for i in xrange(n, 0, -1)] # contains values from n to 1.
# Add head and zeros until tail.
MATRIX = [([j] + zeros + [(n+1)-j]) if j != 1 else head for j in xrange(1, n)]
# Add tail
MATRIX.append(tail)
```

Then, create the smaller `(n/2 + 1) x n`

array:

```
# Split matrix and add last row.
sub_matrix = MATRIX[:(n / 2)] + [tail]
```

Afterwards, a small function called `sub = fill_rows(sub_matrix)`

comes in and takes care of business:

```
def fill_rows(mtrx):
# length of the sub array (= 4)
sub_n = len(mtrx)
# From 1 because 0 = head
# Until sub_n -1 because the sub_n - 1 is = tail (remember, range is exclusive)
for current_row in xrange(1, sub_n - 1):
print "Current row: " + str(current_row)
# -- it gets messy here --
# get values of inner columns and filter out the zeros (matrix[row][n / 2] == 0 evaluates to False)
col_vals_1 = [mtrx[row][n / 2] for row in xrange(0, sub_n) if mtrx[row][(n / 2)]]
col_vals_2 = [mtrx[row][(n / 2) - 1] for row in xrange(0, sub_n) if mtrx[row][(n / 2) - 1]]
col_vals = col_vals_1 + col_vals_2
# print "Column Values = " + str(col_vals)
row_vals = [mtrx[current_row][col] for col in xrange(0, n) if mtrx[current_row][col]]
# print "Row Values = " + str(row_vals)
# Find the possible values by getting the difference of the joined set of column + row values
# with the range from (1 - 6).
possible_values = list(set(xrange(1, n + 1)) - set(row_vals + col_vals))
print "Possible acceptable values: " + str(possible_values)
# Add values to matrix (pop to remove them)
# After removing add to the list of row_values in order to check for the other columns.
mtrx[current_row][(n-1)/2] = possible_values.pop()
row_vals.append(mtrx[current_row][(n - 1) / 2])
mtrx[current_row][(n-1)/2 + 1] = possible_values.pop()
row_vals.append(mtrx[current_row][(n-1) / 2 + 1])
# New possible values for remaining columns of the current row.
possible_values = list(set(xrange(1, n + 1)) - set(row_vals))
print "Possible acceptable values: " + str(possible_values)
# Add values to the cells.
mtrx[current_row][(n - 2)] = possible_values.pop()
mtrx[current_row][1] = possible_values.pop()
# return updated sub-matrix.
return mtrx
```

The only thing left to do now is take those two rows, flip them, reverse them and add the head and tail to them:

```
print '=' * 30 + " Out " + "=" * 30
# Remove first and last rows.
del sub[0]
sub.pop()
# reverse values in lists
temp_sub = [l[::-1] for l in sub]
# reverse lists in matrix.
temp_sub.reverse()
# Add them and Print.
pprint([head] + sub + temp_sub + [tail])
```

This outputs what, I hope, is the right matrix:

```
============================== Out ==============================
[[1, 2, 3, 4, 5, 6],
[2, 3, 6, 1, 4, 5],
[3, 1, 5, 2, 6, 4],
[4, 6, 2, 5, 1, 3],
[5, 4, 1, 6, 3, 2],
[6, 5, 4, 3, 2, 1]]
```

By using this way of solving it the answer to the problem in hand becomes more easy. Viewing the matrix as a combination of these sub-matrices you can tell how many of these combinations might be possible.

As a closing note, it would be a good work-out to modify it a bit in order to allow it to find this array for an arbitrary (** but even**) number of

`n`

.Source (Stackoverflow)

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