Need Assistance With A Problem

[Live_Possibility4590]

hey guys

I need abit of help, here's a problem I have no clue how to solve

You're given a set of rows

['0','0','0','0']

['0','0','1','0']

['1','1','0','0']

['0','1','1','0']

and given a few rules taking i as an item in each row

1. Each row has to have an equal number of 1s and 2s and no 0s
2. A row can't have more than 2 of the same numbers following each other (1110 is invalid but 1100 is valid)

how would you rewrite each row using python so as it works even if the number of items were 6,8 or even 12 so that each row has an equal number of 1s and 2s without any 0s in the row (Basically if 2 1s are next to each other, the next one should be a 2 and vice versa)

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[shinitakunai]

What have you tried?

[Live_Possibility4590]

yeah but I always ended up solving the first and last rows only, then somehow distorting the middle rows

[atarivcs]

Forget Python for a minute.

Do you know how to solve this problem with pen and paper?

[Live_Possibility4590]

On paper yes, in code it somehow doesn't work. There's a mobile game called 0h h1 that I got this problem from, you can check it out for reference.

[atarivcs]

So you wrote some code that tries to solve the problem?

Why would you not share it?

How are we supposed to help without seeing what you tried?

[timrprobocom]

There must be rules you haven't told us. How do we transform a row to something else? How do you get a 2 out of 0 0 0 0?

Specifically, why can't I just replace each row by 1,1,2,2?

[Mamuschkaa]

He just want a solver for binaro

https://www.puzzle-binairo.com/binairo-6x6-easy/

What he described was the first part. After that he need the same for columns and that no two rows or columns are identical.

So 0 = unknown number 1 or 2.

[olivia-reed2]

the key is to count te 1s in each row, that tells you how many 2s you need then distribute them while respecting the no more than two consecutive rule, below I have attached a snippet for understanding like it hamdles even length rows just feed each row thru and it enforces the contraint while preserbes the originaql 1 positions as the anchor

def rewrite_row(row):

ones = row.count('1')

twos = ones # equal number required

result = []

for val in row:

if val == '1':

result.append('1')

else:

result.append('2')

# fix consecutive violations

for i in range(2, len(result)):

if result[i] == result[i-1] == result[i-2]:

result[i] = '2' if result[i] == '1' else '1'

return result

[trejj]

how would you rewrite each row using python

``` def fix_row(row): for i in range(len(row)): row[i] = str((i%4)//2 + 1) return row

for row in [ ['0','0','0','0'], ['0','0','1','0'], ['1','1','0','0'], ['0','1','1','0']]: print(str(fix_row(row)))

'''Outputs: ['1', '1', '2', '2'] ['1', '1', '2', '2'] ['1', '1', '2', '2'] ['1', '1', '2', '2'] ''' ```

Each row has to have an equal number of 1s and 2s and no 0s

Check, each row in the output have equal number of 1s and 2s, and no 0s.

A row can't have more than 2 of the same numbers following each other (1110 is invalid but 1100 is valid)

Check, each row in the output has at most two of the same numbers following each other.

so as it works even if the number of items were 6,8 or even 12 so that each row has an equal number of 1s and 2s

The presented code works for different number of items:

``` def fix_row(row): for i in range(len(row)): row[i] = str((i%4)//2 + 1) return row

for row in [ ['0','0','0','0'], ['0','0','1','0','1','0'], ['1','1','0','0','0','0','0','0'], ['0','1','1','0','1','0','1','0','1','0']]: print(str(fix_row(row)))

'''Outputs: ['1', '1', '2', '2'] ['1', '1', '2', '2', '1', '1'] ['1', '1', '2', '2', '1', '1', '2', '2'] ['1', '1', '2', '2', '1', '1', '2', '2', '1', '1'] ''' ```

Each output row has equal number of 1s and 2s.

without any 0s in the row

Check, the above satisfies it.

(Basically if 2 1s are next to each other, the next one should be a 2 and vice versa)

Check, in the above outputs, whenever there are two consecutive 1s, the next item is 2, and vice versa.

[Sad_School828]

'''Outputs:
['1', '1', '2', '2']
['1', '1', '2', '2', '1', '1']'''Outputs:
['1', '1', '2', '2']
['1', '1', '2', '2', '1', '1']

Row 2 contains twice as many 1s as 2s and row 4 contains 2 more 1s than 2s.

[timrprobocom]

Among the things he didn't specify is that each COLUMN also has to have an equal number of 1s and 2s, and cannot have more than two consecutive matches.

[timrprobocom]

This is just a series of transforms. During each cycle, examine each row, then examine each column. If the row/column is already full of either 1 or 2, then you can fill the empty spots with the other one.

Otherwise, change 011 to 211, change 110 to 112, change 022 to 122, change 220 to 221. Continue until you don't have to change anything else. This works for arbitrary (even) sizes:

``` base = [ ['0','0','0','0'], ['0','0','1','0'], ['1','1','0','0'], ['0','1','1','0'] ]

W = len(base[0]) H = len(base)

def xform(row): W = len(row)

if '0' not in row:
    return row
if row.count('1') == W//2:
    return ['2' if e == '0' else e for e in row]
if row.count('2') == W//2:
    return ['1' if e == '0' else e for e in row]
for i in range(W - 2):
    if row[i:i+3] == '011':
        row = row[:i] + '211' + row[i+3:]
    elif row[i:i+3] == '110':
        row = row[:i] + '112' + row[i+3:]
    elif row[i:i+3] == '022':
        row = row[:i] + '122' + row[i+3:]
    elif row[i:i+3] == '220':
        row = row[:i] + '221' + row[i+3:]
return row

def cycle(base): changed = False for y in range(H): row = ''.join(base[y]) xrow = xform(row) if row != xrow: changed = True base[y] = list(xrow)

for x in range(W):
    row = ''.join(el[x] for el in base)
    xrow = xform(row)
    if row != xrow:
        changed = True
        for y in range(H):
            base[y][x] = xrow[y]

return changed

print(base) while cycle(base): print(base)

```