Another question: I am writing a sudoku solving program. The 'solving' part of is just multiple iterations. It will take random numbers and keep switching it all around until a set of logic statements has been met (ie; all numbers in a row are not equal to each other) ... that's where my question comes in.
Cellboard = my list for storing each row/column's data.
Rather than writing
cellboard[0] is not* (cellboard[1] and cellboard[2] and cellboard[3] and cellboard[4] ... cellboard[8]) cellboard[1] is not (cellboard[0] and cellboard[2] and cellboard[3] and cellboard[4] ... cellboard[8]) etc...
* should this be != ?
the above so that all the data in one row is not equal to each other, is there something I can write to make it simpler? For example, (cellboard[0] is not cellboard[1] is not ... cellboard[8]) only worked for the numbers to the left and right of the cell - is there anyway I can expand this to cover all numbers in a set range?
On 3 Nov 2005 17:01:08 -0800, ale.of.gin...@gmail.com
<ale.of.gin...@gmail.com> wrote: > the above so that all the data in one row is not equal to each other, > is there something I can write to make it simpler? For example, > (cellboard[0] is not cellboard[1] is not ... cellboard[8]) only worked > for the numbers to the left and right of the cell - is there anyway I > can expand this to cover all numbers in a set range?
Python has an operator call 'in', which will allow you to do what you're after.
"1 in (1,2,3)" will be true "4 in (1,2,3)" will be false "not 1 in (1,2,3)" will be false
So you'd be after something along the lines of: not cellboard[0] in (cellboard[1], ...., celboard[8]).
This seems quite tedious to write, maybe you should consider something along the lines of using slicing:
not celboard[0] in cellboard[1:8]
I hope i have given you enough tools to do what you're trying to do.
> cellboard[0] is not* (cellboard[1] and cellboard[2] and cellboard[3] > and cellboard[4] ... cellboard[8]) > cellboard[1] is not (cellboard[0] and cellboard[2] and cellboard[3] and > cellboard[4] ... cellboard[8])
Urgh... the fastest way to check that a list of N numbers has no duplicates is: if len(set(thelist)) == len(thelist): print 'no duplicates'
But if your purpose is to generate N random samples out of a population of M, look at function random.sample (in module random in the Pythons standard library) and you'll do even better!-)
not cellboard[0] in cellboard[1:8] (I knew about ranges/slicing using a colon, can't believe I didn't think of that!)
line, will I have to write that out for each number?
So the line:
not cellboard in ((cellboard[1:8]) and (cellboard[9] and cellboard[18] and cellboard[27] and cellboard[36] and cellboard[45] and cellboard[54] and cellboard[63] and cellboard[72]) and (cellboard[1:2] and cellboard[9:11] and cellboard[18:20]))
will cover all the logic requirements for the number in cell 0 (well, row 1, column 1).
But will I have to copy + paste + edit that for all 81 cells? That isn't complicated, just tedious - thanks though.
>Another question: I am writing a sudoku solving program. The >'solving' part of is just multiple iterations. It will take random >numbers and keep switching it all around until a set of logic >statements has been met (ie; all numbers in a row are not equal to each >other) ... that's where my question comes in.
>Cellboard = my list for storing each row/column's data.
>Rather than writing
>cellboard[0] is not* (cellboard[1] and cellboard[2] and cellboard[3] >and cellboard[4] ... cellboard[8]) >cellboard[1] is not (cellboard[0] and cellboard[2] and cellboard[3] and >cellboard[4] ... cellboard[8]) >etc...
>* should this be != ?
>the above so that all the data in one row is not equal to each other, >is there something I can write to make it simpler? For example, >(cellboard[0] is not cellboard[1] is not ... cellboard[8]) only worked >for the numbers to the left and right of the cell - is there anyway I >can expand this to cover all numbers in a set range?
UIAM if you have a list of items that are comparable and hashable, like integers, you can make a set of the list, and duplicates will be eliminated in the set. Therefore if the resulting set has the same number of members as the list it was made from, you can conclude that the list contains no duplicates. E.g.,
So the test would be >>> len(set(cellboard))==len(cellboard) False And after repairing the list to uniqueness of elements: >>> cellboard[2] = 2 >>> len(set(cellboard))==len(cellboard) True
How do I 'define' set? Is there something to include (like import random)?
while (choice == 3) and len(set(cellboard[0:8]))==len(cellboard[0:8]): # DEFINE TWO RANDOM VARIABLES (ONE FOR ARRAY, ONE FOR NUMBER VALUE) solvingrandom = random.randint(1,9) cellboardrandom = random.randint(0,8) set(cellboard[0:8])
# CHECK TO MAKE SURE THE RANDOMLY ASSIGNED CELL DOES NOT HAVE A VALUE if (cellboard[cellboardrandom] is not ('1' or '2' or '3' or '4' or '5' or '6' or '7' or '8' or '9')): cellboard[cellboardrandom] = solvingrandom
The above is my code (right now it will only work for the first row's numbers). Anything else I need to add?
> while (choice == 3) and len(set(cellboard[0:8]))==len(cellboard[0:8]): > # DEFINE TWO RANDOM VARIABLES (ONE FOR ARRAY, ONE FOR NUMBER > VALUE) > solvingrandom = random.randint(1,9) > cellboardrandom = random.randint(0,8) > set(cellboard[0:8])
> # CHECK TO MAKE SURE THE RANDOMLY ASSIGNED CELL DOES NOT HAVE A > VALUE > if (cellboard[cellboardrandom] is not ('1' or '2' or '3' or '4' > or '5' or '6' or '7' or '8' or '9')): > cellboard[cellboardrandom] = solvingrandom
> The above is my code (right now it will only work for the first row's > numbers). Anything else I need to add?
Simplify your code a bit:
'2' is not ('1' or '2' or '3' or '4' or '5' or '6' or '7' or '8' or '9') evaluates to True '1' is not ('1' or '2' or '3' or '4' or '5' or '6' or '7' or '8' or '9') evaluates to False Somehow I do not believe you want that behavipur.
If cellboard contains characters, you could use: if (cellboard[cellboardrandom] not in '123456789')
for integers, the following should work: if not (1 <= cellboard[cellboardrandom] <= 9)
Using None to code empty cells, you could even have: if (cellboard[cellboardrandom] is None)
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