[Tutor] another for loop question - latin square (was: Tutor Digest, Vol 106, Issue 74)

[Dave Angel]

On 01/01/2013 08:27 PM, Brandon Merritt wrote:
> Sorry, I should have been more clear in what I was trying to accomplish. I
> am trying to build the latin square as follows:
>
> 1234567
> 2345671
> 3456712
> 4567123
> 5671234
> 6712345
> 7123456
>
> So, the "scaleorder" is the n*n scale of the square - in this case, a 7x7.
> The key is that there should be no duplicate number for each row and for
> each column. And the "topleft" variable is the number in the topleft that
> sets how the rest of the square will look - in the example I gave above, it
> is 1.
>
> The link below is the original exercise - I am not in school, purely for my
> own education as I am really trying hard to become a passable Python
> programmer. I've been working at it for 6 month now, and I don't know any
> other languages:
>
> http://www.cse.msu.edu/~cse231/PracticeOfComputingUsingPython/02_Control/LatinSquares/Project03.pdf
>
>
> Thank you,
> Brandon

Please reply to a single message, not to some digest.  This way your new
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>
> On Mon, Dec 31, 2012 at 3:00 AM, <tutor-request at python.org> wrote:
>
>> <snip>
>>
>> When replying, please edit your Subject line so it is more specific
>> than "Re: Contents of Tutor digest..."
>>
>>
>>

So you don't just want a latin-square, you want a particular one.  You
picked the most trivial one, where all the spots are numeric, and each
line is simply sequentially chosen from a ring, 1 through scaleorder.

Assuming scaleorder is less than 10, the following should work.

table = [str(i) for i in range(1, scaleorder+1) * 3]

for row in range(topleft-1, topleft+scaleorder-1):
    line = table[row:row+scaleorder]
    print "".join(line)

If you have to deal with scaleorders bigger than 9, then you'll need a
separator between the numbers.  Probably the easiest place to fix that
would be in constructing the table.  But a q&d fix might be to use  "
".join() on the last line.

-- 

DaveA