help how to sort a list in order of 'n' in python without using inbuilt functions??

[Steven D'Aprano]

On Sun, 26 May 2013 01:41:58 +1000, Chris Angelico wrote:

> On Sun, May 26, 2013 at 12:28 AM, Steven D'Aprano
> <steve+comp.lang.python at pearwood.info> wrote:
>> On Sat, 25 May 2013 19:14:57 +1000, Chris Angelico wrote:
>>
>>> def random_number():
>>>     return 7
>>
>> I call shenanigans! That value isn't generated randomly, you just made
>> it up! I rolled a die *hundreds* of times and not once did it come up
>> seven!
> 
> You've obviously never used a REAL set of dice.

You're right, all my dice are eight-sided and complex:

1+0i
1+1i
1-1i
-1+0i
-1+1i
-1-1i


:-)


But seriously, I have various D&D style gaming dice, d4, d6, d8, d12, d20 
and d30. But I thought the opportunity for a joke was more important than 
pedantic correctness :-)


> Now, I have here with me
> a set used for maths drill (to be entirely accurate, what I have here is
> the company's stock of them, so there are multiples of each of these -
> anyone need to buy dice?) 

Are you serious? What's the cost, posted to Melbourne?


> with everything except the classic 1 through 6 that everyone knows:
> 
> * Six sides, faces marked 7 through 12 
> * Six sides, faces marked "+x-\xf7+" and a "wild" marker 
>   (yes, two of +)

Oh, you mean ÷ (division sign)! Why didn't you say so? :-P

And another thing, shame on you, you mean × not x. It's easy to find too:

py> from unicodedata import lookup
py> print(lookup("MULTIPLICATION SIGN"))
×


> * Ten sides, numbered 0 through 9
> * Eight sides, numbered 1 through 8
> * Twelve sides, as above
> * Twenty sides, as above
> 
> Now, tabletop roleplayers will recognize the latter four as the ones
> notated as d10, d8, d12, and d20, but these are NOT for gameplay, they
> are for serious educational purposes! Honest!
> 
> Anyway, all of those can roll a 7... well, one of them has to roll a
> \xf7, but close enough right? 

I don't think so...

> Plus, if you roll 2d6 (that is, two
> regular six-sided dice and add them up), 7 is statistically the most
> likely number to come up with. Therefore it IS random.

Yes, but if you subtract them the most common is 0, if you multiply the 
most common are 6 or 12, and if you divide the most common is 1. If you 
decide on the operation randomly, using the +-×÷+ die above (ignoring 
wildcards), the most common result is 6. The probability of getting a 7 
is just 1/15.

from collections import Counter
from operator import add, sub, mul, truediv as div
ops = (add, sub, mul, div, add)
Counter(op(i, j) for op in ops for i in range(1, 7) for j in range(1, 7))


-- 
Steven