[Tutor] A bit long, but would appreciate anyone's help, if time permits!

[Blake Winton]

Hee-Seng Kye wrote:
> [[10, 11, 10, 9, 11, 9], [7, 9, 9, 7, 8, 8], [6, 6, 7, 6, 6, 5], [3, 5, 
> 4, 4, 5, 3], [1, 2, 3, 1, 3, 2]]     #r
> 
> How would I find the smallest values of a list r[0], take only those 
> values (r[0][3] and r[0][5]) for further comparison (r[1][3] and 
> r[1][5]), and finally print a p3?

Well, what do you have so far?

Here, I'll help you out a little.  So you want to find the smallest 
values of a list, huh?  Well, this is one way to do it.

 >>> def findSmallestValues( r ):
...   # We'll need somewhere to store the indices of the smallest values.
...   indices = []
...   for index in range(len(r)):
...     if len(indices) == 0 or r[index] < r[ indices[0] ]:
...       indices = [index]
...     elif r[index] == r[ indices[0] ]:
...       indices.append( index )
...   return indices
...
 >>> findSmallestValues([10, 11, 10, 9, 11, 9])
[3, 5]
 >>> r = [10, 11, 10, 9, 11, 9]
 >>> findSmallestValues(r)
[3, 5]
 >>> r[3], r[5]
(9, 9)
 >>>

Or, I think you can make that shorter (but much less efficient) by using 
list comprehensions, and the reduce function.

 >>> def findSmallestNumber(r):
...   return [i for i in range(len(r)) if r[i] == reduce( min, r ) ]
...
 >>> findSmallestNumber(r)
[3, 5]

It would be a little more efficient like this.

 >>> def findSmallestNumber(r):
...   s = reduce( min, r )
...   return [i for i in range(len(r)) if r[i] == s ]
...
 >>> findSmallestNumber(r)
[3, 5]

(Uh, I ran into problems trying to time the various versions of the
  functions, so I'll let someone else do that part of it.)

Later,
Blake.