castironpi wrote:
On Sep 5, 9:20 pm, "Manu Hack" <[EMAIL PROTECTED]> wrote:
On Fri, Sep 5, 2008 at 1:04 PM, castironpi <[EMAIL PROTECTED]> wrote:
On Sep 5, 3:28 am, "Manu Hack" <[EMAIL PROTECTED]> wrote:
On Thu, Sep 4, 2008 at 4:25 PM, castironpi <[EMAIL PROTECTED]> wrote:
On Sep 4, 2:42 pm, [EMAIL PROTECTED] wrote:
David C. Ullrich:
At least in mathematics, the sum of the elements of
the empty set _is_ 0, while the maximum element of the
empty set is undefined.
What do you think about my idea of adding that 'default' argument to
the max()/min() functions?
Bye,
bearophile
For max and min, why can't you just add your argument to the set
itself?
The reason max([]) is undefined is that max( S ) is in S.
It makes sense.
The reason sum([]) is 0 is that sum( [ x ] ) - x = 0.
It doesn't make sense to me.  What do you set x to?
For all x.
But then how can you conclude sum([]) = 0 from there?  It's way far
from obvious.

You can define sum([a1,a2,...,aN]) recursively as
sum([a1,a2,...a(N-1)])+aN.  Call the sum sum([a1,a2,...,aN]) "X", then
subtract aN.

sum([a1,a2,...a(N-1)])+aN=X
sum([a1,a2,...a(N-1)])+aN-aN=X-aN

For N=2, we have:

sum([a1,a2])=X
sum([a1,a2])-a2=X-a2
sum([a1,a2])-a2-a1=X-a2-a1

Since X= a1+ a2, replace X.

sum([a1,a2])-a2-a1=(a1+a2)-a2-a1

Or,

sum([a1,a2])-a2-a1=0

Apply the recursive definition:

sum([a1])+a2-a2-a1=0

And again:

sum([])+a1+a2-a2-a1=0

And we have:

sum([])=0.


This is not necessarily so.

The flaw is that you provide a recursive definition with no start value,
which is to say it is not a recursive definition at all.

A recursive definition should be (for lists where elements
can be added, and ignoring pythonic negative indexing):

Define 'sum(L)' by
a.  sum(L[0]) = L[0]
b.  sum(L[0:i]) = sum(L[0:i-1]) + L[i]  ... if i > 0

From this you can prove the reverse recursion
    sum{L[0:k]) = sum(L[0:k+1]) - L[k+1]
   __only__ if k >= 0

It says nothing about the empty list.

You could add, as part of the definition, that sum{[]) = 0, or any other value.

A rather different approach, not quite simple recursion, would be to
start with

A. a slicing axiom, something like:

  for all non-negative integers, a,b,c with a <=b <= c:

  sum(L[a:c]) = sum(L[a:b]) + sum(L[b:c])

B. a singleton axiom:

  for all integers a where L[a] exists:
 sum(L[a:a]) = L[a]




2a.  sum{
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