```Mauricio Fern?ndez wrote:

> On Wed, Dec 08, 2004 at 08:53:27AM +0900, trans.  (T. Onoma) wrote:
>> On Tuesday 07 December 2004 06:12 pm, Oliver Cromm wrote:
>>| ruby -e'require "matrix";a=[1];(1..10).each{|n| print "
>>| "*(10-n)*3,a.map{|i|\
>>| i.to_s.center(6)}.join,\$/;a.push(0);a=(Vector[*a]+Vector[*a.reverse]).to_a}
>>|'
>>
>> A touch smaller:
>>
>> ruby -e'require"matrix";a=[1];(1..10).each{|n|print" "*(10-n)*3,a.map{|i|\
>> i.to_s.center(6)}.join,\$/;a<<0;a=(Vector[*a]+Vector[*a.reverse]).to_a}'
>
> Some obvious modifications:
>
> ruby -rmatrix -le'a=[1];10.times{|n|print" "*(9-n)*3,"%3d   "*a.size%a;a<<0;a=(
> Vector[*a]+Vector[*a.reverse]).to_a}'

Optimal once you know the stuff - I'm learning!

> There's still potential for further optimization but I prefer to zzzZZ.

a.size = n+1, better yet:

ruby -rmatrix -le'a=[1];1.upto(10){|n|print" "*(10-n)*3,"%3d   "*n%a
a<<0;a=(Vector[*a]+Vector[*a.reverse]).to_a}'

I found the Vector stuff /kakkou ii/, but without, it's shorter.
Accepting the extra indent, my shortest is:

ruby -le'a=[];10.times{|n|a<<1;print" "*(9-n)*3,"%6d"*(n+1)%a
n.downto(1){|i|a[i]+=a[i-1]}}'
--
Oliver C.

```