[scilab-Users] Re: exercise on list

[Calixte DENIZET]

Hi,

Funny exercise... :)

i) N=9;L=zeros(1,N*(N+1)/2 - 1);L(cumsum(1:(N - 1)))=1;L=[1 cumsum(L) + 1]
The first idea is to use a diff on the result to observe that there is 
ones at position 1,3,6,10,.... and to "revert" a diff use a cumsum.

ii) N=9;L=zeros(1,(N+1)*(N+2)/2 - 1);L(cumsum(1:N))=1:N
The idea is to observe that you've a number at position 1,3,6,10,... so 
just put this number at the right position.

Calixte

On 14/04/2012 10:55, philippe wrote:
> Le 08/04/2012 21:04, abd_bela a écrit :
>>
>>
>> without using (loops: for or while)  create  the list of numbers as 
>> follow:
>
> whitout loops it's fun :-)
>
>> 1, 2, 2, 3 3 3  4 4 4 4    5 5 5 5 5    ......       9 9 9 9 9  9 9 9 9
>
> easy : build a 9x9 matrix with A(i,j)=i, take the lower triangular 
> part of A with tril), rewrite it as a one line list (with :), and keep 
> only non zero coefficients :
>
> n=9;
> A=(tril(([1:n]')*ones(1:n)))'; B=A(:); C=B(find(B<>0))'
>
>>
>> 1 0  2 0 0  3 0 0 0    4 0 0 0 0       .....  9  0 0 0 0  0 0 0 0 0
>
> harder, but based on the same ideas :
>
> n=9;A=-ones(n+1,n+1); A(1,2:n+1)=[1:n]; 
> A=triu(A);B=A(:);C=B(find(B<>0))'; C(find(C==-1))=0; C(1)=[]
>
>
> Philippe.
>
>
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