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Re: Re: faster sublist checking


On 21 Jun 2004, at 16:49, JMyers6761 wrote:

> Why is this?
>
> A = {3, 5}; B = {1, 2, 3, 4, 5, 6, 7};
>
> Intersection[B, A] == Sort[A]
> True
>
> but
>
> SublistQ[B_List, A_List] := Intersection[B, A] == Sort[A];
>
> SublistQ[B, A]
> False
>
> Al Myers
>
>
You don't need our help to answer this. Just quit the kernel and run 
evaluate all the above again.

Andrzej Kozlowski
Chiba, Japan
http://www.mimuw.edu.pl/~akoz/


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