On Wed, 31 Oct 2001, Jarett Hailes wrote:
> Mr. Amaral:
>
> For the following sequence:
>
> X = 1 1 1 1 1 1 1 1
> Y = 0 0 1 1 0 0 0 1
> Z = 0 1 0 1 0 0 1 0
a b c d e f g h
at b you got one pair of ones
at c you got two pairs of ones
at d you got three pairs of ones
at e you got four pairs of ones
--- this causes your circuit to reset to zero
thus when leaving e it has not received any one.
at f you have got a single one, but not a pair
af g you got one pair of ones.
> Why Z doesn't output a 1 on its sixth iteration? Since the output reset
> itself in the previous iteration, I thought it would have started to count
> from there, but it looks like it didn't here.
>
> Thanks for your time.
>
> Jarett Hailes
>
--Cheers,
Nelson
/ \ / / Jose Nelson Amaral - amaral@cs.ualberta.ca ) / ( Associate Professor / / \ Dept. of Computing Science - University of Alberta ( / ) Edmonton, Alberta, Canada, T6G 2E8 \ O / Phone: (780)492-5411 Fax: (780)492-1071 \ / http://www.cs.ualberta.ca/~amaral `----'
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