Hi all,

in the meantime, I thought about my last mail and came to the conclusion that
is as simple to calculate the preceeding random number as the following one.

an example:      1 0 0 1 0 1 1 0           the actual random number
                 ~ ~   ~   ~
                 1*0 * 1 * 1      = 1      xor bit7,6,4,2

the next:  1 <-  0 0 1 0 1 1 0 1           shift left the new bit
                 ~   ~   ~     ~
                 0*  1 * 1  *  1   = 1     xor bit 7,5,3,0

the last:        1 0 0 1 0 1 1 0   -> 1    shift right this bit

the xor-ing is like a parity-calculation, the 4 old bits and the new bit always
have even parity. so it is always possible to calculate a missing bit from the
other four.
so if the last position code is known, calculate the following random numbers
out of the last position code until matching to get the relative forward
movement - or if not matching after a limited number of calculations,
calculate the preceeding random numbers until matching to get the relative
backward movement.

This all is a theoretical consideration. I did not test/simulate that yet.


Did everybody understand that?



By the way, what exactly does the TSL213?


Siggi


Siegfried Grob,                                   |
student of electrical engineering,                |
university of ulm, germany                        |
e-mail:  siegfried.grob@student.uni-ulm.de        |
tel&fax: +49 731 25148                            |
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