🚀 go-pugleaf

In article <1992Sep4.174143.13747@hubcap.clemson.edu>, mln@sunkj.larc.nasa.gov (Michael L Nelson) writes:
|> I must be missing something here, because this looks quite simple.
   [....]
It isn't quite this simple :-)
|> 
|> -A & B already know this procedure.
|> 
|> A: says "Bing" to B
|>    B responds with "Bing" to A
|> 
|> this is ABBA time - the total delay
True
|> 
|> A:
|> 1. says "Bong"
I assume you want A to say "Bong" when A hears B's "Bing"
Then when B hears the "Bong" B will have sufficient information to compute
BA + AB == AB + BA, Otherwise B cannot know AB + BA
|> 2. waits, say N minutes (N > ABBA) 
|> 3. says "Bong" again
|> 
|> B already knows what N is, and can then determine
|> What AB is (Bong2 - Bong1 - N)
This is not the case.

	T0 A Goes Bong
	T1 B hears Bong = T0 + AB
	T2 A Goes Bong Second Time = T0 + N
	T3 B Hears Bong = T0 + AB + N
Please note that AB is undetermined (B lacks sufficient information
to determine this).  The problem lies in the fact that A and B do not
share the clock, but each keep their own local clocks, which are not currently
synchronized (as per the given).
|> 
|> B tells A what AB time is.
B does not have the value for AB.
|> 
|> A says "Bang" and waits AB time before zeroing
|> the watch
A does not have the value of AB computed
|> 
|> When B hears "Bang", B immediately zero's the watch
|> 
|> synchronized
Not quite.
|> 
|> Michael Nelson
|> 

Bill
-- 
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|	maniattb@cs.rpi.edu - in real life Bill Maniatty
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