🚀 go-pugleaf

In article <1992Sep3.201927.25562@hubcap.clemson.edu>, leo@nas.nasa.gov
(Leo Dagum) writes:
|> In article <1992Sep3.123226.14205@hubcap.clemson.edu>
steve@hubcap.clemson.edu ("Steve" Stevenson) writes:
|> >[I saw this on sci.logic. I thought it was a nice problem for this group.
|> >  steve]
|> >>From: dougs@tvnews.tv.tek.com (Doug Stevens)
|> >>Message-ID: <1992Sep1.234452.26917@tvnews.tv.tek.com>
|> >
|> >[....]
|> 
|> Unless I'm missing something, this is fairly straightforward.
|> 
|> 1) A zeroes watch, calls B and hangs up.
|> 2) B waits till phone ends ringing then calls A and hangs up.
|> 3) B calls A again and hangs up.
|> 4) A times the first call back, and now knows (AB + BA), when he gets
|>    the second call back, he now knows BA as well (and therefore AB).
|> 5) A sets his watch back by AB, and calls B
|> 6) B picks up the phone and zeroes his watch, they now have (approximately)
|>    synchronized watches.
|> 7) Bob's your uncle.

I have a problem with this - B could place the second call (number 3) at
any time and therefore A is unable to determine the delay BA.  For example,
if B places the second call to A 1 second after the first, then all A will
see is two calls 1 second apart.

Bob isn't my uncle?

 - Simon

[Simon Moore][University of Cambridge, Computer Laboratory][swm11@uk.ac.cam.cl]