🚀 go-pugleaf

In article <1992Sep3.172632.10776@hubcap.clemson.edu> fpst@hubcap.clemson.edu (Steve Stevenson) writes:

>I posted this problem to my news group, comp.parallel. Here are two
>solutions I've received.  ...

>----------------------------------------------------------------------------
>- A Solution -
>----------------------------------------------------------------------------
>Let the local time of A be tA, and the local time of B be tB.
>Further, let tAB be the lag between messages going from A to B,
>and let tBA be the symmetric value. Since both time pieces
>are presumable ticking at the same rate, let tB = tA + C for some
>constant C. If we can determine C, then they can synchronize their
>watches.
>
>1) A asks B the time. As s/he does, s/he notes how long it
>   takes to hear the response from B. If B answers promptly, 
>   A will now know 3 things:
>      (tA)                 (local time)
>      (tAB + tBA)          (total delay)
>      (tA + C + tAB)       (time told to A by B)

TRUE

>2) A subtracts the last value from the first two:
>      (tA) + (tAB + tBA) - (tA + C + tAB) = (tBA - C)

TRUE

>3) B now asks A the time. A hears the request at time
>   tA', notes this time and responds. B now knows:
>      (tA' + tBA)          (time told to B by A)

NO.  A tells B  tA' only, not tA' + tBA.

The  synchroniztation problem can not be solved.  It's easy to see that A and 
B can synchronize their watches in the sense that whenever A sends a message,
A will know what time B will think it is when B receives the message, and
for every message A receives, A will know what time B thought it was when
B sent it, and vice-versa, but  that's as close as they can come to having
synchronized watches.

They can NOT synchronize their watches in the sense that a third person who
is in instant communication with each of them will see that A's watch and
B's watch read the same.



David Petry