Can someone explain why this algorithm is free from deadlock specifically? thanks
+1
A:
Let me guess, you're attending the University of Illinois at Urbana-Champaign (or someplace like it), you're enrolled in CS 425 "Distributed Systems" (or something very similar), and you've just received your second machine problem (or the equivalent assignment).
Am I close?
Chris B.
2010-07-06 05:20:30
+1
A:
Because the need for release messages is removed (assuming all participants are correctly clock-synchronized and you also use timeouts to detect crashed participants). Thank God for Wikipedia.
joe snyder
2010-07-06 05:20:50