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R/W Reductions

R/W Reductions. Eli Gafni UCLA ICDCN06 12/30. Outline. Tasks and r/w impossible task: 2 cons 3 cons NP-completeness R/W reduction “Weakest Unsolvable Task” Thesis Reductions Conclusions/Speculations. 2-cons. Two procs P0 P1 P0 alone output 0 P1 alone output 1

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R/W Reductions

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  1. R/W Reductions Eli Gafni UCLA ICDCN06 12/30

  2. Outline • Tasks and r/w impossible task: • 2 cons • 3 cons • NP-completeness • R/W reduction • “Weakest Unsolvable Task” Thesis • Reductions • Conclusions/Speculations

  3. 2-cons • Two procs P0 P1 • P0 alone output 0 • P1 alone output 1 • P1 and P2 both output either 0 or 1

  4. 2-cons impossible in r/w • The protocol complex is a line • The output is disconnected

  5. 3-cons (election) • Procs P0 P1 P2 • Pi alone output i • All procs output same i

  6. 3-cons impossible in r/w • If 3-cons was possible than 2-cons would be possible, contradiction. • We proved 3-cons impossible by reduction FROM 2-cons

  7. Analogy with NP-completeness • In my Algorithm class students just get the fact that SAT is NP-complete. • Showing that a problem is NP-complete does not require understanding TMs • Showing a Task is r/w impossible should not require knowing Algebraic Topology!

  8. The Quest: An Analog of SAT • In our case it should be the Weakest (W) impossible task • In some sense topology implies no such Weakest exists (upcoming paper: it does for any class of real interest)

  9. R/W reducibility Cont’ed Reducibilty induces a directed graph over tasks • A strongly connected component are tasks which are r/w equivalent • Wishful “Weakest-Thesis:” • There exist a task WEAKEST(n): • WEAKEST(n) is r/w unsolvable. • WEAKEST(n) is reducible to any task which is unsolvable when restricted to participating set of at most n procs.

  10. R/W reducibility Cont’ed • If Weakest-Thesis hold then • “all Maurice can do, we can do better.” • Can think Java and not worry about not knowing Basic. • Can go back to thinking “distributed” rather than “topology.” • Plausibilty All known unsolvable tasks are reducible to SB(n,2n-1) (Symmetry Breaking) SB(n,2n-1): p0,…,p2n-2 procs output 0 or 1 |P|=n not all 0’s and not all 1’s

  11. Have to talk about Task Implementation • 4 procs P0 P1 P2 P4 of which I’ll wake at most 2 • If alone can output any 0 or 1 • If both same parity, one of them output 0 and the other 1, if different parity, then both output same. • (Vassos,Lo) Cannot do anything above r/w

  12. But any implementation will do • If implementable by r/w then a processor in solo execution is apriori decided whether will output 0 or 1. • 2 out of the 3 have same solo say 0 • If same parity, 0 then win (since the other changes to 1) • If no same parity, then there is one parity 0 and other parity 1, the one that changes wins

  13. Instead of Task, any implementation that solves the Task • If the task is on n-procs then commit to some r/w protocol for the max size participating set for which is solvable • For n get the output from the ``oracle.’’

  14. SB(n,2n-1) equivalent to ``comparison’’ • If SB(n,2n-1) know how to rename n ``comparison’’ into 1 to 2n-1 • If ``comparison’’ then can glue together, I.e. the algorithm for 1,2,3 and 1,2,4 when 1,2 work alone no need to worry whether they are going to work with 3 or 4. Invoke object only in ``middle’’ (show example with SB11)

  15. A weakest task for 3 procs

  16. Family of weakest • No task with 0,1 mapping symmetric on the boundary can be r/w colored to avoid all 0’s or all 1’s

  17. R/W Reducibility Between Tasks Task A r/w reducible to B: A R/W B R/W

  18. Reductions: • (3,2) -tst = always at least one proc outsputs 0, not all output 0. • Impossible: Will solve SB11 • Do IS • At level 3 invoke tst, loser goes down • If remain at level 3 do 2-IS to get 0 if alone or 1 if see another

  19. (3,2)-TST=(3,2)-ELECTION • TST implies election: Register in SM Apply to tst If 0 elect yourself If 1, write ``I got 1’’ in SM Scan, elect registerd proc who has not announced 1 The proc who wrote ``got 1’’ first will not be chosen

  20. Lemma: (3,2)-Election = (3,2) Strong Election • Strong election = if elected by any then you elect yourself • Election implies strong: • Phase 1: write (i, elected j) • Scan, if someone elected you elect yourself, if choose else and see someone choose himself choose him • Phase 2 (separate memory): write(I,elected j) • Scan, if someone elected you elect yourself, if choose else and see someone choose himself choose him

  21. Tst=Strong Election (cont’) • Strong election implies TST: • If elect yourself output 0 • Else 1

  22. Conclusions: • Weakest Thesis has ``experimental’’ support. • Not true if ``any’’ task is considered • Find the ``family’’ for which it is the weakest • Missing; Internal reduction 3 SB implies 4 SB

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