R/W Reductions Eli Gafni UCLA Disc Godel Celebration 10/4/04

Outline Tasks R/W reduction “Weakest Unsolvable Task” Thesis Reductions Conclusions/Speculations

Tasks 3-Procs Test-and-Set (TST) =TST(1, 3): {p i } (win) i=0,1,2 {p i,p j } (win,lose), (lose,win) i j {p i,p j,p k }(win,lose,lose),(lose,win,lose), (lose,lose,win) i j k Allowed output tuples for “participating” sets

Tasks Cont’ed UN(n,2n-2) Uniform renaming: p 0,…,p n-1 out(p i ){1,…,2n-2) 0ut(p i )out(p j ) p i P, |P|=k<n out(p i ){1,…,2k-1}

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

R/W reducibility Cont’ed (2,3) set-election task: p 0,p 1,p 2 One or two procs elect themselves, the others elect an elected proc. {pi}(pi) {pi,pj}(pi,pi), (pj,pj), (pi,pj) {pi,pj,pk} (pi,pi,pi),(pi,pj,pi),(pi,pj,pj),…

R/W reducibilty Cont’d (2,3) set-election r/w reducible to 3-tst: pi: ci=di= initially register v i :=tst if v i =win then return p i else c i :=lose (*)scan if exists j s.t. d j =p i then return p i d i :=p j s.t. p j registered, c j = scan if c j = then retrun p j else goto (*)

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.

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): p 0,…,p 2n-2 procs output 0 or 1 |P|=n not all 0’s and not all 1’s

Reductions: SB(n,2n-1) is unsolvable: HS93: Comparison-Renaming CR(n,2n-2) is unsolvable Ramsey-Theorem implies: CR(n,2n-2) unsolvable iff Renaming R(n,2n-1,2n-2) unsolvable R(n,2n-1,2n-2) reducible to SB(n,2n-1): Procs that output 0 Attiya et al rename 1,2,… Procs that output 1 2n-2,2n-3,… Attiya et al k>0 (!) rename into 2k-1 2n n 1 -1=2n-2 if n 0,n 1 >0

Reductions Cont’ed SB(n,2n-1) reducible to R(n,2n-1,2n-2): position 1,…,n-1 output 0 position n,…,2n-2 output 1 TST(n-1,n): p 0,…,p n-1 For all P at least one proc returns “win” Number of “wins” at most n-1 SB(n,2n-1) reducible to TST(n-1,n): Use two TST’s: TST 1 and TST 2 procs p 0,…,p n-1 connected to TST 1 procs p n,…,p 2n-2 TST 2 from TST 1 “win”=1 “lose=0” from TST 2 “win”=0 “lose”=1

Reductions Cont’ed TST(n-1,n) reducible to SB(n,2n-1): I donot think so! (As we heard: SB(n,2n1) is a Torus and Torus is a “manifold” and manifold satisfies Sperner so no Torus van do “real” 2-set “hole”) Currently: SB(n,2n-1) R(n,2n-1,2n-2) TST(n-1,n)

Reductions Cont’ed SE(n-1,n) Set Election: for all P proc elect a proc in P |elected|<n SSE(n-1,n) Strong Set Election: SE with if p i elect p j then p j elects itself TST(n-1,n) reducible to SE(n-1,n): SSE(n-1,n) equivalent to SE(n-1,n): SSE(n-1,n) reducible to SE(n-1,n): (*)c i :=elected(p i ), scan If exits j st d j =p i then elect(p i ):=p i else d i :=follow “arrows” to end or max in cycle scan, if c di = return d i else goto (*)

Reductions Cont’ed SSE(n-1,n) reducible to TST(n-1,n): Same trick as pointing to potential winner and suggesting it elect itself. The first one to announce lose will never elect itself. TST(n-1,n) reducible to SSE(n-1,n): If out(p i )=p i return win else lose.

Reductions Cont’ed TST(n-1,n) reducible to UR(n,2n-2): If return UR(n,2n-2)=k<n return win else lose. |P|=k<n 2k-1-k=k-1<n-1 at least one proc return <n

Reductions Cont’ed UR(n,2n-2) reducible to TST(n-1,n): The BG93 algorithm for UR(n,2n-1): Rename(m,up) Take Imm Snap c i :=S i if p i max id in S i return (m-1)+2|S i |-1 else Rename(2|S i |-2,down) Rename(m,down) Take Imm Snap c i :=S i if p i max id in S i return (m+1)+(2|S i |-1) else Rename(2|S i |,up)

Reductions Cont’ed Example: {7,4} p 7,p 4 {0,2,4,7,3} p 0,p 2,p 3 {6,0,2,4,7,3,5,1} p 1,p 6,p 5 p 7 returns 3, p 4 rename from 2 down p 0,p 2,p 3 rename from 2*5-2=8 down p 1,p 6,p 5 rename from 2*8-2=14 down p 4 will rename at 2 later 8,7,6,5,4 is exactly enough for 3 procs 14,13,12,11,10 is exactly enough for 3 procs

Reductions Cont’ed Variant on the first phase of the BG: TST(n-1,n) Know how by “helping” guarantee that the smallest returned imm snap is all by winners if S i is all made of winners and p i is max then return 2|S i |-1 else Rename(2|S i |-2,down) if S i contain losers and p i max among losers then return 2|S i |-2 else Rename(2|S i |-3,down) There will be loser who will drop off in the first phase from the smallest snap that contains both winners and losers.

Conclusions TST(n-1,n) UR(n,2n-2) SE(n-1,n) SSE(n-1,n) SB(n,2n-1) R(n,2n-1,2n-2)

Conclusion Cont’ed Challenge: Procs p 1,p 2,p 3 q 1,q 2,q 3 Uniform rename to 5 slots when p i and q i can share a slot. Conjecture: need 6. By “Thesis” there should exist a reduction More: Exist min number of rounds for set consensus for synchronous systems by r/w reduction Is all this relations a structure a fluke?