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CPSC 668 Distributed Algorithms and Systems

CPSC 668 Distributed Algorithms and Systems. Spring 2008 Prof. Jennifer Welch. Read/Write Shared Variables. In one atomic step a processor can read the variable or write the variable but not both!. Bakery Algorithm. An algorithm for no lockout mutual exclusion

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CPSC 668 Distributed Algorithms and Systems

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  1. CPSC 668Distributed Algorithms and Systems Spring 2008 Prof. Jennifer Welch Set 7: Mutual Exclusion with Read/Write Variables

  2. Read/Write Shared Variables • In one atomic step a processor can • read the variable or • write the variable • but not both! Set 7: Mutual Exclusion with Read/Write Variables

  3. Bakery Algorithm • An algorithm for • no lockout • mutual exclusion • using 2n shared read/write variables • booleans Choosing[i] : initially false, written by pi and read by others • integers Number[i] : initially 0, written by pi and read by others Set 7: Mutual Exclusion with Read/Write Variables

  4. Bakery Algorithm Code for entry section: Choosing[i] := true Number[i] := max{Number[0], …, Number[n-1]} + 1 Choosing[i] := false for j := 0 to n-1 (except i) do wait until Choosing[j] = false wait until Number[j] = 0 or (Number[j],j) > (Number[i],i) endfor Code for exit section: Number[i] := 0 Set 7: Mutual Exclusion with Read/Write Variables

  5. Bakery Algorithm Provides Mutual Exclusion Lemma (4.5): If pi is in the critical section and Number[k] ≠ 0 (k ≠ i), then (Number[k],k) > (Number[i],i). Proof: Consider two cases: pi 's most recent read of Number[k]; Case 1: returns 0 Case 2: (Number[k],k) > (Number[i],i) pi in CS and Number[k] ≠ 0 Set 7: Mutual Exclusion with Read/Write Variables

  6. pi 's most recent read of Choosing[k], returns false. So pk is not in middle of choosing number. piin CS and Number[k] ≠ 0 pi 's most recent read of Number[k], returns 0. So pk is in remainder or choosing number. pi 's most recent write to Number[i] So pk chooses number in this interval, sees pi 's number, and chooses a larger one. Mutual Exclusion Case 1 Set 7: Mutual Exclusion with Read/Write Variables

  7. Mutual Exclusion Case 2 • Is proved using arguments similar to those for Case 1. Set 7: Mutual Exclusion with Read/Write Variables

  8. Mutual Exclusion for Bakery Algorithm • Lemma (4.6): If pi is in the critical section, then Number[i] > 0. • Proof is a straightforward induction. • Mutual Exclusion: Suppose pi and pk are simultaneously in CS. • By Lemma 4.6, both have Number > 0. • By Lemma 4.5, • (Number[k],k) > (Number[i],i) and • (Number[i],i) > (Number[k],k) Contradiction! Set 7: Mutual Exclusion with Read/Write Variables

  9. No Lockout for Bakery Algorithm • Assume in contradiction there is a starved processor. • Starved processors are stuck at Line 5 or 6 (wait stmts), not while choosing a number. • Let pi be starved processor with smallest (Number[i],i). • Any processor entering entry section after pi has chosen its number chooses a larger number. • Every processor with a smaller number eventually enters CS (not starved) and exits. • Thus pi cannot be stuck at Line 5 or 6. Contradiction! Set 7: Mutual Exclusion with Read/Write Variables

  10. Space Complexity of Bakery Algorithm • Number of shared variables is 2n • Choosing variables are variable • Number variables are unbounded • Is it possible for an algorithm to use less shared space? Set 7: Mutual Exclusion with Read/Write Variables

  11. Bounded 2-Processor ME Algorithm Uses 3 binary shared read/write variables: • W[0] : initially 0, written by p0 and read by p1 • W[1] : initially 0, written by p1 and read by p0 • Priority : initially 0, written and read by both Set 7: Mutual Exclusion with Read/Write Variables

  12. Bounded 2-Processor ME Algorithm with ND • Start with a bounded algorithm for 2 processors with ND, then extend to NL, then extend to n processors. • Some ideas used in 2-processor algorithm: • each processor has a shared boolean W[i] indicating if it wants the CS • p0 always has priority over p1; asymmetric code Set 7: Mutual Exclusion with Read/Write Variables

  13. Bounded 2-Processor ME Algorithm with ND Code for p0 's entry section: • . • . • W[0] := 1 • . • . • wait until W[1] = 0 Code for p0 's exit section: • . • W[0] := 0 Set 7: Mutual Exclusion with Read/Write Variables

  14. Bounded 2-Processor ME Algorithm with ND Code for p1's entry section: • W[1] := 0 • wait until W[0] = 0 • W[1] := 1 • . • if (W[0] = 1) then goto Line 1 • . Code for p1's exit section: • . • W[1] := 0 Set 7: Mutual Exclusion with Read/Write Variables

  15. Discussion of 2-Processor ND Algorithm • Satisfies mutual exclusion: processors use W variables to make sure of this • Satisfies no deadlock (exercise) • But unfair (lockout) • Fix by having the processors alternate having the priority: • shared variable Priority, read and written by both Set 7: Mutual Exclusion with Read/Write Variables

  16. Bounded 2-Processor ME Algorithm Code for entry section: • W[i] := 0 • wait until W[1-i] = 0 or Priority = i • W[i] := 1 • if (Priority = 1-i) then • if (W[1-i] = 1) then goto Line 1 • else wait until (W[1-i] = 0) Code for exit section: • Priority := 1-i • W[i] := 0 Set 7: Mutual Exclusion with Read/Write Variables

  17. W[0] = W[1] = 1, both procs in CS p1 's most recent write of 1 to W[1] (Line 3) p0 's most recent write of 1 to W[0] (Line 3) p0 's most recent read of W[1] (Line 6): must return 1, not 0! Analysis of Bounded 2-Processor Algorithm • Mutual Exclusion: Suppose in contradiction p0 and p1 are simultaneously in CS. Contradiction! Set 7: Mutual Exclusion with Read/Write Variables

  18. No-Deadlock for 2-Proc. Alg. • Useful for showing no-lockout. • If one proc. ever enters remainder for good, other one cannot be starved. • Ex: If p1 enters remainder for good, then p0 will keep seeing W[1] = 0. • So any deadlock would starve both procs. Set 7: Mutual Exclusion with Read/Write Variables

  19. p0 not stuck in Line 2, skips line 5, stuck in Line 6 with W[0] = 1 waiting for W[1] to be 0 p1 skips Line 6, stuck at Line 2 with W[1] = 0, waiting for W[0] to be 0 p0 and p1 stuck in entry, Priority = 0 p0 sees W[1] = 0, enters CS No-Deadlock for 2-Proc. Alg. • Suppose in contradiction there is deadlock. • W.l.o.g., suppose Priority gets stuck at 0 after both processors are stuck in their entry sections. Contradiction! Set 7: Mutual Exclusion with Read/Write Variables

  20. p1 enters entry, gets stuck at Line 2, waiting for W[0] to be 0 p0 stuck at Line 6 with W[0] = 1, waiting for W[1] to be 0 p1 at Line 7; Priority = 0 forever after p0 stuck in entry No-Lockout for 2-Proc. Alg. • Suppose in contradiction p0is starved. • Since there is no deadlock, p1 enters CS infinitely often. • The first time p1 executes Line 7 in exit section after p0 is stuck in entry, Priority gets stuck at 0. Contradiction! Set 7: Mutual Exclusion with Read/Write Variables

  21. Bounded n-Processor ME Alg. • Can we get a bounded space mutex algorithm for n > 2 processors? • Yes! • Based on the notion of a tournament tree: complete binary tree with n-1 nodes • tree is conceptual only! does not represent message passing channels • A copy of the 2-proc. algorithm is associated with each tree node • includes separate copies of the 3 shared variables Set 7: Mutual Exclusion with Read/Write Variables

  22. Tournament Tree 1 2 3 5 6 7 4 p2, p3 p4, p5 p6, p7 p0, p1 Set 7: Mutual Exclusion with Read/Write Variables

  23. Tournament Tree ME Algorithm • Each proc. begins entry section at a specific leaf (two procs per leaf) • A proc. proceeds to next level in tree by winning the 2-proc. competition for current tree node: • on left side, play role of p0 • on right side, play role of p1 • When a proc. wins the 2-proc. algorithm associated with the tree root, it enters CS. Set 7: Mutual Exclusion with Read/Write Variables

  24. More on Tournament Tree Alg. • Code in book is recursive. • pi begins at tree node 2k + i/2, playing role of pi mod 2, where k = log n -1. • After winning node v, "critical section" for node v is • entry code for all nodes on path from v 's parent v/2 to root • real critical section • exit code for all nodes on path from root to v/2 Set 7: Mutual Exclusion with Read/Write Variables

  25. Tournament Tree Analysis • Correctness: based on correctness of 2-processor algorithm and tournament structure: • projection of an admissible execution of tournament alg. onto a particular tree node gives an admissible execution of 2-proc. alg. • ME for tournament alg. follows from ME for 2-proc. alg. at tree root. • NL for tournament alg. follows from NL for the 2-proc. algs. at all nodes of tree • Space Complexity: 3n boolean read/write shared variables. Set 7: Mutual Exclusion with Read/Write Variables

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