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Herman’s algorithm Introduced by T. Herman (1990)

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An odd number of processors are arranged in a circle.

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An odd number of them have tokens. We want to get to a state with exactly one token.

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At each time step, every processor with a token either keeps it or passes it clockwise (independently, 50-50).

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At each time step, every processor with a token either keeps it or passes it clockwise (independently, 50-50).

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At each time step, every processor with a token either keeps it or passes it clockwise (independently, 50-50). When two tokens collide they annihilate each other.

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This eventually reaches a state with only one token. For a fixed initial configuration with N processors, how large can the expected time taken be?

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This eventually reaches a state with only one token. For a fixed initial configuration with N processors, how large can the expected time taken be? Herman (1990): it is O(N 2 logN).

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This eventually reaches a state with only one token. For a fixed initial configuration with N processors, how large can the expected time taken be? Herman (1990): it is O(N 2 logN). McIver and Morgan (2004): it is Θ(N 2 ). Three equally-spaced tokens take time 4N 2 /27. Conjecture: this is the worst case.

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Three equally-spaced tokens take time 4N 2 /27. Conjecture: this is the worst case. Nakata (2005): any configuration has expected time at most N 2 (π 2 –8)/2 (about 6·3 times the conjectured bound).

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Three equally-spaced tokens take time 4N 2 /27. Conjecture: this is the worst case. Nakata (2005): any configuration has expected time at most N 2 (π 2 –8)/2 (about 6·3 times the conjectured bound). We get an upper bound of N 2 (π 2 –8)/12 (about 5% more than the conjectured bound).

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McIver and Morgan showed that if there are three tokens, with the distances between successive tokens being a,b,c, then the expected time is 4abc/N.

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McIver and Morgan showed that if there are three tokens, with the distances between successive tokens being a,b,c, then the expected time is 4abc/N. Here 32/

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McIver and Morgan showed that if there are three tokens, with the distances between successive tokens being a,b,c, then the expected time is 4abc/N. Can this approach be extended to more tokens?

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McIver and Morgan showed that if there are three tokens, with the distances between successive tokens being a,b,c, then the expected time is 4abc/N. Can this approach be extended to more tokens? Yes, sort of. Instead of each step taking time 1, give each step a cost of r if there were 2r+1 tokens at the start of that step. With 3 tokens, cost=time. With more, cost>time.

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We give an exact formula for the expected cost. For r≥1, take variables a 1... a 2r+1. A triple with even gaps is a term of the form a i a j a k with i

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We give an exact formula for the expected cost. For r≥1, take variables a 1... a 2r+1. A triple with even gaps is a term of the form a i a j a k with i

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If r>1, what happens when one variable vanishes? steg(a 1... a 2r–1,0,a 2r+1 )= steg(a 1... a 2r–2, a 2r–1 +a 2r+1 )

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If r>1, what happens when one variable vanishes? steg(a 1... a 2r–1,0,a 2r+1 )= steg(a 1... a 2r–2, a 2r–1 +a 2r+1 ) Suppose we have 2r+1 tokens, and distances between successive tokens are a 1... a 2r+1 in that order, so a 1 + a 2 +···+a 2r+1 =N.

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If r>1, what happens when one variable vanishes? steg(a 1... a 2r–1,0,a 2r+1 )= steg(a 1... a 2r–2, a 2r–1 +a 2r+1 ) Suppose we have 2r+1 tokens, and distances between successive tokens are a 1... a 2r+1 in that order, so a 1 + a 2 +···+a 2r+1 =N. Taking one time step in the algorithm changes these distances to random variables ã 1... ã 2r+1.

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If r>1, what happens when one variable vanishes? steg(a 1... a 2r–1,0,a 2r+1 )= steg(a 1... a 2r–2, a 2r–1 +a 2r+1 ) Suppose we have 2r+1 tokens, and distances between successive tokens are a 1... a 2r+1 in that order, so a 1 + a 2 +···+a 2r+1 =N. Taking one time step in the algorithm changes these distances to random variables ã 1... ã 2r+1. E(steg(ã 1... ã 2r+1 ))= steg(a 1... a 2r+1 )–rN/4.

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So the expected total cost from this starting state is 4steg(a 1... a 2r+1 )/N. How big can this be?

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So the expected total cost from this starting state is 4steg(a 1... a 2r+1 )/N. How big can this be? If x 1... x 2r+1 are non-negative with sum 1 then steg(x 1... x 2r+1 )≤(1-(2r+1) –2 )/24 (the value taken when they are all equal). So for any r, expected cost < N 2 /6.

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We can get a tighter bound on the expected time. Fix a start state with 2r+1 tokens. Write A s for the first configuration with ≤2s+1 tokens, C s for the cost accumulated after that point, and T for the total time.

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We can get a tighter bound on the expected time. Fix a start state with 2r+1 tokens. Write A s for the first configuration with ≤2s+1 tokens, C s for the cost accumulated after that point, and T for the total time. T=∑ s (C s –C s–1 )/s = ∑ s C s (s –1 –(s+1) –1 ) = ∑ s C s s –1 (s+1) –1.

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We can get a tighter bound on the expected time. Fix a start state with 2r+1 tokens. Write A s for the first configuration with ≤2s+1 tokens, C s for the cost accumulated after that point, and T for the total time. T=∑ s (C s –C s–1 )/s = ∑ s C s (s –1 –(s+1) –1 ) = ∑ s C s s –1 (s+1) –1. E(C s )=E(E(C s |A s ))≤(1–(2s+1) –2 )N 2 /6.

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We can get a tighter bound on the expected time. Fix a start state with 2r+1 tokens. Write A s for the first configuration with ≤2s+1 tokens, C s for the cost accumulated after that point, and T for the total time. T=∑ s (C s –C s–1 )/s = ∑ s C s (s –1 –(s+1) –1 ) = ∑ s C s s –1 (s+1) –1. E(C s )=E(E(C s |A s ))≤(1–(2s+1) –2 )N 2 /6. So E(T) ≤ ∑ s s –1 (s+1) –1 (1–(2s+1) –2 )N 2 /6 = ∑ s s –1 (s+1) –1 (4s 2 +4s)(2s+1) –2 N 2 /6 = ⅔N 2 ∑ s (2s+1) –2 = N 2 (π 2 –8)/12.

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