Herman’s algorithm

Introduced by T. Herman (1990)

An odd number of processors are arranged in a circle.

An odd number of processors are arranged in a circle.An odd number of them have tokens.We want to get to a state with exactly one token.

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

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

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

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?

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(N2logN).

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(N2logN).

McIver and Morgan (2004): it is Θ(N2).Three equally-spaced tokens take time 4N2/27. Conjecture: this is the worst case.

Three equally-spaced tokens take time 4N2/27. Conjecture: this is the worst case.

Nakata (2005): any configuration has expectedtime at most N2(π2–8)/2(about 6·3 times the conjectured bound).

Three equally-spaced tokens take time 4N2/27. Conjecture: this is the worst case.

Nakata (2005): any configuration has expectedtime at most N2(π2–8)/2(about 6·3 times the conjectured bound).

We get an upper bound of N2(π2–8)/12(about 5% more than the conjectured bound).

McIver and Morgan showed that if there arethree tokens, with the distances between successivetokens being a,b,c, then the expected time is4abc/N.

McIver and Morgan showed that if there arethree tokens, with the distances between successivetokens being a,b,c, then the expected time is4abc/N.

Here 32/7.

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McIver and Morgan showed that if there arethree tokens, with the distances between successivetokens being a,b,c, then the expected time is4abc/N. Can this approach be extended tomore tokens?

McIver and Morgan showed that if there arethree tokens, with the distances between successivetokens being a,b,c, then the expected time is4abc/N. Can this approach be extended tomore tokens?

Yes, sort of. Instead of each step taking time 1,give each step a cost of r if there were 2r+1tokens at the start of that step. With 3 tokens, cost=time. With more, cost>time.

We give an exact formula for the expected cost.

For r≥1, take variables a1 ... a2r+1.

A triple with even gaps is a term of the formaiajak with i<j<k and k–j, j–i even.

This is preserved by cyclic shifts of the variables.

We give an exact formula for the expected cost.

For r≥1, take variables a1 ... a2r+1.

A triple with even gaps is a term of the formaiajak with i<j<k and k–j, j–i even.

This is preserved by cyclic shifts of the variables.

Write steg(a1 ... a2r+1) for the sum of such triples.

This sum has r(r+1)(2r+1)/6 terms.

If r>1, what happens when one variable vanishes?steg(a1 ... a2r–1,0,a2r+1)= steg(a1 ... a2r–2, a2r–1+a2r+1)

If r>1, what happens when one variable vanishes?steg(a1 ... a2r–1,0,a2r+1)= steg(a1 ... a2r–2, a2r–1+a2r+1)

Suppose we have 2r+1 tokens, and distances betweensuccessive tokens are a1 ... a2r+1 in that order, so

a1+ a2+···+a2r+1=N.

If r>1, what happens when one variable vanishes?steg(a1 ... a2r–1,0,a2r+1)= steg(a1 ... a2r–2, a2r–1+a2r+1)

Suppose we have 2r+1 tokens, and distances betweensuccessive tokens are a1 ... a2r+1 in that order, so

a1+ a2+···+a2r+1=N.

Taking one time step in the algorithm changes these distances to random variables ã1 ... ã2r+1.

If r>1, what happens when one variable vanishes?steg(a1 ... a2r–1,0,a2r+1)= steg(a1 ... a2r–2, a2r–1+a2r+1)

Suppose we have 2r+1 tokens, and distances betweensuccessive tokens are a1 ... a2r+1 in that order, so

a1+ a2+···+a2r+1=N.

Taking one time step in the algorithm changes these distances to random variables ã1 ... ã2r+1.

E(steg(ã1 ... ã2r+1))= steg(a1 ... a2r+1)–rN/4.

So the expected total cost from this starting stateis 4steg(a1 ... a2r+1)/N.

How big can this be?

So the expected total cost from this starting stateis 4steg(a1 ... a2r+1)/N.

How big can this be?

If x1 ... x2r+1 are non-negative with sum 1 then

steg(x1 ... x2r+1)≤(1-(2r+1)–2 )/24

(the value taken when they are all equal).

So for any r, expected cost < N2/6.

We can get a tighter bound on the expected time.Fix a start state with 2r+1 tokens. Write As for the first

configuration with ≤2s+1 tokens, Cs for the cost

accumulated after that point, and T for the total time.

We can get a tighter bound on the expected time.Fix a start state with 2r+1 tokens. Write As for the first

configuration with ≤2s+1 tokens, Cs for the cost

accumulated after that point, and T for the total time.

T=∑s(Cs–Cs–1)/s = ∑sCs(s–1–(s+1)–1) = ∑sCss–1(s+1)–1.

We can get a tighter bound on the expected time.Fix a start state with 2r+1 tokens. Write As for the first

configuration with ≤2s+1 tokens, Cs for the cost

accumulated after that point, and T for the total time.

T=∑s(Cs–Cs–1)/s = ∑sCs(s–1–(s+1)–1) = ∑sCss–1(s+1)–1.

E(Cs)=E(E(Cs|As))≤(1–(2s+1)–2 )N2/6.

We can get a tighter bound on the expected time.Fix a start state with 2r+1 tokens. Write As for the first

configuration with ≤2s+1 tokens, Cs for the cost

accumulated after that point, and T for the total time.

T=∑s(Cs–Cs–1)/s = ∑sCs(s–1–(s+1)–1) = ∑sCss–1(s+1)–1.

E(Cs)=E(E(Cs|As))≤(1–(2s+1)–2 )N2/6.

So E(T) ≤ ∑ss–1(s+1)–1(1–(2s+1)–2 )N2/6

= ∑ss–1(s+1)–1(4s2+4s)(2s+1)–2N2/6

= ⅔N2∑s(2s+1)–2 = N2(π2–8)/12.