Bounds on Herman's algorithm

0.12 MB·

by  John Haslegrave

Bounds on Herman’s algorithm
John Haslegrave
May 21, 2014
4
Abstract
1
0 Herman’s self-stabilisation algorithm allows a ring of N processors
2 having any odd number of tokens to reach a stable state where exactly

n onetokenremains. McIverandMorganconjecturethattheexpectedtime
a taken for stabilisation is maximised when there are three equally-spaced
J tokens. We prove exact results on a related cost function, and obtain a

4 boundon expected time which is very close tothe conjectured bound.
2
Keywords: Randomizedalgorithms;Probabilisticself-stabilization;Herman’s

] algorithm
S
D
1 Introduction
.
s
c
[ Self-stabilisation algorithms were first discussed by Dijkstra [1], and have since

beenwidelystudied(seeeg[2],[10]). Herman’salgorithmprovidesarandomised
1
v self-stabilisation mechanism for N processors connected unidirectionally in a
9 ring, with synchronousupdates. Eachprocessoreither has a token or does not.
0 In aninitial state anunknown, but odd, number ofprocessorsholdtokens, and
2
the system is stable if there is only one token. Herman proposed a scheme
5
where, simultaneously at each time step, each processor which has a token
.
5 decidesindependently atrandombetweenkeepingthetokenorpassingittothe
0
next processor in the ring, choosing each with equal probability. All updates
4
occursimultaneously;iftwotokenscollide(becauseaprocessorkeepingitstoken
1
: receives another from the previous processor) then they annihilate each other.
v
Clearly the number of tokens will remain odd if this procedure is followed,
i
X and will never increase. It is easy to see that from any non-stable state with N
r processorsthereisaprobabilityofatleast2−2N/3thatanannihilationwilloccur
a within 1N steps. Consequently the algorithm stabilises almost surely; in fact
3
thetotaltimetakenhasfiniteexpectation. Thealgorithmwasintroducedin[4],
where Herman showed that the expected time to stabilisation is O(N2logN).
This bound wasimprovedto O(N2) independently, with differentconstants,by
McIver and Morgan [7]; Fribourg, Messika, and Picaronny [3]; and Nakata [8].
McIver and Morgan also conjectured that the expected time is maximised by
a starting state of three equally-spaced tokens; they show that this state has
expected time 4 N2 and that any other three-token configuration has a lower
27
expected time. Kiefer, Murawski, Ouaknine, Wachter and Worrell [5] extend
this result by showing that the probability of stabilisation by time t of any
three-token state is bounded by that of the equally-spaced three-token state
for each t. The conjecture of McIver and Morgan is supported by simulations
1
using the PRISMmodel-checkingsoftware[9]. When there maybe any number
of initial tokens, the best previous upper bound is about 0.64N2, by Kiefer,
Murawski, Ouaknine, Worrell and Zhang [6]. Here we give an upper bound of
about 0.156N2, which is comparatively close to the conjectured value of just
over 0.148N2. Our approach will differ from that of other papers in that we
will in fact prove exact results for the expectation of a different cost function.
A bound of 1N2 will immediately follow from the fact that our cost function
6
is at least the time taken, and this can be slightly improved by considering the
relationship between time and cost more carefully.
2 The steg function and inequality
For any odd m > 3 and variables a ,...a , a triple with even gaps is a term
1 m
which is the product of a , a and a for some i < j < k and k − j,k − i
i j k
both odd (so that the number of unused variables between consecutive used
variables is even). If we consider the variables as indexed by elements of Z
m
then a a a is a triple with even gaps if and only if a a a is.
i+1 j+1 k+1 i j k
Lemma 1. There are 1 m(m+1)(m−1) triples with even gaps on a ,...,a .
24 1 m
Proof. The triples with even gaps containing a are the triples of the form
1
a a a with j < k, j even, and k odd. There is a one-to-one correspondence
1 j k
between these terms and unordered pairs from the set {1,...,m+1}, in which
2
a a a corresponds to the pair j,k+1. Consequently there are (m+1)/2 such
1 j k 2 2 2
triples,and,sinceeachvariableis inthe samenumberoftripleswithevengaps,
there are m (m+1)/2 = 1 m(m+1)(m−1) triples with even ga(cid:0)ps. (cid:1)
3 2 24
Write ste(cid:0)g(a ,..(cid:1).a ) for the sum of all triples with even gaps (note that
1 m
whenm=1therearenotriples,andsosteg(a )≡0). Aswenotedabove,cyclic
1
shifts of the variables do not change which terms are included and so preserve
steg. The motivation for introducing this function is the following reduction
when one of the variables is set to 0; we choose the penultimate variable for
notational convenience.
Lemma 2. For m>3, if am−1 =0 then
steg(a1,...am)=steg(a1,...am−3,am−2+am).
Proof. In steg(a1,...am), all terms with am−1 vanish. No terms include both
am−2 and am apart from am−2am−1am, which vanishes. For i < j < m−2,
aiajam−2 is a triple with even gaps if and only if aiajam is.
Fori<j <k <m−2,a a a isatermintheLHSifandonlyifitis aterm
i j k
in the RHS. The remaining terms on the LHS occur in pairs, with aiajam−2
being such a term if and only if a a a is, and that pair of terms appears in
i j m
the LHS if and only if aiaj(am−2+am) is a term in the RHS, so the two sides
are equal.
We next give an upper bound on steg.
Theorem 3. For any odd m>3, if x ,x ,...,x are non-negative reals such
1 2 m
that x +x +···+x =1 then
1 2 m
1 1
steg(x ,...x )6 1− . (1)
1 m 24 m2
(cid:18) (cid:19)
2
Proof.
1 1 m(m+1)(m−1)
1− = ,
24 m2 24m3
(cid:18) (cid:19)
and, by Lemma 1, steg(x ,...x ) is the sum of m(m+1)(m−1)/24 terms.
1 m
Eachterm takes the value m−3 when x =x ···=x = 1, so the RHS of (1)
1 2 m m
is just the value taken at this point. We will prove that the maximum cannot
be attainedanywhereelse; since we are optimising a continuousfunction overa
compact set, this will be sufficient.
Weuseinductiononm;theresultholdsform=3bytheAM–GMinequality.
Ifm>3andxi =0foranyithensteg(x1,...xm)=steg(y1,...ym−2)forsome
non-negative y summing to 1, by Lemma 2. By the induction hypothesis,
i
1 1
steg(y1,...yl−2)6 24 1− (m−2)2
(cid:18) (cid:19)
1 1
< 1−
24 m2
(cid:18) (cid:19)
1 1
=steg ,..., ,
m m
(cid:18) (cid:19)
and so the maximum is not attained at any such point.
For any point with x >0 for all i, consider the difference
i
steg(x +δ,x ,x +δ,x ,x ,...x )−steg(x ,...x ).
1 2 3 4 5 m 1 m
Certainly any term which does not contain x or x will contribute nothing to
1 3
the difference. For j,k > 2, x x x is a triple with even gaps if and only if
1 j k
x x x is, and in that case x x x +x x x = (x +δ)x x +(x −δ)x x .
3 j k 1 j k 3 j k 1 j k 3 j k
So the only contributions to the difference are from terms of the form x x x ,
1 2 3
x x x (with k odd and greater than 3) and x x x (with k even).
1 2 k 2 3 k
(x +δ)x (x −δ)−x x x =δx x −δx x −δ2x ;
1 2 3 1 2 3 2 3 1 2 2
(x +δ)x x −x x x =δx x ;
1 2 k 1 2 k 2 k
and x (x −δ)x −x x x =−δx x ;
2 3 k 2 3 k 2 k
so modifying x and x in this way increases the value of the function by
1 3
δx [(x +x +x +···)−(x +x +x +···)]−δ2x . Since x ,x ,x >0, if
2 3 5 7 1 4 6 2 1 2 3
(x +x +x +···)6=(x +x +x +···)thenwecanchooseδ suchthatthisis
3 5 7 1 4 6
positive(andx +δ andx −δarenon-negative). Consequentlyifthemaximum
1 3
isattainedatx ,...x thenwemusthave(x +x +x +···)=(x +x +x +···).
1 l 3 5 7 1 4 6
Applyingthesameargumenttox andx showsthatadditionallywemusthave
2 4
(x +x +x +···) = (x +x +x +···), so x = x . Since the function is
2 5 7 1 4 6 2 3
unchanged by a cyclic shift of the variables, we must have x = x for every
i i+1
i∈Z , so the maximum can only be attained when x = 1 for every i.
m i m
Note that many ostensibly similar functions do not satisfy an analogous
inequality. If f(x ,...x ) is a sum of fewer than m3/27 triples then it is not
1 m
maximisedwhenallvariablesareequal,yetsteghasonlyslightlymorethanthis.
Also, if g(x ,...x ) is a sum of fewer than 3m3/64 triples which is maximised
1 m
when all variables are equal then it cannot include three triples from any set of
four variables.
3
Next we consider the average effect of a certain random transformation of
the variables on steg(a ,...,a ). Fix a subset S ⊂ [m] of even size 2h, and
1 m
write S ={i ,...i } with i <i <···<i . The move M+ adds 1 to a for
1 2h 1 2 2h S ik
−
each odd k and subtracts 1 from a for each even k. The move M does the
ik S
opposite. Note that M+ andM− bothleavea ...a unchanged; nevertheless,
∅ ∅ 1 m
we regardthem as different moves. There are then 2m possible moves: 2 moves
for each of the 2m−1 even subsets.
Theorem 4. Writing a=(a ,...,a ), let a˜ be the random vector obtained by
1 m
applying one of the 2m possible moves, chosen uniformly at random, to a. Then
m−1
E(steg(a˜))=steg(a)− a .
k
8 k
X
Proof. Note that
2m steg(a)−E(steg(a˜)) = 2steg(a)−steg M+a −steg M−a ,
S S
S
(cid:0) (cid:1) X(cid:0) (cid:0) (cid:1) (cid:0) (cid:1)(cid:1)
so it will be sufficient to show that the latter is equal to 2m−3(m−1) a .
k k
Fix a triple with even gaps a a a , ordering the variables so as to preserve
i j k
P
the cyclic ordering within Z (that is to say, regarded as members of the set
m
{1,...,m},eitheri<j <k,j <k <i,ork <i<j). Considerthecontribution
of this triple to 2steg(a)−steg M+a −steg M−a ; call this quantity βS .
S S ijk
The move M+ adds η to a , for some η ∈{−1,0,+1}, and M− subtracts the
S i i (cid:0) (cid:1)i (cid:0) (cid:1) S
same amount. Note that η is non-zero if and only if a ∈ S. With the same
i i
notation for j and k,
βS =2a a a −(a +η )(a +η )(a +η )−(a −η )(a −η )(a −η )
ijk i j k i i j j k k i i j j k k
=−2(η η a +η η a +η η a ).
i j k j k i k i j
We will calculate the sum βS ; note that
S ijk
P
2steg(a)−steg M+a −steg M−a = βS ,
S S ijk
!
S t.e.g. S
X(cid:0) (cid:0) (cid:1) (cid:0) (cid:1)(cid:1) X X
where the outer sum is taken over all triples with even gaps.
Write X = {i,...,j}\{i,j} (and similarly define X , X ); recall that
ij jk ki
we chose our ordering of the variables such that k ∈/ X . We distinguish three
ij
cases according to the number of i,j,k that S contains.
Case 1 If S contains at most one of i,j,k then βS =0.
ijk
Case 2 If S contains twoof i,j,k, without lossof generalityi,j ∈S but k ∈/ S,
then βS = −2a if η and η have the same sign, and βS = 2a if they
ijk k i j ijk k
have opposite signs. Write S = S ∩X and S = S ∩(X ∪X ), so that
1 ij 2 jk ki
S = i,j ∪S ∪S . Now |S | ≡ |S | mod 2; η and η have the same sign if
1 2 1 2 i j
and only if these cardinalities are both odd. A non-empty finite set has exactly
half its subsets odd, so if both X and X ∪X are non-empty there are
ij jk ki
2m−5 choices of S for which βS = −2a and 2m−5 for which βS = 2a . If
ijk k ijk k
one of the sets is empty then either i,j are consecutive mod m or j,k,i are
consecutivemod m; ineither ofthese cases eachofthe 2m−4 possible choicesof
S has βS =2a .
ijk k
Case 3 If i,j,k ∈ S then write S = S ∩X , S = S ∩X , S = S ∩X .
1 ij 2 jk 3 ki
|S |+|S |+|S | must be odd, so the possibilities are:
1 2 3
4
(i) all three are odd, when βS =2(−a −a −a );
ijk i j k
(ii) only |S | is odd, when βS =2(a +a −a );
1 ijk i j k
(iii) only |S | is odd, when βS =2(−a +a +a ); and
2 ijk i j k
(iv) only |S | is odd, when βS =2(a −a +a ).
3 ijk i j k
IfX ,X andX areallnon-emptytheneachpossibilityoccursin2m−6ways.
ij jk ki
IfonlyX isemptythen(i)and(ii)areimpossible;therearethenexactly2m−5
ij
waysforeachof(iii)and(iv)to occur. IfX andX areemptythen only(iv)
ij jk
is possible and it occurs in 2m−4 ways.
End of cases
Overall, then, a term which has no two variables consecutive contributes
nothing. A term where exactly two variables, without loss of generality a and
i
a , are consecutive, contributes 2m−3a from choices of S which include i and
j k
j but not k, and another 2m−3a from choices of S which include all three, for
k
a totalof 2m−2a . For eachk, there are m−5 triples with evengapswhich have
k 2
this property, so terms of this form contribute (m−5)2m−3 a in total.
k k
A term where all three variables are consecutive, i.e. one of the form
aj−1ajaj+1, contributes 2m−3(aj−1 −aj +aj+1) from choicPes of S which in-
clude j−1, j and j+1; 2m−3aj−1 from choices of S which include j and j+1
but not j −1; 2m−3a from choices of S which include j −1 and j but not
j+1
j+1; and 2m−3a from choices of S which include j −1 and j+1 but not j.
j
In total, then, such a term contributes 2m−3(2aj−1+2aj+1); since there is one
term of this form for each j, they contribute 4×2m−3 a in total. Thus
k k
P
βS =(m−1)2m−3 a ,
ijk k
!
t.e.g. S k
X X X
as required.
3 Relating Herman’s algorithm to steg
RunHerman’salgorithmwithN processorsfromsomestartingstateA(0) (with
an odd number of tokens), to get a sequence of states (A(t)) . For each time
t>0
step we incur a cost: if there are m tokens attime t the stepfromt to t+1 has
cost m−1. Also write c for the cost accumulated by time t, so that c =0 and
2 t 0
if A(t) has m tokens then c =c + m−1. Note that c is (with probability 1)
t+1 t 2 t
ultimately constant,since it stops increasingonce a stable state is reached. We
shall use the results provedin Section 2 to give an exact value for the expected
total cost.
With each state we associate a vector a(t), whose components are the dis-
tances between consecutive tokens. We define this more precisely as follows: if
A(t) hasmtokensthenwriteb(t) <···<b(t) fortheirpositions,thena(t) isthe
1 m
vector of m components with
b(t) −b(t) for i<m
a(t) = i+1 i
i (N +b(1t)−b(it) for i=m.
5
Note that a(t) =N, the total number of processors. Now define a sequence
i k
of variables (X ) as X =steg(a(t))+ 1Nc .
P t t>0 t 4 t
Theorem 5. The sequence (X ) is a martingale, in the sense that
t t>0
E(X |A(0),...,A(t))=X .
t+1 t
Proof. Herman’s algorithm is a Markov chain so
E(X |A(0),...,A(t))=E(X |A(t)).
t+1 t+1
Fix A(t) and suppose that it has m tokens at positions b(t) < ··· < b(t). Write
1 m
ˆb(t+1),...,ˆb(t+1)forthepositionsofthecorrespondingtokensattimet+1before
1 m
anyannihilationsoccur,anddefinegapsaˆ(t+1),...,aˆ(t+1)asabove. Whilea(t+1)
1 m
is not in general equal to aˆ(t+1), not only because some of the terms may be 0
but also because the latter will be cyclically shifted if a token has moved from
positionN toposition1,weclaimthatsteg(a(t+1))=steg(aˆ(t+1)). Cyclicshifts
do not change steg, so only the collisions need concern us. Note that it is not
possible for aˆ(t+1) and aˆ(t+1) to both be 0, as the former is only possible if the
i i+1
tokenatb(t)+1didnotmoveandthelatterifitdid. Ifthereisasinglecollision,
i
say aˆ(t+1) = 0, then a(t+1) = aˆ(t+1) for j < i−1, a(t+1) = aˆ(t+1)+aˆ(t+1), and
i j j i−1 i−1 i+1
a(t+1) = aˆ(t+1) for j > i+1. By Lemma 2, steg(a(t+1)) = steg(aˆ(t+1)). If
j j+2
therearemultiple collisionswemaycarryouteachoneinturn,andsteg willbe
preserved at each step.
Write S+ for the set of i such that the token at b(t) does not move but the
i
token at b(t) does, and S− for the set of i such that the token at b(t) moves
i+1 i
but the token at b(t) does not. Writing S =S+∪S−, the elements of S must
i+1
alternate between the two types and so |S+|=|S−|. Now
a(t)+1 if i∈S+
i
aˆ(it+1) =a(it)−1 if i∈S−
a(t) otherwise,
i
so, if S is non-empty, aˆ(t+1) =M+(a(t)) if the smallest member of S is in S+,
S
andaˆ(t+1) =M−(a(t))otherwise. Foreachnon-emptyevensetS there aretwo
S
possibilities for (S+,S−), and each of these uniquely determines which of the
tokens moves, so has probability 2−m. If S = ∅ then either all tokens move
or no tokens move; we may think of these as corresponding to M+ and M−
∅ ∅
respectively. So aˆ(t+1) is obtained from a(t) by applying one of the 2m moves,
and each is equally likely. Applying Theorem 4,
N
E(X |A(t))=E(steg(a(t+1))|A(t))+ E(c |A(t))
t+1 t+1
4
N m−1
=E(steg(aˆ(t+1))|a(t))+ +c
t
4 2
(cid:18) (cid:19)
m−1 N m−1
=steg(a(t))− N + +c
t
8 4 2
(cid:18) (cid:19)
=X ,
t
as required.
6
We now use the fact that (X ) is a martingale to deduce the exact value of
t
the expected total cost.
Theorem6. TheexpectedtotalcoststartingfromthestateA(0) is 4 steg(a(0)).
N
Proof. LetT be the earliesttime forwhichA(T) is stable. T isastopping time,
and,asweobservedinSection1,ithasfiniteexpectation. Further,ifa ,...,a
1 m
are non-negative with sum N then
steg(a ,...,a )=N3steg(a /N,...,a /N)
1 m 1 m
1 1
6N3 1−
24 m2
(cid:18) (cid:19)
N3
< ,
24
by Theorem 3, and steg(a ,...,a )>0, so
1 m
|X −X |6|steg(a(t+1))−steg(a(t))|+ 1N(c −c )
t+1 t 4 t+1 t
N3 N2
< + .
24 8
Sincethestoppingtimehasfiniteexpectationandthereisaglobalboundonthe
differencebetweensuccessivevariables,theOptionalStoppingTheorem(see,for
example, [11], p. 100) applies and so E(X )=X . Consequently
T 0
N
E(steg(a(T)))+ E(c )=steg(a(0)).
T
4
Since A(T) is stable, a(T) has only one component, and so steg(a(T))=0 (with
probability 1); also c is the total cost since no further cost is incurred after
T
time T. So we have
4
E(c )= steg(a(0)),
T
N
as required.
Since steg(a(0)) = N3steg(a(0)/N,...,a(0)/N), applying Theorem 3 gives
1 m
the following result.
Corollary 7. The expected total cost from any starting state with 2s+1 tokens
is at most 1−(2s+1)−2 N2/6.
We are(cid:0)now ready to p(cid:1)rove our main result.
Theorem 8. The expected time to stabilisation for Herman’s algorithm from
any start state on N processors is less than (π2−8)N2/12.
Proof. Fix a start state A with 2r + 1 tokens, and let the random variable
T be the total time to stabilisation. For each s > 1, write A for the first
s
configurationwith at most 2s+1 tokens and C for the cost accumulated after
s
that point. Note that A =A =A and C =C for every s>r. Now C =0
s r s r 0
7
and Cs−Cs−1 =0 for all s>r, so for any t>r,
r
T = 1s(Cs−Cs−1)
s=1
X
t
= 1s(Cs−Cs−1)
s=1
X
t
= 1 − 1 C + 1 C
s s+1 s t+1 t
Xs=1(cid:16) (cid:17)
t
= 1 C + 1 C ,.
s(s+1) s t+1 t
s=1
X
Since
lim 1 C =0,
t→∞t+1 t
it follows that
∞
1 C =T.
s(s+1) s
s=1
X
Also,
E(C )=E(E(C |A ))
s s s
1 N2
6 1− ,
(2s+1)2 6
(cid:18) (cid:19)
by Corollary 7, and so
∞
1 1 N2
E(T)6 1−
s(s+1) (2s+1)2 6
s=1 (cid:18) (cid:19)
X
∞
1 4s2+4s N2
=
s(s+1) (2s+1)2 6
s=1 (cid:18) (cid:19)
X
∞
2N2 1
=
3 (2s+1)2
s=1
X
(π2−8)
= N2,
12
as required.
4 Acknowledgements
The author acknowledges support from the European Union through funding
under FP7–ICT–2011–8project HIERATIC (316705).
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8
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