Cut-off theorems for deadlocks and...

Cut-off theorems for deadlocks and serializability

Lisbeth Fajstrup

Department of MathematicsAalborg University

Bedlewo 2013

Lisbeth Fajstrup (AAU) Cut-Off Theorems July 2013 1 / 42

Cut-Off Theorems

General

Property P holds for all n ∈ N if and only P holds for n ≤ M

Think of n as a dimension.Here: Given a computer program T , T n means n copies of T run inparallel.The properties P is deadlock free in one theorem and serializable in theother.


Parallel program: Dining Philosophers


Parallel programs - several models

A non-parallel program is modelled as a graph. Loops, branches.Parallel programs interact. Share resources. May get conflicts. Severalmodels.

Petri Nets

Interleaving semantics

Here: Locks.


The Swiss flag - two dining philosophers.

T1 = PaPbVbVa, T2 = PbPaVaVb

Pa Pb Vb Va

Pb

Pa

Va

Vb

A

B

T1

T2


PV -programs - controlling concurrency through locks

A set of shared resources R - memory, printers,...

A capacity function κ : R → N.

PV programs p are defined by the grammar

p ::= Pa | Va | p.p | p|p | p + p | p∗

Pa is a request to access the resource a, if granted acces, lock it. Va

releases the resource.

A program without the parallel operator is a thread.

Resource r may be locked by at most κ(r) threads at a time.

At a final point, the program has released all resources.

At an initial point, no resources are locked


GeometricallyOne thread is represented by a graph. n threads in parallel is representedby a product of n graphs with “holes” where a resource is locked above itscapacity.

T1 = T2 = Pa.Pb.Va.Pa.Vb.Va κ ≡ 1

Pa Pb Va Pa Vb Va

Pa

Pb

Va

Pa

Vb

Va

A

A

A

A

B

T1

T2


An execution is a directed path

Pa Pb Va Pa Vb Va

Pa

Pb

Va

Pa

Vb

Va

A

A

A

A

B

T1

T2


Deadlock - no time-directed paths leave the point

Pa Pb Va Pa Vb Va

Pa

Pb

Va

Pa

Vb

Va

A

A

A

A

B

T1

T2

The red region is unsafe. No directed path leaves it. The red balls aredeadlock points.


A few words about directed topologyTopological spaces with direction

Definition

Objects of dTop are d-Spaces: (X , ~P(X )) where X ∈ Top, ~P(X ) ⊆ X I ,the dipaths. ~P(X ) is

closed under concatenation,

contains the constant paths

closed under subpath, i.e., composition with f : I → I increasing butnot necessarily surjective.

A d-map f : X → Y is a continuous map satisfyingγ ∈ ~P(X )⇒ f ◦ γ ∈ ~P(Y )

Main examples: Directed paths are paths which are locally increasing wrt.a reasonable local order structure.

In computer science:

X is a state space. ~P(X ) are (partial) executions.


Equivalence of executions

Definition

Two dipaths γ0, γ1 ∈ ~P(X )(p, q) are dihomotopic, if there is a dimapH : I ×~I → X s.t. H(0, t) = γ0(t), H(1, t) = γ1(t), H(I , 0) = p,H(I , 1) = q. (Dipaths in I are constant. Dipaths in ~I are non-decreasing)

The following are equivalent.

Execution paths γ1, γ2 : ~I → X are dihomotopic

They are in the same component of the path space ~P(X )(0, 1)(Compact-Open topology)

They are in the same component of the trace space ~T (X )(0, 1)(Paths modulo directed reparametrization)

Raussens algorithm

Calculates a prod-simplicial model of ~T (X )(0, 1), when X = I n \ F , F is aunion of n-rectangles. RedHom then calculates homology.


Definition

A state x = (x1, . . . , xn) ∈ T1× T2× · · · × Tn is a deadlock if

The only dipaths starting in x are constant.

x is reachable: There is a dipath from 0 to x

x 6= > - not all Ti have finished

If x is a deadlock, then xi = > or xi = Pr(i) and κ(r(i)) other threadshold a lock on r(i) at x


A cut-off theorem for deadlocks

Theorem 1

Let T be a PV thread accessing resources R with capacity κ : R → N.Let T n be n copies of T run in parallel.T n is deadlock free for all n if and only if TM is deadlock free, whereM = Σr∈Rκ(r).


The bound M is sharp.

Theorem 2

Given R = {r1, . . . , rk} with capacity κ : R → N, the threadT = Pr1Pr2Vr1Pr3Vr2 . . .PrkVrk−1Pr1VrkVr1 satisfies:

There is a deadlock in TM (and hence for all n ≥ M)

There are no deadlocks in T n for n ≤ M


The bound M is sharp.

Theorem 2

Given R = {r1, . . . , rk} with capacity κ : R → N, the threadT = Pr1Pr2Vr1Pr3Vr2 . . .PrkVrk−1Pr1VrkVr1 satisfies:

There is a deadlock in TM (and hence for all n ≥ M)

There are no deadlocks in T n for n ≤ M

Proof: The deadlock is x = (x1, . . . , x1, x2, . . . , x2, . . . , xk . . . xk) where fori 6= 1, xi = Pri and xi is repeated κ(ri−1) times. x1 is the last Pr1 and isrepeated κ(rk) times. All resources r are locked κ(r) times.Hence x is a deadlock and 6= >. (Need to check reachability and nodeadlocks in lower dimensions)


Example. When κ ≡ 1

x = (x1, x2, . . . , xk) and M = k :

T1 requests r1 and has a lock on rk

Ti requests ri and has a lock on ri+1

There are no deadlocks in T n for n < M: If y is a deadlock, then there isa yi1 = Pr(i1) and another yi2 with a lock on r(i1), so yi2 = Pr(i1 + 1)(or the last P(r1) if r(i1) = rk .Hence, y is a permutation of (x,>,>, . . . ,>).For κ 6= 1 the argument is similar. Still need to see that x is reachable.


ExampleT = PaPbVaPcVbPaVcVa, κ ≡ 1. T 3 has a deadlock at(6, 2, 4), (2, 4, 6), (6, 4, 2), (2, 6, 4), (4, 2, 6), (4, 6, 2)

Pa Pb Va Pc Vb Pa Vc Va

Pa

Pb

Va

Pc

Vb

Pa

Vc

Va

A

A

A

A

B

C

T1

T2


Proof of Theorem 2 continued

T = Pr1Pr2Vr1Pr3Vr2 . . .PrkVrk−1Pr1VrkVr1

x = (

κ(rk )︷︸︸︷x1, . . . , x1,

κ(r1)︷︸︸︷x2, . . . , x2, . . . ,

κ(rk−1)︷︸︸︷xk . . . xk) = (x1, x2, . . . , xk) is a deadlock:

For i 6= 1, xi = Pri and xi is repeated κ(ri−1) times. Holds κ(ri−1)locks on ri−1

x1 is the last Pr1 and is repeated κ(rk) times. Holds κ(rk) locks on rk


x is reachable

(T = Pr1Pr2Vr1Pr3Vr2 . . .PrkVrk−1Pr1VrkVr1)a dipath from 0 to x is composed of the pieces:

γ0 : 0→ (x1, 0) serially - one coordinate at a time. OBS, now holdκ(rk) locks on rk

γ1 : (x1, 0)→ (x1, 0, xk) one coordinate at a time. Now also κ(rk−1)locks on rk−1

γj : (x1, 0, xk−j+2, . . . , xk)→ (x1, 0, xk−j+1, . . . , xk). Now κ(ri ) lockson rk−j , . . . , rk .

For γ0: ρi (γ0(t)) ≤ 1 for i 6= k ρk(γ0(t)) ≤ κ(rk).

ρi (γ1(t)) ≤ 1 for i 6= k − 1, k. ρk(γ1(t)) = k . ρk−1(γ1(t)) ≤ κ(rk−1).

ρi (γj(t)) ≤ 1 for i ≤ k − j + 1. ρi (γj(t)) = κ(ri ) for i ≥ k − j + 2.ρk−j+1(γj(t)) ≤ κ(rk−j+1).


x is reachable









x is reachable





For j = 2, . . . , k − 1.





x is reachable(T = Pr1Pr2Vr1Pr3Vr2 . . .PrkVrk−1Pr1VrkVr1)a dipath from 0 to x is composed of the pieces:




For j = 2, . . . , k − 1.Need: No resource is locked above its capacity along γ. Let ρi (y) be thenumber of locks held on ri at y ∈ TM .





x is reachable(T = Pr1Pr2Vr1Pr3Vr2 . . .PrkVrk−1Pr1VrkVr1)a dipath from 0 to x is composed of the pieces:




For j = 2, . . . , k − 1.Need: No resource is locked above its capacity along γ. Let ρi (y) be thenumber of locks held on ri at y ∈ TM .





Proof of theorem 1

Theorem 1


Prove: If there is a deadlock in T n, then there is a deadlock in TM .

For n < M: Suppose x = (x1, . . . , xn) is a (reachable) deadlock in T n.Then for all i, either xi = Prj(i) and ρj(i)(x) = κ(rj(i)) or xi = >.

Then x̃ = (x1, . . . , xn,>, . . . ,>) ∈ TM is a (reachable) deadlock in TM ,since

ρi (x̃) = ρi (x) for all i

x̃ is reachable: There is a serial path to (0,>, . . . ,>). Compose with(γ(t),>, . . . ,>), where γ(t) : 0→ x ∈ T n


Proof of theorem 1

Theorem 1


Prove: If there is a deadlock in T n, then there is a deadlock in TM .For n < M: Suppose x = (x1, . . . , xn) is a (reachable) deadlock in T n.Then for all i, either xi = Prj(i) and ρj(i)(x) = κ(rj(i)) or xi = >.





Proof of theorem 1

Theorem 1


Prove: If there is a deadlock in T n, then there is a deadlock in TM .For n < M: Suppose x = (x1, . . . , xn) is a (reachable) deadlock in T n.Then for all i, either xi = Prj(i) and ρj(i)(x) = κ(rj(i)) or xi = >.





Proof of Theorem 1. n > M

x = (x1, . . . , xn) is a (reachable) deadlock in T n.

Construct the directed wait for graph G (x) = (V ,E ).

V = {x1, . . . , xn}

There is an edge E (xi , xk) if xi = Prj and ρj(xk) = 1.Properties of G (x)

If xi = >, then the vertex xi is isolated.

If xi = Prj , then the vertex xi has κ(rj) outgoing edges.

Hence, there are circuits in G (x): Start a walk at a non-isolatedvertex. If xj is the target of an edge, then xj 6= >, so the walkcontinues. G (x) is finite, so there is a circuit L.

Let ↑ L be the vertices reachable from L.Let x̃ = (xi1 , xi2 , . . . , xim) where xij are all the vertices in ↑ L



x = (x1, . . . , xn) is a (reachable) deadlock in T n.Construct the directed wait for graph G (x) = (V ,E ).

V = {x1, . . . , xn}

There is an edge E (xi , xk) if xi = Prj and ρj(xk) = 1.

Properties of G (x)








V = {x1, . . . , xn}









V = {x1, . . . , xn}







x̃ = (xi1 , xi2 , . . . , xim) where xij are all the vertices in ↑ L, future of a circuitin the wait-for-graph.Claim:

1 x̃ is a (reachable) deadlock.

2 m ≤ M

Reachability: Let γ : 0→ x in T n. The restriction to Ti1, . . .Tim is adipath 0→ x̃


The Wait for Graph - the loop

xi1

xi2 xi3

xi4

xi5

xil


The restricted Wait for Graph - future of the loop

xi1

xi2 xi3

xi4

xi5

xil

κ(r(il)) locks

κ(r(i1)) locks

κ(r(i1)) edges

xij 6= >, so xij = P(r(ij)). In G (x), xij = P(r(ij)) had κ(r(ij)) outgoingedges. It still has in G (x̃), since ↑ L contains all targets.

Hence, x̃ is a deadlock.


m ≤ M

In G (x̃), all the m vertices are targets. Hence, they hold a lock on aresource.

The maximal number of locks at an allowed state is M = Σr∈Rκ(r)

Hence, there are at most M vertices in G (x̃) (Less, if some xi hold alock on more than one resource.)


Now to something different: Serializability

Pa Pb Va Pa Vb Va

Pa

Pb

Va

Pa

Vb

Va

A

A

A

A

B

T1

T2

Two executions up to dihomotopy. Equivalent to the serial executionsT1.T2 and T2.T1


SerializabilityT = PaPbVaPcVbPaVcVa is not serializable.

Pa Pb Va Pc Vb Pa Vc Va

Pa

Pb

Va

Pc

Vb

Pa

Vc

Va

A

A

A

A

B

C

T1

T2

4 Executions up to dihomotopy. Two serial executions (green). Twonon-serializable executions (red).


Serializability - a cut-off theorem

Definition

T1|T2 . . . |Tn is serializable if all execution paths are dihomotopic to aserial execution Ti1.Ti2....Tin.

Theorem 3

Let T be a PV -thread accessing resources R all of capacity 1.Let T n be n copies of T run in parallel. Then T n is serializable if and onlyif T 2 is serializable.


In general pairwise serializability is not enough

Example

Let T1 = PcVcPaVa, T2 = PcVcPbVb, T3 = PaVaPbVb,Each pair Ti ,Tj shares one resource.

T1

T2

T3


Schedules as in Raussens Trace algorithm

When κ = 1

All conflict n-rectangles are ×nk=1Ik , Ik = I except for two directions

Ii =]ai , bi [, Ij =]aj , bj [

One choice at a given such n-rectangle “above or below” in theij-plane - i waits or j waits.


T = PaVa, T 3

T

T

T

3! serial executions, pairwise inequivalent.


T n

Each pair of locked intervals ]a1, b1[, ]a2, b2[ in T gives n · (n − 1)forbidden n-rectangles in T n

Ri ,j = ×nk=1Ik

Ii =]a1, b1[, Ij =]a2, b2[, Ik = I otherwise.

The diagonal rectangles in T 2 give ×nk=1Ik Ii =]a, b[, Ij =]a, b[,

Ik = I , n·(n−1)2 n-rectangles. Intersecting in ]a, b[×]a, b[× · · ·×]a, b[.

If T is non-trivial, all the n! serial executions are non-equivalent, since

They have different schedules wrt. the n·(n−1)2 n-rectangles induced

by just one PV -interval.

Equivalently: Their projections to at least one of the T 2 arenon-equivalent.


The symmetric case T 2

k instances of Pa,Va gives k2 forbidden rectangles in T 2

T = PaPbVaPcVbPaVcVa

PaPbVaPc VbPaVc Va

Pa

Pb

Va

Pc

Vb

Pa

Vc

Va

A

A

A

A

B

C

T1

T2

4 rectangles labelled A.


T 2 serializability

If choice at one rectangle implies choice at all other rectangles -below all or above all rectangles

Equivalently: The closure of the forbidden area under addingunreachable and unsafe areas is connected. (Yannakakis,Papadimitriou, Kung,1979)

OBS: Two phase locking is not required.

PaPbVaPcVbPdVcVd

PaPbVaPcVbPdVcVd

A

D

B

C

T1

T2

PaPbVa PaVbVa

Pa

Pb

Va

Pa

Vb

Va

A

A

A

A

B

T1

T2


T 2 serializability

If choice at one rectangle implies choice at all other rectangles -below all or above all rectangles

Equivalently: The closure of the forbidden area under addingunreachable and unsafe areas is connected. (Yannakakis,Papadimitriou, Kung,1979)

OBS: Two phase locking is not required.

PaPbVaPcVbPdVcVd

PaPbVaPcVbPdVcVd

A

D

B

C

T1

T2

A

A

A

A

B

Pd Pd PdVd Vd Vd

Pd

Pd

Pd

Vd

Vd

Vd

D D

DD

D

D DD

D

T1

T2


Proof of Theorem 3

Suppose T 2 is serializable, then

1 Let ]ar , br [ correspond to a lock Pr ,Vr . There are n! schedules forthe rectangles {×n

k=1Ik , Ik =]ar , br [ for k = i , j , i < j ∈ [1 : n]} -corresponding to the serial executions.

2 Fix i , j . There are two schedules - i last for all or j last for all therectangles {I × I × · · ·×]ars , b

rs [×I · · · ×]art , b

rt [×I × · · · × I} where

]ars , brs [×]art , b

rt [ are forbidden rectangles in T 2

3 Choose one of the n! schedules for 1); that fixes all schedules in 2).Hence, there are n! inequivalent schedules.

4 These are all realized by serial executions.


How complicated does T n get?

Answer: Very!

1 Let T1, . . . ,Tk be threads. Compose to get T = T1.T2. . . . .Tk.

2 T1× . . .×Tk ↪−→ T k at [0, 1]× [1, 2]× . . .× [k − 1, k] (and at the k!permutations)

3 All paths on the integer 1-skeleton are allowed (serial execution ofTi).(Ti holds no locks at ⊥ and >)

4 Hence, the lower point (0, 1, 2, . . . , k − 1) is reachable from 0 andthere is a path from (1, 2, . . . , k) to k.

5 So deadlocks, non-serializable paths etc. in T1× . . .× Tk arerealized in T k .


Serializability for κ > 1

There are non-serializable executions.

All serial executions are equivalent (all boundary 2-cells are allowed)

The scheduling algorithm calculates components of the trace space.

More than one component ⇔ non-serializable.


The scheduling algorithm

X = I n \ F , F = ∪li=1Ri

R i =]ai1, bi1[× . . .×]ain, b

in[

Use the Nerve Lemma for a covering of ~P(X )(0, 1), the space of dipaths.Output: A prod-simplicial model of ~P(X )(0, 1). Then RedHom calculateshomology.


X = I n \ F , F = ∪li=1Ri

R i =]ai1, bi1[× . . .×]ain, b

in[

A schedule is an l-tuple J = (J1, . . . , Jl) where Ji ⊂ {1 . . . , n}. Thethreads/directions/coordinates which wait till at least one (other) haspassed R i .J defines a subset of ~P(X )(0, 1):

Extend R i according to Ji - if j ∈ Ji , thenR ij = [0, bi1[× . . .×]aij , b

ij [× . . .× [0, bin[.

FJ = ∪li=1 ∪j∈Ji R ij

XJ = I n \ FJ~P(XJ)(0, 1) is an open subset of ~P(X )(0, 1), either empty or contractible(when Ji 6= ∅ for all i).~P(XJ)(0, 1) ∩ ~P(XJ′)(0, 1) = ~P(XJ∪J′)(0, 1)


A non-serializable κ = 2 example.


Schedule for “two wedges”

J = ({3}, {3}, {2}, {1})Paths following this schedule go through the “hole”.

J1 = ({1, 3}, {3}, {2}, {1}) and J2 = ({3}, {2, 3}, {2}, {1}) also“through the hole”.


















Non-serializability for two wedges



Non-serializable. Since Ji < J̃ ⇒ ~P(XJ̃) = ∅Lisbeth Fajstrup (AAU) Cut-Off Theorems July 2013 41 / 42

More to do

Serializability for κ > 1 - Guess: There is a homological obstruction .This needs to be described specifically for the symmetric case.

As of last night: Proof that κ = 2 has cut-off at n = 3, but thatneeds checking in daylight....

Cut-off for other properties - linearizability?


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