1

Reduction: A Method of Proving Properties of Parallel Programs

By Richard J. LiptonPresented at the Second ACM Symposium of Principles of Programming Languages, Palo

Alto, Calif. 1975

2

Motivation Prove that a Parallel Program does not halt

acq(this) S1

X S2

j=bal S3

Y S4

bal=j+n S5

Z S6

rel(this) S7 S0

3

X T1

Y T2

acq(this) T3

j=bal S4

bal=j+n S5

rel(this) T6

Z S7 S0

acq(this) S1

X S2

j=bal S3

Y S4

bal=j+n S5

Z S6

rel(this) S7 S0

Motivation Prove that a Parallel Program does not halt

4

Goal

• When proving that a system of processes has a given property it is often convenient to assume that a routine is atomic

• The paper presents a reduction that preserves basic properties such as halting.

• Thus correctness proofs of a system of processes can often be greatly simplified

5

Definitions

6

uninterruptible

• A statement is atomic provided it is never interleaved with the rest of the program.– For instance: A statement might be the three

actions:

Assuming it is uninterruptible reduces it to the single action :

X T1

Y T2𝒙→𝒓 T3

inc r S4𝒓→ 𝒙 S5

Z S7 S0

7

Reduction of P by R

• Reduction of P by R is defined to be the parallel program obtained from P by reducing R to one uninterruptible action.

• Notation: P/R• Two ways in which the reduced Q=P/R program

is simpler than P: 1) Q has fewer actions than P2) Assertions about Q are often simpler than

assertions about P.

8

parbegin…parend

• parbegin parend is to interleave the statements in some arbitrary order until no further execution is possible.

• The statement of each form a distinct process

9

A computation

• A computation is a sequence of statements such that is executed first, then is executed, and so on until the last statement is executed. Since an may be a compound statement, m>k is possible.

• For example if is:

then might be the statement or the statement or even “part” of these statements.

10

Indivisible statement

• Notation:– We assume S has a single entry and a single exit.

• The semantics of are:– In a given state of the parallel program, can execute

provided in this state control is ready to enter S and after S is applied control has left S

– In a given state of the parallel program, the effect of the applying provided it can execute, is the same as that of S.

• The key to the definition of is that we can never apply it when we cannot fully complete its execution

11

P(a), V(a)

• P(a) = • V(a) = • Without closing in brackets is it possible to

“lose counts”.– Example:

value of a can be 1 or 2.

12

• is a computation provided is a computation and can execute in the state that results after is executed.

When is a computation?

13

• Example:

When is a computation? – cont.

a=0 T1

B:V(a) T2 S0

a=0 T1

B:V(a) T2

A:P(a) T3 S0

a=0 T1

A:P(a) T2

B:V(a) T3 S0

14

Halt

• Intuitively halt is like deadlock• Usually want to show that a program does not

halt.• A program halts if there is some computation

such that is not a computation for all statements f.

15

What is the relationship between P and P/S?

16

P/S halts iff P halts?• This is false.• Consider:

• This program halts: Let both repeat’s execute their first P’s; then

a=b=0 and the program has halted.

17

P/S halts iff P halts? – cont.• Now consider the following program P/S:

• leaves both a and b fixed.

18

Why is the assertion false?• It is possible to enter S and not to ever be able

to leave it.• This leads to one restriction on statement S:• (R1) If a statement S is ever entered, then it

must be possible eventually to exit S.

19

Is (R1) enough?• No.• Consider:

• The program halts.• Also, the statement satisfies (R1)

20

Is (R1) enough? – cont.• The program P/S is:

• always sets y to 1.• This program does not halt.

21

Why is the assertion false?• This example fails to satisfy assertion because

the effect and when separated and when together is not the same.

• This observation leads to further restriction:• (R2) The effect of the statement in S when

together and separated must be the same.

22

Right Moverb c

c b

b is right mover

23

Right MoverACQ(l,t) c

c ACQ(l,t)

24

Left Moverb c

c b

c is left mover

25

Left Moverb REL(l,t)

bREL(l,t)

26

Right and Left Moversacq(this) S1

X S2

j=bal S3

Y S4

bal=j+n S5

Z S6

rel(this) S7 S0

Red thread holds lock Blue thread does not hold lock operation y does not access balance

( assuming balance protected by lock) operations commute

27

Right and Left Moversacq(this) S1

X S2

j=bal S3

Y S4

bal=j+n S5

Z S6

rel(this) S7 S0

acq(this) S1

X S2

Y T3

j=bal S4

bal=j+n S5

Z S6

rel(this) S7 S0

Red thread holds lock after acquire operation x does not modify lock operations commute

28

Right and Left Moversacq(this) S1

X S2

j=bal S3

Y S4

bal=j+n S5

Z S6

rel(this) S7 S0

acq(this) S1

X S2

Y T3

j=bal S4

bal=j+n S5

Z S6

rel(this) S7 S0

X T1

acq(this) S2

Y T3

j=bal S4

bal=j+n S5

Z S6

rel(this) S7 S0

29

Right and Left Moversacq(this) S1

X S2

j=bal S3

Y S4

bal=j+n S5

Z S6

rel(this) S7 S0

acq(this) S1

X S2

Y T3

j=bal S4

bal=j+n S5

Z S6

rel(this) S7 S0

X T1

acq(this) S2

Y T3

j=bal S4

bal=j+n S5

Z S6

rel(this) S7 S0

X T1

Y T2

acq(this) T3

j=bal S4

bal=j+n S5

Z S6

rel(this) S7 S0

30

Right and Left Movers

X T1

Y T2

acq(this) T3

j=bal S4

bal=j+n S5

rel(this) T6

Z S7 S0

acq(this) S1

X S2

j=bal S3

Y S4

bal=j+n S5

Z S6

rel(this) S7 S0

acq(this) S1

X S2

Y T3

j=bal S4

bal=j+n S5

Z S6

rel(this) S7 S0

X T1

acq(this) S2

Y T3

j=bal S4

bal=j+n S5

Z S6

rel(this) S7 S0

X T1

Y T2

acq(this) T3

j=bal S4

bal=j+n S5

Z S6

rel(this) S7 S0

31

Right and left movers

• f is a right mover provided – for any a computation where f and h lie in different

processes, then is also a computation– The values of all the program variables in and are the

same.• f is a left mover provided– For any a computation where h and g lie in different

processes, then is also a computation – The values of all the program variables in and are the

same.

32

PV parallel program

• A program is a PV parallel program provided there is a distinguished subset of the program variables called semaphores with integer values such that they can be used only in either or .

33

D-reduction • Replacing with is a D-reduction provided, for

some i, are right movers and are left movers ( is unconstrained) and each can always execute.

34

Theorem 1.In any PV parallel program all P(a)’s are right

movers, and all V(a)’s are left movers.Theorem 2.

Suppose that S is a D-reduction in P. Then P halts iff P/S halts.

35

Proof of Theorem 2

• If P/S halts then P halts • This is true because for every scheduling in

which P/S halts, p halts since the same scheduling can be applied to it.

36

Proof of Theorem 2

• If P halts then P/S halts • Proof outline:– Assume P halts– Let be a computation that halts in P.– Assume that – Construct a computation such that all the

program variables agree after and are executed, and always occur atomically in .

– Assume that there are no goto’s in

37

Proof of Theorem 2

• LEMMA 1. Suppose that is a computation in P with i>1. Then where no statement from the process of is in .

• This follows because S has a single entry and no goto’s.

38

Proof of Theorem 2

• LEMMA 2. Suppose that is a computation that halts in P with i<n. Then where no statement from the process of is in .

• This follows because:– If any f occurs in where f is in the process of then the

first such f must be .– Assume that no such f is in . In control must be ready

to enter ; therefore is a computation (because by definition D-reduction can always execute), which is a contradiction because halts.

39

Proof of Theorem 2

• If no is in

then is already in the desired form. (let )

A S1

X S2

𝑩 S3

Y S4

C S5

Z S6

D S7 S0

40

Proof of Theorem 2Thus suppose that some is in .

𝝀 S1

𝑺𝒊−𝟏 S2

𝝁 S3

𝑺𝒊 S4

𝜷 ′ S5 S0

S1

𝜶 ′ S2

𝑺𝒊 S3

𝜷 ′ S4

𝝀 S1 S2

𝝁 S3

𝑺𝒊 S4

𝝀 S5

𝑺𝒊+𝟏 S6

𝝁 ′ S7 S0

By definition of D-reduction𝝀 S1 S2

𝑺𝟏 S3

… S4

𝑺𝒏 S5

𝝀 S6

𝝁 ′ S7 S0

𝝀 S1 S2

𝝁 S3

𝑺𝟐 S4

… S5

𝑺𝒏 S6

𝝁 ′ S7 S0

41

Proof of Theorem 2

• This can be repeated to for the desired computation . Now is a computation where no is in any and and agree on all program variables.

• If halts in P\S then the theorem is proved. • Assume that does not halt in P\S, and that is a

computation in P\S.• Then is a computation in P, since and agree on

all program variables. This is a contradiction.

42

Applications

43

Example 1

• By theorem 1 and 2, the aforementioned halts iff the following program halts:

44

Example 1 – cont.

• Therefore Example 1 halts iff the following halts:

45

Example 1 – cont.• Once again theorem 1 and 2 can be applied;

hence the aforementioned halts iff the following halts:

46

Example 1 – cont.

• Therefore the aforementioned halts iff the following halts:

• This program never halts! Thus Example 1 never halts

47

Example 2

• After applying theorem 1 and 2, Example 2 halts iff the following halts:

48

Example 2 – cont.• The effect of is to decrement a by 1 and

increment b by 1.• The effect of is to decrement b by 1 and

increment a by 1.• Thus a+b is conserved and is always equal to

N.

49

Example 2 – cont.• can execute iff a>0 and can execute iff b>0. • Since a+b=N>0, it is not possible for the

program to halt. • Hence Example 2 does not halt.

50

Conclusion

• In a wide number of nontrivial instances reduction preserves important properties.

• Reduction aids in correctness proof• Note theorem 2 proved that for every computation

that halts in P there’s a computation that halts in P/S that agrees on all program variables.

• Thus if S is a D-reduction the final states of P equal the final states of P/S

• D-reduction then preserves any property that depends only on a program’s final state.

51

Questions?