1 Consensus Hierarchy Part 2
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1
Consensus HierarchyPart 2
2
FIFO (Queue)
FIFO Object
),( xQenq
Q
headtail
x
a
aQdeq )( x
b
ac b
c
bc
3
Special case: empty queue
)(Qdeq
tail head
4
Theorem: A FIFO object has consensusnumber 2
Proof:1. We can solve wait-free consensus using FIFO (and read/write) objects for 2 processes
2. We cannot solve wait-free consensus using FIFO (and read/write) variables for 3 or more processes
5
A Wait-Free Consensus algorithmfor 2 processors using a FIFO object(and read/write objects)
Initially:
tail head
0
Proof-Part 1
6
Shared Memory
Queue
Other Variables
0tail
head
7
0x 1x0p 1p
Local variables
}1,0{, ii yx
:initial values for the consensus problem
0y 1y
ix
iy:resulting values for the consensus problem
8
Shared Memory
Queue
Other Variables
0tail
head
1x
0xPrefer[0]Prefer[1]
Initial values
9
Code for processor ip }1,0{i
If then else
0)( Qdeq
ii xy ]1[ ipreferyi
//am I the first?
//yes, choose my value
//no, choose the other processor’s value
Note: the algorithm uses a FIFO object and read/write objects
ixiprefer ][
10
Example execution:
Shared Memory
Queue0tail
head
01 x
10 xPrefer[0]Prefer[1]
10 x
01 x
0p
1p
0y
1y
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Suppose that accesses first the queue
Shared Memory
Queuetail
head
01 x
10 xPrefer[0]Prefer[1]
10 x
01 x
0p
1p
0y
01 y
1p
)(0 Qdeq
decides on its own value
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Suppose that accesses second
Shared Memory
Queuetail
head
01 x
10 xPrefer[0]Prefer[1]
10 x
01 x
0p
1p
00 y
01 y
2p
decides on the other processor’s value
)(Qdeq
Consensus Reached
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There is no wait-free consensusalgorithm using only FIFO(and read-write) objects for 3n
Consider three processors(the same proof generalizes to more)
Proof-Part 2
We will prove:
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There is a bivalent initial configuration(we proved it before)
We will show that every bivalent configurationhas a processor which is not critical
Therefore, we can construct an infiniteexecution with bivalent configurations whereconsensus is never reached
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Assume for contradiction that all processors are critical
C
0C
1C
1nC
bivalent0p
1p
2p
univalent
univalent
univalent
Possibleexecutions
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bivalent
valent
It cannot be that all have the same valence
v
)1 or 0( v
valentv
valentv
valentv
Contradiction
C
0C
1C
2C
0p
1p
2p
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There must exist two processors with different valences
valent-0
valent-1C
0C
1C
2C
bivalent0p
1p
2p
univalent
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Shared Memory
1QQueue
2QQueue
zQQueue
Read/WriteObjects
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Shared Memory
1QQueue
2QQueue
zQQueue
Read/WriteObjects
valent-0
valent-1
C
0C
1C
2C
bivalent0p
1p
2p
Case: the processors access different objects
0p
1p
Note: if an object was read/write the analysis is the similar
univalent
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valent-0
valent-1
C
0C
1C
2C
bivalent0p
1p
2p
op Q
1
op Q2
op Q2
op Q1
C
C
1p
0p
valent-0
valent-1
CCp
2
Impossible since
Two possible executions
univalent
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Shared Memory
1QQueue
QQueue
zQQueue
Read/WriteObjects
valent-0
valent-1
C
0C
1C
2C
bivalent0p
1p
2p
Case: the processors access same object
0p
1p
Note: if the object was read/write the analysisis the same as in the case with read/write objects
univalent
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valent-0
valent-1
C
0C
1C
2C
bivalent0p
1p
2punivalent
deq(
Q)
deq(Q)
Subcase: deq/deq
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valent-0
valent-1
C
0C
1C
2C
bivalent0p
1p
2punivalent
deq(
Q)
deq(Q)
deq(Q)
deq(Q)
C
C
1p
0p
valent-0
valent-1
CCp
2
Impossible since
Queue Q before operations: bc
a b
a d
a b
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valent-0
valent-1
C
0C
1C
2C
bivalent0p
1p
2punivalent
deq(
Q)
enq(Q,x)
Subcase: deq/enq
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valent-0
valent-1
C
0C
1C
2C
bivalent0p
1p
2punivalent
deq(
Q)
enq(Q,x)
enq(Q,x)
deq(Q)
C
C
1p
0p
valent-0
valent-1
CCp
2
Impossible since
Suppose Q was not empty a bc
a bx
a bx
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valent-0
valent-1
C
0C
1C
2C
bivalent0p
1p
2punivalent
deq(
Q)
enq(Q,x)
enq(Q,x)C 1pvalent-0
1
2
CCp
Impossible since
Suppose Q was empty
x
x
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valent-0
valent-1
C
0C
1C
2C
bivalent0p
1p
2punivalent
enq(
Q,a)
enq(Q,b)
Subcase: enq/enq
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valent-0
valent-1
C
0C
1C
2C
bivalent0p
1p
2punivalent
enq(
Q,a)
enq(Q,b)
enq(Q,b)
enq(Q,a)
C
C
1p
0p
valent-0
valent-1
Suppose Q was not empty yx
yx
yx
a
a
b
b
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valent-1
C 1C
2C
bivalent
0p
1p
2punivalent
enq(
Q,a
)
enq(Q,b) enq(Q,a)C 0p
valent-1
valent-0
0C enq(Q,b)C 1p
yxab b
0p
z w t w
1p0D
yxab z w t w
0p 1p1D
a
valent-0
Dequeue a Dequeue b
Dequeue b Dequeue a
Impossible since
10
2
DDp
30
C
bivalent
0p
enq(
Q,a
)
valent-0
0C enq(Q,b)C 1p
yxab b
0p
a
0 decidesvalent-0
Suppose does not dequeue a
0
A decision will be reached since the consensus algorithm is wait-free
0p
Explanation
0p
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valent-1
C 1Cbivalent
0p
1p
enq(
Q,a
)
enq(Q,b) enq(Q,a)C 0p
valent-0
0C enq(Q,b)C 1p
yxab
0p
yxab
0p
0
0
valent-0
valent-1
The same value will be decided by ,since sees the same shared memory values in both executions
0p
decides 00p
decides 00p
contradiction
0
ba
ba
0p
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In all cases we obtained contradiction;Therefore, there exists a processorwhich is not critical
C
0C
1C
1nC
bivalent0p
1p
1np
univalent
univalent
univalent
kC bivalent(not critical)
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Therefore, we can construct an execution
0C 1C 2C
Initialconfiguration
0ip
1ip
2ip Never
ends
bivalent bivalent bivalent
Consensus can never be reached
End of Theorem Proof
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Compare&Swap
Compare&Swap(X,A,B) { Temp X;
If X==A then X B;Return Temp; }
Shared Memory
X
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Theorem: The consensus number ofthe Compare&Swap object is
Proof:
Given processes, for any , we can solve wait-free consensus using a Compare&Swap object(and read/write objects)
n n
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A Wait-Free Consensus algorithm for n processors using a compare&swap object
Shared Memory
1nx
0x
1np
0p
1ny
0y
First
Local Memory
(compare&swap object)
Initial value
Final value
37
Code for processor ip }1,0{i
If then else
),,(& ii xFirstSwapComparev
ii xy
ii vy
//am I the first?
//yes, choose my value
Note: the algorithm uses a compare&swap and read/write objects
iv
//no, choose the value of the first process which is stored in First
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Shared Memory00 x
0p0y
First
Local Memory
Example execution:
11 x1p
1y
02 x2p
2y
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00 x0p
0y
Local Memory
Suppose executes first
11 x1p
1y
02 x2p
2y
2p
Compare&Swap(First, ,0)
Shared Memory
First
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00 x0p
0y
Local Memory
Suppose executes first
11 x1p
1y
02 x2p
02 y
2p
0
Shared Memory
First
Realizes it is first, decides on its own value
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00 x0p
0y
Local Memory
Suppose executes second
11 x1p
1y
02 x2p
1p
Compare&Swap(First, ,1)
Shared Memory
First0
02 y
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00 x0p
0y
Local Memory
11 x1p
01 y
02 x2p
Shared Memory
First0
0
Realizes is not first, decides on value of First
02 y
Suppose executes second 1p
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00 x0p
00 y
Local Memory
11 x1p
01 y
02 x2p
Shared Memory
First0
0
Realizes is not first, decides on value of First
02 y
Similarly for 0p Consensus has beenreached
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The algorithm is wait-free, since after the completion of the Compare&Swap operation, every processor decides (without consideringwhat the other processors do)
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Consensus HierarchyPart 3
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Consensus Number
Consensus Number of an object type:
The maximum number of processes for which the object can be used tosolve the wait-free consensus problem(together with read/write objects)
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Object Type Consensus Number
Read/Write 1
FIFO 2
Compare&Swap (infinity)
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Simulation:Object Type B
Object Type A
Object Type A
Read/Write Object
Object of type A simulates object of Type B
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Theorem:Objects of type A with consensus numbercannot simulate another object of type B withconsensus number
n
nm
Proof: Since otherwise, object A wouldhave consensus numberm
End of Proof
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Universality
can simulate in a wait-free mannerany other arbitrary object
Universal object:
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Objects with consensus number ncan simulate in a wait-free mannerany other arbitrary object of up toprocessors
n
We can show:
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new-state
Compare&Swap
inv
responsebefore
new-stateinv
responsebefore
initial
head
anchor
Non-Blocking Simulation
53
new-state
Consensus Object
inv
responseafter
new-stateinv
responseafter
initial
Head
anchor
Non-Blocking Simulation
seq seq 1
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new-stateinv
responseafter
Announce
seq
Wait-free Simulation
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