Set 7: Mutual Exclusion with Read/Write Variables
CSCE 668
DISTRIBUTED ALGORITHMS AND
SYSTEMS
CSCE 668
Fall 2011
Prof. Jennifer Welch
1
Read/Write Shared Variables
2

In one atomic step a processor can
 read
the variable or
 write the variable
 but not both!
Set 7: Mutual Exclusion with Read/Write Variables
CSCE 668
Bakery Algorithm
3

An algorithm for
 no
lockout
 mutual exclusion

using 2n shared read/write variables
 booleans
Choosing[i] : initially false, written by pi
and read by others
 integers Number[i] : initially 0, written by pi and
read by others
Set 7: Mutual Exclusion with Read/Write Variables
CSCE 668
Bakery Algorithm
4
Code for entry section:
Choosing[i] := true
Number[i] := max{Number[0], …,
Number[n-1]} + 1
Choosing[i] := false
for j := 0 to n-1 (except i) do
wait until Choosing[j] = false
wait until Number[j] = 0 or
(Number[j],j) > (Number[i],i)
endfor
Code for exit section:
Number[i] := 0
Set 7: Mutual Exclusion with Read/Write Variables
CSCE 668
Bakery Algorithm Provides Mutual
Exclusion
5
Lemma (4.5): If pi is in the critical section and
Number[k] ≠ 0 (k ≠ i), then (Number[k],k) >
(Number[i],i).
Proof: Consider two cases:
pi 's most recent
pi in CS and
read of Number[k];
Number[k] ≠ 0
Case 1: returns 0
Case 2: (Number[k],k) > (Number[i],i)
Set 7: Mutual Exclusion with Read/Write Variables
CSCE 668
Mutual Exclusion Case 1
6
pi 's most recent
read of Choosing[k],
returns false.
So pk is not in middle
of choosing number.
pi 's most recent
write to Number[i]
pi in CS and
Number[k] ≠ 0
pi 's most recent
read of Number[k],
returns 0.
So pk is in remainder
or choosing number.
So pk chooses number in this interval,
sees pi 's number, and chooses a larger one.
Set 7: Mutual Exclusion with Read/Write Variables
CSCE 668
Mutual Exclusion Case 2
7

Is proved using arguments similar to those for
Case 1.
Set 7: Mutual Exclusion with Read/Write Variables
CSCE 668
Mutual Exclusion for Bakery Algorithm
8

Lemma (4.6): If pi is in the critical section, then
Number[i] > 0.
 Proof

is a straightforward induction.
Mutual Exclusion: Suppose pi and pk are
simultaneously in CS.
 By
Lemma 4.6, both have Number > 0.
 By Lemma 4.5,
 (Number[k],k)
> (Number[i],i) and
 (Number[i],i) > (Number[k],k)
Set 7: Mutual Exclusion with Read/Write Variables
CSCE 668
No Lockout for Bakery Algorithm
9






Assume in contradiction there is a starved processor.
Starved processors are stuck at Line 5 or 6 (wait
stmts), not while choosing a number.
Let pi be starved processor with smallest pair
(Number[i],i).
Any processor entering entry section after pi has
chosen its number chooses a larger number.
Every processor with a smaller number eventually
enters CS (not starved) and exits.
Thus pi cannot be stuck at Line 5 or 6.
Set 7: Mutual Exclusion with Read/Write Variables
CSCE 668
Space Complexity of Bakery Algorithm
10




Number of shared variables is 2n
Choosing variables are boolean
Number variables are unbounded
Is it possible for an algorithm to use less shared
space?
Set 7: Mutual Exclusion with Read/Write Variables
CSCE 668
Bounded 2-Processor ME Algorithm
11
Uses 3 binary shared read/write variables:
 W[0] : initially 0, written by p0 and read by p1
 W[1] : initially 0, written by p1 and read by p0
 Priority : initially 0, written and read by both
Set 7: Mutual Exclusion with Read/Write Variables
CSCE 668
12
Bounded 2-Processor ME Algorithm with
ND


Start with a bounded algorithm for 2 processors
with ND, then extend to NL, then extend to n
processors.
Some ideas used in 2-processor algorithm:


each processor has a shared boolean W[i]
indicating if it wants the CS
p0 always has priority over p1 ; asymmetric code
Set 7: Mutual Exclusion with Read/Write Variables
CSCE 668
13
Bounded 2-Processor ME Algorithm with
ND
Code for p0 's entry section:
5
.
.
W[0] := 1
.
.
6
wait until W[1] = 0
1
2
3
4
Code for p0 's exit section:
7
8
.
W[0] := 0
Set 7: Mutual Exclusion with Read/Write Variables
CSCE 668
14
Bounded 2-Processor ME Algorithm with
ND
Code for p1 's entry section:
5
W[1] := 0
wait until W[0] = 0
W[1] := 1
.
if (W[0] = 1) then goto Line 1
6
.
1
2
3
4
Code for p1 's exit section:
7
.
8
W[1] := 0
Set 7: Mutual Exclusion with Read/Write Variables
CSCE 668
Discussion of 2-Processor ND Algorithm
15




Satisfies mutual exclusion: processors use W
variables to make sure of this
Satisfies no deadlock (exercise)
But unfair (lockout)
Fix by having the processors alternate having the
priority:
 shared
variable Priority, read and written by both
Set 7: Mutual Exclusion with Read/Write Variables
CSCE 668
Bounded 2-Processor ME Algorithm
16
Code for entry section:
5
W[i] := 0
wait until W[1-i] = 0 or Priority = i
W[i] := 1
if (Priority = 1-i) then
if (W[1-i] = 1) then goto Line 1
6
else wait until (W[1-i] = 0)
1
2
3
4
Code for exit section:
7
Priority := 1-i
8
W[i] := 0
Set 7: Mutual Exclusion with Read/Write Variables
CSCE 668
17
Analysis of Bounded 2-Processor
Algorithm

Mutual Exclusion: Suppose in contradiction p0
and p1 are simultaneously in CS.
p1 's most
recent write
of 1 to W[1]
(Line 3)
p0 's most
recent write
of 1 to W[0]
(Line 3)
W[0] = W[1] = 1,
both procs in CS
p0 's most recent
read of W[1] (Line 6):
must return 1, not 0!
Set 7: Mutual Exclusion with Read/Write Variables
CSCE 668
No-Deadlock for 2-Proc. Alg.
18


Useful for showing no-lockout.
If one proc. ever enters remainder for good, other
one cannot be starved.
 Ex:
If p1 enters remainder for good, then p0 will keep
seeing W[1] = 0.

So any deadlock would starve both procs.
Set 7: Mutual Exclusion with Read/Write Variables
CSCE 668
No-Deadlock for 2-Proc. Alg.
19


Suppose in contradiction there is deadlock.
W.l.o.g., suppose Priority gets stuck at 0 after
both processors are stuck in their entry sections.
p0 not stuck
p0 and p1
stuck in entry, in Line 2, skips
line 5, stuck in
Priority = 0
Line 6 with W[0] = 1
waiting for
W[1] to be 0
p1 skips
Line 6, stuck
p0 sees
at Line 2 with
W[1] = 0,
W[1] = 0, waiting enters CS
for W[0] to be 0
Set 7: Mutual Exclusion with Read/Write Variables
CSCE 668
No-Lockout for 2-Proc. Alg.
20



Suppose in contradiction p0 is starved.
Since there is no deadlock, p1 enters CS infinitely
often.
The first time p1 executes Line 7 in exit section after
p0 is stuck in entry, Priority gets stuck at 0.
p0 stuck
in entry
p1 at Line 7;
Priority = 0
forever after
p0 stuck at
Line 6 with
W[0] = 1, waiting
for W[1] to be 0
Set 7: Mutual Exclusion with Read/Write Variables
CSCE 668
p1 enters
entry, gets
stuck at
Line 2, waiting
for W[0] to
be 0
Bounded n-Processor ME Alg.
21



Can we get a bounded space mutex algorithm for n
larger than 2 processors?
Yes!
Based on the notion of a tournament tree: complete
binary tree with n-1 nodes


tree is conceptual only! does not represent message
passing channels
A copy of the 2-proc. algorithm is associated with
each tree node

includes separate copies of the 3 shared variables
Set 7: Mutual Exclusion with Read/Write Variables
CSCE 668
Tournament Tree
22
1
2
4
p0, p1
3
5
p2, p3
6
p4, p5
7
p6, p7
Set 7: Mutual Exclusion with Read/Write Variables
CSCE 668
Tournament Tree ME Algorithm
23
Each proc. begins entry section at a specific
leaf (two procs per leaf)
 A proc. proceeds to next level in tree by
winning the 2-proc. competition for current tree
node:

 on
left side, play role of p0
 on right side, play role of p1

When a proc. wins the 2-proc. algorithm
associated with the tree root, it enters CS.
Set 7: Mutual Exclusion with Read/Write Variables
CSCE 668
More on Tournament Tree Alg.
24
Code in book is recursive.
 pi begins at tree node 2k + i/2,
playing role of pi mod 2, where k = log n -1.
 After winning node v, "critical section" for node v
is

 entry
code for all nodes on path from
v's parent v/2 to root
 real critical section
 exit code for all nodes on path from root to v/2
Set 7: Mutual Exclusion with Read/Write Variables
CSCE 668
Tournament Tree Analysis
25

Correctness: based on correctness of 2-processor
algorithm and tournament structure:
projection of an admissible execution of tournament alg.
onto a particular tree node gives an admissible execution
of 2-proc. alg.
 ME for tournament alg. follows from ME for 2-proc. alg.
at tree root.
 NL for tournament alg. follows from NL for the 2-proc.
algs. at all nodes of tree


Space Complexity: 3n boolean read/write shared
variables.
Set 7: Mutual Exclusion with Read/Write Variables
CSCE 668