Parallel algorithms

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Parallel
Algorithms
Shashikant V. Athawale
Assistant Professor ,Computer Engineering
Department AISSMS College of Engineering,
Kennedy Road, Pune , MS, India - 411001

Parallel Algorithms
Parallel: perform more than one operation at a time.
PRAM model: Parallel Random Access Model.
2
p0
p1
pn-1
Shared
memory
Multiple processors connected to a shared memory.
Each processor access any location in unit time.
All processors can access memory in parallel.
All processors can perform operations in parallel.

Concurrent vs. Exclusive AccessFour models
EREW: exclusive read and exclusive write
CREW: concurrent read and exclusive write
ERCW: exclusive read and concurrent write
CRCW: concurrent read and concurrent write
Handling write conflicts
Common-write model: only if they write the same
value.
Arbitrary-write model: an arbitrary one succeeds.
Priority-write model: the one with smallest index
succeeds.
EREW and CRCW are most popular.
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Synchronization and Control
Synchronization:
A most important and complicated issue
Suppose all processors are inherently tightly
synchronized:
 All processors execute the same statements at the same
time
 No race among processors, i.e, same pace.
Termination control of a parallel loop:
Depend on the state of all processors
Can be tested in O(1) time.
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Pointer Jumping –list ranking
Given a single linked list L with n objects,
compute, for each object in L, its distance from the
end of the list.
Formally: suppose next is the pointer field
d[i]= 0 if next[i]=nil
 d[next[i]]+1 if next[i]≠nil
Serial algorithm: Θ(n).
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List ranking –EREW algorithm
 LIST-RANK(L) (in O(lg n) time)
1. for each processor i, in parallel
2. do if next[i]=nil
3. then d[i]←0
4. else d[i]←1
5. while there exists an object i such that next[i]≠nil
6. do for each processor i, in parallel
7. do if next[i]≠nil
8. then d[i]← d[i]+ d[next[i]]
9. next[i] ←next[next[i]]
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7
1
3
1
4
1
6
1
1
1
0
0
5
(a)
3 4 6 1 0 5
(b) 2 2 2 2 1 0
3 4 6 1 0 5
(c) 4 4 3 2 1 0
3 4 6 1 0 5
(d) 5 4 3 2 1 0

List ranking –correctness of EREW algorithm
Loop invariant: for each i, the sum of d values
in the sublist headed by i is the correct
distance from i to the end of the original list L.
Parallel memory must be synchronized: the
reads on the right must occur before the wirtes
on the left. Moreover, read d[i] and then read
d[next[i]].
An EREW algorithm: every read and write is
exclusive. For an object i, its processor reads
d[i], and then its precedent processor reads its
d[i]. Writes are all in distinct locations.
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LIST ranking EREW algorithm running time
O(lg n):
The initialization for loop runs in O(1).
Each iteration of while loop runs in O(1).
There are exactly lg n iterations:
 Each iteration transforms each list into two interleaved lists:
one consisting of objects in even positions, and the other
odd positions. Thus, each iteration double the number of
lists but halves their lengths.
The termination test in line 5 runs in O(1).
Define work =#processors ×running time. O(n lg n).
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Parallel prefix on a list
A prefix computation is defined as:
Input: <x1, x2, …, xn>
Binary associative operation ⊗
Output:<y1, y2, …, yn>
Such that:
 y1= x1
 yk= yk-1⊗ xkfork=2,3, …,n, i.e, yk= ⊗ x1⊗ x2 …⊗ xk.
Suppose <x1, x2, …, xn> are stored orderly in a list.
Define notation: [i,j]= xi⊗ xi+1 …⊗ xj
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Prefix computation LIST-PREFIX(L)
2. do y[i]← x[i]
3. while there exists an object i such that next[i]≠nil
5. do if next[i]≠nil
6. then y[next[i]]← y[i] ⊗ y[next[i]]
7. next[i] ←next[next[i]]
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12
[1,1]
x1
[2,2]
x2
[3,3] [4,4]
x4
[5,5]
x5
[6,6]
x6
(a)
x3
x4
(b)
x1 x2 x5
x6x3
[1,1] [1,2] [2,3] [3,4] [4,5] [5,6]
x1 x2 x5
x6x3
x1 x2 x5
x6x3
(c)
(d)
[1,1] [1,2] [1,3] [1,4] [2,5] [3,6]
[1,1] [1,2] [1,3] [1,4] [1,5] [1,6]

Find root –CREW algorithm
Suppose a forest of binary trees, each node i has a
pointer parent[i].
Find the identity of the tree of each node.
Assume that each node is associated a processor.
Assume that each node i has a field root[i].
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Find-roots –CREW algorithm
 FIND-ROOTS(F)
2. do if parent[i] = nil
3. then root[i]←i
4. while there exist a node i such that parent[i] ≠ nil
6. do if parent[i] ≠ nil
7. then root[i] ← root[parent[i]]
8. parent[i] ← parent[parent[i]]
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Find root –CREW algorithm
Running time: O(lg d), where d is the height of
maximum-depth tree in the forest.
All the writes are exclusive
But the read in line 7 is concurrent, since several
nodes may have same node as parent.
See figure 30.5.
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Find roots –CREW vs. EREW
How fast can n nodes in a forest determine their
roots using only exclusive read?
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Ω(lg n)
Argument: when exclusive read, a given peace of information can only be
copied to one other memory location in each step, thus the number of locations
containing a given piece of information at most doubles at each step. Looking
at a forest with one tree of n nodes, the root identity is stored in one place initially.
After the first step, it is stored in at most two places; after the second step, it is
Stored in at most four places, …, so need lg n steps for it to be stored at n places.
So CREW: O(lg d) and EREW: Ω(lg n).
If d=2(lg n)
, CREW outperforms any EREW algorithm.
If d=Θ(lg n), then CREW runs in O(lg lg n), and EREW is
much slower.

Find maximum – CRCW algorithm Given n elements A[0,n-1], find the maximum.
 Suppose n2
processors, each processor (i,j) compare A[i] and A[j], for 0≤
i, j ≤n-1.
 FAST-MAX(A)
1. n←length[A]
2. for i ←0 to n-1, in parallel
3. do m[i] ←true
4. for i ←0 to n-1 and j ←0 to n-1, in parallel
5. do if A[i] < A[j]
6. then m[i] ←false
7. for i ←0 to n-1, in parallel
8. do if m[i] =true
9. then max ← A[i]
10. return max
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The running time is O(1).
Note: there may be multiple maximum values, so their processors
Will write to max concurrently. Its work = n2
× O(1) =O(n2
).
5 6 9 2 9 m
5 F T T F T F
6 F F T F T F
9 F F F F F T
2 T T T F T F
9 F F F F F T
A[j]
A[i]
max=9

Find maximum –CRCW vs. EREW
If find maximum using EREW, then Ω(lg n).
Argument: consider how many elements “think”
that they might be the maximum.
First, n,
After first step, n/2,
After second step n/4. …, each step, halve.
Moreover, CREW takes Ω(lg n).
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Stimulating CRCW with EREW
Theorem:
A p-processor CRCW algorithm can be no more than O(lg p)
times faster than a best p-processor EREW algorithm for the same
problem.
Proof: each step of CRCW can be simulated by O(lg p)
computations of EREW.
Suppose concurrent write:
 CRCW pi write data xi to location li, (li may be same for multiple pi ‘s).
 Corresponding EREW pi write (li, xi) to a location A[i], (different A[i]’s)
so exclusive write.
 Sort all (li, xi)’s by li’s, same locations are brought together. in O(lg p).
 Each EREW picompares A[i]= (lj, xj), and A[i-1]= (lk, xk). If lj≠ lk or i=0,
then EREW pi writes xj to lj. (exclusive write).
See figure 30.7.
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CRCW vs. EREW
CRCW:
Some says: easier to program and more faster.
Others say: The hardware to CRCW is slower than
EREW. And One can not find maximum in O(1).
Still others say: either EREW or CRCW is wrong.
Processors must be connected by a network, and only
be able to communicate with other via the network, so
network should be part of the model.
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Parallel algorithms

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