instructor neelima gupta [email protected]. table of contents parallel algorithms
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InstructorNeelima Gupta
Table of Contents
Parallel Algorithms
Thanks to: Tejinder Kaur (35, MCS '09)
Instructor: Ms Neelima Gupta
Thanks to: Tejinder Kaur (35, MCS '09)
Solving a problem on multiple processors. S(n) is sequential time to solve a problem. T(n,p) is the parallel time to solve a
problem on p processors. W(n) is the work done by a parallel
algorithm. W(n)=T(n,p) p A parallel algorithm is optimal if the work
done is best of known sequential algorithm. i.e. if W(n)=S(n) Speed up is how much time is gained by
using more processors. speed up = S(n)/T(n,p)
Thanks to: Tejinder Kaur (35, MCS '09)
Take a problem of computing sum of numbers.Sequential time = Θ(n)We have 2 processors p1 and p2 and the numbers are2,3,4,5,1,11,13,10,7,8Initially all the numbers are with p1 and it sends half of them to p2. Both p1 and p2 compute sums and send the sums s1 and s2 to each other. So both have the final sum. p1 p2 2,3,4,5,1 11,13,10,7,8 (s1+s2) (s1+s2) Communication time= Θ(1)Computation time= Θ(n/2)T(n,2)= Θ(n/2)W(n)= n/2 2 =nHence this algorithm is optimal.Speed up = n/ n = 2 2
Thanks to: Tejinder Kaur (35, MCS '09)
PARALLEL MODELSDistributed Computing
Several independent machines are there.They communicate with
each oher by passing messages.The final result comes from all
independent machines.
M1 M2
M3
M5
M4
Thanks to: Tejinder Kaur (35, MCS '09)
SHARED MEMORY MODEL All the processors are reading and writing to the same memory.
There is no communication between them. Can not write at same time but can read at
same time.
SharedmemoryShared
memory
p1
p2
p3
pn
Thanks to: Tejinder Kaur (35, MCS '09)
Models for concurrency in shared memory modelEREW(Exclusive read exclusive write)CREW(Concurrent read exclusive write)CRCW(Concurrent read Concurrent Write)The weakest is EREW.CREW is Better than EREW but weaker thanCRCW.If we go from CRCW to CREW there is a
slowdownof factor of log(n).
Made By : Deepika Kamboj ( Roll No.7, MSc '11 )
Searching for a key Key =
x1 x2 xnx3
p1p
pnp
p3p
p2p
x1= xn=x3==
x2==
…….…
0
COMPARISON
OUTPUT
Thanks to 'PREETI'
xi …….…
…….…
….….…
…….…
…….…
pip
xi==
CRCW Key =
x1 x2 xnx3
p1p
pnp
p3p
p2p
x1= xn=x3==
x2==
…….…
0
COMPARISON
OUTPUT
Thanks to 'PREETI'
xi …….…
…….…
….….…
…….…
…….…
pip
xi==
Match Match found
CRCW Key =
x1 x2 xnx3
p1p
pnp
p3p
p2p
x1= xn=x3==
x2==
…….…
1
COMPARISON
OUTPUT
Thanks to 'PREETI'
xi …….…
…….…
….….…
…….…
…….…
pip
xi==
Match Match found
VERSION 1 OF SEARCHINGTo find the existence of the given KEY.MODEL used
CRCW Common Priority Arbitrary
Thanks to 'PREETI'
example for version1
Key = 7 12 7 3022
p1p
p6p
p3p
p2p
12≠7
0
COMPARISON
OUTPUT
15 7
p4p
p5p
30≠7
7=715≠7
22≠7
7=7
Thanks to 'PREETI'
example for version1
Key = 7 12 7 3022
p1p
p6p
p3p
p2p
12≠7
0
COMPARISON
OUTPUT
15 7
p4p
p5p
30≠7
7=715≠
722≠
77=7
Thanks to 'PREETI'
example for version1
Key = 7 12 7 3022
p1p
p6p
p3p
p2p
12≠7
1
COMPARISON
OUTPUT
15 7
p4p
p5p
30≠7
7=715≠
722≠
77=7
Thanks to 'PREETI'
VERSION 2 OF SEARCHINGTo find the processor id.MODEL used
CRCW Common Priority Arbitrary
Thanks to 'PREETI'
example for version2
Key = 7 12 7 3022
p1p
p6p
p3p
p2p
12≠7
0
COMPARISON
OUTPUT
15 7
p4p
p5p
30≠7
7=715≠7
22≠7
7=7
Thanks to 'PREETI'
example for version2
Key = 7 12 7 3022
p1p
p6p
p3p
p2p
12≠7
0
COMPARISON
OUTPUT
15 7
p4p
p5p
30≠7
7=715≠
722≠
77=7
Thanks to 'PREETI'
example for version2
Key = 7 12 7 3022
p1p
p6p
p3p
p2p
12≠7
p5
COMPARISON
OUTPUT
15 7
p4p
p5p
30≠7
7=715≠
722≠
77=7
P2 or p5
gets
written
Thanks to 'PREETI'
Or
Key = 7 12 7 3022
p1p
p6p
p3p
p2p
12≠7
p2
COMPARISON
OUTPUT
15 7
p4p
p5p
30≠7
7=715≠
722≠
77=7
P2 or p5
gets
written
Thanks to 'PREETI'
VERSION 3 OF SEARCHINGTo find the LEFT MOST OCCURRENCE of
the given KEY.MODEL used
CRCW Common Arbitrary Priority
×
Thanks to 'PREETI'
example for version3
Key = 7 12 7 3022
p1p
p6p
p3p
p2p
12≠7
0
COMPARISON
OUTPUT
15 7
p4p
p5p
30≠7
7=715≠7
22≠7
7=7
Thanks to 'PREETI'
example for version3
Key = 7 12 7 3022
p1p
p6p
p3p
p2p
12≠7
0
COMPARISON
OUTPUT
15 7
p4p
p5p
30≠7
7=715≠
722≠
77=7
Thanks to 'PREETI'
example for version3
Key = 7 12 7 3022
p1p
p6p
p3p
p2p
12≠7
p2
COMPARISON
OUTPUT
15 7
p4p
p5p
30≠7
7=715≠
722≠
77=7
P2 has
highest
priority.
Thanks to 'PREETI'
Thanks to: Tejinder Kaur (35, MCS '09)
SUM PROBLEMFind sum of n numbers and there are n processors. n
processors
n/2 processors
n/4
processors
1 processor
a1 a2 a3 a4 an
Thanks to: Tejinder Kaur (35, MCS '09)
Height of this tree is log n.Each step is taking constant time.Hence this algo takes O(log n) time.W(n)= n log n= nlogn.Speed up=n/log n.This algorithm is not optimal as half of the
processors areidle in first step and number of idle
processors isincreasing in further steps.What if we use n/log n processors.
Thanks to: Tejinder Kaur (35, MCS '09)
As the number of processors is n/log n.Each processor will get log n values.
s1 s2 sm
Take m=n/log nEach processor has n/log n values so sm sums will be generated.
Thanks to: Tejinder Kaur (35, MCS '09)
The height is log m.So it will take log m time <= log nSo T(n,p) <= 2logn = O(log n)W(n)= n=O(S(n))As sequential time is O(n).Hence this algorithm is optimal.
Thanks to: Tejinder Kaur (35, MCS '09)
SORTING Sort n numbers in parallel with n processors. Initially each procesor has an element.
n/2,2 merge
n/4,4 merge
1,n merge
a1 a2 a3 a4 an
Thanks to: Tejinder Kaur (35, MCS '09)
The last step will take n units of time n + n/2 + n/4 + - - - - - - + 2 <= 2nSo it takes O(n) time.W(n)= n2
Thanks to: Surbhi Tripathi (27, MCS '09)
Instructor: Ms Neelima Gupta
Thanks to: Surbhi Tripathi (27, MCS '09)
Definition: Prefix SumsGiven: Set of n values A = {a0,a1…….,an-1}
We want to find the prefix sums S0, S1,………..Sn-1.
Where, S0=a0 S1=a1+a0 | | Sn-1=an-1+…………+a1+a0
Thanks to: Surbhi Tripathi (27, MCS '09)
STEP - IIa0 a1 a2 a3 a4 a5 a6
a7
P1:s1 P2:a2oa3 P3:a4oa5 P4:a6oa7
p2: s3(s1oa2oa3)
p3:a4oa5oa6
p4:a4oa5oa6oa7
p1: s2(s1oa2)
Thanks to: Surbhi Tripathi (27, MCS '09)
STEP - IIIa0 a1 a2 a3 a4 a5 a6 a7
P1:s1 P2:a2oa3 P3:a4oa5 P4:a6oa7
p2: s3(s1oa2oa3)
p3:a4oa4oa6
p4:a4oa5oa6oa7
p1: s2(s1oa2)
p1=s4 (s3oa4)
p4: s7 (s3oa4oa5oa6oa7)
p3: s6 (s3oa4oa5oa6)
p2: s5 (s3oa4oa5)
Thanks to: Surbhi Tripathi (27, MCS '09)
CREW Model
Computations of prefix sums do not require any concurrent writes.
Thanks to: Surbhi Tripathi (27, MCS '09)
TIME COMPLEXITYTo compute prefix sums of n numbersAs,the number of prefix sums computed
doubles at each step.While computing n prefix sums we get a tree of height log n.
Each step takes constant time.
So, computing n prefix sums using n processors in parallel takes log n time