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Interconnection Networks. Direct Indirect Shared Memory Distributed Memory (Message passing). Topology. Diameter: Longest path length between two processors d = max { d min ( v i , v j ) | for all v i , v j V } Node connectivity - PowerPoint PPT Presentation
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Interconnection Networks
• Direct
• Indirect
• Shared Memory
• Distributed Memory (Message passing)
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Topology• Diameter: Longest path length between two
processorsd = max {dmin(vi, vj) | for all vi, vj V}
• Node connectivitymin. # of nodes to be removed for the network to
be disconnected
node connectivity r There are r disjoint paths between every pair of nodes.
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Topology• Fault diameter
Suppose the node connectivity = r, diameter of the network with at most r-1 faulty nodes
• Bisection widthThe # of edges to be removed to separate the
graph into two equal parts.
• Cost1) # of nodes & # of edges
2) Bisection width.
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• cw (Channel width): # of wires connecting two nodes.
• cr (Channel rate): # of bits/sec/wire.
• Channel bandwidth = cw cr
• Bisection bandwidth
= bisection width channel bandwidth
• Cost Bisection bandwidth
bisection bandwidth is constant
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• Hypercube: N = 22n
bisection width = 22n-1
channel width = wh
bisection bandwidth = 22n-1 wh
• 2-D Torus: N = 22n
bisection width = 2 2n
channel width = wd
bisection bandwidth = 2n+1 wd
2n+1 wd = 22n-1 wh
wd / wh = 22n-1 / 2n+1 = 2n-2 = 2n/4 = N /4
• 3-D Torus: w3d / wh = (N)1/3/4
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Network Topology
• Graph Model:Processors Nodes
Wires joining processors Link
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Linear Array
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• # of processors = n
• # of links = n-1
• Connection (i, i+1) for i=1,2, … , n-1
• Diameter(Longest Path Length)= n-1
• Communication delayworst case: diameter
Average case: 1/n2 dij
32 n. . . . .
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Hypercube• Hamming Distance
DH(x, y) = # of positions in which x & y differ
x & y are binary vectors
DH(1100, 0111) = 3
• Nodes are labeled as n-bit binary
• Two nodes, x & y, are adjacent if DH(x, y) =1
001
101 111
011
110
010000100
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Product of graphs• G1 = (V1, E1) & G2 = (V2, E2)
G1 G2 = G = (V, E)= (V1 V2 , E)
Two nodes in G, ((v1, v2), (v3, v4)) are adjacent
1) if v1= v3 and (v2, v4) E2 or
2) if v2= v4 and (v1, v3) E1
Q2 Qn-1= Qn or Qk Qn-k = Qk
001
101 111
011
110
010000100
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Routing: A BA=(an-1, an-2, … , a1, a0)
B =(bn-1, bn-2, … , b1, b0)
an-1, an-2, … , a2, a1, a0
an-1, an-2, … , a2, a1, b0
an-1, an-2, … , a2, b1, b0
an-1, an-2, … , b2, b1, b0
••• an-1, bn-2, … , b2, b1, b0
bn-1, bn-2, … , b2, b1, b0
# of steps required
= DH(A, B)
Diameter of the network
= n = log2N
Average comm. delay
= 1/N (( )+2( ) … n( ))
= n2n-1/N = n2n-1/2n
= n/2
n1
n2
nn
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Connectivity
Let DH(x, y) = d
n : node disjoint paths between x & y
d : paths have length d
n-d : paths have length d+2
Example: 110000 111111, d(110000,111111)=4
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11000
010000 100000 110001 110010 110100 111000
010001 100001 110011 110110 111100 111001
010011 100011 110111 111110 111101 111011
010111 100111
011111 101111
111111
Fault Diameter = n+1
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De Bruijn Network
• # of nodes, N = 2n
• Node A=(an-1, an-2, …, a0 ) is adjacent to
1. an-2, an-3, … , a1, a0, 0
2. an-2, an-3, … , a1, a0, 1
3. 0 an-1, an-2, … , a1
4. 1 an-1, an-2, … , a1
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De Bruijn NetworkN=23 = 8
2n - 4 nodes have degree 4
2 nodes have degree 3
2 nodes have degree 2
# of links 42n / 2 = 2n+1
000
001
100
010
011
101
110
111
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Routing: A B A=(an-1, an-2, … , a1, a0)
B =(bn-1, bn-2, … , b1, b0)
an-1, an-2, … , a2, a1, a0
an-2, an-3, … , a1, a0, bn-1
an-3, an-4, … , a0, bn-1, bn-2
an-4, an-5, … , bn-1, bn-2, bn-3
••• bn-1, bn-2, … , b2, b1, b0
log2N
• Example
A = 1010
B = 1101
1010
0101
1011
0110
1101
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an-1, an-2, … , a1, a0
an-2, … , a1, a0, 0
an-3, … , a0, 0 , 0
… 0, 0, 0, … , 0, 0
b0, 0, 0, … , 0, 0
b1, b0, 0, … , 0, 0…
bn-2, bn-3, … , b0, 0
an-2, … , a1, a0, 1
an-3, … , a0, 1 , 1
… 1, 1, 1, … , 1, 1
b0, 1, 1, … , 1, 1
b1, b0, 1, … , 1, 1 …
bn-2, bn-3, … , b0, 1
bn-1, bn-2, … , b1, b0
2-disjoint paths
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k-ary n-cube
• Multidimensional torus
00 01 02 03
10 11 12 13
20 21 22 23
30 31 32 33
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k-ary n-cube• Given a node (an-1, an-2, … , a1, a0) adjacent
to 2n nodes given by(an-1, an-2, … , a1, a0±1)
(an-1, an-2, … , a1 ±1, a0):
(an-1 ±1, an-2, … , a1, a0)
• k=8, n=3 node (3 2 4) is adjacent to (3, 2, 5), (3, 2, 3)(3, 3, 4), (3, 1, 4)(4, 2, 4), (2, 2, 4)
• N = # of nodes in k-ary n-cube = kn
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00 01 02 03
10 11 12 13
20 21 22 23
30 31 32 33
001
101 111
011
110
010000100
k=4, n=2k=2, n=3
Hypercube
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0• k=8
• Lee Distance
DL ((a3, a2, a1, a0 ), (b3, b2, b1, b0 ))
= min (ai - bi, bi - ai ) mod k
DL (1 2 3, 3 2 1) = min (1-3, 3-1) + min (2-2, 2-2) + min (3-1, 1-3) = 2 + 0 + 2 = 4
4
6 2
1
35
7
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Routing: A BA=(an-1, an-2, … , a1, a0)
B =(bn-1, bn-2, … , b1, b0)
an-1, an-2, … , a2, a1, a0 ±1
an-1, an-2, … , a2, a1, a0 ±2:
an-1, an-2, … , a2, a1, b0
an-1, an-2, … , b2, a1 ±1, b0
:
an-1, an-2, … , a2, b1, b0
:
bn-1, bn-2, … , b2, b1, b0
• Example (k=5)
(4,4,2,1) (2,1,4,4) (4,4,2,1)
(4,4,2,0) (4,4,2,4) (4,4,3,4) (4,4,4,4,) (4,0,4,4)
(4,1,4,4)(3,1,4,4)(2,1,4,4)
# of steps = DL (A, B)
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Torus of size kn-1, kn-2, … , k1, k0
• Mixed radix number system
• Processor (an-1, an-2, … , a1, a0 )
0 ai < ki-1 i = 0, 1, 2, … n-1
Decimal value
= an-1 (kn-2 kn-3, … , k0) + an-2 (kn-3 kn-4, … , k0) + …+ a0
• Two nodes, A=(an-1, an-2, … , a1, a0) &
B =(bn-1, bn-2, … , b1, b0)
are adjacent if DL (A, B) = 1
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Torus of size kn-1, kn-2, … , k1, k0 • Example: (854)
(3,2,1) = 3(54) + 24 + 1 = 60 + 8 +1 = 69 4 69
Decimal to mixed radix 5 17 … 1 69 = (3,2,1) 3 … 2
• Example: (888)(3,2,4) = 382 + 28 + 4 8 212
= 364 + 16 +4 = 212 8 26 … 4 212 = (3,2,4) 3 … 2
• Radix k-number(an-1, an-2, … , a1, a0)= an-1 kn-1 + an-2 kn-2 + … + a0
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Embedding Cycles• Hypercube- Gray codes
0 00 000 00001 01 001 0001
11 011 001110 010 0100
110 0110111 0111101 0101100 0100
110011011111 …
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Mapping (Binary to Gray)
000 000
001 001
010 011
011 010
100 110
101 111
110 101
111 100
f(xn-1xn-2 … x0)
= (gn-1gn-2 … g0)
where
gn-1 = xn-1
gi = xi xi+1
i= n-2, n-3, … ,0
10110 11101
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k-ary n-cube
• f: Radix Gray
f(xn-1xn-2 … x0) = (gn-1gn-2 … g0)
gn-1 = xn-1
gi = xi - xi+1 mod k
for i= n-2, n-3, … ,0
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3-ary 2-cube
Radix Gray0 00 001 01 012 02 023 10 124 11 105 12 116 20 217 21 22 k=58 22 20 2431 2243
00 01 02
10 11 12
20 21 22
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De Bruijn Network
23 - nodesj+3 + sj+1 + sj= 0 sj+3 = sj+1 + sj mod 2
Initial conditions s0 =1, s1 = 0 and s2 = 1
s0 s1 s2 s3 s4 s5 s6 s7 s8 s9 1 0 1 1 1 0 0 1 0 1
101011111110100000001010101
000
001
100
010
011
101
110
111
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How to choose this different equation?
• Take an n-r order difference equation whose characteristic equation is a primitive polynomial of degree r.Primitive Polynomial
x2+x+1 x7+x3+1
x3+x+1 x8+x4+ x3+x2+1
x4+x+1 x9+x4+1
x5+x2+1 x10+x3+1
x6+x2+1 sj+5 = sj+2 + sj
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Disjoint Cycle
• sj+3 = sj+1 + sj+1
Initial conditions s0 =1, s1 = 0 and s2 = 1s0 s1 s2 s3 s4 s5 s6 s7 s8 s9 1 0 1 0 0 0 1 1 0 1
101010100000001011110