EE 351 – Spring 2001-2002 Handout # 21M.J. Narasimha

Homework #4 solutions

Note : a CRS* denotes an MCRS

Problem 2

The last 4 stages of this network can be viewed as an Ω−116 network. We can convert an arbitrary

destination sequence at the input into an Ω−116 -passable sequence as follows:

(i) Set the stage-1 switches to split the given CRS(16) into two CRS(8) systems; the first one isderived from the upper outputs and the second one from the lower outputs.

(ii) Similarly, set the stage-2 switches to split the CRS(8) groups into CRS(4) groups.

(iii) Similarly again, set the stage-3 switches to split the CRS(4) groups into CRS(2) groups.At this point, the sequence is Ω−1

16 -passable. Set the switches in stages 4, 5, 6, and 7 accordingto the self-routing algorithm.

Switch settings for the bit-reversal permutation are as shown in the following figure. All theswitches in the first three stages are straight through; the input sequence to the Ω−1

16 network isclearly Ω−1

16 -passable.

0000 (0)

1000 (8)

0100 (4)

1100 (12)

0010 (2)

1010 (10)

0110 (6)

1110 (14)

0001 (1)

1001 (9)

0101 (5)

1101 (13)

0011 (3)

0111 (7)

1011 (11)

1111(15)

0

4

2

6

1

5

3

7

0

1

2

3

0

1

16-Input Inverse Omega Network

Problem 3

(a) We have to prove that:

d0 d1︸ ︷︷ ︸

d2 d3︸ ︷︷ ︸

CRS(2) CRS(2)︸ ︷︷ ︸

d4 d5︸ ︷︷ ︸

d6 d7︸ ︷︷ ︸

CRS(2) CRS(2)︸ ︷︷ ︸

CRS(4) CRS(4)︸ ︷︷ ︸

CRS(8)

The CRS(2) conditions are obviously satisfied from the the construction in step 3. To proved0, d1, d2, d3 is a CRS(4), note that LSBs of d0 and d2 are equal to 0 by construction (step 3), butthe bits next to the LSBs have to be different since both d0 and d2 originate from the CRS(4) definedby b0, b1, b2, b3 (step 1). Similarly, LSBs of d1 and d3 are equal to 1 by construction (step 3), but

the bits next to the LSBs have to be different as they both originate from the CRS(4) defined byb4, b5, b6, b7 (step 1). Hence d0, d1, d2, d3 is a CRS(4). This implies that d4, d5, d6, d7 is alsoa CRS(4).

(b) Because of the CRS(4) and CRS(2) partitioning, one switch in stage 1 and two switches instage 2 can be nailed up as shown in the figure. Furthermore, all the switches in the fourth stage(the first stage of the Ω−1

8 network) can be nailed up as shown due to the construction in step 3.

d0

d1

d2

d3

d4

d5

d6

d7

c0

c1

c2

c3

c4

c5

c6

c7

b0

b1

b2

b3

b4

b5

b6

b7

a0

a1

a2

a3

a4

a5

a6

a7

(c) See figure.

6

4

2

0

1

7

5

3

5

2

3

7

6

1

4

0

4

0

1

5

6

2

7

3

5

3

6

4

2

7

1

0

5

6

2

1

3

4

7

0

6

1

4

7

5

2

3

0

0

1

2

3

4

5

6

7

1

Problem 4: The following diagram shows an N×N network constructed using two ΩN-1 networks

1x2 switches 2x1 switches

1

2

N

1

2

N

ΩN-1-network

ΩN-1-network

Solution: (a) Show that this network is strictly nonblocking for the case N = 4.

Consider the mapping:

=

3210

3210dddd

π

Set the input switches so that half of the inputs are directed to the upper 14−Ω and the other half to

the lower. Now the upper 14−Ω has to realize the mapping:

=

xdxd 201

3210π

and the lower 14−Ω has to realize:

=

312

3210dxdx

π

Clearly both of these mappings/permutations are 14−Ω passable (they satisfy the CRS conditions).

Furthermore, we can permanently wire the input switches to send d0 and d2 to the upper 14−Ω , and

d1 and d3 to the lower 14−Ω . Hence the network is strictly nonblocking.

(b) Show that this network is rearrangeable for the case N = 8. Briefly delineate your algorithm for realizing an arbitrary permutation of eight inputs with this network. As before, split the original permutation:

=

76543210

76543210dddddddd

π

into two mappings by the input switches, as follows:

=

xdxdxdxd 64201

76543210π

=

75312

76543210dxdxdxdx

π

2

Consider the permutation π1. The destination sequence clearly satisfies the CRS(2) conditions for 1

8−Ω network. If the set xdxd 20 is not a CRS(4), i.e. if the two LSBs of d0 and d2 are

identical, then swap the states of input switches 0 and 1, so that the new mappings are:

=

xdxdxddx 64211

76543210π

=

75302

76543210dxdxdxxd

π

Now both xddx 21 and 30 dxxd are CRS(4). Similar arguments are applicable for the set 30 dxxd . The complete destination sequence π1 and π2 is clearly CRS(8), and hence the network is rearrangeable. The routing algorithm has also been delineated which includes splitting the permutation into two different permutations and then swapping the states of each pair of switches, if required.