Switch and routers architectures: cantor networks
CF
minimum complexity cantor networks
Consider a Cantor network built with m parallel Benes networks.
Let A (κ) (with 1 6 κ 6 log2 (N)) be the number of 2 × 2 modules (in
the worst case) in the Benes network reachable without rearrangement
at stage κ by an input of the Cantor network. Then:
A (1) = m
A (2) = 2 · A (1) − 1
A (3) = 2 · A (2) − 2
A (4) = 2 · A (3) − 4
Thus:
A (κ) = 2 · A (κ − 1) − 2κ−2
κ = 2 , 3 , · · · , log2 (N)
because log2 (N) is the middle stage of the Benes networks. Going on:
A (κ) = 2 · A (κ − 1) − 2κ−2
A (κ − 1) = 2 · A (κ − 2) − 2κ−3
A (κ − 2) = 2 · A (κ − 3) − 2κ−4
Putting all together in the first one:
h
i
A (κ) = 2 · 2 · 2 · A (κ − 3) − 2κ−4 − 2κ−3 − 2κ−2 =
h
i
= 2 · 4 · A (κ − 3) − 2κ−3 − 2κ−3 − 2κ−2 =
= 8 · A (κ − 3) − −2κ−2 − 2κ−2 − 2κ−2 =
= 2y · A (κ − y) − y · 2κ−2
We need that A (κ − y) = A (1), thus:
κ−y = 1
=⇒
y = κ−1
This implies that:
A (κ) = 2κ−1 · A (1) − (κ − 1) · 2κ−2
Since the maximum κ = log2 (N):
A log2 (N) = 2log2 (N)−1 · A (1) − (log2 (N) − 1) · 2log2 (N)−2 =
N
N
· A (1) − (log2 (N) − 1) ·
=
2
4
N
N
=
· m − (log2 (N) − 1) ·
2
4
=
1
Switch and routers architectures: cantor networks
CF
Looking at the network,
from the beginning and from the end, we
can reach A log2 (N) modules; in the middle stage, by construction
of a Benes network, we have m · (N/2) modules: we need to find a free
common module. Therefore we impose:
m·
N
< 2 · A log2 (N)
2
By substituting the value of A log2 (N) computed before:
N
N
N
m·
=
< 2·
· m − (log2 (N) − 1) ·
2
2
4
N
N
m·
< N · m − (log2 (N) − 1) ·
=
2
2
m · N < 2N · m − (log2 (N) − 1) · N =
m · (N − 2N) < −(log2 (N) − 1) · N =
− m · N < −(log2 (N) − 1) · N =
m > log2 (N) − 1
2