170 11. Oblivious Routing Protocols
Proof.
Wit first prove part (a) of the Theorem.
Proof of Part (a): Suppose that
qN
worms, q per input, have to be sent to
random outputs. Let ~ -- min(L, log N}. Then the worms can be given delays
such that the routing can be viewed as divided into [log
N/~J
independent
passes in which every input has to send out at most
q~ = [q/llogN/gJl
worms9 Therfore, we have to show that there exists a rank allocation such
that each pass takes O(-~ (q~.log 1/B N+log N)) rounds, w.h.p., each requiring
2 log N + L time steps9 In the following we do not only show the existence
but also describe a specific allocation that ensures the above routing time.
The ith worm (0 < i <
q')
starting at node
(XlogN_l,... ,
X0) on level
0 is assigned rank i 9 N + (x0,... ,XlogN-1)2. Thus, the rank of a worm is
essentially the bit-reversal number of its input node plus an offset. In the
following we identify a worm with its rank9 (Note that the ranks of all worms
participating in the same phase are distinct.)
For an integer m _> 1 with binary representation (... ,m2, ml,m0), we
define
5(m) = max{i _< logN I
(mi-1,... ,mo) ~-
0/} 9
(i(m) has the nice property that any two worms w and w ~ can not use the
same edge before level log
N - 6(IJ -
w]). Further the following two claims
can be shown that will be used in the further analysis.
Claim 11.1.1.
For any m,k E ~I with k < m, there are at most
(~)/2 i'k
possibilities to choose a subset
{Xl,...,
xk
} C_ {1,..., m}
such that
min{(i(xx),
= i.
Proof.
Let m E IN be an arbitrary constant. According to the definition
of (i(i), for every i E {1,..., LlogmJ} there are at most [m/2iJ numbers
x E {1,...,m} with 6(x) _> i. Therefore there are at most ([m/2"J) choices
for {xl,.
xk
}. Since for all j E {0, , k - 1} it holds that
m/2'-j
< ~ we
9 9 ' " " ~ ~r~--j --
get (%/2'j) < []
Claim 11.1.2.
For any k E g4 and any strictly decreasing function f : ~
I~ it holds that
k k
Z 2'(i)f(i) -< (logU + 1)
Zf(i) .
i=1 i=1
k'
Proof.
Let k E Is/be an arbitrary constant. We first show that Y]~=I 2'(i) ~-
(log N + 1)k' for any k' 6 {1,..., k}. Since there are at most [k'/21J numbers
x E {1,...,k'} with
6(x) = i,
it holds
k' flog ~'] [ k, j 2mi,{i,log N}
i=1 i:I