Next, we draw our attention to bipartite graphs that are related to standard cuckoo hashing. The following theorem shows us, that the structure of this graph is similar to the structure of its non-bipartite counterpart. A detailed discussion of the differences and similarities can be found at the end of this chapter.
6.3 The bipartite Cuckoo Graph
Theorem 6.2. Suppose that ε ∈ (0,1) is fixed and that n = (1−ε)m. Then a la-belled random bipartite multigraph with2×m vertices and nedges satisfies the following properties.
1. The number of unicyclic components with cycle length 2k has in limit a Poisson distribution P o(λk) with parameter
λk= 1
2k(1−ε)2k, (6.58)
and the number of unicyclic components has in limit a Poisson distributionP o(λ), too, with parameter
λ=−1 2log
1−(1−ε)2
. (6.59)
2. Denote the number of tree components with kvertices by tk. Mean and variance of this random variable are asymptotically equal to
mμ= 2mkk−2(1−ε)k−1ek(ε−1)
k! , (6.60)
respectively
mσ2=mμ−2me2k(ε−1)k2k−4(1−ε)2k−3(k2ε2+k2ε−4kε+ 2)
(k!)2 . (6.61)
Furthermore tk satisfies a central limit theorem of the form tk−μ
σ →N(0,1). (6.62)
3. The number of vertices contained in cycles has in limit the distribution with char-acteristic function
φ(s) =
$
1−(1−ε)2
1−e2is(1−ε)2, (6.63) and, hence, expectation is asymptotically given by
(1−ε)2
1−(1−ε)2, (6.64)
and variance by
2(1−ε)2
(1−(1−ε)2)2. (6.65)
4. Furthermore, the expected value of the number of nodes in unicyclic components is asymptotically given by
(1−ε)2
ε(1−(1−ε)2), (6.66)
and its variance by
(1−ε)2(ε2−3ε+ 4)
ε2(1−(1−ε)2)2 . (6.67)
6 The Structure of the Cuckoo Graph
Proof of Theorem 6.2
Similar to the proof of Theorem 6.1, it is sufficient to consider graphs of G◦m,m,n, the set of bipartite graphs without complex components, only. Given a random variable ξ, defined on the setGm,m,n, we denote its restriction toG◦m,m,nbyξ and the corresponding distribution functions byFξ resp. Fξ. Due to Theorem 4.2 and Lemma 6.1, the relation
|Fξ−Fξ| ≤P(Gm,m,n\G◦m,m,n) =O(1/m) (6.68)
holds.
As usual, we define the ratio
ε = 1− n
m = 1−(1−ε)m
m , (6.69)
and consider an infinite series of graphs possessing fixed ε first (cf. Chapter 4).
The further proof is divided into four parts, each of it proves one of the claimed results using a generating function approach. Recall the function
g◦(x, v) = exp 1
v˜t(xv, yv) +1
2log 1
1−t1(x, y)t2(x, y)
= ev1˜t(xv,yv)
1−t1(xv, yv)t2(xv, yv). (6.70) established in Lemma 4.5 that counts graphs without complex components. Again, we use the technique of introducing a new variable to “mark” the parameter of interest.
Number of Cycles
Lemma 6.6. The moment generating function of the limiting distribution of the number of cycles resp. the number of cycles of length 2k in a graph of G◦m,m,n is given by
ψc(s) = exp log
1−(1−ε)2
2 (1−es) 1 +O 1
m
, (6.71)
resp.
ψ2k(s) = exp
−(1−ε)2k
2k (1−es) 1 +O 1
m
. (6.72)
These results hold pointwise for any fixed real number s, as m→ ∞.
Proof. We start considering the number of all cycles, hence we attach w once to each cyclic component, that leads us to the generating function
gc◦(x, y, v, w) = exp 1
vt(xv, yv) +˜ w
2 log 1
1−t1(x, y)t2(x, y)
= exp1
v˜t(xv, yv)
(1−t1(xv, yv)t2(xv, yv))w/2. (6.73) Clearly, the equation gc◦(x, y, v,1) = g◦(x, y, v) is valid. Hence, the moment generating function is given by
ψc(s) = [xmymvn]g◦(x, y, v, es)
[xmymvn]g◦(x, y, v,1). (6.74)
6.3 The bipartite Cuckoo Graph
Again, the number of tree components equals 2m −n, thus the generating function simplifies to We continue using Cauchy’s formula and the double saddle point method as described in Theorem 3.2. Similar to the univariate case, the method is applicable for fixed s. Hence we further obtain, that the equation
ψc(s) = holds, what completes the proof of the first part of the lemma.
The proof of the second part is very similar, we just replace gc◦ by the generating function
g◦k(x, y, v, w) = exp1
v˜t(xv, yv) + (w−1)2k1t1(xv, yv)kt2(xv, yv)k
1−t1(xv, yv)t2(xv, yv) . (6.77) Hereby, w is used to mark cycles of length 2k. Recall that the generating function of a component containing a cycle of length 2kis given by 2k1t1(x, y)kt2(x, y)k, see (4.41). We proceed as usual and yield
xmymvn Finally, the moment generating function equals
ψk(s) = [xmymvn]gk◦(x, y, v, es) and get the claimed results.
Similar to the results obtained for the usual graph, these moment generating functions correspond to Poisson distributions. Note that we may again replaceε by ε, because of the relation
Together with Lemma 6.1, this proves the first statement of Theorem 6.2.
Once more, there exists an additive relation between the parameters, illustrated by the equation
6 The Structure of the Cuckoo Graph
Trees with fixed size
The proof of this result is more complicated, because the parameters depend on m. In what follows, we make use of the generating function of a bipartite tree component that possesses exactlyknodes of both types. Because of Lemma 4.2, we can write this function as
˜tk(x, y) =
m1+m2=k
mm12−1mm21−1xm1 m1!
ym2
m2!. (6.82)
The following lemmata provide more detailed information about this function.
Lemma 6.7.
t˜k(x0, x0) =
k
l=0
lk−l−1(k−l)l−1 xk0
l! (k−l)! = 2kk−2xk0
k!. (6.83)
Proof. We apply Lagrange’s Inversion Formula to obtain the coefficient ofxkin ˜t(x, x) = 2t(x)−t(x)2, wheret(x) denotes the usual tree function that satisfiest(x) =xexp(t(x)).
Because of the previous relation, it is also clear that the number of unrooted bipartite trees possessingk nodes equals twice the number of unrooted (usual) trees of sizek.
Lemma 6.8.
∂
∂u˜tk(x0eu, x0ev)
(0,0)
=xk0
k
l=0
lk−l(k−l)l−1 1
l! (k−l)! =kk−1xk0
k!. (6.84) Proof. The proof of this lemma is a simple application of Abel’s generalisation of the Binomial Theorem,
x−1(x+y+ka)k=
k
l=0
k l
(x+la)l−1(y+ (k−l)a)k−l, (6.85) see, e.g., Riordan [1968]. We set x → k, y → k and a → −1 and obtain the claimed result.
For simplification, we introduce as before the following notation:
Definition 6.2. Let k denote a natural number and suppose that ε ∈ (0,1) is fixed.
Then, we define the numbers
μ= 2kk−2(1−ε)k−1ek(ε−1)
k! , (6.86)
and
σ2 =μ−2e2k(ε−1)k2k−4(1−ε)2k−3(k2ε2+k2ε−4kε+ 2)
(k!)2 . (6.87)
With help of these preliminary results, we are able to prove the following lemma.
6.3 The bipartite Cuckoo Graph
Lemma 6.9. The number of tree components withkvertices of a randomly chosen mem-ber of of G◦m,m,n possesses mean
mμ+O(1) (6.88)
and variance
mσ2+O(1). (6.89)
Proof. We start introducing the variable wto mark trees possessing exactlyknodes and obtain the generating function
g◦t(x, y, v, w) = exp1
vt(xv, yv) + (w˜ −1)˜tk(xv, yv)
1−t1(xv, yv)t2(xv, yv) , (6.90) that allows us to calculate the l−th factorial moment as follows
Ml= We further simplify this expression and obtain the equation
[xmymvn] Now, we use once more Theorem 3.2 to calculate an asymptotic expansion. By using Lemma 6.7, we obtain that the leading term ofMl equals
(2m−n)l Hence, we have completed the proof of the first statement. Moreover, we conclude that the variance is of order O(m) too, thus its calculation requires to determine the next term of the asymptotic expansion. Similar to the “simplified” situation, we do this in a semi-automatic way using Maple and obtain the claimed result. See the corresponding worksheet for further details.
As in previous calculations, we may of course replace ε by ε. Again, it is possible to establish a central limit theorem.
Lemma 6.10. The number of tree components of size k of a randomly selected member of G◦m,m,(1−ε)m minus mμ and divided by
This equation holds pointwise for any fixed real number r, asm→ ∞.
6 The Structure of the Cuckoo Graph
Proof. This result is again obtained using an adopted saddle point method, similar to the proof of Lemma 6.5. In the following, we make again use of the shortened denotation M = 2m−n=m(1 +ε). Similar to (6.44), we obtain that the Taylor expansion
Using this Taylor expansion, we proceed as in the proof of Lemma 6.5 respectively The-orem 3.2.
Nodes in cycles
In this part of the proof, we count the number of nodes contained in cycles, but we do not count the non root nodes of the trees attached to the cycles. Similar to proof of claims concerning the number of cycles, this result is rather easy to obtain. We make use of the generating function
gn◦(x, y, v, w) = exp1
v˜t(xv, yv)
1−w2t1(xv, yv)t2(xv, yv). (6.98) Hence we get the characteristic function
φn(s) = [xmymvn]gn◦(x, y, v, eis) using the double saddle point method, that is again applicable what can be seen as in the univariate case. It is further straightforward to calculate asymptotic mean and variance.
Finally we use the series expansion
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