1 First of all, let $f: \Bool^{{\mathsf{N}}} \rightarrow \Bool^{{\mathsf{N}}}$.
2 It has been shown~\cite[Theorem 4]{bcgr11:ip} that
3 if its iteration graph $\Gamma(f)$ is strongly connected, then
4 the output of $\chi_{\textit{14Secrypt}}$ follows
5 a law that tends to the uniform distribution
6 if and only if its Markov matrix is a doubly stochastic matrix.
9 In~\cite[Section 4]{DBLP:conf/secrypt/CouchotHGWB14},
10 we have presented a general scheme which generates
11 function with strongly connected iteration graph $\Gamma(f)$ and
12 with doubly stochastic Markov probability matrix.
14 Basically, let us consider the ${\mathsf{N}}$-cube. Let us next
15 remove one Hamiltonian cycle in this one. When an edge $(x,y)$
16 is removed, an edge $(x,x)$ is added.
19 For instance, the iteration graph $\Gamma(f^*)$
20 (given in Figure~\ref{fig:iteration:f*})
21 is the $3$-cube in which the Hamiltonian cycle
22 $000,100,101,001,011,111,110,010,000$
26 We first have proven the following result, which
27 states that the ${\mathsf{N}}$-cube without one
29 has the awaited property with regard to the connectivity.
32 The iteration graph $\Gamma(f)$ issued from
33 the ${\mathsf{N}}$-cube where an Hamiltonian
34 cycle is removed is strongly connected.
37 Moreover, if all the transitions have the same probability ($\frac{1}{n}$),
38 we have proven the following results:
40 The Markov Matrix $M$ resulting from the ${\mathsf{N}}$-cube in
42 cycle is removed, is doubly stochastic.
45 Let us consider now a ${\mathsf{N}}$-cube where an Hamiltonian
47 Let $f$ be the corresponding function.
48 The question which remains to solve is
49 can we always find $b$ such that $\Gamma_{\{b\}}(f)$ is strongly connected.
51 The answer is indeed positive. We furtheremore have the following strongest
54 There exist $b \in \Nats$ such that $\Gamma_{\{b\}}(f)$ is complete.
57 There is an arc $(x,y)$ in the
58 graph $\Gamma_{\{b\}}(f)$ if and only if $M^b_{xy}$ is positive
59 where $M$ is the Markov matrix of $\Gamma(f)$.
60 It has been shown in~\cite[Lemma 3]{bcgr11:ip} that $M$ is regular.
61 There exists thus $b$ such there is an arc between any $x$ and $y$.
64 This section ends with the idea of removing a Hamiltonian cycle in the
66 In such a context, the Hamiltonian cycle is equivalent to a Gray code.
67 Many approaches have been proposed a way to build such codes, for instance
68 the Reflected Binary Code. In this one, one of the bits is switched
69 exactly $2^{\mathsf{N}-}$ for a $\mathsf{N}$-length cycle.
73 The function that is built
76 The next section presents how to build balanced Hamiltonian cycles in the
77 $\mathsf{N}$-cube with the objective to embed them into the
78 pseudorandom number generator.