TheTheorem: Schinzel's hypothesis H implies the conjecture is true if you don't insist that the demi-primes are consecutive. Choose
Proof: Choose distinct large primes $q_S$$q_S > 100|G|$ indexed by the 2-element subsets $S$ of $G$. It suffices to construct odd For each $i \in G$, let $Q_i$ be the set of $q_S$ for $S$ such that $i \in S$ and the edge $S$ is not part of $G$. Let $P_i$ be the product of the primes in $p_x$$Q_i$. Let $P = 4 \prod_S q_S^2$.
By the Chinese remainder theorem, for each $x \in G$$i$ we can find a positive integer $a_i$ such that
$p_x$ is not equal to $q_S$ for any $x$ and $S$
$(p_x+1)/2$ is divisible by $q_S$ if and only if $x \in S$ and the edge $S$ is not part of $G$, and
for distinct $x$ and $y$, the numbers $(p_x+1)/2$ and $(p_y+1)/2$ have no common prime factors except for possibly the $q_S$.
$a_i \equiv 1 \bmod{\ell^2}$ for each prime $\ell \le 10|G|$,
In fact$a_i \equiv q-1 \bmod{q^2}$ for each $q \in Q_i$, this is easy: justand
$a_i \equiv 1 \bmod{q_S}$ for each $q_S \notin Q_i$.
Moreover, we can choose the $p_x$ one at a time$a_i$ to be distinct. The conditions imposed on any one Let $p_x$ by 2) and 3) amount$J$ be the set of positive integers up to finitely many congruence conditions with prime moduli$\operatorname{max} a_i$, but excluding all of the $a$'s themselves (i.e., $J$ consists of the numbers in the gaps). For each $j \in J$ choose a prime $s_j$ much larger than all the $a_i$ and all the $q_S$.
Consider the linear polynomials $P n + a_i$ and $(P n + a_i + 1)/(2P_i)$ In $\mathbf{Z}[n]$. For each prime modulus appears$\ell \le 10|G|$ and each $\ell$ of the form $q_S$, all these $2|G|$ polynomials are nonzero mod $\ell$ at most once$n=0$. For each other prime $\ell$, and nonethere exists $n$ such that all these polynomials are askingnonzero mod $p_x$$\ell$, since $n$ needs to beavoid no more than $2|G|$ residue classes mod $\ell$. Furthermore, we can impose the condition that $P n+j$ is divisible by $s_j^2$ for each $j \in J$, and still find $n$ as above. Therefore Schinzel's hypothesis H implies that there exist arbitrarily large positive integers $n$ such that the modulusnumbers $P n+a_i$ and $(P n + a_i + 1)/(2P_i)$ are all prime, so theyand such that $P n+j$ is not prime for $j \in J$. This makes the numbers $p_i:=P n + a_i$ consecutive primes such that $(p_i+1)/2 = P_i r_i$ for some prime $r_i$. If $n$ is sufficiently large, then these primes $r_i$ are satisfiable by Dirichlet's theorem onall distinct and larger than all of the $q_S$. So the greatest common factor of $(p_i+1)/2$ and $(p_j+1)/2$ for $i \ne j$ equals $1$ if there is an edge between $i$ and $j$, and $q_{\{i,j\}}$ otherwise. $\square$
Remark: Given how little is known about consecutive primes in arithmetic progressions, it seems unlikely that the conjecture can be proved unconditionally. But at least now we can be confident that it's true!