= Solution
Consider a walk in the replaced <graph> that starts and ends in $V_0$. Each passage through a triangle enters at one port and leaves at a different port. It cannot revisit that triangle: the first passage uses at least two of its three vertices, whereas a later completed passage would need two previously unused ports. Nor can it immediately return through its entry port, since that would repeat a <graph vertex>. Contracting each passage therefore gives a <self-avoiding walk> in $G$ ending in $V_0$.
Conversely, each length-$2k$ walk of this kind in $G$ visits $k$ distinct vertices of $V_1$. At each one, its incoming and outgoing ports determine exactly two routes through the triangle: the direct internal <edge> or the two internal <edges> through the third port. Including the two external <edges>, these have lengths three and four. The choices at distinct triangles are independent combinatorial choices, so that walk contributes $(x^3+x^4)^k$ to the new <generating function>. This <triangle replacement for self-avoiding walks> gives
$$
\boxed{Z_H^0(x)=\sum_{k\geq0}\sigma_{2k}(x^3+x^4)^k=Z_G^0\bigl(\sqrt{x^3+x^4}\bigr)}.
$$
For $x\geq0$, the square-root argument increases strictly from zero to infinity. Nonnegative coefficients and the radius from part (iii) therefore give the unique positive threshold
$$
\boxed{\rho^3+\rho^4=\mu^{-2}}.
$$
This identifies the radius of the restricted series, without assuming that the new <graph> has equal counts from every <graph vertex>.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/iii/paper-26-triangle-routes.png]
{title=An original two-edge passage and its two triangle replacements, of lengths three and four}
{height=360}
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