 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 givesFor $x\ge 0$, the square-root argument increases strictly from zero to infinity. Nonnegative coefficients and the radius from part (iii) therefore give the unique positive thresholdThis identifies the radius of the restricted series, without assuming that the new graph has equal counts from every graph vertex.