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In a bipartite graph, a self-avoiding walk ends in its starting class exactly when it has even length. If degrees are bounded by $\Delta$, deleting the last edge gives ${\sigma }_{2k+1}\le \Delta {\sigma }_{2k}$ and hence $Z_G^0(x)\le Z_G(x)\le (1+\Delta x)Z_G^0(x)$ for $x\ge 0$. Thus these generating functions have the same radius of convergence.

An exponential generating function encodes a sequence using the displayed factorial denominators. It is especially useful for counting labelled combinatorial structures: taking an unordered set of structures with positive size corresponds to exponentiating their exponential generating function. The rooted-tree generating function illustrates this through $T=z{e}^T$.

A power series encoding a sequence through its coefficients. It turns combinatorial splitting and concatenation rules into algebraic identities. A counting generating function is not necessarily a probability generating function: its coefficients need not sum to one.

A horizontal run has generating function $R(z)=(1+z)/(1-z)$. Decomposing each walk into $H(NH{)}^k$ gives $A(z)=R(z)/(1-zR(z))=(1+z)/(1-2z-z^2)$. Thus $a_0=1,a_1=3$ and $a_n=2a_{n-1}+a_{n-2}$. The exponential growth rate is $1+\sqrt{2}$, furnishing a strict lower bound greater than two for the connective constant of all square-lattice self-avoiding walks.

Count partially directed self-avoiding walks using north, east and west steps. Within one horizontal level, a self-avoiding walk must move consistently east or consistently west. It can change horizontal direction only after a north step. Conversely, these rules guarantee self-avoidance: every visited horizontal level is new, and its horizontal run is monotone.

Let $a_n$ count these walks, including $a_0=1$. A horizontal run has generating functionEvery walk decomposes uniquely into an initial horizontal run followed by zero or more pairs consisting of a north step and a horizontal run. Therefore the generating function isComparing coefficients yields $a_0=1$, $a_1=3$, and $a_n=2a_{n-1}+a_{n-2}$ for $n\ge 2$. Solving this linear recurrence relation givesThese are a subset of all self-avoiding walks, so $c_n\ge a_n$ andAllowing either horizontal direction at successive heights gives the strict gain over the north-east-only family.

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.

The generating function weights each self-avoiding walk by $x$ to the power of its length, including the zero-length walk. Its radius of convergence is the reciprocal of the upper exponential growth rate of the rooted counts, by the Cauchy-Hadamard theorem.

Replace every degree-three vertex of a graph in one class of a bipartite graph by a triangle with one port for each incident edge. A self-avoiding walk whose endpoints remain in the other class cannot make two completed passages through the same triangle: each passage needs two unused ports and only three exist. Each old two-edge passage has exactly two replacements, of lengths three and four. Therefore the restricted generating functions satisfy $Z_H^0(x)=Z_G^0(\sqrt{x^3+x^4})$. If the old radius is $1/\mu$, the new radius obeys ${\rho }^3+{\rho }^4={\mu }^{-2}$.