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In cyclization in step-growth polymerization, compatible functional groups on the same chain react and close it into a ring. They consume groups without merging two molecules. If $R$ ring-closing events and $J$ intermolecular joins occur in an otherwise balanced bifunctional feed, the group conversion is $p=(J+R)/N_0$, while $N=N_0-J=N_0(1-p)+R$. ThusFor example, an $AB$ dimer can close its remaining $A$ and $B$ ends: two group pairings have occurred among two starting monomer molecules, giving $p=1$, but the resulting single cyclic molecule has only two monomeric units. This counterexample shows exactly where the linear Carothers equation fails. Rings have no chain ends, so counting every chain-bearing molecule as two ends also fails.

Use an ideal feed in which every starting molecule has exactly two suitable functional groups: either one $A$ and one $B$, or two $A$ groups mixed with two $B$ groups. Only an $A$–$B$ pairing can form a joining chemical bond. Assume no loss of feed-derived material, no splitting of chains, no competing consumption of the counted functional groups, no branching, and no cyclization in step-growth polymerization. Linear here describes connectivity; a chain can bend in space. Small byproducts of polycondensation do not enter the chain-bearing molecule count.

Let $N_0$ count all initial feed molecules, $J$ the number of intermolecular joins, and $N$ all remaining chain-bearing molecules, including unreacted monomers. Each join turns two molecules into one, so $N=N_0-J$ exactly. If chain $i$ contains $j_i$ feed-derived monomeric units, conservation gives ${\sum }_{i=1}^N{j}_i=N_0$. Consequently the number-average degree of polymerization isThis is a count of both feed species' monomeric units, rather than a count of interior constitutional repeating units. For $AA+BB$ chemistry an interior repeat often contains two feed-derived units, as explained in monomer. Equal functional group reactivities are not needed for this bookkeeping at a specified $J$; predicting a time course or a chain-length probability distribution requires additional assumptions.

For the figure's eight $AB$ feed molecules, three joins can give lengths $(2,3,1,1,1)$, so $N=5$ and $X_n=8/5$. Joining the first two chains gives $(5,1,1,1)$: the incorporated total stays eight, $N$ becomes four, and $X_n=2$. A graph makes the bookkeeping visible: each starting monomer molecule is a vertex of a graph, and each intermolecular joining chemical bond connects two previously distinct connected components of a graph. The connected pieces are the counted molecules.