A fork-noodle pairing computation

All intersections lie on $x=\frac{1}{5}$. In increasing order along $N$ the tine points have heights $(-\frac{3}{20},\frac{1}{25},\frac{7}{100},\frac{1}{10})$ and the parallel points have heights $(-\frac{29}{200},\frac{7}{200},\frac{3}{40},\frac{19}{200})$. Thus the combined order is $z_1,z_1^{\mathrm{\prime }},z_2^{\mathrm{\prime }},z_2,z_3,z_3^{\mathrm{\prime }},z_4^{\mathrm{\prime }},z_4$. The loops ${\xi }_i$ that follow the handle and tine to $z_i$ and return to $d_1$ along $N$ have total puncture windings $(-2,-1,-1,-1)$: the first clockwise loop encloses $p_2,p_3$, while each of the other loops encloses only $p_2$. The hairpin changes none of these windings. The clockwise loop $N$ followed by the lower boundary return has $A_0=-1$. The labelled-loop formula $a_{i,j}=a_i+a_j+A_0$ therefore gives $(a_{i,j})=\left(\begin{array}{cccc}-5& -4& -4& -4\\ -4& -3& -3& -3\\ -4& -3& -3& -3\\ -4& -3& -3& -3\end{array}\right)\mathrm{.}$

Here is an exact ray-crossing calculation of the mutual exponents, rather than an inference from their parities. Let $D_{i,j}$ be the difference of the two labelled tracks along ${\delta }_{i,j}$. Merge the rational segment-time breakpoints of the two tracks; the resulting difference is piecewise affine with rational vertices and never zero. Count signed crossings of the ray in direction $1+\mathrm{i}\mathrm{/}10$: for consecutive difference vertices $(x,y),(X,Y)$ a crossing occurs when $g=y-x\mathrm{/}10$ and $G=Y-X\mathrm{/}10$ have opposite signs and, at $u=-g\mathrm{/}(G-g)$, $x+u(X-x)+(y+u(Y-y))\mathrm{/}10>0$. Its sign is positive when $g<0<G$, negative in the reverse case. No difference vertex lies on this ray. If the labels return, this closed difference path has signed count $-1$, so $b=2(-1)=-2$. If they exchange, concatenate the difference path with its negative; the resulting closed path has signed count $-1$, so $b=-1$. Substitution of the listed vertices gives these counts for every pair. The returning case is exactly $z_i$ preceding $z_j^{\mathrm{\prime }}$ along $N$, yielding $(b_{i,j})=\left(\begin{array}{cccc}-2& -2& -2& -2\\ -1& -1& -2& -2\\ -1& -1& -2& -2\\ -1& -1& -1& -1\end{array}\right)\mathrm{.}$ Together with step 1.1 this specifies every monomial $m_{i,j}=q^{a_{i,j}}{t}^{b_{i,j}}$.

The diagonal exponents are $(-2,-1,-2,-1)$. Applying ${ϵ}_{i,j}=-(-1{)}^{b_{i,i}+b_{j,j}+b_{i,j}}$ to step 2.1 gives $({ϵ}_{i,j})=\left(\begin{array}{cccc}-1& 1& -1& 1\\ -1& 1& 1& -1\\ 1& -1& -1& 1\\ -1& 1& -1& 1\end{array}\right)\mathrm{.}$ The two exponent matrices and this sign matrix list all sixteen labelled contributions; no pair is omitted.

Rows two and three of the signed monomial table cancel entry by entry. In row one, columns two and three cancel; in row four, columns three and four cancel. The four remaining terms give $⟨N,F⟩=-q^{-5}{t}^{-2}+q^{-4}{t}^{-2}-q^{-4}{t}^{-1}+q^{-3}{t}^{-1}=q^{-5}{t}^{-2}(q-1)(1+qt)\mathrm{\ne }0.$ Thus geometrically distinct terms really do cancel, although the collected polynomial is nonzero. The sum of the sixteen signs is zero, so the ordinary algebraic intersection number of the projected surfaces vanishes. The deck-labelled polynomial retains information lost by this unweighted count.