This question is an extension of Who's that Polygon? to arbitrary numbers of sides.
A fundamental polygon for a surface is an polygon with a prescribed pairing for all its \$2n\$ sides, each marked with an arrow, such that identifying the sides according to this pairing and the arrow directions produces the surface (up to homeomorphism).
This pairing can be described as a cyclic fundamental word with two each of \$n\$ distinct symbols, where any symbol may be inverted to represent an arrow pointing opposite to the direction of travel around the polygon. For example, the word \$abab^{-1}\$ describes the Klein bottle:
In light of the classification theorem for surfaces, which states that every closed 3D surface is homeomorphic to either the connect-sum of \$g\ge0\$ tori (including the sphere at \$g=0\$) or the connect-sum of \$k>0\$ cross-caps, it may not be immediately obvious what surface a fundamental polygon represents. Fortunately canonical forms and a canonicalisation process exist – the material below comes from Gerhard Ringel's Map Color Theorem.
Canonical words for surfaces
The canonical fundamental word for the connect-sum of \$g\ge0\$ tori, with genus \$g\$, is $$a_1b_1a_1^{-1}b_1^{-1}a_2b_2a_2^{-1}b_2^{-1}\dots a_gb_ga_g^{-1}b_g^{-1}$$ So the sphere's canonical word is the empty string, the normal torus's is \$aba^{-1}b^{-1}\$ and so on.
The canonical fundamental word for the connect-sum of \$k>0\$ cross-caps, with genus \$-k\$ (although this is incorrect, this definition is used here because then the integers bijectively correspond to surface types) is $$a_1a_1a_2a_2\dots a_ka_k$$ The real projective plane and Klein bottle are the \$k=1\$ and \$k=2\$ cases of this sequence, so their canonical words are \$aa\$ and \$aabb\$ respectively.
Canonisation
The surface represented by a fundamental word is unchanged if
- the word is cyclically rotated: \$aabb\to abba\$
- a symbol is inverted: \$aabb\to aab^{-1}b^{-1}\$
- a symbol and its adjacent inverse are cancelled: \$caba^{-1}b^{-1}c^{-1}\to aba^{-1}b^{-1}\$
The surface is orientable iff each symbol appears once inverted and once uninverted in the word. In that case a "handle" can be extracted by the following word transformation, where \$Q,R,S\$ are not all empty: $$PaQbRa^{-1}Sb^{-1}T\to PSRQTaba^{-1}b^{-1}$$ Repeating this process with intermediate cancellation leads to the canonical form for orientable surfaces described above.
If the two occurrences of a symbol are both inverted or both uninverted, the surface is non-orientable. In this case a cross-cap may be extracted by the following word transformation, where \$Q^{-1}\$ is the group-theoretic inverse (so e.g. \$(abc)^{-1}=c^{-1}b^{-1}a^{-1}\$): $$PaQaR\to aaPQ^{-1}R$$ Do this as many times as possible to get a word of the form \$a_1a_1\dots a_ia_iO\$, where \$O\$ is the fundamental word of an orientable surface. Canonise \$O\$ as above and determine its genus \$g\$; the canonical form of the full non-invertible surface is then \$a_1a_1\dots a_{i+2g}a_{i+2g}\$, because a handle turns into two cross-caps in the presence of another cross-cap.
Example
This picture from one of my MathsSE answers is an embedding of the Johnson graph \$J(5,2)\$:
The star-shaped border represents a fundamental polygon where
- each black (dark blue) arc matches the grey (light blue) arc 5 steps away going counterclockwise
- each golden yellow arc matches the light yellow arc 7 steps away going clockwise
- in all cases the matching is parallel (as hinted by the faded exterior), so the surface is orientable
Thus the corresponding fundamental word is $$ah^{-1}bf^{-1}ca^{-1}db^{-1}ei^{-1}fd^{-1}ge^{-1}hc^{-1}ig^{-1}$$ which reduces as follows: $$(a)h^{-1}(b)f^{-1}c(a^{-1})d(b^{-1})ei^{-1}fd^{-1}ge^{-1}hc^{-1}ig^{-1}$$ $$\to (d)f^{-1}(c)h^{-1}ei^{-1}f(d^{-1})ge^{-1}h(c^{-1})ig^{-1}aba^{-1}b^{-1}$$ $$\to ge^{-1}[hh^{-1}]ei^{-1}[ff^{-1}]ig^{-1}aba^{-1}b^{-1}dcd^{-1}c^{-1}$$ $$\to g[e^{-1}e][i^{-1}i]g^{-1}aba^{-1}b^{-1}dcd^{-1}c^{-1}$$ $$\to[gg^{-1}]aba^{-1}b^{-1}dcd^{-1}c^{-1}$$ $$\to aba^{-1}b^{-1}dcd^{-1}c^{-1}$$ Thus the surface has genus 2, which matched my expectation when I constructed the embedding based on a triangulation with 10 vertices and 36 edges – by Euler's formula such a triangular embedding, if orientable, must live on the genus-2 surface. Indeed (glue the three ends of the open surface below in a Y-shape so that the colours match):
Task
Given a list of nonzero integers where the magnitudes from 1 to \$n\$ appear exactly twice each, representing a fundamental word with a negative number marking an inverted symbol, output the genus of the surface corresponding to that fundamental word, using the definition above where a negative genus \$-k\$ means "connect-sum of \$k\$ cross-caps".
This is code-golf; fewest bytes wins.
Test cases
[] -> 0
[1, -1] -> 0
[1, 1] -> -1
[1, 2, -2, -1] -> 0
[1, 2, -1, -2] -> 1
[1, 2, 1, -2] -> -2
[1, 2, 1, 2] -> -1
[-2, -2, -1, 1] -> -1
[1, 2, 3, -1, -2, -3] -> 1
[-3, 1, 3, -2, 1, 2] -> -2
[1, 2, 3, 4, -1, -2, -3, -4] -> 2
[1, 2, 4, 4, -2, -3, 1, -3] -> -2
[1, -8, 2, -6, 3, -1, 4, -2, 5, -9, 6, -4, 7, -5, 8, -3, 9, -7] -> 2
[1, -8, 2, -9, 3, -1, 4, -2, 5, -3, 6, -4, 7, -5, 8, -6, 9, -7] -> 3
[4, 3, 2, 1, 1, 2, 3, 4] -> -4
[3, 4, 1, 2, 6, 7, -4, 5, -2, -3, -7, -1, -5, -6] -> 3
[1, 2, 3, 3, -1, -2] -> -3
