# Families of graphs derived from classical geometries over finite fields¶

These include graphs of polar spaces, affine polar graphs, graphs related to Hermitean unitals, graphs on nonisotropic points, etc

The methods defined here appear in sage.graphs.graph_generators.

sage.graphs.generators.classical_geometries.AffineOrthogonalPolarGraph(d, q, sign='+')

Returns the affine polar graph $$VO^+(d,q),VO^-(d,q)$$ or $$VO(d,q)$$.

Affine Polar graphs are built from a $$d$$-dimensional vector space over $$F_q$$, and a quadratic form which is hyperbolic, elliptic or parabolic according to the value of sign.

Note that $$VO^+(d,q),VO^-(d,q)$$ are strongly regular graphs, while $$VO(d,q)$$ is not.

For more information on Affine Polar graphs, see Affine Polar Graphs page of Andries Brouwer’s website.

INPUT:

• d (integer) – d must be even if sign is not None, and odd otherwise.

• q (integer) – a power of a prime number, as $$F_q$$ must exist.

• sign – must be equal to "+", "-", or None to compute (respectively) $$VO^+(d,q),VO^-(d,q)$$ or $$VO(d,q)$$. By default sign="+".

Note

The graph $$VO^\epsilon(d,q)$$ is the graph induced by the non-neighbors of a vertex in an Orthogonal Polar Graph $$O^\epsilon(d+2,q)$$.

EXAMPLES:

The Brouwer-Haemers graph is isomorphic to $$VO^-(4,3)$$:

sage: g = graphs.AffineOrthogonalPolarGraph(4,3,"-")
sage: g.is_isomorphic(graphs.BrouwerHaemersGraph())
True


Some examples from Brouwer’s table or strongly regular graphs:

sage: g = graphs.AffineOrthogonalPolarGraph(6,2,"-"); g
Affine Polar Graph VO^-(6,2): Graph on 64 vertices
sage: g.is_strongly_regular(parameters=True)
(64, 27, 10, 12)
sage: g = graphs.AffineOrthogonalPolarGraph(6,2,"+"); g
Affine Polar Graph VO^+(6,2): Graph on 64 vertices
sage: g.is_strongly_regular(parameters=True)
(64, 35, 18, 20)


When sign is None:

sage: g = graphs.AffineOrthogonalPolarGraph(5,2,None); g
Affine Polar Graph VO^-(5,2): Graph on 32 vertices
sage: g.is_strongly_regular(parameters=True)
False
sage: g.is_regular()
True
sage: g.is_vertex_transitive()
True


Return the collinearity graph of the generalized quadrangle $$AS(q)$$, or of its dual

Let $$q$$ be an odd prime power. $$AS(q)$$ is a generalized quadrangle (Wikipedia article Generalized_quadrangle) of order $$(q-1,q+1)$$, see 3.1.5 in [PT2009]. Its points are elements of $$F_q^3$$, and lines are sets of size $$q$$ of the form

• $$\{ (\sigma, a, b) \mid \sigma\in F_q \}$$

• $$\{ (a, \sigma, b) \mid \sigma\in F_q \}$$

• $$\{ (c \sigma^2 - b \sigma + a, -2 c \sigma + b, \sigma) \mid \sigma\in F_q \}$$,

where $$a$$, $$b$$, $$c$$ are arbitrary elements of $$F_q$$.

INPUT:

• q – a power of an odd prime number

• dual – if False (default), return the collinearity graph of $$AS(q)$$. Otherwise return the collinearity graph of the dual $$AS(q)$$.

EXAMPLES:

sage: g=graphs.AhrensSzekeresGeneralizedQuadrangleGraph(5); g
AS(5); GQ(4, 6): Graph on 125 vertices
sage: g.is_strongly_regular(parameters=True)
(125, 28, 3, 7)
AS(5)*; GQ(6, 4): Graph on 175 vertices
sage: g.is_strongly_regular(parameters=True)
(175, 30, 5, 5)

sage.graphs.generators.classical_geometries.CossidentePenttilaGraph(q)

Cossidente-Penttila $$((q^3+1)(q+1)/2,(q^2+1)(q-1)/2,(q-3)/2,(q-1)^2/2)$$-strongly regular graph

For each odd prime power $$q$$, one can partition the points of the $$O_6^-(q)$$-generalized quadrangle $$GQ(q,q^2)$$ into two parts, so that on any of them the induced subgraph of the point graph of the GQ has parameters as above [CP2005].

Directly following the construction in [CP2005] is not efficient, as one then needs to construct the dual $$GQ(q^2,q)$$. Thus we describe here a more efficient approach that we came up with, following a suggestion by T.Penttila. Namely, this partition is invariant under the subgroup $$H=\Omega_3(q^2)<O_6^-(q)$$. We build the appropriate $$H$$, which leaves the form $$B(X,Y,Z)=XY+Z^2$$ invariant, and pick up two orbits of $$H$$ on the $$F_q$$-points. One them is $$B$$-isotropic, and we take the representative $$(1:0:0)$$. The other one corresponds to the points of $$PG(2,q^2)$$ that have all the lines on them either missing the conic specified by $$B$$, or intersecting the conic in two points. We take $$(1:1:e)$$ as the representative. It suffices to pick $$e$$ so that $$e^2+1$$ is not a square in $$F_{q^2}$$. Indeed, The conic can be viewed as the union of $$\{(0:1:0)\}$$ and $$\{(1:-t^2:t) | t \in F_{q^2}\}$$. The coefficients of a generic line on $$(1:1:e)$$ are $$[1:-1-eb:b]$$, for $$-1\neq eb$$. Thus, to make sure the intersection with the conic is always even, we need that the discriminant of $$1+(1+eb)t^2+tb=0$$ never vanishes, and this is if and only if $$e^2+1$$ is not a square. Further, we need to adjust $$B$$, by multiplying it by appropriately chosen $$\nu$$, so that $$(1:1:e)$$ becomes isotropic under the relative trace norm $$\nu B(X,Y,Z)+(\nu B(X,Y,Z))^q$$. The latter is used then to define the graph.

INPUT:

• q – an odd prime power.

EXAMPLES:

For $$q=3$$ one gets Sims-Gewirtz graph.

sage: G=graphs.CossidentePenttilaGraph(3)    # optional - gap_packages (grape)
sage: G.is_strongly_regular(parameters=True) # optional - gap_packages (grape)
(56, 10, 0, 2)


For $$q>3$$ one gets new graphs.

sage: G=graphs.CossidentePenttilaGraph(5)    # optional - gap_packages (grape)
sage: G.is_strongly_regular(parameters=True) # optional - gap_packages (grape)
(378, 52, 1, 8)

sage.graphs.generators.classical_geometries.HaemersGraph(q, hyperoval=None, hyperoval_matching=None, field=None, check_hyperoval=True)

Return the Haemers graph obtained from $$T_2^*(q)^*$$

Let $$q$$ be a power of 2. In Sect. 8.A of [BL1984] one finds a construction of a strongly regular graph with parameters $$(q^2(q+2),q^2+q-1,q-2,q)$$ from the graph of $$T_2^*(q)^*$$, constructed by T2starGeneralizedQuadrangleGraph(), by redefining adjacencies in the way specified by an arbitrary hyperoval_matching of the points (i.e. partitioning into size two parts) of hyperoval defining $$T_2^*(q)^*$$.

While [BL1984] gives the construction in geometric terms, it can be formulated, and is implemented, in graph-theoretic ones, of re-adjusting the edges. Namely, $$G=T_2^*(q)^*$$ has a partition into $$q+2$$ independent sets $$I_k$$ of size $$q^2$$ each. Each vertex in $$I_j$$ is adjacent to $$q$$ vertices from $$I_k$$. Each $$I_k$$ is paired to some $$I_{k'}$$, according to hyperoval_matching. One adds edges $$(s,t)$$ for $$s,t \in I_k$$ whenever $$s$$ and $$t$$ are adjacent to some $$u \in I_{k'}$$, and removes all the edges between $$I_k$$ and $$I_{k'}$$.

INPUT:

• q – a power of two

• hyperoval_matching – if None (default), pair each $$i$$-th point of hyperoval with $$(i+1)$$-th. Otherwise, specifies the pairing in the format $$((i_1,i'_1),(i_2,i'_2),...)$$.

• hyperoval – a hyperoval defining $$T_2^*(q)^*$$. If None (default), the classical hyperoval obtained from a conic is used. See the documentation of T2starGeneralizedQuadrangleGraph(), for more information.

• field – an instance of a finite field of order $$q$$, must be provided if hyperoval is provided.

• check_hyperoval – (default: True) if True, check hyperoval for correctness.

EXAMPLES:

using the built-in constructions:

sage: g=graphs.HaemersGraph(4); g
Haemers(4): Graph on 96 vertices
sage: g.is_strongly_regular(parameters=True)
(96, 19, 2, 4)


sage: g=graphs.HaemersGraph(4,hyperoval_matching=((0,5),(1,4),(2,3))); g
Haemers(4): Graph on 96 vertices
sage: g.is_strongly_regular(parameters=True)
(96, 19, 2, 4)

sage.graphs.generators.classical_geometries.NonisotropicOrthogonalPolarGraph(m, q, sign='+', perp=None)

Returns the Graph $$NO^{\epsilon,\perp}_{m}(q)$$

Let the vectorspace of dimension $$m$$ over $$F_q$$ be endowed with a nondegenerate quadratic form $$F$$, of type sign for $$m$$ even.

• $$m$$ even: assume further that $$q=2$$ or $$3$$. Returns the graph of the points (in the underlying projective space) $$x$$ satisfying $$F(x)=1$$, with adjacency given by orthogonality w.r.t. $$F$$. Parameter perp is ignored.

• $$m$$ odd: if perp is not None, then we assume that $$q=5$$ and return the graph of the points $$x$$ satisfying $$F(x)=\pm 1$$ if sign="+", respectively $$F(x) \in \{2,3\}$$ if sign="-", with adjacency given by orthogonality w.r.t. $$F$$ (cf. Sect 7.D of [BL1984]). Otherwise return the graph of nongenerate hyperplanes of type sign, adjacent whenever the intersection is degenerate (cf. Sect. 7.C of [BL1984]). Note that for $$q=2$$ one will get a complete graph.

For more information, see Sect. 9.9 of [BH2012] and [BL1984]. Note that the page of Andries Brouwer’s website uses different notation.

INPUT:

• m - integer, half the dimension of the underlying vectorspace

• q - a power of a prime number, the size of the underlying field

• sign"+" (default) or "-".

EXAMPLES:

$$NO^-(4,2)$$ is isomorphic to Petersen graph:

sage: g=graphs.NonisotropicOrthogonalPolarGraph(4,2,'-'); g
NO^-(4, 2): Graph on 10 vertices
sage: g.is_strongly_regular(parameters=True)
(10, 3, 0, 1)


$$NO^-(6,2)$$ and $$NO^+(6,2)$$:

sage: g=graphs.NonisotropicOrthogonalPolarGraph(6,2,'-')
sage: g.is_strongly_regular(parameters=True)
(36, 15, 6, 6)
sage: g=graphs.NonisotropicOrthogonalPolarGraph(6,2,'+'); g
NO^+(6, 2): Graph on 28 vertices
sage: g.is_strongly_regular(parameters=True)
(28, 15, 6, 10)


$$NO^+(8,2)$$:

sage: g=graphs.NonisotropicOrthogonalPolarGraph(8,2,'+')
sage: g.is_strongly_regular(parameters=True)
(120, 63, 30, 36)


Wilbrink’s graphs for $$q=5$$:

sage: graphs.NonisotropicOrthogonalPolarGraph(5,5,perp=1).is_strongly_regular(parameters=True) # long time
(325, 60, 15, 10)
sage: graphs.NonisotropicOrthogonalPolarGraph(5,5,'-',perp=1).is_strongly_regular(parameters=True) # long time
(300, 65, 10, 15)


Wilbrink’s graphs:

sage: g=graphs.NonisotropicOrthogonalPolarGraph(5,4,'+')
sage: g.is_strongly_regular(parameters=True)
(136, 75, 42, 40)
sage: g=graphs.NonisotropicOrthogonalPolarGraph(5,4,'-')
sage: g.is_strongly_regular(parameters=True)
(120, 51, 18, 24)
sage: g=graphs.NonisotropicOrthogonalPolarGraph(7,4,'+'); g # not tested (long time)
NO^+(7, 4): Graph on 2080 vertices
sage: g.is_strongly_regular(parameters=True) # not tested (long time)
(2080, 1071, 558, 544)

sage.graphs.generators.classical_geometries.NonisotropicUnitaryPolarGraph(m, q)

Returns the Graph $$NU(m,q)$$.

Returns the graph on nonisotropic, with respect to a nondegenerate Hermitean form, points of the $$(m-1)$$-dimensional projective space over $$F_q$$, with points adjacent whenever they lie on a tangent (to the set of isotropic points) line. For more information, see Sect. 9.9 of [BH2012] and series C14 in [Hub1975].

INPUT:

• m,q (integers) – $$q$$ must be a prime power.

EXAMPLES:

sage: g=graphs.NonisotropicUnitaryPolarGraph(5,2); g
NU(5, 2): Graph on 176 vertices
sage: g.is_strongly_regular(parameters=True)
(176, 135, 102, 108)

sage.graphs.generators.classical_geometries.Nowhere0WordsTwoWeightCodeGraph(q, hyperoval=None, field=None, check_hyperoval=True)

Return the subgraph of nowhere 0 words from two-weight code of projective plane hyperoval.

Let $$q=2^k$$ and $$\Pi=PG(2,q)$$. Fix a hyperoval $$O \subset \Pi$$. Let $$V=F_q^3$$ and $$C$$ the two-weight 3-dimensional linear code over $$F_q$$ with words $$c(v)$$ obtained from $$v\in V$$ by computing

$c(v)=(\langle v,o_1 \rangle,...,\langle v,o_{q+2} \rangle), o_j \in O.$

$$C$$ contains $$q(q-1)^2/2$$ words without 0 entries. The subgraph of the strongly regular graph of $$C$$ induced on the latter words is also strongly regular, assuming $$q>4$$. This is a construction due to A.E.Brouwer [Bro2016], and leads to graphs with parameters also given by a construction in [HHL2009]. According to [Bro2016], these two constructions are likely to produce isomorphic graphs.

INPUT:

• q – a power of two

• hyperoval – a hyperoval (i.e. a complete 2-arc; a set of points in the plane meeting every line in 0 or 2 points) in $$PG(2,q)$$ over the field field. Each point of hyperoval must be a length 3 vector over field with 1st non-0 coordinate equal to 1. By default, hyperoval and field are not specified, and constructed on the fly. In particular, hyperoval we build is the classical one, i.e. a conic with the point of intersection of its tangent lines.

• field – an instance of a finite field of order $$q$$, must be provided if hyperoval is provided.

• check_hyperoval – (default: True) if True, check hyperoval for correctness.

EXAMPLES:

using the built-in construction:

sage: g=graphs.Nowhere0WordsTwoWeightCodeGraph(8); g
Nowhere0WordsTwoWeightCodeGraph(8): Graph on 196 vertices
sage: g.is_strongly_regular(parameters=True)
(196, 60, 14, 20)
sage: g=graphs.Nowhere0WordsTwoWeightCodeGraph(16) # not tested (long time)
sage: g.is_strongly_regular(parameters=True)       # not tested (long time)
(1800, 728, 268, 312)


sage: F=GF(8)
sage: O=[vector(F,(0,0,1)),vector(F,(0,1,0))]+[vector(F, (1,x^2,x)) for x in F]
sage: g=graphs.Nowhere0WordsTwoWeightCodeGraph(8,hyperoval=O,field=F); g
Nowhere0WordsTwoWeightCodeGraph(8): Graph on 196 vertices
sage: g.is_strongly_regular(parameters=True)
(196, 60, 14, 20)

sage.graphs.generators.classical_geometries.OrthogonalDualPolarGraph(e, d, q)

Return dual polar graph on $$GO^e(n,q)$$ of diameter $$d$$. The value of $$n$$ is determined by $$d$$ and $$e$$.

The graph is distance-regular with classical parameters $$(d, q, 0, q^e)$$.

INPUT:

• e – integer; type of the orthogonal polar space to consider; must be $$-1, 0$$ or $$1$$.

• d – integer; diameter of the graph

• q – integer; prime power; order of the finite field over which to build the polar space

EXAMPLES:

sage: G = graphs.OrthogonalDualPolarGraph(1,3,2)
sage: G.is_distance_regular(True)
([7, 6, 4, None], [None, 1, 3, 7])
sage: G = graphs.OrthogonalDualPolarGraph(0,3,3)
sage: G.is_distance_regular(True)
([39, 36, 27, None], [None, 1, 4, 13])
sage: G.order()
1120


REFERENCES:

See [BCN1989] pp. 274-279 or [VDKT2016] p. 22.

sage.graphs.generators.classical_geometries.OrthogonalPolarGraph(m, q, sign='+')

Returns the Orthogonal Polar Graph $$O^{\epsilon}(m,q)$$.

For more information on Orthogonal Polar graphs, see the page of Andries Brouwer’s website.

INPUT:

• m,q (integers) – $$q$$ must be a prime power.

• sign"+" or "-" if $$m$$ is even, "+" (default) otherwise.

EXAMPLES:

sage: G = graphs.OrthogonalPolarGraph(6,3,"+"); G
Orthogonal Polar Graph O^+(6, 3): Graph on 130 vertices
sage: G.is_strongly_regular(parameters=True)
(130, 48, 20, 16)
sage: G = graphs.OrthogonalPolarGraph(6,3,"-"); G
Orthogonal Polar Graph O^-(6, 3): Graph on 112 vertices
sage: G.is_strongly_regular(parameters=True)
(112, 30, 2, 10)
sage: G = graphs.OrthogonalPolarGraph(5,3); G
Orthogonal Polar Graph O(5, 3): Graph on 40 vertices
sage: G.is_strongly_regular(parameters=True)
(40, 12, 2, 4)
sage: G = graphs.OrthogonalPolarGraph(8,2,"+"); G
Orthogonal Polar Graph O^+(8, 2): Graph on 135 vertices
sage: G.is_strongly_regular(parameters=True)
(135, 70, 37, 35)
sage: G = graphs.OrthogonalPolarGraph(8,2,"-"); G
Orthogonal Polar Graph O^-(8, 2): Graph on 119 vertices
sage: G.is_strongly_regular(parameters=True)
(119, 54, 21, 27)

sage.graphs.generators.classical_geometries.SymplecticDualPolarGraph(m, q)

Returns the Symplectic Dual Polar Graph $$DSp(m,q)$$.

For more information on Symplectic Dual Polar graphs, see [BCN1989] and Sect. 2.3.1 of [Coh1981].

INPUT:

• m,q (integers) – $$q$$ must be a prime power, and $$m$$ must be even.

EXAMPLES:

sage: G = graphs.SymplecticDualPolarGraph(6,3); G       # not tested (long time)
Symplectic Dual Polar Graph DSp(6, 3): Graph on 1120 vertices
sage: G.is_distance_regular(parameters=True)            # not tested (long time)
([39, 36, 27, None], [None, 1, 4, 13])

sage.graphs.generators.classical_geometries.SymplecticPolarGraph(d, q, algorithm=None)

Returns the Symplectic Polar Graph $$Sp(d,q)$$.

The Symplectic Polar Graph $$Sp(d,q)$$ is built from a projective space of dimension $$d-1$$ over a field $$F_q$$, and a symplectic form $$f$$. Two vertices $$u,v$$ are made adjacent if $$f(u,v)=0$$.

See the page on symplectic graphs on Andries Brouwer’s website.

INPUT:

• d,q (integers) – note that only even values of $$d$$ are accepted by the function.

• algorithm – if set to ‘gap’ then the computation is carried via GAP library interface, computing totally singular subspaces, which is faster for $$q>3$$. Otherwise it is done directly.

EXAMPLES:

Computation of the spectrum of $$Sp(6,2)$$:

sage: g = graphs.SymplecticPolarGraph(6,2)
sage: g.is_strongly_regular(parameters=True)
(63, 30, 13, 15)
sage: set(g.spectrum()) == {-5, 3, 30}
True


The parameters of $$Sp(4,q)$$ are the same as of $$O(5,q)$$, but they are not isomorphic if $$q$$ is odd:

sage: G = graphs.SymplecticPolarGraph(4,3)
sage: G.is_strongly_regular(parameters=True)
(40, 12, 2, 4)
sage: O=graphs.OrthogonalPolarGraph(5,3)
sage: O.is_strongly_regular(parameters=True)
(40, 12, 2, 4)
sage: O.is_isomorphic(G)
False
sage: graphs.SymplecticPolarGraph(6,4,algorithm="gap").is_strongly_regular(parameters=True) # not tested (long time)
(1365, 340, 83, 85)


Return the collinearity graph of the generalized quadrangle $$T_2^*(q)$$, or of its dual

Let $$q=2^k$$ and $$\Theta=PG(3,q)$$. $$T_2^*(q)$$ is a generalized quadrangle (Wikipedia article Generalized_quadrangle) of order $$(q-1,q+1)$$, see 3.1.3 in [PT2009]. Fix a plane $$\Pi \subset \Theta$$ and a hyperoval $$O \subset \Pi$$. The points of $$T_2^*(q):=T_2^*(O)$$ are the points of $$\Theta$$ outside $$\Pi$$, and the lines are the lines of $$\Theta$$ outside $$\Pi$$ that meet $$\Pi$$ in a point of $$O$$.

INPUT:

• q – a power of two

• dual – if False (default), return the graph of $$T_2^*(O)$$. Otherwise return the graph of the dual $$T_2^*(O)$$.

• hyperoval – a hyperoval (i.e. a complete 2-arc; a set of points in the plane meeting every line in 0 or 2 points) in the plane of points with 0th coordinate 0 in $$PG(3,q)$$ over the field field. Each point of hyperoval must be a length 4 vector over field with 1st non-0 coordinate equal to 1. By default, hyperoval and field are not specified, and constructed on the fly. In particular, hyperoval we build is the classical one, i.e. a conic with the point of intersection of its tangent lines.

• field – an instance of a finite field of order $$q$$, must be provided if hyperoval is provided.

• check_hyperoval – (default: True) if True, check hyperoval for correctness.

EXAMPLES:

using the built-in construction:

sage: g=graphs.T2starGeneralizedQuadrangleGraph(4); g
T2*(O,4); GQ(3, 5): Graph on 64 vertices
sage: g.is_strongly_regular(parameters=True)
(64, 18, 2, 6)
T2*(O,4)*; GQ(5, 3): Graph on 96 vertices
sage: g.is_strongly_regular(parameters=True)
(96, 20, 4, 4)


sage: F=GF(4,'b')
sage: O=[vector(F,(0,0,0,1)),vector(F,(0,0,1,0))]+[vector(F, (0,1,x^2,x)) for x in F]
T2*(O,4); GQ(3, 5): Graph on 64 vertices
sage: g.is_strongly_regular(parameters=True)
(64, 18, 2, 6)

sage.graphs.generators.classical_geometries.TaylorTwographDescendantSRG(q, clique_partition=None)

constructing the descendant graph of the Taylor’s two-graph for $$U_3(q)$$, $$q$$ odd

This is a strongly regular graph with parameters $$(v,k,\lambda,\mu)=(q^3, (q^2+1)(q-1)/2, (q-1)^3/4-1, (q^2+1)(q-1)/4)$$ obtained as a two-graph descendant of the Taylor's two-graph $$T$$. This graph admits a partition into cliques of size $$q$$, which are useful in TaylorTwographSRG(), a strongly regular graph on $$q^3+1$$ vertices in the Seidel switching class of $$T$$, for which we need $$(q^2+1)/2$$ cliques. The cliques are the $$q^2$$ lines on $$v_0$$ of the projective plane containing the unital for $$U_3(q)$$, and intersecting the unital (i.e. the vertices of the graph and the point we remove) in $$q+1$$ points. This is all taken from §7E of [BL1984].

INPUT:

• q – a power of an odd prime number

• clique_partition – if True, return $$q^2-1$$ cliques of size $$q$$ with empty pairwise intersection. (Removing all of them leaves a clique, too), and the point removed from the unital.

EXAMPLES:

sage: g=graphs.TaylorTwographDescendantSRG(3); g
Taylor two-graph descendant SRG: Graph on 27 vertices
sage: g.is_strongly_regular(parameters=True)
(27, 10, 1, 5)
sage: from sage.combinat.designs.twographs import taylor_twograph
sage: T = taylor_twograph(3)                           # long time
sage: g.is_isomorphic(T.descendant(T.ground_set())) # long time
True
sage: g=graphs.TaylorTwographDescendantSRG(5)    # not tested (long time)
sage: g.is_strongly_regular(parameters=True)  # not tested (long time)
(125, 52, 15, 26)

sage.graphs.generators.classical_geometries.TaylorTwographSRG(q)

constructing a strongly regular graph from the Taylor’s two-graph for $$U_3(q)$$, $$q$$ odd

This is a strongly regular graph with parameters $$(v,k,\lambda,\mu)=(q^3+1, q(q^2+1)/2, (q^2+3)(q-1)/4, (q^2+1)(q+1)/4)$$ in the Seidel switching class of Taylor two-graph. Details are in §7E of [BL1984].

INPUT:

• q – a power of an odd prime number

EXAMPLES:

sage: t=graphs.TaylorTwographSRG(3); t
Taylor two-graph SRG: Graph on 28 vertices
sage: t.is_strongly_regular(parameters=True)
(28, 15, 6, 10)

sage.graphs.generators.classical_geometries.UnitaryDualPolarGraph(m, q)

Returns the Dual Unitary Polar Graph $$U(m,q)$$.

For more information on Unitary Dual Polar graphs, see [BCN1989] and Sect. 2.3.1 of [Coh1981].

INPUT:

• m,q (integers) – $$q$$ must be a prime power.

EXAMPLES:

The point graph of a generalized quadrangle (see Wikipedia article Generalized_quadrangle, [PT2009]) of order (8,4):

sage: G = graphs.UnitaryDualPolarGraph(5,2); G   # long time
Unitary Dual Polar Graph DU(5, 2); GQ(8, 4): Graph on 297 vertices
sage: G.is_strongly_regular(parameters=True)     # long time
(297, 40, 7, 5)


Another way to get the generalized quadrangle of order (2,4):

sage: G = graphs.UnitaryDualPolarGraph(4,2); G
Unitary Dual Polar Graph DU(4, 2); GQ(2, 4): Graph on 27 vertices
sage: G.is_isomorphic(graphs.OrthogonalPolarGraph(6,2,'-'))
True


A bigger graph:

sage: G = graphs.UnitaryDualPolarGraph(6,2); G   # not tested (long time)
Unitary Dual Polar Graph DU(6, 2): Graph on 891 vertices
sage: G.is_distance_regular(parameters=True)     # not tested (long time)
([42, 40, 32, None], [None, 1, 5, 21])

sage.graphs.generators.classical_geometries.UnitaryPolarGraph(m, q, algorithm='gap')

Returns the Unitary Polar Graph $$U(m,q)$$.

For more information on Unitary Polar graphs, see the page of Andries Brouwer’s website.

INPUT:

• m,q (integers) – $$q$$ must be a prime power.

• algorithm – if set to ‘gap’ then the computation is carried via GAP library interface, computing totally singular subspaces, which is faster for large examples (especially with $$q>2$$). Otherwise it is done directly.

EXAMPLES:

sage: G = graphs.UnitaryPolarGraph(4,2); G
Unitary Polar Graph U(4, 2); GQ(4, 2): Graph on 45 vertices
sage: G.is_strongly_regular(parameters=True)
(45, 12, 3, 3)
sage: graphs.UnitaryPolarGraph(5,2).is_strongly_regular(parameters=True)
(165, 36, 3, 9)
sage: graphs.UnitaryPolarGraph(6,2)    # not tested (long time)
Unitary Polar Graph U(6, 2): Graph on 693 vertices