Rigid realizations of graphs with few locations in the plane

Adiprasito and Nevo (2018) proved that there exists a set of 76 points in R 3 such that every triangulated planar graph has an infinitesimally rigid realization in which each vertex is mapped to a point in this set. In this paper we show that there exists a set of 26 points in the plane such that every planar graph which is generically rigid in R 2 has an infinitesimally rigid realization in which each vertex is mapped to a point in this set. It is known that a similar result, with a set of constant size, does not hold for the family of all generically rigid graphs in R d , d ≥ 2. We show that there exists a constant c such that for every positive integer n there is a set of c ( √ n ) points in the plane such that every generically rigid graph in R 2 on n vertices has an infinitesimally rigid realization on this set. This bound is tight up to a constant factor. its members on a given subset of R d of constant cardinality? They showed that triangulated planar graphs have such realizations on 76 points in R 3 . They also gave similar results

a b s t r a c t Adiprasito and Nevo (2018) proved that there exists a set of 76 points in R 3 such that every triangulated planar graph has an infinitesimally rigid realization in which each vertex is mapped to a point in this set.
In this paper we show that there exists a set of 26 points in the plane such that every planar graph which is generically rigid in R 2 has an infinitesimally rigid realization in which each vertex is mapped to a point in this set.
It is known that a similar result, with a set of constant size, does not hold for the family of all generically rigid graphs in R d , d ≥ 2. We show that there exists a constant c such that for every positive integer n there is a set of c( √ n) points in the plane such that every generically rigid graph in R 2 on n vertices has an infinitesimally rigid realization on this set. This bound is tight up to a constant factor.

Introduction
Adiprasito and Nevo [1] asked the following question: ''How generic does the realization of a generically rigid graph need to be to guarantee that it is infinitesimally rigid?'' In fact, Adiprasito and Nevo considered a more exact question. Which graph classes have infinitesimally rigid realizations for each of its members on a given subset of R d of constant cardinality? They showed that triangulated planar graphs have such realizations on 76 points in R 3 . They also gave similar results for 3-dimensional realizations of triangulations of closed surfaces. The problem whether a similar statement is true for planar rigid graphs in R 2 was left open in [1].
The first result of this paper is that there exists a set A of 26 points in the plane such that every planar graph which is generically rigid in R 2 has an infinitesimally rigid realization on A. Furthermore, a similar result follows when we change the class of rigid planar graphs to the class of rigid graphs whose members can be embedded in a given closed surface. This implies our second result that states that for every positive integer n there exists a set A n of O( √ n) points of the plane such that every graph on n vertices which is rigid in the plane has an infinitesimally rigid realization on A n . We note that the above question of Adiprasito and Nevo was also considered before by Fekete and Jordán [2] who proved that instead of using generic points one can always find an infinitesimally rigid injective realization on a grid of size ( (1)).
Before introducing the above problems formally, we summarize some basics of rigidity theory.
We refer to [4] for more details. A d-dimensional framework is a pair (G, p), is a graph and p is a map from V to R d . We will also refer to (G, p) (or less precisely to p) as a realization of G and to p(v) as the location of v for a vertex v ∈ V . We assign to (G, p) a matrix, called the rigidity matrix R(G, p) ∈ R |E|×d|V | , which is defined as follows. We assign a row of R(G, p) to each edge uv ∈ E and d columns to each v ∈ V . The row of R(G, p) assigned to uv ∈ E contains the d + d coordinates of p(u) − p(v) and p(v) − p(u) in the d columns assigned to u and in the d columns assigned to v, respectively, while the other entries are zeros.
An infinitesimal motion of a framework (G, p) is an assignment m : V → R d of infinitesimal velocities to the vertices, such that where K V is the complete graph on V . (G, p) is infinitesimally rigid in R d if all of its infinitesimal motions are trivial. We also note that the dimension of the vector space of the trivial infinitesimal motions of a d-dimensional framework is when the underlying graph has at least d vertices.
A set of points A ⊆ R d is said to be generic if the (multi)set of the coordinates of the points in A is algebraically independent over Q. A realization p of G is said to be generic if p is injective and its image is a generic set. It follows by the definition of a generic realization that if the determinant of a square submatrix of R(G, p 0 ) is 0 for a generic realization p 0 , then the determinant of the same submatrix of R(G, p) is also 0 for every other realization p. Thus rank(R(G, p 0 )) = max{rank(R(G, p)) : p : V → R d }. Therefore, the infinitesimal rigidity of frameworks in R d is a generic property, that is, the infinitesimal rigidity of (G, p) depends only on the graph G and not the particular realization p, if (G, p) is generic (see [10]). We say that the graph G is rigid in R d if all (or equivalently, if some) It is easy to see that if G = (V , E) is minimally rigid and p is an infinitesimally rigid realization of G, then the rows of R(G, p) are linearly independent. Let E(X ) denote the set of edges in a graph We have the following necessary conditions for minimal rigidity.
Pollaczek-Geiringer [7] (and Laman [5]) showed that these necessary conditions are also sufficient for minimal rigidity when d = 2. We note that Fekete and Jordán [2] observed that the class of graphs which are rigid on the line (that is, the class of connected graphs) is rigid on the line with 2 locations. A similar result with a constant c does not hold in R 2 by the following result. We prove Theorem 1.6 in Section 3. In Section 4 we obtain the following result by using some observations on graph embeddings in closed surfaces. Note that Proposition 1.5 shows that the above bound on the cardinality of A is sharp up to a constant factor.
Finally, in Section 5, we show the following theorem by using another idea of Fekete and Jordán [2].
This result implies that some slightly weaker statements remain true if we change 'generic' in Theorems 1.4, 1.6, or 1.7 to 'integral'. However, note that, in Theorem 1.8, the image set B G of p ′ depends on the graph G.

Preliminaries
In this section, we list the main lemmas which we use to prove Theorem 1. 6. In what follows, we will say that a set X ⊆ V is tight in a sparse graph G = (V , E) if the subgraph G[X ] induced by X is tight. The following two lemmas follow from the supermodularity of the function i G , see [4]. Lemma 2.1. Let G = (V , E) be a sparse graph and let X , Y ⊂ V be two tight sets in G with |X ∩ Y | ≥ 1.
Our main tool in the proof of Theorem 1.6 is the following generalization of the key lemma of Adiprasito Proof. By deleting some edges of G for which the corresponding row of the rigidity matrix R(G, p) is linearly dependent from the other rows of R(G, p), we can assume that (G, p) is minimally by Theorem 1.1. Let us consider the rigidity matrix R(G, p v ) of another realization p v of G which arises by taking One can look at the polynomials over R with d variables and maximum degree at most c as a ( d+c d ) -dimensional vector space over R whose bases are the monomials with d variables and maximum degree at most c. As (G, p) is infinitesimally rigid, at least one of the polynomials corresponding to the submatrices of R(G, p v ), say P, must be not identically zero.
We claim that no choice of Therefore, we can extend p ′ We note that Lemma 2.3 immediately implies the following.  . Then there exists a Adiprasio and Nevo [1] used Lemma 2.3 and the fact that contraction of an edge uv maintains rigidity in R 3 when u and v have two common neighbors (see [9]) to prove Theorem 1.4 by induction on |V |. Beside other ideas, to use Lemma 2.3, they first showed that the above contraction can be performed in triangulated planar graphs in such a way that one endvertex of the contracted edge has low degree and hence, when the reverse operation of a contraction is performed, the arising new vertex will have low degree. We note that it is enough to prove Theorem 1.6 for planar Laman graphs. Such graphs always have at least two triangle faces and it is known that we always can contract an edge incident to a triangle face by maintaining rigidity and vice versa (see [3]). However, by repeatedly using the operation in Fig. 1 on the 4-faces neighboring the four vertices of the two central triangles, one can see that the degree of each vertex, which is incident with a triangle face, can be arbitrary large. This implies that, when we use the reverse operation to build up our graph we cannot guarantee any upper bound for the degree of the new vertex and hence we cannot use Lemma 2.3 for the induction. Hence, for our proof, we shall use some other operations that preserve the rigidity of frameworks. The Henneberg-0 extension, or simply 0-extension, on G adds a new vertex and connects it to 2 distinct vertices of G. The 1-extension, deletes an edge uw ∈ E, adds a new vertex v and connects it to u, w and one other vertex of G. The following two lemmas show that 0-and 1-extensions preserve rigidity.
. Let G + be a 0-extension of G that arises by adding a new vertex v with two incident edges vv 1 and vv 2 and let us take The following well-known result was a key in the proof of Theorem 1.2 in [5,7].

Lemma 2.7 ([5,7]). A graph is Laman if and only if it arises from K 2 by using 0-and 1-extensions. □
The inverse operation of 1-extension is called a 1-reduction. The following lemma is also well-known.

Lemma 2.8 ([8]). Let G be a Laman graph and v be a vertex of G with exactly
Note that, to use Lemmas 2.5 and 2.6 in our inductive proof, some pairs of vertices must have different location in the realization of the reduced framework. To ensure this property, we introduce a set F of extra edges which denotes the pair of vertices which must have different locations. Since there are infinitely many Laman graphs with constant number of vertices which have degree at most three, we cannot guarantee that (after a sequence of reductions) our graph has a vertex of degree at most three with low ''F -degree'', that is, low degree when restricted to edges in F . Thus, although all Laman graphs can be constructed by using only 0-and 1-extensions, we will also need to use the following operation which is called an X-replacement. Let G = (V , E) and v 1 v 2 , v 3 v 4 ∈ E be two vertex-disjoint edges. The X-replacement deletes v 1 v 2 , v 3 v 4 , adds a new vertex v and connects it to v 1 , v 2 , v 3 and v 4 . The following lemma shows that X-replacement preserves rigidity.   [8] showed that a degree 4 vertex can always be removed from a Laman graph along with adding two, possibly not independent, new edges between its neighbors such that the resulting graph is Laman. The following lemma shows when we can get a Laman graph after an inverse X -replacement.

Tay and Whiteley
Proof. Since the necessity of the condition is obvious, we only prove its sufficiency. Observe that G ′ has 2|V | − 3 − 4 + 2 = 2|V − v| − 3 edges, hence we only need to prove its sparsity. Assume for a contradiction that there is a set Therefore, equality holds in the last inequality, implying that X is tight in G, contradicting the assumption. □ As we have seen before the introduction of X-replacements, the problem with using only 0-and 1-extension in our proof is that it is possible that there are just a constant number of vertices of degree at most three in a Laman graph. The following lemma shows that the number of vertices with degree at most four is much higher. Proof. Let n i (n ≤i , n ≥i , respectively) denote the number of vertices in G with degree i (at most i, at least i, respectively). Then Hence, n = n ≤4 + n ≥5 ≤ 3n ≤4 − 6. Therefore, n/3 + 2 ≤ n ≤4 . □ It is easy to see that we can maintain the planarity of our graph extended with the extra edges in F while we delete degree-two vertices and perform 1-reductions (see Fig. 2 for an example of a 1-reduction). However, if we need to perform the inverse of an X-replacement we may need to add crossing edges to F to ensure condition of Lemma 2.9 that any three element of the set {p(v 1 ), p(v 2 ), p(v 3 ), p(v 4 )} affinely span the plane (see Fig. 3 for an example). To guarantee the low number of the edges in F , we need the following definition.
We say that a graph G ′ continuous curves in the plane such that only edges in F can cross each other and each edge in F can cross at most one other edge in F . It is easy to observe the following property of F -crossing graphs.

then there exists a partition of F into two sets F 1 and F 2 such that both of G
To guarantee the existence of a vertex with maximum degree four and with low F -degree we will need the following property. Proof. Since G is Laman, |E| = 2n −3. By Proposition 2.12, there exists a partition of F into two sets F 1 and F 2 such that both of G ′ 1 = (V , E ∪ F 1 ) and G ′ 2 = (V , E ∪ F 2 ) are planar. As G ′ i is simple planar, we get |E ∪ F i | ≤ 3n − 6 for i = 1, 2. Hence |F 1 | ≤ n − 3 and |F 2 | ≤ n − 3 and thus |F | ≤ 2n − 6.

Rigid planar graphs with few locations
In this section we prove Theorem 1.6. As we observed in Section 2, it is enough to prove Theorem 1.6 for planar Laman graphs. In fact, we will prove a slightly stronger result, as follows. Note that a Laman graph is always simple. Moreover, without loss of generality, we may assume that G ′ is simple since deleting each edge of F which is parallel to an edge in E does not change our statement. By Lemmas 2.11 and 2.13, there exists a vertex v ∈ V with d G (v) ≤ 4 and d F (v) ≤ 11.   Fig. 2). By induction, there exists an infinitesimally rigid realization affinely span the plane. Lemma 2.6 implies that we can define p(v) in such a way that (G, p) is infinitesimally rigid. However, at this point we cannot guarantee that p(v) ∈ A, although, we have p(u) ∈ A for every u ∈ V − v. Now, by Lemma 2.3, we can define a map p ′ : is not equal to the location of any of its F -neighbors since |A| = 26 ≥ ( 5 2 ) + 11. Note that p ′ (v) is not equal to the location of any of the neighbors of v in G ′ since otherwise one of the 2|V | − 6 rows of the rigidity would be 0, contradicting the infinitesimal rigidity of (G, p ′ ). 3 and v 4 , such that this is the order of the outgoing edges in E from v in a fixed F -crossing drawing of G ′ . We have the following two subcases: By relabeling the neighbors of v we can assume that G−v+v 1 v 2 +v 3 v 4 is Laman. It is easy to see that G−v+v 1 v 2 +v 3 v 4 is planar and Fig. 3). By induction, there exists an infinitesimally rigid realization p of Since A is generic, we can use Lemma 2.9 to prove that there exists a placement of p(v) in R 2 such that (G, p) is infinitesimally rigid. However, we need to take it from the set A. By Lemma 2.3, we can define a map p ′ is infinitesimally rigid, and p ′ (v) is not equal to the location of any of its F -neighbors since |A| = 26 ≥ ( 6 2 ) + 11. Note that, as in Case 2, p ′ (v) is not equal to the location of any of its neighbors in G ′ .

Subcase 3.2:
If neither G −v +v 1 v 2 +v 3 v 4 nor G −v +v 1 v 4 +v 2 v 3 is Laman, then, by Lemma 2.10, there exists an i ∈ {1, 2, 3, 4} such that there are tight sets X , 4 and v 5 := v 1 . By relabeling the vertices cyclically we can assume that contradicting the sparsity condition (L2). We will use the following two observations.

Claim 3.2. There exists no tight set Z
Proof. For the sake of contradiction, suppose that  Proof. For the sake of contradiction, suppose that Z ′ contradicting Claim 3.3. By swapping v 2 and v 4 , we can assume that there is no tight set Fig. 4). By induction, there exists an infinitesimally rigid realization p of G − v + v 2 v 4 + v 3 v 4 on A such that the two endvertices of each edge in Next we add v to G − v + v 2 v 4 + v 3 v 4 by a 1-extension on v 2 v 4 along with the edges vv 1 , vv 2 and vv 4 . By using Lemma 2.3 as in the proof of Case 2, we can see that from any 10 points in A we can find at least one, say a, for which the extension p a of p with p a (v) := a is an infinitesimally rigid realization of G − vv 3 + v 3 v 4 . Furthermore, this also implies that from any 11 points in A we can find at least two, say a and b, for which the extensions p a and p b of p with p a (v) := a and p b (v) := b are both infinitesimally rigid realizations of G − vv 3 + v 3 v 4 . As |A| = 26 ≥ 11 + 11 + 4, we can choose such a and b in such a way that p(u) ̸ = a and p(u) ̸ = b both hold for every u ∈ V for which uv ∈ E ∪ F . We shall show that (G, p a ) or (G, p b ) is infinitesimally rigid.

. This fact together with Claims 3.2 and 3.3 imply that
Note that, in an infinitesimally rigid realization of a Laman graph G * on vertex set V , any tight set in G * induces an infinitesimally rigid subframework (since otherwise the corresponding rows of the rigidity matrix are not linearly independent and hence the rigidity matrix has at most 2|V | − 4 linearly independent rows contradicting to the infinitesimal rigidity of G * ). For the tight sets X and Observe that G − vv 3 has only 2|V | − 4 edges and hence neither (G − vv 3 , p a ) nor (G − vv 3 , p b ) is infinitesimally rigid. However, the infinitesimal rigidity of (G − vv 3 + v 3 v 4 , p a ) (and of (G − vv 3 + v 3 v 4 , p b ), respectively) implies that the dimension of the space of the infinitesimal motions of (G − vv 3 , p a ) (and of (G − vv 3 , p b ), respectively) is four.
infinitesimally rigid, we can add a trivial infinitesimal motion to any non-trivial infinitesimal motion of (G − vv 3 , p a ) in such a way that we get a non-trivial infinitesimal motion m 0 of (G − vv 3 , p a ) for which m 0 (u) = 0 holds for each u ∈ X ∪ Y ∪ {v}. Observe that m 0 is also such a non-trivial infinitesimal motion of (G − vv 3 , p b ), since p a and p b only differ on the location of v and the value of m 0 on v and on all its neighbors is 0. The previous dimension constraint implies that such infinitesimal motion m 0 of (G − vv 3 , p a ) (or (G − vv 3 , p b )) is unique up to a constant multiplier. Note also that m 0 (v 3 ) ̸ = 0 since otherwise m 0 is also an infinitesimal motion of the infinitesimally rigid framework (G − vv 3 + v 3 v 4 , p a ) which contradicts its non-triviality.
Assume now that each of (G, p a ) and (G, p b ) has a non-trivial infinitesimal motion, say, m a and m b . Like for m 0 , we may assume without loss of generality that m a (u) = m b (u) = 0 holds for each u ∈ X ∪ Y ∪ {v}. Since m a and m b are also infinitesimal motions of (G − vv 3 , p a ) and (G − vv 3 , p b ), respectively, m a = c a m 0 and m b = c b m 0 must hold for some constants c a , c b ̸ = 0. Furthermore, (1)

Rigid graphs with few locations
In this section we show that Theorem 1.6 can be extended to the class of graphs that can be embedded in a fixed closed surface. Later we use this generalization of Theorem 1.6 to prove Theorem 1.7. We refer to the book of Mohar and Thomassen [6,Chapter 3] for an introduction to the topic of graph embeddings in surfaces.

Graphs on surfaces
Note that in the proof of Theorem 1.6 we used planarity twice: • In Lemma 2.13, we used the edge bound (which follows from Euler's formula) for planar graphs.
• In our reduction steps, we used planarity 'locally' to show that the reduced graphs are also Note that Euler's formula extend for graphs which can be embedded in a given closed surface (by using the Euler characteristic of the surface), furthermore, a closed surface is locally homeomorphic to the plane. Hence we get the following result with the same proof. √ −χ C ) such that for every graph G = (V , E) which has an embedding into C and is rigid in R 2 and for every set A of generic points in R 2 with |A| ≥ k C , there exists an infinitesimally rigid realization p : V → A of G.
Proof of Sketch. Since the proof is just a copy of our proof for the planar case, we only show why . In our proof for the planar case, we used Euler's formula in the proof of Lemma 2.13.
As in the planar case, we say that G ′ = (V , E ∪F ) is F -crossing on C for a closed surface C if E ∩F = ∅ and G ′ can be drawn with continuous curves on C such that only edges in F can cross each other and each edge in F can cross at most one other edge in F . Now Lemma 2.13 can be modified, as follows.

Lemma 4.2.
Let G = (V , E) be a Laman graph on n vertices, let C be a closed surface with Euler characteristic χ C , and let G ′ = (V , E ∪ F ) be F -crossing on C and simple. Then it has less than n/3 vertices of F -degree more than 12 − 36 n (χ C − 1).

Proof.
Since G is Laman, |E| = 2n − 3. Like in the planar case, there exists a partition of F into to sets F 1 and F 2 such that both of G ′ 1 = (V , E ∪ F 1 ) and G ′ 2 = (V , E ∪ F 2 ) can be embedded into C. As G ′ i can be embedded into C that has Euler characteristic χ C , |E ∪ F i | = n + n * i − χ C for i = 1, 2 where n * i is the number of faces of G ′ i embedded into C. Since n * i ≤ 2 3 |E ∪ F i | follows by the simplicity of G i , we get |E ∪ F i | ≤ 3n − 3χ C for i = 1, 2. Hence |F 1 | ≤ n − 3χ C + 3 and |F 2 | ≤ n − 3χ C + 3 and thus |F | ≤ 2n − 6χ C + 6.
For a constant c ∈ R + , let n ′ >c denote the number of vertices in G ′ of F -degree more than c. Now, cn ′ >c < 2|F | ≤ 4n − 12χ C + 12. To prove that n ′ >c < n/3, we need 4n − 12χ C + 12 ≤ cn/3 and hence 12 + 36 n (1 − χ C ) ≤ c. □ As in the planar case, it is enough to prove Theorem 4.1 for Laman graphs. We prove the following slightly stronger result.

Genus of Laman graphs
Next we show the following simple bound on the genus of Laman graphs.

Proof.
It is easy to check that each Laman graph on at most 5 vertices is planar. By Lemma 2.7, each Laman graph can be constructed by 0-and 1-extensions from the complete graph on 2 vertices. If G has an embedding in an orientable closed surface C and G ′ is its 0-extension, then it is easy to see that we can add a new handle to C (with ends close to the location of the two neighbors of the new vertex in G ′ ) in such a way that G ′ is embeddable into this new surface. Similarly, when G ′ is a 1-extension of G, we can add a new handle to C (with ends close to the subdivided edge and to the third neighbor of the new vertex in G ′ ). □ Now we are ready to prove Theorem 1.7.
Proof of Theorem 1.7. Again it is enough to consider the case when G = (V , E) is Laman and hence its genus is at most max(|V | − 5, 0) by Lemma 4.4. It is well-known that an orientable closed surface with genus g has Euler characteristic 2 − 2g (see [6]). Hence our statement follows from Theorem 4.1 (or Theorem 1.6 when g = 0). □

Rigid realizations on few integer points
Fekete and Jordán [2] showed that one can construct an infinitesimally rigid realization of a graph G = (V , E) with integer coordinates from {1, . . . , |V |} by changing the coordinates one-by-one of an infinitesimally rigid realization of G, preserving infinitesimal rigidity. We prove Theorem 1.8 by showing that the coordinates of coincident vertices can be changed simultaneously.