On central-max-point tolerance graphs

Max-point-tolerance graphs (MPTG) were introduced by Catanzaro et al. in 2017. This class has a wide application in genome studies as well as in telecommunication networks. In our paper we have considered central-max-point tolerance graphs (CMPTG) by taking the points of MPTG as center points of their corresponding intervals. We have presented a necessary and a sufficient condition for this class of graphs and we have shown that the class of CMPTG is same as the class of unit-max-tolerance graphs (UMTG). A proper CMPTG (PCMPTG) becomes a proper interval graph. Next we introduce 50\% max-tolerance graphs and separated this class from UMTG which are same for min-tolerance graphs. We also have shown that proper interval graphs are belonging to a subclass of the class of 50\% max-tolerance graphs. We have proved that the class of interval graphs and CMPTG are both incomparable with that of 50\% max-tolerance graphs. We have shown that every chordless cycle is a 50\% max-tolerance graph. We found a close relation between a CMPTG and a central interval catch digraph (CICD). Finally we have introduced the concepts of optimized digraphs, optimized graphs and proved that these classes of graphs merge with the class of CICD and that of proper interval graphs respectively.


Introduction
The class of interval graphs was initially posed by Hajös in 1957 [10] as a study of intersection graphs of intervals on real line. In 1959, the molecular biological scientist Benzer [2] used the model of interval graphs to obtain a physical map from information on pairwise overlaps of the fragments of DNA. Interval graphs were well studied by many people in Computer Science and Discrete Mathematics for its wide application.
Many combinatorial problems have been solved for interval graphs in linear time.
Due to their lot of applications in theories and practical situations the concept was generalized to several variations. In one direction it went in developing concepts of probe interval graphs, circular-arc graphs, interval digraphs. [1], [9] On the other hand in 1982, Golumbic and Monma introduced the concept of mintolerance graphs (commonly known as tolerance graphs). [7] A simple undirected graph G = (V, E) is a min-tolerance graph if each vertex u ∈ V corresponds to a real interval I u and a positive real number t u , called tolerance, such that uv is an edge of G if and only if |I u ∩ I v | min {t u , t v }.
In [8] Golumbic introduced max-tolerance graphs (in brief, MTG) where each vertex u ∈ V corresponds to a real interval I u and a positive real number t u (known as tolerance) such that uv is an edge of G if and only if |I u ∩ I v | max {t u , t v }. An MTG is a unit-max-tolerance graph (in brief, UMTG) if |I u | = |I v | for all u, v ∈ V . Some combinatorial problems like finding maximal cliques were obtained in polynomial time whereas the recognition problem was proved to be NP-hard for max-tolerance graphs in 2006 [11]. Also they have given a geometrical connection of max-tolerance graphs to semi-squares. For further details of tolerance graphs see [8].
Recently max-point-tolerance graphs (in brief, MPTG) are introduced in [5], 1 where each vertex u ∈ V corresponds to a pointed interval (I u , p u ), where I u is an interval on the real line and p u ∈ I u , such that uv is an edge of G if and only if {p u , p v } ⊆ I u ∩I v . The graphs MPTG have a huge number of practical applications in human genome studies and modelling of telecommunication networks [5]. An MPTG G = (V, E) is central if p u is the center point of I u for each u ∈ V . We call this graph by central-max-point-tolerance graph (in brief, CMPTG). This graph class actually matches with the graph defined as c-p-box(1) graph in [18]. They have shown that c-p-box(1) graphs form a subclass of max-tolerance graphs.
In our paper we prove that CMPTG is same as UMTG. Moreover we show that a unit CMPTG (in brief, UCMPTG) is same as a proper CMPTG (in brief, PCMPTG) and also is same as a proper interval graph.
We introduce 50% max-tolerance graphs analogous to the similar concept for min-tolerance graphs. In case of min-tolerance graphs, unit and 50% are defining the same class of graphs [4]. In our paper we show that for max-tolerance graphs they are not same. Also we show that the classes CMPTG and 50% max-tolerance graphs are not comparable although they contain various classes of graphs (for example C n , n ≥ 3, proper intervals graphs) in common. Finally we find a close relation of CMPTG with central interval catch digraphs (in brief, CICD) where the points p v associated to the intervals I v are center points. For more details about interval catch digraph (in brief ICD) see Section 2. Further we introduce the concepts of optimized digraphs, optimized graphs and prove that these classes of graphs merge with the class of CICD and that of proper interval graphs respectively. Also we obtained the adjacency matrix characterization of a CICD which forms a tournament. In conclusion we show the relations between the subclasses of MTG and CMPTG and list major open problems in this area.

Preliminaries
The following characterization of MPTG is known: Theorem 2. 1. [5] Let G = (V, E) be a simple undirected graph. Then G is an MPTG if and only if there is an ordering of vertices of G such that the following condition holds: For any x < u < v < y, xv, uy ∈ E =⇒ uv ∈ E.
Among many characterizations of proper interval graphs we list the following which will serve our purpose.
The reduced graphG is obtained from G by merging vertices having same closed neighborhood. G(n, r) is a graph with n vertices x 1 , x 2 , . . . , x n such that x i is adjacent to x j if and only if 0 < |i − j| ≤ r, where r is a positive integer.

For all
, uv ∈ E. Therefore G is a UMTG with interval representation {T u = [c u , y u ]|u ∈ V } and tolerances {t u |u ∈ V } as defined above.
Conversely, let G = (V, E) be a UMTG with interval representation {T u = [l u , r u ] | u ∈ V } and tolerances Then c u , the center the required condition such that every interval has unit (or, same) length. We call it unit-central-max-point tolerance graph (in brief, UCMPTG).
Theorem 3. 4. Let G be a simple undirected graph. Then the following are equivalent. 1. G is a PCMPTG.

2.
G is a UCMPTG. 3. G is a proper interval graph. Proof.
(1 ⇐⇒ 2) : First from a PCMPTG representation of G, we obtain a UCMPTG representation. When no interval properly includes another, the right end points have the same order as the left end points and so as the center points as well. We process the representation from left to right, adjusting all intervals to length l where l is the length of first interval (.i.e; |I 1 | = l). At each step until all have been adjusted, let be the unadjusted interval that has the leftmost left endpoint among the unadjusted intervals. Now one of the following cases may happen. 1. I x does not contain centre point of any adjusted interval. 2. I x contains center points of some adjusted intervals and c = a + b 2 also belong to those intervals. 3. I x contains center points of some adjusted intervals but c = a + b 2 does not belong to any of them.
Let I i and I j be the adjusted intervals referred in (2) and (3) respectively. We note that I j must occur before I i (i.e; a j < a i , b j < b i where a k , b k denote the leftmost, rightmost endpoint of an adjusted interval I k ). Take α = a if (1) occurs and α = c i if (2) occurs where c i is the leftmost centre point which belong to I x such that c ∈ I i . Note c i belong to an interval that has already been adjusted to length l. It is easy to observe c k ∈ I x and c ∈ I k for all k where i < k < x. Now when (1), (2) does not occur but (3) occurs then we take α = b i where b i is the rightmost endpoint for which c i ∈ I x but c / ∈ I i . We adjust the portion [a, ∞) by shrinking or expanding [a, b] to [α, α + l] and scaling and shifting [b, ∞) to [α + l, ∞).
Note that in second case after adjustment centre point of I x becomes c imply c x / ∈ I j for any j ≤ i. In all of three cases the order of endpoints does not change and the adjacency relation also remains intact with I x , intervals earlier than I x still have length l, and I x also now has length l. Iterating this operation produces the UCMPTG representation.
Conversely, if G is a UCMPTG then all intervals associated to the vertices of G must be of the same length.
Thus none of them contains other properly and so G is a PCMPTG with the same interval representation. Next let u, v ∈ V . Suppose p u = i j and p v = i k . Then u is a copy of v ij and v is a copy of v i k . If uv ∈ E, Thus G is a PCMPTG.
Therefore u i u j ∈ E, as required. Similarly, it can be shown that Thus G is a proper interval graph. Observation 3. 6. Let G be a simple undirected graph. Then following are equivalent: There is an ordering of vertices of G such that for any u < v, u, v ∈ V , uv ∈ E =⇒ uw ∈ E for all w > v or, wv ∈ E for all w < u.   In the following we present a necessary condition for CMPTG.
Theorem 3. 8. Let G = (V, E) be a CMPTG. Then there is an ordering of vertices of G such that the following condition holds: For any x < u < v < y, xv, uy ∈ E =⇒ uv ∈ E and (xu ∈ E or vy ∈ E or xu, vy ∈ E).
We arrange vertices according to the increasing order of center points (i.e; in < C order) of representing intervals. Suppose in this ordering we have x < u < v < y and xv, uy ∈ E. Then c v , c x ∈ I v ∩ I x and c u , c y ∈ I u ∩ I y . Also Again Combining Definition 3. 9. Let G be a CMPTG and C n (n ≥ 4) be an induced cycle of G. Let A = (a 1 , . . . , a n ) be the list of all vertices of C n , n ≥ 4 arranged in a C-order. Then a x ,a y are said to be circularly consecutive in A if they are in consecutive places of the list, or if x = n and y = 1 (or x = 1 and y = n). C n , n ≥ 4 is said to be circularly consecutive ordered if starting from a fixed vertex (say u) we can order all of its vertices in a circularly consecutive way in clockwise (or anticlockwise) direction until u is reached.
Definition 3. 10. An ending edge in a path P n is an edge that contains a pendant vertex. Let P n (n ≥ 4) be an induced path in a CMPTG G. Then P n is said to have vertex consecutive ending edges if vertices corresponding to ending edges of P n occur consecutively in a C-order (up to permutations between them) at least in one end.
Corollary 3. 11. All induced C 4 in CMPTG are circularly consecutive ordered and all induced P 4 in CMPTG have vertex consecutive ending edges. Proof. In all other cases it will violate condition (3.3) follows from the proof of Theorem 3.8.
The following theorem is a sufficient condition that an MPTG to be a CMPTG.
Theorem 3. 12. Let G = (V, E) be a MPTG with n vertices. Let the ordering {v 1 , v 2 , . . . , v n } of vertices of G that satisfies (2.1) and each v i corresponds to a natural number x i such that x 1 < x 2 < · · · < x n and the following conditions hold for all i = 1, 2, . . . , n: where i 1 and i 2 be the least and the highest indices such that Then G is a CMPTG.
Proof. Suppose the conditions hold. Define . . , v n } and this ordering of vertices satisfies (2.1). Suppose v i v j ∈ E. Then by definition Without loss of generality we assume i < j. Suppose j 1 < i and j < i 2 . Then we have Next, following Observation 3.6, we have the following in a similar way: 13. Let G be a simple undirected graph. Then G is a CMPTG if and only if there exists a central interval catch digraph D such that G = D ∩D T where D T is the digraph obtained from D by reversing direction of every arc.
It is proved in [5] that if G is an MPTG with non-adjacent vertices u and v, then is an interval graph. We found the following analogous result for CMPTG.
are arranged according to the increasing order of center points (i.e., in C-order We observe the followings. • Vertices belong to N (u) ∩ N (v), occurs between u, v form a clique. Let x, y be two such vertices such that u < x < y < v. By (3.3) one can conclude that xy ∈ E.
• If there exist a vertex belongs to N (u) ∩ N (v), occurs before u then no vertex can occur after v which belongs to N (u) ∩ N (v) and conversely. On contrary let u < u < v < v such that u , v ∈ N (u) ∩ N (v) then from(3.3) one can find uv ∈ E which is a contradiction.
• Vertices belong to N (u) ∩ N (v), occurs before u forms a clique. Let u < u < u < v such that Again as a u , c u ∈ [a u , c u ] from above we can conclude that b u ≤ b u . But this is a contradiction as Similarly one can show vertices belong to N (u) ∩ N (v), occurs after v forms a clique.
Similarly one can show if there exists vertices of G[N (u) ∩ N (v)], occurs after v in G then with respect to the ordering {x 1 , . . . , ] arranged according to increasing order of right end points (< N say), occurs between u and v in G and {v 1 , . .
The above proposition leads to a construction of the following forbidden graph for the class of CMPTG.
Example 3. 16. By Proposition 3.15 we see that the graph G (see Figure 1) formed by taking K 1,3 together with two non-adjacent vertices (say, u, v) which are adjacent to each vertex of K 1,3 is not a CMPTG whereas [3,15], [11,15] to the pendants of K 1,3 and [1, 21], [1,29] to v and the central vertex of K 1,3 respectively then it is easy to check G \ {u} becomes a CMPTG with this representation.
u v Figure 1: The graph G in Example 3. 16 4 50% max-tolerance graphs all u ∈ V where t u denotes the tolerance associated with the vertex u ∈ V .
We know that classes of unit (min) tolerance graphs and 50% (min) tolerance graphs are same [8]. But the following theorem shows that this is not the case for max-tolerance graphs Theorem 4. 2. CMPTG (i.e., UMTG) and the class of 50% max-tolerance graphs are not comparable.
Proof. We prove this theorem with the help of following Lemmas. will show that c k / ∈ I k for any k > k that will imply v k v k for any k, k ∈ {1, 2, . . . , n} . Let k > k. Here > 0 clearly as k > k ≥ 1 and α > 0.
Lemma 4. 4. K 1,n where n ≥ 8 is a CMPTG but is not a 50% max-tolerance graph.
Proof. It is sufficient to show that K 1,n , n ≥ 8 is not a 50% max-tolerance graph. Then the result will follow using Lemma 4. 3.
We will prove that K We consider the intervals corresponding to any two pendants must be distinct as they are non-adjacent. We note the following observations which will lead us to the proof.
• Claim 1: Number of pendant vertices whose intervals satisfy a i ≤ 0 < 1 ≤ b i is at most 1.
Let [a 1 , b 1 ], [a 2 , b 2 ] be the two intervals associated to two such pendant vertices (say v 1 , v 2 ). Then the As every pendant vertex is adjacent to u and I u is of unit length, so each pendant vertex has tolerance atmost 1 and therefore has length Hence v 1 ↔ v 2 which contradicts the fact that they are pendant vertices.
2 and a 2 < 0) which contradicts the fact b 2 < 1. Hence the three intervals form a well ordered set with inclusion as the ordering.
From the statement above without loss of generality we can assume that [ as v 3 ↔ u). Now b 1 − a 1 ≤ 2b 1 < 2 from (4.1) and the fact b 1 < 1. Hence we are through.
This proof is same as in Claim 2.
• Claim 4: Number of pendant vertices whose intervals satisfy 0 ≤ a i < b i ≤ 1 is at most 2.
On the contrary we assume that there exists three such vertices with representations [a i , b i ] ⊆ [0, 1] for i ∈ {1, 2, 3}. 1. First we will show that for any two intervals, one must not contain the other.
If not, we assume which is a contradiction. 2. Next we will show that no two intervals are disjoint.
If possible let [a 1 , b 1 ] and [a 2 , b 2 ] be disjoint. Without loss of generality we can assume a 1 < a 2 .
Hence since they are disjoint b 1 < a 2 Now without loss of generality we can assume a 1 < a 2 < a 3 . Hence from 1 and 2 we conclude 3 . Under this situation we will show that there exist no choice of three such For the remaining of the proof we split into two cases.
Hence we arrive at a contradiction.
Hence we have established our claim. Now using the above results we will show that K 1,n where n ≥ 8 is not a 50% max-tolerance graph. Let [a, b] be an interval corresponding to a pendant vertex I v . Then I v must belong to one of the four sets, We note that any pendant vertex cannot have [0, 1] as its interval representation. Hence the above sets are mutually exclusive. As every pendant vertex is adjacent to the central vertex u hence it follows that the four sets are also exhaustive.
Lemma 4. 5. C 6 is a 50% max-tolerance graph but is not a CMPTG.
Proof. Let {v 1 , . . . , v 6 } are the vertices occurred in circularly consecutive way (clockwise or anticlockwise order) in C 6 . We assign the following intervals and tolerances for all the vertices so that they satisfy 50% max-tolerance representation in C 6 . I v1 = [0, 20], t v1 = 10, I v2 = [12,24] 6 , v 3 } are circularly consecutive ordered. So without loss of generality we can assume that in its CMPTG representation c 1 < c 4 < c 6 < c 3 . Now we observe the following, But as Combining we get c 1 < a 6 ≤ c 4 .
Thus the proof is complete.
Now from Lemma 4.4 and 4.5 we can easily conclude that CMPTG and 50% max-tolerance graphs are incomparable.
The following theorem shows that the class of proper interval graphs is subclass of the class of 50% maxtolerance graphs.
Proof. Let G = (V, E) be a proper interval graph. Now as we know from Theorem 3.4 that proper interval graphs are same as PCMPTG. We can consider a PCMPTG representation {I u = [a u , b u ]|u ∈ V } of it. We will show that G is a 50% max-tolerance graph with this same interval representation.
as none of them contains other properly. Hence . From this we can conclude that Theorem 4. 7. Any cycle C n , (n ≥ 3) is a 50% max-tolerance graph. Proof. We prove that C n belongs to the class of 50% max-tolerance graphs by constructing a realization • Let n ≥ 6. We prove now in two cases considering n even and odd. We define k = n 2 when n is even, and k = n + 1 2 when n is odd.
n is even . . , n}. * Claim 1: We will show that v 1 is adjacent to only v 2 , v n .
Next to show v 2 v i for i ∈ {4, . . . , k} we note, a 2 = a i = 1 and b Now clearly a k+1 < 1 = a 2 and b k+1 < n = b 2 from (4.4). Hence, Claim 4, 5 can be shown similarly to the previous cases.
Hence the proof is complete.
Remark 4. 8. Combining Theorems 4.6 and 4.7 along with results of [18] we can easily conclude that proper intervals graphs and C n both belong to the intersection class of CMPTG and 50% max-tolerance graphs.

Optimized Digraph
In the following we define a class of directed graphs, namely optimized digraphs. Let R + be the set of all positive real numbers.
, every out-neighbor distance is less than every non-out-neighbor distance from a vertex). such that v i v j ∈ E and v i v k / ∈ E. So from our above definition it follows that G is an optimized digraph.  [11,17]  It is important to note that d(i, j) = d(j, i) and if we arrange vertices according to increasing order of their labels, then we have d(i, k) = d(i, j) + d(j, k) for all i < j < k.
We consider following cases: Case I: i < j < k or, k < j < i. Then d(i, k) = d(i, j) + d(j, k) > d(i, j) for d(j, k) > 0 as p i 's are distinct.
Case II: i < k < j or, j < k < i. These cases are not possible by (2.2) as D is an ICD.
Case III: j < i < k or, k < i < j. Now v i v j ∈ E and v i v k ∈ E imply p j ∈ I i but p k ∈ I i . Let r = |Ii| 2 . Since p i is the central point of Conversely, suppose D is an optimized digraph with a labeling f . Let us arrange vertices according to increasing order of their labels. For each i = 1, 2, . . . , n, define p i = f (v i ). Let i 1 and i 2 be the least and the highest numbers such that for this it is sufficient to prove that D is an ICD.
We verify (2.2) to show that D is an ICD. Let i < j < k and v i v k ∈ E. Now d(i, k) = d(i, j)+d(j, k) > d(i, j).
Note that the digraph G in Example 5.2 is indeed a CICD as it is evident from the matrix in Figure 2  where i 1 and i 2 be the least and the highest numbers such that for each i = 1, 2, . . . , n.
Proof. We first note that it follows from the proof of Theorem 5.3 that D is an optimized digraph with a labeling for each vertex the center point of the corresponding interval in a CICD representation of D. Thus arrangement of vertices according to the increasing sequence center points is same as the arrangement of vertices according to the increasing labeling of vertices when D is considered as an optimized digraph. Let Then d(k, j) < d(k, i) and so v k v j ∈ E. Therefore the ordering satisfies (2.2).
Next suppose there are i < j such that i 1 > j 1 and i 2 > j 2 . Then j 1 < i 1 i < j j 2 < i 2 . Now by (5.1), This necessary condition gives rise to many forbidden digraphs for the class of optimized digraphs. We describe one of them in the following example.
Remark 5. 6. In [12], Maehara conjectured that if an ICD does not contain any of the digraphs in Figure 4 as an induced subgraph, then it is a CICD. Clearly the graph G 1 in Example 5.5 disproves the conjecture.
Note that the digraph G 1 can be obtained from the digraph (a) of Figure 4 by adding the arc 4 to 3.

INTERVAL DIGRAPHS
An interval digraph is a digraph representable by a family of balanced intervals ( = pointed intervals with base points at their centers) on the real line. If D is an interval digraph, then by Theorem 2, the augmented adjacency matrix A*@) has the consecutive 1's property for rows. But the converse is not true. For example, neither of the digraphs given in Figure 1 is an interval digraph, though they are clearly representable by families of pointed intervals. This is seen as follows. Let {(Ii, bi); i = 1, 2 , 3,4} be a family of pointed intervals representing the digraph of (a). Definition 5. 7. Let G = (V, E) be an undirected graph. Let V = {v 1 , v 2 , . . . , v n } be an ordering of vertices and f : V → R + be a distinct labeling of vertices such that d(u, v) < d(u, w) for all u, v, w ∈ V such that uv ∈ E but uw / ∈ E. Then G is called an optimized graph.
Theorem 5. 8. Let G = (V, E) is an undirected graph. Then G is an optimized graph if and only if G is a proper interval graph.
Proof. Let G = (V, E) be a proper interval graph. By Theorem 2.5 the reduced graphG = (Ṽ ,Ẽ) of G is an induced subgraph of G(n, r) = (V n , E ) for some n, r ∈ N with n > r. Let V n = {x 1 , x 2 , . . . , x n } such that For convenience, we write y j = x ij for j = 1, 2, . . . , m. Define f : if u is a k th copy among z copies of y j (for any but a fixed permutation of them). We arrange the vertices of V according to the increasing order of vertices inṼ keeping copies of same vertices together. Let u, v, w ∈ V such that . Therefore d(u, v) < d(u, w) for all u, v, w ∈ V such that u ↔ v and u w. Then G is an optimized graph.
Conversely, let G = (V, E) be an optimized graph with a labeling f . We arrange the vertices of G according to the increasing order of their labels.
Let i < j < k, where i, j, k ∈ {1, 2, . . . , n} and v Then by Definition 5.7, v i ↔ v j and v j ↔ v k . Thus in the given ordering of vertices, (closed) neighbors of any vertex is consecutively ordered. Therefore by Theorem 2.5, G is a proper interval graph.
A Ferrers digraph D = (V, E) is a directed graph whose successor sets are linearly ordered by inclusion, where the successor set of v ∈ V is its set of out-neighbors {u ∈ V | vu ∈ E}. A binary matrix M is a

Conclusion
We have established the following relations between some graph classes related to this paper.
• K 1,3 ∈ 50% Max-Tolerance Graphs Proper Interval Graphs. K 1,3 is an example of a graph which has a 50% max-tolerance representation having interval [1.9, 6 But it is not a proper interval graph.
The following examples lead us to conclude that interval graphs and 50% max-tolerance graphs are incomparable.
• K 1,n , n ≥ 8 ∈ Interval Graphs 50% Max-Tolerance Graphs. K 1,n is an interval graph having interval [1, 2n]  Finally we note the major unsolved problems in this area. 1. Recognition algorithm and forbidden subgraph characterization of max-point tolerance graphs.