Computers versus human brains a cooperative game for scientific discoveries Alain Hertz Polytechnique Montréal Mons, August 23, 2017

The presentation is divided into three parts Extremal carbon graphs for Adriatic indices Forbidden subgraph characterizations Proofs by graph transformations

Extremal carbon graphs for Adriatic indices 1st part Extremal carbon graphs for Adriatic indices

Is there a simple graph 𝐺 Let 𝑀= 𝑚 𝑖𝑗 Is there a simple graph 𝐺 with exactly 𝑚 𝑖𝑗 edges with end-degrees 𝑖 and 𝑗 ? Example 𝑚 12 = 6 𝑚 22 =10 𝑚 𝑖𝑗 = 0 otherwise

Let 𝒓 be the largest degree in 𝐺. Example 𝒎 𝟏𝟐 =𝟔 𝒎 𝟐𝟐 =𝟏𝟎 𝒎 𝒊𝒋 =𝟎 otherwise Let 𝒓 be the largest degree in 𝐺. (i.e., 𝒓 is the largest integer such that 𝑖 with 𝑚 𝑖𝑟 ≥1) We have proved that there is a simple graph 𝐺 that matches the 𝑚 𝑖𝑗 values  the three following conditions hold Condition 1 𝑛 𝑖 = 1 𝑖 𝑗=𝑖 𝑟 𝑚 𝑖𝑗 + 𝑗=1 𝑖 𝑚 𝑗𝑖 is an integer for all 𝑖=1,…,𝑟 ( 𝑛 𝑖 is the number of vertices of degree 𝑖) 𝑛 1 =6 𝑛 2 =13 𝑛 𝑖 =0 otherwise Condition 2 𝑚 𝑖𝑖 ≤ 1 2 𝑛 𝑖 𝑛 𝑖 −1 for 𝑖=1,…,𝑟 (no parallel edges between vertices with the same degree) 𝑚 11 ≤15 𝑚 22 ≤78 Condition 3 𝑚 𝑖𝑗 ≤ 𝑛 𝑖 𝑛 𝑗 for 1≤𝑖<𝑗≤𝑟 (no parallel edges between vertices with different degrees) 𝑚 12 ≤78

NO Let 𝑀= 𝑚 𝑖𝑗 Is there a simple connected graph 𝐺 with exactly 𝑚 𝑖𝑗 edges with end-degrees 𝑖 and 𝑗 ? Example 𝑚 12 = 6 𝑚 22 =10 𝑚 𝑖𝑗 = 0 otherwise 𝑛 1 = 6 𝑛 2 =13 𝑛 𝑖 = 0 otherwise NO

NO Let 𝑀= 𝑚 𝑖𝑗 Is there a simple connected graph 𝐺 with exactly 𝑚 𝑖𝑗 edges with end-degrees 𝑖 and 𝑗 ? Example 𝑚 11 =1 𝑚 22 =3 𝑚 𝑖𝑗 =0 otherwise 𝑛 1 = 2 𝑛 2 = 3 𝑛 𝑖 = 0 otherwise NO

Let 𝒓 be the largest degree in 𝐺. (i.e., 𝒓 is the largest integer such that 𝑖 with 𝑚 𝑖𝑟 ≥1) We have proved that there is a simple connected graph 𝐺 that matches the 𝑚 𝑖𝑗 values  the four following conditions hold Conditions 1, 2 and 3 we have seen before Condition 4 (connectivity constraints) 𝑩 𝟏 =𝟏 𝑩 𝟐 =𝟐 𝑩 𝟑 =𝟓 𝑩 𝟒 =𝟏𝟓 𝑩 𝟓 =𝟓𝟐 𝑩 𝟔 =𝟐𝟎𝟑 𝑩 𝟕 =𝟖𝟕𝟕 𝑩 𝟖 =𝟒𝟏𝟒𝟎 𝑩 𝟗 =𝟐𝟏𝟏𝟒𝟕 The number of connectivity constraints is the number of partitions of the set 1,…,𝑟 , i.e, the 𝑟th Bell number 𝑩 𝒓

This was found with

Remarks The four conditions can be checked with an integer linear program having 2 𝑟−1 +𝑟 𝑟+2 −2 variables and 2 𝑟 + 𝑟 2 𝑟 2 −𝑟+6 −4+ 𝐵 𝑟 constraints The proof that the four conditions are sufficient is algorithmic. In other words, if the four conditions hold the, using a computer, we know how to build a graph that matches the 𝑚 𝑖𝑗 values

In summary, carbon trees are trees with maximum degree 4. There is a clear correspondence between chemical and graph-theoretical notions: atoms are represented by vertices and chemical bonds by edges. The ability of atoms to make chemical bonds, i.e. their valences, are equivalent to the notion of vertex degrees in graph. Alkanes are organic compounds exclusively composed of carbon and hydrogen atoms. Each carbon atom has four chemical bonds and each hydrogen atom has one chemical bond. Therefore, the hydrogen atoms can be removed without losing information about the molecule. The resulting representation of an alkane is a carbon tree that can be represented as a graph by replacing the carbon atoms with vertices. In summary, carbon trees are trees with maximum degree 4. 

Extremal graphs for Adriatic indices Adriatic indices are graph invariants that depend on the 𝑚 𝑖𝑗 . For example, the Randić index is the sum of the contributions 1 𝑑 𝑖 𝑑 𝑗 of all edges {𝑖,𝑗}, where 𝑑 𝑖 denotes the degree of vertex 𝑖. In other words, the Randić index is equal to 𝑖=1 𝑟 𝑗=𝑖 𝑟 𝑚 𝑖𝑗 𝑖𝑗

Which are the carbon trees with minimum Randić index ? Question Which are the carbon trees with minimum Randić index ? We solve this problem as follows We fix 𝒓=4, and for every number 𝑛 of vertices we solve an integer program The objective is to minimize 𝑖=1 𝑟 𝑗=𝑖 𝑟 𝑚 𝑖𝑗 𝑖𝑗 The constraints are conditions 1, 2, 3, 4 plus the two following constraints 𝑖=1 𝑟 𝑛 𝑖 =𝑛 𝑖=1 𝑟 𝑗=𝑖 𝑟 𝑚 𝑖𝑗 =𝑛−1

Independently of 𝑛, this gives a total of 2 𝑟−1 +𝑟 𝑟+2 −2 = 30 variables 2 𝑟 + 𝑟 2 𝑟 2 −𝑟+6 −2+ 𝐵 𝑟 = 65 constraints 2. For every solution 𝑴= 𝒎 𝒊𝒋 , we then construct graphs that correspond to 𝑴. 3. It is not difficult to adapt the integer program so that it finds not only all matrices 𝑴= 𝒎 𝒊𝒋 that reach the minimum value, but also all those that reach the second-minimum, third minimum, etc.

This can be found with

Result

“Edge realizability of connected simple graphs”, For more details : P. Hansen, A. Hertz, C. Sellal, D. Vukičević, M. Aouchiche, G. Caporossi “Edge realizability of connected simple graphs”, to appear in MATCH Communications in Mathematical and in Computer Chemistry.

Forbidden subgraph charaterizations 2nd part Forbidden subgraph charaterizations

Graph Properties Some properties are valid for all graphs Some properties define a particular class of graphs 𝜔 𝐺 =𝜒(𝐺) for all 𝐺′⊆𝐺  𝐺 is perfect and such classes are possibly characterized by a list of forbidden induced subgraphs 𝜔 𝐺 =𝜒(𝐺) for all 𝐺′⊆𝐺 G does not contain an odd hole or an odd anti-hole as induced subgraph (M. Chudnovsky, N. Robertson, P. Seymour, and R. Thomas, 2006)

𝐺 does not contain any of these 9 graphs as induced subgraphs Another example 𝐺 is a line graph 𝐺 does not contain any of these 9 graphs as induced subgraphs

Some characterizations by forbidden subgraphs give sufficient conditions to belong to a particular class of graphs 𝐺 does not contain as induced subgraph 𝐺 is perfect 𝐺 does not contain as induced subgraph 𝐺 does not contain as induced subgraph

Some characterizations by forbidden subgraphs give Necessary conditions to belong to a particular class of graphs 𝐺 is perfect 𝐺 does not contain as induced subgraph

Definition Let 𝐹 be as set of graphs, and let 𝐺 be any graph. 𝐺 is 𝐹-free  𝐺 does not contain any graph of 𝐹 as induced subgraph Given a class 𝑪 of graphs, we want to determine a set 𝐹 of graphs to get : Sufficient conditions for Forbidden Subgraph Characterizations (SFSC) if 𝐺 is 𝐹 -free, then 𝐺 belongs to 𝑪 Necessary conditions for Forbidden Subgraph Characterizations (NFSC) If 𝐺 belongs to 𝑪 then 𝐺 is 𝐹 –free Necessary and sufficient conditions for Forbidden Subgraph Characterizations (FSC) 𝐺 belongs to 𝑪 if and only if 𝐺 is 𝐹 –free

SFSC Let 𝐹 be a set of graphs such that 𝐺 is 𝐹 –free  𝐺 belongs to 𝐶 If 𝐹 contains two graphs 𝐻 and 𝐻′ such that 𝐻′ is a subgraphs of 𝐻, then 𝐺 is (𝐹− 𝐻 )–free  𝐺 belongs to 𝐶 NFSC L𝑒𝑡 𝐹 be a set of graphs such that 𝐺 belongs to 𝐶  𝐺 is 𝐹 –free then 𝐺 belongs to 𝐶  𝐺 is(𝐹− 𝐻 )–free We will therefore limit our searches to minimal forbidden subgraph characterizations.

Procedure for determining an SFSC with at most 𝑵 subgraphs Let 𝐹={ 𝐻 1 ,…, 𝐻 𝑁 } be the set containing the 𝑁 smallest graphs not in 𝑪. Set 𝐿= 𝐻 𝑁 . 2. While there is an 𝐹-free graph 𝐺 not in 𝑪 do Add 𝐺 to 𝐿 and determine the largest common induced subgraph 𝐻 in 𝐿. Set 𝐹={ 𝐻 1 ,…, 𝐻 𝑁−1 ,𝐻} We work in 𝜞 𝑛 which is the set of graphs having at most a fixed number 𝑛 of vertices. We enumerate the graphs in 𝜞 𝑛 using the algorithm proposed by McKay (1998). This algorithm avoids visiting two isomorphic graphs This makes it possible to determine the initial set 𝐹 (at step 1), and To determine the largest common induced subgraph 𝐻 in 𝐿 (at step 2).

A stable set is a set of pairwise non adjacent vertices A dominating set is a set 𝑇 of vertices such that all vertices not in 𝑇 are adjacent to at least one vertex in 𝑇. Let 𝑊 be a set of vertices. A vertex 𝑤∈𝑊 is irredundant in 𝑊 if 𝑤 has no neighbor in 𝑊, or if there is a vertex 𝑣∉𝑊 which has 𝑤 has unique neighbor in 𝑊. A set of vertices is irredundant if all its vertices are irredundant. Invariants of graphs 𝑖(𝐺) is the minimum cardinality of a maximal stable set of 𝐺 𝛾(𝐺) is the minimum cardinality of a dominating set of 𝐺 𝑖𝑟(𝐺) is the minimum cardinality of a maximal irredundant set of 𝐺 It is known that 𝑖𝑟(𝐺)  𝛾(𝐺)  𝑖(𝐺) We are interested in determining when some of these relations are satisfied at equality.

𝑪 is the set of graphs 𝐺 such that 𝑖𝑟(𝐺) = 𝛾(𝐺) 𝑁 =2 of the smallest graphs not in 𝐶 Is it a SFSC ?

NO 𝑪 is the set of graphs 𝐺 such that 𝑖𝑟(𝐺) = 𝛾(𝐺) 𝑁 =2 of the smallest graphs not in 𝐶 NO Is it a SFSC ?

𝑪 is the set of graphs 𝐺 such that 𝑖𝑟(𝐺) = 𝛾(𝐺) Largest common induced subgraph Is it a SFSC ?

𝑪 is the set of graphs 𝐺 such that 𝑖𝑟(𝐺) = 𝛾(𝐺) NO Is it a SFSC ?

𝑪 is the set of graphs 𝐺 such that 𝑖𝑟(𝐺) = 𝛾(𝐺) Largest common induced subgraph Is it a SFSC ?

𝑪 is the set of graphs 𝐺 such that 𝑖𝑟(𝐺) = 𝛾(𝐺) NO Is it a SFSC ?

𝑪 is the set of graphs 𝐺 such that 𝑖𝑟(𝐺) = 𝛾(𝐺) Largest common induced subgraph Is it a SFSC ?

𝑪 is the set of graphs 𝐺 such that 𝑖𝑟(𝐺) = 𝛾(𝐺) It seems so. It is the case in 𝜞 10 Is it a SFSC ?

This can be found with

We have proved this conjecture which therefore becomes a theorem. Theorem (Favaron, 1986): 𝐺 is - - free  𝑖𝑟(𝐺) = 𝛾(𝐺) Conjecture 𝐺 is - - free  𝑖𝑟(𝐺) = 𝛾(𝐺) We have proved this conjecture which therefore becomes a theorem. This was done with

Procedure for determining an NFSC with 1 subgraph Let 𝐻 be the single vertex graphs. Set 𝐿= 2. While there is a graph 𝐺 in 𝑪 that contains 𝐻 do Choose the smallest such graph 𝐺 and add it to 𝐿 Determine the smallest graph 𝐻 which is not an induced subgraph of a graph in 𝐿 Again, we work in 𝜞 𝑛 . This makes it possible to determine the smallest graph 𝐺 in 𝑪 that contains 𝐻 To determine the smallest graph 𝐻 which is not an induced subgraph of a graph in 𝐿

𝑪 is the set of graphs 𝐺 such that 𝑖𝑟(𝐺) < 𝛾(𝐺) Is is a NFSC ?

𝑪 is the set of graphs 𝐺 such that 𝑖𝑟(𝐺) < 𝛾(𝐺) Is is a NFSC ? NO

𝑪 is the set of graphs 𝐺 such that 𝑖𝑟(𝐺) < 𝛾(𝐺) A smallest graph which is not an induced subgraph of Is is a NFSC ? NO

𝑪 is the set of graphs 𝐺 such that 𝑖𝑟(𝐺) < 𝛾(𝐺) A smallest graph which is not an induced subgraph of NO

𝑪 is the set of graphs 𝐺 such that 𝑖𝑟(𝐺) < 𝛾(𝐺) A smallest graph which is not an induced subgraph of Is is a NFSC ?

𝑪 is the set of graphs 𝐺 such that 𝑖𝑟(𝐺) < 𝛾(𝐺) A smallest graph which is not an induced subgraph of NO

𝑪 is the set of graphs 𝐺 such that 𝑖𝑟(𝐺) < 𝛾(𝐺) A smallest graph which is not an induced subgraph of Is is a NFSC ?

𝑪 is the set of graphs 𝐺 such that 𝑖𝑟(𝐺) < 𝛾(𝐺) A smallest graph which is not an induced subgraph of Is is a NFSC ? It seems so. It is the case in 𝜞 8

Let 𝐾 𝑛 be the clique on 𝑛 vertices. We have proved 𝐺 is in 𝜞 8 and 𝑖𝑟(𝐺) < 𝛾(𝐺)  𝐺 is 𝐾 5 -free. With a similar technique, we have proved 𝐺 is in 𝜞 8 and 𝛾(𝐺) < 𝑖(𝐺)  𝐺 is 𝐾 5 -free. Since 𝑖𝑟(𝐺)  𝛾(𝐺)  𝑖 𝐺 , we have proved 𝐺 is in 𝜞 8 and 𝑖𝑟(𝐺) < 𝑖(𝐺)  𝐺 is 𝐾 5 -free. Varying 𝑛 we have proved 𝐺 is in 𝜞 9 and 𝑖𝑟(𝐺) < 𝑖(𝐺)  𝐺 is 𝐾 6 -free. 𝐺 is in 𝜞 10 and 𝑖𝑟(𝐺) < 𝑖(𝐺)  𝐺 is 𝐾 7 -free.

This was done with

This was done with Conjecture Let 𝐺 be a graph with 𝑛 vertices 𝑖𝑟(𝐺) < 𝑖(𝐺)  𝐺 is 𝐾 𝑛−3 -free We have proved this conjecture which therefore becomes a theorem This was done with

C. Desrosiers, P. Galinier, P. Hansen, A. Hertz For more details : C. Desrosiers, P. Galinier, P. Hansen, A. Hertz “Automated generation of conjectures on forbidden subgraph characterization” Discrete Applied Mathematics 162, 177-194, 2014

Proofs by graph transformations 3rd part Proofs by graph transformations

Let 𝐼 1 and 𝐼 2 be two graph invariants Let 𝑪 be a class of graphs We want to prove that 𝐼 1 (𝐺)≤ 𝐼 2 (𝐺) for all graphs in 𝑪 To do this we consider transformations 𝑇 1 ,…, 𝑇 𝑟 so that when applied to a graph 𝐺 in 𝑪 𝑇 𝑖 (𝐺) belongs to 𝑪 for all 𝑖=1,…,𝑟 𝐼 1 (𝐺)≤ 𝐼 1 ( 𝑇 𝑖 (𝐺)) for all 𝑖=1,…,𝑟 𝐼 2 (𝐺)≥ 𝐼 2 ( 𝑇 𝑖 (𝐺)) ror all 𝑖=1,…,𝑟 The aim is to use these transformations repeatedly until we get a graph 𝐺’ for which it is easy to prove that 𝐼 1 (𝐺′)≤ 𝐼 2 (𝐺′) As a consequence, we will have proved 𝐼 1 (𝐺)≤ 𝐼 2 (𝐺) since 𝐼 1 (𝐺)≤ 𝐼 1 (𝐺′) ≤ 𝐼 2 (𝐺′) ≤ 𝐼 2 (𝐺)

For every transformation 𝑇 𝑖 we can use the computer to determine if there exist a graph 𝐺 in 𝑪 such that 𝐼 1 𝐺 > 𝐼 1 ( 𝑇 𝑖 (𝐺)) or 𝐼 2 𝐺 < 𝐼 2 ( 𝑇 𝑖 (𝐺)) We start with a transformation 𝑇 1 and determine the subclass 𝑪 1 of graphs in 𝑪 such that 𝐼 1 𝐺 > 𝐼 1 ( 𝑇 1 (𝐺)) or 𝐼 2 𝐺 < 𝐼 2 ( 𝑇 1 (𝐺)) We then consider a transformation 𝑇 2 and determine the subclass 𝑪 2 of graphs in 𝑪 1 such that 𝐼 1 𝐺 > 𝐼 1 ( 𝑇 2 (𝐺)) or 𝐼 2 𝐺 < 𝐼 2 ( 𝑇 2 (𝐺)) We then consider a transformation 𝑇 3 and determine the subclass 𝑪 3 of graphs in 𝑪 2 such that 𝐼 1 𝐺 > 𝐼 1 ( 𝑇 3 (𝐺)) or 𝐼 2 𝐺 < 𝐼 2 ( 𝑇 3 (𝐺)) We repeat this process, until we get a class 𝑪 𝑟 such that it is easy to prove 𝐼 1 (𝐺)≤ 𝐼 2 (𝐺) for all 𝐺 in 𝑪 𝑟

This has to be done with We need our brains to imagine transformations 𝑇 𝑖 . We need computer assistance to determine the subclasses 𝑪 𝑖

If we succeed in finding 𝑪 𝑟 such that it is easy to prove 𝐼 1 (𝐺)≤ 𝐼 2 (𝐺) for all 𝐺 in 𝑪 𝑟 this suggest the following proof that 𝐼 1 (𝐺)≤ 𝐼 2 (𝐺) for all graphs in 𝑪 Let 𝑇 1 ,…, 𝑇 𝑟 be the transformations used to reach the empty class. Consider the following algorithm. Set 𝐺 ′ =𝐺 While at least one 𝑇 𝑖 is applicable to 𝐺 ′ do Choose such a 𝑇 𝑖 with minimum index 𝑖 and set 𝐺 ′ = 𝑇 𝑖 ( 𝐺 ′ ) We have to prove that 𝐼 1 (𝐺′)≤ 𝐼 1 ( 𝑇 𝑖 (𝐺′)) and 𝐼 2 (𝐺′)≥ 𝐼 2 ( 𝑇 𝑖 (𝐺′)) when 𝑇 1 ,…, 𝑇 𝑖−1 are not applicable 𝐼 1 (𝐺′)≤ 𝐼 2 (𝐺′) when no 𝑇 𝑖 is applicable to 𝐺 ′

This proof must be done with

If 𝐺 is a connected graph, then 𝜇(𝐺) 𝐹(𝐺) 2 Exemple 1 𝐼 1 (𝐺) : average distance in 𝐺 = 𝜇 𝐺 𝐼 2 (𝐺) : half of the order of a maximum induced forest in 𝐺 = 𝐹(𝐺) 2 𝑪 : set of all graphs 𝑪 𝑟 : balanced double kites We have proved that If 𝐺 is a connected graph, then 𝜇(𝐺) 𝐹(𝐺) 2

We have used 6 transformations (edge removal, edge rotation, vertex removal, kite creation, vertex shifting, etc.) adsfasdfsaf For more details : P. Hansen, A. Hertz, R. Kilani, O. Marcotte, D. Schindl “ Average distance and maximum induced forest ” Journal of Graph Theory 60 (1), 31-54, 2008

Exemple 2 Given a graph 𝐺, we want to know if it is possible to color its vertices in black and white such that No two adjacent vertices are white, and Every black vertex has exactly one black neighbor The black vertices induce a maximal matching The white vertices induce a stable set, which implies that the induced matching is dominating. So, the question is : given a graph 𝐺, does it contain a dominating induced matching ? We are interested in solving this feasibility problem in 𝑆 2,2,2 -free graphs

𝐼 1 (𝐺) : constant 1 for all graphs 𝐺 𝐼 2 (𝐺) : 1 if 𝐺 contains a dominating induced matching; 0 otherwise 𝐼 1 (𝐺)≤ 𝐼 2 (𝐺) ? Does 𝐺 contain a dominating induced matching 𝑪 : 𝑆 2,2,2 -free graphs 𝑪 𝑟 : graphs with maximum degree 4 and a very special structure We have used 17 transformations All 𝑇 𝑖 satisfy 𝐼 2 𝐺 = 𝐼 2 ( 𝑇 𝑖 (𝐺))

Example of transformation No green vertex has another neighbor outside this subgraph We replace the left green vertices by the right ones

If the resulting graph 𝐺 ′ satisfies 𝐼 1 (𝐺′)≤ 𝐼 2 𝐺 ′ then 𝐼 1 (𝐺) = 𝐼 1 (𝐺′) ≤ 𝐼 2 𝐺 ′ = 𝐼 2 𝐺 Hence, 𝐺 has a dominating induced matching If the resulting graph 𝐺 ′ satisfies 𝐼 1 𝐺 ′ > 𝐼 2 𝐺 ′ then 𝐼 1 (𝐺) = 𝐼 1 (𝐺′) > 𝐼 2 𝐺 ′ = 𝐼 2 𝐺 Hence, 𝐺 does not contain an dominating induced matching All transformations can be implemented in polynomial time All transformations are performed a polynomial number of times. We can determine in polynomial time if 𝐼 1 (𝐺′)≤ 𝐼 2 𝐺 ′ for the resulting graph 𝐺 ′ This means that the problem of determining if a graph 𝐺 has a dominating induced matching can be solved in polynomial time in the class of 𝑆 2,2,2 -free graphs (It is an NP-complete problem in general)

A. Hertz, V. Lozin, B. Ries, V. Zamaraev, D. de Werra, For more details : A. Hertz, V. Lozin, B. Ries, V. Zamaraev, D. de Werra, “Dominating induced matchings in graphs containing no long claw” Submitted for publication, 2015

In summary Any questions ? human brains and computers are both needed for scientific discovery. It’s a cooperative game. Any questions ?