Your task is to find the common neighbors of two nodes in an undirected academic network. In this network, nodes represent authors and edges represent research collaborations.
Example 1
- Authors in the network: Keng Peng Tee, Toshiyuki Ohtsuka, David Q. Mayne, Darryl DeHaan, Miroslav Krstic, James B. Rawlings, Knut Graichen, Andreas Kugi, Pierre O. M. Scokaert, Christian Ebenbauer
- Research collaborations between these authors: Keng Peng Tee and Miroslav Krstic, Keng Peng Tee and James B. Rawlings, Keng Peng Tee and David Q. Mayne, Keng Peng Tee and Pierre O. M. Scokaert, Keng Peng Tee and Darryl DeHaan, Toshiyuki Ohtsuka and Knut Graichen, Toshiyuki Ohtsuka and Miroslav Krstic, David Q. Mayne and Knut Graichen, David Q. Mayne and James B. Rawlings, David Q. Mayne and Christian Ebenbauer, David Q. Mayne and Pierre O. M. Scokaert, Darryl DeHaan and Knut Graichen, Darryl DeHaan and Christian Ebenbauer, Miroslav Krstic and Andreas Kugi, Miroslav Krstic and Christian Ebenbauer, James B. Rawlings and Knut Graichen, James B. Rawlings and Pierre O. M. Scokaert, James B. Rawlings and Christian Ebenbauer, Knut Graichen and Andreas Kugi, Knut Graichen and Pierre O. M. Scokaert, Pierre O. M. Scokaert and Christian Ebenbauer.
Common neighbors between Keng Peng Tee and Toshiyuki Ohtsuka: [Miroslav Krstic]
Problem to Solve
- Authors in the network: Manfred Schmidt-Schauss, David Sabel, Markus Lohrey, Sebastian Maneth, Conrad Rau, Manfred Schmidt-Schauß, Jordi Levy
- Research collaborations between these authors: Manfred Schmidt-Schauss and Jordi Levy, Manfred Schmidt-Schauss and David Sabel, Manfred Schmidt-Schauss and Manfred Schmidt-Schauß, Manfred Schmidt-Schauss and Markus Lohrey, Manfred Schmidt-Schauss and Sebastian Maneth, Manfred Schmidt-Schauss and Conrad Rau, David Sabel and Conrad Rau, David Sabel and Manfred Schmidt-Schauß, Markus Lohrey and Manfred Schmidt-Schauß, Markus Lohrey and Jordi Levy, Markus Lohrey and Sebastian Maneth, Sebastian Maneth and Manfred Schmidt-Schauß, Sebastian Maneth and Jordi Levy, Conrad Rau and Manfred Schmidt-Schauß, Manfred Schmidt-Schauß and Jordi Levy. Please identify the common neighbors of Manfred Schmidt-Schauss and David Sabel in this network. Present your answer in the following format: [AuthorA, AuthorB, AuthorC, AuthorD, ...].
[Manfred Schmidt-Schauß, Conrad Rau]
[Conrad Rau, Manfred Schmidt-Schauß, Markus Lohrey, Sebastian Maneth, Jordi Levy]
[Conrad Rau, Manfred Schmidt-Schauss]
[Manfred Schmidt-Schauß, Jordi Levy]
[Manfred Schmidt-Schauß, Conrad Rau]
[Conrad Rau, Manfred Schmidt-Schauß]
[Conrad Rau, Manfred Schmidt-Schauß]
[Manfred Schmidt-Schauß]
[Manfred Schmidt-Schauß]
[Manfred Schmidt-Schauß, David Sabel, Markus Lohrey]
You are required to identify all connected components in the given social network and output one representative node from each component. Within a connected component, any node can be reached from any other node through the edges in the graph. Different connected components are isolated from each other.
Example 1
- Names in the network: Barbara Perkins, Jonathon Mitchell, Jenny Trujillo, Adam Hoffman, Devin Nunez.
- Friendship connections: Barbara Perkins and Jonathon Mitchell, Jenny Trujillo and Devin Nunez, Jenny Trujillo and Adam Hoffman, Adam Hoffman and Devin Nunez.
The answer including one representative element from each connected component in the given social network: [Barbara Perkins, Jenny Trujillo]
Problem to Solve
- Names in the network: Veronica Garcia, Katherine Brennan, Angel Chavez, Steven Martin, Brett Johnson, Megan Banks, Julia Dominguez, Rachel Mitchell
- Fiendship connections: Veronica Garcia to Brett Johnson, Veronica Garcia to Megan Banks, Katherine Brennan to Brett Johnson, Katherine Brennan to Megan Banks, Angel Chavez to Megan Banks, Angel Chavez to Rachel Mitchell, Steven Martin to Megan Banks, Brett Johnson to Megan Banks, Megan Banks to Julia Dominguez, Megan Banks to Rachel Mitchell Identify all connected components in this network. Note that for each connected component, you should only output one of its nodes. Present your answer in the following format: [UserA, UserB, UserC, UserD, ...]
[Veronica Garcia, Katherine Brennan, Angel Chavez, Steven Martin, Julia Dominguez]
[Veronica Garcia, Angel Chavez, Steven Martin]
[Veronica Garcia, Katherine Brennan, Angel Chavez, Steven Martin, Brett Johnson, Megan Banks, Julia Dominguez, Rachel Mitchell]
[Veronica Garcia, Katherine Brennan, Angel Chavez, Steven Martin, Megan Banks, Julia Dominguez, Rachel Mitchell]
[Veronica Garcia, Angel Chavez, Steven Martin]
[Veronica Garcia, Angel Chavez]
[Veronica Garcia, Megan Banks, Rachel Mitchell]
In this social network, there are two connected components. The first connected component consists of the nodes Veronica Garcia, Brett Johnson, and Megan Banks. The second connected component includes Katherine Brennan, Angel Chavez, Rachel Mitchell, Julia Dominguez.
The representative nodes from each connected component are Veronica Garcia and Katherine Brennan.
Therefore, the answer is: [Veronica Garcia, Katherine Brennan]
[Veronica Garcia, Katherine Brennan, Angel Chavez, Steven Martin, Brett Johnson, Megan Banks, Julia Dominguez, Rachel Mitchell]
[Veronica Garcia, Katherine Brennan, Angel Chavez, Steven Martin]
You are required to calculate the diameter of an undirected knowledge graph. The diameter of a graph is the maximum distance between any pair of nodes in the graph. To compute this, you need to find the shortest path between all pairs of nodes and then determine the maximum length of these shortest paths.
Example 1
- Entities in this knowledge graph: Village, Niwica Żary County, Prosity, Warmian-Masurian Voivodeship, Dębinka Lubusz Voivodeship, Gmina Trzebiel, Święta Lipka, Poland, Stare Leśno, Wielki Łęck, Gorzów County.
- The relationships between these entities are as follows:
- Village is connected to Wielki Łęck via the relationship type.
- Village is connected to Dębinka Lubusz Voivodeship via the relationship type.
- Village is connected to Niwica Żary County via the relationship type.
- Village is connected to Prosity via the relationship type.
- Village is connected to Święta Lipka via the relationship type.
- Niwica Żary County is connected to Gorzów County via the relationship http://www.w3.org/2000/01/rdf-schema#seeAlso.
- Niwica Żary County is connected to Poland via the relationship country.
- Niwica Żary County is connected to Gmina Trzebiel via the relationship isPartOf.
- Prosity is connected to Poland via the relationship country.
- Prosity is connected to Warmian-Masurian Voivodeship via the relationship isPartOf.
- Warmian-Masurian Voivodeship is connected to Wielki Łęck via the relationship isPartOf.
- Warmian-Masurian Voivodeship is connected to Dębinka Lubusz Voivodeship via the relationship http://www.w3.org/2000/01/rdf-schema#seeAlso.
- Warmian-Masurian Voivodeship is connected to Poland via the relationship country.
- Warmian-Masurian Voivodeship is connected to Święta Lipka via the relationship isPartOf.
- Dębinka Lubusz Voivodeship is connected to Poland via the relationship country.
- Dębinka Lubusz Voivodeship is connected to Gmina Trzebiel via the relationship isPartOf.
- Gmina Trzebiel is connected to Poland via the relationship country.
- Święta Lipka is connected to Poland via the relationship country.
- Poland is connected to Gorzów County via the relationship country.
- Poland is connected to Stare Leśno via the relationship country.
- Poland is connected to Wielki Łęck via the relationship country.
One shortest path corresponding to the diameter in this graph is: [Village, Święta Lipka, Poland, Stare Leśno]
Problem to Solve
- Entities in this knowledge graph: Seine-et-Marne, Montereau-Fault-Yonne, Vittorio De Sica, Saint-Fiacre Seine-et-Marne, France
- The relationships between these entities are as follows:
- Seine-et-Marne is connected to Saint-Fiacre Seine-et-Marne via the relationship department.
- Seine-et-Marne is connected to Montereau-Fault-Yonne via the relationship department.
- Seine-et-Marne is connected to France via the relationship country.
- Montereau-Fault-Yonne is connected to France via the relationship country.
- Vittorio De Sica is connected to France via the relationship deathPlace.
- Saint-Fiacre Seine-et-Marne is connected to France via the relationship country. Please determine the diameter of this network and output the corresponding path in the following format: [Entity1, Entity2, ..., EntityN].
[Seine-et-Marne, Saint-Fiacre Seine-et-Marne, France, Montereau-Fault-Yonne]
To calculate the diameter of the given knowledge graph, we need to find the longest shortest path between any two nodes. The graph consists of the following entities and relationships:
- Seine-et-Marne is connected to Saint-Fiacre Seine-et-Marne via the relationship department.
- Seine-et-Marne is connected to Montereau-Fault-Yonne via the relationship department.
- Seine-et-Marne is connected to France via the relationship country.
- Montereau-Fault-Yonne is connected to France via the relationship country.
- Vittorio De Sica is connected to France via the relationship deathPlace.
- Saint-Fiacre Seine-et-Marne is connected to France via the relationship country.
From these relationships, we can deduce the following paths:
- Seine-et-Marne to Saint-Fiacre Seine-et-Marne (1 step)
- Seine-et-Marne to Montereau-Fault-Yonne (1 step)
- Seine-et-Marne to France (1 step)
- Montereau-Fault-Yonne to France (1 step)
- Vittorio De Sica to France (1 step)
- Saint-Fiacre Seine-et-Marne to France (1 step)
The longest shortest path in this graph is 1 step, as all entities are directly connected to each other or to France. Therefore, the diameter of the graph is 1.
The corresponding path for the diameter is: [Seine-et-Marne, France]
[Seine-et-Marne, Saint-Fiacre Seine-et-Marne, France]
[Seine-et-Marne, France, Vittorio De Sica]
The given knowledge graph consists of the following entities and relationships:
- Entities: Seine-et-Marne, Montereau-Fault-Yonne, Vittorio De Sica, Saint-Fiacre Seine-et-Marne, France
- Relationships:
- Seine-et-Marne is connected to Saint-Fiacre Seine-et-Marne (department).
- Seine-et-Marne is connected to Montereau-Fault-Yonne (department).
- Seine-et-Marne is connected to France (country).
- Montereau-Fault-Yonne is connected to France (country).
- Vittorio De Sica is connected to France (deathPlace).
- Saint-Fiacre Seine-et-Marne is connected to France (country).
First, let's visualize the graph:
-
Seine-et-Marne is directly connected to:
- Saint-Fiacre Seine-et-Marne
- Montereau-Fault-Yonne
- France
-
Montereau-Fault-Yonne is directly connected to:
- Seine-et-Marne
- France
-
Saint-Fiacre Seine-et-Marne is directly connected to:
- Seine-et-Marne
- France
-
Vittorio De Sica is directly connected to:
- France
-
France is directly connected to:
- Seine-et-Marne
- Montereau-Fault-Yonne
- Saint-Fiacre Seine-et-Marne
- Vittorio De Sica
Now, we will determine the shortest paths between all pairs of nodes and find the longest of these shortest paths to determine the diameter of the graph.
The graph can be represented as follows:
- Seine-et-Marne ↔ Saint-Fiacre Seine-et-Marne
- Seine-et-Marne ↔ Montereau-Fault-Yonne
- Seine-et-Marne ↔ France
- Montereau-Fault-Yonne ↔ France
- Saint-Fiacre Seine-et-Marne ↔ France
- Vittorio De Sica ↔ France
Given the connections, the shortest paths between nodes are:
- Seine-et-Marne to Saint-Fiacre Seine-et-Marne: 1 edge
- Seine-et-Marne to Montereau-Fault-Yonne: 1 edge
- Seine-et-Marne to France: 1 edge
- Montereau-Fault-Yonne to France: 1 edge
- Saint-Fiacre Seine-et-Marne to France: 1 edge
- Vittorio De Sica to France: 1 edge
By analyzing the shortest paths, it is clear that the longest shortest path is between Vittorio De Sica and any other node except France, which is 2 edges long, such as:
[Vittorio De Sica, France, Montereau-Fault-Yonne]
Thus, the diameter of the graph is [2], and one of the paths corresponding to the diameter is:
[Vittorio De Sica, France, Montereau-Fault-Yonne]
[Seine-et-Marne, Montereau-Fault-Yonne, France, Vittorio De Sica]
To calculate the diameter of the graph, we need to find the shortest path between all pairs of nodes and then determine the maximum length of these shortest paths.
After analyzing the relationships between the entities in the Seine-et-Marne graph, we can identify the shortest paths between all pairs of nodes:
- Seine-et-Marne to Saint-Fiacre Seine-et-Marne: [Seine-et-Marne, Saint-Fiacre Seine-et-Marne] (distance 1)
- Seine-et-Marne to Montereau-Fault-Yonne: [Seine-et-Marne, Montereau-Fault-Yonne] (distance 1)
- Seine-et-Marne to France: [Seine-et-Marne, France] (distance 1)
- Montereau-Fault-Yonne to France: [Montereau-Fault-Yonne, France] (distance 1)
- Vittorio De Sica to France: [Vittorio De Sica, France] (distance 1)
- Saint-Fiacre Seine-et-Marne to France: [Saint-Fiacre Seine-et-Marne, France] (distance 1)
The maximum distance between any pair of nodes is 1, which means the diameter of the graph is 1.
The corresponding path for the diameter is: [Seine-et-Marne, Saint-Fiacre Seine-et-Marne, France]
The diameter of this network is 3, and the corresponding path is: [Seine-et-Marne, Montereau-Fault-Yonne, France, Vittorio De Sica]
The diameter of this network is 3. One possible shortest path corresponding to the diameter is: [Seine-et-Marne, Vittorio De Sica, France, Saint-Fiacre Seine-et-Marne]
The diameter of this knowledge graph can be calculated by finding the shortest path between all pairs of entities and then finding the maximum length. Here's the path with the maximum length:
[Vittorio De Sica, France, Seine-et-Marne, Saint-Fiacre Seine-et-Marne, France, Montereau-Fault-Yonne, France]
This path corresponds to the longest distance between any pair of entities in the graph. Therefore, the diameter is 6. The output format would be:
[Vittorio De Sica, France, Seine-et-Marne, Saint-Fiacre Seine-et-Marne, France, Montereau-Fault-Yonne, France]
Your task is to identify the shortest path between two specified entities in an undirected knowledge graph, minimizing the number of hops.
Example 1
- Entities in this knowledge graph: Poland, Village, Warmian-Masurian Voivodeship, Dębinka Lubusz Voivodeship
- The relationships between these entities are as follows:
- Poland is connected to Warmian-Masurian Voivodeship via the relationship country.
- Poland is connected to Dębinka Lubusz Voivodeship via the relationship country.
- Village is connected to Dębinka Lubusz Voivodeship via the relationship type.
- Warmian-Masurian Voivodeship is connected to Dębinka Lubusz Voivodeship via the relationship http://www.w3.org/2000/01/rdf-schema#seeAlso.
One shortest path between Dębinka Lubusz Voivodeship and Poland is: [Dębinka Lubusz Voivodeship, Poland]
Problem to Solve
- Entities in this knowledge graph: Senegal, Government of Australia, Niger, Bamako, Greenwich Mean Time, Midfielder (association football), Mali, Burkina Faso, Mali national football team, Defender (association football), Ségou, Ouattagouna, Ansongo Cercle, Araouane, Mopti, Suðuroy, Midfielder
- The relationships between these entities are as follows:
- Senegal is connected to Greenwich Mean Time via the relationship timeZone.
- Bamako is connected to Mali via the relationship capital.
- Greenwich Mean Time is connected to Suðuroy via the relationship timeZone.
- Greenwich Mean Time is connected to Ansongo Cercle via the relationship timeZone.
- Greenwich Mean Time is connected to Ségou via the relationship timeZone.
- Greenwich Mean Time is connected to Mali via the relationship timeZone.
- Mali is connected to Ouattagouna via the relationship country.
- Mali is connected to Ségou via the relationship country.
- Mali is connected to Ansongo Cercle via the relationship country.
- Mali is connected to Mopti via the relationship country.
- Mali is connected to Araouane via the relationship country.
- Ouattagouna is connected to Ansongo Cercle via the relationship isPartOf. Please determine the shortest path between Bamako and Suðuroy in this network. Submit your answer in the format: [Entity1, Entity2, ..., EntityN], where Entity1 and EntityN are the specified start and end entities, and Entity2 through EntityN-1 are the intermediate entities on the shortest path.
[Bamako, Mali, Greenwich Mean Time, Suðuroy]
[Bamako, Mali, Greenwich Mean Time, Suðuroy]
[Bamako, Mali, Ansongo Cercle, Suðuroy]
[ Bamako, Mali, Greenwich Mean Time, Suðuroy]
The shortest path between Bamako and Suðuroy in the given knowledge graph is: [Bamako, Mali, Greenwich Mean Time, Suðuroy].
[Bamako, Mali, Greenwich Mean Time, Suðuroy]
[Bamako, Mali, Greenwich Mean Time, Suðuroy]
[Bamako, Mali, Suðuroy]
The shortest path between Bamako and Suðuroy in this network is: [Bamako, Mali, Greenwich Mean Time, Suðuroy]
The shortest path between Bamako and Suðuroy in this network is: [Bamako, Mali, Greenwich Mean Time, Suðuroy]
You are required to solve the Graph Edit Distance problem between two molecules. Each edit operation (adding or deleting an edge, adding or deleting an isolated node, or relabeling a node) has the identity cost. Your objective is to establish a mapping between the atom IDs from Molecule A to Molecule B, ensuring that each atom ID in Molecule A corresponds to exactly one atom ID in Molecule B. The mapping corresponds to the minimum edit cost between the two graphs.
Example 1
Molecule A:
- Atoms: N (atom 0), O (atom 1), Si (atom 2), O (atom 3), O (atom 4).
- Bonds: 0-1, 1-2, 2-3, 2-4. Molecule B:
- Atoms: F (atom 0), B (atom 1), C (atom 2), N (atom 3), Br (atom 4).
- Bonds: 0-1, 1-2, 1-4, 2-3.
One optimal node mapping: [3, 2, 1, 0, 4].
Problem to Solve
You are given the following two molecules: Molecule A:
- Atoms: N (atom 0), C (atom 1), N (atom 2), F (atom 3).
- Bonds: 0-1, 1-2, 1-3. Molecule B:
- Atoms: O (atom 0), C (atom 1), F (atom 2), F (atom 3).
- Bonds: 0-1, 1-2, 1-3. Represent the node mapping as a list of integers, where the position in the list corresponds to the atom ID in Molecule A and the value at that position indicates the corresponding atom ID in Molecule B. For instance, if atom 0 in Molecule A corresponds to atom 1 in Molecule B, atom 1 in Molecule A corresponds to atom 0 in Molecule B, and atom 2 remains unchanged, the mapping would be represented as [1, 0, 2, ...].
[2, 1, 0, 3]
To solve the Graph Edit Distance problem for the given molecules, we need to find the optimal mapping that minimizes the edit cost. Here are the steps to find the optimal mapping:
-
Identify the atoms and their labels in both molecules:
- Molecule A: N (atom 0), C (atom 1), N (atom 2), F (atom 3)
- Molecule B: O (atom 0), C (atom 1), F (atom 2), F (atom 3)
-
Analyze the bonds in both molecules:
- Molecule A: 0-1, 1-2, 1-3
- Molecule B: 0-1, 1-2, 1-3
-
Find the optimal mapping:
- Atom 0 in Molecule A (N) can map to atom 1 in Molecule B (C) or atom 0 in Molecule B (O). Mapping to atom 1 (C) aligns better with the bond structure.
- Atom 1 in Molecule A (C) can map to atom 0 in Molecule B (C) or atom 1 in Molecule B (C). Mapping to atom 0 (C) aligns with the bond structure.
- Atom 2 in Molecule A (N) can map to atom 2 in Molecule B (F) or atom 3 in Molecule B (F). Mapping to atom 2 (F) aligns with the bond structure.
- Atom 3 in Molecule A (F) can map to atom 2 in Molecule B (F) or atom 3 in Molecule B (F). Mapping to atom 3 (F) aligns with the bond structure.
-
Final mapping:
- The optimal mapping that maintains the bond structure and minimizes the edit cost is: [1, 0, 2, 3]
This mapping ensures that each atom ID in Molecule A corresponds to exactly one atom ID in Molecule B, and the bond structure is preserved as much as possible, minimizing the edit cost.
[1, 0, 2, 3]
The graph edit distance problem can be solved using the Hungarian algorithm to find the optimal node mapping between the two molecules. First, construct the cost matrix where the cost of matching atom i in Molecule A with atom j in Molecule B is the edit distance between the two atoms. Then, use the Hungarian algorithm to find the optimal assignment that minimizes the total cost.
For the given molecules: Molecule A:
- Atoms: N (atom 0), C (atom 1), N (atom 2), F (atom 3).
- Bonds: 0-1, 1-2, 1-3. Molecule B:
- Atoms: O (atom 0), C (atom 1), F (atom 2), F (atom 3).
- Bonds: 0-1, 1-2, 1-3.
The cost matrix is:
| 1 0 1 1 |
| 0 1 0 0 |
| 1 0 1 1 |
| 1 0 1 1 |
Using the Hungarian algorithm, the optimal node mapping is: [1, 2, 0, 3]
For the given problem of finding the Graph Edit Distance between two molecules with the provided data:
Molecule A:
- Atoms: N (atom 0), C (atom 1), N (atom 2), F (atom 3)
- Bonds: 0-1, 1-2, 1-3
Molecule B:
- Atoms: O (atom 0), C (atom 1), F (atom 2), F (atom 3)
- Bonds: 0-1, 1-2, 1-3
We need to find the optimal mapping of atoms from Molecule A to Molecule B that minimizes the edit distance. Considering the costs and the nature of atoms and bonds, the optimal mapping that aligns the most similar atoms and bond structures would be as follows:
[0, 1, 3, 2]
This means:
- Atom 0 in Molecule A (N) maps to atom 0 in Molecule B (O)
- Atom 1 in Molecule A (C) maps to atom 1 in Molecule B (C)
- Atom 2 in Molecule A (N) maps to atom 3 in Molecule B (F)
- Atom 3 in Molecule A (F) maps to atom 2 in Molecule B (F)
So the optimal node mapping is [0, 1, 3, 2]
[0, 1, 2, 3]
[Molecule A: N (atom 0), C (atom 1), N (atom 2), F (atom 3)] [Molecule B: O (atom 0), C (atom 1), F (atom 2), F (atom 3)]
One optimal node mapping: [1, 1, 2, 2].
The node mapping between Molecule A and Molecule B that minimizes the Graph Edit Distance is [1, 0, 3, 2]. Here's the reasoning:
- Atom 0 (N) in Molecule A corresponds to Atom 1 (C) in Molecule B.
- Atom 1 (C) in Molecule A corresponds to Atom 0 (O) in Molecule B.
- Atom 2 (N) in Molecule A corresponds to Atom 3 (F) in Molecule B.
- Atom 3 (F) in Molecule A remains unchanged, as there is no better match in Molecule B.
An optimal node mapping for the given molecules is [2, 1, 0, 3].
The node mapping for the given example is [1, 0, 2, 3].
You are required to solve the Maximum Clique Problem for an undirected academic network. In this network, nodes represent authors and edges represent research collaborations. Your objective is to find the largest subset of nodes such that every pair of vertices in this subset is connected by an edge.
Example 1
- Authors in the network: Keng Peng Tee, Veit Hagenmeyer, Bartosz Käpernick, Karl Henrik Johansson, Darryl DeHaan, James B. Rawlings, Andreas Kugi, Knut Graichen, Tilman Utz, Christian Ebenbauer.
- Research collaborations between these authors: Keng Peng Tee and James B. Rawlings, Keng Peng Tee and Darryl DeHaan, Veit Hagenmeyer and Knut Graichen, Bartosz Käpernick and Andreas Kugi, Bartosz Käpernick and Knut Graichen, Karl Henrik Johansson and Christian Ebenbauer, Darryl DeHaan and Knut Graichen, Darryl DeHaan and Christian Ebenbauer, James B. Rawlings and Knut Graichen, James B. Rawlings and Christian Ebenbauer, Andreas Kugi and Knut Graichen, Knut Graichen and Tilman Utz.
One Maximum Clique: [Knut Graichen, Bartosz Käpernick, Andreas Kugi].
Problem to Solve
- Authors in the network: Manfred Schmidt-Schauss, David Sabel, Manfred Schmidt-Schauß, Guillem Godoy
- Research collaborations between these authors: Manfred Schmidt-Schauss and David Sabel, Manfred Schmidt-Schauss and Manfred Schmidt-Schauß, Manfred Schmidt-Schauss and Guillem Godoy, David Sabel and Manfred Schmidt-Schauß, Manfred Schmidt-Schauß and Guillem Godoy. Identify the clique with the maximum number of authors in this network. Present your answer in the following format: [AuthorA, AuthorB, AuthorC, AuthorD, ...].
[Manfred Schmidt-Schauss, Manfred Schmidt-Schauß, Guillem Godoy]
[Manfred Schmidt-Schauss, David Sabel, Manfred Schmidt-Schauß, Guillem Godoy]
[Manfred Schmidt-Schauss, David Sabel, Guillem Godoy]
[Manfred Schmidt-Schauss, David Sabel, Manfred Schmidt-Schauß, Guillem Godoy]
The Maximum Clique in the given network consists of the following authors:
[Manfred Schmidt-Schauss, David Sabel, Manfred Schmidt-Schauß, Guillem Godoy]
[Manfred Schmidt-Schauss, Manfred Schmidt-Schauß, David Sabel, Guillem Godoy]
The clique with the maximum number of authors in this network is [Manfred Schmidt-Schauss, David Sabel, Manfred Schmidt-Schauß, Guillem Godoy].
[Manfred Schmidt-Schauss, David Sabel, Manfred Schmidt-Schauß, Guillem Godoy]
[Manfred Schmidt-Schauss, David Sabel, Manfred Schmidt-Schauß, Guillem Godoy]
[Manfred Schmidt-Schauß, David Sabel, Manfred Schmidt-Schauß, Guillem Godoy]
You are required to solve the Maximum Common Subgraph problem. Your goal is to identify the common subgraph with the maximum number of atoms shared between the two molecules.
Example 1
Molecule A consists of 8 atoms with the following 9 bonds: 0-1, 0-6, 1-2, 2-3, 3-4, 3-7, 4-5, 5-6, 5-7. Molecule B consists of 7 atoms with the following 7 bonds: 0-1, 1-2, 1-4, 2-3, 3-4, 3-6, 4-5.
One max common subgraph: [2, 3, 4, 5, 7, 6], [0, 1, 2, 3, 4, 6].
Problem to Solve
You are given the following two molecules: Molecule A consists of 4 atoms with the following 3 bonds: 0-1, 1-2, 2-3. Molecule B consists of 4 atoms with the following 3 bonds: 0-1, 1-2, 1-3. Provide the indices of the atoms in the common subgraph for each molecule in the following format: [Node indices in molecular A], [Node indices in molecular B]. For example, if the common subgraph is the subgraph of atom 1, 2, 3 in molecule A and the subgrah of atom 2, 3, 4 in molecule B, you should answer: [1, 2, 3], [2, 3, 4].
[0, 1, 2], [0, 1, 2]
[0, 1, 2], [0, 1, 2]
[1, 2, 3], [2, 3]
[1, 2, 3], [1, 2, 3]
The maximum common subgraph between Molecule A and Molecule B with the given bonds can be determined as follows:
Molecule A:
- Atoms: 0, 1, 2, 3
- Bonds: 0-1, 1-2, 2-3
Molecule B:
- Atoms: 0, 1, 2, 3
- Bonds: 0-1, 1-2, 1-3
The common subgraph can be identified by comparing the structures of both molecules to find the largest matching substructure. The bonds 0-1 and 1-2 in Molecule A and Molecule B are identical and form a linear subgraph.
Thus, the maximum common subgraph is:
- For Molecule A: Atoms 0, 1, 2
- For Molecule B: Atoms 0, 1, 2
Therefore, the solution is: [0, 1, 2], [0, 1, 2]
[0, 1, 2], [0, 1, 2]
[1, 2, 3], [1, 2, 3]
The maximum common subgraph for the given molecules can be obtained by comparing the bonds between atoms in both molecules. By identifying the shared bonds and their connecting atoms, we can derive the common subgraph.
For the given molecules: Molecule A: 0-1, 1-2, 2-3 Molecule B: 0-1, 1-2, 1-3
The maximum common subgraph is: [1, 2], [2, 1, 3]
Explanation:
- Atom 1 in molecule A and atom 2 in molecule B are connected to each other.
- Atom 2 in molecule A and atom 1 in molecule B are connected to each other.
- Atom 2 in molecule A and atom 3 in molecule B are connected to each other through atom 1 in molecule B.
- Atom 3 in molecule A is not connected to any atom in molecule B, but since atom 2 in molecule A is connected to atom 3 in molecule B, atom 3 in molecule A is also part of the common subgraph.
Therefore, the maximum common subgraph is formed by atoms [1, 2] in molecule A and atoms [2, 1, 3] in molecule B.
The common subgraph between Molecule A and Molecule B includes atoms that are connected by a path of bonds that exists in both molecules. In this case, the common subgraph consists of atoms 0, 1, and 2, as these atoms are bonded together in both molecules.
Solution: [0, 1, 2], [0, 1, 2].
[1, 2, 3], [1, 2, 3]
Your task is to solve the Maximum Independent Set problem in the given social network. In this network, each node represents a user, and each edge represents a friendship connection. You need to identify the largest subset of users such that no two users in this subset are friends connected by an edge.
Example 1
- Users in the network: Melinda Vaughan, Mary Thornton, Jeremiah Griffith, Lisa Anderson, Alfred Powell.
- Fiendship connections: Melinda Vaughan and Jeremiah Griffith, Mary Thornton and Jeremiah Griffith, Jeremiah Griffith and Lisa Anderson, Jeremiah Griffith and Alfred Powell. One Maximum Independent Set: [Melinda Vaughan, Lisa Anderson, Alfred Powell, Mary Thornton].
Problem to Solve
- Users in the network: William Lawson, Daniel Shelton, Michelle Lewis, Julie Hayes.
- Fiendship connections: William Lawson and Daniel Shelton, William Lawson and Julie Hayes, Michelle Lewis and Julie Hayes.
Identify the Maximum Independent Set of this network and present your answer in the following format: [UserA, UserB, UserC, UserD, ...].
[William Lawson, Michelle Lewis]
[Michelle Lewis, Julie Hayes]
[William Lawson, Michelle Lewis, Julie Hayes]
[William Lawson, Michelle Lewis]
To solve the Maximum Independent Set problem for the given social network, we need to identify the largest subset of users such that no two users in this subset are friends (i.e., there are no edges between any two nodes in this subset).
Let's list the users and their friendship connections:
Users:
- William Lawson
- Daniel Shelton
- Michelle Lewis
- Julie Hayes
Friendship Connections:
- William Lawson and Daniel Shelton
- William Lawson and Julie Hayes
- Michelle Lewis and Julie Hayes
First, let's create a visual representation of the network:
- William Lawson (connected to Daniel Shelton and Julie Hayes)
- Daniel Shelton (connected to William Lawson)
- Michelle Lewis (connected to Julie Hayes)
- Julie Hayes (connected to William Lawson and Michelle Lewis)
Now, we can see the connections:
William Lawson - Daniel Shelton
William Lawson - Julie Hayes
Michelle Lewis - Julie Hayes
To identify the Maximum Independent Set, we need to select the largest group of users such that no two users are directly connected.
-
Set 1: William Lawson, Michelle Lewis
- William Lawson is not connected to Michelle Lewis.
-
Set 2: Daniel Shelton, Michelle Lewis
- Daniel Shelton is not connected to Michelle Lewis.
-
Set 3: Daniel Shelton, Julie Hayes
- Daniel Shelton is not connected to Julie Hayes.
-
Set 4: Michelle Lewis
- Only one user, trivially independent.
From these sets, Set 2 and Set 3 both include 2 users, making them candidates for the Maximum Independent Set.
Thus, one possible Maximum Independent Set is: [Daniel Shelton, Michelle Lewis]
Alternatively, another valid Maximum Independent Set is: [Daniel Shelton, Julie Hayes]
Therefore, the Maximum Independent Set can be presented as [Daniel Shelton, Michelle Lewis] or [Daniel Shelton, Julie Hayes].
[Michelle Lewis, Daniel Shelton]
The Maximum Independent Set of this network is: [William Lawson, Michelle Lewis, Daniel Shelton].
[William Lawson, Michelle Lewis, Julie Hayes] or [Daniel Shelton, Michelle Lewis, Julie Hayes]
Explanation: Both sets {William Lawson, Michelle Lewis, Julie Hayes} and {Daniel Shelton, Michelle Lewis, Julie Hayes} are maximum independent sets for this network, as they each contain three users with no friendship connections between them.
[William Lawson, Michelle Lewis]
[William Lawson, Michelle Lewis]
Your task is to solve the Minimum Vertex Cover problem in the given social network. In this network, each node represents a user, and each edge represents a friendship connection. You need to identify the smallest subset of users such that every friendship connection has at least one user from this subset.
Example 1
- Users in the network: Julie Harris, David Torres, Vanessa Parker, Shawn Barnett, Karl Dean.
- Fiendship connections: Julie Harris and Vanessa Parker, Julie Harris and David Torres, Julie Harris and Shawn Barnett, Julie Harris and Karl Dean, David Torres and Vanessa Parker, David Torres and Shawn Barnett, David Torres and Karl Dean, Vanessa Parker and Shawn Barnett, Shawn Barnett and Karl Dean.
One Minimum Vertex Cover: [Julie Harris, David Torres, Shawn Barnett].
Problem to Solve
- Users in the network: Pamela Haynes, Kyle Meadows, Adam Nichols, Anna Lowery, Heather Dixon, Matthew Lee, Elizabeth Wood, Stephen Hess.
- Fiendship connections: Pamela Haynes and Stephen Hess, Kyle Meadows and Matthew Lee, Kyle Meadows and Stephen Hess, Kyle Meadows and Adam Nichols, Adam Nichols and Stephen Hess, Adam Nichols and Heather Dixon, Anna Lowery and Stephen Hess, Heather Dixon and Stephen Hess, Matthew Lee and Stephen Hess, Elizabeth Wood and Stephen Hess. Identify the Minimum Vertex Cover of this network and present your answer in the following format: [UserA, UserB, UserC, UserD, ...].
[Stephen Hess, Kyle Meadows, Adam Nichols]
[Stephen Hess, Kyle Meadows, Adam Nichols]
[Pamela Haynes, Kyle Meadows, Adam Nichols, Stephen Hess]
[Stephen Hess, Kyle Meadows, Adam Nichols]
The Minimum Vertex Cover for the given social network with the specified users and friendship connections is [Stephen Hess, Adam Nichols, Kyle Meadows].
[Adam Nichols, Kyle Meadows, Stephen Hess]
The Minimum Vertex Cover of this network is: [Stephen Hess, Adam Nichols, Kyle Meadows].
[Pamela Haynes, Kyle Meadows, Anna Lowery, Matthew Lee] or [Pamela Haynes, Kyle Meadows, Adam Nichols, Elizabeth Wood] or [Pamela Haynes, Kyle Meadows, Heather Dixon, Matthew Lee] or [Pamela Haynes, Kyle Meadows, Heather Dixon, Elizabeth Wood]
Explanation: The minimum vertex cover for this network is 4. There are multiple possible solutions, and I have provided 4 of them. Each solution includes 4 users such that every friendship connection has at least one user from this subset. The friendship connections for each solution are covered as follows:
-
[Pamela Haynes, Kyle Meadows, Anna Lowery, Matthew Lee]:
- Pamela Haynes and Stephen Hess
- Kyle Meadows and Matthew Lee
- Kyle Meadows and Stephen Hess
- Kyle Meadows and Adam Nichols
- Adam Nichols and Stephen Hess
- Anna Lowery and Stephen Hess
- Heather Dixon and Stephen Hess
- Matthew Lee and Stephen Hess
-
[Pamela Haynes, Kyle Meadows, Adam Nichols, Elizabeth Wood]:
- Pamela Haynes and Stephen Hess
- Kyle Meadows and Matthew Lee
- Kyle Meadows and Stephen Hess
- Kyle Meadows and Adam Nichols
- Adam Nichols and Stephen Hess
- Adam Nichols and Heather Dixon
- Elizabeth Wood and Stephen Hess
- Matthew Lee and Stephen Hess
-
[Pamela Haynes, Kyle Meadows, Heather Dixon, Matthew Lee]:
- Pamela Haynes and Stephen Hess
- Kyle Meadows and Matthew Lee
- Kyle Meadows and Stephen Hess
- Kyle Meadows and Adam Nichols
- Adam Nichols and Stephen Hess
- Heather Dixon and Stephen Hess
- Matthew Lee and Stephen Hess
-
[Pamela Haynes, Kyle Meadows, Heather Dixon, Elizabeth Wood]:
- Pamela Haynes and Stephen Hess
- Kyle Meadows and Matthew Lee
- Kyle Meadows and Stephen Hess
[Stephen Hess, Kyle Meadows, Adam Nichols]
[Stephen Hess, Kyle Meadows, Anna Lowery, Matthew Lee]
You are required to solve the Travelling Salesman Problem for an undirected flight route network. Your objective is to determine the shortest possible route that visits each of the listed airports exactly once and returns to the starting point.
Example 1
- Airports to visit: VPY, AAE, BGA, YWB.
- Travel distances (in kilometers) between each pair of airports: VPY to YWB: 16285 VPY to AAE: 9488 VPY to BGA: 13255 AAE to YWB: 7807 AAE to BGA: 9332 BGA to YWB: 6575
One shortest route: [VPY, AAE, YWB, BGA, VPY].
Problem to Solve
- Airports to visit: YNT, TEB, CFC, VIE, NSH, ROV
- Travel distances (in kilometers) between each pair of airports: YNT to VIE: 7962 YNT to ROV: 6644 YNT to NSH: 6176 YNT to CFC: 18668 YNT to TEB: 12263 TEB to VIE: 6987 TEB to ROV: 8529 TEB to NSH: 10389 TEB to CFC: 8846 CFC to VIE: 10716 CFC to ROV: 12244 CFC to NSH: 13193 VIE to ROV: 1736 VIE to NSH: 3406 NSH to ROV: 3000 Please calculate the shortest tour and format your answer as follows: [Airport A, Airport B, Airport C, ..., Airport A]
[YNT, TEB, CFC, VIE, NSH, ROV, YNT]
[YNT, NSH, ROV, VIE, TEB, CFC, YNT]
[YNT, CFC, TEB, NSH, ROV, YNT]
[YNT, NSH, VIE, ROV, TEB, CFC, YNT]
The Travelling Salesman Problem (TSP) aims to find the shortest possible route that visits each airport exactly once and returns to the starting point. Given the distances between the airports, we can determine the shortest route by examining the various permutations and selecting the one with the minimum distance. Here is the detailed process:
Airports to visit: YNT, TEB, CFC, VIE, NSH, ROV
Travel distances between each pair of airports:
- YNT to VIE: 7962 km
- YNT to ROV: 6644 km
- YNT to NSH: 6176 km
- YNT to CFC: 18668 km
- YNT to TEB: 12263 km
- TEB to VIE: 6987 km
- TEB to ROV: 8529 km
- TEB to NSH: 10389 km
- TEB to CFC: 8846 km
- CFC to VIE: 10716 km
- CFC to ROV: 12244 km
- CFC to NSH: 13193 km
- VIE to ROV: 1736 km
- VIE to NSH: 3406 km
- NSH to ROV: 3000 km
To find the shortest tour, we need to consider all permutations of the routes and calculate their total distances, then select the one with the minimum distance.
After evaluating all possible permutations, the shortest route with the minimum total distance is:
[YNT, NSH, ROV, VIE, TEB, CFC, YNT]
This route ensures each airport is visited exactly once, and it returns to the starting point (YNT), minimizing the total travel distance.
[YNT, NSH, ROV, VIE, CFC, TEB, YNT]
After analyzing the given travel distances, I found the shortest possible route that visits each of the listed airports exactly once and returns to the starting point. Here is the solution:
[YNT, TEB, VIE, NSH, ROV, CFC, YNT].
This route has a total distance of 73,311 kilometers.
The shortest tour is: [YNT, NSH, ROV, VIE, TEB, YNT] with a total distance of 48292 kilometers.
[TEB, VIE, YNT, ROV, NSH, CFC, TEB]
The shortest route visiting each airport exactly once and returning to the starting point (YNT) would be:
[YNT, TEB, CFC, VIE, NSH, ROV, YNT]