Question
Task 4.(1.5 points)There are several students in the class and some pairs are friends (the friendship relationship is symmetrical). Every student has at least one friend. The whole class came to MatFyz, where each student chooses exactly one of the lectures: either mathematics or computer science. Can you prove that (in each such class) the students can divide themselves into two lectures so that each studentZhad at least one friend who attended a different lecture than the student himself Z.
201
likes1006 views
Answer to a math question Task 4.(1.5 points)There are several students in the class and some pairs are friends (the friendship relationship is symmetrical). Every student has at least one friend. The whole class came to MatFyz, where each student chooses exactly one of the lectures: either mathematics or computer science. Can you prove that (in each such class) the students can divide themselves into two lectures so that each studentZhad at least one friend who attended a different lecture than the student himself Z.
Ali
4.492 Answers
1. Consider the students and friends as a graph where each student is a vertex, and an undirected edge exists between two vertices if both students are friends.
2. The problem requires proving that this graph is bipartite.
3. A graph is bipartite if and only if it has no odd-length cycles.
4. Start with any vertex and perform BFS or DFS to try to color the graph using two colors, such that no two adjacent vertices share the same color.
5. If successful, you have a bipartite graph, meaning that nodes can be divided into two groups, each corresponding to one lecture.
6. If at any point you find a vertex with the same color during the graph traversal across an edge, there exists an odd-length cycle.
7. The presence of such a cycle would contradict the criteria as friendship is symmetrical, ensuring each cycle detected can be colored alternately, validating bipartiteness.
8. Since each student has at least one friend, and that ensures the graph has no isolated points, every path will connect properly into two sets.
9. Thus, the students can always be divided into two lectures maintaining the condition.
Answer: The students can always be divided in a manner where each student has at least one friend attending a different lecture.
2. The problem requires proving that this graph is bipartite.
3. A graph is bipartite if and only if it has no odd-length cycles.
4. Start with any vertex and perform BFS or DFS to try to color the graph using two colors, such that no two adjacent vertices share the same color.
5. If successful, you have a bipartite graph, meaning that nodes can be divided into two groups, each corresponding to one lecture.
6. If at any point you find a vertex with the same color during the graph traversal across an edge, there exists an odd-length cycle.
7. The presence of such a cycle would contradict the criteria as friendship is symmetrical, ensuring each cycle detected can be colored alternately, validating bipartiteness.
8. Since each student has at least one friend, and that ensures the graph has no isolated points, every path will connect properly into two sets.
9. Thus, the students can always be divided into two lectures maintaining the condition.
Answer: The students can always be divided in a manner where each student has at least one friend attending a different lecture.
Frequently asked questions (FAQs)
Math question: Find the limit as x approaches 0 of (sin x) / x using L'Hospital's rule.
+
What is the value of 'x' in the equation 2x - 5 = 3x + 1?
+
What is the derivative of f(g(x)) using the power rule variant?
+
New questions in Mathematics