Graph Theory and Combinatorics Winter 2022 GTU Paper Solution | 3171611

In graph theory, a tree is an undirected graph in which any two vertices are connected by exactly one path. A tree has the following properties:

1. It is an undirected graph with no cycles.
2. There is exactly one path between any two nodes.
3. It is connected and acyclic.
4. Adding an edge between any two vertices will create a cycle.
5. Every edge is a bridge.
6. A tree with n vertices has exactly n-1 edges.
7. The degree of any internal node (a node with degree greater than 1) is at least 2.
8. The degree of the root node (a node with degree 1) is 1.
9. The sum of the degrees of all the vertices is 2(n-1), where n is the number of vertices in the tree.
10. A tree can be represented as a rooted tree with a designated root node.

In graph theory, graph coloring is the assignment of colors to the vertices of a graph such that no adjacent vertices share the same color. The minimum number of colors required to color a graph is called the chromatic number of the graph. Graph coloring has applications in various fields such as scheduling, map coloring, register allocation in compilers, etc.

Greedy coloring is a simple and widely used heuristic algorithm for coloring a graph. The algorithm works by iteratively selecting an uncolored vertex and assigning it the lowest possible color that is not already assigned to any of its neighbors. The algorithm continues this process until all vertices are colored.

The steps of the Greedy coloring algorithm are as follows:

1. Initialize an array of colors with all colors set to unassigned.
2. Pick an uncolored vertex and assign it the lowest numbered color that is not already assigned to any of its neighbors.
3. Repeat step 2 for all remaining uncolored vertices.

Greedy coloring is a simple and easy-to-implement algorithm. However, it may not always produce the optimal solution, i.e., the minimum number of colors required to color the graph. In some cases, the Greedy coloring algorithm may use more colors than necessary. Nonetheless, it is a useful algorithm for practical applications and can be improved upon using various heuristics and techniques.

The Königsberg Bridge Problem is a famous problem in graph theory, named after the city of Königsberg (now Kaliningrad, Russia). The city was situated on both sides of the Pregel River and included two large islands connected to each other and the mainland by seven bridges. The problem was whether it was possible to take a walk through the city, crossing each bridge once and only once, and returning to the starting point.

The problem was first posed by the mathematician Leonhard Euler in 1736, who proved that it was impossible to find such a walk. Euler approached the problem using the concept of a graph, where the bridges are represented as edges and the landmasses as vertices. By analyzing the degree of each vertex (the number of edges connected to it), Euler showed that in order to have a walk that crosses each bridge once and only once, each vertex must have an even degree.

In the case of Königsberg, however, there were four vertices with an odd degree, meaning that it was impossible to find such a walk. Euler’s solution to the problem laid the foundation for the field of graph theory and the development of the concept of the Eulerian path and circuit. The Königsberg Bridge Problem remains a classic example of the practical application of graph theory.

In combinatorics, the rook polynomial is a generating function that counts the number of ways to place non-attacking rooks on a chessboard. A rook is a chess piece that can move any number of squares vertically or horizontally. The rook polynomial is a polynomial in two variables: one variable for the size of the chessboard and the other for the number of rooks.

Let R(m,n) be the rook polynomial of an m×n chessboard. The expansion formula for R(m,n) is:

R(m,n) = ∑(-1)^i C(m,i) C(n,i) (m – i)^n

where C(m,i) is the binomial coefficient “m choose i”.

The rook polynomial counts the number of ways to place k non-attacking rooks on an n×n chessboard. The coefficient of the kth degree term in the rook polynomial gives the number of such configurations. The rook polynomial is useful in solving combinatorial problems involving non-attacking rooks, such as the maximum number of non-attacking rooks that can be placed on a chessboard with no two rooks attacking each other.

The stable matching problem is a well-known problem in the field of mathematics, economics, and computer science. The problem is essentially a matching problem where a set of men are to be matched with a set of women based on their preferences.

The problem is as follows: There are n men and n women, and each individual has a ranked preference list of the opposite sex. The goal is to find a stable matching between the men and women. A matching is considered stable if there is no pair of individuals who would both prefer each other to their current partners. In other words, a matching is stable if there are no “rogue couples” who would be happier with each other than their current partners.

One algorithm to solve the stable matching problem is the Gale-Shapley algorithm. This algorithm proceeds as follows:

1. Initially, all men are free and have not been matched to any woman.
2. Each man proposes to the woman he prefers most on his list who has not yet rejected him.
3. Each woman considers the proposal from the man she is currently paired with (if any) and the proposal from the man who just proposed to her. She keeps the man she prefers and rejects the other.
4. Each rejected man proposes to his next preferred woman who has not yet rejected him.
5. The process repeats until every man has been paired with a woman.

The Gale-Shapley algorithm guarantees a stable matching, and it is also efficient as it runs in O(n^2) time.

To determine the number of farmers that farm beetroot, yams, and radish, we need to use the inclusion-exclusion principle.

First, we find the total number of farmers that farm each crop:

• |B| = 260 (number of farmers that farm beetroot)
• |Y| = 100 (number of farmers that farm yams)
• |R| = 70 (number of farmers that farm radish)

Next, we find the number of farmers that farm two crops:

• |B ∩ R| = 40 (number of farmers that farm both beetroot and radish)
• |Y ∩ R| = 40 (number of farmers that farm both yams and radish)
• |B ∩ Y| = 30 (number of farmers that farm both beetroot and yams)

Finally, we find the number of farmers that farm all three crops:

• |B ∩ Y ∩ R| = ?

To find |B ∩ Y ∩ R|, we need to use the inclusion-exclusion principle:

|B ∩ Y ∩ R| = |B| + |Y| + |R| – |B ∪ Y ∪ R| – |B ∩ R| – |Y ∩ R| – |B ∩ Y|

where |B ∪ Y ∪ R| is the number of farmers that farm at least one of the crops.

|B ∪ Y ∪ R| = |B| + |Y| + |R| – |B ∩ Y| – |B ∩ R| – |Y ∩ R| + |B ∩ Y ∩ R|

Plugging in the values we know, we get:

|B ∪ Y ∪ R| = 260 + 100 + 70 – 30 – 40 – 40 + |B ∩ Y ∩ R|

|B ∪ Y ∪ R| = 320 + |B ∩ Y ∩ R|

Substituting back into the first equation, we get:

|B ∩ Y ∩ R| = |B| + |Y| + |R| – |B ∪ Y ∪ R| – |B ∩ R| – |Y ∩ R| – |B ∩ Y| |B ∩ Y ∩ R| = 260 + 100 + 70 – 320 – 40 – 40 – 30 |B ∩ Y ∩ R| = 0

Therefore, there are 0 farmers that farm beetroot, yams, and radish.

Hall’s Marriage Theorem is a fundamental result in graph theory that provides a necessary and sufficient condition for the existence of a perfect matching in a bipartite graph. It states that a bipartite graph with bipartition (X, Y) has a matching that covers all vertices in X if and only if for every subset S of X, the neighborhood N(S) of S (i.e., the set of all vertices in Y that are adjacent to at least one vertex in S) has size greater than or equal to |S|, i.e., |N(S)| >= |S|.

In other words, for a perfect matching to exist, there must be no “blocking sets”, i.e., subsets of X for which the corresponding neighborhood in Y is too small to contain a matching.

The theorem provides a condition for a set of men to be able to each marry a woman, where each woman can only be married to one man. It says that such a matching is possible if and only if every subset of men has at least as many women available to them as there are men in the subset. The theorem has many applications in computer science, economics, and other fields.

This theorem has applications in various fields, including computer science, economics, and social choice theory, where it is used to model matching problems such as stable marriages, job matching, and roommate assignment.

1. An example of a graph that has an Eulerian circuit but no Hamiltonian circuit is the following:

   a --- b
   |     | \
   |     |  \
   c --- d   e

This graph has an Eulerian circuit a-b-d-e-c-a, which is a closed path that traverses every edge of the graph exactly once. However, it does not have a Hamiltonian circuit, which is a closed path that visits every vertex of the graph exactly once.

2. An example of a graph that has a Hamiltonian circuit but no Eulerian circuit is the following:

  a ----- b
  | \   / |
  |   x   |
  | /   \ |
  c ----- d

This graph has a Hamiltonian circuit a-b-d-c-a, which visits every vertex of the graph exactly once. However, it does not have an Eulerian circuit, since there is no closed path that traverses every edge of the graph exactly once.

In combinatorics and graph theory, the Exponential Generating Function (EGF) is a tool used to count the number of objects with respect to some property or structure. It is a formal power series that encodes the information about the number of objects of different sizes or types.

Suppose that we have a sequence of objects {a_n}, where a_n is the number of objects of size n. The EGF of this sequence is defined as:

​Here, the coefficient of $x^n/n!$ counts the number of objects of size $n$. The EGF has some useful properties, such as linearity, product rule, and composition rule, that make it a powerful tool for solving combinatorial problems.

For example, suppose we want to count the number of labeled graphs with $n$ vertices. We can represent a graph as an adjacency matrix, where the $(i,j)$ entry is 1 if there is an edge between vertices $i$ and $j$, and 0 otherwise. The number of labeled graphs with $n$ vertices can be expressed as the coefficient of $x^n/n!$ in the EGF of the sequence {a_n}, where $a_n$ is the number of labeled graphs with $n$ vertices. Using the product rule, we can write the EGF of {a_n} as:

The coefficient of $x^n/n!$ in the above expression is $\binom{n}{2}$, which is the number of ways to choose 2 vertices from $n$ vertices. Hence, the number of labeled graphs with $n$ vertices is $\binom{n}{2}$, which is well-known result.

In summary, the EGF is a powerful tool for counting combinatorial objects and solving combinatorial problems.

The Binomial Theorem is a formula for expanding a binomial raised to a positive integer power. Specifically, it states that:

(x + y)^n = C(n,0)x^n*y^0 + C(n,1)x^(n-1)y^1 + C(n,2)x^(n-2)y^2 + … + C(n,n-1)x^1y^(n-1) + C(n,n)x^0y^n

where C(n,k) denotes the binomial coefficient “n choose k” and is defined as:

C(n,k) = n! / (k! * (n-k)!)

The formula can also be written more compactly as:

(x + y)^n = ∑ C(n,k) * x^(n-k) * y^k

The Binomial Theorem is useful in combinatorics, probability theory, and other fields that deal with discrete mathematics. It allows us to quickly calculate the coefficients of a binomial expansion, which can represent the outcomes of many different types of events.

Catalan numbers are a sequence of natural numbers that appear in many combinatorial problems, particularly those involving the arrangement of objects. They are named after the Belgian mathematician Eugène Charles Catalan, who discovered them in the mid-19th century.

The Catalan numbers can be defined recursively or algebraically. The recursive definition is:

C_0 = 1 C_n+1 = sum(C_i * C_n-i) from i=0 to n, for n >= 0

The algebraic definition is:

C_n = (2n choose n)/(n+1), for n >= 0

Here are some examples of problems that are solved using Catalan numbers:

1. Parenthesization of expressions: How many ways are there to parenthesize an expression with n terms? For example, the expression a+b+c can be parenthesized in five ways: ((a+b)+c), (a+(b+c)), (a+b)+c, a+(b+c), and (a+b+c). The answer is the nth Catalan number.
2. Triangulation of polygons: How many ways are there to triangulate a polygon with n sides using non-intersecting diagonals? The answer is the (n-2)nd Catalan number.
3. Mountain ranges: A mountain range is a sequence of up and down steps. How many mountain ranges of length 2n are there that never go below the starting point? The answer is the nth Catalan number.
4. Dyck words: A Dyck word of length 2n is a string of n X’s and n Y’s such that no initial segment has more Y’s than X’s. How many Dyck words are there? The answer is the nth Catalan number.

Catalan numbers have many other applications in mathematics and computer science, including the analysis of algorithms, the enumeration of graphs and trees, and the study of pattern avoidance in permutations.

A recurrence relation is an equation that defines a sequence recursively in terms of its previous terms. In other words, it is a way of describing a sequence of numbers by giving a formula for each term in the sequence based on the previous terms. There are several types of recurrence relations, including linear, homogeneous, non-homogeneous, and constant coefficient recurrence relations.

1. Linear recurrence relation: A linear recurrence relation is a recurrence relation in which the current term is a linear combination of the previous terms. The general form of a linear recurrence relation is:

an = c1an-1 + c2an-2 + … + ckank

where c1, c2, …, ck are constants and k is the order of the recurrence relation.

2. Homogeneous recurrence relation: A homogeneous recurrence relation is a recurrence relation in which the right-hand side of the equation is zero. In other words, the recurrence relation only depends on the previous terms in the sequence.
3. Non-homogeneous recurrence relation: A non-homogeneous recurrence relation is a recurrence relation in which the right-hand side of the equation is non-zero. In other words, the recurrence relation depends on both the previous terms in the sequence and some external input or function.
4. Constant coefficient recurrence relation: A constant coefficient recurrence relation is a recurrence relation in which the coefficients c1, c2, …, ck are constant for all n. These types of recurrence relations can be solved using various methods, such as the characteristic equation, generating functions, or matrix methods.

To construct an optimal binary prefix code, we need to follow the Huffman coding algorithm. The steps are as follows:

1. Arrange the weights in ascending order.
2. Combine the two smallest weights and create a parent node with the combined weight.
3. Repeat step 2 until all weights are combined into a single tree.
4. Assign 0 to the left branch and 1 to the right branch at each node.
5. The code for each weight is the sequence of 0’s and 1’s obtained by traversing the tree from the root to the corresponding leaf node.

Here is the construction process:

Step 1: Arrange the weights in ascending order.

5, 7, 8, 15, 35, 40

Step 2: Combine the two smallest weights and create a parent node with the combined weight.

5+7=12: 5,7 -> 0; 12
8,12 -> 1; 20
15,20 -> 1; 35
35,40 -> 0; 75

Step 3: Repeat step 2 until all weights are combined into a single tree.

5+7=12: 5,7 -> 0; 12
8,12 -> 1; 20
15,20 -> 1; 35
35,40 -> 0; 75

Step 4: Assign 0 to the left branch and 1 to the right branch at each node.

          75
         /  \
        0    1
      35      \
             20
            /  \
           1    0
         15     8
               / \
              1   0
             5   7

Step 5: The code for each weight is the sequence of 0’s and 1’s obtained by traversing the tree from the root to the corresponding leaf node.

5:  110
7:  111
8:  10
15: 01
35: 00
40: 11

Therefore, the optimal binary prefix code is:

5:  110
7:  111
8:  10
15: 01
35: 00
40: 11

The summation operator is used in generating functions to represent the coefficients of the power series. Given a generating function, the coefficient of a certain term can be found by taking the derivative of the generating function and evaluating it at the appropriate point.

For example, if f(x) is a generating function and we want to find the coefficient of x^n in the power series expansion of f(x), we can use the following formula:

[ x^n ] f(x) = (1/n!) * [ d^n/dx^n ] f(x)

where [ x^n ] f(x) represents the coefficient of x^n in the power series expansion of f(x), and [ d^n/dx^n ] f(x) represents the nth derivative of f(x) with respect to x.

In other words, the coefficient of x^n is equal to the nth derivative of f(x) divided by n!. This allows us to easily find coefficients of power series without having to explicitly compute each term.

To solve the recurrence relation an = an-1 + n with the initial term a0 = 4, we can use backward substitution method.

First, we have:

a1 = a0 + 1 = 4 + 1 = 5

a2 = a1 + 2 = 5 + 2 = 7

a3 = a2 + 3 = 7 + 3 = 10

a4 = a3 + 4 = 10 + 4 = 14

and so on.

Alternatively, we can use the formula for the nth term:

an = a0 + 1 + 2 + … + (n-1)

This is the sum of the first n natural numbers, which is n(n-1)/2. Substituting a0 = 4, we get:

an = 4 + n(n-1)/2

Therefore, the solution to the recurrence relation an = an-1 + n with the initial term a0 = 4 is:

an = 4 + n(n-1)/2, for n ≥ 1.