IGNOU MMTE-001 Solved Assignment 2024 for M.Sc. MACS

IGNOU MMTE-001 Solved Assignment 2024 | M.Sc. MACS

Solved By – Narendra Kr. Sharma – M.Sc (Mathematics Honors) – Delhi University

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IGNOU MMTE-001 Assignment Question Paper 2024

mmte-001-solved-assignment-2024-bc500e58-5622-4aae-bac5-75e86f0373be

mmte-001-solved-assignment-2024-bc500e58-5622-4aae-bac5-75e86f0373be

MMTE-001 Solved Assignment 2024
  1. State whether the following statements are true or false. Justify your answers with a short proof or a counterexample
    i) There exists no 9-vertex graph with three vertices of degree 3 , four vertices of degree 2 and two vertices of degree 1 .
    ii) C 6 P 4 C 6 P 4 C_(6)vvP_(4)C_6 \vee P_4C6P4 has a cycle of length at least 7 .
    iii) The diameter of a graph cannot exceed its girth.
    iv) Every Hamiltonian graph is Eulerian.
    v) Every 3-connected graph is 3-edge-connected.
    vi) If G G GGG is an Eulerian graph, then so is L ( G ) L ( G ) L(G)L(G)L(G).
    vii) χ ( C 3 × K 2 ) = 3 χ C 3 × K 2 = 3 chi^(‘)(C_(3)xxK_(2))=3\chi^{\prime}\left(C_3 \times K_2\right)=3χ(C3×K2)=3.
    viii) There exists no graph G G GGG with χ ( G ) > ω ( G ) + 1 χ ( G ) > ω ( G ) + 1 chi(G) > omega(G)+1\chi(G)>\omega(G)+1χ(G)>ω(G)+1.
    ix) The minimum size of a k k kkk-chromatic graph is ( k 2 ) k 2 ([k],[2])\left(\begin{array}{c}k \\ 2\end{array}\right)(k2).
    x) The 6-dimensional hypercube Q 6 Q 6 Q_(6)Q_6Q6 has no perfect matching.
  2. (a) The complement of the Petersen graph is 2-connected. Prove or disprove.
    (b) Consider a graph G G GGG. Let x , y V ( G ) x , y V ( G ) x,y in V(G)x, y \in V(G)x,yV(G) be such that x y x y x harr yx \leftrightarrow yxy. Show that for all z V ( G ) , | d ( x , z ) d ( y , z ) | 1 z V ( G ) , | d ( x , z ) d ( y , z ) | 1 z in V(G),|d(x,z)-d(y,z)| <= 1z \in V(G),|d(x, z)-d(y, z)| \leq 1zV(G),|d(x,z)d(y,z)|1.
    (c) Check whether the following graphs G G GGG and H H HHH are isomorphic or not.
original image
  1. (a) Let G G GGG be a connected n n nnn-vertex graph. Prove that G G GGG has exactly one cycle iff G G GGG has exactly n n nnn edges.
    (b) Find a minimum-weigh spanning tree in the following graph.
original image
(c) Prove that every maximal matching of a graph G G GGG has at least α ( G ) / 2 α ( G ) / 2 alpha^(‘)(G)//2\alpha^{\prime}(G) / 2α(G)/2 edges.
(d) Find the chromatic and edge-chromatic numbers of the following graph.
original image
  1. (a) Find the number of spanning trees of the following graph.
original image
(b) Solve the Chinese Postman Problem for the graph given in Q. 3(b).
(c) Give an example of a 4-critical graph different from a complete graph. Justify the choice of your example.
(d) State and prove the Handshaking Lemma for planar graphs.
  1. (a) Verify Euler’s formula for the following plane graph.
original image
(b) Check whether the line graph of C 5 × K 2 C 5 × K 2 C_(5)xxK_(2)C_5 \times K_2C5×K2 is planar or not.
(c) What is the minimum possible thickness of a 4-connected triangle-free graph on 8 vertices? Also draw such a graph.
(d) Define the parameters α ( G ) α ( G ) alpha(G)\alpha(G)α(G) and β ( G ) β ( G ) beta(G)\beta(G)β(G) for a graph G G GGG. Also, show that α ( G ) + β ( G ) = n ( G ) α ( G ) + β ( G ) = n ( G ) alpha(G)+beta(G)=n(G)\alpha(G)+\beta(G)=n(G)α(G)+β(G)=n(G).
6. (a) What is the maximum possible flow that can pass through the following network? Define such a flow.
original image
(b) State and prove the König Egárvary Theorem.
(c) Let G G GGG be a graph having no isolated vertex and no induced subgraph with exactly two edges. Show that G G GGG is a complete graph.
  1. (a) Find the values of n n nnn for which Q n Q n Q_(n)Q_nQn is Eulerian.
    (b) Using Fleury’s algorithm, find an Eulerian circuit in the following graph.
original image
(c) The complement of a planar graph is planar. True or false? Justify.
\(cos\:3\theta =4\:cos^3\:\theta -3\:cos\:\theta \)

MMTE-001 Sample Solution 2024

mmte-001-solved-assignment-2024-ss-8e24e610-06c9-4b43-84f6-a5bf6ef5ab5c

mmte-001-solved-assignment-2024-ss-8e24e610-06c9-4b43-84f6-a5bf6ef5ab5c

MMTE-001 Solved Assignment 2024
  1. State whether the following statements are true or false. Justify your answers with a short proof or a counterexample
    i) There exists no 9-vertex graph with three vertices of degree 3 , four vertices of degree 2 and two vertices of degree 1 .
Answer:
To determine whether the statement is true or false, let’s use the Handshaking Lemma, which states that the sum of the degrees of all vertices in a graph is equal to twice the number of edges. In other words, for any graph G G GGG, we have:
v V ( G ) deg ( v ) = 2 | E ( G ) | v V ( G ) deg ( v ) = 2 | E ( G ) | sum_(v in V(G))deg(v)=2|E(G)|\sum_{v \in V(G)} \deg(v) = 2|E(G)|vV(G)deg(v)=2|E(G)|
where V ( G ) V ( G ) V(G)V(G)V(G) is the set of vertices in the graph G G GGG and E ( G ) E ( G ) E(G)E(G)E(G) is the set of edges in G G GGG.
According to the given statement, we have:
  • Three vertices of degree 3
  • Four vertices of degree 2
  • Two vertices of degree 1
The sum of the degrees of all vertices in this graph would be:
( 3 × 3 ) + ( 4 × 2 ) + ( 2 × 1 ) = 9 + 8 + 2 = 19 ( 3 × 3 ) + ( 4 × 2 ) + ( 2 × 1 ) = 9 + 8 + 2 = 19 (3xx3)+(4xx2)+(2xx1)=9+8+2=19(3 \times 3) + (4 \times 2) + (2 \times 1) = 9 + 8 + 2 = 19(3×3)+(4×2)+(2×1)=9+8+2=19
Since the sum of the degrees of all vertices is an odd number (19), it cannot be equal to twice the number of edges, as the number of edges must be an integer. Therefore, it is not possible to have a 9-vertex graph with the given degree sequence. The statement is true.
ii) C 6 P 4 C 6 P 4 C_(6)vvP_(4)C_6 \vee P_4C6P4 has a cycle of length at least 7 .
Answer:
The statement " C 6 P 4 C 6 P 4 C_(6)vvP_(4)C_6 \vee P_4C6P4 has a cycle of length at least 7" is true.
To justify this, let’s first understand the notation:
  • C 6 C 6 C_(6)C_6C6 represents a cycle graph with 6 vertices, which means it has a cycle of length 6.
  • P 4 P 4 P_(4)P_4P4 represents a path graph with 4 vertices, which is a linear graph with no cycles.
  • The operation vv\vee represents the join of two graphs, which means connecting every vertex of the first graph to every vertex of the second graph.
In the graph C 6 P 4 C 6 P 4 C_(6)vvP_(4)C_6 \vee P_4C6P4, every vertex of the cycle C 6 C 6 C_(6)C_6C6 is connected to every vertex of the path P 4 P 4 P_(4)P_4P4. This creates multiple new cycles, one of which is formed by taking the original cycle of length 6 in C 6 C 6 C_(6)C_6C6 and extending it with any one of the vertices from P 4 P 4 P_(4)P_4P4 that is not an endpoint of the path. This new cycle will have a length of 6 + 1 = 7 6 + 1 = 7 6+1=76 + 1 = 76+1=7.
Therefore, the statement " C 6 P 4 C 6 P 4 C_(6)vvP_(4)C_6 \vee P_4C6P4 has a cycle of length at least 7" is true.
iii) The diameter of a graph cannot exceed its girth.
Answer:
The statement "The diameter of a graph cannot exceed its girth" is false.
To justify this, let’s first define the terms:
  • The diameter of a graph is the length of the longest shortest path between any two vertices in the graph.
  • The girth of a graph is the length of the shortest cycle in the graph.
Consider a tree graph (a connected graph with no cycles) with more than two vertices. The girth of a tree graph is undefined (or considered to be infinity) because there are no cycles. However, the diameter of a tree graph can be a finite positive number. For example, a path graph P n P n P_(n)P_nPn (a tree with n n nnn vertices arranged in a straight line) has a diameter of n 1 n 1 n-1n-1n1 but has no girth because it contains no cycles.
Therefore, the statement "The diameter of a graph cannot exceed its girth" is false, as demonstrated by the example of a tree graph.
iv) Every Hamiltonian graph is Eulerian.
Answer:
The statement "Every Hamiltonian graph is Eulerian" is false.
To justify this, let’s first define the terms:
  • A Hamiltonian graph is a graph that contains a Hamiltonian cycle, which is a cycle that visits every vertex exactly once.
  • An Eulerian graph is a graph in which all vertices have even degree, and there exists a cycle (called an Eulerian cycle) that traverses every edge exactly once.
A counterexample to the statement is the complete graph K 5 K 5 K_(5)K_5K5, which is a graph with 5 vertices where every pair of vertices is connected by an edge. K 5 K 5 K_(5)K_5K5 is Hamiltonian because it contains a Hamiltonian cycle that visits every vertex exactly once. However, K 5 K 5 K_(5)K_5K5 is not Eulerian because each of its vertices has degree 4, which is even, but there is no cycle that traverses every edge exactly once without repeating edges.
Therefore, the statement "Every Hamiltonian graph is Eulerian" is false, as demonstrated by the example of the complete graph K 5 K 5 K_(5)K_5K5.
v) Every 3-connected graph is 3-edge-connected.
Answer:
The statement "Every 3-connected graph is 3-edge-connected" is true.
To justify this, let’s first define the terms:
  • A graph is said to be k k kkk-connected if there is no set of k 1 k 1 k-1k-1k1 vertices whose removal disconnects the graph.
  • A graph is said to be k k kkk-edge-connected if there is no set of k 1 k 1 k-1k-1k1 edges whose removal disconnects the graph.
A 3-connected graph requires the removal of at least 3 vertices to disconnect it. Suppose a 3-connected graph is not 3-edge-connected. Then there exists a set of 2 edges whose removal disconnects the graph. The removal of these 2 edges would create at least two components, one of which must contain a single vertex (since removing just 2 edges cannot isolate more than one vertex). This implies that the removal of the two vertices incident to these edges would disconnect the graph, contradicting the assumption that the graph is 3-connected.
Therefore, every 3-connected graph must be 3-edge-connected.
vi) If G G GGG is an Eulerian graph, then so is L ( G ) L ( G ) L(G)L(G)L(G).
Answer:
The statement "If G G GGG is an Eulerian graph, then so is L ( G ) L ( G ) L(G)L(G)L(G)" is true.
To justify this, let’s first define the terms:
  • An Eulerian graph is a graph in which there exists a closed walk that visits every edge exactly once.
  • L ( G ) L ( G ) L(G)L(G)L(G) represents the line graph of G G GGG, which is a graph whose vertices represent the edges of G G GGG and where two vertices in L ( G ) L ( G ) L(G)L(G)L(G) are adjacent if and only if their corresponding edges in G G GGG are incident (share a common vertex) in G G GGG.
If G G GGG is an Eulerian graph, it means that every vertex in G G GGG has an even degree (since the Eulerian cycle enters and exits each vertex an equal number of times). In the line graph L ( G ) L ( G ) L(G)L(G)L(G), the degree of each vertex (which corresponds to an edge in G G GGG) is equal to the sum of the degrees of the two endpoints of that edge in G G GGG minus 2 (since the edge itself is not counted in L ( G ) L ( G ) L(G)L(G)L(G)). Since the sum of two even numbers is even, each vertex in L ( G ) L ( G ) L(G)L(G)L(G) also has an even degree.
Therefore, L ( G ) L ( G ) L(G)L(G)L(G) satisfies the necessary and sufficient condition for a graph to be Eulerian (every vertex has an even degree), so if G G GGG is an Eulerian graph, then L ( G ) L ( G ) L(G)L(G)L(G) is also Eulerian.
vii) χ ( C 3 × K 2 ) = 3 χ C 3 × K 2 = 3 chi^(‘)(C_(3)xxK_(2))=3\chi^{\prime}\left(C_3 \times K_2\right)=3χ(C3×K2)=3.
Answer:
The statement " χ ( C 3 × K 2 ) = 3 χ C 3 × K 2 = 3 chi^(‘)(C_(3)xxK_(2))=3\chi^{\prime}\left(C_3 \times K_2\right)=3χ(C3×K2)=3" is true.
To justify this, let’s first define the terms:
  • χ ( G ) χ ( G ) chi^(‘)(G)\chi^{\prime}(G)χ(G) represents the edge chromatic number of a graph G G GGG, which is the minimum number of colors needed to color the edges of G G GGG such that no two adjacent edges share the same color.
  • C 3 C 3 C_(3)C_3C3 is the cycle graph with 3 vertices, also known as a triangle.
  • K 2 K 2 K_(2)K_2K2 is the complete graph with 2 vertices, which is essentially a single edge.
  • C 3 × K 2 C 3 × K 2 C_(3)xxK_(2)C_3 \times K_2C3×K2 represents the Cartesian product of C 3 C 3 C_(3)C_3C3 and K 2 K 2 K_(2)K_2K2, which results in a graph where each vertex of C 3 C 3 C_(3)C_3C3 is connected to each vertex of K 2 K 2 K_(2)K_2K2, forming a hexagon.
The graph C 3 × K 2 C 3 × K 2 C_(3)xxK_(2)C_3 \times K_2C3×K2 is a hexagon, which is a 2-regular graph (each vertex has degree 2). For a 2-regular graph, the edge chromatic number is equal to 2 if the graph has an even number of vertices (since it can be divided into two perfect matchings), and it is equal to 3 if the graph has an odd number of vertices (since it cannot be divided into two perfect matchings without leaving one edge uncolored).
Since C 3 × K 2 C 3 × K 2 C_(3)xxK_(2)C_3 \times K_2C3×K2 has 6 vertices (an even number), it might seem that χ ( C 3 × K 2 ) χ C 3 × K 2 chi^(‘)(C_(3)xxK_(2))\chi^{\prime}\left(C_3 \times K_2\right)χ(C3×K2) should be 2. However, C 3 × K 2 C 3 × K 2 C_(3)xxK_(2)C_3 \times K_2C3×K2 is also a cycle graph, and for cycle graphs with an even number of vertices, the edge chromatic number is equal to the maximum degree of the graph plus 1. In this case, the maximum degree of C 3 × K 2 C 3 × K 2 C_(3)xxK_(2)C_3 \times K_2C3×K2 is 2, so χ ( C 3 × K 2 ) = 2 + 1 = 3 χ C 3 × K 2 = 2 + 1 = 3 chi^(‘)(C_(3)xxK_(2))=2+1=3\chi^{\prime}\left(C_3 \times K_2\right) = 2 + 1 = 3χ(C3×K2)=2+1=3.
Therefore, the statement " χ ( C 3 × K 2 ) = 3 χ C 3 × K 2 = 3 chi^(‘)(C_(3)xxK_(2))=3\chi^{\prime}\left(C_3 \times K_2\right)=3χ(C3×K2)=3" is true.
viii) There exists no graph G G GGG with χ ( G ) > ω ( G ) + 1 χ ( G ) > ω ( G ) + 1 chi(G) > omega(G)+1\chi(G)>\omega(G)+1χ(G)>ω(G)+1.
Answer:
The statement "There exists no graph G G GGG with χ ( G ) > ω ( G ) + 1 χ ( G ) > ω ( G ) + 1 chi(G) > omega(G)+1\chi(G)>\omega(G)+1χ(G)>ω(G)+1" is true.
To justify this, let’s first define the terms:
  • χ ( G ) χ ( G ) chi(G)\chi(G)χ(G) represents the chromatic number of a graph G G GGG, which is the minimum number of colors needed to color the vertices of G G GGG such that no two adjacent vertices share the same color.
  • ω ( G ) ω ( G ) omega(G)\omega(G)ω(G) represents the clique number of G G GGG, which is the size of the largest complete subgraph (clique) in G G GGG.
According to the Brooks’ Theorem, for any connected graph G G GGG that is neither a complete graph nor an odd cycle, the chromatic number χ ( G ) χ ( G ) chi(G)\chi(G)χ(G) is less than or equal to the maximum degree Δ ( G ) Δ ( G ) Delta(G)\Delta(G)Δ(G) of the graph. Since the maximum degree of a graph is always greater than or equal to the size of its largest clique (minus one), we have:
Δ ( G ) ω ( G ) 1 Δ ( G ) ω ( G ) 1 Delta(G) >= omega(G)-1\Delta(G) \geq \omega(G) – 1Δ(G)ω(G)1
Therefore, for such graphs:
χ ( G ) Δ ( G ) + 1 ω ( G ) χ ( G ) Δ ( G ) + 1 ω ( G ) chi(G) <= Delta(G)+1 <= omega(G)\chi(G) \leq \Delta(G) + 1 \leq \omega(G)χ(G)Δ(G)+1ω(G)
For complete graphs and odd cycles, the chromatic number is equal to the clique number (for complete graphs) or the clique number plus one (for odd cycles). In either case, the chromatic number does not exceed the clique number plus one:
χ ( G ) ω ( G ) + 1 χ ( G ) ω ( G ) + 1 chi(G) <= omega(G)+1\chi(G) \leq \omega(G) + 1χ(G)ω(G)+1
Therefore, there exists no graph G G GGG with χ ( G ) > ω ( G ) + 1 χ ( G ) > ω ( G ) + 1 chi(G) > omega(G)+1\chi(G) > \omega(G) + 1χ(G)>ω(G)+1, making the statement true.
ix) The minimum size of a k k kkk-chromatic graph is ( k 2 ) k 2 ([k],[2])\left(\begin{array}{c}k \\ 2\end{array}\right)(k2).
Answer:
The statement "The minimum size of a k k kkk-chromatic graph is ( k 2 ) k 2 ([k],[2])\left(\begin{array}{c}k \\ 2\end{array}\right)(k2)" is true.
To justify this, let’s first define the terms:
  • A k k kkk-chromatic graph is a graph whose chromatic number χ ( G ) χ ( G ) chi(G)\chi(G)χ(G) is equal to k k kkk.
  • The size of a graph refers to the number of edges in the graph.
  • ( k 2 ) k 2 ([k],[2])\left(\begin{array}{c}k \\ 2\end{array}\right)(k2) is the binomial coefficient, representing the number of ways to choose 2 elements from a set of k k kkk elements, which is also equal to k ( k 1 ) 2 k ( k 1 ) 2 (k(k-1))/(2)\frac{k(k-1)}{2}k(k1)2.
The minimum size of a k k kkk-chromatic graph is achieved by the complete graph K k K k K_(k)K_kKk, which is a graph with k k kkk vertices where every pair of distinct vertices is connected by a unique edge. The complete graph K k K k K_(k)K_kKk is k k kkk-chromatic because no two adjacent vertices can share the same color, and there are k k kkk vertices, so k k kkk colors are needed.
The size of the complete graph K k K k K_(k)K_kKk is equal to the number of edges, which is ( k 2 ) k 2 ([k],[2])\left(\begin{array}{c}k \\ 2\end{array}\right)(k2) because each edge corresponds to a pair of vertices, and there are ( k 2 ) k 2 ([k],[2])\left(\begin{array}{c}k \\ 2\end{array}\right)(k2) distinct pairs of vertices in a set of k k kkk vertices.
Therefore, the minimum size of a k k kkk-chromatic graph is ( k 2 ) k 2 ([k],[2])\left(\begin{array}{c}k \\ 2\end{array}\right)(k2), making the statement true.
x) The 6-dimensional hypercube Q 6 Q 6 Q_(6)Q_6Q6 has no perfect matching.
Answer:
The statement "The 6-dimensional hypercube Q 6 Q 6 Q_(6)Q_6Q6 has no perfect matching" is false.
To justify this, let’s first define the terms:
  • A hypercube Q n Q n Q_(n)Q_nQn is a graph that represents an n n nnn-dimensional cube. It has 2 n 2 n 2^(n)2^n2n vertices, each corresponding to a unique n n nnn-bit binary string. Two vertices are adjacent if and only if their binary strings differ in exactly one bit.
  • A perfect matching in a graph is a set of edges such that every vertex in the graph is incident to exactly one edge in the set.
For a hypercube Q n Q n Q_(n)Q_nQn, it can be shown that a perfect matching exists if and only if n n nnn is even. This is because each vertex in Q n Q n Q_(n)Q_nQn has degree n n nnn, and a necessary condition for a perfect matching to exist in a graph is that the graph has an even number of vertices (which is true for Q n Q n Q_(n)Q_nQn as 2 n 2 n 2^(n)2^n2n is even) and that the graph is k k kkk-regular with even k k kkk (which is true for Q n Q n Q_(n)Q_nQn when n n nnn is even).
Since Q 6 Q 6 Q_(6)Q_6Q6 is a 6-dimensional hypercube, it has 2 6 = 64 2 6 = 64 2^(6)=642^6 = 6426=64 vertices and each vertex has degree 6. Therefore, Q 6 Q 6 Q_(6)Q_6Q6 satisfies the conditions for the existence of a perfect matching. Hence, the statement "The 6-dimensional hypercube Q 6 Q 6 Q_(6)Q_6Q6 has no perfect matching" is false.

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