Cover of IGNOU MMTE-001 Solved Assignment 2023 for MSC MACS

IGNOU MMTE-001 Solved Assignment 2023 | M.Sc. MACS

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

₹365.00

Get Instant Access

Access via our Android App Only

Please read the following points before ordering :

Share with your Friends
Details For MMTE-001 Solved Assignment

IGNOU MMTE-001 Assignment Question Paper 2023

untitled-document-15-c4a14609-db5c-41e1-99f0-81aab3fdac8f
MMTE-001 Question Paper 2023
1.State whether the following statements are true or false. Justify your answers with a short proof or a counterexample
i) Every tree has a perfect matching.
ii) Every 2-connected bipartite graph is Hamiltonian.
iii) For some nonnegative integers d 1 , d 2 , … , d n d 1 , d 2 , … , d n d_(1),d_(2),dots,d_(n)d_1, d_2, \ldots, d_nd1,d2,…,dn the sequence ( max { d 1 , n } , d 2 , d 3 , … , d n ) max d 1 , n , d 2 , d 3 , … , d n (max{d_(1),n},d_(2),d_(3),dots,d_(n))\left(\max \left\{d_1, n\right\}, d_2, d_3, \ldots, d_n\right)(max{d1,n},d2,d3,…,dn) is graphic.
iv) Grötzsch graph is Eulerian.
v) K 4 K 4 K_(4)K_4K4, as a plane graph, is self-dual.
vi) The vertex-covering number of an odd cycle is 1 more than its independence number.
vii) The complement of a disconnected graph is connected.
viii) If κ ( G ) < κ ′ ( G ) κ ( G ) < κ ′ ( G ) kappa(G) < kappa^(‘)(G)\kappa(G)<\kappa^{\prime}(G)κ(G)<κ′(G), then δ ( G ) ≥ 4 δ ( G ) ≥ 4 delta(G) >= 4\delta(G) \geq 4δ(G)≥4.
ix) The line graph of the Petersen graph has 30 edges.
x) There exists a complete binary tree on 15 vertices.
2 (a) Does there exist a 3 -edge-colourable graph on 10 vertices and having 20 edges? Justify.
(b) Prove or disprove that the height of an n n nnn-vertex complete k k kkk-ary tree is at least log k ⁡ ( n + 1 ) − 1 log k ⁡ ( n + 1 ) − 1 log _(k)(n+1)-1\log _k(n+1)-1logk⁡(n+1)−1.
(c) Find a minimum-weigh spanning tree in the following graph.
original image
3(a) An n n nnn-vertex forest with n / 2 n / 2 n//2n / 2n/2 edges has exactly n / 2 n / 2 n//2n / 2n/2 trees as its components. True or false? Justify.
(b) Draw the complement of the following graph.
original image
Is the complement Hamiltonian? Justify your answer.
(c) Verify the König Egarváry Theorem for the following graph.
original image
4(a) Draw a diagram, as nice as possible, of the line graph of the Petersen graph. Write the number of vertices, the number of edges, the minimum and maximum degrees of it.
(b) There exists a self-complementary graph on 2023 vertices. True of false? Justify your answer.
(c) Every 3-colourable graph contains an odd cycle. True or false? Justify.
5(a) If G G GGG is a k k kkk-connected graph having n n nnn vertices, what is the minimum size of G G GGG ? Justify.
(b) Draw the dual of the following plane graph.
original image
Does the dual have any cut-vertex? Justify.
(c) Define a flow on the following network, having value at least 5.
original image
6(a) Check whether the sequence ( 4 , 4 , 4 , 3 , 2 , 2 , 1 , 1 , 1 ) ( 4 , 4 , 4 , 3 , 2 , 2 , 1 , 1 , 1 ) (4,4,4,3,2,2,1,1,1)(4,4,4,3,2,2,1,1,1)(4,4,4,3,2,2,1,1,1) is graphic or not. If yes, draw a graph realising this degree sequence.
(b) 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.
7(a) Draw an (8,15)-graph G G GGG with χ ′ ( G ) = 5 χ ′ ( G ) = 5 chi^(‘)(G)=5\chi^{\prime}(G)=5χ′(G)=5.
(b) Explain the difference between a maximal and a maximum matching, with the help of an example.
(c) Find the thickness of the line graph of K 4 K 4 K_(4)K_4K4.
8(a) Starting with the cycle ( v 1 , v 2 , v 3 , v 4 , v 5 , v 1 ) v 1 , v 2 , v 3 , v 4 , v 5 , v 1 (v_(1),v_(2),v_(3),v_(4),v_(5),v_(1))\left(v_1, v_2, v_3, v_4, v_5, v_1\right)(v1,v2,v3,v4,v5,v1) in the following weighted K 5 K 5 K_(5)K_5K5 perform the reduction step twice to get a Hamiltonian cycle with smaller weight.
original image
(b) Let G G GGG be a planar graph with at least 11 vertices. Show that G ¯ G ¯ bar(G)\bar{G}G¯ is nonplanar.
  1. (a) Let G G GGG and H H HHH be any graphs such that L ( G ) ≅ L ( H ) L ( G ) ≅ L ( H ) L(G)~=L(H)L(G) \cong L(H)L(G)≅L(H). Is it necessary that G ≅ H G ≅ H G~=HG \cong HG≅H ? Justify.
(b) For the graph given in Q. 3(b), find the number of ( v 2 , v 5 ) v 2 , v 5 (v_(2),v_(5))\left(v_2, v_5\right)(v2,v5)-walks of length 3.
\(b=c\:cos\:A+a\:cos\:C\)

MMTE-001 Sample Solution 2023

untitled-document-15-c4a14609-db5c-41e1-99f0-81aab3fdac8f
MMTE-001 Solved Assignment 2023

Question:-01

1.State whether the following statements are true or false. Justify your answers with a short proof or a counterexample
i) Every tree has a perfect matching.
Answer:
let’s address the statement:
“Every tree has a perfect matching.”
This statement is false.
Justification:
A perfect matching in a graph is a set of edges such that every vertex of the graph is adjacent to exactly one edge in the set, and no two edges in the set share a common vertex.
For a tree to have a perfect matching, it must have an even number of vertices. This is because each edge in the matching covers exactly two vertices, so if there’s an odd number of vertices, at least one vertex will be left unmatched.
Counterexample:
Consider a simple tree with 3 vertices and 2 edges (a path of length 2). This tree does not have a perfect matching because one of the three vertices will always be left unmatched.
Thus, the statement “Every tree has a perfect matching” is false.
ii) Every 2-connected bipartite graph is Hamiltonian.
Answer:
The statement is:
“Every 2-connected bipartite graph is Hamiltonian.”
This statement is false.
Justification:
A graph is Hamiltonian if it has a Hamiltonian cycle, which is a cycle that visits every vertex exactly once and returns to the starting vertex. A bipartite graph is one whose vertices can be divided into two disjoint sets such that no two vertices within the same set are adjacent.
For a bipartite graph to be Hamiltonian, it must satisfy the following condition: both sets of vertices (let’s call them A and B) must have the same number of vertices. This is because a Hamiltonian cycle in a bipartite graph will alternate between vertices in A and B. If one set has more vertices than the other, it’s impossible to form a Hamiltonian cycle.
Counterexample:
Consider a bipartite graph with sets A and B where | A | = 3 | A | = 3 |A|=3|A| = 3|A|=3 and | B | = 2 | B | = 2 |B|=2|B| = 2|B|=2. The graph is 2-connected, but it cannot have a Hamiltonian cycle because the two sets have different numbers of vertices. Specifically, imagine three vertices on the left (set A) and two on the right (set B). Even if every vertex is connected to every other vertex across the sets, you can’t form a cycle that visits every vertex exactly once and returns to the starting vertex.
Thus, the statement “Every 2-connected bipartite graph is Hamiltonian” is false.
iii) For some nonnegative integers d 1 , d 2 , … , d n d 1 , d 2 , … , d n d_(1),d_(2),dots,d_(n)d_1, d_2, \ldots, d_nd1,d2,…,dn the sequence ( max { d 1 , n } , d 2 , d 3 , … , d n ) max d 1 , n , d 2 , d 3 , … , d n (max{d_(1),n},d_(2),d_(3),dots,d_(n))\left(\max \left\{d_1, n\right\}, d_2, d_3, \ldots, d_n\right)(max{d1,n},d2,d3,…,dn) is graphic.
Answer:
The statement is:
“For some nonnegative integers d 1 , d 2 , … , d n d 1 , d 2 , … , d n d_(1),d_(2),dots,d_(n)d_1, d_2, \ldots, d_nd1,d2,…,dn the sequence ( max { d 1 , n } , d 2 , d 3 , … , d n ) max d 1 , n , d 2 , d 3 , … , d n (max{d_(1),n},d_(2),d_(3),dots,d_(n))\left(\max \left\{d_1, n\right\}, d_2, d_3, \ldots, d_n\right)(max{d1,n},d2,d3,…,dn) is graphic.”
To determine if this statement is true or false, we need to understand what it means for a sequence to be graphic. A sequence of nonnegative integers is said to be graphic if there exists a simple, undirected graph with no loops or multiple edges such that the degrees of its vertices correspond to the numbers in the sequence.
The statement is suggesting that for some sequences of degrees, if we replace d 1 d 1 d_(1)d_1d1 with the maximum of d 1 d 1 d_(1)d_1d1 and n n nnn, the resulting sequence is still graphic.
This statement is false.
Justification:
Consider the sequence d 1 , d 2 , … , d n d 1 , d 2 , … , d n d_(1),d_(2),dots,d_(n)d_1, d_2, \ldots, d_nd1,d2,…,dn where d 1 = 0 d 1 = 0 d_(1)=0d_1 = 0d1=0 and all other d i d i d_(i)d_idi values are 1. This sequence is graphic because it represents a star graph with n − 1 n − 1 n-1n-1n−1 edges and n n nnn vertices.
Now, let’s replace d 1 d 1 d_(1)d_1d1 with the maximum of d 1 d 1 d_(1)d_1d1 and n n nnn. Since d 1 = 0 d 1 = 0 d_(1)=0d_1 = 0d1=0, the new d 1 d 1 d_(1)d_1d1 value will be n n nnn. The new sequence becomes n , 1 , 1 , … , 1 n , 1 , 1 , … , 1 n,1,1,dots,1n, 1, 1, \ldots, 1n,1,1,…,1. This sequence is not graphic. The reason is that the first vertex is supposed to have n n nnn edges connected to it, but there are only n − 1 n − 1 n-1n-1n−1 other vertices available, each of which can only provide one edge.
Thus, the statement “For some nonnegative integers d 1 , d 2 , … , d n d 1 , d 2 , … , d n d_(1),d_(2),dots,d_(n)d_1, d_2, \ldots, d_nd1,d2,…,dn the sequence ( max { d 1 , n } , d 2 , d 3 , … , d n ) max d 1 , n , d 2 , d 3 , … , d n (max{d_(1),n},d_(2),d_(3),dots,d_(n))\left(\max \left\{d_1, n\right\}, d_2, d_3, \ldots, d_n\right)(max{d1,n},d2,d3,…,dn) is graphic” is false.
iv) Grötzsch graph is Eulerian.
Answer:
The statement is:
“Grötzsch graph is Eulerian.”
A graph is Eulerian if it contains an Eulerian cycle, which is a cycle that visits every edge exactly once and returns to the starting vertex. For a graph to be Eulerian, every vertex must have an even degree.
The Grötzsch graph is a triangle-free graph with 11 vertices and 20 edges.
To determine if the Grötzsch graph is Eulerian, we need to check the degrees of its vertices.
Justification:
The Grötzsch graph is known to have the following degree sequence: 5, 5, 5, 5, 4, 4, 4, 4, 4, 4, 4.
From the degree sequence, it’s clear that there are four vertices of degree 5, which is odd. Since an Eulerian graph must have all vertices of even degree, the Grötzsch graph cannot be Eulerian.
Thus, the statement “Grötzsch graph is Eulerian” is false.
v) K 4 K 4 K_(4)K_4K4, as a plane graph, is self-dual.
Answer:
The statement is:
“ K 4 K 4 K_(4)K_4K4, as a plane graph, is self-dual.”
A graph is self-dual if its planar embedding is isomorphic to its dual graph. The dual graph of a planar graph has a vertex for each face of the original graph and an edge for each edge in the original graph that separates the corresponding faces.
Justification:
The graph K 4 K 4 K_(4)K_4K4 is the complete graph on 4 vertices. When drawn in the plane without any edges crossing, K 4 K 4 K_(4)K_4K4 divides the plane into 4 regions: 3 triangular regions and 1 outer region.
The dual graph of this planar embedding of K 4 K 4 K_(4)K_4K4 will have 4 vertices (one for each region). Since each triangular region in K 4 K 4 K_(4)K_4K4 is bounded by 3 edges, each vertex in the dual graph will have degree 3. This means the dual graph is also K 4 K 4 K_(4)K_4K4.
Thus, the planar embedding of K 4 K 4 K_(4)K_4K4 is isomorphic to its dual, making K 4 K 4 K_(4)K_4K4 self-dual when considered as a plane graph.
Therefore, the statement “ K 4 K 4 K_(4)\boldsymbol{K_4}K4, as a plane graph, is self-dual” is true.
\(c^2=a^2+b^2-2ab\:Cos\left(C\right)\)
Get The AppClick Here
Get Instant Access

Frequently Asked Questions (FAQs)

You can access the Complete Solution through our app, which can be downloaded using this link:

App Link 

Simply click “Install” to download and install the app, and then follow the instructions to purchase the required assignment solution. Currently, the app is only available for Android devices. We are working on making the app available for iOS in the future, but it is not currently available for iOS devices.

Yes, It is Complete Solution, a comprehensive solution to the assignments for IGNOU. 

Yes, the Complete Solution is aligned with the requirements and has been solved accordingly.

Yes, the Complete Solution is guaranteed to be error-free.The solutions are thoroughly researched and verified by subject matter experts to ensure their accuracy.

As of now, you have access to the Complete Solution for a period of 1 Year after the date of purchase, which is sufficient to complete the assignment. However, we can extend the access period upon request. You can access the solution anytime through our app.

The app provides complete solutions for all assignment questions. If you still need help, you can contact the support team for assistance at Whatsapp +91-9958288900

No, access to the educational materials is limited to one device only, where you have first logged in. Logging in on multiple devices is not allowed and may result in the revocation of access to the educational materials.

Payments can be made through various secure online payment methods available in the app.Your payment information is protected with industry-standard security measures to ensure its confidentiality and safety. You will receive a receipt for your payment through email or within the app, depending on your preference.

The instructions for formatting your assignments are detailed in the Assignment Booklet, which includes details on paper size, margins, precision, and submission requirements. It is important to strictly follow these instructions to facilitate evaluation and avoid delays.

\(2\:cos\:\theta \:sin\:\phi =sin\:\left(\theta +\phi \right)-sin\:\left(\theta -\phi \right)\)

Terms and Conditions

  • The educational materials provided in the app are the sole property of the app owner and are protected by copyright laws.
  • Reproduction, distribution, or sale of the educational materials without prior written consent from the app owner is strictly prohibited and may result in legal consequences.
  • Any attempt to modify, alter, or use the educational materials for commercial purposes is strictly prohibited.
  • The app owner reserves the right to revoke access to the educational materials at any time without notice for any violation of these terms and conditions.
  • The app owner is not responsible for any damages or losses resulting from the use of the educational materials.
  • The app owner reserves the right to modify these terms and conditions at any time without notice.
  • By accessing and using the app, you agree to abide by these terms and conditions.
  • Access to the educational materials is limited to one device only. Logging in to the app on multiple devices is not allowed and may result in the revocation of access to the educational materials.

Our educational materials are solely available on our website and application only. Users and students can report the dealing or selling of the copied version of our educational materials by any third party at our email address (abstract4math@gmail.com) or mobile no. (+91-9958288900).

In return, such users/students can expect free our educational materials/assignments and other benefits as a bonafide gesture which will be completely dependent upon our discretion.

Categories , , Tags , , , , , , , , , , , , , ,

Related products

Our educational materials are solely available on our website and application only. Users and students can report the dealing or selling of the copied version of our educational materials by any third party at our email address (abstract4math@gmail.com) or mobile no. (+91-9958288900).

In return, such users/students can expect free our educational materials/assignments and other benefits as a bonafide gesture which will be completely dependent upon our discretion.

All Policies

Account

Company

Affiliate Program

Shipping Policy

Privacy Policy

Return Policy

Terms & Conditions

Cancellation Policy

Registration & Security

Refund Policy

Copyright & DMCA Notice

Privacy Policy

Content Copyright

Dispute Resolution

DMCA.com Protection Status

Terms of Services

Contact Us

Address: L-1/2, Uttam Nagar, Delhi, 110059
Email ID: support@abstractclasses.com 
WhatsApp No.: +91 9958288900

Online Classes available . More than 120 students joined last Month.
Contact+91-9958288900

Home

UPSC

IGNOU

My Account

Scroll to Top
Scroll to Top