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Assignment Brief
Student Name /ID Number
A.A.Lahiru Shamika / KUR00064943
Unit Number and Title
Unit 18 :Discrete Mathematics
Academic Year
2021/22
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Discrete mathematics in Computing
Assignment Title
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Submission Date
27/07/2022
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Unit Learning Outcomes:
LO1
Examine set theory and functions applicable to software engineering.
LO2
Analyse mathematical structures of objects using graph theory.
LO3
LO4
Investigate solutions to problem situations using the application of Boolean algebra.
Explore applicable concepts within abstract algebra.
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Assignment Brief and Guidance:
Activity 01
Part 1
1. Perform algebraic set operations in the following formulated mathematical problems.
i. Let A and B be two non-empty finite sets. If cardinalities of the sets A, B, and are 72, 28 and 13
respectively, find the cardinality of the set.
ii.
If n()=45, n()=110 and n()=15, then find n(B).
iii.If n(A)=33, n(B)=36 and n(C)=28, find n().
Part 2
1. Write the multisets (bags) of prime factors of given numbers.
i. 160
ii. 120
iii. 250
2. Write the multiplicities of each element of multisets (bags) in Part 2-1(i,ii,iii) separately.
3. Determine the cardinalities of each multiset (bag) in Part 2-1(i,ii,iii).
Part 3
1. Determine whether the following functions are invertible or not and if a function is invertible,
then find the rule of the inverse using appropriate mathematical technique.
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Part 4
1.
Formulate corresponding proof principles to prove the following properties about defined sets.
i. .
ii. De Morgan’s Law by mathematical induction.
iii. Distributive Laws for three non-empty finite sets A, B, and C.
Activity 02
Part 1
1. Model two contextualized problems using binary trees both quantitatively and qualitatively.
Part 2
1. State the Dijkstra’s algorithm for a directed weighted graph with all non-negative edge
weights.
2. Use Dijkstra’s algorithm to find the shortest path spanning tree for the following weighted
directed graph with vertices A, B, C, D, and E given. Consider the starting vertex as E.
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Part 3
1. Assess whether the following undirected graphs have an Eulerian and/or a Hamiltonian cycle.
i.
ii.
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iii.
Part 4
1. Construct a proof of the five color theorem for every planar graph.
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Activity 03
Part 1
1. Diagram two real world binary problems in two different fields using applications of Boolean
Algebra.
Part 2
1. Produce truth tables and its corresponding Boolean equation for the following scenarios.
i. If the driver is present and the driver has not buckled up and the ignition switch is on,
then the warning light should turn on.
ii. If it rains and you don't open your umbrella, then you will get wet.
2. Produce truth tables for given Boolean expressions.
i.
ii.
Part 3
1. Simplify the following Boolean expressions using algebraic methods.
i.
ii.
iii.
iv.
Part 4
1. Consider the K-Maps given below. For each K- Map
i.
Write the appropriate standard form (SOP/POS) of Boolean expression.
ii. Design the circuit using AND, NOT and OR gates.
iii. Design the circuit only by using
 NAND gates if the standard form obtained in part (i) is SOP.
 NOR gates if the standard form obtained in pat (i) is POS.
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(a)
AB/C
0
1
00
0
0
01
0
1
11
0
1
10
1
0
(b)
AB/CD 00
01
11
10
00
1
0
0
1
01
0
1
0
1
11
1
1
1
0
10
1
1
1
1
(c)
AB/C
0
1
00
1
0
01
1
1
11
1
0
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10
0
1
Activity 04
Part 1
1. Describe the distinguishing characteristics of different binary operations that are performed on the
same set.
Part 2
1. Determine the operation tables for group G with orders 1, 2, 3 and 4 using the elements a, b, c, and e
as the identity element in an appropriate way.
2.
i. State the relation between the order of a group and the number of binary operations that can be
defined on that set.
ii. How many binary operations can be defined on a set with 4 elements?
3.
i. State the Lagrange’s theorem of group theory.
ii. For a subgroup H of a group G, prove the Lagrange’s theorem.
iii. Discuss whether a group H with order 6 can be a subgroup of a group with order 13 or not.
Clearly state the reasons.
Part 3
1.
Validate whether the set is a group under the binary operation ‘*’defined as for any two elements.
Part 4
1.
Prepare a presentation for ten minutes to explore an application of group theory relevant to your
course of study. (i.e. in Computer Sciences)
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Acknowledgement
I would like to make this a moment to thank our dearest lecturer Mr Tharanga who gave his
fullest support in teaching this subject to us and also I would like to thank my family and
friends who gave me their fullest support in completing this assignment.
A.A.L.S.Amarasinghe.
Esoft Metro Campus
Kurunegala
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Contents
Acknowledgement................................................................................................................21
List of figures.......................................................................................................................22
List of tables.........................................................................................................................23
Activity 01....................................................................................................25
Part 1....................................................................................................................................25
Part 2....................................................................................................................................27
Part 3....................................................................................................................................29
Part 4....................................................................................................................................34
Activity 02....................................................................................................38
Part 1....................................................................................................................................38
Activity 02....................................................................................................38
Part 1....................................................................................................................................38
Part 2....................................................................................................................................42
Part 4....................................................................................................................................45
Activity 03....................................................................................................50
Part 1....................................................................................................................................50
Part 2....................................................................................................................................52
Part 3....................................................................................................................................54
Part 4....................................................................................................................................57
Activity 04....................................................................................................62
Part 1....................................................................................................................................62
Part 2....................................................................................................................................63
Part 3....................................................................................................................................67
Part 4....................................................................................................................................67
Bibliography..................................................................................................72
List of figures
Figure 1.....................................................................................................................................37
Figure 2.....................................................................................................................................38
Figure 3.....................................................................................................................................38
Figure 4.....................................................................................................................................41
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Figure 5.....................................................................................................................................41
Figure 6.....................................................................................................................................42
Figure 7.....................................................................................................................................44
Figure 8.....................................................................................................................................46
Figure 9.....................................................................................................................................47
Figure 10...................................................................................................................................47
Figure 11...................................................................................................................................49
Figure 12...................................................................................................................................49
Figure 13...................................................................................................................................50
Figure 14...................................................................................................................................50
Figure 15...................................................................................................................................51
Figure 16...................................................................................................................................51
Figure 17...................................................................................................................................62
Figure 18...................................................................................................................................62
Figure 19...................................................................................................................................63
Figure 20...................................................................................................................................63
Figure 21...................................................................................................................................64
Figure 22...................................................................................................................................71
Figure 23...................................................................................................................................72
Figure 24...................................................................................................................................72
Figure 25...................................................................................................................................73
Figure 26...................................................................................................................................73
Figure 27...................................................................................................................................74
Figure 28...................................................................................................................................75
Figure 29...................................................................................................................................76
List of tables
Table 1......................................................................................................................................31
Table 2......................................................................................................................................54
Table 3......................................................................................................................................55
Table 4......................................................................................................................................56
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Table 5......................................................................................................................................56
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Activity 01
Part 1
1. Perform algebraic set operations in the following formulated mathematical problems.
I. Let A and B be two non-empty finite sets. If cardinalities of the sets A, B, and are 72, 28
and 13 respectively, find the cardinality of the set.
P(A) = 72
P(B) = 28
P() =13
()=P(A)+P(B) - P()
=72 + 28 – 13
=87
II. If n()=45, n()=110 and n()=15, then find n(B).
P()= P()=P(A) - P()= 45
P()=110
P()= 15
P(A)= P() = 45
P(A)= 60
P(A)= 45 + P()
P(A)= 45 + 15= 60
P()=P(A)+P(B) - P()
110= 60 + P(B) – 15
110= 60 – 15 + P(B)
110 – 45 = P(B)
P(B) = 65
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II. If n(A)=33, n(B)=36 and n(C)=28, find n().
P()= ?
P(A)= 33
P(B)= 36
P(C)= 28
P()=P(A)+ P(B)+ P(C) - P() - P()-P()
P(A)= 10+ a + 5+b = 33
15+a+b = 33
a + b = 18
1
15+a+5+c = 36
20+ a+ c =36
a + c = 16
2
13+ 5+b+c = 28
b + c= 28 – 18
b + c = 10
3
a + b = 18
1
b + c= 160
2
b + c = 16
3
1+3
a + b – b-c = 8
a - c =8
4
2+4
2a = 24
a= 12
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b=6
c=4
P()
= 33+ 36+28- 17-11-9
= 97 – 37
= 60
Part 2
I. Write the multisets (bags) of prime factors of given numbers.
i.
160 = {2, 2, 2, 2, 2, 5}
ii.
120 = {2, 2, 2, 2, 3, 5}
iii.
250 = {2, 5, 5, 5}
II. Write the multiplicities of each element of multisets (bags) in Part 2-1(i,ii,iii) separately.
i.
160={2, 2, 2, 2, 2, 5}
Multiplicity of 2 = 5
Multiplicity of 5 = 1
ii.
120={2, 2, 2, 2, 3, 5}
Multiplicity of 2 = 3
Multiplicity of 3 = 1
Multiplicity of 5 = 5=1
iii.
250={2, 5, 5, 5}
Multiplicity of 2 = 2
Multiplicity of 5 = 3
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III. Determine the cardinalities of each multisets (bag) in Part 2-1(i,ii,iii).
i.
Cardinality of multi set = 5 + 1 = 6
ii.
Cardinality of multi set = 3 + 1 +1 = 5
iii.
Cardinality of multi set = 2 + 3 = 5
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Part 3
I. Determine whether the following functions are invertible or not and if a function is
invertible, then find the rule of the inverse using appropriate mathematical technique.
N = {1,2,3,4,5,}
Z = {…. -3, -2, -1,0,1,2,3,4, 5….}
Z+ = {1,2,3,4, 5….}
Z- = {…. -3, -2, -1,}
I.
f(x) = 2x
f (-3) = 2*(-3) = (-6) →f2 * (-6) = (-12)
f (-2) = 2*(-2) = (-4) →f2 * (-4) = (-8)
f (-1) = 2*(-1) = (-2) →f2 * (-2) = (-4)
f (0) = 2*(0) = (0) →f2 * (0) = (0)
f (1) = 2*(1) = (2) →f2 * (2) = (4)
f (2) = 2*(2) = (4) →f2 * (4) = (8)
f (3) = 2*(3) = (6) →f2 * (6) = (12)
II.
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X1, x2 be two different numbers from domain
→①
→②
①-②
F(x1) = f(x2)
x1=x2
1-1 function
F(x) = y
Y= 1/x
X=1/y
X→y→x
Y= 1/x (inverse function)
=
x
1
2
4
5
25
X2
1
4
16
25
625
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0.1
0.0
0.01
0.000
1
Table 1
1
X1, x2 be two different numbers from domain
→①
→②
①=②
X12= X22
Take the square root
x1 = +x2
x1 = x2
then 1 – 1 function
inverse exit.
f(x) = y
y = x2
Take the square root
x=y&y=x
(inverse)
=
III.
= - sin =(-1)
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sin (-x) = - sin (x)
= -sin
=sin
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Sin 0 = 0 onto function
X1, x2 be two different numbers
f (x1) = sin (x1) →①
f (x2) = sin (x2) →②
①=②
sin (x1) = sin (x2)
f (x) = sin (x)
y = f (x)
y = sin (x)
sin -1(y) = x
x→y&y→x
y = sin -1(x)
= sin -1(x)
IV.
= - sin =(-1)
=sin = 1
sin (-x) = - sin (x)
= -sin
=sin
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Sin 0 = 0 onto function
X1, x2 be two different numbers
f (x1) = sin (x1) →①
f (x2) = sin (x2) →②
①=②
sin (x1) = sin (x2)
f (x) = sin (x)
y = f (x)
y = sin (x)
sin -1(y) = x
x→y&y→x
y = sin -1(x)
= sin -1(x)
Part 4
I. Formulate corresponding proof principles to prove the following properties about defined sets.
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.
If , it must be (maximum; A<B or minimum ; A=B);
B
A
But for become both of above condition, it must be A=B;
Then it will be;
It proof that
De Morgan’s Law by mathematical induction.
De Morgan’s Law is
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Let P = (A U B)' and Q = A' ∩ B'
Let x be an arbitrary element of P then x ∈ P ⇒ x ∈ (A U B)'
⇒ x ∉ (A U B)
⇒ x ∉ A and x ∉ B
⇒ x ∈ A' and x ∈ B'
⇒ x ∈ A' ∩ B'
⇒x∈Q
Therefore, P ⊂ Q …………….. (i)
Again, let y be an arbitrary element of Q then y ∈ Q ⇒ y ∈ A' ∩ B'
⇒ y ∈ A' and y ∈ B'
⇒ y ∉ A and y ∉ B
⇒ y ∉ (A U B)
⇒ y ∈ (A U B)'
⇒y∈P
Therefore, Q ⊂ P …………….. (ii)
Now combine (i) and (ii) we get; P = Q i.e. (A U B)' = A' ∩ B'
Distributive Laws for three non-empty finite sets A, B, and C.
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Figure 1
Distributive Laws for three non-empty finite sets A, B, and C
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Figure 2
Activity 02
Part 1
Figure 3
Activity 02
Part 1
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I. Model two contextualized problems using binary trees both quantitatively and qualitatively.
A binary tree is a tree data structure in which each node can have up to two child nodes, which form
the tree's branches. The left and right nodes are the names given to the two offspring. Child nodes may
include references to their parents, whereas parent nodes may contain references to their children.
We broaden the definition of linked data structures to include nodes with multiple self-referenced fields. A
binary tree is made up of nodes, each of which has a "left" and "right" reference as well as a data element.
The root refers to the tree's topmost node.
More tree terminology:
1.
The depth of a node is the number of edges from the root to the node.
2.
The height of a node is the number of edges from the node to the deepest leaf.
3.
The height of a tree is a height of the root.
4.
A full binary tree is a binary tree in which each node has exactly zero or two children.
5.
A complete binary tree is a binary tree, which is completely filled, with the possible exception of the
bottom level, which is filled from left to right.
A complete binary tree is very special tree, it provides the best possible ratio between the number of nodes
and the height. The height h of a complete binary tree with N nodes is at most O(log N). We can easily
prove this by counting nodes on each level, starting with the root, assuming that each level has the
maximum number of nodes:
Advantages of trees
Trees are so useful and frequently used, because they have some very serious advantages:
1.
Trees reflect structural relationships in the data
2.
Trees are used to represent hierarchies
3.
Trees provide an efficient insertion and searching
4.
Trees are very flexible data, allowing to move sub trees around with minimum effort
Binary Search Trees
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We consider a particular kind of a binary tree called a Binary Search Tree (BST). The basic idea behind this
data structure is to have such a storing repository that provides the efficient way of data sorting, searching
and retrieving.
A BST is a binary tree where nodes are ordered in the following way:
1.
each node contains one key (also known as data)
2.
the keys in the left sub tree are less than the key in its parent node, in short L < P;
3.
the keys in the right sub tree are greater the key in its parent node, in short P < R;
4.
Duplicate keys are not allowed.
In the following tree all nodes in the left sub tree of 10 have keys < 10 while all nodes in the right sub tree
> 10. Because both the left and right sub trees of a BST are again search trees; the above definition is
recursively applied to all internal nodes: [ CITATION bin21 \l 1033 ]
Implementation
We implement a binary search tree using a private inner class BSTNode. In order to support the binary
search tree property, we require that data stored in each node is Comparable:
Figure 4
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Binary Trees quantitatively and qualitatively;
Quantitatively mean we can use tree data structure for valuable examples. Let’s focus on following
situation.
For example;
There is a school with two classes and need to know about number of girls and boys on these classes.
Figure 5
Qualitatively mean we can use tree data structure for quality measurement examples. Let’s focus on
following situation.
For example;
A recursive partitioning tree showing survival of passengers on the Titanic ("sibsp" is the number of
spouses or siblings aboard). The figures under the leaves show the probability of survival and the
percentage of observations in the leaf.
[ CITATION Bin21 \l 1033 ]
Figure 6
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Part 2
I. State the Dijkstra’s algorithm for a directed weighted graph with all non-negative edge weights.
Dijkstra's algorithm is one method for determining the shortest path from a beginning node to a target node
in a weighted network. The technique builds a tree of shortest routes from the source vertex to every other
point in the graph.
Dijkstra's method, named after its developer, Dutch computer scientist "Edsger Dijkstra," was released in
1959 and may be used on a weighted graph. There are two types of graphs: directed and undirected. One
requirement for utilizing the technique is that every edge in the graph has a nonnegative weight.
1. Make a set called sptSet to keep track of the vertices in the shortest path tree.
2. For instance, whose minimal distance from the source has been determined and finalized. This
collection is initially empty.
3.
Assign a distance value to each of the input graph's vertices. All distance values should be set to
INFINITE. Assign the source vertex a distance value of 0 to ensure that it is chosen first.
4. sptSet does not contain all vertices.
5. Choose a vertex u that is not in sptSet and has the smallest distance value.
6. Add u to the sptSet.
7. Update the distance between all of u's neighboring vertices. Iterate over all neighboring vertices to
update the distance values. If the total of the distance value of u and the weight of edge u-v is less
than the distance value of v for each neighboring vertex v, then update the distance value of v.
[ CITATION Dij211 \l 1033 ]
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II. Use Dijkstra’s algorithm to find the shortest path spanning tree for the following weighted directed
graph with vertices A, B, C, D, and E given. Consider the starting vertex as E.
Figure 7
A to B = 5 units
A to C = 3 units
A to D = 7units
A to E = 7 units.
First Strat from E.
Then;
A – B = 5.
A– D = 3.
Select D.
Then;
D – B = 2.
D – B = 6.
Select B.
Then;
B– E = 4.
Select E.
End is C.
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Shortest path is A – D – B – E – C.
Part 3
I. Assess whether the following undirected graphs have an Eulerian and/or Hamiltonian cycle.
i.
Figure 8
This is not Eulerian: “E” vertices cannot cover because there is only one edge to visit “E”.
But this has Hamilton circuit.
ii.
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Figure 9
This is not Eulerian: Because one edge is needed to complete the route
So, this has Hamilton circuit.
iii.
Figure 10
This is not Eulerian: Because one edge is needed to complete the route.
So this has Hamilton circuit.
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Part 4
1. Construct a proof of the five color theorem for every planar graph.
The five color theorem is a graph theory conclusion that states that given a plane divided into regions, such
as a political map of a state's counties, the areas may be colored using no more than five colors, with no
two neighboring parts receiving the same color.
Contradiction serves as proof.
Let G be the smallest planar structure (in terms of vertices) that can't be colored with more than five colors.
Let v be the vertex in G with the highest degree. The corollary to Euler's formula tells us that deg(v) = 6.
1.Case #1: deg(v) ≤ 4. G-v can be colored with five colors.
G-v may be colored with five different colors in case #1: deg(v) = 4.
The neighbors of v have been painted in a maximum of four colors. Then v. is available in at least one
color.
As a result, G may be colored in five different ways, which is a contradiction.
Figure 11
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Deg(v) = 5 in case #2. G-v can be colored in five different ways.
There is a color accessible for v if two of its neighbors are colored with the same color. As a result, we may
suppose that all vertices next to v are colored in clockwise sequence with colors 1,2,3,4,5. Assume that
colors 1 and 3 are used to color all of the vertices (and all the edges among them).
Figure 12
We can flip the colors 1 and 3 in the component with v1 if this subgraph G is unconnected and v1 and v3
are in separate components.
Figure 13
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G-v will still be colored in five colors. Furthermore, in this new 5-coloring, v1 is colored with color 3 and
v3 is still colored with color 3. Color 1 would be accessible for v, which is incongruent. As a result, v1 and
v3 must be in the same component in that sub graph, i.e. there must be a path from v1 to v3 that is colored
with either color 1 or color 3.
Figure 14
Assume that all of the vertices are colored in colors 2 and 4. (and all the edges among them). If v2 and v4
are not linked components, we can swap the colors in the chain starting at v2 and utilize the leftover color
for v.
Figure 15
If they are on the same linked component, a path exists from v2 to v4 along which every vertex has either
color 2 or color 4.
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Figure 16
This implies that there must be two intersecting edges. This violates the graph's planarity, bringing the
evidence to a close.[ CITATION pro21 \l 1033 ]
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Activity 03
Part 1
I. Diagram two real world binary problems in two different fields using applications of Boolean algebra.
A sequence of logic gates are coupled together to turn on and off each part of a calculator's display.
Consider only the lower right-hand section. If we're showing the numerals 0 (binary 00), 1 (01), 3 (11), 4
(100), 5 (101), 6 (110), 7 (111), 8 (1000), and 9 (1000), we'll need to turn this segment on (1001)
But not if the number 2 is displayed (10). By wiring up three OR gates and one NOT gate like this, you can
make the segment turn on and off appropriately for the digits 1–10.
If you enter the binary number patterns into the four inputs on the left, each section will turn on and off
appropriately.
Representing numbers in binary;
Computers and calculators have been created using a variety of switching devices that can be in one of two
positions over the previous century or more.
It just contains “one (1)” or “zero (0)” in it. As a result, computers and calculators employ binary code to
store and process numbers, which requires only two symbols (0 and 1) to represent any number.
So in binary code, the number 18 is written 10010,
Which means; (1 × 16) + (0 × 8) + (0 × 4) + (1 × 2) + (1 × 0)
=16+2
=18
Using logic gates with binary
Someone wants to do the sum 3 + 2 = 5.
A calculator solves a problem like this by converting the two integers to binary, resulting in 101 (5 in
binary = 1 4 + 0 2 + 1 1) by adding 11 (which is 3 in binary = 1 2 + 1 1) and 10 (2 in binary = 1 2 + 0 1).
So, how does the calculator calculate the total? It employs logic gates to compare the active switch pattern
and generate a new switch pattern in its place.
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A logic gate is a basic electrical circuit that compares two numbers and generates a third number based on
the original numbers' values.
There are 4 very common types of logic gates.
1. OR
2. AND
3. NOT
4. XOR
An OR gate has two inputs and it produces an output of 1 if either of the inputs is 1; it produces a zero
otherwise.
An AND gate also has two inputs, but it produces an output of 1 only if both inputs are 1.
A NOT gate has a single input and reverses it to make an output. So if feed it a zero, it produces a 1.
An XOR gate gives the same output as an OR gate, but switches off if both its inputs are one.
[ CITATION rea21 \l 1033 ]
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Part 2
I. Produce truth tables and its corresponding Boolean equation for the following scenarios.
i. If the driver is present and the driver has not buckled up and the ignition switch is on,
then the warning light should turn on.
P: - The driver is present
Q: -The driver has buckled up
R: - The ignition switch is on
S: - The warning light should turn ON
Pᴧ QᴧR=S
Boolean equalization = P.Ǭ.R
P
Q
R
Q
T
T
T
T
T
F
T
F
F
T
F
T
F
F
F
F
Table 2
T
F
T
F
T
F
T
F
F
F
T
T
F
F
T
T
Pᴧ Q
F
F
T
T
F
F
F
F
Pᴧ QᴧR=S
F
F
T
F
F
F
F
F
ii. If it rains and you don't open your umbrella, then you will get wet.
P: - It rains
Q: - you open your umbrella
R: - you will get wet
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(P ᴧ Q) → R
P
Q
R
Q
T
T
T
T
T
F
T
F
F
T
F
T
F
F
F
F
Table 3
T
F
T
F
T
F
T
F
F
F
T
T
F
F
T
T
A.A.L.S.Amarasinghe
Pᴧ Q
F
F
T
T
F
F
F
F
Pᴧ Q→R
T
T
T
F
T
T
T
T
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II.
Produce truth tables for given Boolean expressions.
I.
A B
0 0
0 0
0 1
0 1
1 0
1 0
1 1
1 1
Table 4
C
0
1
0
1
0
1
0
1
1
1
1
1
0
0
0
0
1
1
0
0
1
1
0
0
1
0
1
0
1
0
10
0
1
1
1
1
1
1
1
1
1
0
1
1
1
1
1
0
1
1
1
1
1
1
1
1
1
1
1
1
0
1
1
0
1
1
0
1
0
0
1
II.
A
0
0
0
0
1
1
1
1
B
0
0
1
1
0
0
1
1
C
0
1
0
1
0
1
0
1
1
1
1
1
0
0
0
0
1
1
0
0
1
1
0
0
1
0
1
0
1
0
11
0
1
1
0
1
1
1
0
1
1
1
1
1
1
1
1
1
1
1
1
1
0
1
1
0
1
0
1
1
0
1
1
following Boolean expressions using algebraic methods.
I.
= AA + AB + BB + BC + CC + CA
= A + AB + B + BC +C + CA
Table 5
= A + B (A + 1) + C (B + 1) + CA
= B + C + A (1 + C)
=A+B+C
II.
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Part 3
I. Simplify the
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= AB + AC + B`B + B`C + AC + A`A + BC + BA`
= AB + AC + 0 + B`C + AC + 0 + BC + BA`
= AB + AC + C (B` + B) + BA`
= B (A + A`) + AC + C
= B + C (A + 1)
=B+C
III.
=AAC + AAC` + BAC + BAC` + AB + B
= A (C + C`) + BA (C + C`) + (A +1) B
= A + BA + B
= A + AB + B + AB
= A (1 + B) + B (1 + A)
=A+B
IV.
= AA` + A`B + BA + BB` + AA + AB`
= A`B + AB + A + AB`
= (A` + A) B + A (1 + B`)
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= B+A
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Part 4
I. Consider the K-Maps given below. For each K- Map
I.
Write the appropriate standard form (SOP/POS) of Boolean expression.
The sum-of-products (SOP) form is a technique (or form) for reducing logic gates' Boolean statements. The
variables are ANDed (summed or added) together in this SOP form of Boolean function representation to
produce a product term, and all of these product terms are ORed (summed or added) together to make the
final function.
Part a) is for the left K-map, Part b) is for the right K-map.
AB/C
0
00
0
01
AB/C
1
00
0
01
1
11
10
0
11
D
00
0
1
10
0
1
10
01
1
0
01
0
1
11
1
1
1
0
10
1
1
1
1
SOP=A′BC+ABC+AB′C′
POS=(A'B'+C')(A'B'+C)(A'B+C')(AB+C')(AB'+C)POS=(A′B′+C′)(A′B′+C)(A′B+C′)(AB+C′)(AB′+C)
(A'B'+C')=(A'+C')(B'+C')=(A'+C'+B')(A'+C'+B)\times(A′B′+C′)=(A′+C′)(B′+C′)=(A′+C′+B′)(A′+C′+B)×
\times (B'+C'+A')(B'+C'+A)=(A'+C'+B')(A'+C'+B))(B'+C'+A)×(B′+C′+A′)(B′+C′+A)=(A′+C′+B′)(A′+C′
+B))(B′+C′+A)
(A'B'+C)=(A'+C)(B'+C)=(A'+C+B)(A'+C+B')\times(A′B′+C)=(A′+C)(B′+C)=(A′+C+B)(A′+C+B′)×
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\times(B'+C+A)(B'+C+A')=(A'+C+B)(A'+C+B')(B'+C+A)×(B′+C+A)(B′+C+A′)=(A′+C+B)(A′+C+B′)(B′
+C+A)
(A'B+C')=(A'+C')(B+C')=(A'+C'+B)(A'+C'+B')\times(A′B+C′)=(A′+C′)(B+C′)=(A′+C′+B)(A′+C′+B′)×
\times(B+C'+A)(B+C'+A')=(A'+C'+B)(A'+C'+B')(B+C'+A)×(B+C′+A)(B+C′+A′)=(A′+C′+B)(A′+C′+B′)
(B+C′+A)
(AB+C')=(A+C')(B+C')=(A+C'+B)(A+C'+B')\times(AB+C′)=(A+C′)(B+C′)=(A+C′+B)(A+C′+B′)×
\times(B+C'+A)(B+C'+A')=(A+C'+B)(A+C'+B')(B+C'+A')×(B+C′+A)(B+C′+A′)=(A+C′+B)(A+C′+B′)
(B+C′+A′)
(AB'+C)=(A+C)(B'+C)=(A+C+B)(A+C+B')\times(AB′+C)=(A+C)(B′+C)=(A+C+B)(A+C+B′)×
\times(B'+C+A)(B'+C+A')=(A+C+B')(A+C+B')(B'+C+A')×(B′+C+A)(B′+C+A′)=(A+C+B′)(A+C+B′)(B′
+C+A′)
Standard form:
POS=(A'+C'+B')(A'+C'+B))(B'+C'+A)(A'+C+B)(A'+C+B')\timesPOS=(A′+C′+B′)(A′+C′+B))(B′+C′+A)
(A′+C+B)(A′+C+B′)×
\times(B'+C+A)(B+C'+A)×(B′+C+A)(B+C′+A)
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II.
Design the circuit using AND, NOT and OR gates.
Figure 17
III.
Design the circuit only by using
 NAND gates if the standard form obtained in part (i) is SOP.
 NOR gates if the standard form obtained in pat (i) is POS.
OR gate using NAND gate:
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Figure 18
AND gate using NAND gate:
Figure 19
THEN;
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Figure 20
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B)
i.
SOP=A′B′C′D′+A′B′CD′+A′BC′D+A′BCD′+ABC′D′+
+ABC'D+ABCD+AB'C'D'+AB'C'D+AB'CD+AB'CD'+ABC′D+ABCD+AB′C′D′+AB′C′D+AB′CD+AB
′CD′
POS=(A'B'+C'D)(A'B'+CD)(A'B+C'D')(A'B+CD)(AB+CD')POS=(A′B′+C′D)(A′B′+CD)(A′B+C′D′)(A
′B+CD)(AB+CD′)
II.
Figure 21
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Activity 04
Part 1
I. Describe the distinguishing characteristics of different binary operations that are performed on the same
set.
Just like when two numbers are combined, subtracted, multiplied, or divided, we obtain a number. Any two
members of a set can be linked using binary operations. The resultant of the two is in the same set as the
resultant of the two. Calculations that combine two items of a set (called operands) to generate another
element of the same set are known as binary operations on a set.
The binary operations * on a non-empty set A are functions from A × A to A. The binary operation, *: A ×
A → A. It is an operation of two elements of the set whose domains and co-domain are in the same set.
Addition, subtraction, multiplication, division, exponential is some of the binary operations.
Types of Binary Operations
Commutative
A binary operation * on a set A is commutative if a * b = b * a, for all (a, b) ∈ A (non-empty set). Let
addition be the operating binary operation for a = 8 and b = 9, a + b = 17 = b + a.
Associative
The associative property of binary operations hold if, for a non-empty set A, we can write (a * b) *c = a*(b
* c). Suppose N be the set of natural numbers and multiplication be the binary operation. Let a = 4, b = 5 c
= 6. We can write (a × b) × c = 120 = a × (b × c).
Distributive
Let * and o be two binary operations defined on a non-empty set A. The binary operations are distributive
if a*(b o c) = (a * b) o (a * c) or (b o c)*a = (b * a) o (c * a). Consider * to be multiplication and o be
subtraction. And a = 2, b = 5, c = 4. Then, a*(b o c) = a × (b − c) = 2 × (5 − 4) = 2. And (a * b) o (a * c)
= (a × b) − (a × c) = (2 × 5) − (2 × 4) = 10 − 6 = 2.
Identity
If A be the non-empty set and * be the binary operation on A. An element e is the identity element of a
∈ A, if a * e = a = e * a. If the binary operation is addition (+), e = 0 and for * is multiplication (×), e = 1.
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
Inverse
If a binary operation * on a set A which satisfies a * b = b * a = e, for all a, b ∈ A. a-1 is invertible if for a
* b = b * a= e, a-1 = b. 1 is invertible when * is multiplication.
[ CITATION dis21 \l 1033 ]
Part 2
I.
Determine the operation tables for group G with orders 1, 2, 3 and 4 using the elements a, b, c, and
e as the identity element in an appropriate way.
Multiplication is a binary operation on each of the sets of Natural numbers (N), Integer numbers (Z),
Rational numbers (Q), Real Numbers (R), and Complex numbers (C) (C). On the set of Natural numbers
(N), integers (Z), Rational numbers (Q), Real Numbers (R), and Complex numbers (C), division is not a
binary operation (C). As Binary Operation Properties On the set of Natural numbers, subtraction is not a
binary operation (N). The binary operations on each of the sets of Natural numbers (N), Integers (Z),
Rational numbers (Q), Real Numbers (R), and Complex numbers (C) are known as additions (C).
As Properties of Binary Operation Exponential operation (x, y) → xy is a binary operation on the set of
Natural numbers (N) and not on the set of Integers (Z).
II.
State the relation between the order of a group and the number of binary operations that can be
defined on that set.
Because the set has n items and binary operations (i.e., two operations) may be performed on each
of them in relation, the total number of possible combinations is n.
2n.2n−1...22.2=21+2+...n=22n(n−1) .
III.
How many binary operations can be defined on a set with 4 elements?
The formula is 2^{n(n-1)/2}2n(n−1)/2 where n is number of elements
Therefore binary operations can be defined on a set with 4 elements
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= 2^{4(4-1)/2} = 2^6 = 64=24(4−1)/2=26=64
iv. State the Lagrange’s theorem of group theory.
Lagrange's theorem is a statement in group theory that may be thought of as an extension of Euler's
theorem's number theoretical result. It's a crucial lemma in group theory for proving more complex
conclusions.
V. For a subgroup H of a group G, prove the Lagrange’s theorem.
If G is a finite group and H is a subgroup of G, then ∣H∣ divides ∣G∣.
As an immediate corollary, we get that if G is a finite group and g ∈G, then o(g) divides ∣G∣. In
particular, ∣G∣=e for all g ∈G.
ii.
Before proving Lagrange’s Theorem, state and prove three lemmas.
Lemma 1. If G is a group with subgroup H, then there is a one to one correspondence between H and any
coset of H.
Proof. Let C be a left coset of H in G. Then there is a g ∈ G such that C = g ∗ H. 1 Define f : H → C by
f(x) = g ∗ x.
f is one to one.
If x1 6= x2, then as G has cancellation, g ∗ x1 6= g ∗ x2. Hence, f(x1) 6= f(x2).
f is onto.
If y ∈ C, then since C = g ∗ H, there is an h ∈ H such that y = g ∗ h. It follows that f(h) = y and as y was
arbitrary, f is onto.
This completes the proof of Lemma 1.
Lemma 2. If G is a group with subgroup H, then the left coset relation, g1 ∼ g2 if and only if g1 ∗ H = g2
∗ H is an equivalence relation.
Proof. The essence of this proof is that ∼ is an equivalence relation because it is defined in terms of set
equality and equality for sets is an equivalence relation. The details are below.
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• ∼ is reflexive. Let g ∈ G be given. Then, g ∗H = {g ∗h : h ∈ H} and as this set is well defined, g ∗H =
g ∗H.
1We use “∗” to represent the binary operation in G.
• ∼ is symmetric.
Let g1, g2 ∈ G with g1 ∼ g2. Then by the definition of ∼, g1 ∗H = g2 ∗H. That is, {g1 ∗h : h ∈ H} =
{g2 ∗ h : h ∈ H} and as set equality is symmetric, {g2 ∗ h : h ∈ H} = {g1 ∗ h : h ∈ H}. Hence, g2 ∼ g1
and as g1 and g2 were arbitrary, ∼ is symmetric.
• ∼ is transitive.
Let g1, g2, g3 ∈ G with g1 ∼ g2 and g2 ∼ g3. Then, g1 ∗ H = {g1 ∗ h : h ∈ H} = {g2 ∗ h : h ∈ H} =
g2 ∗ H
And
g2 ∗ H = {g2 ∗ h : h ∈ H} = {g3 ∗ h : h ∈ H} = g3 ∗ H.
As set equality is transitive, it follows that
g1 ∗ H = {g1 ∗ h : h ∈ H} = {g3 ∗ h : h ∈ H} = g3 ∗ H, or g1 ∗ H = g3 ∗ H. That is, g1 ∼ g3, and as
g1, g2, g3 ∈ G are arbitrary, ∼ is transitive.
This complete the proof of the lemma.
Lemma 3. Let S be a set and ∼ be an equivalence relation on S. If A and B are two equivalence classes
with A ∩ B 6= ∅, then A = B.
Proof. To prove the lemma, we show that A ⊂ B and B ⊂ A. As A and B are arbitrarily labeled, it suffices
to show the former.
Let a ∈ A. As A ∩ B 6= ∅, there is a c ∈ A ∩ B. As A is an equivalence class of ∼ and both a and a c are
in A, it follows that a ∼ c. But as a ∼ c, c ∈ B and B is an equivalence class of ∼, it follows that a ∈ B.
Armed with these three lemmas we proceed to the main result.
Theorem 1. [Lagrange’s Theorem] If G is a finite group of order n and H is a subgroup of G of order k,
then k|n and n k is the number of distinct cosets of H in G.
Proof. Let ∼ be the left coset equivalence relation defined in Lemma 2. It follows from Lemma 2 that ∼ is
an equivalence relation and by Lemma 3 any two distinct cosets of ∼ are disjoint. Hence, we can write
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G = (g1 ∗ H) ∪ (g2 ∗ H) ∪ · · · ∪ (g` ∗ H)
Where the gi ∗ H, i = 1, 2, . . . , ` are the disjoint left cosets of H guaranteed by Lemma 3.
By Lemma 1, the cardinality of each of these cosets is the same as the order of H, and so
This completes the proof.
VI. Discuss whether a group H with order 6 can be a subgroup of a group with order 13 or not. Clearly
state the reasons.
Group H with order 6 can be a subgroup of a group with order 13 or not; however, as 6 does not divide 13,
a group with order 13 elements cannot have a subgroup of 6 members. Because only this number has
divisors of 13, it might have subgroups with 13 components.
Part 3
I. Validate whether the set is a group under the binary operation ‘*’defined asfor any two elements.
For Check whether the set is a group under the binary operation ‘*’defined as for any two elements ;
need to think about group axioms.
Assume that a * b = (-1) for this a = 1 and b = -1 or a = -1 and b = 1.
If a = 1 and b = -1;
LHS = -1 and RHS = -1
But .
There for the set is a group under the binary operation ‘*’defined as for any two elements.
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Part 4
Prepare a presentation for ten minutes to explore an application of group theory relevant to your course of
study. (i.e. in Computer Sciences)
Figure 22
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Figure 23
Figure 24
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Figure 25
Figure 26
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Figure 27
Figure 28
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Figure 29
Bibliography
binary trees . (2021). Retrieved from https://www.geeksforgeeks.org/binary-tree-data-structure/
Binary Trees quantitatively and qualitatively. (2021). Retrieved from https://www.javatpoint.com/discretemathematics-binary-trees
Dijkstra’s algorithm . (2021). Retrieved from https://www.geeksforgeeks.org/dijkstras-shortest-pathalgorithm-greedy-algo-7/
distinguishing
characteristics
of
different
binary
operations
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(2021).
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https://www.toppr.com/guides/maths/relations-and-functions/binary-operations/
proof
of
the
five
color
theorem
.
(2021).
Retrieved
from
http://cgm.cs.mcgill.ca/~athens/cs507/Projects/2003/MatthewWahab/5color.html#:~:text=5%2Dcolor
%20theorem%20%E2%80%93%20Every%20planar,that%20has%20the%20maximum%20degree.
real world binary problems in two different fields using applications of Boolean algebra. . (2021).
Retrieved
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https://oneclass.com/homework-help/mathematics/6552407-the-question-is-attached-
below.en.html
w3schools. (2019). w3schools. Retrieved from w3schools: https://www.w3schools.in/data-structurestutorial/binary-trees/
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www.cs.cmu.edu.
(2018).
www.cs.cmu.edu.
Retrieved
from
www.cs.cmu.edu:
https://www.cs.cmu.edu/~adamchik/15-121/lectures/Trees/trees.html
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Grading Rubric
Grading Criteria
Achieved
Feedback
LO1 : Examine set theory and functions applicable to software
engineering.
P1 Perform algebraic set operations in a formulated mathematical
problem.
P2 Determine the cardinality of a given bag (multiset).
M1 Determine the inverse of a function using appropriate
mathematical technique.
D1 Formulate corresponding proof principles to prove properties
about defined sets.
LO2 : Analyse mathematical structures of objects using graph
theory.
P3 Model contextualized problems using trees, both quantitatively and
qualitatively.
P4 Use Dijkstra’s algorithm to find a shortest path spanning tree in a
graph.
M2 Assess whether an Eulerian and Hamiltonian circuit exists in an
undirected graph.
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D2 Construct a proof of the Five colour theorem.
LO3 : Investigate solutions to problem situations using the
application of Boolean algebra.
P5 Diagram a binary problem in the application of Boolean Algebra.
P6 Produce a truth table and its corresponding Boolean equation from
an applicable scenario.
M3 Simplify a Boolean equation using algebraic methods.
D3 Design a complex system using logic gates.
LO4 : Explore applicable concepts within abstract algebra.
P7 Describe the distinguishing characteristics of different binary
operations that are performed on the same set.
P8 Determine the order of a group and the order of a subgroup in
given examples.
M4 Validate whether a given set with a binary operation is indeed a
group.
D4 Explore with the aide of a prepared presentation the application of
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group theory relevant to your course of study
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