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Hi! This is my third post about my International Diploma in Computing assignment. Now I going to post about my C1003 Mathematics for Computing module. I do this module in Term 3 2007 (August – December 2007).

So, lets begin..!!

SECTION A

A1

a) Perform the following base conversions, showing all workings.

i)        3345 Hexadecimal to Denary

3		3		4		5
X		X		X		X
163		162		161		160
1228	+	768	+	64	+	5	=	13125

3345_{16 =} 13125₁₀

ii)      5714 Octal to Hexadecimal

First step: Octal to Binary

5	7	1	4
▼	▼	▼	▼
101	111	001	100

5714₈ = 1011 1100 1100₂

Second step: Binary to Hexadecimal

1011	1100	1100
▼	▼	▼
B	C	C

1011 1100 1100₂ = BCC₁₆

5714₈ = BCC₁₆

iii)    5971 Denary to Binary

2	5971
2	2985				►	1
2	1492				►	1
2		746			►	0
2		373			►	0
2		186			►	1
2			93		►	0
2			46		►	1
2			23		►	0
2			11		►	1
2				5	►	1
2				2	►	1
				1	►	0

5971₁₀ = 1 0111 0101 0011₂

b) Evaluate F9D ₁₆ + 56A ₁₆, showing your working.

	1	1
	F	9	D
	5 +	6 +	A +
	21	16	23
	16 +	16 -	16 +
1	5	0	7

F9D₁₆ + 56A ₁₆ = 1507₁₆

A2

For each of the following statement, select one correct answer from those given.

a)      A(n) truncation error occurs when the least significant digits from a number are discarded.

i)                    overflow

ii) truncation

iii)                transposition

b)      Errors that exist by themselves in the measurement scale is known as inherent errors.

i) inherent

ii)                  induced

iii)                propagation

c)      Absolute error is the difference between the true value and the reported value.

i) absolute

ii)                  relative

iii)                transcription

d)     Error bound is the difference between the minimum and maximum possible reported value.

i)                    absolute error

ii)                  induced error

iii) error bound

A3 State which word from the list below would be most suitable in the place of each of the letters (a) to (f) in the text. (You may, if you wish, use a word more than once, but you do not have to use them all.)

Truncation	Iteration	Serial	overflow
Sequential	Unconditional	Offset	data transmission
Conditional	Direct	Base	data transfer
Arithmetic	Indexed	control transfer	indirect

(a)Direct addressing gives the address of the operand to be used, whereas (b)indirect addressing is where the instruction word gives the address of the address of the operand. Besides, the two addressing techniques, another kind of addressing techniques is known as (c)Indexed addressing. There are several types of machine instruction, which include (d)Arithmetic, string and (e)control transfer. In addition, (f)data transfer instructions move data and addresses between registers and memory locations or I/O ports.

A4 An 8 bit register stores numbers in binary form. Using the following bit patterns:

A = 11010110                         B = 10011010

Write your answer as bit patterns:

i)

X = A’.B

A	B	A’	A’. B
1	1	0	0
1	0	0	0
0	0	1	0
1	1	0	0
0	1	1	1
1	0	0	0
1	1	0	0
0	0	1	0

X = A’.B= 00001000

ii)

X = A’.B’

A	B	A’	B’	A’.B’
1	1	0	0	0
1	0	0	1	0
0	0	1	1	1
1	1	0	0	0
0	1	1	0	0
1	0	0	1	0
1	1	0	0	0
0	0	1	1	1

X = A’.B’ = 00100001

iii)

X = A + B’

A	B	B’	A + B’
1	1	0	1
1	0	1	1
0	0	1	1
1	1	0	1
0	1	0	0
1	0	1	1
1	1	0	1
0	0	1	1

X = A + B’ = 11110111

A5

Suppose that the universal set U is given by

U = {a, b, c, d, e, f, g} and that the sets P and Q are defined by

P = {a, c, e}, Q = {b, c, d, e}.

a)      Calculate :

i)        n(P Ç Q)

P = {a, c, e}, Q = {b, c, d, e}, U = {a, b, c, d, e, f, g}

P Ç Q = {c, e}

n(P Ç Q) = 2

ii)      n(P’ È Q)

P = {a, c, e}, Q = {b, c, d, e}, U = {a, b, c, d, e, f, g}

P’= {b, d, f, g}

P’ È Q = {b, c, d, e, f, g}

n(P’ È Q) = 6

b)      List the elements of the following sets :

i)        (P È Q)’

P = {a, c, e}, Q = {b, c, d, e}, U = {a, b, c, d, e, f, g}

P È Q = {a, b, c, d, e}

(P È Q)’ = {f, g}

ii)      P’ Ç Q

P = {a, c, e}, Q = {b, c, d, e}, U = {a, b, c, d, e, f, g}

P’= {b, d, f, g}

P’ Ç Q = {b, d}

A6

Suppose that numbers are to be held in an 8-bit computer using two’s complement form.

a) Show how -56 would be evaluated by the computer.

i) Convert -56 to binary

-56 = – 0111000

ii) Insert it to 8-bit register

00111000

▼

11000111

00000001₊

11001000

So, -56 will be stored as 11001000

b) Convert 10011010 to denary from stored using the above format.

10011010

00000001_-

10011001

▼

01100110

0		1		1		0		0		1		1		0
X		X		X		X		X		X		X		X
27		26		25		24		23		22		21		20
0	+	64	+	32	+	0	+	0	+	4	+	2	+	0	=	102

So, the original number is -102₁₀

SECTION B

Answer ALL questions in this section.

B1

a) A binary tree is represented in a computer by the following table:

Node	Content	Left Pointer	Right Pointer
1	Heather	2	3
2	Coriander	5	-1
3	Olive	4	6
4	Magnolia	-1	-1
5	Clytia	7	-1
6	Rosette	-1	8
7	Blossom	-1	-1
8	Suzette	-1	-1

i) Draw a binary tree which this table represents.

ii) Construct a table to show the back and trace pointer, to represent the binary tree below.

Node	Content	Back Pointer	Trace Pointer
1	Guava	-1	6
2	Blue berry	1	4
3	Watermelon	1	-1
4	Citrus	2	1
5	Avocado	2	2
6	Pear	3	7
7	Rumbutan	6	3

b) From the tree in part(a)(ii) above, state the order in which the nodes would be visited if they were traversed in :

i)        Preorder

Guava(1), Blue berry(2), Avocado(5), Citrus(4), Watermelon(3), Pear(6), Rumbutan(7)

ii)      Inorder

Avocado(5), Blue berry(2), Citrus(4), Guava(1), Pear(6), Rumbutan(7), Watermelon(3)

iii)    Postorder

Avocado(5), Citrus(4), Blue berry(2), Rumbutan(7), Pear(6), Watermelon(3), Guava(1)

c) Briefly explain the difference between a stack and a queue.

The basic difference between stack and queue is:

Ø  Stack : Use LIFO (Last In First Out). So, the first item enter the stack will be the last item to exit the stack.

Ø  Queue : Use FIFO (First In First Out). So, the first item enter the queue will be the first item to exit the queue.

d)  Explain the term binary tree.

Binary tree (T) is a finite set of elements called nodes. It contains a distinguished node (R ) called as the root of T. The other nodes will form the left and the right subtrees of the root.

e) Suppose that a computer stores numbers in an 8-bit register using two’s complement form. How would the computer evaluate 33 – 45? Give your answer in binary form.

i) Convert each 33 and – 45 to binary, and then store it in 8-bit register using two’s complement form

*) 33:

2	33
2	16		►	1
2	8		►	0
2		4	►	0
2		2	►	0
2		1	►	0

33 = 100001

Stored in 8-bit register = 00100001

*)- 45:

2	45
2	22		►	1
2	11		►	0
2		5	►	1
2		2	►	1
2		1	►	0

- 45 = – 101101

Stored in 8-bit register =    –   00101101

▼

11010010

00000001₊

11010011

ii) Add together

00100001

11010011₊

11110100

So, 33₁₀ – 45₁₀ will be evaluated as 11110100₂

B2

a) Lucy likes singing. She sings Chinese(A), English(B) and Japanese(C) songs. Write Boolean expressions to represent the following statements:

i)        She sings Chinese and English songs, or she does not sing Chinese and Japanese songs.

X = A.B + (A.B)’

ii)      She sings English or Japanese songs, but not Chinese songs.

X = (B+C).A’

iii)    It is not true that she sings Chinese but not Japanese songs.

X = (A.C’)’

b) Suppose that X = (A.(A.B)’)’ and Y =(B.(A.B)’)’.

i)        Produce two separate truth tables to represent X and Y respectively.

• for X:

A	B	A.B	(A.B)’	A.(A.B)’	X = (A.(A.B)’)’
0	0	0	1	0	1
0	1	0	1	0	1
1	0	0	1	1	0
1	1	1	0	0	1

• for Y:

A	B	A.B	(A.B)’	B.((A.B)’)	Y =(B.(A.B)’)’
0	0	0	1	0	1
0	1	0	1	1	0
1	0	0	1	0	1
1	1	1	0	0	1

ii)      Draw a logic circuit diagram to represent (X.Y)’ with inputs A and B.

c) Simplify the following expression

A’.B.C + A.B’.C’ + A.B.C’ + A’.B.C’ + A’.B’.C’

= A’.B.C + (A. C’).(B’+ B) + (A’.C’).(B’+ B)

= A’.B.C + (A. C’).( 1 ) + (A’.C’).( 1 )

= A’.B.C + C’.(A + A’)

= A’.B.C + C’

d) John has 5 red discs and 4 white discs in a bag. He takes 2 discs at random from the bag one after the other, and are not replaced. Expressing your answer as fractions, find the probability that

i)        the first disc is red and the second is white.

5/9 * 4/8 = 20/72 = 5/18

ii)      the two discs have different colours.

5/9 * 4/8 = 20/72 = 5/18

Or

4/9 * 5/8 = 20/72 = 5/18 So, 5/18 + 5/18 = 10/18 = 5/9

iii)    the second disc chosen is white.

5/9 * 4/8 = 20/72 = 5/18 or  4/9 * 3/8 = 12/72 = 3/18

So, 5/18 + 3/18 = 8/18 = 4/9

Do you have any comment on this? maybe you have something to ask me or want to discuss with me about this?Or maybe you are IDIC student and taking this module also?
Feel free to use the comment box below (click here if you cannot see the comment box).

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