BIRKBECK
(University of London)
BSc/FD Examination for Internal Students
Department of Computer Science and Information Systems
Introduction to Computer Systems
SUMMARY ANSWERS
(BUCI008H4 - value: 15 credits)
Date and time of Examination: Wednesday 5th June 2013 at 10.00 am
Duration of Paper: 2 hours
Calculators are not allowed
Answer ALL TEN questions. Each question is worth 10 marks.
The paper is divided into two sections: Section A (40 marks) and Section
B (60 marks). Answer each section in a different Answer Book. Do not
include in the same Answer Book answers to questions in two or more
sections.
Where appropriate, include a justification or an explanation for each
answer.
This paper is not prior disclosed.
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Section A: Boolean Operations and Number Representations (40
marks)
1. Show your working in all parts of this question.
a) Add the two binary numbers 1001 and 1101. Show the result in binary
form.
(2 marks)
Answer: 10110. One mark for the correct answer with no working.
b) Multiply the two binary numbers 1101 and 11. Show the result in
binary form.
(2 marks)
Answer: 100111. One mark for the correct answer with no working.
c) Convert the following two binary numbers into hexadecimal numbers:
11101010 and 111010.
(2 marks)
Answer: EA and 3A. One mark for the correct answers with no working.
d) Express the following two decimal numbers in three bit excess
notation: 3 and -2.
(2 marks)
Answer: 111 and 010. One mark for the correct answers with no working.
e) Add the following two hexadecimal numbers: 7 and 8. Show the result
in hexadecimal form.
(2 marks)
Answer: F. One mark for the correct answer with no working.
2. (a) A number is a binary fraction if it can be written in the form a/b
such that a is an integer and b is a power of 2. For example 5 is a binary
fraction for which a = 5 and b = 1 = 2^(0). Which of the following
numbers are binary fractions and which are not binary fractions?
i) Decimal 37
ii) Binary 10.1
iii) Decimal 7/3
iv) Binary 0.11
(2 marks)
Answer: yes, yes, no, yes. ½ mark for each correct answer.
b) The Brookshear representation for a binary fraction x consists of 8 bits,
labelled s, e1, e2, e3, m1, m2, m3, m4 from left to right. If x is zero then
all 8 bits are zero. Next, suppose that x is not zero. The bit s is 1 if x is
strictly negative and 0 if x is strictly positive. To obtain the remaining
bits, x is written in the form
2^(r) * 0.t
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where r is an integer and t is a bit string with the left most bit equal to 1.
Then e1, e2, e3 specify the three bit excess notation for r and m1, m2,
m3, m4 are the left most four bits of t.
i) Obtain the Brookshear floating point representation for the decimal
integer 7. Show your working.
(3 marks)
Answer: 01111110. Two marks for the correct answer with no
working.
ii) Obtain the decimal number that has the Brookshear floating point
representation 01101010. Show your working.
(3 marks)
Answer: 2+(1/2). Two marks for the correct answer with no working.
c) Find the largest positive decimal number that can be expressed exactly
in Brookshear floating point notation, i.e. without any truncation error.
Explain how you obtained the answer.
(2 marks)
Answer: 7+(1/2). One mark for the correct answer with no explanation.
3. a) Consider the following four truth tables in which 0 stands for False
and 1 stands for True.
A
0
0
1
1
B
0
1
0
1
T1(A,B)
0
0
0
1
A
0
0
1
1
B
0
1
0
1
T2(A,B)
0
1
1
0
A
0
0
1
1
B
0
1
0
1
T3(A,B)
1
1
0
0
A
0
0
1
1
B
0
1
0
1
T4(A,B)
1
0
0
0
The four truth tables are for the Boolean operations NOT(A), A XOR B
(XOR is exclusive or), A AND B and NOT(A OR B). State which table
T1, T2, T3, T4 corresponds to which operation.
(4 marks)
Answer: A AND B, A XOR B, NOT(A), NOT(A OR B). One mark for each
correct item.
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b) Consider the following truth table for the Boolean expression
A IMPLIES B.
A
0
0
1
1
B
0
1
0
1
A IMPLIES B
1
1
0
1
Use the Boolean operations NOT, AND and the Boolean variables A, B
to construct a Boolean expression which has the same truth table as A
IMPLIES B.
Hint: look at the third row of the truth table for A IMPLIES B, and note
that it is the only row for which the value of A IMPLIES B is 0.
(2 marks)
Answer: NOT(A AND NOT B). The expression A AND NOT(B) has the
value 1 if and only if A = 1 and B = 0. It follows that NOT(A AND
NOT(B)) has the value 0 if and only if and only if A = 1 and B = 0. Thus
it matches the truth table for A IMPLIES B. One mark for the correct
answer with no explanation.
c) Evaluate the following two Boolean expressions, given that A = 0
(False), B = 1 (True) and C = 1 (True).
i) (A AND B) OR (B AND C)
(2 marks)
Answer: 1 (True). One mark for the correct answer with no working.
ii) (NOT(A) OR NOT(C)) AND B
(2 marks)
Answer: 1 (True). One mark for the correct answer with no working.
In both answers, show your working.
4. a) Evaluate the following expressions, under the assumption that x = 4
and y = 7. Note that = = is a test for equality.
i) x > y
ii) (x < y) AND (x > 0)
iii) x = = y
iv) NOT(x = = 0)
Answer: False, True, False, True. One mark each.
(4 marks)
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b) Let A1, A2 be binary bits. The sum of A1 and A2 is expressed as a
single bit B and a carry bit C. The table for B is as follows.
A1
0
0
1
1
A2
0
1
0
1
B
0
1
1
0
Write out the table for the carry bit C. Show how the truth tables for B
and C can be used to find the result of adding the binary numbers 11 and
1.
(6 marks)
Answer:
A1 A2
C
0 0
0
0 1
0
1 0
0
1 1
1
11+1=100. To obtain the right most bit use the table for B with A1=1 and
A2=1, to obtain right most bit = 0. To obtain the right most carry use the
table for C with A1=1 and A2=1, to obtain 1. Add the right most carry
and the left most bit of 11 in the same way to obtain 0 carry 1. The 0 is
the next bit of the answer and this second carry is then the left most bit of
the answer. Two marks for the table and 4 marks for the explanation.
Section B: Programming and Structure of a Computer (60 marks)
5. a) Convert the following two decimal integers to binary integers: 11,
42.
(2 marks)
Answers: 1011, 101010. One mark for the correct answers with no
working.
b) Convert the following two binary integers to decimal integers: 11,
1101.
(2 marks)
Answers: 3, 13. One mark for the correct answer with no working.
c) Consider the following pseudo code algorithm for finding the binary
number corresponding to a non-negative decimal integer m. Comments
are in the form /* comment */. Prepend places a bit at the beginning of a
bit string, for example Prepend[0, {1, 1}] has the value {0, 1, 1}.
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Input: non-negative integer m
Output: bit string b that specifies the binary representation of m.
1. If[m = = 0, Output {0}; Halt]
2. b = {};
/* empty bit string*/
3. While (m > 0)
4.
Divide m by 2 to give quotient q and remainder r;
5.
m = q;
6.
b = Prepend[r, b];
/* r followed by b */
7. EndWhile;
8. Output b;
9. Halt;
Describe in detail the action of the pseudo code algorithm for m = 3.
(4 marks)
Answer: Test to see if m equals 0. The test returns False. Initialise b with
the empty list {}. Test to see if m is strictly greater than 0. The test returns
True. Enter the while loop. Divide m by 2 to obtain quotient q = 1 and
remainder r = 1. Prepend r to b to obtain b = {1}. Assign the value 1 of q
to m. Test to see if m is strictly greater than 0. The test returns True.
Enter the while loop. Divide m by 2 to obtain quotient q = 0 and
remainder r = 1. Prepend r to b to obtain b = {1, 1}. Assign the value 0
of q to m. Test to see if m is strictly greater than 0. The Test returns
False. Exit the while loop, output b and halt.
d) Describe how the algorithm in part (c) of this question can be modified
to produce a new algorithm that finds the hexadecimal number
corresponding to a non-negative decimal integer m. Explicit calculations
with hexadecimal digits are not required.
(2 marks)
Answer: replace the lines within the while loop with
4. Divide m by 16 to give quotient q and remainder r
5. Convert r to the appropriate hexadecimal integer h
6. m = q;
7. b = Prepend[r, b];
6. The following table describes instructions of length 16 bits, made by
concatenating an op-code and an operand. The first 4 bits record the opcode. The remaining 12 bits record the operand. Four bits are required to
specify a register R and 8 bits are required to specify a memory location
XY. Each register holds 8 bits and each memory location holds 8 bits.
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Op- Operand Description
code
1
RXY
Load register R with the bit pattern at memory cell
XY.
2
RXY
Load register R with the bit pattern XY.
3
RXY
Store the bit pattern in register R at memory cell XY.
4
0RS
Move the bit pattern in register R to register S.
5
RST
Add (two’s complement) the bit patterns in registers R
and S. Put the result in register T.
6
RST
Add (floating point) the bit patterns in registers R and
S. Put the result in register T.
7
RST
Or the bit patterns in registers S and T. Put the result in
register R.
8
RST
And the bit patterns in registers S and T. Put the result
in register R.
9
RST
Exclusive Or the bit patterns in registers S and T. Put
the result in register R.
A
R0X
Rotate the bit pattern in register R one bit to the right
X times.
B
RXY
Jump to the instruction in memory cell XY if the bit
pattern in register R is equal to the bit pattern in
register 0.
C
000
Halt.
Each 16 bit instruction is coded by four hexadecimal digits. For example,
the four hexadecimal digits 37A9 specify an instruction with op-code 3,
in which the “7” refers to register 7 and “A9” refers to the memory cell
A9. The registers are numbered in hexadecimal from 0 to F.
All memory addresses in this question are given in hexadecimal notation.
a) Memory cells 90 and 91 each hold an integer in 8 bits two’s
complement notation. The following program halts with 1 in register 1 if
the two integers are equal and it halts with 0 in register 1 otherwise.
25C0
3592
2500
3593
2101
1290
1091
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B292
2100
C000
Use the table at the beginning of this question to describe the action
carried out by each instruction. For example, 25C0 loads the bit pattern
C0 in register 5 and 3592 stores the contents of register 5 in memory cell
92.
(4 marks)
Answer
25C0
3592
2500 load the bit pattern 00 in register 5
3593 store the contents of register 5 at memory cell 93
2101 load the bit pattern 01 in register 1
1290 load the contents of memory cell 90 in register 2
1091 load the contents of memory cell 91 in register 0
B292 Jump to the instruction stored at memory cells 92 and 93 if register
2 and register 0 have the same contents.
2100 load register 1 with bit pattern 00
C000 halt
½ mark for each correct line.
b) Describe in words the action of the program shown in part (a) of this
question. A high level description is required, suitable for a reader who
knows about registers, memory and instructions but who finds the list of
instructions in part (a) of this question un-illuminating.
(6 marks)
Answer: The instruction C000 for Halt is stored in memory cells 92 and
93. The bit pattern 01 is loaded into register 1. The two numbers to be
compared are loaded into registers 2 and 0 respectively. The contents of
registers 2 and 0 are compared. If the contents are equal then the value
in register 1 is correct and the next instruction to be carried out is C000
(Halt). If the contents are not equal, then 00 is loaded into register 1 and
the program halts.
7. a) Every integer strictly greater than 0 can be written as a sum of
distinct powers of 2. For example, the decimal integers 1,4 and 7 can be
expressed as sums of powers of 2 in the following ways:
i) 1 = 2^(0)
ii) 4 = 2^(2)
iii) 7 = 2^(2)+2^(1)+2^(0)
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Explain how this property ensures that every integer strictly greater than
0 can be expressed in binary form. What is the connection between the
binary digits for an integer and the expression of that integer as a sum of
powers of 2. Illustrate your answer using the number 7.
(4 marks)
Answer: A bit string specifies a set of powers of 2 in the following way. A
1 in the nth place, reading from the left, corresponds to the power 2^(n1). A 0 in the nth place indicates that the power 2^(n-1) is absent. The
sum of the powers of 2 associated with a bit string yields the integer
represented by the bit string. In the case of the number 7 the bit string is
111, which specifies 2^(0), 2^(1) and 2^(2), on reading the bit string from
right to left. The sum of these powers of 2 is equal to 7 thus 7 is
represented by 111. Four marks for any reasonable answer. Two marks
for a good illustration of the answer using the integer 7.
b) Let n be an integer strictly greater than 0. It is required to find an
integer k such that
2^(k) <= n and 2^(k+1) > n.
For example, if n = 5, then k = 2 because 4 <= 5 and 8 > 5. An algorithm
to find k is specified as follows: begin with k = 0. Test to see if 2^(k) <=
n. If the test returns True then increase k by 1 and repeat the test. If the
test returns False then output k-1 and halt. Write out a pseudo code
version of this algorithm that includes a while loop.
(4 marks)
Answer
1. Input n;
2. k = 0;
3. While (2^(k) <= n)
4. k = k+1;
5. EndWhile;
6. Output k-1;
Four marks for any reasonable version. No marks if there is no while
loop.
c) Explain how the algorithm in part (b) of this question can be used to
find the length of the binary representation of a strictly positive integer.
The length is by definition the number of bits in the representation,
assuming that the left most bit is 1. For example the length of 1101 is 4.
(2 marks)
Answer: k is the greatest integer such that 2^(k)<=n and the least integer
such that 2^(k+1)>=n. Suppose that n has a binary representation with r
bits. It follows that 2^(r) > n, thus r >= k+1. The smallest number with r
bits is 2^(r-1) thus 2^(r-1) <= n and hence r-1<=k. It follows that r =
k+1. One mark for the correct answer with no supporting argument.
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8. (a) Define a pointer.
(2 marks)
Answer: a pointer is a storage area containing the address at which a
piece of information is stored.
(b) It is required to store a list of items in memory. The items are of
varying size, and it may be necessary to add or remove items at a later
date. Give one reason why it is inefficient to insist that the items be stored
side by side in memory.
(2 marks)
Answer: if items are to be stored side by side by side then the gap left by
any deleted item must be filled by moving the remaining items in memory.
This could take a lot of time, especially if the items are large.
(c) Suppose that it is no longer required to store the items in the list
referred to in part (b) of this question side by side in memory. Explain
how the items can be stored in memory as a linked list. Include in your
explanation an example with a list consisting of three items. Give explicit
values for the pointers and the addresses of the items in your example.
(4 marks)
Answer: a pointer is associated with each item. The items are stored at
any convenient locations in memory. The pointer associated with a given
item is initialised with the address of the next item in the list. The pointer
associated with the last item has a special null value. The location of the
first item is indicated by a head pointer associated with the list as a
whole. Two marks for the description, two marks for a reasonable
diagram.
(d) Describe the process of deleting the first item from your example of a
linked list in part (c) of this question.
(2 marks)
Answer: The value of the head pointer is set equal to the position of the
second item.
9. (a) Draw a labelled diagram showing the CPU and the main memory of
a computer. The diagram should show the high level structure of the CPU
and the link between the CPU and the main memory.
(4 marks)
Answer: the diagram should show the control unit, the arithmetic and
logic unit, the registers, the programme counter, the instruction register,
the bus and the main memory, all labelled correctly.
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b) Describe the machine cycle, i.e. the sequence of three actions carried
out repeatedly by the CPU.
(6 marks)
Answer: the three actions are as follows. Fetch: load the instruction
register with an instruction from the main memory using the address in
the program counter. Update the program counter with the next memory
location. Decode: find the op code of the instruction, and the actions
specified by the instruction. Execute: carry out he instruction. Two marks
for each action.
10. Consider the following problem: a man wishes to ferry a wolf, a
sheep and a bale of hay across a river. His boat will only hold one item at
a time. The wolf cannot be left with the sheep and the sheep cannot be
left with the hay. How can he ferry the three items?
The state of the man is represented by a bit. The bit is 0 if the man is on
the near side (original side) of the river and 1 if the man is on the far side
(final side) of the river. The states of the wolf, the sheep and the hay are
represented in the same way. The four bits are concatenated. For
example, if the state is 0011 then the man and the wolf are on the near
side of the river and the sheep and the hay are on the far side of the river.
a) What is the original or first state? What is the desired final state?
(2 marks)
Answers: 0000, 1111. One mark each.
b) Suppose that the current state is 0001. Describe in words the locations
of the participants.
(2 marks)
Answers: the man, the wolf and the sheep are on the near side of the
river. The hay is on the far side of the river. ½ mark for each correct
location.
c) List all the states, both allowed and forbidden, that can follow from
0001 after one crossing of the river. Explain your answer.
(4 marks)
Answers: 1001, 1101, 1011. One mark each and one for the explanation.
d) List all the allowed states that can follow from 0001 after one crossing
of the river.
(2 marks)
Answers: 1101, 1011. One mark each.
(End of the question paper)
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