Practice Exam Name Personal Number Question 1: We had discussed magnetic tape data storage in previous lecture. In
mainframe computers, IBM tape drives product number 3592, was introduced
under the nickname Jaguar. The next drive was the TS1120, also having the
nickname Jaguar. As of June, 2011, the latest and current drive is the TS1140.
1) Above figure shows TS1140 drive, cartridge, tapes, and format for one track.
Now, we know there is a tape has been formatted for TS1140, having 500,000 bits
per inch on one track, 2560 tracks, 880 meters long, please calculate the capacity
for the tape in terms of byte( 1 inch = 2.54 centimeter) (5pts)
Capacity = density * length * num_tracks
=
=
(500, 000 bits/inch. track) * ( (880 * 100 / 2.54 ) inch) * (2560 tracks )
44,346,456,692,913.385
bits
The capacity is about 44TB, so in terms of byte, it is about 5.5 TB.
2) Please first briefly explain similarity between tape drives and hard drives, and
based on following data: Seagate Barracuda ST380021A, 8 platters, 16 headers,
number of cylinder is 158816, 63 sectors, Block size is 512.
calculate capacity (5pts)
Similar to the previous question Capacity = density * length * num_tracks = ((63 sectors*(512bytes/sector))/2pi * (2pi radians) ( 1between
58816 chard
ylinders/header 16 header) 2) Please first 2)briefly
Please
explain
firstradians) similarity
2)briefly
Please
explain
between
first
briefly
similarity
tape
explain
drives
between
similarity
and* tape
hard
drives
drives,
and
and
tape
drives
drives,
andand
hard drives,*and
2) Please first briefly explain
between
tape
drives
andBarracuda
hard
and
basedsimilarity
on following
based
data:
on following
Seagate
based
Barracuda
data:
on following
Seagate
ST380021A,
data:
Seagate
8drives,
platters,
ST380021A,
Barracuda
16 headers,
8ST380021A,
platters, 16 8headers,
platters, 16 headers,
= 81,964,302,336 bytes approximately 80GB based on following data: Seagate
8 platters,
16
number ofBarracuda
cylinder
number
is 158816,
ofST380021A,
cylinder
number
63 sectors,
is 158816,
of cylinder
Block
63size
sectors,
is 158816,
is 512.
Block
63headers,
sectors,
size is 512.
Block size is 512.
2)
Please
first
2)
briefly
Please
explain
first
similarity
briefly between
explaintape
similarity
drives and
between
hard drives,
tapeanddrives and hard drives, and
number
of cylinder is 158816, 63 sectors, Block size is 512.
based(5pts)
on following
baseddata:
on following
Seagate Barracuda
data: Seagate
ST380021A,Barracuda
8 platters, 16
ST380021A,
headers, 8 platters, 16 headers,
calculate capacity
calculate
(5pts) capacity
calculate
(5pts)capacity
calculate
capacity
(5pts)
of cylinder
number
is 158816,
of cylinder
63 sectors,
is 158816,
Block size63and
issectors,
512. each
Block size is 512.
3) IBM S54 frame support up to 1,320 number
TS1140
LTO
cartridges,
!
!
!
!
!
!
!
!
!
!
!
!
!
cartridge has 4TB. Please use Egyptian Hieroglyphics system to calculate S54
calculate capacity
calculate
(5pts) capacity (5pts)
frame’s capacity
! in terms
! of TB.
! (5pts)
!
!
!
!
!
!
3) IBM S54 frame
3) IBM
support
S54 up
frame
3)to 1,320
IBM
support
S54
TS1140
frame
up toLTO
1,320
support
cartridges,
TS1140
up to 1,320
LTO
and each
cartridges,
TS1140 LTOandcartridges,
each and each
3) IBM S54 frame support up to 1,320 TS1140 LTO cartridges, and each
cartridge has 4TB.
cartridge
Pleasehas
use4TB.
cartridge
Egyptian
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Hieroglyphics
use4TB.
Pleasesystem
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use
to calculate
Hieroglyphics
system
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systemS54
to calculate S54
! Egyptian
! Egyptian
has 4TB.similarity
Please use
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Hieroglyphics
to calculate
explain
Please firstcartridge
briefly
between
drives andsystem
hard drives,
and S54
frame’s
capacity
frame’s
in
terms
capacity
of
TB.
frame’s
in
(5pts)
terms
capacity
of
TB.
in
(5pts)
terms
of
TB.
(5pts)
sed on following
data:
Seagate
Barracuda
ST380021A, 8 platters, 16 headers,
frame’s
capacity
in terms
of TB. (5pts)
3) IBM S54 frame
3) IBM
support
S54upframe
to 1,320support
TS1140 up
LTOtocartridges,
1,320 TS1140
and eachLTO cartridges, and each
mber of cylinder is 158816, 63 sectors, Block size is 512.
cartridge hascartridge
4TB. Pleasehas
use 4TB.
Egyptian
Please
Hieroglyphics
use Egyptian
system toHieroglyphics
calculate S54 system to calculate S54
culate capacity (5pts)
frame’s capacity
frame’s
in terms
capacity
of TB. (5pts)
in terms of TB. (5pts)
!
!
!
!
!
!
!
!
!
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!
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!
!
!
!
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!
!
!
IBM S54 frame support up to 1,320 TS1140 LTO cartridges, and each
!
!
tridge has 4TB.
Please use Egyptian
system
to calculate S54
! Hieroglyphics
!
!
!
me’s capacity in terms of TB. (5pts)
!
1320 = 4 = !
!
!
!
!
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!
!
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2
333nine = 3ten ! 9 ten
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So we can represent the -273 number using base-2 using two’s complement
number:
invert(0100010001two ) +1two = 1011101110 two +1two = 1011101111two Question 2: Please read following C program and Perl regular expression: Character
Description
.
a single character
\s
a whitespace character (space, tab, newline)
\S
non-whitespace character
\d
a digit (0-9)
\D
a non-digit
\w
a word character (a-z, A-Z, 0-9, _)
\W
a non-word character
[aeiou]
matches a single character in the given set [^aeiou]
matches a single character outside the given set (foo|bar|baz)
matches any of the alternatives specified
#include<stdio.h>
int Fibonacci(int);
main()
{
int n, i = 0, c;
scanf("%d",&n);
printf("Fibonacci series\n");
for ( c = 1 ; c <= n ; c++ )
{
printf("%d\n", Fibonacci(i));
i++;
}
return 0;
}
int Fibonacci(int n)
{
if ( n == 0 )
return 0;
else if ( n == 1 )
return 1;
else
return ( Fibonacci(n-1) + Fibonacci(n-2) );
} 1) Following Perl program is a lexer, which group tokens and assign the group’s
name. Please fill out the program to print out two groups: firstly, all type
declarations, such as “int:=type”; secondly, all so called ‘if else’ clause, e.g.,
“if:=if clause” (5pts)
# Read the file one line at a time.
while(my $line = <FILE>) {
@tokens = split(/\s+/, $line );
foreach
{
my $token (
@tokens )
# print $token;
if ($token eq "int" || $token =~ m/void/ )
{
print $token;
print ":=type\n";
}
if ($token =~ m/else\s+if/ )
{
print $token;
print ":=else if clause\n";
}
elsif ($token =~ m/if/)
{
print $token;
print ":=if clause\n";
}
elsif ($token =~ m/else/)
{
print $token;
print ":=else clause\n";
}
}
}
2) Please construct a Definite Finite Automata(DFA) according to above
question. (7pts)
Any char other
Any char other
than t
than n
any char other
Any char other
than \s
n
than i, v, e
i
t
f
Epslon
d
v
o
i
e
e
\s +
i
\s
Any char other
than l
l
Any char
other than s
Any char other
than e
s
Epslon
Epslon
3) Answer the following Perl Regular Expression questions (9pts)
(a) does following program print yes?
my $mystring;
$mystring = “Hello world!”;
if ($mystring =~ m/World/) {print “Yes”}
No, Perl is Case Sensitive
Any char
other than l
Any char
other than \s
(b) does following program print “The first digit is $1” ?
$mystring = “[2013/06/25] The date of this exam paper.”;
if
($mystring = ~ m/(\d)/) {
print “The first digit is $1.”
}
Prints "The first digit is 2." In order to designate a pattern for extraction,
one places parenthesis around the pattern. If the pattern is matched, it is
returned in the Perl special variable called $1. If there are multiple
parenthesized expressions, then they will be in variables $1, $2, $3, etc.
(c ) What is the output of following program?
$mystring = “The start text always precedes the end of the end text.”
If ($mystring =~ m/start(.*)end/) {
print $1;
}
Prints “text always precedes the end of the”. The pattern .* is two different
metacharacters that tell Perl to match everything between the start and end. Specifically,
the metacharacter . means match any symbol except new line. The pattern
quantifier * means match zero or more of the preceding symbol. By default, the
quantifiers are greedy. This means that when you say .*, Perl matches every character
(except new line) all the way to the end of the string, and then works backward until it
finds end. To make the pattern quantifier miserly, you use the pattern quantifier limiter ?.
This tells Perl to match as few as possible of the preceding symbol before continuing to
the next part of the pattern.
Question 3: Please refer to the C grammar program in question2 (Fibonacci.c) .
1) Please parse following statement using parse tree(Notice the leave element
could be expression) (6pts)
if ( n == 0 )
return 0;
else if ( n == 1 )
return 1;
else
return ( Fibonacci(n-1) + Fibonacci(n-2) );
Selection-­‐Statement if expression n == 0
Statement else Statement return 0
Selection-­‐statement if !
expression Statement else statement n == 1
return 1
jump-­‐statement Notice: we will not expand all return
statement further
Freturn Fibonacci(n-1) + Fibonacci(n-2)
2) Please show how the tree is parsed using right-derivation using Push-Down
Automata (Stack)
Selection-­‐stmt stmt Pop it, Pop it, Push new rule Push new rule Pop it, Push new rule stmt Pop it, Jump-­‐stmt Push new rule Notice: we will not expand jump-statement further
Selection-­‐stmt 3) Usually, in order to generate assembly code, the grammar-tree or parse-tree
need to be gone through several times. Briefly explain how MIPS assembly code
is generated for the question3 part 1, and write the code below.
If_label1: J finish Else_label1: If_label2: J finish Else_label2: finish: Question 4:
1) Briefly explain lazy evaluation and eager evaluation in lambda-calculus
In programming language theory, lazy evaluation or call-by-need is an evaluation
strategy which delays the evaluation of an expression until its value is needed
(non-strict evaluation) and which also avoids repeated evaluations (sharing).
The sharing can reduce the running time of certain functions by an exponential
factor over other non-strict evaluation strategies, such as call-by-name.
In eager evaluation, an expression is evaluated as soon as it is bound to
a variable
culus (20 points):
Solution:
olution:
(λa b.b) Use
((λz.z
z)((λf.f)(λg.g)))
→β evaluation order to reduce the follo
(a) (5 points)
β-reduction
with am lazy
β
a b.b) ((λz.z z)((λf.f)(λg.g)))
→β
(λb.b)
mmevaluation
→
fully
possible:
) Use β-reduction
with
a as
lazy
order to reduce the following term as
β
b.b) m →
m
ossible:
2) Use lazy evaluation order to reduce the
term as
fully as possible:
(λafollowing
b.b) ((λz.z
z)((λf.f)(λg.g)))
m
(λa b.b) ((λz.z z)((λf.f)(λg.g))) m
Solution:
n:
(λa b.b) ((λz.z z)((λf.f)(λg.g))) m →β
β β
(λb.b)m m
((λz.z z)((λf.f)(λg.g)))
→→
→β
m
(b) (5 points) Use β-reduction with an eager evaluation order to reduce the followi
as fullyanaseager
possible:
points) Use 3)
β-reduction
evaluation
to reduce
theasfollowing
term
Use eagerwith
evaluation order
to reduceorder
the following
term
fully as possible:
fully as possible:
(λf.f g h) (((λa.a a) (λb.b)) (λc d.d))
(λf.f g h) (((λa.a a) (λb.b)) (λc d.d))
Solution:
(λf.f g h) (((λa.a a) (λb.b)) (λc d.d)) →β
olution:
(λf.f g h) (((λb.b)(λb.b)) (λc d.d)) →β
f.f g h) (((λa.a a) (λb.b)) (λc d.d)) →β
(λf.f g h) ((λb.b)
(λc d.d)) →β
β
f.f g h) (((λb.b)(λb.b))
(λc d.d))Use
→ β-reduction
(b) (5 points)
with an eager evaluation order to reduce the fo
(λf.fβ g h)
(λc d.d) →β
)f.f
Use
β-reduction
an eager
evaluation order to reduce the following term
g h)
((λb.b) (λcwith
d.d))
as fully
as possible:
(λc→d.d)
g h →β
β
sf.f
possible:
g h) (λc d.d) →
(λd.d) h →β
β
(λf.f g h) (((λa.a a) (λb.b)) (λc d.d))
c d.d) g h →
h
(λf.f g h) (((λa.a a) (λb.b)) (λc d.d))
d.d) h →β
4) calculate following formula:
Solution:
(let ((x 2) (y 3))
n:
(λf.f
a) (λb.b)) (λc d.d)) →β
(let
((x 7)g h) (((λa.a
β
) (((λa.a a) (λb.b))(λf.f
(λc d.d))
→
(z
y)))
g (+
h)xβ(((λb.b)(λb.b))
(λc d.d)) →β
) (((λb.b)(λb.b)) (λc
d.d))g h)
→ ((λb.b) (λc d.d)) →β
(* z x)))
(λf.f
β
) ((λb.b) (λc d.d)) (λf.f
→ g h) (λc d.d) →β
β
) (λc d.d) →
(λc35d.d) g h →β
β
gh →
(λd.d) h →β
β
→
h
University of Illinois at Urbana-Champaign
Illinois at Urbana-Champaign
Netid:
Netid:
Question 5:
Please refer to the C grammar program in question2 (Fibonacci.c) .
1) Please use Scheme to implement the same function recursively.
define (fib n)
(if (< n 2)
n
(+ (fib (- n 1)) (fib (- n 2)))))
2) Please answer if the stack’s change is the same as Fibonacci.c ? Sketch
stack’s change of the both program, and if they are the same, sketch one figure;
otherwise sketch both figures.
Ans: Yes
Address Variable Value
Variable
fib(3)
Address
fib(3)
0x1000 n
3
0x1004 fibn
?
0x1008 fibn1
?
Step 1
Value
Variable
Value
fib(3)
0x1000
n
3
0x1000
n
3
0x1004
fibn
?
0x1004
fibn
?
0x1008
fibn1
?
0x1008
fibn1
?
fib2(3, 0x1004, 0x1008)
fib2(3, 0x1004, 0x1008)
0x1010
n
3
0x1010
n
3
0x1014
pfibn
0x1004
0x1014
pfibn
0x1004
0x1018
pfibn1
0x1008
0x1018
pfibn1
0x1008
0x101c
fibn2
?
0x101c
fibn2
?
fib2(2, 0x1008, 0x101c)
Step 2
0x1020
n
2
0x1024
pfibn
0x1008
0x1028
pfibn1
0x101c
0x102c
fibn2
?
Step 3
Address
Variable
Value
Address
fib(3)
0x1000
n
0x1004 fibn
0x1008
fibn1
Variable
Value
Address
fib(3)
Variable
Value
fib(3)
3
0x1000
n
3
?
0x1000
n
3
0x1004
fibn
?
?
0x1004
fibn
?
0x1008
fibn1
?
0x1008
fibn1
?
fib2(3, 0x1004, 0x1008)
fib2(3, 0x1004, 0x1008)
0x1010
n
3
0x1010
n
3
0x1014
pfibn
0x1004
0x1010
n
3
0x1014
pfibn
0x1004
0x1018
pfibn1
0x1008
0x1014
pfibn
0x1004
0x1018
pfibn1
0x1008
0x101c
fibn2
?
0x1018
pfibn1
0x1008
0x101c
fibn2
1
0x101c
fibn2
1
fib2(3, 0x1004, 0x1008)
fib2(2, 0x1008, 0x101c)
fib2(2, 0x1008, 0x101c)
0x1020
n
2
0x1020
n
2
0x1024
pfibn
0x1008
0x1020
n
2
0x1024
pfibn
0x1008
0x1028
pfibn1
0x101c
0x1024
pfibn
0x1008
0x1028
pfibn1
0x101c
0x102c
fibn2
?
0x1028
pfibn1
0x101c
0x102c
fibn2
1
0x102c
fibn2
1
fib2(1, 0x101c, 0x102c)
fib2(1, 0x101c, 0x102c)
0x1030
n
2
0x1030
n
2
0x1034
pfibn
0x101c
0x1034
pfibn
0x101c
0x1038
pfibn1
0x102c
0x1038
pfibn1
0x102c
0x103c
fibn2
?
0x103c
fibn2
?
Step 4
Step 5
fib2(2, 0x1008, 0x101c)
Step 6
Address
Variable
Value
fib(3)
0x1000
n
3
0x1004
fibn
?
0x1008
fibn1
2
Address
Variable
Value
Address
fib(3)
Variable
Value
fib(3)
0x1000 n
3
0x1000
n
3
fib2(3, 0x1004, 0x1008)
0x1004 fibn
?
0x1004
fibn
3
0x1010
n
3
0x1008 fibn1
2
0x1008
fibn1
2
0x1014
pfibn
0x1004
fib2(3, 0x1004, 0x1008)
fib2(3, 0x1004, 0x1008)
0x1018
pfibn1
0x1008
0x1010 n
3
0x1010
n
3
0x101c
fibn2
1
0x1014 pfibn
0x1004
0x1014
pfibn
0x1004
fib2(2, 0x1008, 0x101c)
0x1018 pfibn1
0x1008
0x1018
pfibn1
0x1008
0x1020
n
2
0x101c fibn2
1
0x101c
fibn2
1
0x1024
pfibn
0x1008
0x1028
pfibn1
0x101c
0x102c
fibn2
1
Step 7
Step 8
Step 9
Address Variable Value
fib(3)
0x1000 n
3
0x1004 fibn
3
0x1008 fibn1
2
Step 10
3) If we implement the same program using Fortran 77, is the stack’s change
same as Fibonacci.c ?
No
4) Please convert (compile) your Scheme into MIPS assembly code
fib:
cont:
slti $t0, $a0, 2 # if i < 2 (i.e i == 1)
beq $t0, $zero, cont
# if i >= 2 go to cont
addi $v0, $zero, 1
# else make the return value 1
jr $ra
# OPERATION 1: MAKE SPACE IN STACK
addi $sp, $sp, -16
# make space for 3 elements in the stack
sw $ra, 0($sp)
# save the return address
sw $a0, 4($sp)
# OPERATION 2: RECURSIVE CALLS
addi $a0, $a0, -1
# calculate n - 1
jal fib
# calculate fib(n - 1)
sw $v0, 8($sp)
# save into the stack result of fib(n - 1)
lw $a0, 4($sp)
# load the value of n
addi $a0, $a0, -2
# calculate n - 2
jal fib
sw $v0, 12($sp)
# save into the stack the result of fib(n-2)
lw $ra, 0($sp)
# load the return address
lw $t0, 8($sp)
# load the value of fib(n - 1)
lw $t1, 12($sp)
# load the value of fib(n - 2)
addi $sp, $sp, 16
# pop from stack # OPERATION 4: OPEARATION
add $v0, $t0, $t1
# fib(n - 1) + fib(n - 2)
jr $ra
5) Briefly explain how Scheme functions handle parameter-list ?
Recall in lecture 10
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