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Automata and Compiler
(1) [25 marks] Let N be the last two digits of your student number. Design a finite automaton that accepts the language of strings that end with the last four bits of the binary expansion of N. (1.1) Make a regular expression ? of this language. For example the set of strings that end with 101 is expressed by a regular expression (0+1)*101. (1.2) Make an NFA that accepts this expression ?. You should remove any ?-moves that can be done trivially by inspection. (1.3) Make a subset automaton that accepts the language. (1.4) Perform state minimization on the above automaton.
(2) [25 marks] A CFG is given by S ? aSbS, S ? bSaS, S ? c
(2.1) Draw a syntax chart for this grammar. [5]
(2.2) Write a Python program for the recursive descent parser Trace the parser using two strings of at least 10 symbols, one for an accepted case and one for an unaccepted case. Do the trace using the style in the notes. [20]
(3) [25 marks] A sample program for computing the greatest common divisor by recursive call and its object program are given below. Some sample comments are given.
const a=75, b=55;
var x, y;
procedure gcd;
var w;
begin
if y>0 then begin
w:=y;
y:=x ? (x/y)*y;
x:=w;
call gcd;
end;
x:=a; y:=b;
write(x);
end.
0 jmp 0 21 Jump to 21, start of main
1 jmp 0 2
2 inc 0 4
3 lod 1 4
4 lit 0 0 Load literal 0
5 opr 0 12 Test if y>0
6 jpc 0 20 Jump to 20 if false
7 lod 1 4 Load y
8 sto 0 3 Store in w
9 lod 1 3
10 lod 1 3
11 lod 1 4
12 opr 0 5
13 lod 1 4
14 opr 0 4
15 opr 0 3
16 sto 1 4
17 lod 0 3
18 sto 1 3
19 cal 1 2
20 opr 0 0
21 inc 0 5
22 lit 0 75
23 sto 0 3
24 lit 0 55
25 sto 0 4
26 cal 0 2
27 lod 0 3
28 wrt 0 0 Write stack top
29 opr 0 0
Let L1 and L2 be CGF. We show that L1 ∩ L2 is CFG too. Let M1 be a decider for L1 and M2 be a decider for L2 . Consider a 2-tape TM M: "On input x: 1. copy x on the sec
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The upper string r ∈ Q+ is the sequence of states visited by the automaton as it scans the lower string w ∈ Σ*. We will refer to this string over Q as the run of A on w. The automa
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Proof (sketch): Suppose L 1 and L 2 are recognizable. Then there are DFAs A 1 = (Q,Σ, T 1 , q 0 , F 1 ) and A 2 = (P,Σ, T 2 , p 0 , F 2 ) such that L 1 = L(A 1 ) and L 2 = L(
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