Test: Syntax Directed Translation- 1 - Computer Science Engineering (CSE) MCQ

► Expression without any calls in it ⇒ 1+2*3-4

► Expression with embedded calls ⇒ 1 + fun1(a,b,c) *fun2(3.4,58) - fun3(x,yz);

First we can convert Infix to Postfix using single stack (Using it as operator stack)

Then we can evaluate that expression using Single stack.

at left  leafs
+   -----> (1,1)=2         intermediate + ------->  2-(-1)=3    -  -------->(0,1)=-1
at right leafs
-  minus ------->(1,0)=1       intermediate   +    ------> 1+2=3 +     ------------->(1,1)=2
at root   + -------> 3+3=6

Option (b) is free from left recursion. No direct left recursion. No indirect left recursion.

Set minimum element as root (i.e 1), now 6 are remaining and left subtree will have 3 elements, each left subtree combination can be permuted in 2! ways.

Total ways to design min-heap with 7-elements =  *2! * 2! = 20*2*2 = 80

Alternative approach – Total number of min or max heap tree with 1 to N elements are using recurrence relation:

• T(N) = (N-1)Ck * T(k) * T(N-k-1), were k = number of nodes on left subtree
•  T(1) = 1 T(2) = 1 T(3) = 2 T(4) = 3C2 * T(2) * T(1) = 3 T(5) = 4C3 * T(3) * T(1) = 8 T(6) = 5C3 * T(3) * T(2) = 20 T(7) = 5C3 * T(3) * T(3) = 80

So, answer is 80.

Since CFG's are used to show syntactic rules while designing compiler, and syntactic rules don't check for meaningful things such as if a variable has been declared before its use or not. Such things are meant to be handled by Semantic Analysis phase (requires power of a context sensitive grammar).
For D, a regular expression does not restrict the string length. Languages have restriction for variable name length for storing purpose like in symbol table.
For A, if then else is inherently ambiguous. But CFG can represent inherently ambiguous languages just that there are more than one parse trees possible for some strings.

S∝ ->

This violates the conditions of context-free and hence the grammar becomes context-sensitive.

 if a then if b then c: = d else c:= f  has two parse trees as follows:

1. if a then (if b then c:= d) else c:= f 2. and if a then (if b then c:=d else c:= f)

It will be A. For multiple ' ', the derivation is possible only via ' ' which is on left side of ' ' in the production. Hence it is left associative. 

For multiple '* ', the derivation is possible only via '* ' which is on the right side of ' ' in the production. Hence it is right associative. 

If both left and right derivations were possible, the grammar would have been ambiguous and we couldn't have given associativity.

Let grammar is like this :

This grammar is left as well as right recursive but still unambiguous.. A is useless production but still part of grammar..
A grammar having both left as well as right recursion may or may not be ambiguous ..

Let's take a grammar say 

Now, according to the link https://en.wikipedia.org/wiki/Formal_grammar

For a grammar G, if we have L(G) then all the strings/sentences in this language can be produced after some finite number of steps .

But, for the grammar

Can we produce any string after a finite number of steps ? NO, and hence language of this grammar is empty set {} .

Hence, If a grammar is having both left and right recursion, then grammar may or may not be ambiguous .

Operator which is at lower level in the grammar is termed to have higher precedence.

(A) We can not have different Left Most Derivation and Right Most Derivation parse trees BUT we can certainly have different LMD and RMD for a given string.(LMD and RMD here refers to the order of various productions used for derivation which could be different.)

(D) is wrong w.r.t. question because IT IS TRUE that any LR(k) IS NEVER AMBIGUOUS and so an ambiguous can never be an LR(K) for any k, no matter how large k becomes.

(B) and (C) can not be the answer because LL(1) is TOP-DOWN parser and LALR is powerful than SLR. So both are TRUE.

Let us make the parse tree for 9+5+2 in top down manner, left first derivation.

Steps:

1. Exapnd S-> TR
2. apply T-> Num...
3. apply R -> +T...
4. apply T->Num...
5. apply R-> +T..
6. apply T-> Num..
7. apply R-> epsilon

After printing through the print statement in the parse tree formed you will get the answer as 9 5 + 2 +.

We can do it in a simple manner.

Here # is multiplication and & is addition by semantics rules given in the question and by observation of productions.

1. here &(+) is higher precedence than #(*). bcoz & is far from starting symbol.
2. and both &,# are left associative so we can solve the express in this way ((2*(3+5))*(6+4)) =160 so ans should be (C).

because operator grammer  does not contain  1)  nullable variable  2) 2 adjacent non-terminal on rhs of production

above CFG generate string b, aaa.. b will eliminate options A and D aaa eliminate options B.
C is answer i.e. number of a = 3k, k = 0,1,2....

n+n*n

E+n*n

E*n

E

string in red indicates handle here

1.G is ambiguous. E.g. the string  has multiple derivation trees like

II. False.  does not produce all strings with equal no. of `s and `s. ( cannot be generated).

III. True. The given grammar G generates the language , which is Regular and therefore also DCFL. So, a DPDA can be designed for G .

as A will generate a or aa or aaa... i,e a+ ,  so S→ AC will generate a⁺aⁿbⁿ , i.e aⁱb^j i>j

A  LL(1) grammar doesn’t give to multiple entries in a single cell of its parsing table. It has only single entry in a single cell, hence it should be unambiguous. 

Option A is wrong. Grammar is not left recursive. For a grammar to be left recursive a production should be of form A->Ab, where A is a single Non-Terminal and b is any string of grammar symbols.

Option B is wrong. Because a right recursive grammar has nothing to do with LL(1).

Option D is wrong. Because the given grammar is clearly a Context Free Grammar. A grammar is CFG if it has productions of the form A->(V∪ T)* , where A is a single non-terminal and V is a set of Non-terminals and T is a set of Terminals. 

Hence Option C should be the correct one. i.e. the grammar is ambiguous.

But let’s see how the grammar is ambiguous.

If the grammar is ambiguous then it should give multiple entry in a cell while making its parsing table. And Parse table is made with the aid of two functions : FIRST and FOLLOW.

A parsing table of a grammar will not have multiple entries in a cell( i.e. will be a LL(1) grammar) if and only if the following conditions hold for each production of the form A->α|β

1) FIRST(α)  ∩ FIRST(β) =   Φ

2) if FIRST(α) contains ‘ ε ‘ then FIRST(α) ∩ FOLLOW (A) =  Φ and vice-versa.

Now,

• For the production , S->iCtSS1|a, rule 1 is satisfied, because FIRST(iCtSS1) ∩ FIRST(a) = {i} ∩ {a} = Φ
• For the production, S1->eS|ε, rule 1 is satisfied, as FIRST(eS) ∩ FIRST(ε) = {e} ∩ {ε} = Φ   .  But here due to ‘ε’ in FIRST, we have to check for rule 2. FIRST(eS)  ∩ FOLLOW(S1) = {e} ∩ {e, $} ≠ Φ .  Hence rule 2 fails in this production rule. Therefore there will be multiple entries in the parsing table, hence the grammar is ambiguous and not LL(1).