Compiler Design Preparation Resources for GATE CSE

July 29, 2015 by Arjun Suresh 12 Comments

Compiler Design Syllabus

Lexical analysis, parsing, syntax-directed translation.
Runtime environments.
Intermediate code generation.
Local optimization, Data flow analyses: constant propagation, liveness analysis, common subexpression elimination.

” The optimization part is newly added. But these were already included before GATE 2016 in the form of “Basics of Code Optimization”

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$(uv)^R = v^Ru^R,$ for all strings $u,v.$ If $u = \epsilon,$ then $u^R = \epsilon$ and hence $(uv)^R = v^R = v^Ru^R.$ If $v = \epsilon,$ then $v^R = \epsilon$ and hence $(uv)^R = u^R = v^Ru^R.$ Now, suppose $u = u_1u_2 \dots u_m$ and $v = v_1v_2 \dots v_n$, with $m,n \geq 1.$ Then, $(uv)^R = (u_1u_2 \dots u_mv_1v_2 \dots v_n)^R = v_n \dots v_2v_1x_m \dots x_2x_1 = v^Ru^R.$ Detailed Video Solution [...]

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Answered: Kenneth Rosen Edition 7 Exercise 6.1 Question 41 (Page No. 397)

Ans: $2^{\lceil n/2 \rceil}$ APPROACH: for even length strings: 1st bit and nth bit are same. similarly, 2nd and (n-1)th bits are same. Now, 2 bits for every position, and we are filling n/2 positions. therefore $2^{n/2}$ strings for odd-length strings 1st bit and nth bit are same. similarly, 2nd and (n-1)th bits are same. but {(n+1)/2}th positions has 2 more ways to be filled Now, 2 bits for every position, and we are filling n/2 positions. therefore $2^{n+1/2}$ strings. We can also write it as $2^{\lceil n/2 \rceil}$. [...]

Answered: #nptel

C. To be able to infer the flow of control from the program source code … The main goal of structured Programming is to get an understanding about the flow of control in the given program text. In structure programming various control structures such as switch case. If then – else, while, etc. Allows a programmer to decode the flow of the program easily… [...]

Answered: GATE CSE 2008 | Question: 40

Say we need to find the number of occurence of a number k. If k has to occur atleast n/2 times (that is half of the array size), then k must be there in the middle. There is no way that there are atleast n/2 occurences of k and k is not there in array[middle]. BUT, vice-versa is not true. Maybe what we can do here is to first look in the condition C1: (array[middle] == k) If C1 is False, go home. But if C1 is True we may have something to work with. Now the n/2 appearances of… [...]

Answered: Kenneth Rosen Edition 7 Exercise 6.1 Question 40 (Page No. 397)

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