# GATE CS 2010

Question 1 |

Let G = (V,E) be a graph. Define ξ(G) = Σd id x d, where id is the number of vertices of degree d in G. If S and T are two different trees with ξ(S) = ξ(T),then

|S| = 2|T| | |

|S| = |T|-1 | |

|S| = |T| | |

|S| = |T|+1 |

**GATE CS 2010**

**Graph Theory**

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Question 1 Explanation:

The expression ξ(G) is basically sum of all degrees in a tree. For example, in the following tree, the sum is 3 + 1 + 1 + 1.

a / | \ b c dNow the questions is, if sum of degrees in trees are same, then what is the relationship between number of vertices present in both trees? The answer is, ξ(G) and ξ(T) is same for two trees, then the trees have same number of vertices. It can be proved by induction. Let it be true for n vertices. If we add a vertex, then the new vertex (if it is not the first node) increases degree by 2, it doesn't matter where we add it. For example, try to add a new vertex say 'e' at different places in above example tee.

Question 2 |

Newton-Raphson method is used to compute a root of the equation x

^{2}-13=0 with 3.5 as the initial value. The approximation after one iteration is3.575 | |

3.676 | |

3.667 | |

3.607 |

**GATE CS 2010**

**Numerical Methods and Calculus**

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Question 2 Explanation:

In Newton-Raphson's method, We use the following formula to get the next value of f(x). f'(x) is derivative of f(x).

f(x) = x^{2}-13 f'(x) = 2x Applying the above formula, we get Next x = 3.5 - (3.5*3.5 - 13)/2*3.5 Next x = 3.607

Question 3 |

What is the possible number of reflexive relations on a set of 5 elements?

2 ^{10} | |

2 ^{15} | |

2 ^{20} | |

2 ^{25} |

**GATE CS 2010**

**Combinatorics**

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Question 3 Explanation:

Question 4 |

Consider the set S = {1, ω, ω2}, where ω and w

^{2}are cube roots of unity. If * denotes the multiplication operation, the structure (S, *) formsA group | |

A ring | |

An integral domain | |

A field |

**GATE CS 2010**

**Set Theory & Algebra**

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Question 4 Explanation:

A group is a set of elements together with an operation that combines any two of its elements to form a third element also in the set while satisfying four conditions called the group axioms, namely closure, associativity, identity and invertibility. (Source: http://en.wikipedia.org/wiki/Group_(mathematics)
(S, *) is a group with identity as 1

Question 5 |

What is the value of Lim

_{n->∞}(1-1/n)^{2n}?0 | |

e ^{-2} | |

e ^{-1/2} | |

1 |

**GATE CS 2010**

**Numerical Methods and Calculus**

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Question 5 Explanation:

The value of e (mathematical constant) can be written as following
And the value of 1/e can be written as following.

Question 6 |

The minterm expansion of f(P, Q, R) = PQ + QR' + PR' is

m2 + m4 + m6 + m7 | |

m0 + m1 + m3 + m5 | |

m0 + m1 + m6 + m7 | |

m2 + m3 + m4 + m5 |

**GATE CS 2010**

**Digital Logic & Number representation**

**Logic functions and Minimization**

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Question 7 |

A main memory unit with a capacity of 4 megabytes is built using 1M × 1-bit DRAM chips. Each DRAM chip has 1K rows of cells with 1K cells in each row. The time taken for a single refresh operation is 100 nanoseconds. The time required to perform one refresh operation on all the cells in the memory unit is

100 nanoseconds | |

100×2 ^{10} nanoseconds | |

100×2 ^{20} nanoseconds | |

3200×2 ^{20} nanoseconds |

**GATE CS 2010**

**Computer Organization and Architecture**

**Cache and main memory**

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Question 7 Explanation:

Number of chips required for 4MB MM = (4 * 2^20 * 8) / (1 * 2^20) = 32 chips
In a refresh cycle, a whole row of a memory chip is refreshed at once. This implies the given time of 100 ns for one refresh operation refreshes one row of memory chip. Since there are 1K=2^10 such rows, time for refreshing a whole chip would be: 2^10 * 100 ns.
Second question arises, how to arrange these chips as there can be many possible arrangements. There is a logical arrangement provided in the problem statement itself as "1M x 1 bit chip". This indicates that to make a "1M x 32 bits" MM, we need to arrange all 32 chips in a line. It is to be noted that a row in all chips in series can be refreshed in one refresh cycle. This makes the total time to refresh the 4MBytes of memory as same as that of one chip. Hence, time required to refresh MM = 100 * 2^10 ns.
So, option (B) is correct.

Question 8 |

P is a 16-bit signed integer. The 2's complement representation of P is (F87B)

_{16}.The 2's complement representation of 8*P(C3D8) _{16} | |

(187B) _{16} | |

(F878) _{16} | |

(987B) _{16} |

**GATE CS 2010**

**Number Representation**

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Question 8 Explanation:

P = (F87B)

_{16}is -1111 1000 0111 1011 in bianry Note that most significant bit in the binary representation is 1, which implies that the number is negative. To get the value of the number perform the 2's complement of the number. We get P as -1925 and 8P as -15400 Since 8P is also negative, we need to find 2's complement of it (-15400) Binary of 15400 = 0011 1100 0010 1000 2's Complement = 1100 0011 1101 1000 = (C3D8)_{16}Question 9 |

(P(XOR)Q(XOR)R)' | |

P(XOR)Q(XOR)R | |

(P+Q+R)' | |

P+Q+R |

**GATE CS 2010**

**Digital Logic & Number representation**

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Question 9 Explanation:

For 4 to 1 mux truth table
SEL INPUT O/P

p’q’r+p’qr’+pq’r’+pqr
pXORqXORr

Q | P | R | R’ | R’ | R | F |

0 | 0 | X | X | X | 1 | 1 |

0 | 1 | X | X | 1 | X | 1 |

1 | 0 | X | 1 | X | X | 1 |

1 | 1 | 1 | X | X | X | 1 |

Question 10 |

What does the following program print?

#include void f(int *p, int *q) { p = q; *p = 2; } int i = 0, j = 1; int main() { f(&i, &j); printf("%d %d n", i, j); getchar(); return 0; }

2 2 | |

2 1 | |

0 1 | |

0 2 |

**C Pointer Basics**

**GATE CS 2010**

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Question 10 Explanation:

See below comments for explanation.

/* p points to i and q points to j */ void f(int *p, int *q) { p = q; /* p also points to j now */ *p = 2; /* Value of j is changed to 2 now */ }

There are 64 questions to complete.