Introduction to Artificial Intelligence Questions and Answers

Q: The defining length and order of the schema 1***********1 is :

The defining length is 12 and the order is 2: δ(H) = 12, O(H) = 2.The schema 1***********1 has 13 positions: a fixed 1 at position 1, eleven don't-care symbols (*), and a fixed 1 at position 13.The order O(H) is the number of fixed positions, here 2. The defining length δ(H) is the distance between the first and last fixed positions, 13 - 1 = 12.Option (A) gives the string length 13 as the defining length, and options (B) and (D) give 13 as the order, which counts every position instead of only the fixed ones.

Q: The following table shows attributes of biological and artificial neurons respectively. Match the following according to the functions performed by their counterpart :List IList II(i) Cell/soma(a) Neuron/Node(ii) Synapse(b) Weights(iii) Dendrites(c) Net input(iv) Axon(d) Output

The correct matching is (i)-(a), (ii)-(b), (iii)-(c), (iv)-(d).Cell body (soma)Neuron or nodeSynapseWeightsDendritesNet inputAxonOutputThe soma, where signals are combined, corresponds to the artificial neuron or node itself. The strength of a synapse, which scales the signal passing from one neuron to another, corresponds to a connection weight. The dendrites carry incoming signals to the cell, the counterpart of the inputs summed into the net input. The axon carries the neuron's signal onward, as the node's output does.Each of the other options rotates the pairs, for example matching the axon to the weights.

Q: The probability of 1-point crossover, with crossover rate Pc, to break the schema of length 'l' given by :

The probability is \(\dfrac{P_c \cdot D(H)}{l - 1}\).In Holland's schema theorem a chromosome of length l has l - 1 possible cut points for one-point crossover. A schema H with defining length D(H), the distance between its first and last fixed positions, is broken when the cut falls between those positions, which happens at D(H) of the l - 1 points. Crossover takes place with probability \(P_c\), so\[ P_{break} = P_c \cdot \frac{D(H)}{l - 1} \]Replacing D(H) by a second \(P_c\), or dividing by l instead of the l - 1 cut points, gives the wrong options.

Q: In LISP the function that provides the initial element of a list is :

The function is car.In LISP, (car '(a b c)) returns a, the first element of the list, and cdr returns the rest of the list, (b c). The names come from the IBM 704 register fields used in the original implementation.set assigns a value to a symbol. second returns the second element. Common Lisp also defines first as a synonym, but the primitive LISP function the question has in mind is car.

Q: Which of the following is not a correct encoding technique used in genetic algorithms ?

The technique that is not a genetic algorithm encoding is premier encoding.Standard chromosome encodings are binary encoding (strings of bits), value encoding (real numbers, characters or objects), permutation encoding (orderings, as for the travelling salesman problem) and tree encoding (used in genetic programming).'Premier encoding' is not among them; the similar-sounding real technique is permutation encoding.

101.

The defining length and order of the schema 1***********1 is :

A.δ(H) = 13, O(H) = 02
B.δ(H) = 10, O(H) = 13
C.δ(H) = 12, O(H) = 2
D.δ(H) = 12, O(H) = 13

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: C

The defining length is 12 and the order is 2: δ(H) = 12, O(H) = 2.

The schema 1***********1 has 13 positions: a fixed 1 at position 1, eleven don't-care symbols (*), and a fixed 1 at position 13.

The order O(H) is the number of fixed positions, here 2. The defining length δ(H) is the distance between the first and last fixed positions, 13 - 1 = 12.

Option (A) gives the string length 13 as the defining length, and options (B) and (D) give 13 as the order, which counts every position instead of only the fixed ones.

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102.

The following table shows attributes of biological and artificial neurons respectively. Match the following according to the functions performed by their counterpart :

List I	List II
(i) Cell/soma	(a) Neuron/Node
(ii) Synapse	(b) Weights
(iii) Dendrites	(c) Net input
(iv) Axon	(d) Output

A.(i)-(a), (ii)-(b), (iii)-(c), (iv)-(d)
B.(i)-(b), (ii)-(c), (iii)-(d), (iv)-(a)
C.(i)-(c), (ii)-(d), (iii)-(a), (iv)-(b)
D.(i)-(d), (ii)-(a), (iii)-(b), (iv)-(c)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: A

The correct matching is (i)-(a), (ii)-(b), (iii)-(c), (iv)-(d).

Cell body (soma)	Neuron or node
Synapse	Weights
Dendrites	Net input
Axon	Output

The soma, where signals are combined, corresponds to the artificial neuron or node itself. The strength of a synapse, which scales the signal passing from one neuron to another, corresponds to a connection weight. The dendrites carry incoming signals to the cell, the counterpart of the inputs summed into the net input. The axon carries the neuron's signal onward, as the node's output does.

Each of the other options rotates the pairs, for example matching the axon to the weights.

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103.

What features amongst the following would be required for a computer program to pass the Turing test ?

(i) Natural language processing
(ii) Knowledge representation
(iii) Automated reasoning
(iv) Computer vision
(v) Machine learning
(vi) Problem solving

A.(i), (ii) and (iii)
B.(i), (ii), (iii) and (iv)
C.(i), (ii), (iii) and (v)
D.(i), (ii), (iv) and (vi)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: C

The features are (i), (ii), (iii) and (v), so the answer is (i), (ii), (iii) and (v).

In the standard account, a program would need four capabilities to pass Turing's test: natural language processing to communicate, knowledge representation to store what it knows or hears, automated reasoning to answer questions and draw conclusions, and machine learning to adapt and detect patterns.

Computer vision (with robotics) is needed only for the 'total Turing test', in which the interrogator can also test perception and handle objects, so options with (iv) are wrong. Problem solving is not listed as a separate requirement.

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104.

The probability of 1-point crossover, with crossover rate P_c, to break the schema of length 'l' given by :

A.\(\dfrac{P_{c} * P_{c}}{l - 1}\)
B.\(\dfrac{P_{c} * D(H)}{l - 1}\)
C.\(\dfrac{P_{c}}{l}\)
D.\(\dfrac{P_{c} * D(H)}{l}\)

Exam Year(s)

MH SET - 2021, SET Exam - 2021,

Answer: B

The probability is \(\dfrac{P_c \cdot D(H)}{l - 1}\).

In Holland's schema theorem a chromosome of length l has l - 1 possible cut points for one-point crossover. A schema H with defining length D(H), the distance between its first and last fixed positions, is broken when the cut falls between those positions, which happens at D(H) of the l - 1 points. Crossover takes place with probability \(P_c\), so

\[ P_{break} = P_c \cdot \frac{D(H)}{l - 1} \]

Replacing D(H) by a second \(P_c\), or dividing by l instead of the l - 1 cut points, gives the wrong options.

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105.

In LISP the function that provides the initial element of a list is :

A.Car
B.Set
C.Second
D.First

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

The function is car.

In LISP, (car '(a b c)) returns a, the first element of the list, and cdr returns the rest of the list, (b c). The names come from the IBM 704 register fields used in the original implementation.

set assigns a value to a symbol. second returns the second element. Common Lisp also defines first as a synonym, but the primitive LISP function the question has in mind is car.

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106.

Which of the following is not a correct encoding technique used in genetic algorithms ?

A.Premier encoding
B.Tree encoding
C.Binary encoding
D.Value encoding

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

The technique that is not a genetic algorithm encoding is premier encoding.

Standard chromosome encodings are binary encoding (strings of bits), value encoding (real numbers, characters or objects), permutation encoding (orderings, as for the travelling salesman problem) and tree encoding (used in genetic programming).

'Premier encoding' is not among them; the similar-sounding real technique is permutation encoding.

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