Q1 – In the content of Alpha Beta pruning in game trees which of the following statements are correct regarding cut off procedures?
(a) Alpha Beta pruning can eliminate subtrees with certainly when the value of a node exceeds both the alpha and beta bonds.
(b) The primarily purpose of Alpha-Beta pruning is to save computation time by searching fewer nodes in the same tree.
(c) Alpha Beta pruning guarantees the optimal solution in all cases by exploring the entire game tree.
(d) Alpha and Beta bonds are initialized to negative and positive infinity respectively at the root note.
(a), (c), (d) only
(b), (c), (d) only
(a), (b), (d) only
(c), (b) only
(UGC NET DEC 2023)
Ans – (3)
Explanation – Alpha-Beta pruning is, in short, an optimization technique of the Minimax algorithm in decision-making and game theory. It reduces the number of nodes that are evaluated, thus providing an efficient way to reach the same optimal move as in Minimax.
(a) If a node’s value goes beyond the current threshold of either alpha (α) or beta (β), it means that the subtree will not be further explored. This statement is true.
(b) Alpha-Beta pruning is capable of eliminating as many as half of the nodes that would be referred to the Minimax algorithm, thus improving efficiency considerably. Hence this statement is also true.
(c) Alpha-Beta pruning may not go through the complete game tree but passes only through the nodes that are important to get sufficient results. Therefore, this statement is false.
(d) α = -∞, which is the worst case for the maximizer
β = +∞, which is the worst case for the minimizer
So, option 3 is (a), (b) and (d).
Q2 – In a genetic algorithm optimization problem, the fitness function is defined as F(x) = x2 – 4x + 4
Given a population of four individuals with values of x: {1.5, 2.0, 3.0, 4.5}
What is the fitness value of the individual that will be selected as the parent for reproduction in one generation?
2.25
6.0
0.0
6.25
(UGC NET DEC 2023)
Ans – (4)
Explanation – A function F(x) = x2 − 4x + 4 indicates how “fit” or “good” an individual is within the framework of optimization problems.
The function F(x) must be evaluated for every x in order to scale their fitness values. Higher fitness values mean a greater chance of being selected for mating in Genetic Algorithms.
A population of four individuals with values of x: {1.5, 2.0, 3.0, 4.5}
F(1.5) = (1.5)2 – 4(1.5) + 4 = 2.25 – 6.0 + 4 = 0.25
F(2.0) = (2.0)2 – 4(2.0) + 4 = 4.0 – 8.0 + 4 = 0.0
F(3.0) = (3.0)2 – 4(3.0) + 4 = 9.0 – 12 + 4 = 1.0
F(4.5) = (4.5)2 – 4(4.5) + 4 = 20.25 + 18 + 4 = 6.25 (highest)
Thus, x = 4.5 with fitness value 6.25 will be selected as the parent.
Q3 – In a feed forward neural network with the following specifications:
Input layer has 4 neurons, hidden layer has 3 neurons and output layer has 2 neurons using the sigmoid activation function for given input values [0.5, 0.8, 0.2, 0.6] as well as the initial weights for the connections.
Input layer to hidden layer weights
W1: [0.1, 0.3, 0.5, 0.2]
W2: [0.2, 0.4, 0.6, 0.2]
W3: [0.3, 0.5, 0.7, 0.2]
Hidden layer to output layer weights
W4: [0.4, 0.1, 0.3]
W5: [0.5, 0.2, 0.4]
What is the output of the output layer when the given input values are passed through neural network? Round the answer to two decimal places:
[0.62, 0.68]
[0.72, 0.78]
[0.82, 0.88]
[0.92, 0.98]
(UGC NET DEC 2023)
Ans – (1)
Explanation – The question asks what will be the final output of the neural network after passing the given inputs through the hidden layer and then to the output layer.
We need to calculate the weighted sum at each layer, apply the sigmoid activation function, and find the final output values at the output layer.
Finally, we round the result to two decimal places.
Input values – [0.5, 0.8, 0.2, 0.6]
The weighted sum (net input) for each hidden neuron is
W1: [0.1, 0.3, 0.5, 0.2]
Hidden Neuron 1 (h1) = 0.5*0.1 + 0.8*0.3 + 0.2*0.5 + 0.6*0.2
= 0.05 + 0.24 + 0.10 + 0.12 = 0.51
W2: [0.2, 0.4, 0.6, 0.2]
Hidden Neuron 2 (h2) = 0.5*0.2 + 0.8*0.4 + 0.2*0.6 + 0.6*0.2
= 0.10 + 0.32 + 0.12 + 0.12 = 0.66
W3: [0.3, 0.5, 0.7, 0.2]
Hidden Neuron 3 (h3) = 0.5*0.3 + 0.8*0.5 + 0.2*0.7 + 0.6*0.2
= 0.15 + 0.40 + 0.14 + 0.12 = 0.81
We apply sigmoid activation function – σ(x) = 1/(1 + e-x)
For h1 = 0.51, σ(0.51) = 1/(1 + e-0.51) = 0.6248
For h2 = 0.51, σ(0.66) = 1/(1 + e-0.66) = 0.6592
For h3 = 0.51, σ(0.81) = 1/(1 + e-0.81) = 0.6921
The weighted sum for each output neuron is
W4: [0.4, 0.1, 0.3]
Output Neuron 1 (O1) = 0.6248*0.4 + 0.6592*0.1 + 0.6921*0.3
= 0.24992 + 0.06592 + 0.2076 = 0.5234
W5: [0.5, 0.2, 0.4]
Output Neuron 2 (O2) = 0.6248*0.5 + 0.6592*0.2 + 0.6921*0.4
= 0.3124 + 0.1318 + 0.2768 = 0.7210
Now, we apply sigmoid activation function – σ(x) = 1/(1 + e-x)
For O1 = 0.5234, σ(0.5234) = 1/(1 + e-0.5234) = 0.6279
For O2 = 0.7210, σ(0.7210) = 1/(1 + e-0.7210) = 0.6728
Option 1 is right [0.62, 0.68]
Q4 – Arrange the following encoding strategies used in Genetic Algorithms (GAs) in the correct sequence starting from the initial step and ending with the final representation of solution:
(a) Binary Encoding
(b) Real valued Encoding
(c) Permutation Encoding
(d) Gray coding
Choose the correct answer from the options given below:
(d), (b), (a), (c)
(b), (d), (a), (c)
(c), (d), (a), (b)
(b), (c), (a), (d)
(UGC NET DEC 2023)
Ans – (3)
Explanation – In GAs, we must encode the solutions in some way understandable to the computer.
(c) Permutation Encoding
This is just like making a list.
Example – For Traveling Salesman Problem, if we know the order in which we will be visiting the cities, we can simply list them.
(d) Gray Code
Gray codes are used to minimize the risk of errors caused by changes occurring between adjacent numbers.
Gray coding changes only one bit at any given time when representing a number.
Example – Binary 1100 → Gray Code 1010
(a) Binary Encoding
Binary refers to the process of converting information into 0s and 1s (since computers understand binary).
Most commonly used by genetic algorithms.
(b) Real-Valued Encoding
Converts binary back into real values for final output.
Useful when, in problems, decimal values are required as solutions.
Q5 – Arrange the following steps in the correct sequence for applying an unsupervised learning technique such as K-means clustering is to a data set:
(a) Randomly initialise cluster centroids.
(b) Assign each data point to nearest cluster centroid.
(c) Update the cluster centroids based on the mean of data points assigned to each cluster.
(d) Specify the number of clusters (K) to partition the data into
(e) Repeat steps B and C until convergence criteria are met.
Choose the correct answer from the options given below:
(d), (a), (b), (c), (e)
(a), (b), (c), (d), (e)
(c), (b), (a), (d), (e)
(d), (c), (a), (b), (e)
(UGC NET DEC 2023)
Ans – (1)
Explanation – (d) Specify the number of clusters (K) to partition the data into – Decide how many groups you want, in other words, choose a number K, which is the number of clusters.
(a) Randomly initialise cluster centroids – Decide on initial locations for each of the K clusters. Randomly choose K points to serve as the initial centres of clusters.
(b) Assign each data point to nearest cluster centroid – Assign each of the collected data points into a cluster. For each collected data point, determine the closest centre, thus the data point gets assigned to that cluster.
(c) Update the cluster centroids based on the mean of data points assigned to each cluster – Calculate the new centre for each of the clusters. For each cluster, the new centre is the average of all data points assigned to that cluster.
(e) Repeat steps B and C until convergence criteria are met –
Repeat Assignment and Update. Continue reassigning points and recalculate centres until the groups stay constant or cease to change effectively.
So, option 1 is the answer.
Q6 – A __________ point of fuzzy set A is a point x ԑ X at which μA(x) = 0.5.
Core
Support
Crossover
α – cut
(UGC NET DEC 2023)
Ans – (3)
Explanation – Assume you have a fuzzy set having an element x, which is assigned a degree of membership from 0 to 1 according to how much it belongs to the set. The meaning of the scores can be summarized as follows –
1 – The element completely belongs to the set.
0 – The element does not belong in any way.
Between 0 and 1 – The element belongs partially.
Now the crossover point is that one special point where the membership score is exactly 0.5. Thus, an element is halfway in and halfway out of the set.
Core – The score that is 1 (fully belonging to the set).
Support – A set of all points with scores more than 0.
α–cut – This is a set of all points where the score is at least α.
Q7 – Match List-I with List-II
List-I | List-II |
(a) Hill Climbing | I. O(b^d) |
(b) Best first search | II. O(bd) |
(c) A* Search | III. O(1) |
(d) Depth first Search | IV. O(b^m) |
Choose the correct answer from the options given below:
(a) | (b) | (c) | (d) | |
1. | III | I | IV | II |
2. | II | I | IV | III |
3. | II | IV | I | III |
4. | I | III | II | IV |
(UGC NET DEC 2023)
Ans – (1)
Explanation – Hill Climbing is a heuristic approach that moves towards the best neighbour at each step. So, its time complexity is O(1).
Best First Search (BFS) is a graph search algorithm that selects the promising node first according to a heuristic. BFS expands those nodes that have the best (lowest) heuristic value, using a priority queue in the process.
The time complexity is heavily dependent on the branching factor (b) and depth (d). In the worst-case scenario, the complexity is O(b^d), meaning it can become exponentially increasing with more alternatives.
The A* search is an informed search algorithm that takes the best of both uniform cost search and greedy best-first search methods to find the shortest path. The time complexity of A* search, in the worst case, is given by O(b^m), where b is the branching factor (the average number of possible child nodes per tree node) and m is the maximum depth of the search tree. For example, if the heuristic function is poor or misleadingly informative, A* may expand cells, for instance, two or three steps below the actual solution, leading to exponential growth in the number of expanded cells with respect to m, rather than stopping at the depth d of the optimal solution.
Depth First Search (DFS) explores as deep as possible before backtracking. So, the time complexity is O(bd) where b is the branching factor and d is depth.
Q8 – Match List – I with List – II.
List – I
(A) Greedy Best first search
(B) A*
(C) Recursive best first search
(D) SMA*
List – II
(I) The space complexity as O(d) where d = depth of the deepest optimal solution
(II) Incomplete even if the search space is finite
(III) Optimal if optimal solution is reachable otherwise return the best reachable optimal solution
(IV) Computation and space complexity is two light
Choose the correct answer from the options given below :
(A)-(II), (B)-(IV), (C)-(I), (D)-(III)
(A)-(II), (B)-(III), (C)-(I), (D)-(IV)
(A)-(III), (B)-(II), (C)-(IV), (D)-(I)
(A)-(III), (B)-(IV), (C)-(II), (D)-(I)
(UGC NET DEC 2023)
Ans – (1)
Explanation – Greedy Best First Search is an algorithm that selects the next step based only on how close it seems to the goal. While it is fast and does not make heavy memory demands, it is not always reliable due to its tendency to get caught up in misleading paths. This allows it to fit statement (II) because it is not guaranteed to be complete.
An A* is a more intelligent search because it looks at both the cost so far and an estimate of the distance that is left to the goal. An A* is likely to find the best path but needs a lot of memory and processing power. This fits statement (IV).
(Note: the word “two light” is a bit unusual, here it is meant to indicate the heavy resource usage of A* relative to the other algorithms.)
Recursive Best-First Search (RBFS) designs simplified memory requirements relative to A*. This approach needs the space of only the maximal depth of solutions (d) and uses space powerfully. It matches statement (I).
Simplified Memory-Bounded A* (SMA*) is yet another version of A* that works with bounded memory. The A* quality of approximation assures a good solution if there is sufficient memory; otherwise, the best possible solution with available resources will return. It matches statement (III).
Q9 – What is the generic structure of Multi Agent System (MAS) ?
Single agent with multiple objectives
Multiagents with a single objectives
Multiagents with diverse objectives and communication abilities
Multiagent with two objectives
(UGC NET DEC 2023)
Ans – (3)
Explanation – A multi-agent system represents a system in which agents cooperate together toward a goal for which they have different objectives, yet they communicate and coordinate with each other. These agents can be either independent or cooperative; levels of cooperation vary with respect to the environment shared by the agents for the achievement of separate or common objectives.