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Complement of fuzzy set:

$\mu_{A'}(x) = 1 - \mu_A(x)$

Compute:

$x_1$: $1 - 0.3 = 0.7$

$x_2$: $1 - 0.7 = 0.3$

$x_3$: $1 - 0.4 = 0.6$

Thus,

$A' = {(x_1,0.7),(x_2,0.3),(x_3,0.6)}$

Solution

Common properties of social networks:

• Scale-free degree distribution → True

• Small world property → True

• Short average path length → True

However, most social networks are not necessarily directed; many are modeled as undirected networks.

Thus the statement that is not a common property is:

Solution

Morphological analysis, semantic analysis and pragmatic analysis are standard components of natural language understanding.

Meaning analysis is not a standard separate component because meaning is covered under semantic analysis.

Solution

In a production system, conflict resolution can be based on the rule matched, the object matched, or the action of the matched rule.

But assigning preference based on the action that the matched object would perform is not a standard conflict resolution approach.

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Complexity increases as follows:

Rule-based system → simplest

Shallow neural network → moderate complexity

CNN image classification → more complex

Autonomous driving → highly complex system

Thus increasing order:

A → B → C → D

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In uniform crossover, a mask (binary mask) determines which parent contributes each gene.

Thus mask-based crossover = uniform crossover.

Solution

Solution

In fuzzy set theory, cardinality is the sum of membership values:

$|A| = \sum \mu_A(x)$

The result depends on membership values and may vary.

Solution

In genetic algorithms, Actual Count refers to the expected number of copies of a chromosome in the next generation, used in selection (fitness proportionate selection).

Solution

The right parts of the parents are exchanged:

Parent₁: 10110 | 010

Parent₂: 10101 | 111

After crossover:

Child₁: 10110 | 111

Child₂: 10101 | 010

This is exactly what is shown in figure (2).

Thus it represents one-point crossover.

Solution

Predicate logic → symbolic syntactic structure

Slot-and-filler → symbolic frame representation

Conceptual dependency → semantic

Nonmonotonic systems → reasoning method

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The root node is a $MAX$ node.

In minimax method:

$MAX$ node selects the maximum value from its children.

$MIN$ node selects the minimum value from its children.

From the left subtree, the minimum level gives values such that the final value of the left side becomes

$7$

From the right subtree, the available minimum values make the right side value less than or equal to $5$.

Now, root is a $MAX$ node, so it selects maximum of left and right subtree values.

$\max(7,\le 5)=7$

Hence, the value of the root node is

$7$

Solution

Greedy best-first search selects the node with minimum heuristic value.

So,

$(a)\rightarrow(ii)$

Lowest cost-first search selects the node with minimum path cost.

So,

$(b)\rightarrow(iii)$

$A^*$ algorithm selects the node using cost plus heuristic value.

So,

$(c)\rightarrow(i)$

Therefore, the correct matching is

$(a)-(ii);\ (b)-(iii);\ (c)-(i)$

Solution

STRIPS is used in AI planning.

In STRIPS, actions are represented using:

Preconditions
Add list
Delete list

So, it mainly describes how actions change the state of the world.

Therefore, STRIPS representation is an action-centric representation.

Solution

For fuzzy conjunction and fuzzy disjunction to be dual, they must satisfy De Morgan type relation.

The duality condition is:

$t(x,y)=1-s(1-x,1-y)$

Solution

In fuzzy sets, a higher $\alpha$-cut gives a smaller or equal set.

Since,

$\alpha_1<\alpha_2$

the cut at $\alpha_1$ will contain all elements of the cut at $\alpha_2$.

So,

$A_{\alpha_1}\supseteq A_{\alpha_2}$

Solution

Defuzzification converts fuzzy output into a crisp value.

Mean of maximum is a defuzzification method.

Centre of area is also a defuzzification method.

Height method is also used for defuzzification.

Therefore, all $M1,M2$ and $M3$ are defuzzification methods.

Solution

Genetic algorithms are based on the idea of natural evolution.

They use selection to choose better individuals.

They use reproduction to create new generations.

They use mutation to introduce variations.

So, evolution, selection, reproduction and mutation are all related to genetic algorithms.

Solution

Using perceptron learning rule: Δw = η (t − y) x, where y = 1 if y_in > 0 else 0.

S1 (0,0,t=0): y_in=0 ⇒ y=0 ⇒ no change ⇒ (w1,w2)=(0,0)

S2 (0,1,t=1): y_in=0 ⇒ y=0 ⇒ Δw=0.1(1−0)(0,1) ⇒ (w1,w2)=(0,0.1)

S3 (1,0,t=1): y_in=0 ⇒ y=0 ⇒ Δw=0.1(1−0)(1,0) ⇒ (w1,w2)=(0.1,0.1)

S4 (1,1,t=1): y_in=0.1+0.1=0.2>0 ⇒ y=1 ⇒ no change

Final Weights: w1=0.1, w2=0.1

Answer: (1)

Solution

Solution

In probability theory,

$P(A)+P(\neg A)=1$

But in Dempster-Shafer theory, belief values need not add up to $1$ because some uncertainty may remain unassigned.

So,

$Bel(A)+Bel(\neg A)\le 1$

Solution

In the given game tree:

$\square$ represents MAX node

$\circ$ represents MIN node

So, we evaluate the tree from bottom to top using the minimax rule.

At MIN node, minimum value is selected.

At MAX node, maximum value is selected.

From the left side of the tree:

$\min(17,19)=17$

After applying MAX and MIN levels, the left subtree gives value:

$17$

From the right side of the tree:

The available minimum values include $15$ and $14$, so the right subtree gives a value less than $17$.

At the root node, we have MAX node.

So root will select maximum value:

$\max(17,14)=17$

Therefore, the value of the root node is:

$17$

Solution

Additive rule models combine the contribution of rules in an additive manner.

Takagi-Sugeno-Kang model is an additive rule model.

Kosko's Standard Additive Model $($SAM$)$ is also an additive rule model.

Mamdani model is not generally considered an additive rule model in this classification.

So, examples of additive rule model are:

$M_2$ and $M_3$

Solution

Given:

$t(x,y)=1-s(1-x,1-y)$

So,

$s(x,y)=1-t(1-x,1-y)$

Now,

$t(1-x,1-y)=\dfrac{(1-x)(1-y)}{(1-x)+(1-y)-(1-x)(1-y)}$

Numerator:

$(1-x)(1-y)=1-x-y+xy$

Denominator:

$(1-x)+(1-y)-(1-x)(1-y)$

$=2-x-y-(1-x-y+xy)$

$=1-xy$

So,

$t(1-x,1-y)=\dfrac{1-x-y+xy}{1-xy}$

Now,

$s(x,y)=1-\dfrac{1-x-y+xy}{1-xy}$

$s(x,y)=\dfrac{1-xy-1+x+y-xy}{1-xy}$

$s(x,y)=\dfrac{x+y-2xy}{1-xy}$

Solution

In gradient descent, weights are updated in the opposite direction of the gradient.

The general rule is:

$New\ Weight=Old\ Weight-\alpha\times Gradient$

Here,

$\alpha$ is the learning rate.

So,

$W_{ij}(t+1)=W_{ij}(t)-\alpha\dfrac{\partial E}{\partial W_{ij}}$

Solution

In genetic algorithm, the order of a schema is the number of fixed positions.

Fixed positions are symbols other than $?$

For schema $?10?101?$:

Fixed symbols are:

$1,\ 0,\ 1,\ 0,\ 1$

So, order is:

$5$

For schema $???0??1$:

Fixed symbols are:

$0,\ 1$

So, order is:

$2$

Solution

Solution

Convert chromosomes into decimal form:

$(01101)_2=13$

$(11000)_2=24$

$(01000)_2=8$

$(10011)_2=19$

Now calculate $f(x)=x^2$

For $13$:

$f(13)=13^2=169$

For $24$:

$f(24)=24^2=576$

For $8$:

$f(8)=8^2=64$

For $19$:

$f(19)=19^2=361$

Now,

$\sum f(x)=169+576+64+361=1170$

Strength of fitness for chromosome $(11000)$:

$Sf(24)=\dfrac{576}{1170}$

$Sf(24)=0.4923$

In percentage:

$0.4923\times 100=49.23%\approx 49.2$

Solution

Solution

Solution

Explanation:

• A → IV → Intersection → $\min(\mu_A(x), \mu_B(x))$
• B → III → Bounded Sum → $\min(1, \mu_A(x) + \mu_B(x))$
• C → II → Bounded Difference → $\max(0, \mu_A(x) - \mu_B(x))$
• D → I → Algebraic Sum → $\mu_A(x) + \mu_B(x) - \mu_A(x)\mu_B(x)$

Solution

Hill climbing algorithm has some common limitations.

It may get trapped at a local maximum.

It may reach a plateau where all neighboring states have almost same value.

It may face difficulty while traversing along a ridge.

So, all three are shortcomings of hill climbing algorithm.

Solution

Using best first search, we always choose the next node with the minimum available cost.

From $A$:

$A \to C = 2$

$A \to F = 7$

So, choose $C$.

From $C$:

$C \to F = 3$

$C \to B = 10$

So, choose $F$.

From $F$:

$F \to D = 1$

$F \to E = 6$

So, choose $D$.

From $D$:

$D \to E = 2$

$D \to Z = 7$

So, choose $E$.

From $E$:

$E \to Z = 4$

Hence, the traversal order before reaching $Z$ is:

$A \to C \to F \to D \to E$

Correct option is:

$\boxed{(a)\ A \to C \to F \to D \to E}$

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For a fuzzy set $A$, the $\lambda$-cut is defined as: $$A_\lambda = \{x \mid \mu_A(x) \ge \lambda\}$$

• Union Property: $(A \cup B)_\lambda = A_\lambda \cup B_\lambda$ ✅
• Intersection Property: $(A \cap B)_\lambda = A_\lambda \cap B_\lambda$ ✅
• Monotonicity: If $\lambda \le \beta$, then $A_\beta \subseteq A_\lambda$ ✅
• Complement: $\mu_{\overline{A}}(x) = 1 - \mu_A(x)$ ⇒ $(\overline{A})_\lambda = \{x \mid 1 - \mu_A(x) \ge \lambda\} = \{x \mid \mu_A(x) \le 1 - \lambda\}$ ❌

Hence, $(\overline{A})_\lambda$ is not equal to $A_\lambda$, even when $\lambda = 1$. Therefore, the statement “except when $\lambda = 1$” is incorrect.

✅ Final Answer: Option 3 is not valid.

Solution

Solution

Solution

$A^*$ algorithm uses the evaluation function:

$f(n)=g(n)+h(n)$

where,

$g(n)$ is the cost from start node to current node.

$h(n)$ is the estimated cost from current node to goal node.

$A^*$ is optimal when the heuristic function $h$ is admissible.

Admissible means:

$h(n)$ never overestimates the actual cost.

So, $A^*$ always finds the solution with the lowest total cost when $h$ is admissible.

Solution

The question asks about understanding associations and relationships between objects and events.

In Artificial Intelligence, symbolic representation is used to represent facts, objects, events, and their relationships.

Here, "Relative symbolism" means representing relationships among different objects and events using symbols.

Heuristic processing is used for problem-solving using rules of thumb.

Cognitive science studies human thinking and intelligence.

Pattern matching is used to identify similar patterns.

So, the technique related to associations and relationships is Relative symbolism.

Solution

In alpha-beta pruning, two values are maintained:

$ \alpha $ = best value found so far for MAX player

$ \beta $ = best value found so far for MIN player

These values are updated while moving along the search path in the game tree.

So, alpha and beta values do not update only at the initial state or only at the end.

They are updated along the path of search.

Solution

Solution

In a genetic algorithm, solution representation refers to how the potential solutions, or individuals in the population, are encoded. The given options can be categorized as follows:

• Binary valued: Represents solutions using binary encoding, where each gene is typically represented as either 0 or 1.

• Real valued: Represents solutions using real numbers, allowing for a more continuous representation.

• Permutation: Represents solutions as permutations of elements, where the order of elements matters.

• Combinations: This is not a standard representation in genetic algorithms. While permutations represent ordered arrangements, combinations represent unordered selections of elements. Genetic algorithms typically use representations that capture the structure and relationships among elements, and combinations may not be suitable for many optimization problems.

Solution

A Genetic Algorithm is based on the idea of natural selection and evolution.

It improves solutions using operations like:

Selection
Crossover
Mutation

So, the AI system that mimics the evolutionary process is:

III. Genetic algorithm

Solution

Solution

Text planning is a part of Natural Language Generation.

So,

a → II

Sentence planning is also a part of Natural Language Generation.

So,

b → II

Sentence generation is also a part of Natural Language Generation.

So,

c → II

Mapping the input to useful representations is a part of Natural Language Understanding.

So,

d → I

Correct matching:

a-II, b-II, c-II, d-I

Solution

For fuzzy sets, the intersection operation is defined as:

$\mu_{A \cap B}(x) = \min(\mu_A(x), \mu_B(x))$

Now compute for each element:

$x_1$ : $\min(0.3, 0.5) = 0.3$
$x_2$ : $\min(0.7, 0.2) = 0.2$
$x_3$ : $\min(0.4, 0.5) = 0.4$

Thus,

$A \cap B = {(x_1,0.3),(x_2,0.2),(x_3,0.4)}$