GATE DA Supervised Learning – Data Science & AI | Set 1

Original solution basis:
Distance = sqrt((4 - 1)^2 + (6 - 2)^2) = sqrt(3^2 + 4^2) = sqrt(9 + 16) = sqrt(25) = 5.

Deep Dive
The coordinate differences are 4-1=3 and 6-2=4. Euclidean distance is √(3²+4²)=√25=5. This is the ordinary straight-line distance in the two-dimensional plane.

Key Points to Remember:
• Euclidean distance is the L2 norm of the difference vector.
• d=√$\Sigma (xᵢ-yᵢ)²$.
• The 3-4-5 right-triangle identity gives distance 5.

Exam tip:
Keep full precision during intermediate calculations and round only at the final step, as requested.

Original solution basis:
Manhattan distance = |4 - 1| + |6 - 2| = 3 + 4 = 7.

Deep Dive
Manhattan distance sums absolute coordinate differences: |4-1|+|6-2|=3+4=7. It corresponds to movement along axis-aligned paths rather than the straight-line path used by Euclidean distance.

Key Points to Remember:
• Manhattan distance is the L1 norm.
• d=$\Sigma |xᵢ-yᵢ|$.
• For the given points, the coordinate differences sum to 7.

Exam tip:
Keep full precision during intermediate calculations and round only at the final step, as requested.

Original solution basis:
Distance = |3(0) + 4(0) - 12| / sqrt(3^2 + 4^2) = |-12| / 5 = 12 / 5 = 2.4.

Deep Dive
For a line ax₁+bx₂+c=0, the perpendicular distance from (x₀,y₀) is |ax₀+by₀+c|/√(a²+b²). Here a=3, b=4, c=-12, and the point is (0,0). Thus distance=|-12|/√(9+16)=12/5=2.4.

Key Points to Remember:
• Point-to-line distance = |ax₀+by₀+c|/√(a²+b²).
• The denominator is the norm of the normal vector (a,b).
• The distance is nonnegative because of the absolute value.

Exam tip:
Keep full precision during intermediate calculations and round only at the final step, as requested.

Original solution basis:
Training fold size = (5 - 1)/5 * 1000 = 4/5 * 1000 = 800 instances.

Deep Dive
In K-fold cross-validation, one fold is held out for validation and the other K-1 folds are used for training. With K=5, the training fraction is 4/5. Therefore 1000$\times 4/5=800 instances are used$ for training in each fold (assuming equal fold sizes).

Key Points to Remember:
• Training fraction per fold = (K-1)/K.
• Validation fraction per fold = 1/K.
• For 5 folds and 1000 examples, training size is 800 and validation size is 200.

Exam tip:
Keep full precision during intermediate calculations and round only at the final step, as requested.

Original solution basis:
Always predicting positive means FP = 80, TN = 0. FPR = FP / (FP + TN) = 80 / 80 = 1.0.

Deep Dive
The classifier predicts Positive for every example. All 20 actual positives are true positives, while all 80 actual negatives are false positives; there are no true negatives. FPR=FP/(FP+TN)=80/(80+0)=1, or 100%.

Key Points to Remember:
• FPR is calculated only over actual negatives.
• Here FP=80 and TN=0.
• Always predicting positive gives FPR=1.

Exam tip:
Keep full precision during intermediate calculations and round only at the final step, as requested.