ITPEC FE Subject B April 2025 Question 15

Math & Numbers — Minkowski distance calculation with varying exponent n

The function calcDistance(p1, p2, n) computes the Minkowski distance of order n:

(Σ|p1[i] - p2[i]|^n)^(1/n)

Given p1 = {3, 1, 5, 2} and p2 = {4, 6, 2, 3}:

• •Absolute differences: |3-4|=1, |1-6|=5, |5-2|=3, |2-3|=1

For n=1 (Manhattan distance):
- distance = 1 + 5 + 3 + 1 = 10
- ex = 1/1 = 1
- result = 10^1 = 10 → A = 10

For n=2 (Euclidean distance):
- distance = 1 + 25 + 9 + 1 = 36
- ex = 1/2 = 0.5
- result = 36^0.5 = 6 → B = 6

Why not others:
- A=4 options (a–d): incorrect sum for n=1; forgetting to include all four terms or miscalculating
- B=10 or B=18: wrong exponent application or summing without taking the root
- B=4: would imply √16, but the sum of squares is 36, not 16

Key rule: As n → ∞, Minkowski distance converges to the Chebyshev distance = max(|p1[i] - p2[i]|). Here max(1,5,3,1) = 5, confirming the problem statement.

AI-generated — may contain errors