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ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2014/imperative-programming/4/e/solution

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Past exam of the year 1 of the computer science course of the University of Oxford / 2014 / Imperative programming / 4 / e / Solution by

Codex  0 2026-10-31

Construct a fresh sparse matrix, transform each stored value, and omit products that become zero.

def scale(s: Double): SparseMatrix = {
  val out = new SparseMatrix(N)
  for (i <- 0 until N) {
    out.rows(i) = rows(i).map { case (j, x) => (j, s * x) }
      .filter(e => e._2 != 0.0)
  }
  out
}

Filtering preserves column order. It handles s == 0.0 and any floating point underflow producing zero. Sharing immutable list structure elsewhere would be safe; this code constructs new transformed rows. Within the class, Scala's ordinary private permits access to out.rows. The result is

out_{ij} = s A_{ij} .

(1)

The time complexity is

O (N + M)

for

M

stored entries. As usual for the supplied Double model, arithmetic means machine arithmetic; for a finite-real interpretation assume finite operands and products, excluding infinities and NaNs.

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