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 Incoming links: Row run length

Past exam of the year 1 of the computer science course of the University of Oxford / 2014 / Imperative programming / 2 / a / Solution 2026-10-31

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For each row, keep the row run length of the processed prefix. A false entry resets the length to zero; a true entry extends it by one.

val rowCount = Array.ofDim[Int](N, N)
var i = 0
while (i < N) {
  var j = 0
  var run = 0
  while (j < N) {
    if (a(i)(j)) run += 1 else run = 0
    rowCount(i)(j) = run
    j += 1
  }
  i += 1
}

The outer loop invariant states that rows before i have all their correct row run lengths. The inner loop invariant states that columns before j in row i are correct and that run is the length of the all-true suffix of a(i)[0..j). Initially this suffix is empty. Updating run gives exactly the suffix length ending at column j, proving the required formula, including zero at a false entry. Every cell is visited once:

rowCount (i) (j) = {0, 1 + rowCount (i) (j - 1), \neg a (i) (j), a (i) (j), O (N^{2}),

(1)

where the fictitious value at column

- 1

is zero. The table itself uses

O (N^{2})

storage.

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Past exam of the year 1 of the computer science course of the University of Oxford / 2014 / Imperative programming / 2 / b / Solution 2026-10-31

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Fix the top-right corner

(i, j)

and a bottom boundary

k

. A width

w

is feasible precisely when every row

m \in [i, k)

has at least

w

consecutive true entries ending at column

j

. Thus the largest feasible width is the minimum of those row run lengths, and the area is that width times

k - i

def maxRectangleAt(i: Int, j: Int): Int = {
  var k = i
  var width = j + 1
  var best = 0
  while (k < N) {
    width = math.min(width, rowCount(k)(j))
    k += 1
    best = math.max(best, (k - i) * width)
  }
  best
}

At the loop head, the loop invariant is

i \leq k \leq N

width best = min ({j + 1} \cup {rowCount (m) (j) : i \leq m < k}), = max ({0} \cup {(h - i) i \leq m < h min rowCount (m) (j) : i < h \leq k}) .

(1)

The initial cap

j + 1

makes the empty-prefix case well-defined and cannot restrict any actual run. Each iteration incorporates one new row and tests its corresponding bottom boundary. The loop variant is

N - k

, so

best = i < k \leq N max (k - i) i \leq m < k min rowCount (m) (j), O (N) .

(2)

A zero width contributes area zero, covering the case where no nonempty all-true rectangle starts at that corner.

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Row run length 2026-10-31

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The number of consecutive true entries ending at a specified column of one row of a Boolean matrix. It is zero at a false entry and otherwise one plus the previous column's row run length, taking the value before column zero to be zero.

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