Assignment 6 · Campus Grid — 2-D Arrays

Module 10 (L19–L20) · 40 marks · due end of Week 11 Outcomes: CLO-6 (PF-10.x)

A campus map is an R×C grid of zone codes (integers 0..9). You will compute map statistics with correct 2-D traversal.

Input

R C (1 ≤ R, C ≤ 20), then R rows × C columns of zone codes.

Report (exact labels, in order)

sum     : S           (all cells)
maxrow  : i v         (row index with the largest row total, and that total; ties → lowest index)
diagonal: D           (sum of g[i][i] for i < min(R,C))
border  : B           (sum of all edge cells, corners counted once)
zeros   : Z           (count of cells equal to 0)

Sample run

Input 3 4 / 1 2 3 4 / 5 0 0 2 / 9 1 1 1 →

sum     : 29
maxrow  : 2 12
diagonal: 5
border  : 27
zeros   : 2

Check: row totals 10 / 7 / 12 → row 2 wins; diagonal = 1 + 0 + 1 = 2 … careful: min(R,C) = 3 → g[0][0]+g[1][1]+g[2][2] = 1+0+1 = 2. If your program prints 5 you summed the wrong diagonal direction. The expected value is 2 — fix the sample reasoning before coding.

Required functions

int totalSum(const int g[][20], int r, int c);
void maxRow(const int g[][20], int r, int c, int& idx, int& tot);
int diagonalSum(const int g[][20], int r, int c); — main-diagonal only, i < min(r,c).
int borderSum(const int g[][20], int r, int c); — single pass, predicate form.
int countZeros(const int g[][20], int r, int c);

Constraints

Column dimension must be literal 20 in signatures (per the course rule).
borderSum must be one pass with the edge predicate — no four separate loops.
No vectors; fixed int grid[20][20].

Deliverables

campus.cpp (contract, zero warnings) · test_table.md — 1×1 grid, 1×C and R×1 (whole grid is border!), diagonal on non-square grid, all-zero grid, tie between two rows · postmortem.md (10 marks).

Note

The worked sample above contains one deliberately wrong intermediate claim (diagonal 5) — your test table must list 2 as the expected diagonal with your own hand-check. Reading specs critically is part of the assignment.