Past exam of the computer science course of the University of Oxford / 2015 / Object oriented programming 2026-10-31
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The solutions use the original seven-page paper, including its corrected Question 3.[ref]
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Past exam of the computer science course of the University of Oxford / 2015 / Object oriented programming / 1 / e / Solution 2026-10-31
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Use the weekday number as an array index. Before adding dates, add one empty entry to each weekday before start; otherwise the first date in those columns would appear a week too early. For day , the slot number is , whose remainder modulo 7 selects the column. The source's example makes clear that start counts empty weekday cells, rather than individual printed space characters.[ref]
With import java.util.*;, the remaining Java code is:
final class Column extends ArrayList<Data> {
    private static final long serialVersionUID = 1L;
}

static String month(int start, int numDays) {
    if (start < 0 || start >= 7 || numDays < 1 || numDays > 31)
        throw new IllegalArgumentException("invalid month parameters");
    String[] days = {"Su", "Mo", "Tu", "We", "Th", "Fr", "Sa"};
    int ncols = days.length;
    List<Column> output = new ArrayList<Column>(ncols);
    for (int c = 0; c < ncols; ++c) {
        Column column = new Column();
        column.add(new StringData(days[c], true));
        output.add(column);
    }
    for (int c = 0; c < start; ++c)
        output.get(c).add(new StringData("", true));
    for (int d = 1; d <= numDays; ++d) {
        int c = (start + d - 1) % ncols;
        output.get(c).add(new IntegerData(d, false));
    }
    return printData(output);
}
The two-character headers force every column to width two, and right justification aligns one-digit dates with the units digits of two-digit dates. Date appears in data row , below the header. For month(1,30) the visible result is
Su Mo Tu We Th Fr Sa
    1  2  3  4  5  6
 7  8  9 10 11 12 13
14 15 16 17 18 19 20
21 22 23 24 25 26 27
28 29 30
The last physical line may also carry spaces for its empty trailing cells. Every physical line ends with a newline. In particular,
(1)
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Past exam of the computer science course of the University of Oxford / 2015 / Object oriented programming / 2 / f / Solution 2026-10-31
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For each sample, include both its marker and its bar. A bar's lower vertical limit is , including negative bars below the horizontal axis. Its horizontal half-width is , rather than . Its marker extends by in both directions. Therefore that sample's axis-aligned bounding box is
(1)
Taking successive bounding unions includes every sample. Finally unite with to include the stipulated axes. The printed upper-corner formula has one extra closing parenthesis; both the lower-corner formula and the picture confirm the coordinatewise interpretation used here.[ref]
The following commented Java object method may be placed in a graph-view utility software class. It uses Rect from part (a) and one iterator snapshot from part (e).
static Rect barBounds(DataList data, double C, double M) {
    if (!Double.isFinite(C) || !Double.isFinite(M) || C <= 0 || M <= 0)
        throw new IllegalArgumentException("positive finite margins required");
    List<Point> p = new ArrayList<Point>();
    for (Point point : data) p.add(point);
    int n = p.size();
    if (n < 2) throw new IllegalArgumentException("at least two points required");

    // This central square supplies the required minimum axis extents.
    Rect bounds = new Rect(new Point(-M, -M), new Point(M, M));
    for (int i = 0; i < n; ++i) {
        double x = p.get(i).getX(), y = p.get(i).getY();
        double width;
        if (i == 0) {
            width = p.get(1).getX() - x;
        } else if (i == n - 1) {
            width = x - p.get(i - 1).getX();
        } else {
            width = Math.min(x - p.get(i - 1).getX(),
                             p.get(i + 1).getX() - x);
        }
        // Include whichever is wider: marker or half of the bar.
        double half = Math.max(C, width / 2.0);
        // The bar reaches the baseline; the marker reaches y +/- C.
        Rect item = new Rect(new Point(x - half, Math.min(0.0, y - C)),
                             new Point(x + half, Math.max(0.0, y + C)));
        bounds = bounds.Union(item);
    }
    return bounds;
}
This uses the usual assumption that coordinate differences and expanded bounds are representable as finite double values. Rect rejects overflowed bounds; numerical range extension is not part of the requested geometry. As with all floating-point arithmetic, the calculation approximates the corresponding exact real number formulas.
Every side produced by the coordinatewise extrema is required by some bar, marker enclosure or axis extent. The result is therefore the smallest guaranteed rectangle for the specified marker enclosures and axes. If denote the graph-only bounds, the final answer is
(2)
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Past exam of the computer science course of the University of Oxford / 2015 / Object oriented programming / 3 / c / Solution 2026-10-31
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For an integer cell coordinate, occupancy is an existential predicate:
(1)
The first occupied unit cell of brick is at , and its last is at . Thus the required support predicate, with the baseplate explicitly included, is
(2)
For width one, the two support cells coincide. For a wider brick, its middle may bridge an empty gap; the question only requires support beneath its two ends. Checking would incorrectly examine a cell beyond the brick's right end. The potentially infinite baseplate supplies every required support at height zero.[ref]
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