pdqContext

pub fn pdqContext(a: usize, b: usize, context: anytype) void {
    // slices of up to this length get sorted using insertion sort.
    const max_insertion = 24;
    // number of allowed imbalanced partitions before switching to heap sort.
    const max_limit = std.math.floorPowerOfTwo(usize, b - a) + 1;

    // set upper bound on stack memory usage.
    const Range = struct { a: usize, b: usize, limit: usize };
    const stack_size = math.log2(math.maxInt(usize) + 1);
    var stack: [stack_size]Range = undefined;
    var range = Range{ .a = a, .b = b, .limit = max_limit };
    var top: usize = 0;

    while (true) {
        var was_balanced = true;
        var was_partitioned = true;

        while (true) {
            const len = range.b - range.a;

            // very short slices get sorted using insertion sort.
            if (len <= max_insertion) {
                break sort.insertionContext(range.a, range.b, context);
            }

            // if too many bad pivot choices were made, simply fall back to heapsort in order to
            // guarantee O(n*log(n)) worst-case.
            if (range.limit == 0) {
                break sort.heapContext(range.a, range.b, context);
            }

            // if the last partitioning was imbalanced, try breaking patterns in the slice by shuffling
            // some elements around. Hopefully we'll choose a better pivot this time.
            if (!was_balanced) {
                breakPatterns(range.a, range.b, context);
                range.limit -= 1;
            }

            // choose a pivot and try guessing whether the slice is already sorted.
            var pivot: usize = 0;
            var hint = chosePivot(range.a, range.b, &pivot, context);

            if (hint == .decreasing) {
                // The maximum number of swaps was performed, so items are likely
                // in reverse order. Reverse it to make sorting faster.
                reverseRange(range.a, range.b, context);
                pivot = (range.b - 1) - (pivot - range.a);
                hint = .increasing;
            }

            // if the last partitioning was decently balanced and didn't shuffle elements, and if pivot
            // selection predicts the slice is likely already sorted...
            if (was_balanced and was_partitioned and hint == .increasing) {
                // try identifying several out-of-order elements and shifting them to correct
                // positions. If the slice ends up being completely sorted, we're done.
                if (partialInsertionSort(range.a, range.b, context)) break;
            }

            // if the chosen pivot is equal to the predecessor, then it's the smallest element in the
            // slice. Partition the slice into elements equal to and elements greater than the pivot.
            // This case is usually hit when the slice contains many duplicate elements.
            if (range.a > a and !context.lessThan(range.a - 1, pivot)) {
                range.a = partitionEqual(range.a, range.b, pivot, context);
                continue;
            }

            // partition the slice.
            var mid = pivot;
            was_partitioned = partition(range.a, range.b, &mid, context);

            const left_len = mid - range.a;
            const right_len = range.b - mid;
            const balanced_threshold = len / 8;
            if (left_len < right_len) {
                was_balanced = left_len >= balanced_threshold;
                stack[top] = .{ .a = range.a, .b = mid, .limit = range.limit };
                top += 1;
                range.a = mid + 1;
            } else {
                was_balanced = right_len >= balanced_threshold;
                stack[top] = .{ .a = mid + 1, .b = range.b, .limit = range.limit };
                top += 1;
                range.b = mid;
            }
        }

        top = math.sub(usize, top, 1) catch break;
        range = stack[top];
    }
}