Hellish Bricks DevBlog 2

This is an update on what I have recently done to Hellish Bricks.

Progression Curves

The game got progression curves for running speed and initial jump speed.

Speed Curve

Jump Speed Curve

These are important so that the character evolves, creating more gameplay possibilities as the game progresses.

Colored Lights

I’ve also enabled lantern and ambient light commands so that the player can change the lantern and the ambient light colors during the game or during initialization, through the boot script.

Default Boot Script

# Set lantern to a light yellow.
lantern #EEEEBB
# Set ambient to a dark red.
ambient #443322

White and Colored Lights Comparison

White Lights

Colored Lights

Hellish Bricks DevBlog 1

This is the first of what may end up being a series of posts regarding the development of my closed-source game Hellish Bricks.

The game is heavily inspired on Devil Daggers (2016), but intends to focus on low-poly and minimalist graphics instead of retro ones.

Hellish Bricks is written using modern C++ and requires at least OpenGL 3.3 and just about 128 MiB of free memory for now.

I intend to write better and more visually appealing shaders in the future and start working on some fancier graphical effects, such as particles and shadows.

Gameplay Video

Playable Demo

You need at least Windows Vista to run the game and you may also need to download and install the Microsoft Visual C++ Redistributable for Visual Studio 2017 before being able to execute the game.

New versions of Wine are also able to run the abovementioned executable.

In the future I may release Linux, BSD and maybe even macOS demo builds.

Bitwise And of Range

Recently I had to solve a problem which asked you to determine the bitwise and of a range of nonnegative numbers. There is an obvious linear solution to this problem which simply computes the bitwise and of the range.

Bitwise and of [4, 10] = 4 & 5 & 6 & 7 & 8 & 9 & 10

However, after thinking about how the anding ends up “erasing” bits permanently I figured out the following logarithmic solution:

def bitwise_and_of_range(begin, end):
    if begin == end:
        return begin
        return bitwise_and_of_range(begin >> 1, end >> 1) << 1

Essentially, if you have at least two numbers in the range, the last bit will be zero, so you can compute the bitwise and of the prefixes and append a zero to the result (by shifting it to the left).

Sliding Maximum

In this post I will present a simple and somewhat generic solution for the sliding maximum problem.

Find the maximum element for every K consecutive elements.

Note that the sliding minimum can be seen as a negated sliding maximum problem, just like the maximum spanning tree can be seen as the minimum spanning tree of the negated graph.


Below is a generic C++ sliding window I implemented that takes a comparator as a template parameter. This allows it to be instantiated for both the sliding maximum and the sliding minimum.

template <typename T, typename Comparator>
class SlidingWindow {
  struct Block {
    Block(T v, size_t w) : value(v), width(w) {}
    T value;
    size_t width;
  Comparator comp;
  deque<Block> data;

  void push(const T t) {
    size_t width = 1;
    while (!data.empty() && comp(data.back().value, t)) {
    data.emplace_back(t, width);
  T get() const { return data.front().value; }
  void pop() {
    // Either reduce the width of the best block (front), or drop it.
    if (data.empty()) {
    if (data.front().width > 1) {
    } else {

This solution is amortized \(O(1)\) for all operations, making it \(O(N)\) for \(N\) elements. By using the standard library trees directly we cannot do better than \(O(N \log N)\).

Sliding Maximum Example

Using 20 terms and a window of width 4, we have the following table:

Current Maximum Minimum
16 16 16
1 16 1
2 16 1
13 16 1
9 13 1
11 13 2
15 15 9
14 15 9
4 14 4
8 8 4
3 8 3
7 7 3
6 7 3
12 12 6
5 12 5
18 18 5
19 19 5
17 19 17
20 20 17
10 20 10


This post is an update on a small side project I just started. It is called BigWord and it aims to find the biggest word in a dictionary based on a multiset of letters.

Sample Usage

For instance, it finds out that the biggest English words made with the letters from “linuxwars” are “urinals” and “insular” and that the biggest English words made with “applepies” are “pappies” and “applies”.

$ ./bigword linuxwars
> urinals
> insular
$ ./bigword applepies
> pappies
> applies

Implementation Notes

The existing code is written using C++14 and should run under any system which has reasonable C++14 compiler. The program relies on the public domain word list shipped with Fedora, but you may use any plain text dictionary you want.

It loads all words which may be constructed with the provided letter multiset, ignoring words which are too big in order to improve performance, and then traverses the word vector from longest to shortest trying to match as many as possible of the biggest subsets of the provided multiset.

Even though it is performant enough for a fairly big word list, I think there must be a more efficient way to find the biggest subsets. Currently, the first abovementioned example finishes after 200 ms on a list of 479,000 words.

Ideally, I would like even the worst-case queries to finish in less than 100 ms for word lists with less than one million words, as suggested in Usability Engineering by Jakob Nielsen.

Some possible improvements are dumping the processed word vector after the first execution and reusing it in the next executions and sorting the input text file by word length so IO can be stopped earlier, but these possibilities add undesired complexity to the setup the user would need.

The following code snippet shows how letter counts are currently compared.

  static bool is_contained(const LetterCount &a, const LetterCount &b) {
    if (a.letter_count > b.letter_count) {
      return false;
    size_t remaining = b.letter_count - a.letter_count;
    for (size_t i = 0; i < alphabet_size; i++) {
      if (a.counters[i] > b.counters[i]) {
        return false;
      // By catching excessive differences we may fail early.
      const size_t difference = b.counters[i] - a.counters[i];
      if (difference > remaining) {
        return false;
      remaining -= difference;
    return true;

There may be a faster way to do it, but I am failing to spot it.

You can find the project source code (licensed under the ISC license) at the GitHub repository.