So, currently the Illuminate\Pipeline\Pipeline is used for Route Middleware and Job Middleware. In both cases, the Pipeline is used once: you new up an Instance and pass it the Request or Job, then use the result. It is clearly optimized for that use case.

I have a use-case in which I would like to call a single Pipeline (consisting of the same pipes) over and over again. When I used the Pipeline as is, I was surprised to learn that each time I run the pipeline with ->then(), I would get new instances of my Middleware. An example:

$pipeline = resolve(Pipeline::class)
    ->through($pipes);

$handle = fopen('logs.txt', 'r');
while ($line = fgets($handle)) {
    $result = $pipeline
        ->send($line)
        ->then(fn($line) => $line);
    // ...
}

This will create new instances of all $pipes for each line in the file. This is undesirable if you want to do something with the lines based on a state of the pipe. As a silly example:

class LogRowCounter
{
    protected $counter = 0;

    public function handle($line, $next)
    {
        $this->counter++;

        return $next($this->counter . ' - ' . $line);
    }
}

Apart from resolving all the $pipes, a lot of checks are also done while running the pipeline, meaning it happens in the while loop of my example. Some of those things could be done beforehand.

Reusable Pipeline

To make the Pipeline reusable, I would propose a new method called prepare(). (I was also thinking about compile(), but ‘prepare’ is used in some protected methods of Pipeline already.) It would extract running the array_reduce() from then() and return the resulting closure, and then() can call it.

    public function prepare(Closure $destination)
    {
        $pipeline = array_reduce(
            array_reverse($this->pipes()), $this->carry(), $this->prepareDestination($destination)
        );

        return $pipeline;
    }

    public function then(Closure $destination)
    {
        $pipeline = $this->prepare($destination);

        return $pipeline($this->passable);
    }

This would still resolve all the pipes on ‘runtime’ though, because of how the closures are structured. Here is a new version of carry(), which resolves the pipes outside of the returned closure. The outer closure returned is meant for array_reduce(), like the current version; didn't want to refactor that.

    protected function carry()
    {
        return function ($stack, $pipe) {
            if (is_string($pipe)) {
                [$name, $parameters] = $this->parsePipeString($pipe);

                $pipe = $this->getContainer()->make($name);
            } else {
                $parameters = [];
            }

            return function ($passable) use ($stack, $pipe, $parameters) {
                try {
                    if (is_callable($pipe)) {
                        return $pipe($passable, $stack);
                    }

                    $parameters = array_merge([$passable, $stack], $parameters);

                    $carry = method_exists($pipe, $this->method)
                                    ? $pipe->{$this->method}(...$parameters)
                                    : $pipe(...$parameters);

                    return $this->handleCarry($carry);
                } catch (Exception $e) {
                    return $this->handleException($passable, $e);
                } catch (Throwable $e) {
                    return $this->handleException($passable, new FatalThrowableError($e));
                }
            };
        };
    }

All Pipeline tests pass, and the following new test would pass too:

    public function testPipelinePrepared()
    {
        $pipeline = (new Pipeline(new Container))
                    ->through([
                        PipelineTestPipeOne::class,
                        PipelineTestStatefulPipe::class
                    ])
                    ->prepare(function ($piped) {
                        return $piped;
                    });

        $result = $pipeline('foo');

        $this->assertSame('foo', $result);
        $this->assertSame('foo', $_SERVER['__test.pipe.one']);
        $this->assertSame([], $_SERVER['__test.pipe.state']);

        $result = $pipeline('bar');

        $this->assertSame('bar', $result);
        $this->assertSame('bar', $_SERVER['__test.pipe.one']);
        $this->assertSame([0, 1], $_SERVER['__test.pipe.state']);

        unset($_SERVER['__test.pipe.one']);
        unset($_SERVER['__test.pipe.state']);
    }
}

class PipelineTestStatefulPipe
{
    protected $count = 0;

    public function handle($piped, $next)
    {
        $_SERVER['__test.pipe.state'][] = $this->count++;

        return $next($piped);
    }
}

Things to note

First off: I’ve taken the liberty to rename the ! is_object($pipe) check to is_string($pipe), because the very next thing we do is call $this->parsePipeString($pipe) with it.

Because of the way the closure is bound, the following code would work, and it seems like I can put it to use, but it looks wonky and weird too:

$filename = 'logs.txt';
$pipeline = resolve(Pipeline::class)
    ->through($pipes);

$run = $pipeline->prepare(fn($line) => $line);

// Call all `init($filename, $next)` methods
$pipeline->via('init');
$run($filename);

// Set method back to 'handle'
$pipeline->via('handle');

$handle = fopen($filename, 'r');
while ($line = fgets($handle)) {
    // Call all `handle($line, $next)` methods
    $result = $run($line);

    // ...
}

… that syntax makes me feel like Pipeline should store the Closure as a property somewhere, so prepare() returns $this, and then add an __invoke() or run() to Pipeline, so you can at least use the same variable instead of $pipeline and $run.

Alternative solutions

While we could do the above, I also have a different solution: singletons in the Container.

I originally rejected this (hence the work above) because I don’t want my stateful pipes to be global singletons. I want the pipes to belong to the Pipeline: if the Pipeline goes, so go the references to the pipes.

Think for example of the above file-handling example, but as a queued Job: I don’t want the state of the pipes from one run to be around when I process the next log-file.

But: this last point can also be resolved by creating a new Container instance per Job, and giving that to the Pipeline while constructing it. You can bind singletons into that separate container, which will belong to the pipeline.

Note that in that case – if we don’t change the Pipeline class – you’re still making the calls to the Container on ‘runtime’, which is not optimal. On the other hand, it might not be a very big deal.

I don't really like the way I have to new-up or resolve the Pipeline in this case, but it is doable.

---

I’d be happy to work this into a PR, but I thought this might be a better start, especially since I’m still debating whether or not I got a point here.

Thanks for reading, I’m curious about your thoughts.

tl;dr: This is nerding out on Tetris scores, and quite frankly, I don't think the conclusion holds. But there are nice tables along the way.

I've been watching Classic Tetris for a few months now, and as a result, I'm also playing it. I am noticeably better after I watched some pro's scoring Tetris after Tetris at level 19, so I keep this pace of watching, playing, watching playing.

There is a problem though: at first I deemed playing at level 18 myself impossible, it's just too fast. Then I noticed how my mind could keep up with the players on the screen, but not with the pieces falling on my screen. Of course, I need more practice, but I have an excuse: I live in Europe.

Turns out old European TVs have a different refresh-rate than Americans. Therefore Tetris on the European PAL systems gets faster in earlier levels than the American NTSC systems. There is a nice article by Tetris Finland about this and other differences between the two versions.

Despite the faster speeds of the PAL version, I found no evidence of the scoring system being different. This means the PAL version is just harder, you just get less points for your hard work. (Yea, I will come back at this.)

I am playing Tetris on a real NES, brought ±15 years ago on a flee market at the local community playground for the insane price of 10 euros. The cartridge of Tetris has cost me more. But to compare myself with the pro's on screen I would have to spend a bit more to import an American NES.

Isn't there a way to calculate the NTSC score based on a PAL score?

Yea, is there a way to calculate NTSC score?

My current, not too impressive high-score on PAL is 174_731 (and yes I will use underscores here). How much is that worth on NTSC?

My first attempt was to look at the maximum scores. Players have been known to 'max out' the game on NTSC. A max out score is a score that is higher than 999_999, at which point the score will freeze on those six nines.

On the PAL version, however, we still have a world record, which is currently set to 758_360 by Joseph Saelee. This video is also a demonstration of how much harder the PAL game gets. (Although Joseph has a talent of making Tetris look easy on all versions.)

With some tweaks in the software you can get Tetris to report past 999_999, so higher scores are known. But for simplicity, let's take the world record and equate it with a NTSC max out score. This gives us ntsc = pal / 758_360 * 999_999, which will give me a NTSC score of 230_406.

But is that accurate?

It's always comparing apples with oranges, but I feel like this formula is a bit too loose. Let's bring in the information of the aforementioned article of Tetris Finland, which gives a nice table that compares the different frame rates, and calculates per level the time it takes, in seconds, to drop 20 spaces. Here is a table with that data:

Level	NTSC	PAL
9	1.996 sec	1.999 sec
10-12	1.663 sec	1.599 sec
13-15	1.331 sec	1.199 sec
16-18	0.998 sec	0.7998 sec
19-28	0.6655 sec	0.39994 sec
29+	0.33278 sec	as above

Now, the scoring stays the same in both versions, but changes per level. Let's bring that down to a single number per level, let's agree on a perfect score.

The best way of scoring points is to only score tetrises (clear 4 lines at once). Each tetromino is made up of 4 blocks, and the board is 10 blocks wide. A perfect score would be if the Random Number Generator (RNG) gives us only lines and we can stand them up over the full width of the board, resulting in a tetris and an all-clear board each 10 pieces. This is not really achievable, but this would be the best score possible.

According to the Tetris Wiki we get 1200 * (level + 1) for each tetris, and scores are calculated based on the level after the line clear. Let's ignore push-down points. Let's also say that if the drop speed goes below 0.5 sec for 20 spaces the game is over (this makes levels 29 for NTSC and 19 for PAL the kill screens).

Then there is the concept op the start level. When you start at level 1, you go up to level 2 after 10 lines. But when you start at level 18, you get 130 lines before going to level 19, after which you get a new level ever 10 lines again. The Tetris Wiki gives the formula of "(startLevel × 10 + 10) or max(100, (startLevel × 10 - 50)) lines, whatever comes first". This means 100 lines for levels 9-15, then 110, 120 and 130 for 16, 17 and 18.

So I wrote this odd-looking piece of Javascript (because it's my day off and I can do whatever) to list out the various perfect scores for the various level-starts.

const killscreen = 19 // or 29

tetrisScore = level => 1200 * (level + 1)

startlevelLines = level => Math.min(
  (level * 10 + 10),
  Math.max(100, (level * 10 - 50))
)

scoreForLevel = (level, lines) => ({
  score: (Math.floor(lines / 4) * tetrisScore(level)),
  restlines: lines % 4
})

restlevelScore = (level, rest) => {
  if (level >= killscreen) return tetrisScore(level)
  const {score, restlines} = scoreForLevel(level, 10 + rest)
  return score + restlevelScore(level + 1, restlines)
}

perfectScore = level => {
  const {score, restlines} = scoreForLevel(level, startlevelLines(level))
  return score + restlevelScore(level + 1, restlines)
}

const levels = [...Array(killscreen).keys()]
levels.map(level => console.log(`| ${level} | ${perfectScore(level)}`))

I was kind of surprised by the results: as you can see, the world record by Joseph is higher than the perfect score. This is because he had a 100% tetris rate, burned one line in level 18 for 760 points, then tetrised into level 19, and then made a few singles and doubles in level 19. My definition of 'kill screen' seems too strict, but at least it's even on both sides of the table.

Start level	NTSC perfect	PAL perfect
0	1_332_000	588_000
1	1_334_400	590_400
2	1_340_400	596_400
3	1_348_800	604_800
4	1_360_800	616_800
5	1_375_200	631_200
6	1_393_200	649_200
7	1_413_600	669_600
8	1_437_600	693_600
9	1_464_000	720_000
10	1_478_400	728_400
11	1_454_400	710_400
12	1_462_800	712_800
13	1_432_800	688_800
14	1_435_200	685_200
15	1_399_200	655_200
16	1_429_200	685_200
17	1_461_600	717_600
18	1_497_600	753_600
19	1_536_000
20	1_578_000
21	1_622_400
22	1_670_400
23	1_720_800
24	1_774_800
25	1_831_200
26	1_891_200
27	1_953_600
28	2_019_600

Given that my high score was a level-5-start score of 174_731, I should use ntsc = pal / 631_200 * 1_375_200, giving me a NTSC score of 380_687, a lot more than my previously calculated 230_406. But is this fair now?

Score per second

The table above is not by any means a fair comparison. It does account for the higher scoring potential in NTSC, but it still compares levels regardless of the speed they are played at. Let's combine the previous two tables into one.

So in the PAL version, you get less points, but also less time to think about how to score them. In that regard, the PAL version is harder, and the points should be worth more.

In the next table, I've taken the perfect score per level, and divided it by the thinking time you have for it (the seconds of the earlier table times 240, the number of lines before transition you get when you start in level 29). The result is a series of numbers that say how 'easy' it was to get that number of points.

Start level	NTSC score	sec / 20 lines	modifier
9	1_464_000	1.996	3.0561122244
10	1_478_400	1.663	3.7041491281
11	1_454_400	1.663	3.6440168370
12	1_462_800	1.663	3.6650631389
13	1_432_800	1.331	4.4853493614
14	1_435_200	1.331	4.4928625094
15	1_399_200	1.331	4.3801652893
16	1_429_200	0.998	5.9669338677
17	1_461_600	0.998	6.1022044088
18	1_497_600	0.998	6.2525050100
19	1_536_000	0.665	9.6240601504
20	1_578_000	0.665	9.8872180451
21	1_622_400	0.665	10.1654135338
22	1_670_400	0.665	10.4661654135
23	1_720_800	0.665	10.7819548872
24	1_774_800	0.665	11.1203007519
25	1_831_200	0.665	11.4736842105
26	1_891_200	0.665	11.8496240602
27	1_953_600	0.665	12.2406015038
28	2_019_600	0.665	12.6541353383

Start level	PAL score	sec / 20 lines	modifier
9	720_000	1.999	1.5007503752
10	728_400	1.559	1.9467607441
11	710_400	1.559	1.8986529827
12	712_800	1.559	1.9050673509
13	688_800	1.199	2.3936613845
14	685_200	1.199	2.3811509591
15	655_200	1.199	2.2768974145
16	685_200	0.799	3.5732165207
17	717_600	0.799	3.7421777222
18	753_600	0.799	3.9299123905

This would say an NTSC max-out on a level 18 start is 999_999 / 6.25250501, so has a difficulty of 159_935. The world record on PAL is 758_360 / 3.9299123905, so has a difficulty of 192_971. To me this feels believable, because Joseph has had many beyond-max-out games. We don't know their scores, because the game does not tell us, but their difficulty might also approach 200k.

And my score? Well, as I mentioned somewhere it was a level-5-start score, and as you can see in the first table: there is not much difference between level 9 on PAL and NTSC, and level 5 is not even listed. I should really stop whining and just practice more.

Apples and oranges

If you actually read the article by Tetris Finland you already would know: the speed is not the only difference. There are also differences in the way DAS works in PAL vs NTSC. The article concludes that these games are therefore very different, so different that you might consider them entirely different titles, despite looking the same.

It was fun figuring out these numbers and formula's, but in the end, you really can't compare the two. Gotta fix myself an NTSC somewhere if I want to compare my progress to the pro's.

Note that this difficulty number is a thing I invented in this post, and it is in itself not comparable to Tetris scores. But you can compare it to other difficulty numbers, if you calculate them the same way. Actually, I don't really think this whole thing is reliable at all, but I didn't want to throw a whole morning of calculating and writing out of the window. Some of the tables bear more value than others.

So, I've been addicted to Tetris for the last week or so. Last night I decided to do something else, and when I came across the Phoenix LiveView contest, I thought: why not try and make Tetris? #myelixirstatus

I was surprised by how far I got. I have watched videos of people coding games before, have been coding myself for years and years, but somehow games felt out of my league. But here, after a (long) evening, I got a game! I should've streamed it myself.

I mean I'm standing of shoulders of giants. I got so much for free from Erlang/Elixir, LiveView to hook it up, @chris_mccord's Snake example gave me the basic idea of using just <div>'s for blocks, the browser is doing the drawing and key-repeats for me.

But it felt magical, once the game got playable. I was hooked again to my own creation. And so many "features" from NES Tetris like sliding, tucking and spinning "just worked" in my first implementation (probably because I was close to their implementation).

It all started with this HTML based board, using flexbox to to the hard works of blocks for me. I later wrapped the board in a Game-struct, and added bindings for the scores.

How to get a game loop in Elixir? Just send yourself a message after, say, 500ms. The LiveView component has a handle_info/2, which queues the next tick and moves the current Tetromino down.

Thanks to the phx-keydown="keydown" in the template, we get messages in handle_event/3 for each key. Just delegating to my Game module. And yes, I use Vim, so I need those H, J and L. (Dropping with K is not implemented.)

My Game.move/2 got a bit complicated, some refactoring is in place. But I started with a Tetromino struct, with :x and :y keys and a :color. To move it down, you paint the current :x and :y :white, and then paint :y + 1 to the :color. Code below reduces for all four points:

To obtain all the points I got this wonderful Tetromino.points/1 function. It's ugly and was a pain to write out and get right, but it works like a charm.

Colors and rotations work similar, with Tetromino.rotate/1. (color_for_type/1 gets called in new/1, and it never changes once it is created.)

Then the messy Game.move/1: can you move down? Move down. Otherwise, call Board.clear_lines/1 to remove full lines and obtain the score, get a new random Tetromino and put it on the board. The game-over handling is a bit buggy still, I had only one evening.

Clearing the board is simple: reject rows of which all cells are not empty and use counts to add new rows to the top. The Board.color_at/2 gives the color for that particular cell. The guards against negative numbers prevent List.pop_at/2 to get items from the back of the list :)

I think that's mostly it! Again, I wish I had streamed it, it was very weird watching myself doing it, but also very rewarding. Keep coding, people, you can do this too!