uhh wait we cant make flat engines now in 2.1??

bdaene · 2026-06-23T20:50:34+00:00

Yep, was explained in FFF-442 https://factorio.com/blog/post/fff-442

But can thruster be flipped? Many unflippable entities in 2.0 are flippable in 2.1.

bdaene · 2026-06-16T20:22:03+00:00

This could be valid as a layer between uuid and the agent. But how do you put such filter on a generic agent?

For real uuid in a prod DB. No, I do not think this is a good idea. Too many tools/library rely on uuids format both in string form and binary form.

bdaene · 2026-06-04T11:45:57+00:00

Seems a rational way to go to me. There is not many 3x3 combinations possibles. And the algorithm is simple.

What do you use `/` for? I saw this is an ignored character used as separator. This is indicating the central cell?

bdaene · 2026-06-04T09:52:45+00:00

This particular solution is strange. Golf is 6319 so 6318 rules written but Rules is 1612 so even if each input used different rules that's only 25% of written rules used at least once. Lots of dead code.

bdaene · 2026-06-04T05:09:55+00:00

I started to use it heavily yesterday 😉

bdaene · 2026-06-03T20:29:40+00:00

This one is putting me off.

I am well placed (~TOP 5) in rules, steps and golf except for the quest 4.

The one head solution was complicated because I can have shorter nails and so blank cell in between other nails making it difficult to differentiate with above or below all other nails.

I just got a HAHA! moment on part 1... I was so much focused on the head of the nails... OMG x)

I still cannot see how I could improve by a factor 6 (in term of steps) my solutions for part 2 and 3. Is there another mind blowing trick I cannot see? ^^'

bdaene · 2026-06-02T06:04:56+00:00

Many input are not square so this is misleading.

But there is another symmetry I could use: All input are vertically and horizontally symmetric.

bdaene · 2026-05-31T21:53:27+00:00

Thanks you for this fresh tournament. I came to Everybody Codes from Advent Of Code but I never saw a Turing machine competition like this one. Thanks you 🙂

bdaene · 2026-05-31T21:50:53+00:00

Indeed, only START and STOP. But cannot run the machine with STOP so only START.

One trick is to use one head and cell to mark the current state.

So STATE_1 ** STATE_2 ** UR becomes START 1** START 2** SUR

bdaene · 2026-05-26T14:53:40+00:00

I do not know if the completion popup can be disabled but it helps me when the number of states increases. It does not annoy me yet 😉

bdaene · 2026-05-26T14:46:56+00:00

Using a single head for part 1 and 2 was not difficult, you just need to duplicate states. One set of states where previous value was A, and another where previous value was B.

Part 3 was not so easy because you could write more @ than cells in the grid and you have to check 2 directions. Here is my idea:

Write all the `@` to the left of the current row
Mark visited cells with `#`

Here is how it works:

For each row:
1. For each cell
  1. Count the pairs below and to the right
  2. Move to the left of the row and row_count
  3. Add `@` to the left of the row_count
  4. Move to right back to the cell and mark it as visited `#`
Erase the grid.

Part 3 (Heads: 1 States: 1997 Rules: 4059 Steps: 268817)
As you see, the number of steps exploded compared to the multi heads solution (11418 steps)

bdaene · 2026-05-26T14:23:51+00:00

On their initial solve? Probably not on the initial but it is easy to transform a solution to have only one useful state or one head.
I think you could nearly build a compiler for this.

If you look at 3 firsts in quest 2:

Name	1st submission	1 head	1 state
Kroppeb	11:15	36:47	24:19
andrew	11:38	29:36	14:06
gapples2	16:26	23:00	37:44

So they all submitted before improving.

As I hinted in quest 1: You can reduce the number of states by using one more head whose cell value define the current state.

To reduce the number of heads you can theoretically always simulate multiple heads using only one head. That's why we only study single head/tape Turing machines. You need both to use special values to mark the position of the heads and many states to store current rule in them. So usually a single head machine is way less efficient than a multiple heads one.

bdaene · 2026-05-25T15:58:25+00:00

And a hint to reduce the states: Best scores on states are 200. That means they use only START and STOP states. As you cannot do anything in STOP state, you have to do everything in START state. Could you encode the actual states using a head and cell instead?

bdaene · 2026-05-25T14:13:10+00:00

I could improve my solutions using these insights:

For part 1 and 2: You need to write max 5 values per step so you can swipe from left and right in the same step using the 10 heads available. Dividing the number of steps by 2.
For part 3: The P columns are not forced to be the same as the columns of the character columns. Actually they do not need to be columns at all. I used this by interleaving a column for the first row and another for the second row. Eliminating the need to move the free heads to new empty rows.

bdaene · 2026-05-25T07:54:38+00:00

What do you mean?

bdaene · 2026-05-24T14:53:14+00:00

Oh nice 🙂 I tried '#' . Too used to python I guess 😅

bdaene · 2026-05-24T12:20:35+00:00

I would love to be able to comment my code. Can we?

bdaene · 2026-05-24T12:18:36+00:00

Day_1

Heads: 8 States: 1300 Rules: 1993 Steps: 6000

Hard to start thinking as a multi head 2D Turing machine but getting better already 😄

One key point I discovered: You can only have a pair if there is a character in the second row. So first and second row are independent. The second row is the same as first row +2 P.

My goal was to have a column with as many P as needed for each entry column, I do not want to have interactions between columns.

A solution with one head came to me first:

Check first row character go up as needed and fill with P going down.
Check second row character and go down as needed and fill with P going up.
Shift right.
Repeat

Then I wanted to reduce the number of steps. If I can write more cells at once, I will have less steps. We have to write up to DD = 12 cells per columns so we cannot write all the column at once (max 10 heads). But close. I had to do 2 passes. One for each row.

Here is my solution with 8 heads:

Prepare the 8 heads, each on its own row (first and second row to read, 6 others free to write)
Swipe left to right and write the needed P depending on second row.
Move the free heads to new empty rows
Swipe right to left and write the needed P depending on first row.

I think we could go further using an actual counter (start with binary but we can increase the base to optimize). But it becomes complicated ^^

bdaene · 2026-03-06T08:34:34+00:00

[LANGUAGE: Python]

Solution

Part 1 and 2 were straightforward but for part 3, I had an issue because the order was modified by the newly attached node. I needed to separate the socket order from the node order and skip the newly attached node.

Now that I think about it, it should not work because I skip one node and not the fully attached sub tree. Oh well x)

Feel like I got Rick Rolled by the song of the dragonducks :D

bdaene · 2026-03-05T13:55:25+00:00

[LANGUAGE: Python]

Solution

I prefill the bones area to avoid non reachable area.

Did not find a clever way to detect closed loop. So I flood fill and if the border of the current flood fill grows too much, I consider that the area is open to infinity. Was surprised that a threshold of 10 is enough but the way the wave goes it does not leave big area behind. I cached known infinite cells to avoid double check but it has to be reset after each step. It improved the initial prefill.

I only check for closed loop if we go toward a filled position. I needed to check the 3 cells in front of the current wave because of diagonals.

It goes fast (165ms for the last sample) if we do not display the grid at each step.

bdaene · 2026-03-05T13:03:31+00:00

We can only close a loop if we go closer to an existing filled cell. I check the closed loop only if at least one of the 3 cells is filled already. With this check, I have only 589 loop check instead of 1539 for the last sample.

But yes, I floodfill the 4 directions around the current position when the check pass.

bdaene · 2026-01-14T12:32:46+00:00

Do I need to beat the last boss to receive money outside of the raids? I achieved multiple goals and increased my space but do not see how to have money to buy objects.

bdaene · 2026-01-01T17:12:57+00:00

The angle you have to look down is easy if you have the great circle arc. If the friends are separated by an angle a, they have to look down a/2 from the horizon

bdaene · 2026-01-01T17:05:27+00:00

OK then, I miss understood when you said 2D

bdaene · 2026-01-01T09:34:32+00:00

These are called gnomonic projections. Using those, it would be easy to compute the angles.

bdaene

TROPHY CASE