===== 'I HATE THE NUMBER 4 EVEN MORE' =====
 0.	🜔🜔🜔🜔🜔	 => 	Gold-Copper
 1.	🜔🜔🜔🜔🜂	 => 	Silver-Tin
 2.	🜔🜔🜔🜂🜔	 => 	Iron-Copper
 3.	🜔🜔🜔🜂🜂	 => 	Tin-Tin
 4.	🜔🜔🜂🜔🜔	 => 	Gold-Tin
 5.	🜔🜔🜂🜔🜂	 => 	Copper-Gold
 6.	🜔🜔🜂🜂🜔	 => 	Iron-Tin
 7.	🜔🜔🜂🜂🜂	 => 	Lead-Gold
 8.	🜔🜂🜔🜔🜔	 => 	Gold-Iron
 9.	🜔🜂🜔🜔🜂	 => 	Silver-Lead
10.	🜔🜂🜔🜂🜔	 => 	Iron-Iron
11.	🜔🜂🜔🜂🜂	 => 	Tin-Lead
12.	🜔🜂🜂🜔🜔	 => 	Gold-Lead
13.	🜔🜂🜂🜔🜂	 => 	Copper-Silver
14.	🜔🜂🜂🜂🜔	 => 	Iron-Lead
15.	🜔🜂🜂🜂🜂	 => 	Lead-Silver
16.	🜂🜔🜔🜔🜔	 => 	Silver-Gold
17.	🜂🜔🜔🜔🜂	 => 	Copper-Copper
18.	🜂🜔🜔🜂🜔	 => 	Tin-Gold
19.	🜂🜔🜔🜂🜂	 => 	Lead-Copper
20.	🜂🜔🜂🜔🜔	 => 	Silver-Copper
21.	🜂🜔🜂🜔🜂	 => 	Copper-Tin
22.	🜂🜔🜂🜂🜔	 => 	Tin-Copper
23.	🜂🜔🜂🜂🜂	 => 	Lead-Tin
24.	🜂🜂🜔🜔🜔	 => 	Silver-Silver
25.	🜂🜂🜔🜔🜂	 => 	Copper-Iron
26.	🜂🜂🜔🜂🜔	 => 	Tin-Silver
27.	🜂🜂🜔🜂🜂	 => 	Lead-Iron
28.	🜂🜂🜂🜔🜔	 => 	Silver-Iron
29.	🜂🜂🜂🜔🜂	 => 	Copper-Lead
30.	🜂🜂🜂🜂🜔	 => 	Tin-Iron
31.	🜂🜂🜂🜂🜂	 => 	Lead-Lead

“i believe the best encoding is the friends we made along the way” - bist

due to recent happenings in the community that i’m sure you all are aware of, i didn’t spend as much time on this puzzle as i wanted to (sorry haxton). still, i’m quite proud of what i’ve made, especially when i compare it to the mess that was my week 4 solution. i can feel just how much i’ve improved since week 1. huge thanks to haxton for a great tourney and also to the whole community for just being such amazing, friendly people. i couldn’t have asked for a better experience as a newcomer. good luck to kazyan and sergyd as they battle for the top spot, and see you all in the weeklies (hopefully)!

explanation below. i’d tell you not to read it but i know you will anyway.

here are my encoding details: the first thing to note is that i decided to make salt 1 and fire 0, as this made disposal easier. the values of each bit, from right to left, are 8, 18, 2, 1, 4. 18 is used instead of 16 as it is easier to add in base 6, and the 4 extra output cases allow us this leeway. the ordering of these bits is important for later.

at the bottom, an arm breaks off one atom at a time from the input, starting from the right. it is read by attempting to triplex it, and if the triplex succeeds, it is wanded away to be disposed. when a salt is read, it is used to wand over quicksilver to add to the base 6 number represented by metals on the left. in base 6, these are the values of the bits in the order they are read: 12, 30, 2, 1, and 4. they are ordered like this so that the first four bits are guaranteed to require no carrying. the 12 (8) bit is placed first because it needs both digits to be incremented, so we can place 2 quicksilver before the lead is ready, reducing the machine’s period. despite this, some cases still collided, so a second arm takes the first quicksilver from the 12 bit and holds it until it can be safely used.

now comes the reason for the solution title. when the 4 bit is added, we may have to perform a carry operation. after each quicksilver is placed on the projector, we check if it’s still there. if it’s gone, we use whatever metal was on the projector to suppress the lead input before putting it back to where it was. (i learned that trick from week 4!) but if it’s still there, we must carry. to do so, the quicksilver is used to wand away the gold before being used on the upper digit. this time there is nothing to suppress the lead input, so a lead will be grabbed, resetting the lower digit and completing the carry operation. after doing this four times, we can bond the metals together and output. if we made a waste gold, it goes into a waste chain because bonders are cheaper than arms. fortunately i managed to tuck the waste into the area used by the disposal arm, so it doesn’t cost extra area.