fire is 0, salt is 1. 00000 = Pb-Pb 00001 = Cu-Pb 00010 = Sn-Fe 00011 = Ag-Fe 00100 = Pb-Fe 00101 = Cu-Fe 00110 = Sn-Ag 00111 = Ag-Ag 01000 = Pb-Sn 01001 = Cu-Sn 01010 = Sn-Cu 01011 = Ag-Cu 01100 = Pb-Cu 01101 = Cu-Cu 01110 = Sn-Au 01111 = Ag-Au 10000 = Pb-Ag 10001 = Cu-Ag 10010 = Fe-Pb 10011 = Au-Pb 10100 = Sn-Pb 10101 = Ag-Pb 10110 = Fe-Fe 10111 = Au-Fe 11000 = Pb-Au 11001 = Cu-Au 11010 = Fe-Sn 11011 = Au-Sn 11100 = Sn-Sn 11101 = Ag-Sn 11110 = Fe-Cu 11111 = Au-Cu oh wait the computation mod does it for you? well that was a waste of time. anyway, notes below: 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 proud of what i’ve made, especially when i compare it to the mess that was my previous computation solution. i can feel how much i’ve improved since week one. huge thanks to haxton for a great tourney and also to the whole community for being amazing, friendly people. i couldn’t have asked for a better experience as a newcomer. good luck to kazyan and sergyd fighting for the top spot, congrats to biggie for winning week 8, and see you all in the weeklies! explanation below. i’d tell you not to read it but i know you will anyway. my encoding is similar but not identical to a binary to base 6 conversion. the first thing to note is that i decided to make salt 1 and fire 0, as this made placing disposal easier. the first bit on the right has a value of 18 rather than 16; this is 3*6 making it easier to add, and we are afforded this leeway by the 4 spare outputs. the remaining bits from right to left are valued 8, 2, 1, and 4. this ordering is important. each time a salt is read at the bottom, it is used to wand over quicksilver to add its bit value to the base 6 number represented by metals on the left. in base 6, the values of the bits in order are 30, 12, 2, 1, and 4. they are ordered like this so that the first four bits are guaranteed to require no carrying. when the 4 bit is read, after each quicksilver is placed on the projector we check if it’s still there. if it’s gone, we move whatever metal was on the projector onto the lead input before putting it back to where it was, but if it’s still there, we have gold in the lower digit and we must carry. to do so, the quicksilver is used to wand away the gold before instead being applied to the upper digit. as the gold has been wanded away, there is nothing to suppress the lead input, so a lead will be grabbed, resetting the lower digit. after doing this four times, the string has been encoded. all that’s left is to bond them together and output. if we made a waste gold, it’ll go into a waste chain because it’s not worth the area to send it to disposal. oh yeah there’s also a salt input just chilling there i guess.