Topic: General Gaming /
The Codex of Alchemical Engineering; you'll waste hours
Wow, I’m really late coming to this party; how did that happen? This game is absolutely brilliant, and I love Lightbot et al, and I took almost three years to come across this game. I guess I mostly browse newer games and missed this one when it first debuted … but anyway.
My approach to optimization was quite different from that in this thread, focusing on execution time and minimum period above all else. A solution that produces output after 70 cycles and every 6 cycles thereafter will complete five copies in 100 cycles, while a solution that produces output after 30 cycles and every 12 cycles thereafter will complete 90 copies in 90 cycles. However, the former solution will prove to be much more efficient the longer the machine runs, were it allowed to do so, and this “long-term” execution time was the basis of my models.
Put differently, I want to use every spawned atom as soon as it is available.
It’s possible and relatively easy to calculate the minimum period for each puzzle. Every Glyph of Spawning produces a new atom every two cycles. If there are two spawning glyphs and the finished object requires six atoms to produce (including any needed for calcination/transmutation/projection), then each glyph will have to produce three atoms for every copy of the compound, meaning the machine at best will produce a completed compound every six cycles (once it’s “primed”).
Compounds which require an odd number of atoms are rather trickier. With two spawning glyphs and a compound which requires five atoms, for example, you can use a period of six and “waste” one atom every six cycles, or you can have period of 10 and produce two different copies in that timeframe. The latter is of course optimal but I ran into a lot of problems with space on the board with most of the actual problems in the game.
Finally, for compounds with different types of spawning glyphs, particularly with metals involved, the minimum period for each ‘type’ of atom are calculated separately (the most obvious example is Gold, which makes use of only one metal atom for every five Mercury atoms, and there’s no way to improve on that)
So, without further adieu, here are some of my solutions, many still being worked on:
(symbols/cycles/period)
Aqua Vitae (16/13/2)
m,3,12,0,3,clor;m,11,12,270,3,olcr;m,3,4,0,3,orcl;m,11,4,90,3,crol;
Sal Ammoniac (40/17/2)
b,6,8;b,7,10;m,6,12,0,3,clor;m,7,12,180,2,crol;m,7,6,90,1,oucd;m,4,4,0,1,orcl;m,4,8,0,1,clor;m,7,13,270,3,cdou;m,12,3,90,3,crol;m,4,9,0,2,orcl;m,8,6,0,3,orcl;m,10,10,180,1,olcr;
Cinnabar (40/17/2)
b,7,11;b,6,9;t,12,8;t,3,5;m,7,14,270,3,cdou;m,7,13,180,2,crol;m,6,13,0,3,clor;m,10,11,180,1,olcr;m,4,10,0,2,orcl;m,7,7,90,1,oucd;m,3,9,0,2,clor;m,3,2,0,2,orcl;m,8,7,0,3,orcl;m,12,3,90,3,crol;
Litharge (84/21/2)
b,4,11;b,9,11;b,1,7;b,12,7;m,9,13,270,1,olcr;m,14,11,180,2,cuod;m,4,7,180,1,olcr;m,0,11,0,3,odcu;m,5,7,90,3,crol;m,2,5,90,1,clor;m,1,11,270,3,owc2;m,4,6,90,3,cdou;m,10,6,90,2,oucd;m,3,13,180,1,crol;m,4,13,270,2,olcr;m,5,13,0,1,clor;m,11,13,270,1,orcl;m,10,13,0,2,wwwjclor;m,9,7,0,3,orcl;m,13,11,270,3,c1ow;m,10,7,90,1,clor;m,12,5,180,1,olcr;m,6,5,0,3,clor;m,8,5,270,3,olcr;
Oil of Vitriol (108/23/2)
b,3,11;b,10,11;t,3,4;t,1,2;c,1,8;t,7,11;t,9,9;b,7,7;c,5,9;c,13,8;t,11,4;m,3,2,180,1,rorrcr;m,6,0,180,2,wc2lor;m,1,9,0,3,orwwcl;m,7,9,180,1,clllol;m,3,13,180,1,wwcrol;m,4,13,180,2,cdrolu;m,10,13,0,2,cdloru;m,11,13,0,1,wwclor;m,10,7,180,1,lucdro;m,13,10,90,3,rdolcu;m,13,11,270,2,1olwcr;m,1,11,270,2,2orwcl;m,2,6,0,1,rcrror;m,12,6,180,1,lcllol;m,5,7,0,2,2lorwc;m,11,0,90,2,u1owcd;m,10,2,180,2,olwwcr;m,14,4,180,3,wwcuod;
Haematite (76/37/4)
p,9,11;b,3,11;p,6,7;b,10,6;m,9,9,90,1,clor;m,7,10,0,2,clor;m,5,13,270,1,orcl;m,3,13,270,1,olcr;m,7,11,180,1,cuod;m,3,8,0,2,orcl;m,13,9,270,1,olcr;m,13,5,90,1,crol;m,10,8,270,1,oucd;m,11,8,270,1,oucd;m,9,8,180,2,crol;m,14,6,180,2,crol;m,12,2,270,1,orcl;m,12,3,270,2,clor;m,11,2,180,1,owc1;m,9,13,180,1,crol;m,14,11,180,3,odcu;m,13,11,90,1,crol;m,10,13,270,2,orcl;
Aqua Regis (72/59/8)
This is only half of the solution; the completed solution will have period 4 and 43 cycles.
b,3,11;b,4,9;b,5,7;c,2,5;t,1,2;m,1,11,90,1,cluodrww;m,3,13,180,1,wwcrolww;m,4,13,270,3,dlwwcuro;m,1,9,0,3,orwwwwcl;m,6,11,270,2,ldwwcruo;m,5,5,90,3,cddouuww;m,3,7,270,1,orwcl211;m,0,5,0,2,cl1212or;m,3,3,270,2,r2ollwcr;
Gold (108/34/5)
p,9,10;p,3,10;m,10,13,0,2,clor;m,13,10,180,2,olcr;m,9,13,180,1,cruoldwwww;m,7,10,90,2,wwcurodlww;m,3,13,180,1,cuodwwwwww;m,4,13,180,2,wwcrolcrol;m,1,10,0,2,orwwclorcl;m,3,12,270,1,wwwuodcudw;m,9,6,90,3,wrrorrcwww;m,3,7,90,1,wwwwclorww;m,0,10,90,3,wlordcuwww;m,5,5,270,1,rrwwwwcrro;
??? (15) (188/73/10)
p,10,11;b,6,7;p,3,10;b,2,4;t,13,5;b,11,2;t,1,6;m,7,11,0,2,drwwwwcluo;m,10,13,180,2,cdroluwwww;m,11,13,270,1,orcl;m,13,11,90,1,crol;m,10,7,90,3,crolwwcrol;m,11,5,180,1,cl1ludollw;m,11,0,180,2,dloruuwc1d;m,3,0,90,3,clorw;m,0,4,0,2,wcdlrouwww;m,3,6,270,1,lwcro;m,3,7,180,2,rollwwcdur;m,1,9,90,3,clorclorww;m,1,10,90,3,rrorrdwwcu;m,3,13,0,3,wwcldoruww;m,4,13,180,2,wwcrolcrol;m,7,9,270,1,olcrolwwcr;m,6,5,0,1,orcl2;m,7,5,0,1,orwcl;m,8,5,180,1,2rocl;m,10,5,0,2,11ollcdllu;m,10,4,270,1,wclor;m,14,2,180,1,wwwuolwcrd;m,14,3,90,1,wwclllolww;
