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Correct me if I'm wrong, but it seems that the androids level can only be solved up to certain size inputs due to space restrictions.
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I missed something critical: The input strings never contain green or yellow -- these are yours to use in calculations or state transitions. I've been solving for potentially 4-color inputs and it is much, much harder (sometimes impossible because you can't mark the end of a string in a way that might not occur randomly).
The game is much easier knowing only blue and red are heading my way. :)
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hehe obviously I'm not since there are some better solutions like ThirdParty's :) I guess I could get rid of most of these conveyers but I like that way ^^
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@restoshammyman - the random tests are randomly generated, to prevent people building a simple machine that just checks for the fixed test strings, and spits out the right answer. With the random tests, you have to solve the problem properly.
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Two suggestions:
1. Allow rewinding while testing design.
2. Allow break points. Make it so that you can pause the machine when the robot hits a certain state.
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Well, for Orphanim the easiest way is still to use binary substraction (compare strings 'upwards' from the last bit, and keep track of the carry.
By the way people, if you're interested in sharing solution, go to this thread:
http://www.kongregate.com/forums/3-general-gaming/topics/90661-manufactoria-walkthrough-optimization-thread?page=1
ThirdParty has posted a very nice solution for Orphanim there that runs in 37 seconds, and I believe many of you have something to share too. Keep this comment up if possible.
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Very challenging, addictive, fairly amusing -- 5/5. If this ever gets badges you would want goals for less than x parts/time used on individual levels, or sets of levels, etc. You're tracking this info in-game, but if you're also using the Kong API you should make it visible (not just the one highscore).
For that matter: don't wait for badges! Add in-game achievements like the above to give players additional goals to shoot for, and feedback on how good their solutions are.
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finally completed ophanim! :D man that was hard. kinda proud of the solution though: "only" 74 parts, 52 seconds.
?lvl=30&code=g12:2f3;y12:3f3;c12:4f3;c12:5f3;c12:6f3;p12:7f3;q11:7f1;q13:7f5;b11:6f2;r13:6f0;c12:8f3;q15:6f7;b15:7f2;c16:7f1;r17:7f0;g14:7f3;c14:8f2;c15:8f2;c16:8f1;g14:6f1;c16:4f0;c15:4f0;c13:4f0;p16:6f5;c16:5f2;q14:5f5;y13:5f1;y15:5f1;c14:4f0;c16:3f3;q17:3f0;c18:4f1;c18:3f0;c16:2f3;g10:7f3;c10:8f0;c9:8f0;c8:8f1;c8:7f1;g10:6f1;c9:4f2;c11:4f2;q9:6f3;q7:6f7;g6:6f1;q17:6f3;c17:5f3;g18:6f1;c18:5f1;c6:5f1;b9:7f0;r7:7f2;p8:6f1;q10:5f1;c8:5f0;c7:5f3;c10:4f2;q7:3f2;c6:4f1;c6:3f2;c8:2f3;c8:3f3;q12:9f0;p12:10f3;c13:10f0;q12:11f0;q12:12f0;g9:5f1;y11:5f1;g17:4f0;y17:2f0;g7:2f2;g7:4f2;c8:4f2;
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Here are some custom ones they aren't to hard but kinda fun.
?ctm=4_red;Accept_only_4_reds_than_any_number_of_blue.;bbbrr:x|rrrrbbbb:*|rrrrbr:x|rrrbbbbbrr:x|rrrr:*;5;3;0;
?ctm=Fun_stuff;Accept_only_blue_then_add_a_red_at_the_beggining_and_end.;:rr|brbrb:rbbbr|bbbrb:rbbbbr|rrrrr:rr|br:rbr;13;3;0;
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I just finished the game. I totally loved it!! Here i share my metatron solution:
?lvl=31&code=q8:7f2;p8:8f4;q8:9f6;r9:7f3;c9:8f0;b9:9f1;g10:8f0;b11:7f2;q11:8f1;g12:6f3;c12:7f3;p12:8f3;r13:7f0;q13:8f5;g14:8f2;r15:2f3;p15:3f4;b15:4f1;r15:5f1;r15:7f3;c15:8f2;b15:9f1;c16:3f0;p16:5f5;g16:6f1;q16:7f4;p16:8f2;q16:9f0;g16:10f3;p16:11f7;c17:3f0;r17:4f1;r17:5f1;c12:4f3;g12:2f3;r12:3f3;c12:5f3;q14:3f6;c14:4f0;c13:4f0;r15:11f3;b15:12f3;b17:11f3;r17:12f3;c15:13f2;c16:13f2;c17:13f2;c18:13f1;c18:12f1;c18:11f1;c18:10f1;c18:9f1;c18:8f1;c18:7f1;c18:6f1;c18:5f1;c18:4f1;c18:3f0;g8:10f3;g8:6f1;p8:5f5;p8:11f7;b9:11f3;b9:12f3;b7:11f3;r7:12f3;r9:5f1;b9:4f1;b7:5f1;r7:4f1;c9:13f0;c8:13f0;c7:13f0;c6:13f1;c6:12f1;c6:11f1;c6:10f1;c6:9f1;c6:8f1;c6:7f1;c6:6f1;c6:5f1;c6:4f1;c6:3f2;c7:3f2;c8:3f2;p9:3f2;r9:2f3;q10:3f6;c10:4f2;c11:4f2;q12:9f6;q12:10f6;q12:12f6;p12:11f3;c17:8f1;c17:7f0;c7:7f2;c7:8f1;c14:10f1;c14:9f1;c13:11f0;b11:11f3;r11:13f1;p11:12f6;c13:10f2;
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@voodooattack: I solved Orphanim in a similar way: read the last bit of the first string, read the last bit of the second string, write the difference (second-first), using yellows to represent borrows. If one string was shorter than the other (read a green rather than a red or blue), the missing bit was treated as a red. I then worked out the borrow bits in the difference string. If all the yellows merged out (blue then yellow = red, red then yellow = yellow blue), then the difference was positive and B >= A, so it was rejected. If the first bit in the difference string ended up yellow, then the difference was negative so A>B and it was accepted. I marked all three strings (A, B, difference) with greens and just had to keep track of where I was and what the next green meant. I haven't tried very hard to optimize it, but I used 99 parts, 1:23.
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@bachan: I'm going for a bitwise-AND approach for Orphanim, kind of struggling with the odd ends and special situations at the moment, and the size of the board isn't helping much; but I hope it works out.
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@snorlax6: actually your solution only works for words having 3 consecutive blues, although words like BBRB should be accepted.
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my idea for engineer: mark the beginning and end of the string with green, then bubble two yellows toward each other, checking to make sure the colors in between the yellow and the green are the same at each step. if they are, remove the blue and the blue or the red and the red. if they arent, discard the robot.
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@ antso101001: you made things a little too complicated.
?lvl=3&code=p12:5f3;p11:5f3;p10:5f3;c13:5f0;c9:5f3;c9:6f3;c9:7f3;c9:9f3;c9:8f3;c9:10f2;c10:10f2;c11:10f2;
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not the best, but for Teachers:
?lvl=21&code=p12:3f3;g11:3f0;c10:3f1;c10:2f2;c11:2f2;c12:2f3;y13:3f2;c13:2f0;c14:3f1;c14:2f0;c12:4f3;c11:4f2;c13:4f0;c15:5f1;c15:4f0;b14:4f0;c9:4f2;r10:4f2;c8:5f1;c8:4f1;c8:3f2;c9:3f2;c16:5f1;c16:4f1;c16:3f0;c15:3f0;c12:10f3;c12:12f3;c12:5f3;c12:6f3;q12:9f7;q14:9f2;q14:8f5;q10:9f0;c11:9f0;c13:9f2;c13:8f0;c11:8f2;c12:8f3;c12:7f3;q10:8f5;c10:10f0;c9:10f0;c8:10f1;c8:9f1;c8:8f1;c8:6f2;c9:6f2;c10:6f2;c11:6f2;c13:6f0;c14:6f0;c15:6f0;c16:6f0;c14:10f2;c15:10f2;c16:9f1;c16:8f1;c16:10f1;r16:7f1;b8:7f1;p12:11f3;y13:11f2;g11:11f0;c10:11f0;c9:11f0;c8:11f0;c7:11f1;c7:10f1;c7:9f1;c7:8f1;c7:7f1;c7:6f1;c7:5f2;c14:11f2;c15:11f2;c16:11f2;c17:11f1;c17:10f1;c17:9f1;c17:8f1;c17:7f1;c17:6f1;c17:5f0;c9:8f1;c9:7f1;c15:8f1;c15:7f1;
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@CeruleanDragon: If you put your mouse over the control, it will give you a description of what it is/how to use it--it takes about 3-4 seconds to show up.
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29 parts ?lvl=14&code=p12:10f3;r13:10f0;b11:10f2;c12:11f3;q12:12f2;g12:2f3;y12:3f3;c12:4f3;c12:9f3;c10:6f2;b10:7f1;c11:6f2;q11:7f0;r11:8f2;c12:6f3;p12:7f3;c12:8f3;q13:7f2;r14:7f1;c12:5f3;c14:6f0;i13:6f2;g13:5f1;b13:4f2;p14:4f3;r15:4f0;y15:6f0;q15:5f2;c14:5f2;
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I'm working on an algorithm for orphanim. It's a bit complicated and I'm hoping there's a tricky way to slim it down, but from what I've seen all tricks to make it smaller fail at powers of two. Step 1. shift off leading reds from both numbers, place yellows at ends of both numbers. Step 2. Develop two machines- one for when A is currently >B, one when A<=B. Each machine shifts of the last bit to determine if you should switch machines, stay in this machine, fail (A<=B), or pass (A>B)
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Generals 1:10 32 parts ?lvl=14&code=p12:10f3;r13:10f0;b11:10f2;c12:11f3;q12:12f2;g12:2f3;y12:3f3;c12:4f3;c12:9f3;c10:6f2;b10:7f1;c11:6f2;q11:7f0;r11:8f2;c12:6f3;p12:7f3;c12:8f3;q13:7f2;r14:7f1;c12:5f3;c14:6f0;i13:6f2;g13:5f1;b13:4f2;b15:5f2;p16:5f3;q16:6f3;r17:5f0;y15:6f0;c14:4f2;c15:4f2;c16:4f3;
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If you're having trouble with excessive lag while testing your solutions, you probably have had the game open for too long; exit the level saving changes and refresh the page.
Doing this solves this issue for me, just thought I'd mention it.
Technobabble: Looks like the Flash Player's garbage collector isn't that efficient after all.
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@freezit4 : The algorithm for Teachers is to read a blue first (reject the red side). The use a red/blue switch with a blue writer only to dupe blues until you hit a red. The red arrow goes to another red/blue switch which dupes reds until you hit another blue. You then dupe blues until the end of the chain, rejecting the part if you hit another red. Then rewrite the green dot at the end and recycle to the beginning, until you either run out of spots (which means equal numbers of blue, then red, then blue) or an excess of one color shows up and it gets rejected. I've managed to massage it down to 19 parts, 0:49.
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My own lvl:
?ctm=Greenish_red!;Replace_all_red_with_green!;brbbrrbrb:bgbbggbgb|rbrbrbrb:gbgbgbgb|rrrrrrrrrr:gggggggggg|bbbbbbbbbr:bbbbbbbbbg;5;3;0;
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just cleaned the last one !
my solutions are not beatiful, not optimized, but works no matter what you throw at them, and that's not so bad ;)
btw : thx for the addictive game (even if I was a bit tortured at some point, like there was a big step to overcome midgame)
5/5
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@djpawl
My solution for Police: 3:01, 43 parts.
Using green to keep track of string endings and moving two yellows towards eachother from both ends. Your solution sounds like it could be quicker tho if executed correctly.
?lvl=19&code=g12:2f3;y12:3f3;r11:4f2;p12:4f7;b13:4f0;y12:6f3;q12:5f0;b9:8f2;c9:9f1;c9:10f1;c9:11f1;c9:12f1;c9:13f1;q10:7f5;p10:8f5;y10:9f2;q10:10f6;p10:11f4;q10:12f2;y10:13f0;c11:9f1;r11:10f3;c11:11f0;b11:12f1;c12:7f2;b12:8f2;b12:10f3;y12:11f0;r12:12f2;g13:7f3;p13:8f3;q13:9f2;p13:10f3;i13:11f6;p13:12f7;q13:13f1;r14:8f0;r14:10f3;c14:11f0;b14:12f0;y14:13f0;r11:8f0;c11:7f2;
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@PleasingFungus: please add the possibility to rotate pasted elements (with shift+V) using WASD/arrows keys :)
This would make optimizations & rearrangements easier!
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my solution for level 22
?lvl=22&code=b6:2f3;c7:1f3;p7:2f3;c8:1f0;r8:2f3;c6:3f3;c6:4f2;b6:5f3;c6:6f3;c6:7f2;b6:8f3;c6:9f3;c7:4f3;p7:5f3;c7:7f3;p7:8f3;c8:3f3;c8:4f0;r8:5f3;c8:6f3;c8:7f0;r8:8f3;c8:9f3;q10:3f4;p10:4f1;r9:4f2;b11:4f0;g9:1f0;c10:2f1;c10:1f0;c12:2f0;c11:2f0;q11:8f0;r10:8f1;c10:7f2;c11:7f2;c12:7f3;p14:8f6;q14:9f3;r14:10f0;c13:10f1;c13:9f1;c13:8f2;c14:7f1;p14:6f1;p12:8f7;c12:12f3;c9:10f1;c9:9f1;c9:8f1;c9:7f2;c12:6f0;c11:6f0;c10:6f1;c10:5f1;p7:11f7;r6:11f2;b8:11f0;c6:10f2;c8:10f0;c7:10f3;q7:12f3;g8:12f2;c9:12f1;c9:11f1;q16:6f2;b17:6f3;c17:7f0;c16:7f0;g16:5f2;q13:6f4;c13:5f1;i15:6f7;c15:7f0;c18:5f3;c18:6f3;c18:7f3;c18:8f3;c18:9f3;c18:10f3;c18:11f3;c18:12f0;c17:12f0;c16:12f0;c15:12f0;c14:12f0;c13:12f0;q14:5f5;g15:5f3;c13:4f2;i14:4f3;p16:3f1;g16:4f1;c15:4f2;r15:3f1;b17:3f1;i17:2f4;c15:2f1;c17:1f0;c16:1f0;c15:1f0;c14:1f3;c14:2f3;c14:3f0;c13:3f0;c12:3f3;c12:4f3;c12:5f3;c16:2f2;c18:2f3;c18:4f3;c18:3f3;
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my robo-children in 29 parts (0:36, works in all cases): ?lvl=18&code=y12:3f3;y9:4f2;b9:5f1;c9:6f1;c9:7f1;c10:4f2;c10:7f0;c11:4f2;p11:5f3;c11:7f0;c12:4f3;p12:5f3;q12:6f2;c13:4f0;p13:5f3;c13:7f2;c14:4f0;r14:5f1;c14:7f2;y15:4f0;r15:5f1;c15:6f1;c15:7f1;c12:8f3;c12:9f3;c12:10f3;c12:11f3;p12:7f3;b10:5f1;
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I had more or less the same machine on generals and officers. Search for the last red or blue, and then switch that one and the following symbols.
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I could never even imagine that I will once see a game that can rapee my brain harder than Codex of Alchemical Engineering... :(
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Robochildren, 12 parts :
?lvl=18&code=p12:3f3;y11:3f2;y13:3f0;q12:4f3;q11:4f7;c12:5f3;c12:6f3;c12:7f3;c12:8f3;c12:9f3;c12:10f3;c12:11f3;
Thanks a lot Roboduck.
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Does anyone have a short solution for Police? Mine is 67 parts and 3:17, but I can't find a shorter routine (well, my current setup can be optimized a little by rearranging the design, but I believe there are much faster algorithms out there).
?lvl=19&code=c12:12f3;r8:3f2;p9:3f1;g9:4f1;r9:5f2;b10:3f0;q10:4f1;p10:5f1;b11:5f0;c12:5f3;y12:2f3;g12:3f3;c8:1f0;c7:1f3;c7:2f3;c7:3f3;c7:4f3;c7:5f3;c7:6f3;c7:7f3;c7:8f3;c7:9f3;c7:10f3;c7:11f3;c7:12f2;p8:12f2;b8:13f1;r8:11f3;q9:12f3;c10:12f2;c11:12f2;q9:2f3;q10:2f6;c11:3f3;c11:4f2;c11:2f3;y10:1f0;c9:1f0;c10:6f1;c10:7f1;c10:8f1;b10:9f1;y10:10f1;c11:6f0;b11:7f3;g11:8f2;b11:9f2;q11:10f5;i12:6f6;p12:7f3;c12:8f3;c12:9f3;p12:10f3;c13:6f0;r13:7f3;g13:8f0;r13:9f0;q13:10f1;c14:6f0;c14:7f1;c14:8f1;r14:9f1;y14:10f1;c12:11f3;q12:4f7;q13:4f6;y13:5f0;
Basically, each loop moves a green dot forward, and a yellow dot backwards; the center is detected when both colors meet.