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Incredible game with so much potential for a sequel! This would especially get coding geeks and nerds out there especially excited! Needs: more space available for certain levels and/or level editor, online level sharing instead of text copying, and most of all A SEQUEL!
And for all those super geeks out there who feel up for a SUPER challenge (not sure if its possible though, as I do not have the coding power to even test it!) try this:
?ctm=Adv_Cryptography;Go_to:
http://www._xkcd.com/153
You_will_find_what_to_do_there!;rbbrbrbr:rbrbrrbb|bbrrbrrbb:rrrbbrbbr|:|b:r|brb:rrb;13;3;1;
If you don't feel like going to the website here are the simple instructions:
1: move 1st sticker to end
2: flip red and blue colors
3: reverse the whole chain (i.e. R-B-B would become B-B-R)
In shorter words: take it down, flip it, and reverse it!
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MagiMaster, here is your level with better tests:
?ctm=3-in-a-row;Accept_if_the_tape_does_NOT_contain_three_of_the_same_color_in_a_row.;bbbr:x|rbrr:*|brrr:x|rrbb:*|:*|rrbrrbbrbbrbbrb:*|bbrbrrbrbbrbbbr:x|brrbrbbrbrbrbbr:*;7;3;0;
The problem with your tests is that a circuit with 5 branches in line is enough to pass them:
?lvl=32&code=p12:5f7;p11:5f7;p10:5f7;p13:5f7;p14:5f7;c10:6f2;c11:6f2;c14:6f0;c13:6f0;c12:6f3;c12:7f3;c12:8f3;c12:9f3;&ctm=3-in-a-row;Accept_if_the_tape_does_NOT_contain_three_of_the_same_color_in_a_row.;brbbr:*|bbrrbbbr:x|rrbbrbrrrb:x|brbrr:*|:*;7;3;0;
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A somewhat simple level:
?ctm=3-in-a-row;Accept_if_the_tape_does_NOT_contain_three_of_the_same_color_in_a_row.;brbbr:*|bbrrbbbr:x|rrbbrbrrrb:x|brbrr:*|:*;7;3;0;
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@Gotan: Looks like you've been ignoring the help screen too; Shift+C copies, Shift+V pastes.
Rotation has been implemented for the next version yet to be uploaded (per PleasingFungus' response on feature requests), mirroring/flipping would be great to have but will have to wait, I guess.
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Very nice game!
Some Ideas:
I found out about some editing capabilities (like flipping components) only from comments, only now i realised, that i was constantly ignoring the explanation line at the bottom of the screen.
A mouseover of the "select" function says, that you can also copy parts, but i have yet to find out how, atm i can only move. It would be nice if you could also rotate and mirror "selected" parts. A clipboard for storing often used aggregates might also be a nice thing, but maybe that would be overdoing it.
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...I love this. Simple visuals, solid design, but the level design... My God, the level design. It's gorgeous. People in the industry can learn a lot from you, my good man. A blue-red-blue to you.
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I have waited until I completed the game to post in here. Simply putting it this game is AWESOME; my head had to work quite a lot to figure out all of the levels. Art is simple but efficient; levels are challenging and with an awesome difficulty curve. It's smarter than lightbot, more appealing and immediate than krispykrem's games. Also (even if no one cares) educational, implementing concepts as "finite state machines" in a cute visual environment. This game will get even more awesome if the author will add the kong built-in level sharing.
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I don't very much like having to press shift to make bridges. Very much of the controls are all over the place anyway and hard to use efficiently.
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i retract my previous statement... but was still better the other way. everyone give + to the comment about using shift
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Ok, it appears now that with the new update (1.13) in order to make a BRIDGE with conveyors you must hold down the SHIFT key when you place the new conveyor.
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Umm. Is it just me or are bridges no longer working? If i place a vertical conveyor on a horizontal conveyor, for example, it just replaces it now... I don't think some levels are possible without this :/
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Androids is not that complicated:
1. Mark the end (write green/yellow).
2. If no blue or red, accept; if red, discard; if blue, go to 3.
3. If blue, write blue; if red go to 4.
4. If red, write red; if blue discard, if no blue or red, read green/yellow, go to 1.
E.g.
?lvl=17&code=p12:5f7;c11:4f2;c10:9f1;c10:8f1;c10:6f1;c10:5f1;c10:7f1;g12:4f3;c14:5f3;c13:6f3;c13:7f3;p13:8f3;p14:6f7;r14:8f0;b15:6f0;c13:9f0;i12:9f1;i12:8f6;c12:6f3;c12:7f3;c13:5f2;c12:10f3;q10:4f5;c11:9f0;
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Huge tip: Press the spacebar to flip the sorters, so you can keep your machines symetrical.
Why isn't that in the instructions? Good Grief.
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Androids solution for any pattern:
?lvl=17&code=p12:5f3;g11:5f0;p10:5f0;y10:4f2;c11:4f2;c12:4f3;g13:5f2;y8:6f3;c8:7f3;p8:8f3;c9:7f0;g9:8f2;y10:7f0;p10:8f2;c10:6f0;c9:6f0;c15:5f2;c15:6f1;g15:7f1;p16:5f6;y16:6f3;p16:7f7;c14:5f2;c14:11f0;c13:11f0;c8:9f3;c8:10f2;c9:10f2;c10:10f2;c11:10f2;c12:9f1;i13:8f7;c14:8f2;c12:10f1;c12:8f2;c15:11f0;c12:6f2;c13:6f3;c13:7f3;c13:9f3;c13:10f3;q16:9f7;q15:9f7;c15:8f2;c16:8f3;c16:10f3;c16:11f0;
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missingno: If you do androids in two parts, you can solve for any string length (my solution checks that the string is all blues followed by reds first, then checks that there's an equal number of reds and blues)
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missingno: solution for general input
?lvl=17&code=c8:8f1;c8:9f1;c8:10f1;b9:9f3;p9:10f0;r9:11f1;c10:10f0;b11:8f3;p11:9f0;c11:10f0;c12:8f3;p12:9f3;c12:10f3;r13:8f3;p13:9f2;c13:10f2;c14:10f2;r15:9f3;p15:10f2;b15:11f1;c16:8f1;c16:9f1;c16:10f1;c8:7f2;c9:7f2;c10:7f2;q11:7f3;g12:7f3;q13:7f1;c15:7f0;c16:7f0;c12:4f3;c12:5f3;c12:6f3;c14:7f0;
I also got one which works for things like "bbbrrrbbbbb", as in only the first set of blues has to match the first set of reds, but I think it is meant to accept when every set of blue is followed by the the same amount of reds
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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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Just finished ophanim:
1) Clear all leading red
2) Pop off the first digit of both strings, you don't have to look at what you pop off unless it's the last digit of the first number
3) If the first number is exhausted first, do a quick compare to what's left of the second, and if the second number is exhausted first accept right away
Also, maybe I'm dumd but I can't figure out how to fit judiciary >.> there are two other similar puzzles (the yellow in the middle one and the identical strings seperated by green one), and it would be the combination of those two.
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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;