Topic by nitsuja

Posted: 6/20/2010 9:10 PM

Quote

nitsuja

Emulator Coder, Skilled player (1300)

Joined: 12/21/2004
Posts: 2687

Engineers in 21 parts: ?lvl=24&code=c12:8f3;c12:9f3;c12:10f3;c12:11f3;c12:12f3;g9:3f2;c9:4f1;c10:2f2;p10:3f5;q10:4f7;r10:5f2;c11:2f3;c11:3f3;p11:4f3;q11:5f5;g12:2f0;q12:3f4;b12:4f1;p12:6f3;c12:7f3;q12:5f3;

Posted: 6/20/2010 9:18 PM

Quote

Bisqwit

Editor, Active player (296)

Joined: 3/8/2004
Posts: 7469
Location: Arzareth

No Ophanim yet. I ran out of field space. First time this happens.

I tried to make this machine without a separate program file. Hmm... (There is duplication on the bottom. The two sides should not be identical, because one is > and one is >=, but I ran out of space.) P.S. Did I mention I hate these conveyer belts?

Posted: 6/20/2010 9:23 PM
(Edited: 6/23/2010 6:48 PM)

Quote

Scepheo

Player (142)

Joined: 7/16/2009
Posts: 686

TAH-DAH! A list! Robotoast! ACCEPT: Move robots from the entrance (top) to the exit (bottom)! Least parts: 3 Robocoffee! If a robot's string starts with blue, accept. Otherwise, reject! Least parts: 3 Robolamp! ACCEPT: if there are three or more blues! Least parts: 8 Robofish! ACCEPT: if a robot contains NO red! Least parts: 4 Robobugs! ACCEPT: if the tape has only alternating colors! Least parts: 11 (TheRandomPie_IV) Robocats! ACCEPT: if the tape ends with two blues! Least parts: 11 Robobears! ACCEPT: Strings that begin and end with the same color! Least parts: 15 (flagitious) RC Cars! OUTPUT: The input, but with the first symbol at the end! Least parts: 7 Robocars! OUTPUT: Replace blue with green, and red with yellow! Least parts: 7 Robostilts! OUTPUT: Put a green at the beginning, and a yellow at the end! Least parts: 9 Milidogs! ACCEPT: With blue as 1 and red as 0, accept odd binary strings! Least parts: 8 Soldiers! OUTPUT: With blue as 1 and red as 0, multiply by 8! Least parts: 7 Officers! OUTPUT: With blue as 1 and red as 0, add 1 to the binary string! Least parts: 18 (flagitious) Generals! OUTPUT: Subtract 1 from the binary string! (Input >= 1) Least parts: 17 (flagitious) Robotanks! OUTPUT: With blue as 1 and red as 0, accept binary string > 15! Least parts: 25 (TheRandomPie_IV) Robospies! OUTPUT: With blue as 1 and red as 0, accept natural powers of 4! Least parts: 8 Androids! ACCEPT: Some number of blue, then the same number of red! Least parts: 12 (nitsuja) Robochildren! ACCEPT: An equal number of blue and red, in any order! Least parts: 17 (nitsuja) Police! OUTPUT: Put a yellow in the middle of the (even-length) string! Least parts: 32 (klmz) Judiciary! ACCEPT: (Even-length) strings that repeat midway through! Least parts: 46 (klmz) Teachers! ACCEPT: X blue, then X red, then X more blue, for any X! Least parts: 19 (flagitious) Politicians! ACCEPT: If there are exactly twice as many blues as red! Least parts: 23 (Scepheo) Academics! OUTPUT: Reverse the input string! Least parts: 25 (Pointless Boy) Engineers! ACCEPT: Perfectly symmetrical strings! Least parts: 21 (nitsuja) Roborockets! OUTPUT: Swap blue for red, and red for blue! Least parts: 7 Roboplanes! OUTPUT: All of the blue, but none of the red! Least parts: 7 Rocket Planes! OUTPUT: The input, but with all blues moved to the front! Least parts: 14 (klmz) Robomecha! OUTPUT: The input, but with the last symbol moved to the front! Least parts: 19 (Pointless Boy) Seraphim ACCEPT: Two identical strings, separated by a green! Least parts: 21 (Gunty) Ophanim ACCEPT: Read the tape as two numbers, A and B, split by a green: accept if A > B! Least parts: 51 (TheRandomPie_IV) Metatron OUTPUT: Read the tape as two numbers, A and B, split by a green: output A + B! Least parts: 84 (Nitrodon)

Posted: 6/21/2010 4:12 AM

Quote

Nitrodon

Active player (283)

Joined: 3/4/2006
Posts: 341

Just to get this level started, here's Metatron in 84 parts (1:40). I haven't tried optimizing it yet. ?lvl=31&code=g12:2f2;y13:2f3;c9:11f3;r9:12f2;q10:11f3;p10:12f7;q10:13f3;c11:11f3;b11:12f0;q11:13f3;b7:5f2;y7:6f1;c7:7f1;c7:8f1;c7:9f1;c7:10f1;q8:3f5;c8:4f1;p8:5f5;q8:6f1;c8:7f1;g8:8f1;c8:10f0;c9:3f2;r9:5f0;y9:6f1;r9:7f2;q9:8f1;c10:3f2;g10:5f3;c10:6f3;c10:7f3;p10:8f7;i10:10f6;c11:3f2;r11:4f2;q11:5f1;c11:6f1;b11:7f0;q11:8f5;c11:10f0;g12:3f3;c12:4f3;p12:5f7;c12:6f0;g12:8f3;c12:9f3;q12:10f3;c13:3f0;b13:4f0;q13:5f5;c13:6f1;r13:7f2;q13:8f1;c13:9f1;c13:10f2;c14:3f0;g14:5f3;c14:6f3;c14:7f3;p14:8f7;c14:9f0;c14:10f2;c15:3f0;b15:5f2;g15:6f1;b15:7f0;q15:8f5;c15:10f2;q16:3f1;c16:4f1;p16:5f5;q16:6f1;c16:7f1;g16:8f1;c16:10f2;r17:5f0;g17:6f1;c17:7f1;c17:8f1;c17:9f1;c17:10f1;c9:10f0;q10:9f0;

Posted: 6/21/2010 5:58 AM

Quote

Bisqwit

Editor, Active player (296)

Joined: 3/8/2004
Posts: 7469
Location: Arzareth

Scepheo wrote:

Finally! You guys stopped updating while I was working on this!

I like Flagitious's list better, it lists authors.

Posted: 6/21/2010 8:24 AM
(Edited: 6/21/2010 2:47 PM)

Quote

Scepheo

Player (142)

Joined: 7/16/2009
Posts: 686

When I have time left, I'll add them. See if I can keep track of them. Some would require "author: trivial" or I would claim them =P There's only one non-trivial one that has my name on it, sadly (Politicians!). EDIT: Bummer, seems TheRandomPie_IV had a better one I missed... EDITED EDIT: Mwuhaha! Managed to beat that one. 23 parts now, see list on last page.

Posted: 6/21/2010 11:16 AM
(Edited: 6/21/2010 1:29 PM)

Quote

TheRandomPie_IV

Active player (482)

Joined: 2/19/2007
Posts: 114

Ophanim in 51, just a slight rearrangement of my previous solution. ?lvl=30&code=y12:2f2;c13:1f3;p13:2f2;c13:3f2;g14:1f0;q14:2f2;i14:3f7;r15:1f3;p15:2f4;b15:3f1;b12:4f3;p12:5f6;r12:6f1;c13:4f2;q13:5f0;g13:6f3;c14:4f2;c14:5f0;c14:6f1;y15:4f3;p15:5f7;g15:6f3;q16:5f6;c16:6f0;r17:4f3;p17:5f4;b17:6f1;c13:7f3;c13:8f2;q13:13f3;c14:7f1;r14:9f2;y14:10f3;q14:13f7;p15:7f3;c15:8f3;p15:9f7;q15:10f5;c15:11f1;c15:12f1;c16:7f3;c16:8f0;r16:9f0;g16:10f3;p16:11f7;c16:12f0;p14:11f3;c13:11f3;i14:12f5;c13:12f2;p14:8f2; Edit: A small adjustment to Gunty's solution gives Police! in 33: ?lvl=19&code=g12:3f3;y12:2f3;y12:6f3;c12:4f3;c12:5f3;r13:8f0;c12:8f3;p12:9f3;q13:9f1;c14:5f0;c15:9f1;r15:8f1;p15:7f0;c14:6f1;b15:6f3;b13:6f0;r13:4f0;p13:5f4;i12:7f7;q14:7f6;g14:8f1;y14:9f2;q12:10f2;p12:11f3;b11:11f2;r13:11f0;q12:12f0;q11:9f0;b10:9f1;c10:8f2;i11:8f3;y11:7f2;b13:7f2; and Judiciary! in 59 as Bisquit's solution had the police solution nested in it: ?lvl=20&code=p12:12f3;p11:12f4;p13:12f2;p10:10f6;p9:10f1;p14:10f0;p15:10f1;r11:11f3;r13:11f3;r14:11f1;r10:11f1;c9:11f1;c11:10f2;c13:10f0;q12:10f3;c15:11f1;b11:13f1;b13:13f1;b10:9f3;b14:9f3;g12:11f3;y9:12f1;y15:12f1;q10:12f3;q14:12f7;b11:9f2;p12:9f3;r13:9f0;q12:8f2;y12:4f3;i12:5f7;c12:6f3;p12:7f3;r13:6f0;q13:7f1;g12:3f3;y12:2f3;c15:4f1;q15:5f6;g15:6f1;b16:4f3;p16:5f0;r16:6f1;c14:5f2;c13:4f0;c13:2f3;c13:3f3;b14:4f0;r14:2f0;p14:3f4;c15:3f0;y14:7f1;r14:6f1;q11:7f0;b10:7f1;i11:6f3;y11:5f2;b13:5f2;c10:6f2;

Posted: 6/21/2010 12:11 PM
(Edited: 6/21/2010 11:05 PM)

Post subject: Metatron in ...

Quote

Bisqwit

Editor, Active player (296)

Joined: 3/8/2004
Posts: 7469
Location: Arzareth

Suddenly, this game is no longer so stressful! Here is my Metatron (and its textual flowgraph). ;)

117 parts. Verified to work properly, barring transcription errors. Tags: Sacrifices space for entertainment.

begin
|
YG loop1a
   |
   0G --> G loop1b_0
            |                    _
            0x --> 0 <loop1b_0>    \_ transcribe until Y
            1x --> 1 <loop1b_0>   _/
            1Y --> Y1 <01Y-to-01Y transcriber> G <loop1a>   -- 0+1 = 1
            0Y --> Y0 <01Y-to-01Y transcriber> G <loop1a>   -- 0+0 = 0
            Y  --> Y0 <01Y-to-01Y transcriber> G <loop1a>   -- 0+nil = 0
   1G --> G loop1b_1
            |                    _
            0x --> 0 <loop1b_1>    \_ transcribe until Y
            1x --> 1 <loop1b_1>   _/
            1Y --> YY <01Y-to-01Y transcriber> G <loop1a>   -- 1+1 = Y (Y=2)
            0Y --> Y1 <01Y-to-01Y transcriber> G <loop1a>   -- 1+0 = 1
            Y  --> Y1 <01Y-to-01Y transcriber> G <loop1a>   -- 1+nil = 1
   0x --> 0 <loop1a>
   1x --> 1 <loop1a>
   G  --> G loop1b_nul
            |                     _
            0x --> 0 <loop1b_nul>   \_ transcribe until Y
            1x --> 1 <loop1b_nul>  _/                    
            0Y --> Y0 <01Y-to-01Y transcriber> G <loop1a>   -- nil+0 = 0
            1Y --> Y1 <01Y-to-01Y transcriber> G <loop1a>   -- nil+1 = 1
            Y  --> Y0 <01Y-to-01Y transcriber> G <eat G> <eat Y> <loop2>
loop2
|got0
||                _
0x --> 0 <loop2>   \_ transcribe until G/Y
1x --> 1 <loop2>  _/
0G --> 0 <end>                    -- end of loop, no Y remain
1G --> 1 <end>                    -- end of loop, no Y remain
0Y --> 1 <got0>                   -- 0Y = 10
1Y --> Y0 <01Y-to-01Y transcriber> G <loop2>  -- 1Y = Y0 (Y remains, rescan)

Posted: 6/21/2010 7:31 PM

Quote

henke37

Editor

Joined: 3/10/2010
Posts: 899
Location: Sweden

How did you manage to move the exit?

Posted: 6/21/2010 7:49 PM
(Edited: 6/22/2010 8:57 AM)

Quote

TheRandomPie_IV

Active player (482)

Joined: 2/19/2007
Posts: 114

Feel a little less guilty of my steal of Judiciary now. I found a "proper" improvement bringing it down to 49 (Police! followed by Seraphim): ?lvl=20&code=q12:8f2;y12:4f3;i12:5f7;c12:6f3;p12:7f3;r13:6f0;q13:7f1;g12:3f3;y12:2f3;c15:4f1;q15:5f6;g15:6f1;b16:4f3;p16:5f0;r16:6f1;c14:5f2;c13:4f0;c13:2f3;c13:3f3;b14:4f0;r14:2f0;p14:3f4;c15:3f0;y14:7f1;r14:6f1;q11:7f0;b10:7f1;i11:6f3;y11:5f2;b13:5f2;c10:6f2;b10:11f3;p10:12f6;r10:13f1;p11:10f5;g11:11f1;q11:12f4;g12:9f3;c12:10f3;c12:11f3;p12:12f7;p13:10f5;g13:11f1;q13:12f2;b14:11f3;p14:12f0;r14:13f1;c11:13f1;c13:13f1; EDIT: Judiciary in 47: ?lvl=20&code=p11:10f5;c12:11f3;p12:12f7;p13:10f5;q13:12f2;b14:11f3;p14:12f0;r14:13f1;c13:13f1;q12:9f2;g12:10f3;c13:11f1;g11:2f3;y12:2f0;c10:7f2;b10:8f1;c11:3f2;y11:6f2;i11:7f3;q11:8f0;c12:3f3;c12:4f3;y12:5f3;i12:6f7;c12:7f3;p12:8f3;r13:3f0;p13:4f4;b13:5f0;b13:6f2;r13:7f0;q13:8f1;c14:4f0;c14:5f1;q14:6f6;g14:7f1;y14:8f2;b15:5f3;p15:6f0;r15:7f1;c15:8f1;c11:11f1;q11:12f4;c11:13f1;b10:11f3;p10:12f6;r10:13f1;

Posted: 6/22/2010 2:12 AM

Quote

flagitious

Player (200)

Joined: 7/6/2004
Posts: 511

Didn't have time to fully optimize this but I'll post what I got (a couple more may be shaved using this method, but it might be harder to figure out what this method actually is versus figuring it out from scratch) 37 metatron http://pleasingfungus.com/?lvl=31&code=r10:3f2;p11:3f3;b12:3f0;g12:2f3;q11:4f3;g12:4f3;b9:7f2;g10:6f2;q10:7f5;r10:8f1;c11:5f2;p11:6f2;b11:7f3;q11:8f7;b11:9f2;r12:5f3;c12:6f3;c12:7f3;p12:8f3;q12:9f1;r13:5f2;g13:6f0;r13:7f0;q13:8f5;p14:5f7;q14:6f1;y14:7f1;g14:8f1;b15:5f0;y15:6f0;q13:9f7;c14:9f0;q12:11f0;p12:10f7;r13:10f0;b11:10f2;p12:12f2; Also this is the correct but slower version, a shorter version is faster but will fail in some extreme cases (that are never tested for though)

g,o,p,i=1e4,a[10001];main(x){for(;p?g=g/x*p+a[p]*i+2*!o: 53^(printf("%.4d",o+g/i),p=i,o=g%i);a[p--]=g%x)x=p*2-1;}

Posted: 6/22/2010 7:51 AM

Quote

Bisqwit

Editor, Active player (296)

Joined: 3/8/2004
Posts: 7469
Location: Arzareth

flagitious wrote:

37 metatron

begin_flagitious
|
G <01-to-01 transcriber> G0 loop2
                            |    
                            0 --> got0
                                  |   
                                  G --> GY <01Y-to-01Y transcriber> G <loop2>
                                  x --> 0 <loop2>
                            1 --> got1
                                  |   
                                  x --> 1 <loop2>
                                  G --> 0, loop3
                                           |
                                           Y -> Y  <loop3> 
                                           G -> 1  <got1> 
                                           1 -> G1 <got1> 
                                           0 -> G0 <loop2> 
                            Y  --> <eat Y> <01-to-10 transcriber> <eat G> <eat 1> <end>

Looks nice. You've even incorporated Nitsuja's bit there. I like particularly how you layouted the exit condition from the main loop.

Posted: 6/22/2010 10:03 AM

Post subject: Challenge: RoboticHay!

Quote

Bisqwit

Editor, Active player (296)

Joined: 3/8/2004
Posts: 7469
Location: Arzareth

I've got a challenge that I think might be rather difficult in Manufactoria: ?ctm=RoboticHay!;Input:_string1_green_string2._Accept_if_string1_is_a_substring_of_string2.;bgb:*|rbgrrr:x|rrbbgrrbbbr:*|bbrbrgbrbr:x|bbrrbgrrbbrrrbbrbb:x|rrbrrgbbbrbbrrbrrbrb:*|rbrbbgbbrbrbrbb:*;13;3;0; EDIT: Renamed; name already used by the first poster.

Posted: 6/22/2010 10:53 AM
(Edited: 6/22/2010 12:14 PM)

Post subject: Re: Henke37's custom puzzles

Quote

Bisqwit

Editor, Active player (296)

Joined: 3/8/2004
Posts: 7469
Location: Arzareth

henke37 wrote:

I did on the other hand manage to make a few levels of my own.

Cool!

?ctm=RoboNeedles;Only_legal_input_color_robots.;:*|bry:x|rb:*|ygy:x|rbbbyrb:x|rbrb:*;5;3;0;

Trivial; 5 parts.

?ctm=RoboMixers;Remove_one_from_each_end.;brb:r|rrbb:rb|rbbrb:bbr|br:|rrrbrbbrbrb:rrbrbbrbr;13;3;0;

19 parts, 1:16.

?ctm=RoboWorms;Remove_every_even_color;bbrr:br|brbr:bb|rrbbrrbb:rbrb|rrrbbbrrbb:rrbrb|rrrrrrrrrrbbbbbbbbbb:rrrrrbbbbb;9;3;0;

19 parts, 2:48

?ctm=RoboGuns;OUTPUT:_Remove_all_sequences_longer_than_one_color.;bbbbb:|brbr:brbr|bbr:r|rbrbrbbr:rbrbrr|brrbrrbrrbrrbrrbrr:bbbbbb;13;3;0;

Boy, that's a sneaky challenge. Here's my result, 2:08, 44 parts. ?lvl=36&code=c11:3f3;c11:4f3;i11:5f7;c11:6f0;i12:3f6;i12:5f7;i12:7f6;c13:3f0;c13:6f2;i13:7f2;c12:4f3;g12:2f3;c12:12f3;c13:4f0;c14:3f0;c14:5f1;c15:7f3;c12:11f3;c11:9f2;c11:8f2;c12:10f3;c13:8f1;c14:9f0;c13:9f0;c9:7f3;c9:8f3;c10:7f3;c10:8f2;c10:9f2;c15:8f3;p14:4f4;p14:6f6;p12:6f7;p12:8f7;p10:6f0;r9:9f2;b15:9f0;c9:6f3;c15:6f3;r13:5f3;c10:5f2;b11:7f1;c14:7f0;q12:9f6;&ctm=RoboGuns;OUTPUT:_Remove_all_sequences_longer_than_one_color.;bbbbb:|brbr:brbr|bbr:r|rbrbrbbr:rbrbrr|brrbrrbrrbrrbrrbrr:bbbbbb;13;3;0;

&ctm=RoboModems;ACCEPT:_Only_odd_parity!;brbbr:*|rbrbb:*|rrbbrbb:x|:x|b:*|r:x|bbbrbbbbbrbrbrbrrrb:x|rbbbr:*;7;3;0;

8 parts, 0:52; refreshingly trivial.

?ctm=RoboCalculators;Accept:
Only_evenly_divideable_by_three!;bb:*|br:x|brr:x|bbbbr:*|bbbrbrb:*|bbbrbbb:x|bbrrbrbbbbrbb:x;9;3;1;

Oh dear. That requires thinking!

Posted: 6/22/2010 11:36 AM

Quote

henke37

Editor

Joined: 3/10/2010
Posts: 899
Location: Sweden

I happen to have a shorter one from #zsnes by "zinx": 2:32/41

?lvl=36&code=c12:8f3;c12:9f3;c12:10f3;c12:11f3;c12:12f3;g12:2f3;p12:4f3;c12:7f3;q12:6f6;i12:3f5;i12:5f1;c13:4f2;c8:6f2;i10:4f2;i14:4f7;c10:5f1;c10:3f2;c13:3f3;c14:3f3;c14:5f0;c13:5f0;c16:6f0;c14:6f0;c13:6f0;c10:6f2;c11:6f2;c11:3f2;c11:4f0;p15:5f3;p15:4f6;r15:3f0;i16:5f6;i8:5f7;p9:5f3;p9:4f4;b9:3f2;c11:5f1;b8:4f3;c9:6f2;c15:6f0;r16:4f3;&ctm=RoboGuns;Remove_all_sequences_longer_than_one_color.;bbbbb:|brbr:brbr|bbr:r|rbrbrbbr:rbrbrr|brrbrrbrrbrrbrrbrr:bbbbbb;13;3;0;

Also, new challenge (hopefully I didn't screw the testcases up):

?ctm=RoboAbacuses;In:_two_numbers_separated_by_green.
Out:_absolute_difference_of_the_numbers.;brgb:b|bbgbrb:br|brbrgbbbr:brr|bbrrbrrbgbbbrbr:brrrbbbb|bbbbbrgbbbrbrrr:brbrbrbr;13;3;1;

Posted: 6/22/2010 12:02 PM

Post subject: Re: Henke37's custom puzzles

Quote

Pointless_Boy

Banned User

Joined: 6/18/2010
Posts: 183

Sorry I haven't put up any more optimization walkthroughs, I've been feeling really lazy lately. I really like all the custom levels you guys have been posting. But what does "only legal input color robots" mean? Here's a challenge of my own, it shouldn't be too hard :)

?ctm=Machine_Guns!;Convert_a_binary_string_to_a_yellow_unary_string!;:|b:y|br:yy|brbb:yyyyyyyyyyy|bbrrr:yyyyyyyyyyyyyyyyyyyyyyyy;13;3;1;

Bisqwit wrote:

?ctm=RoboCalculators;Accept:
Only_evenly_divideable_by_three!;bb:*|br:x|brr:x|bbbbr:*|bbbrbrb:*|bbbrbbb:x|bbrrbrbbbbrbb:x;9;3;1;

Oh dear. That requires thinking!

Here's a hint that might help: the divisibility rule for 11 in base 10 also works for 11 (3) in base 2!

Posted: 6/22/2010 12:07 PM

Quote

henke37

Editor

Joined: 3/10/2010
Posts: 899
Location: Sweden

Legal means: Only red and blue allowed.

Posted: 6/22/2010 12:18 PM

Quote

Pointless_Boy

Banned User

Joined: 6/18/2010
Posts: 183

henke37 wrote:

Legal means: Only red and blue allowed.

So if a robot has any number of yellows or greens, reject, otherwise accept?

Posted: 6/22/2010 12:21 PM

Quote

henke37

Editor

Joined: 3/10/2010
Posts: 899
Location: Sweden

Yeah, there is a reason why it is using the smallest board size. It's trivial.

Posted: 6/22/2010 12:23 PM

Post subject: Re: Henke37's custom puzzles

Quote

Bisqwit

Editor, Active player (296)

Joined: 3/8/2004
Posts: 7469
Location: Arzareth

Pointless Boy wrote:

?ctm=RoboCalculators;Accept:
Only_evenly_divideable_by_three!;bb:*|br:x|brr:x|bbbbr:*|bbbrbrb:*|bbbrbbb:x|bbrrbrbbbbrbb:x;9;3;1;
Oh dear. That requires thinking!
Here's a hint that might help: the divisibility rule for 11 in base 10 also works for 11 (3) in base 2!

Ah! That is an useful clue. (A google search revealed what's special about that.) It still requires thinking though. :)

Posted: 6/22/2010 12:56 PM

Post subject: Re: Henke37's custom puzzles

Quote

Bisqwit

Editor, Active player (296)

Joined: 3/8/2004
Posts: 7469
Location: Arzareth

Bisqwit wrote:

?ctm=RoboCalculators;Accept:
Only_evenly_divideable_by_three!;bb:*|br:x|brr:x|bbbbr:*|bbbrbrb:*|bbbrbbb:x|bbrrbrbbbbrbb:x;9;3;1;
Ah! That is an useful clue. (A google search revealed what's special about that.) It still requires thinking though. :)

Lesson learned: Don't trust everything you read. Counting the difference between the sums of even and odd bits respectively got me nowhere. I got it now. 0:43, 11 parts.

Posted: 6/23/2010 7:19 AM

Post subject: More challenges

Quote

Bisqwit

Editor, Active player (296)

Joined: 3/8/2004
Posts: 7469
Location: Arzareth

New challenges:

http://pleasingfungus.com/?ctm=Sawdust;Remove_all_one-color_sequences_that_are_just_one_bit_long.;br:|bbrbb:bbbb|brrbbbbbrrbr:rrbbbbbrr|rrbb:rrbb|brrbrrbrbbrbb:rrrrbbbb|b:|brbrbrbrbrbrbrbrbrbrbrbrb:;9;3;0;

It resembles Henke's similar challenge.

http://pleasingfungus.com/?ctm=Sticks;Accept_if_has_one-color_sequence_of_the_same_length_at_the_begin_and_at_the_end.;bbbbrrrr:*|bbr:x|rrrbrrbbb:*|brrrb:*|brbrbrbr:*|bbbbbbrrrrrrb:x|bbbbbbbbrrrrrrrrbrrrrrrrr:*|bbrbb:*;11;3;0;

http://pleasingfungus.com/?ctm=Stakes;Accept_if__input_made_entirely_of_same-length_one-color_strings_of_*any*_color;bryg:*|yyyrrrgggbbb:*|rrr:*|bbyrrgg:x|brry:x|ggggbbbb:*|yyrryy:*|rrrbbbbbrr:x;13;3;0;

http://pleasingfungus.com/?ctm=Rainbow;Accept_if_input_contains_all_four_colors.;rgbgbrbrbgbrbgrg:x|ygbrg:*|ggrrrbyg:*|ybbbrrbyb:x|:x|brgy:*;13;3;0;

This one might be impossible!

Posted: 6/23/2010 7:54 AM

Post subject: Re: More challenges

Quote

Scepheo

Player (142)

Joined: 7/16/2009
Posts: 686

Bisqwit wrote:

New challenges:

http://pleasingfungus.com/?ctm=Rainbow;Accept_if_input_contains_all_four_colors.;rgbgbrbrbgbrbgrg:x|ygbrg:*|ggrrrbyg:*|ybbbrrbyb:x|:x|brgy:*;13;3;0;

This one might be impossible!

It's possible, but I'm quite sure you don't have enough space.

Posted: 6/23/2010 9:47 AM

Post subject: Re: More challenges

Quote

Bisqwit

Editor, Active player (296)

Joined: 3/8/2004
Posts: 7469
Location: Arzareth

Scepheo wrote:

Bisqwit wrote:
New challenges:
http://pleasingfungus.com/?ctm=Rainbow;Accept_if_input_contains_all_four_colors.;rgbgbrbrbgbrbgrg:x|ygbrg:*|ggrrrbyg:*|ybbbrrbyb:x|:x|brgy:*;13;3;0;
This one might be impossible!
It's possible, but I'm quite sure you don't have enough space.

Alternative reply 1: Hence, impossible. That is what I meant. Alternative reply 2: Well, d'oh! That's what I meant. Here's my attempt, below!

Posted: 6/23/2010 9:56 AM

Quote

henke37

Editor

Joined: 3/10/2010
Posts: 899
Location: Sweden

My idea is to just eat all of each color until there is nothing of the current color.