Timeline for Dancing arrays. Output them with some forward-backward moves
Current License: CC BY-SA 3.0
12 events
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| Mar 21, 2017 at 16:01 | history | edited | Dead Possum | CC BY-SA 3.0 |
added 81 characters in body
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| Mar 20, 2017 at 16:01 | comment | added | hyperneutrino♦ | Oh never mind, I didn't notice the missing end part. Whoops. | |
| Mar 20, 2017 at 15:56 | comment | added | Dead Possum | @HyperNeutrino Nope, it does not, check proof | |
| Mar 20, 2017 at 15:52 | comment | added | hyperneutrino♦ |
It still behaves as expected either way... I'm running it with s*m<l-F+1 and it still works properly.
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| Mar 20, 2017 at 15:05 | comment | added | Dead Possum |
@HyperNeutrino No, I can't do that as (s*m<l-F) should evaluates to 0 or 1. If I remove brackets, than it would evaluates as (s*m<l)-F
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| Mar 20, 2017 at 14:48 | comment | added | hyperneutrino♦ |
You can save 2 bytes by changing 1+(s*m<1-F) to s*m<1-F+1 because that way you don't need the brackets.
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| Mar 20, 2017 at 14:38 | comment | added | Dead Possum | @fəˈnɛtɪk I re-read post again and brought a little fix. Indeed it seems that program should not output anything in case of endless loop, until whole array fits in one step | |
| Mar 20, 2017 at 14:36 | history | edited | Dead Possum | CC BY-SA 3.0 |
added 66 characters in body
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| Mar 20, 2017 at 14:13 | comment | added | Dead Possum | @fəˈnɛtɪk "Whole array can be printed in first iteration, run till the end is working, even if F and B form endless-loop condition." Doen't that counts? | |
| Mar 20, 2017 at 14:11 | history | edited | Dead Possum | CC BY-SA 3.0 |
added 966 characters in body
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| Mar 20, 2017 at 14:04 | comment | added | fəˈnɛtɪk | "If endless loop or backward progression requested, program should print nothing and exit" You are printing the array once on endless loop... | |
| Mar 20, 2017 at 14:01 | history | answered | Dead Possum | CC BY-SA 3.0 |