Timeline for Which really big number is bigger?
Current License: CC BY-SA 4.0
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| Jun 17, 2020 at 9:04 | history | edited | CommunityBot |
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| May 18, 2019 at 8:26 | history | edited | user58988 | CC BY-SA 4.0 |
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| May 18, 2019 at 6:13 | history | edited | user58988 | CC BY-SA 4.0 |
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| May 18, 2019 at 4:53 | history | edited | user58988 | CC BY-SA 4.0 |
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| May 18, 2019 at 4:48 | history | edited | user58988 | CC BY-SA 4.0 |
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| May 17, 2019 at 15:47 | history | edited | user58988 | CC BY-SA 4.0 |
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| May 17, 2019 at 15:04 | comment | added | Simply Beautiful Art |
Perhaps it fails 9, 10, 10, 10, 10, 10, 9, 10, 10, 10, which should return 1, but s(9,10,10,10,10) < s(10,9,10,10,10).
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| May 17, 2019 at 13:39 | history | edited | user58988 | CC BY-SA 4.0 |
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| May 17, 2019 at 12:39 | history | edited | user58988 | CC BY-SA 4.0 |
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| May 17, 2019 at 12:33 | comment | added | user58988 | @SimplyBeautifulArt the test 10 10 2 3 10 z 10 10 8 3 9 return 0 | |
| May 17, 2019 at 12:20 | comment | added | user58988 | @SimplyBeautifulArt in that case Apl return infinity both side, and so code would use the s() function... | |
| May 17, 2019 at 12:17 | history | edited | user58988 | CC BY-SA 4.0 |
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| May 17, 2019 at 12:09 | comment | added | Simply Beautiful Art |
I'm not really sure how your code works, so I'm looking forward to that explanation. Perhaps it fails a test case like 10, 10, 2, 3, 10, 10, 10, 8, 3, 9, which should return 0.
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| May 17, 2019 at 12:05 | comment | added | Simply Beautiful Art |
You said you calculated each by taking log(log()), but for that test case, the difference between log(log(10^10^10^10^10)) and log(log(9^10^10^10^10)) would require an absurd amount of accuracy to pick up on. You'd need to have a floating point with about 2e10 base 10 digits of accuracy. And this is ignoring the fact that both sides are approximately as large as 10^10^10, which I find it hard to believe you were able to compute.
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| May 17, 2019 at 11:59 | comment | added | user58988 | @SimplyBeautifulArt 10 10 10 10 10 z 9 10 10 10 10 return 1 so the left > right... What does it mean"logging twice"? | |
| May 17, 2019 at 11:52 | history | edited | user58988 | CC BY-SA 4.0 |
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| May 17, 2019 at 11:50 | comment | added | Simply Beautiful Art |
Only logging twice leaves you vulnerable to fail test cases like 10, 10, 10, 10, 10, 9, 10, 10, 10, 10.
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| May 17, 2019 at 11:28 | comment | added | Kevin Cruijssen | If all test cases are correct, then +1 from me. Looking forward seeing an explanation of your code. :) | |
| May 17, 2019 at 11:26 | comment | added | user58988 | @KevinCruijssen Here your test , if exclude the one above seems ok... | |
| May 17, 2019 at 11:26 | comment | added | Kevin Cruijssen | Ah oops, that one is indeed supposed to be true. Copy-paste error.. My bad. All the other test cases should be correct. | |
| May 17, 2019 at 11:25 | history | edited | user58988 | CC BY-SA 4.0 |
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| May 17, 2019 at 10:58 | comment | added | user58988 | @KevinCruijssen the test 5 4 3 2 1 z 1 2 3 4 5 would result true n>1 your test would be wrong on that case | |
| May 17, 2019 at 10:24 | comment | added | Kevin Cruijssen | The tests in the challenge description are lacking some edge cases. Could you verify that it's also working for all these test cases? | |
| May 17, 2019 at 10:07 | history | edited | user58988 | CC BY-SA 4.0 |
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| May 17, 2019 at 9:41 | history | edited | user58988 | CC BY-SA 4.0 |
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| May 17, 2019 at 9:36 | history | answered | user58988 | CC BY-SA 4.0 |