Timeline for Is there a single straight line in the matrix?
Current License: CC BY-SA 4.0
24 events
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| Jun 19, 2021 at 18:17 | history | edited | theorist | CC BY-SA 4.0 |
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| Jun 19, 2021 at 9:28 | history | edited | theorist | CC BY-SA 4.0 |
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| Jun 19, 2021 at 9:17 | history | edited | theorist | CC BY-SA 4.0 |
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| Jun 19, 2021 at 7:27 | history | edited | theorist | CC BY-SA 4.0 |
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| Jun 19, 2021 at 6:51 | history | edited | theorist | CC BY-SA 4.0 |
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| Jun 19, 2021 at 6:47 | comment | added | theorist | @Nitrodon Nice catch! I've modified my code to account for the type of case you identified. | |
| Jun 19, 2021 at 6:45 | history | edited | theorist | CC BY-SA 4.0 |
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| Jun 17, 2021 at 17:17 | history | edited | theorist | CC BY-SA 4.0 |
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| Jun 17, 2021 at 13:55 | comment | added | Nitrodon |
Fails on {{1, 1, 0, 0}, {1, 1, 0 ,0}, {0, 0, 0, 0}, {0, 0, 0, 0}, which has operator norm 2.
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| Jun 17, 2021 at 12:27 | comment | added | ZaMoC | Nice answer! I'm glad I helped. (no need for credits..) ;) | |
| Jun 17, 2021 at 12:10 | history | edited | theorist | CC BY-SA 4.0 |
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| Jun 17, 2021 at 12:08 | comment | added | theorist | @ZaMoC Thanks, I wasn't quite sure how to set that up with tio.run. If you don't mind, I'll add your link to my answer, crediting you. | |
| Jun 17, 2021 at 12:04 | comment | added | ZaMoC | here are all test cases: Try it online! | |
| Jun 17, 2021 at 12:00 | comment | added | MarcMush |
right sorry, it seems the correct equivalent in Julia would be opnorm
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| Jun 17, 2021 at 11:56 | comment | added | theorist |
Flatten@SingularValueDecomposition@m = {1/Sqrt[2], 0, -(1/Sqrt[ 2]), 1/Sqrt[2], 0, 1/Sqrt[2], 0, 1, 0, Sqrt[2], 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1}; hence Norm@m returns the largest value, which is Sqrt[2]. Thus (Norm@m)^2 = 2. But m is 3x3, and 2!=3, so my funciton returns False.
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| Jun 17, 2021 at 11:56 | comment | added | theorist |
@MarcMush I tried your test matrix, and it correctly returned False. "norm" seems to means something different in Julia—according to discourse.julialang.org/t/performance-of-norm-function/14709/9, it instead computes the Frobenius norm. In Mathematica, when Norm is applied to a matrix, it does a singular value decomposition, and then selects the element with the maximum value. So, for your matrix m= {{1, 0, 0}, {1, 0, 0}, {0, 1, 0}}, [continued]
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| Jun 17, 2021 at 11:28 | comment | added | MarcMush |
I don't have mathematica but it seems like this only counts if there is the right amount of 1s Try it online with Julia!
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| Jun 17, 2021 at 8:26 | history | edited | theorist | CC BY-SA 4.0 |
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| Jun 17, 2021 at 6:43 | history | edited | theorist | CC BY-SA 4.0 |
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| Jun 17, 2021 at 6:29 | comment | added | Bubbler | I think it should be fine to strip the if statement entirely (giving true or false). | |
| Jun 17, 2021 at 6:26 | history | edited | theorist | CC BY-SA 4.0 |
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| Jun 17, 2021 at 6:04 | history | edited | theorist | CC BY-SA 4.0 |
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| Jun 17, 2021 at 5:51 | history | edited | theorist | CC BY-SA 4.0 |
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| Jun 17, 2021 at 5:40 | history | answered | theorist | CC BY-SA 4.0 |