Timeline for What is the point at infinity on secp256k1 and how to calculate it?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Jan 1, 2022 at 17:27 | history | edited | fgrieu♦ | CC BY-SA 4.0 |
Minor fix and polish
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| Jul 16, 2020 at 11:02 | history | edited | kelalaka | CC BY-SA 4.0 |
added 52 characters in body
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| Feb 18, 2020 at 10:12 | history | edited | kelalaka | CC BY-SA 4.0 |
polish
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| Feb 16, 2020 at 13:31 | vote | accept | PouJa | ||
| Jan 7, 2019 at 18:33 | history | edited | kelalaka | CC BY-SA 4.0 |
added the image
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| Jan 6, 2019 at 11:23 | history | edited | kelalaka | CC BY-SA 4.0 |
polish
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| Jan 6, 2019 at 11:17 | comment | added | kelalaka | The points are forming an additive group with identity $O$ so $P+(-P) = O$. Wikipedia also says the opposite of a point $P$ is $-P$. For the programming part, unfortunately, off-topic here. Please ask a question on StackOverflow. | |
| Jan 6, 2019 at 10:57 | comment | added | PouJa | @kelalaka the Wikipedia link you have supplied is saying that if P and Q are opposites of each other, we define $P + Q = O$. Why we cannot conclude $P+(-P)=O$ is always true? By the way what is the random looking output value is that my computer program in python is giving as output? finally is it possible to calculate $O$ so that for any randomly given $P$ the computer program can calculate $P+O=P$? | |
| Jan 6, 2019 at 10:18 | history | edited | kelalaka | CC BY-SA 4.0 |
typo thanks to dave_thompson_085
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| Jan 6, 2019 at 10:13 | comment | added | dave_thompson_085 | In 2 you mean P + Q = O (or P + -P) | |
| Jan 5, 2019 at 18:55 | history | edited | kelalaka | CC BY-SA 4.0 |
polish
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| Jan 5, 2019 at 17:24 | history | answered | kelalaka | CC BY-SA 4.0 |