Timeline for Quantum Phase Estimation Circuit and Modular Exponentiaton
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 26, 2021 at 9:01 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| May 29, 2021 at 8:09 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jan 29, 2021 at 8:03 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Oct 1, 2020 at 7:06 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jun 3, 2020 at 7:38 | comment | added | DaftWullie | There are different conventions for the ordering of the bits when converting to binary. For example, "big endian" or "little endian", which is worth keeping an eye on. | |
| Jun 3, 2020 at 7:00 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| May 3, 2020 at 22:28 | answer | added | usercs | timeline score: 1 | |
| May 3, 2020 at 22:24 | comment | added | usercs | Isn't it in the reverse order? $j=j_t 2^{t-1} + \dots + j_1 2^0$? This is where I am confused. | |
| May 3, 2020 at 15:45 | comment | added | glS♦ | $j_i$ are by definition the base-2 digits of $j$, thus $j=j_1 2^{t-1} + j_2 2^{t-2} + ... + j_t 2^0$. Is this what you are asking? | |
| May 3, 2020 at 10:30 | history | asked | usercs | CC BY-SA 4.0 |