Timeline for Simplify the boolean function $$Z=A\bar B \bar{C_i} + \bar A B \bar{C_i} + \bar A\bar B {C_i} + A B {C_i}$$
Current License: CC BY-SA 4.0
25 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Sep 12, 2020 at 20:06 | history | bounty ended | CommunityBot | ||
| S Sep 12, 2020 at 20:06 | history | notice removed | CommunityBot | ||
| Sep 9, 2020 at 7:22 | vote | accept | Ski Mask | ||
| Sep 7, 2020 at 23:22 | answer | added | jcaron | timeline score: 4 | |
| Sep 7, 2020 at 18:51 | comment | added | sergiu reznicencu | Or better. Just assume the final result contains all inputs used once with one operator in between. This is very much possible | |
| Sep 7, 2020 at 18:49 | comment | added | sergiu reznicencu | Brute force would be cool. I wonder if some optimization algorithm could be found for minimizing the number of operators used. Some kind of derivative for this problem? Perhaps AI? | |
| Sep 7, 2020 at 17:57 | comment | added | Sean | @Ski Mask, I was actually asking out of my own interest, and I'm still interested in the answer! | |
| Sep 7, 2020 at 15:00 | history | tweeted | twitter.com/StackElectronix/status/1302984991903756289 | ||
| Sep 7, 2020 at 11:19 | comment | added | Ski Mask | @Sean Unfortunately this is from a written question from a past paper. So I have to show all my working out and the laws that I used. | |
| Sep 7, 2020 at 10:15 | answer | added | schnedan | timeline score: 0 | |
| Sep 5, 2020 at 22:04 | comment | added | Sean | Are there any solver tools that can brute-force (or use other heuristics to obtain) a solution to this sort of problem? | |
| Sep 5, 2020 at 14:37 | review | Suggested edits | |||
| Sep 5, 2020 at 16:32 | |||||
| Sep 5, 2020 at 13:49 | answer | added | Shashank V M | timeline score: 5 | |
| Sep 4, 2020 at 21:50 | comment | added | cjferes | Seems like $$A\equiv B$$ is the same as $$\overline{(A\oplus B)}$$. In that case, $$\begin{align} Z&=\overline{C_i} (A \oplus B) + C_i (A\equiv B)\\ &=\overline{C_i} (A \oplus B) + C_i\overline{(A \oplus B)} \\ &= C_i \oplus (A\oplus B)\\ &=A\oplus B\oplus C_i \end{align}$$ | |
| Sep 4, 2020 at 19:10 | comment | added | edmz | By \$X\equiv Y \$, what do you mean exactly? I've never seen this notation so far. | |
| S Sep 4, 2020 at 18:22 | history | bounty started | Ski Mask | ||
| S Sep 4, 2020 at 18:22 | history | notice added | Ski Mask | Authoritative reference needed | |
| Sep 3, 2020 at 12:44 | comment | added | Ski Mask | Wouldn't it be (NOT X AND Y) OR (X AND Y)? | |
| Sep 3, 2020 at 12:07 | comment | added | jcaron | It's another instance of (X AND NOT Y) OR (NOT X AND Y) = X XOR Y, with X being Ci here and Y being A XOR B. | |
| Sep 3, 2020 at 12:02 | comment | added | Ski Mask | @jcaron Yes but I'm trying to figure out why $$\bar C_i(A \oplus B) + C_i(A \equiv B)=A \oplus B \oplus C_i$$. | |
| Sep 2, 2020 at 21:05 | comment | added | jcaron | $$A \equiv B$$ is the same as NOT $${(A \oplus B)}$$ (sorry not very good at formulas) | |
| Sep 2, 2020 at 20:11 | history | edited | Ski Mask | CC BY-SA 4.0 |
[Edit removed during grace period]
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| Sep 2, 2020 at 19:34 | comment | added | DKNguyen | You need a double dollar sign for your title. | |
| Sep 2, 2020 at 19:34 | history | edited | DKNguyen | CC BY-SA 4.0 |
edited title
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| Sep 2, 2020 at 18:14 | history | asked | Ski Mask | CC BY-SA 4.0 |