Timeline for Finite State Machine validity [duplicate]
Current License: CC BY-SA 3.0
27 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 17, 2020 at 9:04 | history | edited | CommunityBot | Commonmark migration | |
| Jan 15, 2016 at 0:39 | history | closed | xnor CommunityBot | Duplicate of Golf Me An OOP! | |
| Jan 15, 2016 at 0:38 | comment | added | xnor | I think this is a dupe. While I think a pure canonical is-there-a-path question would distinguish itself from the many other, this one is yet another challenge to extract the graph from a somewhat inconveninent input format and then do pathfinding. While I have to choose a single question when specifying a dupe, there's many reachability questions that cover similar ground | |
| Jan 14, 2016 at 23:24 | history | edited | Askirkela | CC BY-SA 3.0 | Added a test case that doesn't start with A |
| Jan 14, 2016 at 23:18 | comment | added | Askirkela | In that case, I'll add one. | |
| Jan 14, 2016 at 23:17 | comment | added | user81655 | Test cases are meant for testing if a program follows the specs or not. Hardcoding A as the initial state would save 6 bytes in my solution, but technically make it invalid even though it passes all the test cases. The only way anyone would know is by studying the code. | |
| Jan 14, 2016 at 23:08 | comment | added | Askirkela | @user81655 It doesn't matter as I said The first character of the input is the FSM's initial state and the last character is the FSM's final state. | |
| Jan 14, 2016 at 23:07 | history | edited | Askirkela | CC BY-SA 3.0 | Changed third test case |
| Jan 14, 2016 at 23:06 | answer | added | user81655 | timeline score: 3 | |
| Jan 14, 2016 at 23:03 | comment | added | user81655 | You should also include a test case that doesn't start with A. | |
| Jan 14, 2016 at 23:03 | comment | added | Askirkela | @MegaTom You're totally right. I edit that right away | |
| Jan 14, 2016 at 22:54 | comment | added | MegaTom | The third test case is wrong as: In this challenge, the FSM is considered valid when you have at least one path from the initial state to the final state. or is there something I do not understand? | |
| Jan 14, 2016 at 21:34 | comment | added | Askirkela | @feersum I would say that [A-Z] are valid names to keep it simple. | |
| Jan 14, 2016 at 21:31 | comment | added | feersum | Which characters are valid names for states? | |
| Jan 14, 2016 at 21:30 | review | Close votes | |||
| Jan 15, 2016 at 0:43 | |||||
| Jan 14, 2016 at 21:27 | history | edited | Askirkela | CC BY-SA 3.0 | Took advices from comments |
| Jan 14, 2016 at 21:19 | comment | added | Askirkela | Thank you. I'll edit the post to take @MartinBüttner 's advices in account. If you think this is a dupe, I'm sorry but since I haven't found a challenge close to what I had in mind, I posted this one. | |
| Jan 14, 2016 at 21:14 | comment | added | Peter Taylor | @feersum, pretty much everything in path-finding is close to a dupe, but I can't find such a minimalistic reachability question before, so I will hold off on the supervote. | |
| Jan 14, 2016 at 20:55 | comment | added | Digital Trauma | @FryAmTheEggman I think this is supposed to illustrate an answer with only the second bonus | |
| Jan 14, 2016 at 20:53 | comment | added | feersum | Darn.. I looked in Peter Taylor's graph theory challenge index for a dupe target, but the entry under reachability was 'too many to mention'. | |
| Jan 14, 2016 at 20:52 | comment | added | Digital Trauma | Welcome to PPCG! With respect to bonusses, check out this meta answer. 4 votes is certainly not to be taken as consensus, but IMO your bonuses - especially the first one - add little to the challenge | |
| Jan 14, 2016 at 20:50 | comment | added | feersum | This is not that related to finite state machines. It's just reachability in a directed graph. | |
| Jan 14, 2016 at 20:47 | comment | added | Addison Crump | Never assume that we know. :P I know I don't know what a Finite State Machine is. | |
| Jan 14, 2016 at 20:46 | comment | added | Martin Ender | It would also be a good to have a truthy test case where some state other than the final state is not reachable. | |
| Jan 14, 2016 at 20:46 | comment | added | Martin Ender | You should probably clarify that the transition pairs are not overlapping. So ABCD represents only two transitions A->B and C->D and not B->C. Otherwise the challenge would be trivial. | |
| Jan 14, 2016 at 20:42 | review | First posts | |||
| Jan 14, 2016 at 21:13 | |||||
| Jan 14, 2016 at 20:42 | history | asked | Askirkela | CC BY-SA 3.0 |