Newest at the top
| 2026-03-12 20:34:07 +0100 | <monochrom> | Given that non-determinism cannot be used with mutable variables, you have a multiple-world semantics, such that there is no communication between two sibling worlds. |
| 2026-03-12 20:34:05 +0100 | ski | . o O ( "Concepts, Techniques, and Models of Computer Programming" by Peter Van Roy,Seif Haridi in 2003-06-05 at <https://webperso.info.ucl.ac.be/~pvr/book.html>,<https://webperso.info.ucl.ac.be/~pvr/VanRoyHaridi2003-book.pdf> ) |
| 2026-03-12 20:31:29 +0100 | target_i | (~target_i@user/target-i/x-6023099) target_i |
| 2026-03-12 20:31:10 +0100 | <monochrom> | Yeah I'm thinking "what's there to handle?" because the only thing left to do is collecting results. |
| 2026-03-12 20:31:08 +0100 | AlexNoo | (~AlexNoo@5.139.232.240) |
| 2026-03-12 20:30:58 +0100 | AlexNoo | (~AlexNoo@5.139.232.240) (Ping timeout: 248 seconds) |
| 2026-03-12 20:30:37 +0100 | ski | was (re)reading it recently (didn't get to that part yet, though) |
| 2026-03-12 20:30:13 +0100 | <ski> | (the CTM book talks about this, in the chapter about constraint programming) |
| 2026-03-12 20:29:50 +0100 | <ski> | presumably the alternation acts as a barrier, where information can flow into the "engine", but not out of (except at the end of the computation, collecting results) |
| 2026-03-12 20:29:28 +0100 | <monochrom> | I don't know. |
| 2026-03-12 20:28:49 +0100 | <ski> | how does it handle trying to communicate a result non-locally out of an alternative ? |
| 2026-03-12 20:27:25 +0100 | <monochrom> | Yeah |
| 2026-03-12 20:27:11 +0100 | <ski> | breadth-first ? |
| 2026-03-12 20:26:51 +0100 | <monochrom> | There is a Curry compiler, Curry2Go, which compiles non-determinism to Go multi-threading so all alternatives are literally tried concurrently. |
| 2026-03-12 20:26:42 +0100 | AlexNoo__ | (~AlexNoo@5.139.232.240) (Ping timeout: 248 seconds) |
| 2026-03-12 20:26:33 +0100 | AlexNoo | (~AlexNoo@5.139.232.240) |
| 2026-03-12 20:26:25 +0100 | <ski> | oh, and yes, the system would ensure that both are not defined at the same time |
| 2026-03-12 20:25:59 +0100 | <EvanR> | similar to lub |
| 2026-03-12 20:25:52 +0100 | <ski> | (and so the recursion pattern of them both must align) |
| 2026-03-12 20:25:40 +0100 | <ski> | they are tried, simultaneously |
| 2026-03-12 20:25:38 +0100 | AlexNoo_ | (~AlexNoo@5.139.232.240) (Ping timeout: 248 seconds) |
| 2026-03-12 20:25:31 +0100 | <ski> | they are not independent operations, one tried after the other |
| 2026-03-12 20:25:29 +0100 | <monochrom> | <3 unordered alternation |
| 2026-03-12 20:25:13 +0100 | <EvanR> | is the commutativity guaranteed somehow by the compiler |
| 2026-03-12 20:25:13 +0100 | <ski> | operational semantics should be as if you executed `(-?<) :: Nat -> Nat -> Either () Nat' |
| 2026-03-12 20:25:06 +0100 | AlexNoo | (~AlexNoo@5.139.232.240) (Ping timeout: 248 seconds) |
| 2026-03-12 20:24:41 +0100 | <EvanR> | oh |
| 2026-03-12 20:24:34 +0100 | <ski> | and `|' is commutative |
| 2026-03-12 20:24:25 +0100 | <ski> | types have to match |
| 2026-03-12 20:23:02 +0100 | <EvanR> | and the types don't have to match immediately |
| 2026-03-12 20:22:25 +0100 | <EvanR> | so the Succ ((m - n) / n) will be "tried" first and if it's undefined go to the thing right of | |
| 2026-03-12 20:22:09 +0100 | AlexNoo__ | (~AlexNoo@5.139.232.240) |
| 2026-03-12 20:21:25 +0100 | AlexNoo_ | (~AlexNoo@5.139.232.240) |
| 2026-03-12 20:21:22 +0100 | AlexNoo__ | (~AlexNoo@5.139.232.240) (Ping timeout: 248 seconds) |
| 2026-03-12 20:20:51 +0100 | <ski> | (that does not terminate for `n = 0'. dependently typed version i hinted at above is total, though) |
| 2026-03-12 20:20:51 +0100 | Ranhir | (~Ranhir@157.97.53.139) (Ping timeout: 246 seconds) |
| 2026-03-12 20:20:42 +0100 | AlexNoo | (~AlexNoo@5.139.232.240) |
| 2026-03-12 20:20:20 +0100 | <ski> | m / n = Succ ((m - n) / n) | let () = m ?< n in Zero ; m % n = (m - n) % n | let () = m ?< n in m |
| 2026-03-12 20:20:18 +0100 | AlexNoo_ | (~AlexNoo@5.139.232.240) (Ping timeout: 248 seconds) |
| 2026-03-12 20:19:46 +0100 | AlexNoo | (~AlexNoo@5.139.232.240) (Ping timeout: 248 seconds) |
| 2026-03-12 20:17:19 +0100 | <EvanR> | what does applying both operators look like |
| 2026-03-12 20:17:02 +0100 | <ski> | (if you're just using one of them, it would be partial, yes) |
| 2026-03-12 20:16:49 +0100 | AlexNoo__ | (~AlexNoo@5.139.232.240) |
| 2026-03-12 20:16:49 +0100 | <ski> | nope, because you're supposed to pair up an application of `-', with an application of `?<' so that one of them will be defined, succeed |
| 2026-03-12 20:16:05 +0100 | AlexNoo_ | (~AlexNoo@5.139.232.240) |
| 2026-03-12 20:16:02 +0100 | AlexNoo__ | (~AlexNoo@5.139.232.240) (Ping timeout: 248 seconds) |
| 2026-03-12 20:15:38 +0100 | <EvanR> | so the program will crash or |
| 2026-03-12 20:15:24 +0100 | <EvanR> | Zero - Succ _ = undefined |
| 2026-03-12 20:15:22 +0100 | AlexNoo | (~AlexNoo@5.139.232.240) |
| 2026-03-12 20:14:58 +0100 | AlexNoo_ | (~AlexNoo@5.139.232.240) (Ping timeout: 248 seconds) |