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Programming Language Checklist | |
by Colin McMillen, Jason Reed, and Elly Fong-Jones, 2011-10-10. | |
For the Dusa programming language, https://dusa.rocks | |
You appear to be advocating a new: | |
[ ] functional [ ] imperative [ ] object-oriented [ ] procedural [ ] stack-based | |
[ ] "multi-paradigm" [ ] lazy [x] eager [ ] statically-typed [x] dynamically-typed | |
[x] pure [ ] impure [ ] non-hygienic [ ] visual [ ] beginner-friendly | |
[ ] non-programmer-friendly [x] completely incomprehensible |
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first_word | second_word | first_much_more_familiar | first_more_familiar | about_the_same | second_more_familiar | second_much_more_familiar | |
---|---|---|---|---|---|---|---|
a | axion | 1 | 0 | 0 | 0 | 0 | |
a | chase | 1 | 0 | 0 | 0 | 0 | |
a | decal | 1 | 0 | 0 | 0 | 0 | |
a | desire | 1 | 0 | 0 | 0 | 0 | |
a | dowry | 0 | 0 | 1 | 0 | 0 | |
a | drying | 1 | 0 | 1 | 0 | 0 | |
a | grandfathers | 1 | 0 | 0 | 0 | 0 | |
a | grate | 2 | 0 | 0 | 0 | 0 | |
a | halve | 1 | 0 | 0 | 0 | 0 |
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#### First proof I wrote #### | |
Given ~A & ~B | |
[ Suppose A | B | |
[ Suppose A | |
~A # By conjunction elimination left with (2) | |
F ] # By negation elimination with (5) and (4) | |
[ Suppose B | |
~B # By conjunction elimination right with (2) | |
F ] # By negation elimination with (8) and (7) | |
F ] # By disjunction elimination with (3), (4-6), (7-9) |
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import { hammock } from "@calculemus/oli-hammock"; | |
//import * as widgets from "@calculemus/oli-widgets"; | |
import { Set } from "immutable"; | |
type box = { value: string; correct: boolean }; | |
const emp = { value: "", correct: false }; | |
function renderBox(box: box, id: string) { | |
$(`#${id}`).val(box.value); | |
if (box.value === "" || box.correct) { |
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// A non-empty rope is either a leaf (string) or a node with two children | |
// that are both non-empty ropes | |
interface Tagged<T> { | |
tag: T; | |
} | |
type RopeNode = | |
| Tagged<"node"> & { | |
left: RopeNode; | |
right: RopeNode; |
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CREATE TABLE `feedback` ( | |
`job` int NOT NULL, | |
`task` char(20) NOT NULL, | |
`payload` longtext NOT NULL, | |
`timestamp` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP, | |
UNIQUE KEY `job` (`job`, `task`), | |
CONSTRAINT `feedback_job` FOREIGN KEY (`job`) REFERENCES `job` (`job`) ON DELETE CASCADE | |
); |
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Original puzzle: https://twitter.com/tausbn/status/251686364788170752 | |
First observation: | |
If G ->* {0} implies G =>* {0}, it must be the case that if | |
G -> G' and G' =>* {0} then G =>* {0}. | |
Proof: Given G -> G' and G' =>* {0}, we have G' ->* {0} | |
immediately (->* contains =>*) and G ->* {0} by direct | |
composition (G -> G' ->* {0}). Then we have G =>* {0} by the |
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\documentclass[12pt,openany]{article} | |
\begin{document} | |
\setcounter{page}{51} | |
Basically just read \cite{And01}. | |
\begin{thebibliography}{WCPW02} |
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structure A = | |
struct | |
val say = fn () => print "A\n" | |
end | |
structure B = | |
struct | |
val say = fn () => print "B\n" | |
end |
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datatype foo = A | B | C | D | |
(*[ datasort ab = A | B ]*) | |
(*[ datasort bc = B | C ]*) | |
(*[ datasort abc = A | B | C ]*) | |
(* This works *) | |
(*[ val x: ab & bc ]*) | |
val x = B | |
(* This, however, does not... |
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