Jason Orendorff jorendorff

This was written in response to: https://twitter.com/hillelogram/status/1227118562373963776

People mix the predicate "If a directed graph is a tree, it's a DAG" with the proposition "For all directed graphs, if a directed graph is a tree, it's a DAG".

First, for simplification: let's assume we're talking here about all graphs in the world with up to 10 nodes to be able to enumerate them. Also, note that we're talking about directed, not undirected, graphs.

To prove the proposition "For all..." is true, which is "(For graph G1, if it's a tree, it's a DAG) AND (For graph G2, if it's a tree, it's a DAG) AND...", since it's ANDs, we need to prove (For graph G1, if...) is true and we need to prove (For graph G2, if...) is true, ... etc.

	# Some more privileged methods, with some private data.
	class Availability:
	def __init__(self):
	self._is_open = False

	def open(self):
	self._is_open = True

	def close(self):
	self._is_open = False

	extern crate termion;
	use std::io::{Write, Stdout, stdout, stdin};
	use termion::event::{Key, Event};
	use termion::input::TermRead;
	use termion::raw::{IntoRawMode, RawTerminal};
	use termion::screen::AlternateScreen;


	pub struct Editor {
	screen: AlternateScreen<Stdout>,

	import random

	random.seed()

	def flip():
	return random.choice([True, False])

	def run():
	b = flip()
	n = 1

	use std::fmt::Debug;

	struct Reader<'a> {
	buf: &'a [u8],
	}

	impl<'a> Reader<'a> {
	fn read<'b>(&'b mut self, len: usize) -> Result<&'a [u8], ()> {
	if len <= self.buf.len() {
	let (read, remains) = self.buf.split_at(len);