LightNews — News, ranked by scholars

ProgrammerHumor.io

@programmerhumor-io.bsky.social

ChatGPT Code Leaked

#programming #python #machinelearning #ifstatement #Programminghumor

https://programmerhumor.io/ai-memes/chatgpt-code-leaked-xw65

October 21, 2026 at 3:16 PM

Epstein Web

@epsteinweb.bsky.social

Parts 1-2/2 — EFTA01139475.jpg
#epsteinweb #efta01139475
https://epsteinweb.org
Available in the iOS app store now!
https://apps.apple.com/us/app/epstein-web/id6758880661

EFTA01139453 | EFTA01139475
V
1 •

Figure 7: Boolean-value gadget: Dashed edges are in Et and non-dashed are in Eq.
where 1 ≤ a, k ≤ nr, val is either true or false, each cnj is either a Boolean variable be or its negation Ige. An ifstatement is satisfied if each of the cnj is true (i.e., cnj is true if one of the following holds: if crt, is be and be is true,
or cnj is bi and bt is false).
Concurrent-if system. A concurrent-if system consists of a set B
, b2
b„} of n Boolean variables and a set
s„,} of rn if-statements over the Boolean variables in B. The set of statements defines an execution
P = {at, 82,
from an initial setting of the Boolean variables as follows: repeatedly, a satisfied if-statement Mcn
, cnk)
:= val is selected and then bi is set to val. If the first Boolean variable Eh is eventually true, then the execution is
accepted. If at each point of an execution there is at most one satisfied if-statement, then we say that the execution is
deterministic.
The decision problem. Given a concurrent-if system, the associated decision problem is as follows: Given a set B
of Boolean variables, an initial setting of the variables of B, and a set P of concurrent if-statements, such that the
execution e from the initial setting is deterministic, whether e is accepting.
5.2.1 Reduction of the concurrent-if problem to evolutionary games on graphs
We first describe how we encode the Boolean variables and the if-statements of a concurrent-if system in evolutionary
games on graphs. Later we show how to construct a graph such that if the fixation of the mutant happens, then the
fixation happens according to three phases which are as follows.
I. First phase: The setup phase (to initialize the Boolean variables).
2. Second phase: The execution of the concurrent-if system.
3. Third phase: The fixation phase.
The fixation can only happen if the execution of the concurrent-if system accepts.
Density constraint Again our lower bound result will be for a special case of LBF,

where we have constant fitness
with density constraints (recall Remark l). Our construction will be for Oft = Bit = 0, but a similar construction will
work for any choice of OR,Oni E [0, 1). The thresholds Oft = b yt = 0 indicates that a vertex v can reproduce precisely
as long as all its successors in El are of the opposite type of v, because of the density constraint.
Ideas and gadgets behind the reduction. We first introduce some key ideas and gadgets behind the reduction.
• States which are nearly always a mutant/resident: Similar to the previous lower bounds, we have a vertex vs
without any predecessor in Eq. Thus, if vs is not made a mutant at the start, then it cannot become a mutant.
Hence we will only consider the case when vs is a mutant in the beginning and stays a mutant forever. We
will also have a vertex
and our construction will ensure that it stays a resident until all other vertices are
mutants and then (after a few more steps) all vertices become mutants, and we get fixation. We will use the
vertices vs and it s to ensure that a given vertex has a desired type, and otherwise the vertex cannot reproduce.
Our construction will ensure (using the density constraint) the following properties:
— A vertex v with 6, as a successor under El can only reproduce if it is a mutant (using the density constraint
and re, is a resident). Similarly, a vertex v with vs as a successor under Ef can only reproduce if it is a
resident.

EFTA01139475

May 10, 2026 at 3:10 PM

Mark J. Nelson

@mm-jj-nn.bsky.social

I've done a version of this demo at least twice in the past and never hit this, I assume because I just never happened to try to use a camelCase rule name like ifStatement in the past, but this year I did.

September 28, 2023 at 6:23 PM

sai

@texoport.in

big fixes coming to parserator

(repost bec red arrow was fucked earlier)

const ifStatement: Parser<Statement> = parser(function* () {
yield* keyword("if");
yield* commit();

yield* token(char("(")).expect("opening parenthesis after 'if'");
const test = yield* expression.expect("condition expression");
yield* token(char(")")).expect("closing parenthesis");

const consequent = yield* statement.expect("if body");

const alternate = yield* optional(
parser(function* () {
yield* keyword("else");
yield* commit();
return yield* statement.expect("else body");
})
);

return { type: "if" as const, test, consequent, alternate };
});

$Error at line 1, column 3: > 1 | if x > 0 { } ^ Expected: opening parenthesis after 'if'$

June 4, 2025 at 12:49 AM

JFM Digital Works

@jfmdigitalworks.bsky.social

What’s an if statement? It’s how your code makes decisions.
If something’s true ➜ do a thing.
Short post, simple explanation, quick challenge to try!
#javascript #learnprogramming #codingforall #webdev #ifstatement

jamesfmcgrath.org/what-is-an-i...

What Is an If Statement in JavaScript? A Beginner-Friendly Guide

Learn how if statements work in JavaScript, including how to write them, why they're useful, and how to use them to control program flow. Includes examples, analogies, and a simple coding challenge fo...

jamesfmcgrath.org

July 24, 2025 at 6:43 AM

James McGrath

@jamesmcgrath.bsky.social

What’s an if statement? It’s how your code makes decisions.
If something’s true ➜ do a thing.
Short post, simple explanation, quick challenge to try!
#javascript #learnprogramming #codingforall #webdev #ifstatement

jamesfmcgrath.org/what-is-an-i...