Readit News“My point today is that, if we wish to count lines of code, we should not regard them as ‘lines produced’ but as ‘lines spent’: the current conventional wisdom is so foolish as to book that count on the wrong side of the ledger.”
https://www.cs.utexas.edu/~EWD/transcriptions/EWD10xx/EWD109...
But why not both?
That's an assumption increasingly false, unfortunately. The spirit of collegiality has been beaten back.
Far better to hone logical skills that sift between fact and error than to rely on social reputation. Ironically we're discussing a text designed to do exactly that.
The savvy LLM user already knows to be on the lookout for falsehood, if not bad pedagogy. That's a benefit, not a drawback of LLMs.
Susanna Epp's Discrete Mathematics With Applications is also a really good option
Following Dijkstra’s EWD1094, here’s a way to solve the hairs-on-heads problem eschewing the language of pigeonholes and employing the fact that the mean is at most the maximum of a non-empty bag of numbers.
We are given that Boston has 500,000 non-bald people. The human head has at most 200,000 hairs. Show that there must be at least 3 people in Boston who have the same number of hairs on their head.
Each non-bald Bostonian must have a hair count between 1 and 200,000. The average number of such people per hair count is 500,000 / 200,000 = 2.5. The maximum is at least that; moreover, it must be a round number. So the maximum >= 3. QED.
Wait! Crap!
We can’t sanction Russia - if we do it might destabilize the Russian dictator and if he goes out a worse authoritarian regime might come to power!
> By that logic the US shouldn’t get involved in any other foreign entanglement or global police action because of unintended consequences.
Strawman. No-one is claiming that.
The EU doesn’t even recognize Maduro as the legitimate leader of Venezuela and he forced the democratically elected president into hiding!
Where is the right you're seeing?
Problem: A plane has every point colored red, blue, or yellow. Prove that there exists a rectangle whose vertices are the same color.
Solution: Consider a grid of points, 4 rows by 82 columns. There are 3^4=81 possible color patterns of columns, so by the Pigeonhole Principle, at least two columns have the same color pattern. Also by the Pigeonhole Principle, each column of 4 points must have at least two points of the same color. The two repeated points in the two repeated columns form a rectangle of the same color. QED.
The Pigeonhole Principle is very neat. It would be hard not to use it for the proof.
Partly that article argues against proof by contradiction which does seem to be overused.
For fun, try strengthening the result to a square.
https://codeberg.org/inetutils/inetutils/commit/fa3245ac8c28...
One of the changes is:
What is the reason for a rename these days? If I saw that in a code review I’d immediately get annoyed (and probably pay more attention)The present fix is to sanitize user input. Does it cover all cases?