ryancdotorg

#!/usr/bin/env python

import os
import sys
import gzip
import itertools

from struct import pack, pack_into, unpack, unpack_from, calcsize
from binascii import hexlify, unhexlify
from subprocess import Popen, PIPE

tdunning

We want to show that any divisor $d | M(n) = 2^n-1$ is a Fermat pseudo-prime for base $2^{n/2}$. That would mean that $\big( 2^{(n/2)} \big)^{d-1} \equiv 1 \pmod d$. Assume $n$ is even and $n>4$.

1. $M(n) \equiv 2^n - 1 \equiv 0 \pmod d$ (given)
2. $2^n \equiv 1 \pmod d$ (restatement of step (1))
3. $\left( 2^{(n/2)}\right)^{d-1} \equiv 2^{\left(\frac {n(d-1)} 2 \right)} \equiv \left(2^n\right)^{\left( \frac {d-1} 2 \right)} \equiv 1 \pmod d$ (step (2), $d-1$ is even)
4. $(2^{n/2})^{d-1} \equiv 1 \pmod d \rightarrow$ $d$ is pseudo-prime in base $2^{n/2}$ (definition)

Thus, $d$ is a Fermat pseudo-prime for base $2^{n/2}$

thesamesam

FAQ on the xz-utils backdoor (CVE-2024-3094)

This is a living document. Everything in this document is made in good faith of being accurate, but like I just said; we don't yet know everything about what's going on.

Update: I've disabled comments as of 2025-01-26 to avoid everyone having notifications for something a year on if someone wants to suggest a correction. Folks are free to email to suggest corrections still, of course.

Background

darkerego

#! /bin/bash
# Modified to run both (redudantly, yes, I know, I am paranoid, you should be too) checks
#
set -eu

# find path to liblzma used by sshd
path="$(ldd $(which sshd) | grep liblzma | grep -o '/[^ ]*')"
echo 'Check one: does it even exist?'
# does it even exist?
if [ "$path" == "" ]

RobbedColek

from scapy.all import PPPoED, Ether, sniff, sendp, srp1, hexdump

source = b"\xXX\xXX\xXX\xXX\xXX\xXX" # MAC address of your adapter on PC
destination = b"\xXX\xXX\xXX\xXX\xXX\xXX" # MAC address of LAN on your PS4
interface = "Realtek PCIe 2.5GbE Family Controller #2" # get via "ipconfig /all" or eth0 or similiar on Linux

packet = sniff(iface=interface, filter="pppoed", count=1)
tag_value = packet[PPPoED][0].tag_list[1].tag_value
payload = destination + source + b"\x88\x63\x11\x07\x00\x00\x00\x0c\x01\x03\x00\x08" + tag_value
sendp(payload, iface=interface)

a1k0n

#include <stdint.h>
#include <stdio.h>
#include <string.h>
#include <unistd.h>
#include <math.h>

#define debug(...)
//#define debug printf

// torus radii and distance from camera

mlaprise

Training open-source LLMs on ChatGPT output is a really bad idea.

Everyone is now racing to create open-source alternatives to compete with GPT3.5/GPT4. A common shortcut used by some teams to bootstrap their effort is to fine-tune their model on ChatGPT output. I used to think it was a good idea and totally fair play to do this. Actually, I still think it’s fair play. OpenAI effectively distilled the entire web into its models. They are saying themself that they are using publicly accessible information (mostly). So distilling their model is, in effect, distilling the public open web, so small Term of Service details aside, I don’t see major ethical problems with that. Right? Well, it’s not entirely true and I realized now that, even when ignoring the ethical considerations, using their output is a really bad idea.

First of all, from a purely technical point of view, as @yoavgo is explaining it beautifully in his recent post, there is no way to align LLMs correctly without the RLHF component. I encourag

juliusgeo

20 million digits of pi in under a minute with Julia

I recently discovered a relatively obscure algorithm for calculating the digits of pi: https://en.wikipedia.org/wiki/Gauss–Legendre_algorithm. Well, at least obscure compared to Chudnovsky's. Wikipedia notes that it is "memory-intensive" but is it really? Let's compare to the MPFR pi function:

function gauss_legendre(prec)
    setprecision(BigFloat, prec, base=10)
    GC.enable(false)

juliusgeo

n=256;q, t, m, e, p, c, f=(range(n),16,lambda x
      ,y,       r=0:m((h:=x   <<1,h^283)[
     h&n!=      0],y>>      1,(r,r    ^x)
    [y&1])if    y else       r,lambda
   a,w= 1,p=n   -2:e(m(a,       a),(w,m
  (w,a))[p&1]   ,p>>               1)if p
 else       w,  lambda    b:         list(
map      (print,["%.2x "*t%(*b[r:r+t],)for
r in [*q][::t]])),lambda a,i: (a<<i|a>>8-i)&255
,lambda a: (a^c(a,1)^c(a,2)^c(a,3)^c(a,4))^99);

crypto-cypher

##### COURTESY
# @CuratedIntel
# https://curatedintel.org

##### BACKGROUND
# Initally shared on RAMP ransomware forum
# Last shared on Groove ransomware extortion website
# Publicized by Bleeping Computer, which led to this post being issued to help blue teamers
# https://www.bleepingcomputer.com/news/security/hackers-leak-passwords-for-500-000-fortinet-vpn-accounts/