nlitsme

Demonstrate how to calculate the messagehash for the two signatures in this transaction

See ecdsa_demo.py for code showing how to use this to crack the bitcoin secret key.

These are the values extracted from the example transaction below:

pk="04 db d0 c6 15 32 27 9c f7 29 81 c3 58 4f c3 22 16 e0 12 76 99 63 5c 27 89 f5 49 e0 73 0c 05 9b 81 ae 13 30 16 a6 9c 21 e2 3f 18 59 a9 5f 06 d5 2b 7b f1 49 a8 f2 fe 4e 85 35 c8 a8 29 b4 49 c5 ff"
r="d4 7c e4 c0 25 c3 5e c4 40 bc 81 d9 98 34 a6 24 87 51 61 a2 6b f5 6e f7 fd c0 f5 d5 2f 84 3a d1"
s1="44 e1 ff 2d fd 81 02 cf 7a 47 c2 1d 5c 9f d5 70 16 10 d0 49 53 c6 83 65 96 b4 fe 9d d2 f5 3e 3e"

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from z3 import *
# https://www.youtube.com/watch?v=9ATC_uBF8ow
# Normal sudoku rules apply.
#  - ALL horizontally and vertically neighbouring digits with the sum 10 are marked with X;
#  - ALL horizontally and vertically neighbouring digits with the sum 5 are marked with V.

# these pairs sum to 10:
#   (0,2) + (1,2),  (0,4) + (1,4), (0,6) + (1,6)
#   (1,1) + (2,1), (1,3) + (2,3), (1,5) + (2,5), (1,7) + (2,7)
#   (2,2) + (3,2), (2,4) + (3,4), (2,6) + (3,6)

nlitsme

Marching Band problem

In this document I describe a solution to the 'marchingband' problem as stated on think-maths:

What is the fewest number of performers you require for your marching band to
have 64 marching options? (Only whole positive numbers will be accepted)

Where a 'marchingband' can march only in a rectangular shape.

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EMRO	AF	Afghanistan	647500	37171921	World;Asia;NonCKIR
EURO	AL	Albania	28748	2876591	World;Europe;NonEU
AFRO	DZ	Algeria	2381740	34178188	World;Africa;North Africa
WPRO	AS	American Samoa (United States)	199	55465	World;Oceania;Polynesia
EURO	AD	Andorra	468	77281	World;Europe;NonEU
AFRO	AO	Angola	1246700	12799293	World;Africa;Central Africa
AMRO	AI	Anguilla (United Kingdom)	91	14731	World;North America
AMRO	AG	Antigua and Barbuda	442	96286	World;North America
AMRO	AR	Argentina	2766890	44361150	World;South America
EURO	AM	Armenia	29743	2951745	World;Europe;NonEU

nlitsme

This gist describes my solution to Matt Parker's puzzle #8:

• http://think-maths.co.uk/foldingchallenge
• https://www.youtube.com/watch?v=GpClxF41ugg

My solution involves looking at the pattern of folds when viewing the folded stack from the sides. And then using some python code to generate all possible ways the folds can match up in a way that you will still end up with two rows of 4 sheets.

The list below shows all ways to fold a piece of paper, including ways where the top-left ends up on the inside of the folded stack.

nlitsme

#!/bin/bash
#
# Script which scans all your mounted timemachine volumes, and apfs snapshots for a specific file.
#
# Willem Hengeveld <itsme@xs4all.nl>
#

if [[ $(id -u) -ne 0 ]]; then
    sudo $0 "$@"
    exit $?

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#!/bin/bash
# query.sh from BreachCompilation - magnet:?xt=urn:btih:7ffbcd8cee06aba2ce6561688cf68ce2addca0a3
# works with older bash versions.
# patch by Willem <itsme@xs4all.nl>
dir=$( cd "$( dirname "${BASH_SOURCE[0]}" )" && pwd )

if [ "$1" != "" ]; then
    word1=$(echo $1 | tr A-Z a-z)
	letter1=$(echo $word1|cut -b1)
	if [[ $letter1 == [a-zA-Z0-9] ]]; then

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"""
Python script for finding a still available pretty phonenumber on the sha2017 POC network.

Register your phone number on: https://poc.sha2017.org/

Author: Willem Hengeveld <itsme@xs4all.nl>

This code is released under the Beerware license.
"""
from __future__ import print_function

nlitsme

"""
Example how to decrypt whatsapp msgstore backups with extension .crypt12.

Author: Willem Hengeveld <itsme@xs4all.nl>
"""
from Crypto.Cipher import AES
import zlib
import sys

datafile = keyfile = None

nlitsme

# Linux kernel kallsyms unpacker
# Version 0.2
# Copyright (c) 2010-2013 Igor Skochinsky
#
# This software is provided 'as-is', without any express or implied
# warranty. In no event will the authors be held liable for any damages
# arising from the use of this software.
#
# Permission is granted to anyone to use this software for any purpose,
# including commercial applications, and to alter it and redistribute it