nlitsme

"""
PT/CT pairs

Key       expanded    Plaintext                      Ciphertext
AAAAAAAA  AAAAAAAA    AAAAAAAAAA                     VTPL?DL?HV
AAAAAAAA  AAAAAAAA    ABCDEFGHIJKLMNOPQRSTUVWXY      VMHTL?QFPCBKSZCJGRZM-EEEZ
AAAA      AAAAAAAA    BBBB                           OTBZ
AAAAAAAA  AAAAAAAA    BBBBBBBBBB                     OTBZNRZNVA
AAAAAAAA  AAAAAAAA    BBBBBBBBBBBBBBBB               OTBZNRZNVAJRJFVR
AAA       AAAAAAAA    BBBBBBBBBBBBBBBBBBBBBBBBB      OTBZNRZNVAJRJFVRVVOTVBBJB

nlitsme

"""
Author: itsme@xs4all.nl
"""
import html.parser
import html
import urllib.request
import json
from datetime import datetime, timedelta
import re
import os

nlitsme

	Secp256k1	EdDSA
message	m = hash(M)	M, given
privkey	.	given
private-key	x, given	x = low32(sha512(privkey))
signing-secret	k, given	k = sha512(high32(sha512(privkey)) // M)
privkey-to-pubkey	Y = G * x	Y = G * x
calc-R	R = G * k	R = G * k
r-value	r = R.x	r = SHA512(R // Y // M)
s-value	s = (m + x * r) / k	s = k + x * r

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"

nlitsme

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.

nlitsme

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 $?

nlitsme

#!/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