David Blume dblume

## openpgp.md

      
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                dblume
                / openpgp.md
            
            
              Created
              May 27, 2023 08:29
            
              
                Proof for keyoxide
              
          
      View openpgp.md
  
      
    openpgp4fpr:1CBD499C61C7A3A7DED3935F5E21C0D40ED9EB54

  
## qrcodes.sh
#!/usr/bin/env bash
#
# Spoiler:
# $ ./qrcodes.sh | sort | head
# 416 http://7c.dblu.me
# 416 https://j.dblu.me
# 416 http://wc.dblu.me
# 420 http://2E.dblu.me
# 420 http://73.dblu.me
#

## playlists_from_itunes_xml.py
#!/usr/bin/env python3
import libpytunes  # https://github.com/liamks/libpytunes


if __name__ == '__main__':
    l = libpytunes.Library('iTunes Music Library.xml')
    with open('playlists.txt', 'w', encoding='utf-8') as f:
        for p in l.getPlaylistNames():
            if p not in ('Downloaded', 'Audiobooks', 'Voice Memos', 'Not One Star Rating'):
                f.write(p)

## city_names.py
#!/usr/bin/env python
# From https://hackernoon.com/how-to-lose-an-it-job-in-10-minutes-3d63213c8370
#
# Find cities whose names are rotated versions of other cities.
#
# Given: ['Tokyo', 'London', 'Rome', 'Donlon', 'Kyoto', 'Paris']
#
# Return:
#
#[[ 'Tokyo', 'Kyoto' ],

## fathers_day_puzzle2.py
#!/usr/bin/env python
# Father's Day Card Puzzle #2
# A number's persistence is the number of steps to reduce it to a single
# digit by multiplying all its digits to obain a second number, then
# multiplying all the digits of that number to obtain a third number, and so
# on until a one-digit number is obtained.
#
# For example 77 has a persistence of four because it requires four steps to
# reduce it to one digit: 77 -> 49 -> 36 -> 18 -> 8. The smallest number of
# persistence one is 10. The smallest number of persistence two is 25. The

## fathers_day_puzzle3.py
#!/usr/bin/env python
#
# Father's Day card puzzle #3:
# In the figure below, fill in each of the sixteen numbers from 1 to 16 such
# that the four rows and three columns add up to 29.
#
#                      (  )---(  )---(  )
#                       |             |
# (  )---(  )---(  )---(  )          (  )
#         |             |             |

## confluence_markup.lua
-- confluence_markup.lua
-- A Pandoc Atlassian Confluence Markup Writer
--
-- Invoke with: pandoc -to confluence_markup.lua
--
-- Based on pandoc/data/sampsdfsle.lua and zhzhxtrrk's "pandoc confluence markup writer"
-- Fixes tables and lists of lists

local image_index = 0

## first_attempt.py
def ways_to_break(amount, coins):
    """
    :param amount: the monetary value to break into some number of coins
    :param coins: a container of descending coin denominations
    :return: the number of different ways to break amount into coins
    """
    this_coin = coins[0]

    # If this is the only coin, there's one way to break it.
    if len(coins) == 1:

## gist:f1356788354e4fe684d0
Verifying I am +dblume on my passcard. https://onename.com/dblume

## acid_write_file.py
import platform
import tempfile
import ntpath
import filelock  # https://github.com/dmfrey/FileLock

# It's interesting to see what's different between Windows and Linux.
# On Windows, you can't rename to an existing file, and you can't
# use os.fdatasync().
# Also see: http://blog.gocept.com/2013/07/15/reliable-file-updates-with-python/
	#!/usr/bin/env bash
	#
	# Spoiler:
	# $ ./qrcodes.sh \| sort \| head
	# 416 http://7c.dblu.me
	# 416 https://j.dblu.me
	# 416 http://wc.dblu.me
	# 420 http://2E.dblu.me
	# 420 http://73.dblu.me
	#
	#!/usr/bin/env python3
	import libpytunes # https://github.com/liamks/libpytunes


	if __name__ == '__main__':
	l = libpytunes.Library('iTunes Music Library.xml')
	with open('playlists.txt', 'w', encoding='utf-8') as f:
	for p in l.getPlaylistNames():
	if p not in ('Downloaded', 'Audiobooks', 'Voice Memos', 'Not One Star Rating'):
	f.write(p)
	#!/usr/bin/env python
	# From https://hackernoon.com/how-to-lose-an-it-job-in-10-minutes-3d63213c8370
	#
	# Find cities whose names are rotated versions of other cities.
	#
	# Given: ['Tokyo', 'London', 'Rome', 'Donlon', 'Kyoto', 'Paris']
	#
	# Return:
	#
	#[[ 'Tokyo', 'Kyoto' ],
	#!/usr/bin/env python
	# Father's Day Card Puzzle #2
	# A number's persistence is the number of steps to reduce it to a single
	# digit by multiplying all its digits to obain a second number, then
	# multiplying all the digits of that number to obtain a third number, and so
	# on until a one-digit number is obtained.
	#
	# For example 77 has a persistence of four because it requires four steps to
	# reduce it to one digit: 77 -> 49 -> 36 -> 18 -> 8. The smallest number of
	# persistence one is 10. The smallest number of persistence two is 25. The
	#!/usr/bin/env python
	#
	# Father's Day card puzzle #3:
	# In the figure below, fill in each of the sixteen numbers from 1 to 16 such
	# that the four rows and three columns add up to 29.
	#
	# ( )---( )---( )
	# \| \|
	# ( )---( )---( )---( ) ( )
	# \| \| \|
	-- confluence_markup.lua
	-- A Pandoc Atlassian Confluence Markup Writer
	--
	-- Invoke with: pandoc -to confluence_markup.lua
	--
	-- Based on pandoc/data/sampsdfsle.lua and zhzhxtrrk's "pandoc confluence markup writer"
	-- Fixes tables and lists of lists

	local image_index = 0
	def ways_to_break(amount, coins):
	"""
	:param amount: the monetary value to break into some number of coins
	:param coins: a container of descending coin denominations
	:return: the number of different ways to break amount into coins
	"""
	this_coin = coins[0]

	# If this is the only coin, there's one way to break it.
	if len(coins) == 1:
	import platform
	import tempfile
	import ntpath
	import filelock # https://github.com/dmfrey/FileLock

	# It's interesting to see what's different between Windows and Linux.
	# On Windows, you can't rename to an existing file, and you can't
	# use os.fdatasync().
	# Also see: http://blog.gocept.com/2013/07/15/reliable-file-updates-with-python/