Colin Kelley ColinDKelley

Solution to Three Gods Logic Puzzle

Ask god A: "Would the other non-random god answer 'Ja' to the question 'Does the random god sit to your left¹?'"

=> If the answer is 'Ja', then it's false (or random), therefore NonRandom1 = C; else it's true (or random), therefore NonRandom1 = B

Ask gods[NonRandom1]: "Would the other non-random god answer 'Ja' to the question 'Does the random god sit to your left¹?"

Answer

The tallest prisoner counts the white hats in front of him. If odd, he calls out "white"; if even he calls out "black".

Then each prisoner down the line makes the same calculation. If the color they compute is different from what was called out before them, they know their color is "white"; if it's the same, their color is "black". They call out their color.

	destination_address =
	if call_flow.sip_integration_if_enabled
	if advertiser.feature_enabled?(:sip_integration_destination_in_display_format) \|\|
	!(destination_phone_number = PhoneNumberWithExtension.new(destination)).valid?
	destination.presence
	else
	destination_phone_number
	end
	else
	destination_phone_number(destination)

	orig = File.read("King Bee emails from Rachel.txt")

	cleaned = orig
	.force_encoding('BINARY')
	.gsub("\xCA".force_encoding('BINARY'), '')
	.gsub(/\r+/, '; ')
	.gsub("\"; <", "\" <")
	.gsub(/; */, "\n")
	.gsub(/^"/, '')
	.gsub("\" ", ' <')

	tiles = {}
	STDIN.readlines.map.with_index do \|line, y\|
	x = 0
	line.chomp.split(".").each do \|segment\|
	unless segment.empty?
	tile = [x, segment.size, segment[0]]

	if value = tiles[tile]
	tiles[tile] = [value.first, value.last + 1]
	else

	result =
	retry_on_exception(Timeout::Error, total_tries: 2) do
	http_fe.post(...)
	end

	module ParseFigLeaf
	def parse(*args)
	arg0 = args.first
	arg0.is_a?(String) && arg0 =~ /\A\d\d[-\/]\d\d[-\/]\d\d/ and raise "Ambiguous parse format in #{arg0.inspect}"
	super
	end
	end

	class << Date
	prepend ParseFigLeaf

	Race 5 horses. Take the fastest 3 and race them against 2 chosen from the remainder. Repeat until there are are no remaining.
	The final 3 from the last race are the fastest of the group.
	races = 1 + (25 - 5)/2 = 11

	require 'mysql2'

	module Invoca
	module Mysql2AdapterKillOnTimeout
	def kill(thread_id)
	@connection.query("KILL #{thread_id}")
	end

	def execute(sql, name = nil)

	class Secrets
	def initialize(secrets_path)
	@hash = YAML.load_file(secrets_path)
	end

	def [](key)
	@hash[key]
	end

	def set_instance

Colin Kelley ColinDKelley

Solution to Three Gods Logic Puzzle

Answer

Reasoning