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The program is run with a positive integer as an argument, e.g.,

  ruby entry.rb 27

It has been confirmed to be run on

  ruby 1.9.3p385 (2013-02-06 revision 39114) [x86_64-darwin11.4.2]
  ruby 2.0.0p481 (2014-05-08 revision 45883) [universal.x86_64-darwin13]
  ruby 2.2.3p173 (2015-08-18 revision 51636) [x86_64-linux]


The program prints a Collatz sequence started with a given number, that is, it repeatedly outputs numbers obtained by applying the following Half-Or-Triple-Plus-One (HOTPO) process to the previous number:

If the number is even, divide it by two, otherwise, multiply it by three and add one.

until the number becomes 1. Collatz conjectured that no matter from the process starts it always eventually terminates. This is still an open problem, hence the program may not terminate for some numbers. It is known that there is no such exception below 260.


The source code does not contain either conditional branch or arithmetic operation. The trick shall be revealed step by step.

First, the code is obfuscated by using %-notations, *(String#join), %-formatting, restructuring, and so on. Here is an equivalent readable program:

n = ARGV[0].to_i
  # do nothing
end while begin
  puts n
  n = (/(.)...\1=/ =~ eval('[",,,,,"'+ '",'*n + '  ?=].join#"].join("3x+1?")'))

The line

  n = (/(.)...\1=/ =~ eval('[",,,,,"'+ '",'*n + '  ?=].join#"].join("3x+1?")'))

performs the HOTPO process. The eval expression here returns a string as explained in detail later. Since regex=~str returns index of first match of regex in str, the regular expression (.)...\1 must match the string at index n/2 if n is even and at 3*n+1 if n is odd greater than 1. The match must fail in the case of n = 1 so that it returns nil.

The key of simulating the even-odd conditional branch on n in the HOTPO process is an n-length sequence of the incomplete fragments ", where the double-quote " changes its role of opening/closing string literals alternately. If n is even, the string in the eval expression is evaluated as

  => '[",,,,,"'+ '",' + '",' + '",' + ... + '",' + '  ?=].join#...'
  => '[",,,,,"",",",...",  ?=].join#...'

where the last double-quote " is closing hence the code after # is ignored as comments. Note that "ab""cd" in Ruby is equivalent to "abcd". Therefore the eval expression is evaluated into


where the number of commas is 5+n/2. As a result, the regular expression (.)...\1= matches ,,,,,= at the end of string, that is, at index 5+n/2-5 = n/2.

If n is odd, the string in the eval expression is evaluated as

  => '[",,,,,"'+ '",' + '",' + '",' + '",' + ... + '",' + '  ?=].join#"].join("3x+1?")'
  => '[",,,,,"",",",",...,",  ?=].join#"].join("3x+1?")'

where the last element in the array is ", ?=].join#". Threfore the eval expression is evaluated into

  ",,,,,,3x+1?,3x+1?,...,3x+1?,  ?=].join#"

where the number of ,3x+1? is (n-1)/2. As a result, the regular expression (.)...\1= matches ?, ?= at the almost end of string, that is, at index 5+(n-1)/2*6-1 = 3n+1, provided that the match fails in the case of n = 1 because the symbol ? occurs only once in the string.

One may notice that the string 3x+1 in the code could be other four-character words. I chose it because the Collatz conjecture is also called the 3x+1 problem.


The Collatz conjecture is equivalently stated as,

no matter from the HOTPO process starts, it always eventually reaches the cycle of 4, 2, and 1

instead of termination of the process at 1. This alternative statement makes the program simpler because we do not have to care the case of n = 1. It can be obtained by replacing the regular expression is simply /=/ and removing a padding ",,,,,". The program no longer terminates, though.


The implementation requires to manipulate long strings even for some small starting numbers. For example, starting from 1,819, the number will reach up to 1,276,936 which causes SystemStackError on Ruby 1.9.3. The program works on Ruby 2.0.0 and 2.2.3, though.