From: Kelly Jones <kelly.terry.jones@...>
Date: 2010-04-02T02:31:21+09:00
Subject: Re: Ruby solve classic math problem: make 56 from four 4s

On 4/1/10, Robert Klemme <shortcutter@googlemail.com> wrote:
> 2010/3/31 Kelly Jones <kelly.terry.jones@gmail.com>:
>> A classic math problem asks "using only four 4s and any
>> mathematical operation, come up with the number
>> 56". "4!+4!+4+4" is one of probably several answers [1].
>>
>> I want to write a ruby program that solves the
>> generalization: "given X number of the digit Y and a
>> set of permitted mathematical operations/functions, can
>> you come up with the number Z?"
>
> As long as you do not limit the number of unary operations that
> program is not guaranteed to terminate - especially if there is no
> solution for given X, Y and Z.  Am I missing something?
>
> It should be interesting to see what solution strategy you are picking.

Thanks, and thanks to everyone else who replied.

You're right: I hadn't thought of this.

If I start with 4 and allow sqrt, I get sqrt(4),
sqrt(sqrt(4)), sqrt(sqrt(sqrt(4))) and so on. Of
course, these aren't integers after the first sqrt, but
they could theoretically combine with other 4
combinations later to form an integer.

I incorrectly thought one iteration of a unary operator would suffice.

I originally got interested in this problem because I
thought factorial was a cheat. Some of the solutions on
http://mathforum.org/ruth/four4s.puzzle.html use the
gamma function, integer 4th root, etc. Where do you
draw the line? If you allow constant functions (eg,
f(x) = 56), the solution is trivial.

My new goal is to solve the simpler problem, very
similar to http://www.rubyquiz.com/quiz7.html (thanks,
Brian!).

Given X copies of the digit Y and the 5 mathematical
operators plus, minus, multiply, divide, and exponent,
along with concatenation and decimals (see below), can
you construct the number Z?

My approach for 5 copies of the digit 7 (example):

 % With one 7, you have {7, 0.7} (the latter because we
 allow decimals-- but not 0.07)

 % With two 7s, we union two sets:

  % {77, 7.7, .77} (from decimals and concatenation)

  % Apply + - * / ^ to every ordered pair of elements
  in the resultset for one 7 (including "pairs" like
  {7,7}). Not showing the results, but you get the
  idea.

 % We then recurse. For n 7s, we union:

  % {777...[n times], 777...[n-1 times].7, 777...[n-2 times].77, etc}

  % Applying + - * / ^ to every ordered pair of the
  resultset for n-1.

It might still be interesting to create a website that does this.

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