       Re: Google's aptitude test equation

• To: mathgroup at smc.vnet.net
• Subject: [mg55030] Re: [mg55004] Google's aptitude test equation
• From: DrBob <drbob at bigfoot.com>
• Date: Thu, 10 Mar 2005 05:24:30 -0500 (EST)
• References: <200503091134.GAA06997@smc.vnet.net>
• Reply-to: drbob at bigfoot.com
• Sender: owner-wri-mathgroup at wolfram.com

Here's a minor rewrite:

prob = "wwwdot" - "google" == "dotcom";
eqn = prob /. s_String :> FromDigits@ToExpression@Characters@s;
vars = Variables[Subtract @@ eqn];
ineqs = And @@ Thread[0 <= vars <= 9];
Timing[sol = Reduce[eqn && ineqs, vars, Integers];]
Length[sol]
Timing[Select[sol, Unequal @@ #[[All,2]] & ]]

{8.875 Second,Null}
449
{0.*Second, (c == 4 && d == 5 &&
e == 3 && g == 1 &&
l == 0 && m == 6 &&
o == 8 && t == 9 &&
w == 7) || (c == 4 &&
d == 5 && e == 6 &&
g == 1 && l == 0 &&
m == 3 && o == 8 &&
t == 9 && w == 7)}

If we impose e < m (since e and m can be interchanged), it goes faster:

(sol = Reduce[eqn && ineqs && e < m, vars, Integers];) // Timing
Length@sol
Select[sol, Unequal @@ #[[All, 2]] &] // Timing

{5.3910000000000196*Second, Null}
204
{0.*Second, c == 4 && d == 5 &&
e == 3 && g == 1 && l == 0 &&
m == 6 && o == 8 && t == 9 &&
w == 7}

Bobby

On Wed, 9 Mar 2005 06:34:29 -0500 (EST), Lorenzo Castelli <gcastelli at NOSPAMracine.ra.it> wrote:

> Hi everybody, I have a little question about Mathematica (version 5.1).
> Please note that I just began to use this program, so sorry if I'm missing
> some trivial points.
>
> As an exercise I've tried to resolve the Google's equation that appeared on
> the web some time ago, using this page as a reference:
>
> The text of the exercise is:
> 1. Solve this cryptic equation, realizing of course that values for M and E
> could be interchanged. No leading zeros are allowed.
> WWWDOT - GOOGLE = DOTCOM
>
> I've implemented the solution in Mathematica in this way:
>
> - Setup the problem:
> prob = "wwwdot" - "google" == "dotcom"
>
> - Setup the equation associated with the problem:
> eqn = prob /. s_String :> FromDigits[ToExpression[Characters[s]], 10]
>
> - Extract the variables of the equation:
> vars = Union[Cases[eqn, _Symbol, Infinity]]
>
> - Setup the constraints:
> ineqs = And @@ (0 <= # <= 9 & /@ vars)
>
> - Resolve the equation:
> (sol = Reduce[eqn && ineqs, vars, Integers]) // Timing
>
> This operation gives 449 solutions, calculated in roughly 13 seconds.
> Only two of these solutions are correct, since I have to remove the
> solutions which yield more than one variable to have the same value.
>
> Obviously this task is pretty easy, so I should expect it to take just a
> short time to compute.
> And that is actually the case if use a code like this:
>
> Select[sol, Unequal @@ (Part[#, 2] & /@ Cases[#, _Equal, 1]) &] // Timing
>
> The two solutions are extracted in a fraction of a second.
>
> On the contrary if I try to do that in the following way, Mathematica starts
> a long session of evaluation and eventually crashes after some minutes:
>
> Reduce[sol && Unequal @@ vars] // Timing
>
> I understand that Reduce might be too general to perform this operation in
> an efficient way, but I haven't been able to find a more suitable function.
>
> Also I've noticed that if I change the constraints on the single variables
> for the first Reduce in order to explicity consider that no leading zeros
> are allowed (so just replacing the 0<=x<=9 with 0<x<=9 for the variables w,
> g and d), the evaluation time jumps from 13 seconds to 40 seconds.
> That's odd, as in this case the new solutions are obviously just an easily
> computable subset of the previous ones.
>
> So I don't really know if Reduce is actually the best function even in the
> first usage.
>
> Can you help me out?
>
> Thanks,
>
> Lorenzo Castelli
> E-Mail: gcastelli at racine.ra.it
>
>
>
>

--
DrBob at bigfoot.com

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