Coffeehouse Thread

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Nicely done, Channel 9 (posted by Dr Herbie).

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  • User profile image
    Charles

    Nicely done, Niners!

     

    To be clear, I had little to do with this clever prank. Dan and Duncan came up with the basic idea, I recruited the best two people in the world IMO, Erik and Brian, to come up with the complex expressions used in the gate (Erik provided the Church numeral encoding(nice job, W3bbo!), Brian the beautiful mathematical expression, of course). Geoff, a recent stellar addition to our dev team, implemented this in record time. Rock star.

     

    So, nicely done Dan, Duncan, Geoff, Erik and Brian.

     

    Happy April Fools,

     

    C

  • User profile image
    brianbec

    CreamFilling512 said:

    I put the second problem into Wolfram alpha:

     

    http://www.wolframalpha.com/input/?i=3+%2B+(product+((1+%2B+1/k)^3)/(1+%2B+3/k)+k%3D1+to+infinity)">http://www.wolframalpha.com/input/?i=3+%2B+(product+((1+%2B+1/k)^3)/(1+%2B+3/k)+k%3D1+to+infinity)

    Now, for a little secret: type the same thing into Wolfram Alpha, just replacing "3" with "n" (and, for even more secrets, leave off the first "n + " -- just type in the big infinite product

  • User profile image
    Richard.Hein

    giovanni said:
    CreamFilling512 said:
    *snip*

    That was easy! Your solution is much faster than mine! Excel is not the best tool for this kind of problems, but I didn't have anything handy at the office.

     

    Still not sure what the first equation meant...

    This is my naive solution to the second problem, it approaches 9 as k approaches infinity.

     

    static void Main(string[] args) { 
        Func<double, double> f = (k) => (Math.Pow((1 + (1 / k)), 3)) / (1 + (3 / k)); 
        double z = f(1); 
        for (double k = 2; k < double.MaxValue; k++) { 
            z *= f(k);
            Console.WriteLine(3 + z);
        } 
    }

  • User profile image
    mstefan

    Completely wasn't thinking of what day it was ... nice job Charles and co.! Wink

  • User profile image
    daSmirnov

    I viewed the page source, got the URL for this thread.... Smiley Nice and simple.

  • User profile image
    JoshuaAllen

    giovanni said:
    CreamFilling512 said:
    *snip*

    That was easy! Your solution is much faster than mine! Excel is not the best tool for this kind of problems, but I didn't have anything handy at the office.

     

    Still not sure what the first equation meant...

    @giovanni -- the first formula was the easiest: just count the f's!  Smiley

  • User profile image
    Charles

    brianbec said:
    CreamFilling512 said:
    *snip*

    Now, for a little secret: type the same thing into Wolfram Alpha, just replacing "3" with "n" (and, for even more secrets, leave off the first "n + " -- just type in the big infinite product

    Awesome! I have to say, Wolfram Alpha is rocking and rolling, man.

     

    C

  • User profile image
    Richard.Hein

    brianbec said:
    CreamFilling512 said:
    *snip*

    Now, for a little secret: type the same thing into Wolfram Alpha, just replacing "3" with "n" (and, for even more secrets, leave off the first "n + " -- just type in the big infinite product

    The second problem is n + the gamma function, Γ(n+1), with n = 3.

     

    Here's a graph of the absolute value of the Gamma function on the complex plane, from Wikipedia:

     

    Generic Forum Image

     

  • User profile image
    Charles

    JoshuaAllen said:
    giovanni said:
    *snip*

    @giovanni -- the first formula was the easiest: just count the f's!  Smiley

    Nice job, Joshua. Smiley
    C

  • User profile image
    Charles

    daSmirnov said:

    I viewed the page source, got the URL for this thread.... Smiley Nice and simple.

    Also a clever solution. Smiley
    C

  • User profile image
    giovanni

    W3bbo said:
    Sven Groot said:
    *snip*

    I got it, but only by coincidence: I had a series of classes on lambda calculus last week and I recognised the Church encoding of the natural numbers as lambda functions (as contextfree said). Simply count the level of recursion to get the number, in this case it was 9-deep, so it represents '9'.

     

    I recognised the other question as product notation. Going by the frequent use of '3' and how 3*3 == 9 I figured the answer was the same.

     

    Note that I actually advocate implementing something similar to this as an actual captcha in future. Keep it just as difficult, I like the idea of there being a forum with just me on it Tongue Out

    Actually that was 3+product()

  • User profile image
    giovanni

    JoshuaAllen said:
    giovanni said:
    *snip*

    @giovanni -- the first formula was the easiest: just count the f's!  Smiley

    Yes, it is easy once you know what the notation mean Smiley

  • User profile image
    Charles

    brianbec said:
    CreamFilling512 said:
    *snip*

    Now, for a little secret: type the same thing into Wolfram Alpha, just replacing "3" with "n" (and, for even more secrets, leave off the first "n + " -- just type in the big infinite product

    Try WolframAlpha on the Church numeral encoding  λf.λx.f(f(f(f(f(f(f(f(f(x))))))))))

    The computed result isn't correct.

     

    C

  • User profile image
    brian.​shapiro

    I just did a Google search on λf.λx.f(x and got the wiki entry on Church notation.

     

    Before that, I tried just entering random things a couple of times to see if it would eventually let you in. After 9 times of entering bad answers you get a subtle hint.

  • User profile image
    turrican

    yey! I passed.

  • User profile image
    brianbec

    Richard.Hein said:
    brianbec said:
    *snip*

    The second problem is n + the gamma function, Γ(n+1), with n = 3.

     

    Here's a graph of the absolute value of the Gamma function on the complex plane, from Wikipedia:

     

    Generic Forum Image

     

    A+

  • User profile image
    raptor3676

    Richard.Hein said:
    giovanni said:
    *snip*

    This is my naive solution to the second problem, it approaches 9 as k approaches infinity.

     

    static void Main(string[] args) { 
        Func<double, double> f = (k) => (Math.Pow((1 + (1 / k)), 3)) / (1 + (3 / k)); 
        double z = f(1); 
        for (double k = 2; k < double.MaxValue; k++) { 
            z *= f(k);
            Console.WriteLine(3 + z);
        } 
    }

    I did it with PowerShell.  Anyway, way to go Channel9!

  • User profile image
    sushovande

    Ion Todirel said:
    mastermine said:
    *snip*

    I didn't bother to look, I was like "knowing the C9 devs they probably did server side validation" Big Smile

    http://www.wolframalpha.com/input/?i=product+of+((1+%2B+1/k)^3+/+(1+%2B+3/k))+for+k+%3D+1+to+inf">http://www.wolframalpha.com/input/?i=product+of+((1+%2B+1/k)^3+/+(1+%2B+3/k))+for+k+%3D+1+to+inf

    http://www.wolframalpha.com/input/?i=product+of+((1+%2B+1/k)^3+/+(1+%2B+3/k))+for+k+%3D+1+to+inf">

    Although I solved it by counting the number of f's in the first formula, then realizing that the number '9' was special to C9.

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