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

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

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

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); } }

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

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

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!

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

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:

JoshuaAllen said:giovanni said:*snip*
@giovanni  the first formula was the easiest: just count the f's!
Nice job, Joshua.
C 
daSmirnov said:
I viewed the page source, got the URL for this thread.... Nice and simple.
Also a clever solution.
C 
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 9deep, 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
Actually that was 3+product()

JoshuaAllen said:giovanni said:*snip*
@giovanni  the first formula was the easiest: just count the f's!
Yes, it is easy once you know what the notation mean

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

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.

yey! I passed.

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!

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"
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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