Multivariable Calculus and Encryption Challenge

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Multivariable Calculus and Encryption Challenge

Post by xfelix on Sat Nov 28, 2009 1:21 pm
([msg=30829]see Multivariable Calculus and Encryption Challenge[/msg])

Challenge:
Part 1
f(x,y) = (x^2)*(2^y)
Determine the magnitude of the gradient vector of the given function f(x,y) at the point P where x=1337 and y=8.
Symbol Notation:|grad(f(1337,8))|
Truncate (floor) the decimal point once you have found the magnitude of the gradient vector (i.e. the key)

What the hell is a gradient?
This much for a hint I will give. The gradient vector arises in many situations as finding the tangent plane to a surface at a given point, Determining the Normal vector (i.e. orthogonal vector) at a given point, etc. The most basic article of information provided by the gradient vector of a function is the direction it points is always to the maximum of the function. If you follow the direction of the gradient it will lead you to the local maximum of the function.

Part 2
The answer that you've discovered in part1 is also the key to a blowfish encryption algorithm. Here is the encrypted message
ca1c622b4cde4c59afaaf5e018d121b5
d142079cb7a7765bafbb8bfaa254f068
210f7a8f30af2febf287d466b18702b8
3c1e3c5a034908d2

Part 2 mission is to crack the message with the determined answer of part 1 8-)

Rulez:
For full credit the rulez must apply
1. Show you're work (this means providing a substantial mathematical calculations)
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Re: Multivariable Calculus and Encryption Challenge

Post by Goatboy on Sat Nov 28, 2009 3:04 pm
([msg=30832]see Re: Multivariable Calculus and Encryption Challenge[/msg])

Looks interesting, but are you asking us to do your homework? =P
Assume that everything I say is or could be a lie.
1UHQ15HqBRZFykqx7mKHpYroxanLjJcUk
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Re: Multivariable Calculus and Encryption Challenge

Post by xfelix on Sat Nov 28, 2009 4:55 pm
([msg=30834]see Re: Multivariable Calculus and Encryption Challenge[/msg])

nahh I was skimming over some stuff so I wouldn't forget, and decided to make a challenge. definitely not a bad idea though :mrgreen:
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Re: Multivariable Calculus and Encryption Challenge

Post by runninggee57 on Fri Dec 11, 2009 7:38 pm
([msg=31266]see Re: Multivariable Calculus and Encryption Challenge[/msg])

HAHA finally, a use for my calculus skills in the real world!!

OK, here we go!

f(x,y) = x^2 * 2^y
grad(f(x,y)) = df/dx * i + df/dy * j
= 2x * 2^y * i + ln(2) * x^2 * 2^y * j
=<2x * 2^y, ln(2) * x^2 * 2^y> written as a vector
x = 1337 , y = 8
grad(f(x,y)) = <2 * 1337 * 2^8, ln(2) * 1337^2 * 2^8>
= <684544, 457617664 * ln(2)>

|grad(f(x,y))| = sqrt(684544^2 + (457617664 * ln(2))^2)
= 317197132.235...
truncate the decimal places

key = 317197132

Put the encrypted message into webnet77.com/cgi-bin/helpers/blowfish.pl with key and get the message:

"Good Job, You win youve got some 1337 skillz"



Thanks for giving me something academically worthwhile for a few minutes!!
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Re: Multivariable Calculus and Encryption Challenge

Post by xfelix on Fri Dec 11, 2009 8:35 pm
([msg=31268]see Re: Multivariable Calculus and Encryption Challenge[/msg])

lmao runninggee57 that's exactly the correct magnitude of the gradient vector and is legit answer with work shown. I'm definitely not kidding with the decrypted text. I wasn't sure if anyone would solve this puzzle, good job!! :D
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