challenge

[kryptech.net]

any good ideas, I want to start a challenge... any ideas, sans, jemfinch, stuka, anybody???

[zachogden]

i've always wanted to challenge someone. like; someone posts a problem, and you have 24 hours to solve it in any language you know. I'd like to challenge you, krytech. As soon as someone posts a problem, we setup a start time. Sound like a plan?

[kryptech.net]

awsome, hope I can hold my ground... :)

[zachogden]

All we need now is a problem.... one that can be solved in 24 hours.

[kryptech.net]

hmm.. that eliminates a lot. :)

[kryptech.net]

anyone, a problem?

[GnuVince]

Write a function that returns the perimeter of a rectangle.

[zachogden]

Thst's not much of a challenge. lol.

[kryptech.net]

oh, yeah. even I can do that one. :) (I'm just hoping that was a joke)

[woodwind]

So do it, guys. It can't hurt you.

[MJPhill]

Expanded upon: Write a function that, upon being given the perimeter of a rectangle, gives the dimensions of the rectangle with the greatest area as well as the one with the smallest area using the given perimeter.

[woodwind]

Assuming you want integral dimensions, here is a messy brute-force solution:


import math

# Returns a tuple of tuples containing ((max_w, max_l), (min_w, min_l))
# p is the perimeter
def find_dims(p):
r = int(math.ceil(p/2))
max_w = 0
max_l = 0
for w in range(1, r+1):
l = r-w
if w*l > max_w*max_l:
max_w = w
max_l = l
min_w = max_w
min_l = max_l
for w in range(1, r):
l = r-w
if w*l < min_w*min_l:
min_w = w
min_l = l

return ((max_w, max_l), (min_w, min_l))


Am I right?

[woodwind]

Here's a formulaic method. Maybe hard to understand, but it's short and it works:

import math

def find_dims2(p):
max_w = int(math.floor(p/4))
max_l = max_w
if 2*(max_w+max_l+1) <= p:
max_l += 1
min_l = int(math.floor(p/2)) - 1
return ((max_w, max_l), (1, min_l))

[Strike]

Well the simple answer is that for any given perimeter, the dimensions with the max area are equal. That is, for any given rectangle perimeter p, p/4 is the length of the side of the rectangle which will have the maximum area. The dimensions of the rectangle with the smallest area are where one of the dimensions is epsilon and the other dimension is (p/2) minus epsilon (okay, minus two epsilon ;)).

[kryptech.net]

okay.... so the smallest area is (very, very, very near) zero, and the max is p/4 squared

[woodwind]

Yeah, it's a lot easier with non-integral dimensions. You have a square for max and you have an infinitely small rectangle (which is cool to think about!) for min. The brute-force solution requires a bit of programming, though, so I figured that's what he meant.

[iDxMan]

Originally posted by woodwind
So do it, guys. It can't hurt you.

Notice that neither have posted any code towards a solution. Gee...

-r

[kryptech.net]

WTF.. you want me to make a function that returns a perimeter or a rectangle... like that's hard...
int perimiter(int width, int length) {
return width * 2+ length * 2;
}

[iDxMan]

I don't really care what you do. There seems to be lots of talk and no challenge.

At the least view http://acm.uva.es/problemset/ and pick something that fits. I believe several of the challenges have been suggested here in the past. Perhaps you'll find something that's not too easy for you....

-r

[kryptech.net]

most things are not too easy for me... ;-)