I have reduced a problem I am trying to solve to something simpler though rather difficult: Four side lengths and an angle uniquely describe a quadrilateral, solve one of the angles adjacent to the one given. No matter what algebraic / trigonometric manipulations I apply I cannot get a...
'A' could be with any of the other 15 letters, so the chances of it being with B are 1/15, for that to happen n times is (1/15)^n
EDIT: And for all the groups being the same it is not 1/120, it is \frac{2^8.8!}{16!} = 1/2027025
No, the first person's guess is irrelevant. No matter what he chooses the second person has a 1/10000 chance of matching it. What you described is the chances that they both pick the number 1 but what you are forgetting is that they could have both gotten 2 or 3 or any other number so you must...
Thanks vsage. The speed is a huge factor, I will be having shapes of all kinds with no limit to the number of vertices, no convex constraints, and it will be possible to have shapes cut out of them [see these examples]. At the moment I was thinking this:
1. For each polygon edge:
2. If the...
I have a series of points on a 2 dimensional plane that form a polygon.
Somewhere inside the polygon I have another point and an angle that defines a ray. What algorithm would I use to find which interval is first interstected with the ray?
If the point happens to be outside the polygon then I...
The guy who asked the question differs from the other because he was not involved in the selection. Imagine it this way, there are a thousand people and all but one are guilty, I could ask the guard which 998 of the others are guilty and it would narrow it down to me and one other guy, obviously...
Actually posting that just compounds your stupidity (joking). For anybody who cannot see, the main man still has a 1/3 chance of survival, the known guilty guy has a 0 chance and the other a 2/3 chance. It has already been explained so I won't bother.
There is only one problem with that, there is an implied 10. I would definantly accept log in base 4 and maby the natural logarithm of an expression but that (albeit quite novel) just seems to violate in my opinion.
If I don't ever find a better solution I think I will settle for that though...
(n+\tfrac{1}{2})! = \sqrt{\pi} \prod_{k=0}^{n}\frac{2k+1}{2}
I know that the gamma function can be used to solve factorials for any number but are there any other special rules like this one?
I have read that astronomers are responsible for much of the advances in trig hundreds of years ago but never seem to find who exactly pioneered this field of research.
My logarithm knowledge is kind of sketchy but I think this is right:
x = log_3 25
3^x = 25
If x were 3 then RHS would be 27, if it were 2 then RHS would be 9, since 25 is between these values then so is x.
Q2:
I think there is a rule that goes something like this (someone please confirm)...
That condition is not accurate, consider this function:
f(x) = \frac{x^2-1}{x-1}
The limit above and below f(1) is equal to 2 though it is undefined at that point.
Aha, I didn't read the question carefully enough, thanks for pointing that out Ceptimus. Whichever moderator deleted I think should have instead pointed out my error or at least PM'd me, rather rude I feel.
I posted the solution to this the other day but for some reason it has been removed, here it is again, not sure if it is the same as Ceptimus' as his attatchment is still pending.
LarrrSDonald: I like your ideas! Your repetition idea got me a whole heap of answers including 37! And this latest one, 79 is great lateral thinking, Keep em coming.
ArielGenesis: If I were to include square root signs then it might as well be called the four 4 or 2s challenge, it would make it...
Oh right, I don't think I will be able to help with this question but I think that you will need another variable for how many people use the elevators.
Thanks for 71 ceptimus!
The others I am not as interested in because (as I probably should have stated) I am not considering square roots because they require a 2, I would however consider 4th roots so long as the 4 is counted. No triggonometric functions either, basically whatever techniques I...