Any demoscener from Mensa?
category: residue [glöplog]
Except that's a bad proof, isn't it. \o/
iq, it looks like doom solution is valid, and also my last solution (both look similar)
iq. Those three vectors i proposed are all perpendicular to your original vector, and they can't all be 0. The question is if two of them can be proportional if the third is 0. And i'm at work, and they "don't pay me to think". So. Anyway I'm pretty sure a linear combination of several possible solutions is at least most of the answer.
I suppose real proof you want to look at the length of teh vector.
f(Y)=(y2+y3)^2+(y3-y1)^2+(y1+y2)^2
So you differentiate and let f'(Y)=(0,0,0). OH NOES /o\ calculus :'(. But I'm guessing there's a global minimum at Y=(0,0,0) which is yer proof.
f(Y)=(y2+y3)^2+(y3-y1)^2+(y1+y2)^2
So you differentiate and let f'(Y)=(0,0,0). OH NOES /o\ calculus :'(. But I'm guessing there's a global minimum at Y=(0,0,0) which is yer proof.
Actually, f(Y) = |Y|^2 >= 0, so if you can show that Y=(0,0,0) is the only critical point that's enough. So it's not too bad. Still too busy watching the clock tick towards 4pm though. Getting very close now.
"If x is an even number, then why is x mod 10 = (6 x) mod 10?"
x=2n
2n*6=10n+2n
10n mod 10=0
... I suppose it is enough...
x=2n
2n*6=10n+2n
10n mod 10=0
... I suppose it is enough...
*for n integer
Texel: (y+z, -x, -x) doesn't work either. Try w = (0,1,-1)
Doom: I don't know if I follow what you mean. I assume y1,y2,y3 = my x,y,z. Then you are proposing u=(y+z, z-x, -x-y) what doesn't work for (-1,1,-1), so I guess that's not what you mean.
So still no answer unless I misundertood something...
Doom: I don't know if I follow what you mean. I assume y1,y2,y3 = my x,y,z. Then you are proposing u=(y+z, z-x, -x-y) what doesn't work for (-1,1,-1), so I guess that's not what you mean.
So still no answer unless I misundertood something...
Actually, I just realized that its impossible to construct "u" from linear combinations of the components of "w" -x,y,z -. So we will have to look to degree >=2 things.
true iq...
Pah. My guesses are better than proof.
face it Doom, you still don't dominate the world. you are trying hard, i know. :D
texel: Yes, it is enough. Well done.
whoa, the second problem by iq is evil...
it is the evil f*ckin cosh(x) = (e^x + e^-x) /2 that i can't solve, no way...
at least i can tell you that for t = cosh(x) = (e^x + e^-x) /2
the 2 solutions may be:
x = (1/4 * (t + - sqrt(-16 + t^2)))
(says mathematica...)
anyone got info on how to get there?
it is the evil f*ckin cosh(x) = (e^x + e^-x) /2 that i can't solve, no way...
at least i can tell you that for t = cosh(x) = (e^x + e^-x) /2
the 2 solutions may be:
x = (1/4 * (t + - sqrt(-16 + t^2)))
(says mathematica...)
anyone got info on how to get there?
iq: for the first problem, if you count using min, step, ramp like functions as branching then it means all discontinuous functions are prohibited. hairy ball theorem states that you can't find a continuous tangent function on a unit sphere.
ramp wasn't one them actually. i can't find one with a ramp anyway :)
About iq problem 2, I have no idea of calculus, but the formula looks like catenary curve... so maybe it is a way for searching in google :D
Stingray: Well, I have a backup plan. It involves poison and beer. So see you at Breakpoint. \o/
iq: I guess it boils down to finding an explicit formula for one arbitrary vector that is not proportional to a given w, but for any w. So keeping in mind that the hairy balls are not necessarily unit spheres (e.g. mine are rather massive), anus is right. There's no continuous vector field that does what you need.
iq: I guess it boils down to finding an explicit formula for one arbitrary vector that is not proportional to a given w, but for any w. So keeping in mind that the hairy balls are not necessarily unit spheres (e.g. mine are rather massive), anus is right. There's no continuous vector field that does what you need.
So apparently unless someone that studied some maths says the contrary we conclude I cannot solve my first problem... (shit).
For the second, it's a catenary yes, but in this case the variable to isolate is also outside the cosh(), and that makes standard techniques fail as you can see. I wonder if I have a special talent to find non-solvable problems.... I hope not.
For the second, it's a catenary yes, but in this case the variable to isolate is also outside the cosh(), and that makes standard techniques fail as you can see. I wonder if I have a special talent to find non-solvable problems.... I hope not.
calculus problems are so fucking ugly...
iq: if your first problem is iterative, you can always store the last u you got. As soon as you calculate a new w', create a rotation quaternion that represents the minimal rotation from w to w' (as described here) and use that to rotate u.
In Sweden we do this kind of math in elementary school. This thread kindof proves my point that our elementary school is a damn playground.
Kosmiset Avaruus Sienet
You all know the famous problem of the two guys and two doors, one guy who always say the truth and other who always lie. It appeared in the film Labrinth, one of your favourite films.
There is a variation over it really interesant.
There are 2 doors, the one you have to cross and the one you don't. There are 3 guys, with 3 t-shirts numbered, one has number 1, other number 2, other number 3. One of the guys always say the truth, other always lie and the third one *arbitrary* say the truth or lie. They all know wich is the door you have to cross and they know themselves (they know if they say the truth, or always lie or arbitrary lie).
They can answer you only with "yes" or "no". You have to ask only with questions than can be answered by "yes" or "no".
In two questions, you have to choose the right door.
Whose are these questions?
*Note: Even if this problem also uses doors, there is no relation with the problem I posted last time.
There is a variation over it really interesant.
There are 2 doors, the one you have to cross and the one you don't. There are 3 guys, with 3 t-shirts numbered, one has number 1, other number 2, other number 3. One of the guys always say the truth, other always lie and the third one *arbitrary* say the truth or lie. They all know wich is the door you have to cross and they know themselves (they know if they say the truth, or always lie or arbitrary lie).
They can answer you only with "yes" or "no". You have to ask only with questions than can be answered by "yes" or "no".
In two questions, you have to choose the right door.
Whose are these questions?
*Note: Even if this problem also uses doors, there is no relation with the problem I posted last time.
question 1: do you know the door i have to cross?
question 2: do i have to cross door 1?
there will be at least one 'yes' and one 'no' to question 1.
if only one guy says 'yes', he's mr right. choose accordingly to his answer to question 2.
if only one guy says 'no' to question 1. he's mr. wrong. do the opposite of whatever he says to question 2.
question 2: do i have to cross door 1?
there will be at least one 'yes' and one 'no' to question 1.
if only one guy says 'yes', he's mr right. choose accordingly to his answer to question 2.
if only one guy says 'no' to question 1. he's mr. wrong. do the opposite of whatever he says to question 2.