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My "Thats all" means: so you can't get "B" from "A" using that rules.

px: ABBA is never a desired result.

texel: That is correct. Congratulations, you are now a mensa member. As such, you are licensed to play chess and read books with no pictures in them, of any thickness. Your membership card and special mensa cap will arrive by post within 2 business days along with a formal welcome letter. Your membership fee for the first year is €1,995.00, due on 8. December 2006. Please specify a private email address or fax number so I can supply you with the necessary details.

Doom: I am not worthy... Oh well... I've been already a Mensa member and I don't want to be again. Good problem anyway. Do you know the source of it? I would like to add it to my collection

texel: It's the MU puzzle, but cleverly disguised using different symbols. And I removed the irrelevant M. I call it the "B puzzle".

Thanks. I should read Gödel, Escher and Bach. It has been in my list of to read books for some years...

An easier but interesting one:

2^2 = 4 = 1 + 3
3^2 = 9 = 1 + 3 + 5
4^2 = 16 = 1 + 3 + 5 + 7
...

Why?

Too silly problem to even write the solution Adok.

I have two "problems" I would like you to help me solve. These are real problems I found these week at my job.

1] given a (unit)vector w, could you give a othogonal/orthonormal vector u (so I can build a basis) WITHOUT USING contidonals? The usual trick of select a random "up" vector has the problem of the casual aligment with w, so you need to check the dot product first and select another up vector. This is an IF. I don't accept the "bah, the chance of your up vector to be aligned with w is zero", and I don't accept takking the dot, multiplying by something BIG, saturating to 1, and using that to lerp between two predefined up vectors. No programming tricks please. Just some math.

2] could someone help me isolate x in

t = x*cosh(s/x)-x

(I didn't spend much time on it, but Laplace transform doesn't seem to help). Of course, I want some maths, no a bisection or newton-raphson code. Thanks.

(adok+1)^2 == adok^2 +(2adoks+1) ?

I am interested in iq's first question too..

What means "contidonals"?

the_Ye-Ti: Well done!

Another easy one. Mensans ought to be able to solve this within 10 seconds:

"If x is an even number, then why is x mod 10 = (6 x) mod 10?"

Sorry. "Conditionals" means something like

u = { cross(up1,w) if dot(up1,w)==1
{ cross(up2,w) otherwise

(read "conditional branching" if you were to programm this)

iq: Why do you need this? Why can't you work with if clauses?

1)

In 2D is easy: vector V (x, y) - rotate 90 degrees - V' (-y, x)

In 3d, you have infinite orthonormal vectors to V (x, y, z). If you just want *any* of these and be sure it is orthonormal (this is how I've understood your question) without conditionals (I suppose these are "contidonals"), just rotate 2 times. 2 rotations of 90 degrees will do it, just pick 2 coordinates for rotation, for example x and then y. So, since cos(90)=0, and sin(90)=1:

x' = y;
y' = -z;
z' = -x;

Well... not sure if this is what you are asking or not...

Maybe what I wrote has no sense at all...

Oh, I don't deserve being mensa anymore...

Let Y be your original vector.
Let X=(y2+y3,y3-y1,-y1-y2)
Then X/|X| is your answer.

It's the most sum of these three vectors that are all perpendicular to Y:

(y2,-y1,0)
(y3,0,-y1)
(0,y3,-y2)

To test it, let any combination of y1,y2,y3 be 0 or 1, and find the cases in which Z=(0,0,0). If I did it right you'll find it only happens when y1=y2=y3=0.

x'=y+z;
y'=-x;
z'=-x;

This one looks better, but it doesn't give you a normalized v'

Not the "most sum" though, just the sum. :)

Yes, easy, z1=0 iff y2=y3=0, and z2=0 iff y3=y1. There. Proof fer ya.

Uhm.

:) Indeed, (x',y',z') = (y,-z,-x) is not perpendicular to (x,y,z)

(-z,0,x) would be... UNLESS w=(0,1,0) of course. This is the kind of exceptions I want to avoid.

Someone can try to solve (or show it's impossible to solve?)

Doom, we did exactly the same shit.