You have 100 pennies. Exactly 50 of them a heads-up, but you don't know which ones. If you're blindfolded, how would you divide the pennies up into two groups with an equal number of heads? Repeat the same exercise when exactly 10 of them are heads-up.

By feeling the pennies. I probably couldn't do it, but my blind friend Kevin probably could easily.

Where is DD's solution? It's early but I don't see it in the thread.

2. PM answers, add posts to ask questions.

Whats the point of it if we can already see solutions?

And I agree with rabbit and object to nslay's thing being a puzzle, it is just a pure math problem you have to work backwards to get the solution from, given the conditions. Sidoh's original puzzles don't require any complex math, just simple reasoning and deduction to get a solution. In order to solve the math problem you would have to know some advanced math to understand the notations used to describe the problem, not to mention a knowledge of linear algebra to understand matrices and their transpose and properties of them. O yea and integration also, not to mention it just isn't even interesting to solve! Therefore I demand it to be removed thanks

On the contrary, it is very interesting in the sense of optimization. If you had a Hessian of that form (which I did), positive definiteness would imply strict convexity and therefore a unique global optimum.

The reasoning probably isn't as simple as you think it is. There's a twist and you probably didn't realize it.

EDIT: Oh, I misread. You were talking about Sidoh's puzzle. Nah, this really does boil down to really simple math. You're just intimidated by linear algebra and integration (which really are basic math). But as I said, there is a *twist*.

Since it contains real values wouldn't f(x)f(x)^{T} be positive-semidefinite? and in order to be positive-definite f(x) needs to be linearly independent. Probably not correct but thats all I can remember from linear algebra.

I'm not intimidated by math, just that I don't find pure math interesting. Now if you put that into a problem (optimization) that's a different story. And if linear algebra and calculus are considered basic math, then what is addition, subtraction, algebra? Dumbass math?

Indeed, you do get positive semidefinite for free. However, you're still only half right about linear independence (and this got me too!).

Here is an example of linearly independent functions where said integral isn't strictly positive definite.

[latex]

f_1(x) = \sin(x) \\

f_2(x) = \begin{cases}

\sin(x) & x \neq 0

1 & x = 0

\end{cases}

[/latex]

So

[latex]

\begin{bmatrix} 1 & -1 \end{bmatrix} \int_{-\pi}^{\pi} \mathbf{f}(x) \mathbf{f}(x)^T d x \begin{bmatrix} 1 \\ -1 \end{bmatrix} = \begin{bmatrix} 1 & -1 \end{bmatrix} \int_{-\pi}^0 \mathbf{f}(x) \mathbf{f}(x)^T dx \begin{bmatrix} 1 \\ -1 \end{bmatrix} + \begin{bmatrix} 1 & -1 \end{bmatrix} \int_0^{\pi} \mathbf{f}(x) \mathbf{f}(x)^T dx \begin{bmatrix} 1 \\ -1 \end{bmatrix} = 0

[/latex]

One has to additionally say that the functions are

*linearly dependent only on a set with zero measure*. I totally ruined it, but that's the twist!

I asked this question because it is both counter-intuitive (integral of rank 1 matrix somehow giving full rank matrix result) and it's also deceptively obvious once you do exploit linear independence (because one has to remember the

*twist*).