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Hmm... I don't think that it's a good idea to merge all riddle threads. If we run out of riddles or loose the interest, the threads will quickly disappear out of sight. If on the other hand we keep...

CornedBee already has.

Greets,
Philip

Grmbl. I already dismissed the parity approach. Should have thought it through...

Anyway, very nice riddle!

Greets,
Philip

It may even work without making pauses. Are the dwarves allowed to call out their presumed hat color more than once?

Greets,
Philip

If the dwarves are allowed to move, I have another solution:

The last dwarf (the one who can't see any other dwarf) moves forward one step. The next dwarf walks right next to him. Then comes the...

Oops, you are right. I didn't get it in the first place, so I wrote down this table:



BBB
WBB
BBW B
BWB
BWW A
WBW B

I have a solution, but I think it involves cheating:

The first dwarf simply says the color of the hat directly in front of him. The 2nd dwarf thus knows its own hat color. If the third dwarf has a...

Because the situations WBB and BBB are indistinguishable for all dwarves.

Greets,
Philip

I doubt that. If all get a black sticker, there's no way to tell.

But I can think of at least some ways that will make them survive:
If A sees two white stickers, he can immediately answer...

I already had, but it's good to see you join. :)

There are a few questions remaining:

Do the dwarves know in advance the value of n, i.e. the total number of dwarves?
Do the dwarves know in...

Right!

It may sound silly, but I feel happy now.

Greets,
Philip

As there are infinitely many natural numbers, the two orderings ... < 2 < 1 and 1 < 2 < ... are distinct because they have different properties (there's a clear difference between successor and...

Yes, the order has to be total and strict. Note that an element without a specific predecessor is not necessarily a smallest element (although a smallest element certainly doesn't have a...

The natural numbers are usually ordered in the following way:

1 < 2 < 3 < 4 < ...

There is exactly one element (the first) which has no predecessor. Note that we are free to invent our own...

For those who don't see it right away, here's the full proof:

Suppose that "barish" is barish. Then it denotes a property that it has itself, hence it is fooish. Contradiction.
Suppose that...

Maybe. This used to be true for CS students at my university (that's why I knew about it). Now, the corresponding lecture isn't mandatory anymore. It seems to me that the focus has shifted from logic...

I can see some ugly bit twiddling here. Any volunteers? :))



Nothing easier than that if you know that double negation doesn't change the value:
a OR b == !(!a AND !b)

I once had to show...

I introduce two new adjectives fooish and barish.

A word is fooish if it denotes a property that it has itself.
A word is barish if it doesn't.

Examples for fooish words:
- "short" is a short...

Right!

Can someone else explain it?

Greets,
Philip

And I was inspired by this: Labyrinth Puzzle.

Yes.



return (a >b)?1:(a<b)?-1:0;


Nice try.

I love this solution. My little brother suggested killing the dwarves until one of them decides to tell the truth anyway, but this solution suffers in a similar way.

And after all, a 50% chance is...

Oops, I skipped #4, so here it is.

Consider the following program:



int cmp(int a, int b)
{
if(a > b) {
return 1;

Right. I already had several discussions about that. The problem arises due to the inexactness of most natural languages. The most obvious solution is to ask a more concise (and more awful) question....

There are seven dwarfs, each one wearing a unique hat. A wind blows off the hats. The dwarfs start running after their hats and each dwarf puts on the first hat that he manages to catch. Eventually...