This appears to be Elian Script. I'm not sure I can read the writer's handwriting entirely (and they seem to have added some nonstandard things like a zigzag for T), but the first few lines read:
?HE C PROGRAMMING LANGUAGE!
I REALLY WANT MY PEN!
Edit by OP: I spend some time to fully decipher it but I think Deusovi deserves the ...
An answer using the knife as a measure has already been given, but you can also use the cake itself:
I'm going to assume the cake cannot exist in two places at once, so we have to assume it has whipped cream or some other messy stuff on the sides (the candle suggests it's a birthday cake, so that should be a given), so that it leaves that annoying circle of ...
I think the history is something like this.
Start with 8 colours because 8 is a small power of 2. These are simple: bit 0 for red, bit 1 for green, bit 2 for blue, but 000 means grey rather than black because you're only controlling the foreground and not the background.
Then add another 8 colours. Actually, in the image here it looks as if 8-15 may be the ...
I was able to find one of the solutions using "paper and pencil". I stopped searching for more solutions after that. It is certainly very very time consuming.
In my explanation I name rows as A,B,C,D,E - columns as 1, 2, 3, 4, 5. My search strategy is based on an observation that
As you can see this gave me only 4 possible values for $B1$
Page 2 ...
To a real-life problem I had to give a real-life answer:
But you asked for an actual tiling, without gaps, so here it is.
PS: there is a simpler pattern where pairs disassemble with a single translation:
Consider a hexagon filling up the area of the cake. A hexagon is made up of 6 equilateral triangles.
So, each edge of the triangle would be the same as the radius of the cake (candle to edge).
Mark the knife to indicate the radius of the cake going from one point to the candle.
Now, make one cut from the candle to the edge. Then, use the marking on the ...
Avi and DEEM came close to the right answer; the effect is actually
It's caused by a combination of two factors: one,
Surprisingly the combination of these two factors is enough for the effect to be clearly visible.
[EDIT: After this was answered, the puzzle was restricted to $8$ players. In this case, $8+3+3-3=11$ games are required in the worst case.]
The following method requires up to
Suppose $n=2^k$. First,
In total, at most
games are played.
Suppose we add Alice so that $n=2^k+1$. Modifying the above method:
Several optical things can result in this illusion.
First, most roads are slightly elevated in the middle so the driver side (in US for example) is slightly elevated
Secondly, windshields are slightly curved. So looking straight at the left side gives a different perspective than looking at the right side tires
Thirdly, if you draw a line (of ...
Florian F's 2nd pattern is far and away my favorite, but if anyone was curious, I'll post my answers.
First I wanted to show an example of something that doesn't work but really seems like it should:
It's just like the third example from the question, but it uses joinery such that the pieces slide together at an angle. It comes apart in the same way, ...
This is a way more hairy subject than it appears on the surface, so you are not getting a puzzle answer, but a real-world one instead.
TL;DR: The given requirements are logically incompatible with the statement "round-robin tournaments can end in a draw", so OP's question doesn't have a clear correct answer.
The difficulties arise, because you need a model ...
EDIT: Just to let you know before you read this, this actually isn't the solution at all. I've made lots of assumptions and serious mistakes! :)
The way to think about this problem is
My algorithm is really simple:
This algorithm is generalizable
From this it's easy to see that
And that's it.
Very partial solution
Suppose there are an even number of people: say 2n. (Call them A1..An and B1..Bn.) Then as per Brandon_J's conjecture
We can get a lower bound on the number of meeting periods needed from
Here's a construction that's not too bad, though in general it's far from optimal.
[EDITED to add:] No, wait, I'm not sure $m-1$ is ...