What is the best (whatever this means) answer and why?
[Math] answer to iq test with colored squares
puzzlerecreational-mathematics
Related Solutions
I asked this question on MathOverflow and got a great answer there.
For $k = 2$ I can do it with ${\lceil \log_2 N \rceil + 2 \choose 2} - 1$ servants. In particular for $N = 1000$ I can do it with $65$ servants. The proof is somewhat long, so I don't want to post it until I've thought about the problem more.
I haven't been able to improve on the above result. Here's how it works. Let $n = \lceil \log_2 N \rceil$. Let me go through the algorithm for $k = 1$ so we're all on the same page. Number the wines and assign each of them the binary expansion of their number, which consists of $n$ bits. Find $n$ servants, and have servant $i$ drink all the wines whose $i^{th}$ bit is $1$. Then the set of servants that die tells you the binary expansion of the poisoned wine.
For $k = 2$ we need to find $n$ butlers, $n$ maids, and ${n \choose 2}$ cooks. The cooks will be named $(i, j)$ for some positive integers $1 \le i < j \le n$. Have butler $i$ drink all the wines whose $i^{th}$ bit is $1$, have maid $i$ drink all the wines whose $i^{th}$ bit is $0$, and have cook $(i, j)$ drink all the wines such that the sum of the $i^{th}$ bit through the $j^{th}$ bit, inclusive, mod 2, is $1$. This is how the casework breaks down for butlers and maids.
- If both butler $i$ and maid $i$ die, then one of the poisoned wines has $i^{th}$ bit $0$ and the other has $i^{th}$ bit $1$.
- If only butler $i$ dies, then both of the poisoned wines have $i^{th}$ bit $1$.
- If only maid $i$ dies, then both of the poisoned wines have $i^{th}$ bit $0$.
The second two cases are great. The problem with case 1 is that if it occurs more than once, there's still ambiguity about which wine has which bit. (The worst scenario is if all the butlers and maids die.) To fix the issue with case 1, we use the cooks.
Let $i_1 < ... < i_m$ be the set of bits where case 1 occurs. We'll say that the poisoned wine whose $(i_1)^{th}$ bit is $1$ is wine A, and the other one is wine B. Notice that the sum of the $(i_1)^{th}$ through $(i_2)^{th}$ bits of wine A mod 2 is the same as the sum of the $(i_1)^{th}$ through $(i_2)^{th}$ bits of wine B mod 2, and we can determine what this sum is by looking at whether cook $(i_1, i_2)$ died. The value of this sum determines whether the $(i_2)^{th}$ bit of wine A is 1 or 0 (and the same for wine B). Similarly, looking at whether cook $(i_j, i_{j+1})$ died tells us the remaining bits of wine A, hence of wine B.
One last comment for now. The lower bound is not best possible when $k$ is large compared to $N$; for example, when $k = N-1$ it takes $N-1$ servants. The reason is that any servant who drinks more than one wine automatically dies, hence gives you no information.
Although not needing deep mathematical knowledge...
- A blindfolded man is handed a deck of 52 cards and told that exactly 10 of these cards are facing up. How can he divide the cards into two piles (possibly of different sizes) with each pile having the same number of cards facing up?
An old-fashioned implication one:
- Mrs Claus always sneezes just before it starts snowing. She just sneezed. “This means that it’s going to start snowing”, thinks Santa. Is he correct?
Robbed these from http://math.alamzy.com/wp-content/uploads/2012/10/Handbook.pdf
Best Answer
Well, that's funny, 'cause you can get an answer basing only on several conventions (resonable ones). But they are not unique, so guess you can get different answers. Let's try.
In the left column we have initial states (I've placed an axis, running along the colored blocks from blue to red).
In the middle colum we have the state that shows us the direction of transformations (rotations). I've placed an axis along the colored in gray blocks in such a way that it always points upwards. Plus I've added an azure line which coincides in direction with the initial state and has two parts: the inital and mirrored one.
In the right column we have final states which are obtained by the rotation of the initial state along the direction of transformations.
So following this logic I guess the answer is $\mathrm{G}$.