Question

I have a integer number M in my mind, a number between 1 and N where N is a big number. Chances for M to be any integer between 1 and N are the same. A friend tries to guess this number by asking me to compare M with another number, and I'll answer "your number is bigger", "smaller" or "correct". Another constraint is, his number can be bigger than or equal to M for up to 2 times. What is the strategy to figure out M with least questions?

A variation of this question is, in a N stories building, with 2 eggs, find the lowest floor from where egg breaks when it drops to ground.

Example answers

Follow this idea, assume the first egg is tried every X stories, then the total number of attempts is:

f(X) = N/2X + X/2.

When f(x) is minimized, df(x)/dx = 0. Which means: 1 - N/(X^2) = 0, therefore X=N^(1/2).

Enhance difficulty

What if there are 3 eggs? Assume the 1st egg is tried for every y stories, seconds is tried every x stories, and y=g(x), then the total number of attempts is:

f(x) = (N/g(x) + g(x)/x + x)/2.

When f(x) is minimized, its differential expression should be 0, which means:

1 + g'(x)/x - (g(x))^2/(x^2) - N * g'(x) / (g(x) ^ 2) = 0

When g(x) = x ^ 2 and N = x ^ 3, the equation stands. Therefore the answer is N^(2/3) stories for the first egg, N^(1/3) stories for the second egg and every story for the third one.

Next question

On top of what we found, what if we also what the worse case to be as good as possible? In previous analysis, the worse cases are 2 * N^(1/2) for two eggs and 3 * N^(1/3) for 3 eggs. The average number of guess can't be improved any more, since already made the differential result 0. But with some small adjustment, the worst case can be improved dramatically.

I have a integer number M in my mind, a number between 1 and N where N is a big number. Chances for M to be any integer between 1 and N are the same. A friend tries to guess this number by asking me to compare M with another number, and I'll answer "your number is bigger", "smaller" or "correct". Another constraint is, his number can be bigger than or equal to M for up to 2 times. What is the strategy to figure out M with least questions?

A variation of this question is, in a N stories building, with 2 eggs, find the lowest floor from where egg breaks when it drops to ground.

Example answers

- With only one egg, I can try the 1st floor, the 2nd, 3rd ... until the egg break. This strategy works but it's the worst.
- Improved answer is, try 2, 4, 6, 8...until one egg breaks. Then use the other egg figure out answer.
- Or, try 10, 20, 30, 40.... If egg breaks on 60, try 51, 52, 53... This is slightly better than previous answer, but might not be the best.

Follow this idea, assume the first egg is tried every X stories, then the total number of attempts is:

f(X) = N/2X + X/2.

When f(x) is minimized, df(x)/dx = 0. Which means: 1 - N/(X^2) = 0, therefore X=N^(1/2).

Enhance difficulty

What if there are 3 eggs? Assume the 1st egg is tried for every y stories, seconds is tried every x stories, and y=g(x), then the total number of attempts is:

f(x) = (N/g(x) + g(x)/x + x)/2.

When f(x) is minimized, its differential expression should be 0, which means:

1 + g'(x)/x - (g(x))^2/(x^2) - N * g'(x) / (g(x) ^ 2) = 0

When g(x) = x ^ 2 and N = x ^ 3, the equation stands. Therefore the answer is N^(2/3) stories for the first egg, N^(1/3) stories for the second egg and every story for the third one.

Next question

On top of what we found, what if we also what the worse case to be as good as possible? In previous analysis, the worse cases are 2 * N^(1/2) for two eggs and 3 * N^(1/3) for 3 eggs. The average number of guess can't be improved any more, since already made the differential result 0. But with some small adjustment, the worst case can be improved dramatically.

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