Playing poohsticks in the Styx

Ну что, пошёл я на пресловутый chat.openai.com и спросил у него:

Ackermann(4, 4)=?

На что получил вполне разумный ответ

The Ackermann function is a mathematical function that is used to test the computational capabilities of computers. It is a very complex function that grows very quickly, even for relatively small input values.

In general, the Ackermann function is defined as follows:

Ackermann(m, n) =

• n + 1, if m = 0
  
• Ackermann(m - 1, 1), if m > 0 and n = 0
  
• Ackermann(m - 1, Ackermann(m, n - 1)), if m > 0 and n > 0
  

So, in the case of Ackermann(4, 4), we have:

Ackermann(4, 4) = Ackermann(3, Ackermann(4, 3))

Since Ackermann(4, 3) is itself a very large number, the value of Ackermann(4, 4) is extremely large and is not practical to calculate by hand. In general, the Ackermann function grows so quickly that it is not feasible to calculate its value for even moderately large input values.

Замечательно, думаю. А если что посложнее?

What is the expected number of people such that the probability of every day in a year being somebody's birthday is greater than 50%?

Дальше решайте сами, иметь этот вумный как вутка интеллект или не иметь (особо см. выделенное):

The probability that any given day is not somebody's birthday is 1 minus the probability that it is somebody's birthday. For a group of size N, the probability that a given day is not somebody's birthday is (365-1)/365 * (364-1)/364 * ... * (365-N)/365. So the probability that a given day is not somebody's birthday is 1 when the following equation is true:

(365-1)/365 * (364-1)/364 * ... * (365-N)/365 = 1

To solve this equation, we can take the natural logarithm of both sides to get:

ln((365-1)/365 * (364-1)/364 * ... * (365-N)/365) = 0

Since the natural logarithm is a monotonically increasing function, we can apply it to both sides of the inequality to get:

ln(1 - 1/365) + ln(1 - 2/365) + ... + ln(1 - N/365) < 0

We can then rearrange the terms and use the fact that the natural logarithm is an increasing function to get:

N * ln(1 - 1/365) + (N-1) * ln(1 - 2/365) + ... + 1 * ln(1 - N/365) < 0

From this equation, we can see that the value of N that satisfies the inequality is the smallest value of N such that the sum of the natural logarithms on the left-hand side is less than 0.

To find this value of N, we can simply try different values of N and calculate the sum of the natural logarithms until we find the smallest value of N that satisfies the inequality. For example, if we try N = 1, we have:

1 * ln(1 - 1/365) = -0.0027

This value is not less than 0, so we know that N must be greater than 1. If we try N = 2, we have:

2 * ln(1 - 1/365) + ln(1 - 2/365) = -0.0147

This value is also not less than 0, so we know that N must be greater than 2. If we try N = 3, we have:

3 * ln(1 - 1/365) + 2 * ln(1 - 2/365) + ln(1 - 3/365) = -0.0296

This value is less than 0, so we know that N = 3 is the smallest value of N that satisfies the inequality.

Therefore, the expected number of people such that the probability of every day in a year being somebody's birthday is greater than 50% is 3.

Note that this is just an approximation, and the actual value of N may be slightly different depending on the precision of the calculations. Also, this calculation assumes that all birthdays are equally likely, which may not be the case in reality.

Такие дела.