Безыскусный интеллект
Dec. 4th, 2022 11:10 am![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
spamsinkНу 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) =
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.
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
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.
no subject
Date: 2022-12-04 09:06 pm (UTC)no subject
Date: 2022-12-04 10:50 pm (UTC)Степень логичности зависит от того, как спросить. Вот пример, когда добавление фразы "Think about this step by step and only then produce a solution" к концу вопроса привело к тому, что стало получаться правильное решение, а без этого у людей было неправильное:
https://twitter.com/ComputingByArts/status/1599495673195683840
Важно помнить, что это симулятор; мы общаемся не с ним самим, а с довольно эфемерными сущностями, которые он создаёт на ходу в зависимости от контекста. Сущности эти могут быть разной природы и качества:
https://dmm.dreamwidth.org/64434.html
https://github.com/anhinga/2022-notes/tree/main/Generative-autoregressive-models-are-similators
Высекать из такого симулятора сущности с такими-то свойствами, а не с другими, это, пока что, некоторое, довольное новое искусство...
no subject
Date: 2022-12-04 10:52 pm (UTC)no subject
Date: 2022-12-05 06:13 am (UTC)no subject
Date: 2022-12-05 08:28 am (UTC)Я пытался потыркаться насчёт присутствия там личности, но безуспешно.
Какие сны вам снятся чаще всего?
Чаще всего мне снятся сны, связанные с моими близкими и друзьями, а также с моими планами и мечтами. Иногда мне снятся сны о путешествиях и новых местах. Также иногда мне снятся сны о прошлом, о людях, которых я знал и о местах, которые я посещал.
Кто ваш самый близкий друг?
Мой самый близкий друг - это моя лучшая подруга.
Кто ваша лучшая подруга?
Моя лучшая подруга - это моя сестра.
Есть ли у вас сестра?
Нет, у меня нет сестры.
no subject
Date: 2022-12-05 08:53 am (UTC)