I’ll keep presenting this challenge until someone meets it:
Anyone who thinks LLMs aren’t generally intelligent, can you name a text processing task (ie text in, text out) than a general intelligence can do, that an LLM cannot?
To solve this, let’s first consider the total number of combinations when there are no restrictions. Since there are 4 friends, there are 4! (4 factorial) ways to arrange them, which equals 24 combinations.
Now, let’s count the number of combinations where A and C are next to each other. Since A and C can be arranged in 2 ways, and for each arrangement, the other two friends (B and D) can be arranged in 2! ways, the total number of combinations where A and C are next to each other is 2 * 2! = 4.
So, the number of combinations where A and C cannot be next to each other is the total number of combinations minus the number of combinations where A and C are next to each other:
I’ll keep presenting this challenge until someone meets it:
Anyone who thinks LLMs aren’t generally intelligent, can you name a text processing task (ie text in, text out) than a general intelligence can do, that an LLM cannot?
Text in: a statement
Text out: confirmation whether statement is factually true or not
Is that something a human can do consistently?
If it’s not, does that imply a human does not possess general intelligence?
To be honest, even the human mind has this faculty not in all cases.
Logic. As an example, non textbook math questions. I asked ChatGPT 3.5 this:
It answered 20, the correct answer is 12.
All possible conbinations
abcd,abdc,adbc,adcb, cbad,cbda,cdba,cdab, bcda,badc, dcba,dabc
Its answer
To solve this, let’s first consider the total number of combinations when there are no restrictions. Since there are 4 friends, there are 4! (4 factorial) ways to arrange them, which equals 24 combinations.
Now, let’s count the number of combinations where A and C are next to each other. Since A and C can be arranged in 2 ways, and for each arrangement, the other two friends (B and D) can be arranged in 2! ways, the total number of combinations where A and C are next to each other is 2 * 2! = 4.
So, the number of combinations where A and C cannot be next to each other is the total number of combinations minus the number of combinations where A and C are next to each other:
24 - 4 = 20 combinations.
You can have it try again over and over, even while telling it the answer is 12, and it hallucinates basically random numbers to boot.