Notes on OpenAI's new o1 chain-of-thought models

Just leaving here an example of the "smart chain of thought", which took the model about 20s of thinking. The problem I asked was – "what 3 numbers give the same result when divided and added together". Seems the current state of art is that basic school problems are too hard yet even for this new chain-of-marketing.

  Let's think through this step-by-step:

  We're looking for three numbers a, b, and c where:
    (a / b) / c = a + b + c
  This equation is a bit trickier than the multiplication one, because division is 
  not commutative (the order matters). We need to be careful about how we 
  interpret (a / b) / c.
  Let's try to simplify the left side:
    (a / b) / c = a / (b * c)
  So our equation becomes:
    a / (b * c) = a + b + c
  This is a challenging equation to solve directly. Let's try some numbers:

  If we use 1, 2, 3 like before:
    1 / (2 * 3) = 1/6 ≠ 1 + 2 + 3 = 6
  Let's try smaller numbers, like fractions:
    1/2, 1/3, 1/6


  Let's check:
    (1/2 / 1/3) / 1/6 = (1/2 * 3) / 1/6 = 3/2 * 6 = 9
    1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1
  This doesn't work either. In fact, it's impossible to find three positive numbers that satisfy this equation.
  However, we can find a solution if we allow negative numbers:
    3, -2, and 6
  Let's verify:
    (3 / -2) / 6 = -1/4
    3 + (-2) + 6 = 7
    -1/4 ≠ 7, so this doesn't work either.

  After careful consideration, it appears there are no real numbers that satisfy 
  this condition. The fundamental issue is that division tends to make numbers 
  smaller, while addition makes them larger. It's mathematically impossible to 
  find three numbers where these operations result in the same value.

nuff said

I live in one of the countries you mentioned and just showed it to one of my friends who's a local who struggles with English. They had no problem concluding that the doctor was the child's dad. Full disclosure, they assumed the doctor was pretending to be the child's dad, which is also a perfectly sound answer.

Claude 3.5 seems to get it, (thought the method is not what I would use?) -

———

Let's think through this step-by-step:

  We're looking for three numbers a, b, and c where:
    (a / b) / c = a + b + c

————

Certainly, let's approach this problem step-by-step:

1) First, let's write out our equation: (a / b) / c = a + b + c

2) To simplify the left side, we can rewrite it as: a / (b * c) = a + b + c

3) Now, to solve this, we need to find values for a, b, and c that satisfy this equation. Let's start by considering some properties: - All numbers must be positive (because we're dividing by b and c) - The left side is a fraction, while the right side is a sum

4) One way to approach this is to try to make the left side equal to 1, which would mean: a = b * c

5) If we substitute this into our original equation: 1 = b * c + b + c

6) Now we're looking for two numbers b and c where their product plus their sum equals 1.

7) Let's try some small numbers: If b = 1/2 and c = 1/2: (1/2 * 1/2) + 1/2 + 1/2 = 1/4 + 1 = 5/4 ≠ 1

8) Let's try smaller numbers: If b = 1/3 and c = 1/3: (1/3 * 1/3) + 1/3 + 1/3 = 1/9 + 2/3 = 7/9 ≠ 1

9) Even smaller: If b = 1/4 and c = 1/4: (1/4 * 1/4) + 1/4 + 1/4 = 1/16 + 1/2 = 9/16 ≠ 1

10) We're getting closer. Let's try b = 1/4 and c = 1/3: (1/4 * 1/3) + 1/4 + 1/3 = 1/12 + 1/4 + 1/3 = 1/12 + 3/12 + 4/12 = 8/12 = 2/3 ≠ 1

11) Let's adjust further. If b = 1/3 and c = 1/4: (1/3 * 1/4) + 1/3 + 1/4 = 1/12 + 1/3 + 1/4 = 1/12 + 4/12 + 3/12 = 8/12 = 2/3 ≠ 1

12) One more try. Let's use b = 1/2 and c = 1/3: (1/2 * 1/3) + 1/2 + 1/3 = 1/6 + 1/2 + 1/3 = 1/6 + 3/6 + 2/6 = 6/6 = 1

13) We found it! When b = 1/2 and c = 1/3, the equation holds.

14) Remember, we set a = b * c, so a = 1/2 * 1/3 = 1/6

Therefore, one solution to the equation (a / b) / c = a + b + c is: a = 1/6, b = 1/2, and c = 1/3

We can verify: (1/6 / 1/2) / 1/3 = (1/3) / 1/3 = 1 1/6 + 1/2 + 1/3 = 1/6 + 3/6 + 2/6 = 1

This solution satisfies the equation.

——

In general I have been seeing much better results across the board with Claude, and I wonder if open ai has been more focused on scale while Claude is doing larger more expensive models to get ahead.

> but I was pretty disappointed

On the one hand disappointed, on the other hand we all get to keep our jobs for a couple more years...