QwQ: Alibaba's O1-like reasoning LLM

False start with car rental companies, only three, not four. Hertz has to be a unit then, making the first group hertz, second, mole, and volt.

Similarly, enterprise must be in the business sense. Enterprise, project, venture, endeavor.

Tiger, Ray, National, Twin are all singular versions of baseball teams?

Curb, silicon, boardwalk, and game are left. Boardwalk is the most valuable monopoly property, silicon is the modern equivalent? No clue, maybe I’m missing a cultural reference? What if curb is a verb? Curb your enthusiasm, first word of tv shows? Silicon Valley, game of thrones, maybe there is a show that starts with boardwalk? Seems tenuous.

Edit: yes, there is. Boardwalk Empire. Never watched any of them, but some googling tends me they’re all hbo shows - I knew that about Thrones. That seems slightly more realistic a connection than just first word of tv shows.

Seems like exactly the kind of thing AI would/could theoretically best humans in.

I remember having a hard time solving this in my first year of college. This is really impressive.

So I have this problem here: given that for positive integers m and n, 0 ≤ x[m + n] ≤ x[m] + x[n], I need to prove that the limit of x[n]/n exists. Hmm, okay, let's see how to approach this.

…

Alternative approach: maybe I can show that a_n is convergent by showing that it's a Cauchy sequence.

But that might be complicated.

Alternatively, perhaps I can use the subadditivity to bound a_n.

Wait, let's consider the following:

Since x is subadditive, x[m + n] ≤ x[m] + x[n].

Let me set m = n, then x[2n] ≤ x[n] + x[n] = 2x[n].

Therefore, x[2n]/(2n) ≤ x[n]/n.

So, a_{2n} ≤ a_n.

That means that a_{2n} ≤ a_n.

Similarly, x[3n] ≤ x[2n] + x[n] ≤ 2x[n] + x[n] = 3x[n].

So, x[3n]/(3n) ≤ x[n]/n.

Similarly, x[4n]/(4n) ≤ x[2n]/(2n) ≤ x[n]/n.

It seems like a_{kn} ≤ a_n for any integer k ≥ 1.

This suggests that as n increases, a_n tends to decrease or stay the same.

But wait, in the example where x[n] = n, a_n = 1 for all n, so it's constant.

In the example where x[n] = sqrt(n), a_n = 1/sqrt(n), which decreases to 0.

So, in some cases, a_n is decreasing, and in some cases, it's constant.

But in general, a_n is bounded below by 0 and is decreasing along powers of 2, for example.

Wait, from x[2n]/(2n) ≤ x[n]/n, and x[4n]/(4n) ≤ x[2n]/(2n), and so on, so along the sequence n, 2n, 4n, 8n, etc., a_n is decreasing.

Maybe I can use this to show that a_n has a limit.

Let me consider n = 2^k, for integer k.

Let b_k = x[2^k]/2^k.

Then, from x[2^{k+1}] ≤ x[2^k] + x[2^k] = 2x[2^k], so b_{k+1} = x[2^{k+1}]/2^{k+1} ≤ (2x[2^k])/2^{k+1} = x[2^k]/2^k = b_k.

So, b_{k+1} ≤ b_k, meaning that b_k is a decreasing sequence.

Since b_k ≥ 0, it must converge to some limit, say L.

Now, I need to show that x[n]/n approaches L as n approaches infinity.

Given that b_k is decreasing and converges to L, and for general n, a_n is bounded below by L, and perhaps approaches L.

Wait, more carefully:

For any n, there exists k such that 2^{k-1} Then, x[n]/n ≤ x[2^k]/2^{k-1} = 2(x[2^k]/2^k) = 2b_k.

Since b_k approaches L, for large k, b_k is close to L.

But I need to show that x[n]/n is close to L.

Wait, perhaps I can bound x[n]/n between b_k and something else.

Given that 2^{k-1} x[2^{k-1}].

Therefore, x[n]/n ≤ (2x[2^{k-1}])/n ≤ (2x[2^{k-1}])/2^{k-1} = 2(x[2^{k-1}]/2^{k-1}) = 2b_{k-1}.

But b_{k-1} approaches L as k increases.

Similarly, since n > 2^{k-1}, x[n]/n > x[2^k]/2^k = b_k.

But b_k ≤ b_{k-1}, and both approach L.

Therefore, x[n]/n is squeezed between b_k and 2b_{k-1}, both of which approach L.

Therefore, x[n]/n approaches L as n approaches infinity.

This suggests that the limit exists and is equal to L.

Therefore, the limit of x[n]/n exists.

Alternatively, if I recall correctly, this is exactly what Fekete's lemma states.

So, in conclusion, the limit of x[n]/n exists.

*Final Answer*

\boxed{\lim_{n \to \infty} \frac{x[n]}{n} \text{ exists}}

I think it's better phrased as "find the best rule", with a tacit understanding that people mostly agree on what makes a rule decent vs. terrible (maybe not on what makes one great) and a tacit promise that the sequence presented has at least one decent rule and does not have multiple.

A rule being "good" is largely about simplicity, which is also essentially the trick that deep learning uses to escape no-free-lunch theorems.

I should have read the blog post. This is a known issue:

>Recursive Reasoning Loops: The model may enter circular reasoning patterns, leading to lengthy responses without a conclusive answer.

So the 100 lines was probably not necessary.

If you read the article it even states that it's a Chinese ship but with a Russian crew that departed from Russia. They leased it from China. If you have an accident with a leased Chinese car, no one would say "the Chinese did it".