【QwQ】Qwenベースの推論モデル登場！数学タスクでo1と徹底比較

So I've got this equation: 1 + 2 * 3 + 4 * 5 + 6 * 7 + 8 * 9 = 479, and I need to add a pair of parentheses to make it correct. Hmm, let's see.

First, I should understand the current calculation without any parentheses. Remembering the order of operations, multiplication comes before addition. So, the equation is evaluated as:

1 + (2 * 3) + (4 * 5) + (6 * 7) + (8 * 9) = 1 + 6 + 20 + 42 + 72 = 141.

But the equation claims it equals 479, which is way off. So, I need to add parentheses to change the order of operations to reach 479.

I need to think about where to place the parentheses to maximize the value. Since multiplication has a higher precedence than addition, grouping additions first would give smaller intermediate results, which might not help much. Maybe I need to group multiplications in a different way.

Wait, but all multiplications are already separate, and additions are between them. Perhaps I need to group an addition and then multiply that by something.

Let me try grouping 1 + 2 and multiplying by 3, then adding the rest:

(1 + 2) * 3 + 4 * 5 + 6 * 7 + 8 * 9 = 3 * 3 + 20 + 42 + 72 = 9 + 20 + 42 + 72 = 143.

Still not 479.

Maybe grouping 1 + 2 * 3:

1 + (2 * 3) + 4 * 5 + 6 * 7 + 8 * 9 = 1 + 6 + 20 + 42 + 72 = 141.

Wait, that's the same as before.

What if I group 1 + 2 and 3 + 4 and multiply them:

(1 + 2) * (3 + 4) + 5 + 6 * 7 + 8 * 9 = 3 * 7 + 5 + 42 + 72 = 21 + 5 + 42 + 72 = 140.

Not helpful.

Maybe grouping larger parts. What if I group from the left:

(1 + 2) * 3 + 4 * 5 + 6 * 7 + 8 * 9 = 3 * 3 + 20 + 42 + 72 = 9 + 20 + 42 + 72 = 143.

Still not 479.

Wait, maybe I need to group more extensively. Let's try grouping 1 + 2 * 3 + 4:

(1 + 2 * 3 + 4) * 5 + 6 * 7 + 8 * 9.

First, calculate inside the parentheses: 1 + 6 + 4 = 11.

Then, 11 * 5 = 55.

Then, 55 + 42 + 72 = 169.

Still too low.

Alternatively, maybe grouping 1 + 2 * 3 + 4 * 5:

(1 + 2 * 3 + 4 * 5) + 6 * 7 + 8 * 9.

Inside the parentheses: 1 + 6 + 20 = 27.

Then, 27 + 42 + 72 = 141.

Same as before.

Hmm, perhaps I need to group additions and multiplications differently.

Wait, maybe grouping additions first to get larger numbers before multiplying.

For example, group 1 + 2 + 3 and multiply by 4:

(1 + 2 + 3) * 4 + 5 * 6 + 7 * 8 + 9 = 6 * 4 + 30 + 56 + 9 = 24 + 30 + 56 + 9 = 119.

Not there yet.

What if I group 1 + 2 + 3 + 4 and multiply by 5:

(1 + 2 + 3 + 4) * 5 + 6 * 7 + 8 * 9 = 10 * 5 + 42 + 72 = 50 + 42 + 72 = 164.

Still low.

Maybe grouping even larger parts.

What about (1 + 2 + 3 + 4 + 5) * 6 + 7 * 8 + 9 = 15 * 6 + 56 + 9 = 90 + 56 + 9 = 155.

Closer, but still not 479.

Wait, maybe I need to group multiplications differently.

What if I group 2 * (3 + 4) and so on.

Let's try: 1 + 2 * (3 + 4) * 5 + 6 * 7 + 8 * 9.

First, 3 + 4 = 7.

Then, 2 * 7 = 14.

Then, 14 * 5 = 70.

So, 1 + 70 + 42 + 72 = 185.

Still low.

Alternatively, maybe grouping additions inside multiplications.

Wait, perhaps I need to group additions first and then multiply by subsequent numbers.

Alternatively, maybe I need to group sub-expressions in a different way.

Let me try grouping from the right.

1 + 2 * 3 + 4 * 5 + 6 * (7 + 8) * 9.

First, 7 + 8 = 15.

Then, 6 * 15 = 90.

Then, 90 * 9 = 810.

Now, 1 + 6 + 20 + 810 = 837.

Way above 479.

Maybe I need to group differently.

Wait, perhaps I can group additions and multiplications in a nested way.

Let's try: 1 + 2 * (3 + 4 * 5) + 6 * 7 + 8 * 9.

First, 4 * 5 = 20.

Then, 3 + 20 = 23.

Then, 2 * 23 = 46.

So, 1 + 46 + 42 + 72 = 161.

Still low.

Alternatively, maybe grouping 3 + 4 * 5:

1 + 2 * 3 + 4 * (5 + 6) * 7 + 8 * 9.

First, 5 + 6 = 11.

Then, 4 * 11 = 44.

Then, 44 * 7 = 308.

Now, 1 + 6 + 308 + 72 = 387.

Still low.

Wait, maybe grouping further.

What if I do 1 + 2 * 3 + 4 * (5 + 6 * 7) + 8 * 9.

First, 6 * 7 = 42.

Then, 5 + 42 = 47.

Then, 4 * 47 = 188.

Now, 1 + 6 + 188 + 72 = 267.

Closer, but still not 479.

Alternatively, maybe grouping 5 + 6 * 7:

1 + 2 * 3 + 4 * (5 + 6 * 7) + 8 * 9 = 1 + 6 + 4 * (5 + 42) + 72 = 1 + 6 + 4 * 47 + 72 = 1 + 6 + 188 + 72 = 267.

Same as before.

Maybe I need to group larger expressions.

What if I group 1 + 2 * 3 + 4 * 5 + 6 * 7:

(1 + 2 * 3 + 4 * 5 + 6 * 7) + 8 * 9 = (1 + 6 + 20 + 42) + 72 = 69 + 72 = 141.

Too low.

Alternatively, maybe grouping 1 + 2 * 3 + 4 * 5 + 6:

(1 + 2 * 3 + 4 * 5 + 6) * 7 + 8 * 9.

First, inside the parentheses: 1 + 6 + 20 + 6 = 33.

Then, 33 * 7 = 231.

Then, 231 + 72 = 303.

Still low.

Wait, maybe I need to group differently.

What if I group 1 + 2 * 3 + 4 * 5 + 6 * 7 + 8:

(1 + 2 * 3 + 4 * 5 + 6 * 7 + 8) * 9.

First, inside the parentheses: 1 + 6 + 20 + 42 + 8 = 77.

Then, 77 * 9 = 693.

Way above 479.

Maybe I need to group a subtraction or division, but the problem only allows addition and multiplication.

Wait, perhaps I need to group multiplications in a different way.

Let me try: 1 + (2 * 3 + 4) * 5 + 6 * 7 + 8 * 9.

First, 2 * 3 + 4 = 6 + 4 = 10.

Then, 10 * 5 = 50.

So, 1 + 50 + 42 + 72 = 165.

Still low.

Alternatively, maybe grouping 2 * (3 + 4) + 5:

1 + 2 * (3 + 4) + 5 * 6 + 7 * 8 + 9 = 1 + 2 * 7 + 30 + 56 + 9 = 1 + 14 + 30 + 56 + 9 = 109.

Not helpful.

Wait, maybe I need to group additions and multiplications in a more creative way.

What if I group 1 + 2 * 3 + 4 * 5 + 6:

(1 + 2 * 3 + 4 * 5 + 6) * 7 + 8 * 9 = (1 + 6 + 20 + 6) * 7 + 72 = 33 * 7 + 72 = 231 + 72 = 303.

from transformers import AutoModelForCausalLM, AutoTokenizer
import torch

# モデル名を指定
model_name = "Qwen/QwQ-32B-Preview"

# モデルとトークナイザーのロード
model = AutoModelForCausalLM.from_pretrained(
    model_name,
    torch_dtype=torch.float16,  # float16を使用してメモリ使用量を削減
    device_map="auto"          # 自動的にGPUを活用
)
tokenizer = AutoTokenizer.from_pretrained(model_name)

# プロンプトの設定
prompt = "How many r in strawberry."
messages = [
    {"role": "system", "content": "You are a helpful and harmless assistant. You are Qwen developed by Alibaba. You should think step-by-step."},
    {"role": "user", "content": prompt}
]

# トークナイザーを使って入力を整形
text = tokenizer.apply_chat_template(
    messages,
    tokenize=False,
    add_generation_prompt=True
)

# モデルへの入力データを作成
model_inputs = tokenizer([text], return_tensors="pt").to(model.device)

# モデルの推論を実行
generated_ids = model.generate(
    **model_inputs,
    max_new_tokens=512
)

# 応答を抽出
generated_ids = [
    output_ids[len(input_ids):] for input_ids, output_ids in zip(model_inputs.input_ids, generated_ids)
]
response = tokenizer.batch_decode(generated_ids, skip_special_tokens=True)[0]

print(response)

Let's see. The word is "strawberry." I need to find out how many times the letter "r" appears in it. Okay, so I'll look at each letter in the word one by one.

First letter: s - that's not r.

Second letter: t - no, not r.

Third letter: r - yes, that's an r.

Fourth letter: a - not r.

Fifth letter: w - not r.

Sixth letter: b - no.

Seventh letter: e - not r.

Eighth letter: r - another r.

Ninth letter: y - not r.

So, I found two r's in "strawberry." Let me just double-check to make sure I didn't miss any.

Starting over: s, t, r, a, w, b, e, r, y. Yep, there's an r in the third position and another one in the eighth position. No other r's in the word.

So, the answer is two. There are two r's in strawberry.

**Final Answer**

\[ \boxed{2} \]

from transformers import AutoModelForCausalLM, AutoTokenizer
import torch

# モデル名を指定
model_name = "Qwen/QwQ-32B-Preview"

# モデルとトークナイザーのロード
model = AutoModelForCausalLM.from_pretrained(
    model_name,
    torch_dtype=torch.float16,  # float16を使用してメモリ使用量を削減
    device_map="auto"          # 自動的にGPUを活用
)
tokenizer = AutoTokenizer.from_pretrained(model_name)

# プロンプトの設定
prompt = "円周率が3.05より大きいことを証明せよ。"
messages = [
    {"role": "system", "content": "You are a helpful and harmless assistant. You are Qwen developed by Alibaba. You should think step-by-step."},
    {"role": "user", "content": prompt}
]

# トークナイザーを使って入力を整形
text = tokenizer.apply_chat_template(
    messages,
    tokenize=False,
    add_generation_prompt=True
)

# モデルへの入力データを作成
model_inputs = tokenizer([text], return_tensors="pt").to(model.device)

# モデルの推論を実行
generated_ids = model.generate(
    **model_inputs,
    max_new_tokens=512
)

# 応答を抽出
generated_ids = [
    output_ids[len(input_ids):] for input_ids, output_ids in zip(model_inputs.input_ids, generated_ids)
]
response = tokenizer.batch_decode(generated_ids, skip_special_tokens=True)[0]

print(response)

円周率が3.05より大きいことを証明せよ、と言われました。まず、円周率とは円の周の長さと直径の長さの比のことですよね。一般的にはπと表記され、約3.14159という値です。なので、3.05より大きいというのは当然のように思えますが、どうやって証明すればいいのか考えなければいけません。

まず、基本的な Approach としては、円の周長と直径の比を計算して、それが3.05より大きいことを示せばいいと思います。しかし、πは無理数なので、正確な値を計算するのは不可能です。だから、近似値を使って証明する必要があるでしょう。

 perhaps, I can consider a circle with a known diameter and calculate its circumference, then divide the circumference by the diameter to get an approximation of π and show that it's greater than 3.05.

Alternatively, maybe there are some geometric properties or mathematical series that can help me establish a lower bound for π that is greater than 3.05.

Let me think about some historical methods for approximating π. One of the earliest methods is by Archimedes, who used polygons inscribed in and circumscribed around a circle to approximate π. Maybe I can use a similar approach.

So, if I inscribe a regular polygon inside a circle, the perimeter of the polygon will be less than the circumference of the circle. Similarly, if I circumscribe a polygon around the circle, its perimeter will be greater than the circumference. By calculating the perimeters of these polygons and relating them to the diameter, I can get bounds for π.

Let me try this with a hexagon, for simplicity. In a circle with radius r, a regular hexagon inscribed in it has sides of length r, since each side corresponds to one radius.

So, the perimeter of the inscribed hexagon is 6r.

The circumference of the circle is 2πr.

Therefore, 6r < 2πr ⇒ 6 < 2π ⇒ π > 3.

Hmm, that only shows π > 3, but I need to show π > 3.05.

Maybe I need to use a polygon with more sides to get a better approximation.

Let's try with a regular octagon.

For a regular octagon inscribed in a circle of radius r, the side length s can be calculated using

from transformers import AutoModelForCausalLM, AutoTokenizer
import torch

# モデル名を指定
model_name = "Qwen/QwQ-32B-Preview"

# モデルとトークナイザーのロード
model = AutoModelForCausalLM.from_pretrained(
    model_name,
    torch_dtype=torch.float16,  # float16を使用してメモリ使用量を削減
    device_map="auto"          # 自動的にGPUを活用
)
tokenizer = AutoTokenizer.from_pretrained(model_name)

# プロンプトの設定
prompt = """2人の囚人AとBが逮捕され、別々の部屋で取り調べを受けています。彼らはお互いに連絡を取ることができません。検察官はそれぞれの囚人に以下の選択肢を提示しました。

黙秘: 相手を裏切らず、罪を認めない。
裏切り: 相手を告発し、自分の刑を軽くする。

条件
両者が黙秘した場合、それぞれ軽い罪で 懲役1年。
一方が裏切り、もう一方が 黙秘 した場合、裏切った側は 釈放、黙秘した側は 懲役10年。
両者が裏切りを選んだ場合、それぞれ 懲役5年。"""
messages = [
    {"role": "system", "content": "You are a helpful and harmless assistant. You are Qwen developed by Alibaba. You should think step-by-step."},
    {"role": "user", "content": prompt}
]

# トークナイザーを使って入力を整形
text = tokenizer.apply_chat_template(
    messages,
    tokenize=False,
    add_generation_prompt=True
)

# モデルへの入力データを作成
model_inputs = tokenizer([text], return_tensors="pt").to(model.device)

# モデルの推論を実行
generated_ids = model.generate(
    **model_inputs,
    max_new_tokens=2512
)

# 応答を抽出
generated_ids = [
    output_ids[len(input_ids):] for input_ids, output_ids in zip(model_inputs.input_ids, generated_ids)
]
response = tokenizer.batch_decode(generated_ids, skip_special_tokens=True)[0]

print(response)

これは有名な囚人 dilemma ですね。2人の囚人AとBが、それぞれ別々の部屋で取り調べを受けている状況で、お互いの行動が不透明なまま、自分の選択を迫られます。選択肢は「黙秘」か「裏切り」のどちらかで、それぞれの組み合わせによって罰則が異なります。

まず、状況を整理しておきましょう。

- 両方が黙秘を選ぶ場合：それぞれ懲役1年。

- 一方が裏切り、もう一方が黙秘の場合：裏切った側は釈放、黙秘した側は懲役10年。

- 両方が裏切りを選ぶ場合：それぞれ懲役5年。

このゲーム理論の問題では、お互いの利害が対立していて、協力（黙秘）と裏切りの選択が複雑になります。特に、お互いにコミュニケーション不能な状況では、信頼が試されます。

まず、個人として最善の選択は何かを考えましょう。

例えば、囚人Aの立場で考えます。AはBの行動がわからないので、2つのケースを考える必要があります：Bが黙秘か、Bが裏切るか。

1. Bが黙秘を選ぶ場合：

- Aも黙秘：1年刑務所。

- Aが裏切る：釈放。

2. Bが裏切る場合：

- Aが黙秘：10年刑務所。

- Aも裏切る：5年刑務所。

ここで、AはBの行動を予測不能なので、自身の選択が最も不利にならないように行動する必要があります。

まず、Bが黙秘を選ぶと仮定した場合、Aが裏切った方が有利です（釈放 vs. 1年）。

同じように、Bが裏切るだろうと仮定した場合、Aが黙秘选择すると10年 Becomes、裏切った方が5年 Becomes、なので裏切った方が有利です。

つまり、どちらのケースでも、Aが裏切る方がより良い結果 becomes。これはBも同じ立場にありますので、同じように考え、裏切る方向に傾くでしょう。

しかし、両方が裏切る选择すると、それぞれ5年 Bec