【QwQ-32B-preview】Qwenベースの推論モデル登場！数学タスクでo1と徹底比較

Okay, so I need to figure out where to add parentheses to this equation to make it equal 479. The original equation is 1 + 2 * 3 + 4 * 5 + 6 * 7 + 8 * 9 = 479. Hmm, right now, without any parentheses, the operations would follow the standard order of operations, which is multiplication before addition. Let me first calculate what the original equation equals so I can see how far off it is from 479.

Starting with the original expression:

1 + 2 * 3 + 4 * 5 + 6 * 7 + 8 * 9

First, do all the multiplications:

23 = 6, 45=20, 67=42, 89=72.

So substituting those in:

1 + 6 + 20 + 42 + 72.

Now add them all up step by step:

1+6 =7,

7+20=27,

27+42=69,

69+72=141.

So the original equation equals 141, but we need it to be 479. That's a big difference. So adding parentheses must change the order of operations to make the result much larger. Parentheses can group parts of the equation to prioritize their evaluation, so maybe grouping some additions before multiplications?

The target is 479, which is way bigger. Let me think about how parentheses can increase the value. Since multiplication is involved, perhaps grouping a larger number in a way that when multiplied, it gives a bigger result. Let's see.

Let me consider different possibilities. Maybe grouping the first few terms?

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

Calculating this:

(3)*3 + 20 + 42 +72 = 9 + 20 + 42 +72 = 143. Still too low.

Option 2: 1 + (2 * 3 + 4) 5 +67+8*9. Hmm, maybe?

Wait, let me try that step by step. Let's see:

First, (2*3 +4) would be 6+4=10, then multiplied by 5 gives 50.

So the equation becomes 1 + 50 + 42 +72 = 1+50=51, +42=93, +72=165. Still way too low.

Hmm, maybe grouping more terms. Let's see:

What if I group a larger section. Let me think of grouping the first part as (1+2)3+45+67+89. Wait, that was option 1 which gives 143.

Alternatively, perhaps grouping two multiplications together? Like 1 + 2*(3 +4)*5 + ... but that might be overcomplicating.

Alternatively, maybe (1 + 2 * 3 + 4 * 5 + 6 *7 +8)*9? Wait, but that would make the entire left side multiplied by 9, but the equation would then be (1+6+20+42+8)9. Wait, 1+6=7, +20=27, +42=69, +8=77. 779=693, which is way over.

Hmm, maybe not that. Let me think again.

Alternatively, perhaps grouping a part where an addition is done first before a multiplication, leading to a much larger number. For instance, if I can make a term like (something) * (something else) where both are large numbers. Let's see.

Looking at the original expression, maybe grouping the terms after the first few multiplications?

Wait, perhaps if I group 1 + 2*(3 +45 +67 +8*9). Let's see:

First, inside the parentheses: 3 + (45) + (67) + (89). Wait, but without parentheses, that would still do the multiplications first. So inside the big parentheses, it would be 3 +20 +42 +72 = 137. Then multiply by 2: 2137=274. Then add 1 gives 275. Still not enough.

Alternatively, maybe group something like 1 + (2*3 +4)*5 + ... Let's see:

(23 +4) =6+4=10, 5=50. Then the rest: 1 +50 + (67 +89). Wait, but if I group only part of it:

Wait, let me try:

1 + (23 +45 +6*7 +8)*9. Wait, similar to before but different grouping. Let's calculate:

Inside the parentheses: 23=6, 45=20, 67=42, so adding those: 6+20+42+8 = 76. Then times 9 is 684, plus 1? Wait, no, the parentheses would be (23 +45 +67 +8) and then multiplied by 9? Wait, original equation is 1 + [rest]. Wait, maybe:

If the equation is 1 + (23 +45 +6*7 +8)*9, then that would be 1 + (6+20+42+8)*9 =1 + (76)*9= 1 +684=685. Still too big.

Hmm, maybe a different approach. Let me think of the desired result is 479, which is between 400 and 500. Let me see if I can break down 479 into possible multiplicative components.

Alternatively, maybe grouping a longer addition before a multiplication. For example, if I can have a large number added then multiplied.

Suppose I group (1+2)(3+4)(5+6)(7+8)9? Wait, but that might not be allowed because the equation is linear. The original equation is 1 +23+45+67+89. So I can only add parentheses to the existing structure, not rearrange terms. So parentheses can only group existing terms in the sequence.

Wait, the problem says "add a pair of parentheses", so only one pair? Or a pair meaning two parentheses? Maybe the user means adding parentheses in one place, but perhaps more than one pair? Wait, the wording says "a pair of parentheses", which usually means one opening and one closing, so a single pair. Hmm, but maybe two pairs? Let me check the original problem again.

The user says: "Please add a pair of parentheses to the incorrect equation...". So a pair is one pair, so inserting one opening and one closing parenthesis somewhere. So the equation has to have exactly two parentheses added (one pair) to make it correct.

Hmm, so only one pair. So inserting ( and ) somewhere in the equation. Let me think again.

Let me try different placements:

Option: 1 + 2*(3 +4)5 +67+8*9. Let's compute:

First, 3+4=7, multiplied by 2 and 5: 275. Wait, but order matters. It would be 2*(7)*5=70. Then the total would be 1 +70 +42 +72= 1+70=71+42=113+72=185. Still low.

Alternatively, (1 +2)3 +45 +67+89=3*3 +20+42+72=9+20+42+72=143. No.

What if I group the last terms? Like 1 +23 +45 +6*(7+8)*9. Let's see:

Compute 7+8=15, then 615=90, then 9? Wait, no, the grouping would be 6(7+8) which is 615=90, then the rest: 1 +6 +20 +90 +72? Wait, no:

Wait the equation would be 1 +23 +45 +6*(7+8)9. Wait, the original equation is up to +89. Wait, inserting parentheses as 6*(7+8)*9. Let me parse it step by step:

Original expression with parentheses: 1 + 23 +45 +6*(7+8)*9.

Calculating:

First, compute inside the parentheses (7+8)=15. Then 615=90, then multiplied by 9 gives 810? Wait, no. Wait, the expression is 6(7+8)9. Since multiplication is left-associative, so 6(15)9 = 615=90, then 90*9=810. So adding that term:

1 + (23) + (45) + (6*(7+8)*9). So:

1 +6 +20 +810 = 1+6=7+20=27+810=837. That's way over 479.

Hmm, too much. Maybe grouping a smaller part.

What if I group 1 + (23 +4)5 +6789? Wait, but that would require multiple parentheses. Also, the original equation is up to +89, so maybe:

Wait, perhaps grouping (6*7+8)*9? Let's see:

Original equation with parentheses: 1 +23 +45 +67 +89 becomes 1 +23 +45 + (6*7+8)*9.

Compute inside the parentheses: 67=42 +8=50. Then 509=450.

So total equation becomes 1 +6 +20 +450 = 1+6=7+20=27+450=477. Close to 479, but off by 2. Hmm, almost there.

Wait, so 477 vs 479. Maybe close but not quite. Let me check again:

Wait, (6*7 +8)*9:

6*7 is 42, plus 8 is 50, times 9 is 450. Then adding the previous terms:

1 +23=6, so 1+6=7; 45=20, so total so far 7+20=27; then adding the 450 gives 477, plus the remaining terms? Wait no, the original equation after adding parentheses would be:

1 +23 +45 + (67 +8)9. Wait, but in the original, after 67, there's +89. So inserting parentheses around 67 +8 would replace the +89 with (6*7 +8)*9. Wait, perhaps the user wants to keep the structure but insert parentheses. Let me parse it correctly.

Wait, the original equation is:

1 + 2 * 3 + 4 * 5 + 6 * 7 + 8 * 9

If I group the last three terms as (6 * 7 + 8) * 9, then the equation becomes:

1 + 23 +45 + (6*7 +8)*9

Which computes as:

1 +6 +20 + (42+8)9 → 27 + (509) → 27 + 450 = 477. Still 2 short.

Hmm, so maybe that's close but not quite. Let's see where the 2 difference comes from. Maybe grouping a different part.

Alternatively, what if I group (67+89)? Then it would be 42 +72=114. But that's just addition, so multiplying by something? Not sure.

Alternatively, perhaps grouping (1+2)3 +45 +67+89. Wait, we tried that earlier gives 143.

Hmm. Maybe grouping something earlier. Let's see:

If I group (1

!pip install transformers accelerate

!pip install torch torchvision torchaudio --index-url https://download.pytorch.org/whl/cu118

from transformers import AutoModelForCausalLM, AutoTokenizer

import torch

# モデル名を指定

model_name = "Qwen/QwQ-32B"

# 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

問1.2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。

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