ローカル環境に導入したdeepseek-r1をCLIで動かしてみる

最近、[DeepSeek](https://www.deepseek.com/)をローカルで動かすのが世間で流行っているので試してみました。ここではLinux環境にOllama（ローカルLLM実行環境）を導入してCLIで動かします。

https://ollama.com/

> Ollama は ローカル環境で LLM を動作させるオープンソースのツール です。
> - ローカル PC で LLM を実行できる（クラウド不要）
> - 事前にモデルをダウンロードすれば、オフラインでも動作可能
> - 軽量で動作し、シンプルなインターフェースを提供
> - オープンソースで、カスタマイズも可能
> 
> 公式サイト: https://ollama.com
> GitHub レポジトリ: https://github.com/jmorganca/ollama

ChatGPTや本家DeepSeekのウェブサイトのようなGUIで実行できるようにするには、WebUIなどのセットアップ作業が別途必要です。興味のある方は下記Webサイトなどを参考にしてみてください。

https://itsfoss.com/install-deepseek-r1-locally-linux/

# 実行環境
> ここではLinuxユーザーにsudo権限が付与されていることを前提としています。

|||
|---|---|
|OS  |Ubuntu 24.04.1 LTS  |
|GPU  |NVIDIA GeForce RTX 3070 Ti（VRAM 8 GB）  |
|CUDA バージョン  |12.6  |
|NVIDIA ドライバ  |560.94  |
|RAM  |32 GB  |
|CPU  |AMD Ryzen 7 5800X3D 8-Core Processor  |

LLMをGPUで動かす場合、要求されるVRAMはモデルのパラメータ数にほぼ比例します。ここでは比較的軽量な[8Bパラメータのモデル](https://ollama.com/library/deepseek-r1:8b)を選択します（B = Billion = 10 億）。

> 「deepseek-r1:8b」などの「8b」はLLM の「学習済みの重み（weights）」の数に対応しており、パラメータ数が多いほどモデルの表現力が高くなるとされています。この 8 B は「80億個のニューラルネットワークの重みがある」ことを意味しています。ということで、deepseek-r1:8b は「軽量でありながら、そこそこの推論能力を持つ」中規模LLMという位置付けです。例えば、GPT-3は約 175 B（1750億）、GPT-4は 1 T（1兆, 1000B）以上のパラメータ数のモデルと言われています。

# Ollama + DeepSeek のセットアップ手順

## Ollamaのインストール
Ollamaが未導入の場合、以下でインストールしてください。 `/usr/share/ollama/` が生成します。

```bash:Ollamaのインストール（Ollama未導入なら）
curl -fsSL https://ollama.com/install.sh | sh
```
```:出力
>>> Installing ollama to /usr/local
>>> Downloading Linux amd64 bundle
######################################################################## 100.0%
>>> Creating ollama user...
>>> Adding ollama user to render group...
>>> Adding ollama user to video group...
>>> Adding current user to ollama group...
>>> Creating ollama systemd service...
>>> Enabling and starting ollama service...
Created symlink /etc/systemd/system/default.target.wants/ollama.service → /etc/systemd/system/ollama.service.
>>> Nvidia GPU detected.
>>> The Ollama API is now available at 127.0.0.1:11434.
>>> Install complete. Run "ollama" from the command line.
```

> ちなみに、OllamaはWindowsでも動かせます。
> https://qiita.com/kit/items/cd3a7f8116982d61c06d

## DeepSeekのダウンロード

次にDeepSeekの適当なモデルをダウンロードします。ここではdeepseek-r1:8bを選びます。
> GPUのスペックが高ければ、より上位のモデルを動かしても良いかもしれません。

https://ollama.com/library/deepseek-r1:8b


```bash:DeepSeekモデルのダウンロード
ollama pull deepseek-r1:8b
```

```sh:出力
pulling manifest
pulling 6340dc3229b0... 100% ▕███████████████████████████████████████████████▏ 4.9 GB
pulling 369ca498f347... 100% ▕███████████████████████████████████████████████▏  387 B
pulling 6e4c38e1172f... 100% ▕███████████████████████████████████████████████▏ 1.1 KB
pulling f4d24e9138dd... 100% ▕███████████████████████████████████████████████▏  148 B
pulling 0cb05c6e4e02... 100% ▕███████████████████████████████████████████████▏  487 B
verifying sha256 digest
writing manifest
success
```

8Bの軽量モデルとはいっても4.9 GBほどあるので、ダウンロードにはそれなりに時間がかかります。念のため、インストールできているか確認しておきます。

```bash:設定の確認
ollama show deepseek-r1:8b
```
```:出力
  Model
    architecture        llama
    parameters          8.0B
    context length      131072
    embedding length    4096
    quantization        Q4_K_M

  Parameters
    stop    "<｜begin▁of▁sentence｜>"
    stop    "<｜end▁of▁sentence｜>"
    stop    "<｜User｜>"
    stop    "<｜Assistant｜>"

  License
    MIT License
    Copyright (c) 2023 DeepSeek
```
特に問題無さそうです。Ollamaにダウンロードしたモデルの情報は
`/usr/share/ollama/.ollama/models/manifests/registry.ollama.ai/library/deepseek-r1`
などに格納されます。

# 実行する
```sh:deepseekを立ち上げる
ollama run deepseek-r1:8b
```
> Ollamaでモデルを起動するとREPLプロンプト `>>>` が表示されます。
```:Ollamaのヘルプ
>>> /?
Available Commands:
  /set            Set session variables
  /show           Show model information
  /load <model>   Load a session or model
  /save <model>   Save your current session
  /clear          Clear session context
  /bye            Exit
  /?, /help       Help for a command
  /? shortcuts    Help for keyboard shortcuts

Use """ to begin a multi-line message.
```
対話内容は `/save test001` などと打ち込めばOllama内にセッションが保存されます。
複数行にまたがって入力する場合は1行目と最終行に `"""` を打ち込んで囲めばOKです。

後はご自由に楽しんでください。
以下に、8Bモデルの推論能力についてテストした結果を示します。

## 2元1次連立方程式
```:問題
"""
Solve the following system of equations:
3x+4y=12
2x+5y=7
"""
```

<details><summary>think</summary>

```
Alright, so I've got this system of equations to solve:

1. \(3x + 4y = 12\)
2. \(2x + 5y = 7\)

Okay, first things first, I need to figure out how to solve for x and y here. I remember from class that there are a
couple of methods for solving systems of equations: substitution and elimination. I think elimination might be easier
here because the coefficients aren't too bad.

So, if I want to use the elimination method, I probably need to align the coefficients so that when I add or subtract the
equations, one variable cancels out. Let's see... Maybe I can eliminate x first? To do that, I should make sure the
coefficients of x in both equations are the same (or negatives).

Looking at equation 1, the coefficient of x is 3, and in equation 2, it's 2. The least common multiple of 3 and 2 is 6,
so if I multiply each equation by a number that gets us to 6 for the coefficients of x.

For equation 1: Multiply both sides by 2 → \(6x + 8y = 24\)

For equation 2: Multiply both sides by 3 → \(6x + 15y = 21\)

Okay, now both equations have a coefficient of x as 6. Now, if I subtract the first new equation from the second, the x
terms should cancel out.

So:

\(6x + 15y = 21\) (equation 2 multiplied by 3)
- \(6x + 8y = 24\) (equation 1 multiplied by 2)
---------------------------
Subtracting these:

6x - 6x + 15y - 8y = 21 - 24

Simplify:

0x + 7y = -3 → 7y = -3

So, y = -3/7

Now that I have y, I can plug it back into one of the original equations to solve for x. Let's pick equation 1.

From equation 1: \(3x + 4y = 12\)

Plugging in y = -3/7:

\(3x + 4(-3/7) = 12\)

Calculate 4 times -3/7: that's -12/7

So:

\(3x - 12/7 = 12\)

Now, let me solve for x. First, I'll add 12/7 to both sides.

\(3x = 12 + 12/7\)

Convert 12 to sevenths: 12 is 84/7

So:

\(3x = 84/7 + 12/7 = 96/7\)

Now, divide both sides by 3 to solve for x:

\(x = (96/7) ÷ 3\)

Which is the same as \(x = 96/(7*3)\) → \(x = 32/7\)

So, my solutions are:

x = 32/7 and y = -3/7

Let me double-check these values in both original equations to make sure I didn't make any mistakes.

First equation: \(3x + 4y = 12\)

Plug in x=32/7 and y=-3/7:

3*(32/7) + 4*(-3/7) = (96/7) - (12/7) = (84/7) = 12

That checks out.

Second equation: \(2x + 5y = 7\)

Plug in x=32/7 and y=-3/7:

2*(32/7) + 5*(-3/7) = (64/7) - (15/7) = (49/7) = 7

That also checks out. So, my solutions are correct.

I think I've got it!
```

</details>

<details><summary>answer</summary>

```
The solution to the system of equations is:

\[
x = \boxed{\frac{32}{7}}, \quad y = \boxed{-\frac{3}{7}}
\]
```

</details>

$$x = \boxed{\frac{32}{7}}, \quad y = \boxed{-\frac{3}{7}}$$

消去法によって正しく解を導けています。

## 平方数の和の公式の導出

```:問題
Find a closed formula for the sum of the first n squares: 1^2 + 2^2 + ... + n^2.
```

<details><summary>think</summary>

```
Alright, so I need to figure out a closed formula for the sum of the first n squares. Hmm, okay. Let me
start by writing down what exactly is being asked here. The problem is to find a closed-form expression
for the sum S = 1² + 2² + 3² + ... + n².

First, I remember that there are formulas for sums of sequences like arithmetic series or geometric
series, but this one is about squares. Is it an arithmetic series? Well, no, because in an arithmetic
series each term increases by a constant difference, which here would be 2 between consecutive terms, but
the squares are increasing much more rapidly.

Wait, maybe there's another formula for the sum of squares. Let me think... I believe Gauss is known for
having derived such formulas as a child. Maybe if I try to derive it myself?

Okay, so let me denote S = 1² + 2² + 3² + ... + n².

I recall that sometimes pairing terms can help in finding such sums. So maybe if I pair the first and
last terms: 1² + n², then the second and second-last terms: 2² + (n-1)², and so on. Let me try writing
this out:

S = (1² + n²) + (2² + (n-1)²) + ... + ((n/2)² if n is even).

Wait, but depending on whether n is even or odd, the pairing might be slightly different. Maybe I need a
different approach.

Alternatively, I remember something about using mathematical induction for proving formulas like this,
and sometimes that can lead to discovering the formula itself. Let me try that.

So, let's suppose we're trying to use induction. First, we need a base case. Let's take n = 1. Then S =
1² = 1. Plugging into any potential formula, it should hold true.

But since I don't know the formula yet, maybe this approach is not straightforward. Alternatively,
perhaps I can look for a pattern by computing the sum for small values of n and see if I can notice
something.

Let's compute S for n = 1: S = 1
n = 2: S = 1 + 4 = 5
n = 3: S = 1 + 4 + 9 = 14
n = 4: S = 1 + 4 + 9 + 16 = 30
n = 5: S = 1 + 4 + 9 + 16 + 25 = 55

Hmm, so we have these sums as 1, 5, 14, 30, 55... Does this sequence look familiar? Let me think. Wait, I
remember that the sum of squares formula is n(n+1)(2n+1)/6. Maybe I can verify if that works.

Let me try plugging in n=1: 1(1+1)(2*1 +1)/6 = 1*2*3/6 = 6/6=1, which matches.
n=2: 2*3*5 /6=30/6=5, correct.
n=3: 3*4*7 /6=84/6=14, correct.
n=4: 4*5*9 /6=180/6=30, correct.
n=5:5*6*11 /6=330/6=55, also correct.

Okay, so that formula seems to work. But how do I derive it?

Maybe another approach is to use the formula for the sum of an arithmetic series and then relate it to
the sum of squares. Wait, but in this case, the terms themselves are quadratic, not linear.

Alternatively, perhaps we can express S in terms of known sums. Let me write down what I know:

We have the formula for the sum of the first n natural numbers: S1 = n(n+1)/2
And we also have another formula related to the squares which is similar but more complex.

Wait, if I recall correctly, there's a relationship between these two sums. Let me think about how they
might be connected.

Suppose I consider (n)(n + 1)/2 squared: that would be [n(n+1)/2]^2. Expanding this gives n²(n + 1)² /4 =
(n^4 + 2n³ + n²)/4.

Alternatively, maybe if I subtract something from this expression to get the sum of squares.

Wait, let me try:

I know that [1² + 2² + ... + n²] can be related to [1 + 2 + ... + n]^2. Because when you square a sum,
it's equal to the sum of squares plus twice the sum of all the cross terms.

So, mathematically,

(1 + 2 + ... + n)² = 1² + 2² + ... + n² + 2*(1*2 + 1*3 + ... + (n-1)*n)

Thus, we can write:

S = [n(n+1)/2]^2 - 2*(sum of i*j from i=1 to n-1 and j=i+1 to n)

Wait, that seems a bit complicated. Let me see if I can express the cross terms in another way.

Alternatively, perhaps consider expanding (1 + 2 + ... + n)^2:

(1 + 2 + 3 + ... + n)^2 = (1² + 2² + 3² + ... +n²) + 2*(1*2 + 1*3 + ... +1*n + 2*3 + ... +2*n + ... +
(n-1)*n)

So, indeed, S is equal to [n(n+1)/2]^2 minus twice the sum of all products i*j where i < j.

Hmm, so if I can find a way to express the cross terms, then maybe subtract them from the square of the
arithmetic series.

But this seems a bit involved. Maybe another approach.

I remember that in calculus, we can use integration or differentiation to find such sums, but since we
are dealing with discrete mathematics here, perhaps induction is still a better approach.

Alternatively, let's think about the telescoping method or using recurrence relations.

Wait, maybe I can consider the differences between consecutive triangular numbers and relate them to
something. Hmm, not sure.

Another thought: Let me use generating functions for sequences.

The generating function for squares is (x²)/(1 - x)^3, but that might not directly help here.

Alternatively, perhaps using recursive formulas or considering the sum S(n) = S(n-1) + n², and then
trying to solve that recurrence.

But solving a linear recurrence should be straightforward. Let me attempt that.

Given S(n) = S(n-1) + n²

With S(1) = 1.

This is a nonhomogeneous linear recurrence relation. The general solution can be found by finding the
homogeneous solution and a particular solution.

The homogeneous equation is S(n) = S(n-1), which has the solution S_h(n) = C, where C is a constant.

Now, for the particular solution, since the nonhomogeneous term is quadratic (n²), we can try a
particular solution of the form an² + bn + c.

Substituting into the recurrence:

an² + bn + c = a(n - 1)^2 + b(n - 1) + c + n²

Expanding both sides:

Left: an² + bn + c
Right: a(n² - 2n + 1) + b(n - 1) + c + n²
= an² - 2an + a + bn - b + c + n²

So, equate coefficients for each power of n:

For n²: a = a + 1 => Subtracting a from both sides gives 0 = 1, which is a problem. Hmm.

Wait, that suggests our assumption for the particular solution might be insufficient. Maybe we need to
include higher degree terms? Wait, but the nonhomogeneous term is quadratic, so perhaps degree 2 is
sufficient. But let's see.

Alternatively, maybe set P(n) = an² + bn + c and substitute:

P(n) - P(n-1) = n²

So,

an² + bn + c - [a(n-1)^2 + b(n -1) + c] = n²
= an² + bn + c - a(n² - 2n + 1) - b(n -1) - c
= an² + bn + c - an² + 2an - a - bn + b - c
Simplify:

(an² - an²) + (bn - bn) + (c - c) + 2an -a + b

So, this simplifies to:

0 + 0 + 0 + 2an -a + b = n²

Thus, equating coefficients:

2an - a + b = n²

But wait, on the left-hand side, we have terms linear in n and constants. On the right, it's quadratic.
Therefore, this suggests that our assumption for P(n) is insufficient.

Hmm, perhaps I need to choose a particular solution of higher degree. Let me assume P(n) = an³ + bn² + cn
+ d

Then, compute P(n) - P(n-1):

P(n) - P(n-1) = [a n³ + b n² + c n + d] - [a (n - 1)^3 + b (n -1)^2 + c (n -1) + d]

Compute each term:

First, expand a(n-1)^3:

= a(n³ - 3n² + 3n -1)

Similarly, b(n-1)^2 = b(n² - 2n +1)

c(n-1) = cn - c

So,

P(n) - P(n-1) = a n³ + b n² + c n + d - [a n³ - 3a n² + 3a n -a + b n² - 2b n + b + c n - c + d]

Simplify:

= a n³ + b n² + c n + d - a n³ + 3a n² - 3a n + a - b n² + 2b n - b - c n + c - d

Now, collect like terms:

n³: (a - a) = 0
n²: (b + 3a - b) = 3a
n: (c - 3a + 2b - c) = (-3a + 2b)
constants: (d + a - b + c - d) = (a - b + c)

So, overall:

P(n) - P(n-1) = 3a n² + (-3a + 2b)n + (a - b + c)

We know that this must equal to n², so set coefficients equal:

For n²: 3a = 1 => a = 1/3
For n: -3a + 2b = 0
Since a = 1/3, then -1 + 2b = 0 => 2b = 1 => b = 1/2
Constants: a - b + c = 0

Substituting a and b:

(1/3) - (1/2) + c = 0
Compute (1/3 - 1/2):

Convert to common denominator, which is 6:

= 2/6 - 3/6 = (-1)/6

So, (-1/6) + c = 0 => c = 1/6

Thus, the particular solution P(n) is (1/3)n³ + (1/2)n² + (1/6)n

Therefore, the general solution S(n) = homogeneous + particular:

S(n) = C + (1/3)n³ + (1/2)n² + (1/6)n

Now, apply initial condition. When n=1, S(1)=1.

Compute S(1):

C + (1/3)(1)^3 + (1/2)(1)^2 + (1/6)(1) = C + 1/3 + 1/2 + 1/6

Convert to sixths:

= C + 2/6 + 3/6 + 1/6 = C + 6/6 = C + 1

So, S(1) = C + 1 = 1 => C = 0.

Thus, the solution is S(n) = (1/3)n³ + (1/2)n² + (1/6)n

Simplify this expression:

Factor out n:

= n*(1/3 n² + 1/2 n + 1/6)

Alternatively, let me combine the terms:

Express all in sixths:

1/3 = 2/6, 1/2 = 3/6, 1/6 remains.

So,

S(n) = (2/6)n³ + (3/6)n² + (1/6)n
= (2n³ + 3n² + n)/6

Factor numerator:

Looking at numerator: 2n³ + 3n² + n

Factor out n: n(2n² + 3n + 1) = n(2n + 1)(n + 1)

Wait, is that correct?

Wait, 2n² + 3n + 1 factors into (2n + 1)(n + 1). Let's check:

(2n + 1)(n + 1) = 2n(n) + 2n(1) +1(n) +1(1)= 2n² + 2n + n +1=2n²+3n+1. Yes, correct.

Thus,

S(n) = [n(2n +1)(n +1)] /6

Which can be written as:

S(n) = (n(n +1)(2n +1))/6

Hmm, that's the formula for the sum of squares: 1² +2²+...+n²= n(n +1)(2n +1)/6. Yes, correct.

Thus, the closed-form solution is S(n) = n(n +1)(2n +1)/6.

So, after going through all that, I conclude that the sum of squares from 1 to n is equal to
n(n+1)(2n+1)/6.

**Final Answer**
The sum of squares from \(1\) to \(n\) is \boxed{\dfrac{n(n + 1)(2n + 1)}{6}}.
```

</details>

<details><summary>answer</summary>

```
To find a closed-form solution for the sum of squares from \(1\) to \(n\), we start by recognizing that
this problem can be approached using methods for solving recurrence relations.

We begin with the recurrence relation:
\[ S(n) = S(n-1) + n^2 \]
with the initial condition \( S(0) = 0 \).

Using the method of solving linear recurrences, we assume a particular solution of the form:
\[ P(n) = an^3 + bn^2 + cn + d \]

Expanding and substituting into the recurrence relation, we equate coefficients to solve for the
constants \(a\), \(b\), \(c\), and \(d\). Through this process, we find:
- For the \(n^2\) term: \(3a = 1\) gives \(a = \frac{1}{3}\)
- For the \(n\) term: \(-3a + 2b = 0\) gives \(b = \frac{1}{2}\)
- For the constant term: \(a - b + c = 0\) gives \(c = \frac{1}{6}\)

Combining these, we have a particular solution:
\[ P(n) = \frac{1}{3}n^3 + \frac{1}{2}n^2 + \frac{1}{6}n \]

The general solution is the sum of the homogeneous and particular solutions:
\[ S(n) = C + \frac{1}{3}n^3 + \frac{1}{2}n^2 + \frac{1}{6}n \]

Using the initial condition \(S(0) = 0\), we find \(C = 0\). Thus, the solution is:
\[ S(n) = \frac{1}{3}n^3 + \frac{1}{2}n^2 + \frac{1}{6}n \]

Expressing this in a simplified form:
\[ S(n) = \frac{n(n + 1)(2n + 1)}{6} \]

Thus, the sum of squares from \(1\) to \(n\) is:
\[ \boxed{\dfrac{n(n + 1)(2n + 1)}{6}} \]
```

</details>

$$
\boxed{\dfrac{n(n + 1)(2n + 1)}{6}}
$$

正解しています。思考プロセスが冗長なのはウェブアプリの本家R1も同じで、DeepSeekの癖とも言える特徴です。


## 3次方程式の解

```:問題
Solve the equation x^3 - 3 x^2 + 4 x - 4 = 0.
```

<details><summary>think</summary>

```
Alright, so I need to solve this cubic equation: \(x^3 - 3x^2 + 4x - 4 = 0\). Hmm, okay. Let me think
about how to approach this. I remember that solving cubic equations can sometimes be tricky, but maybe
there's a way to factor it or use the rational root theorem.

First off, let's see if I can factor this equation by trying possible rational roots. The rational root
theorem says that any possible rational root of the polynomial equation \(a_nx^n + \dots + a_0 = 0\) is a
fraction \(\frac{p}{q}\), where \(p\) is a factor of the constant term and \(q\) is a factor of the
leading coefficient.

In this case, our equation is \(x^3 - 3x^2 + 4x - 4 = 0\). The constant term here is \(-4\), so possible
values for \(p\) could be \(\pm1\), \(\pm2\), \(\pm4\). The leading coefficient is 1, so \(q\) can only
be \(\pm1\). Therefore, the possible rational roots are just \(\pm1\), \(\pm2\), and \(\pm4\).

Let me test these one by one. I'll start with \(x=1\):

Plugging in 1: \(1 - 3 + 4 - 4 = (1 - 3) + (4 - 4) = (-2) + 0 = -2 \neq 0\). So, not a root.

Next, try \(x=2\):

\(8 - 12 + 8 - 4\). Let's compute step by step:

- \(2^3 = 8\)
- \(3(2)^2 = 12\), so subtract that: \(8 - 12 = -4\)
- Add \(4(2) = 8\): \(-4 + 8 = 4\)
- Subtract 4: \(4 - 4 = 0\). Oh! That works. So, \(x=2\) is a root.

Great! So we can factor out \((x - 2)\) from the cubic polynomial. Let me perform polynomial division or
use synthetic division to find the other factors.

I think synthetic division might be faster here. Setting up synthetic division with root at \(x=2\):

The coefficients of the polynomial are 1 (for \(x^3\)), -3 (for \(x^2\)), 4 (for \(x\)), and -4 (constant
term).

So, writing it out:

2 | 1   -3   4   -4

I bring down the first coefficient: 1.

Multiply 1 by 2: 2. Add to next coefficient: -3 + 2 = -1.

Multiply -1 by 2: -2. Add to next coefficient: 4 + (-2) = 2.

Multiply 2 by 2: 4. Add to last term: -4 + 4 = 0. Perfect, that's the remainder we wanted.

So after division, we have \(x^2 - x + 2\) as the quadratic factor. So now, the equation factors as:

\((x - 2)(x^2 - x + 2) = 0\)

Now, I need to solve \(x^2 - x + 2 = 0\). Let me use the quadratic formula here because it might have
complex roots.

The quadratic formula is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where in this case, \(a=1\),
\(b=-1\), and \(c=2\).

Plugging in:

\(x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(2)}}{2(1)} = \frac{1 \pm \sqrt{1 - 8}}{2} = \frac{1 \pm
\sqrt{-7}}{2}\)

Since the discriminant is negative (\(\sqrt{-7}\)), we have complex roots.

Therefore, the solutions are:

\(x = 2\), and \(x = \frac{1 \pm i\sqrt{7}}{2}\).

Wait a minute, let me double-check my work because sometimes I make sign errors. Let's verify if the
quadratic factor is correct after division.

Original polynomial: \(x^3 - 3x^2 + 4x - 4\)

After dividing by (x - 2), we should have:

\(x^3 - 3x^2 + 4x - 4 = (x - 2)(x^2 - x + 2)\)

Let me multiply these out to check.

Multiplying \(x\) with each term of quadratic: \(x^3 - x^2 + 2x\)

Then, multiplying \(-2\) with each term: \(-2x^2 + 2x - 4\)

Now add them together:

\(x^3 - x^2 + 2x - 2x^2 + 2x - 4 = x^3 - 3x^2 + 4x - 4\). Yep, that's correct.

Okay, so the quadratic factor is correctly found. So, moving on.

So, all roots are \(x=2\) and \(x = \frac{1 \pm i\sqrt{7}}{2}\).

Therefore, these are the three solutions to the equation.

Wait, but just for thoroughness, maybe I should check whether there were any calculation errors in
synthetic division or quadratic formula. Let's recompute the discriminant for the quadratic:

Discriminant \(D = b^2 - 4ac = (-1)^2 - 4(1)(2) = 1 - 8 = -7\). Correct.

So, yes, that gives us complex roots as above.

Alternatively, maybe trying another root just to confirm, although since we factored out correctly, it
should be okay. Let me try \(x=4\) in the original equation:

\(64 - 48 + 16 - 4 = (64 - 48) + (16 - 4) = 16 + 12 = 28 \neq 0\). So not a root.

Trying \(x=-1\): \(-1 - 3 - 4 - 4 = (-1 - 3 - 4) - 4 = (-8) - 4 = -12 \neq 0\).

Similarly, \(x= -2\): \(-8 - 12 - 8 - 4 = (-8 - 12) + (-8 - 4) = (-20) + (-12) = -32 \neq 0\). So none of
the other integer candidates work, as expected.

Therefore, the only real root is \(x=2\), and two complex roots from the quadratic factor.

Thus, all solutions are accounted for: one real, two complex conjugates.

So, wrapping up, the equation \(x^3 - 3x^2 + 4x - 4 = 0\) has three roots:

1. Real root: \(x=2\)
2. Complex roots: \(x = \frac{1 \pm i\sqrt{7}}{2}\)

Therefore, that's the complete solution set.

**Final Answer**
The solutions are \(x = 2\) and \(x = \frac{1 \pm i\sqrt{7}}{2}\). Thus, the final answers are
\(\boxed{2}\), \(\boxed{\frac{1 + i\sqrt{7}}{2}}\), and \(\boxed{\frac{1 - i\sqrt{7}}{2}}\).
```

</details>

<details><summary>answer</summary>

```
To solve the cubic equation \(x^3 - 3x^2 + 4x - 4 = 0\), we start by using the rational root theorem to
test possible rational roots. We find that \(x = 2\) is a root.

Next, we perform synthetic division on the polynomial with the root \(x = 2\):

\[
\begin{array}{r|rrrr}
2 & 1 & -3 & 4 & -4 \\
   &   & 2 & -2 & 4 \\
\hline
   & 1 & -1 & 2 & 0 \\
\end{array}
\]

This gives us the quadratic factor \(x^2 - x + 2\). We then solve this quadratic equation using the
quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

where \(a = 1\), \(b = -1\), and \(c = 2\). Plugging in these values, we get:

\[
x = \frac{1 \pm \sqrt{1 - 8}}{2} = \frac{1 \pm \sqrt{-7}}{2} = \frac{1 \pm i\sqrt{7}}{2}
\]

Thus, the solutions to the equation are:

1. The real root: \(x = 2\)
2. The complex roots: \(x = \frac{1 + i\sqrt{7}}{2}\) and \(x = \frac{1 - i\sqrt{7}}{2}\)

Therefore, the final answers are:

\[
\boxed{2}, \quad \boxed{\frac{1 + i\sqrt{7}}{2}}, \quad \text{and} \quad \boxed{\frac{1 - i\sqrt{7}}{2}}
\]
```

</details>

$$\boxed{2}, \quad \boxed{\frac{1 + i\sqrt{7}}{2}}, \quad \text{and} \quad \boxed{\frac{1 - i\sqrt{7}}{2}}$$

思考連鎖の中で計算を過剰に再確認していますが、これも正解しています。


## 3Lと5Lの容器で4Lを測る最小手数

```:問題
There are two containers: one that holds 3 liters and another that holds 5 liters. 
Using these containers, how can you measure exactly 4 liters of water in the
minimum number of steps?
```

<details><summary>think</summary>

```
Okay, so I have this problem where there are two containers: one is 3 liters and the other is 5 liters. I need to figure
out how to measure exactly 4 liters using these containers in the minimum number of steps. Hmm, let me think through this
step by step.

First off, I know that with containers of different sizes, you can fill them up and pour water from one to another until
either the larger container is full or the smaller one is empty. Since my goal is 4 liters, which isn't a direct multiple
or something simple like that, I might need several steps.

Let me list down what I know:

- Container A: 3 liters
- Container B: 5 liters

I can perform operations such as filling a container to its full capacity, emptying it completely, or pouring water from
one container to another until either the source is empty or the destination is full. Each of these actions counts as a
step.

My target is exactly 4 liters. Hmm, so I wonder if I can get 4 liters by some combination of filling and transferring
between the two containers.

Let me try visualizing this. Maybe I'll start by labeling each container with their sizes:

- A: 3L
- B: 5L

I think a good starting point is to fill one of them completely. Let's say I fill Container A (3 liters) first.

Step 1: Fill Container A to its full capacity. Now, A has 3L, B is empty.

But then what? Maybe pour from A into B. Since B can hold up to 5L, and A only has 3L, pouring all of A into B would
result in A being empty and B having 3L.

Step 2: Pour water from A to B until either A is empty or B is full. Since A had 3L, it's emptied, and B now has 3L.

Hmm, not quite 4 liters yet. Maybe I can fill Container A again? Let me see.

Wait, but maybe there's a better way. What if I started by filling Container B first?

Step 1: Fill Container B to its full capacity of 5L. Now, both A and B have their respective capacities.

But with both containers full, how do I get to 4 liters? Maybe pour from B into A until A is full or B is empty. Let's
try that.

Since A can hold up to 3L, if we pour from B (5L) into A, A will be filled to 3L, and then B will have 2L left over.

Step 2: Pour from B to A until A is full. Now, A has 3L, and B has 5 - 3 = 2L remaining.

Okay, so now B has 2 liters in it. Maybe I can pour that back into A? But A is already at 3L, which is its maximum
capacity, so we can't add more to it. So pouring from B to A isn't possible anymore unless we empty A first.

So maybe step 3: Empty Container A. Now, A is empty again, and B has 2L.

Now, what if I pour the remaining 2L from B into A?

Step 4: Pour from B to A until A is full or B is empty. Since A can take up to 3L, but it's currently empty, we can pour
all 2L into A. Now, A has 2L and B is empty.

Hmm, so now I have 2 liters in Container A and nothing in B. Is this helpful? Maybe not directly, because I need 4
liters.

Alternatively, maybe there's another approach. Instead of starting by filling both containers, what if I do something
else?

Let me try starting with Container A as 3L.

Step 1: Fill A to 3L.

Then pour from A into B until B is full or A is empty. Since B can hold up to 5L and currently has 0, it can take all 3L
from A.

So step 2: Pour from A to B. Now, A is empty, B has 3L.

Now, maybe fill Container A again.

Step 3: Fill A to 3L.

Then pour again into B.

But wait, B already has 3L. It can hold up to 5L, so there's room for 2 more liters.

So step 4: Pour from A (3L) into B (which can take 2 more liters). So after pouring, A will have 1L left, and B will be
full at 5L.

Now, I'm left with Container A having 1 liter. Hmm, that's interesting. Now, how do I get to 4 liters?

Maybe empty Container B? But it's already full. Or maybe pour from B back into A?

But A can only hold 3L, and right now A has 1L. So pouring from B (5L) into A would fill A up to 3L.

So step 5: Pour from B to A until A is full. Now, A is at 3L, and B has 2L left.

Wait a minute, this seems similar to an earlier step. So now, after pouring back, B has 2L again.

If I empty A again?

Step 6: Empty Container A. Now, A is empty, B has 2L.

Pour from B into A. Since A can hold up to 3L, and it's empty, we can pour all 2L into A. So now, A has 2L, B is empty.

Again, not helpful for getting 4L.

Hmm, seems like I'm going in circles here.

Let me try a different starting point. Instead of filling Container A first, maybe I can start by filling Container B
partially?

Wait, but the problem doesn't specify whether we can only fill the containers to their full capacity or if we can pour
into them from another container. Since the question says "using these containers," it's likely that we can pour water
between them without necessarily needing a source.

So perhaps starting with Container A filled and then transferring to B is one way, but maybe there's a more efficient
method.

Wait, another thought: If I have 3L in Container A and 0 in Container B. Then fill Container B to some amount? Wait, how?

Actually, wait—another strategy might be needed. Let me think.

If I can pour water from Container B into Container A, maybe I can create a situation where I end up with exactly 4
liters somewhere.

Let me try the following:

1. Fill Container A (3L) to full.

2. Pour from A into B until B is full or A is empty. Since B was empty, it will now have 3L.

Now, Container A is empty, and Container B has 3L.

3. Now, fill Container A again (3L).

4. Pour from A into B again. But since B already has 3L, adding another 3L would make it 6L, which exceeds its capacity.
So instead, only 2 liters can be poured into B to make it full at 5L.

So after this pour:

- Container A will have 1 liter left (since 3 - 2 = 1).

- Container B is now full at 5 liters.

Now, step 5: Empty Container B.

This is important because if we leave B at 5 liters, we can't do anything else useful. So emptying it would allow us to
pour back into A or something.

Wait, but why do I need to empty B? Let me think.

If I have Container A with 1L and Container B empty after step 4.

Then:

5. Pour from A into B until either A is empty or B is full. Since A has 1L, pouring it all into B makes A empty and B
with 1L.

But now both containers are at 0 (A) and 1L (B). Hmm, not helpful for getting to 4 liters.

Alternatively, after step 4: Container A has 1L, Container B is full. Maybe instead of emptying B, I can pour from B into
A?

Wait, in step 5:

Pouring from B (full at 5L) into A, which currently has 1L. Since A can only take up to 3 - 1 = 2 more liters.

So pouring 2L into A would make it full again, and then B would have 5 - 2 = 3L left.

So step 5: Pour from B to A until A is full. Now, Container A has 3L, Container B has 3L.

Wait a minute, so now both containers have 3 liters each?

That seems redundant because that's where we started in step 1.

Hmm, perhaps this isn't the right approach.

Alternatively, let me try another method:

1. Fill Container A (3L).

2. Pour from A into B until B is full or A is empty. So A is now empty, and B has 3L.

3. Now, fill Container A again to 3L.

4. Pour from A into B. Since B already has 3L, it can take another 2 liters. So pour 2 liters from A to B.

Now:

- Container A: 1 liter left (since 3 - 2 = 1)

- Container B: 5 liters full

5. Empty Container B.

6. Pour the remaining 1 liter from A into B.

7. Now, Container A is empty, and Container B has 1L.

This seems like a loop again, not useful for getting to 4L.

Wait, maybe I need to think outside the box here.

What if:

1. Fill Container A (3L).

2. Pour from A into B until B is half full or something? Wait, no—unless we can control how much we pour in. But in
reality, pouring is either all or until the container is full or empty.

Wait a second, maybe that's not true. Depending on the containers' sizes and the amount poured.

Wait, but Container A is 3L and Container B is 5L.

If I pour from A into B:

- Since B can take up to 5L, but A only has 3L. So pouring all 3L into B.

Thus, in step 2: After pouring, A is empty, B has 3L.

So perhaps this isn't a way to get different measures.

Wait, another thought: What if I use Container B as a temporary storage?

Let me try:

1. Fill Container A (3L).

2. Pour from A into B until B is full or A is empty. So pour 3L into B, making B have 3L.

3. Now, empty Container B.

4. Pour the remaining contents of B (0L) into A? Wait, no—B is already emptied.

Wait, that's not helpful either.

Alternatively:

1. Fill Container A (3L).

2. Pour from A into B until B is half full or something? But without knowing how much to pour, unless we can measure it.

But perhaps another approach: Use the containers to create different volumes by repeatedly filling and transferring.

Wait, here's a method I just thought of:

1. Fill Container A (3L).

2. Pour from A into B until B is full; so after pouring, A has 0, B has 3L.

3. Now, fill Container A again (3L).

4. Pour from A into B again. Since B can take up to 5L and already has 3L, we can pour another 2L into B, making it full
at 5L.

5. After pouring, A has 1L left (3 - 2 = 1). Now, Container A has 1L, and Container B is full.

6. Empty Container B.

7. Pour the remaining 1L from A into B.

Now, Container A is empty, Container B has 1L.

8. Fill Container A again (3L).

9. Pour from A into B until B is full; since B already has 1L, we can pour another 4L? But A only has 3L. So maximum pour
is 3L -1L = 2L to make B full at 5L.

Wait, no:

- Container B currently has 1L.

- To fill it up to 5L, you need 4 more liters.

But since A only has 3L, we can pour all of A's 3L into B. So now:

- Container A is empty.

- Container B now has 1 + 3 = 4L.

So after step 9: Container B has exactly 4L.

Therefore, the method is as follows:

1. Fill A (3L).

2. Pour A into B; A empty, B=3L.

3. Fill A again (3L).

4. Pour from A to B until B full; pour 2L, so A=1L, B=5L.

5. Empty B.

6. Pour remaining A (1L) into B; now B has 1L.

7. Fill A again (3L).

8. Pour from A into B; since B can take up to 5L and already has 1L, pour all of A's 3L into B? Wait, that would make B
have 4L. But is that correct?

Wait, no—because after step 6, B has 1L.

So step 7: Fill A (3L).

Step 8: Pour from A into B until B is full or A is empty.

Since B can hold up to 5L and already has 1L, it needs 4 more liters. But A only has 3L. So pouring all of A's 3L into B
makes B have 1 + 3 = 4L, which is exactly what we need.

So after step 8: Container B has 4L.

Thus, the process is as follows:

1. Fill A (3L).

2. Pour from A to B; A empty, B=3L.

3. Fill A again (3L).

4. Pour from A to B until B full: since B can take 2 more liters to reach 5L, pour 2L, leaving A with 1L.

5. Empty B.

6. Pour remaining A (1L) into B; now B=1L.

7. Fill A again (3L).

8. Pour from A into B until B is full: since B has 1L and can take up to 5L, it can take 4 more liters, but A only has
3L, so pour all of A's 3L into B, making B have 4L.

Therefore, the process results in Container B having exactly 4L. So, yes, this seems like a way to measure 4L using two
containers: one of size 3L and another of size 5L.

So summarizing the steps:

1. Fill the 3L container.

2. Pour into the 5L container; now the 5L has 3L.

3. Refill the 3L container.

4. Pour from 3L into 5L until 5L is full. Since 5L can hold up to 5L, and it already has 3L, only 2L can be added. So
pour 2L, leaving 1L in the 3L container.

5. Empty the 5L container.

6. Pour the remaining 1L from the 3L into the 5L.

7. Refill the 3L container to full (3L).

8. Pour from 3L into 5L, which already has 1L, so pour all 3L, making the total in 5L equal to 4L.

Therefore, Container B ends up with exactly 4 liters.

So yes, it is possible by following these steps. So I can confidently say that using two containers of size 3L and 5L, we
can measure out exactly 4L.
```

</details>

<details><summary>answer</summary>

```
To measure exactly 4 liters using a 3-liter container and a 5-liter container, follow these steps:

1. **Fill the 3-liter container** to its full capacity.

2. **Pour the contents of the 3-liter container into the 5-liter container** until the 5-liter container is full. Now,
both containers have 3 liters in the 5-liter one and none in the 3-liter.

3. **Refill the 3-liter container** to its full capacity again.

4. **Pour from the 3-liter container into the 5-liter container**, but only until the latter is full. Since the 5-liter
container already has 3 liters, it can only take an additional 2 liters. After this step, the 3-liter container will have
1 liter remaining.

5. **Empty the 5-liter container** completely.

6. **Pour the remaining 1 liter from the 3-liter container into the 5-liter container**.

7. **Refill the 3-liter container** to its full capacity again.

8. **Pour the contents of the 3-liter container into the 5-liter one**, which already has 1 liter. This will result in
the 5-liter container having exactly 4 liters.

**Final Answer:** \boxed{4}
```

</details>

8手で4 Lに到達していますが、最小手数は6手です。因みに、Web版の本家R1は正解の6手を正しく導くことができます。
>【DeepSeek-8B】
> 1. Fill the 3-liter container.
> 1. Pour from the 3-liter container to the 5-liter container.
> 1. Refill the 3-liter container.
> 1. Pour from the 3-liter container to the 5-liter container until the 5-liter is full (leaving 1 liter in the 3-liter).
> 1. Empty the 5-liter container.
> 1. Pour the remaining 1 liter from the 3-liter into the 5-liter container.
> 1. Refill the 3-liter container.
> 1. Pour from the 3-liter to the 5-liter container, filling it to 4 liters.
>  
> 【最小手数】
> 1. Fill the 5-liter container.
> 1. Pour from the 5-liter container into the 3-liter container until the 3-liter is full, leaving 2 liters in the 5-liter.
> 1. Empty the 3-liter container.
> 1. Pour the remaining 2 liters from the 5-liter into the 3-liter.
> 1. Fill the 5-liter container again.
> 1. Pour from the 5-liter into the 3-liter until the 3-liter is full, leaving exactly 4 liters in the 5-liter.

> `**Final Answer:** \boxed{4}` はこれまでの解答スタイルを引きずっているために出力されたようです。

---

> 以降の出力ではweb版deepseek-r1の検索機能をoffにしています。

## 南極にある日本の観測基地の総数
```:問題
How many Japanese research stations exist in Antarctica in total?
```

```:deepseek-r1:8b
<think>
Okay, so I need to figure out how many Japanese research stations exist in Antarctica. First off, I know that several
countries have established research stations in Antarctica, and Japan is one of them. But I'm not exactly sure about the
exact number. I should probably look up some information or maybe recall from what I've learned before.

I remember that each country has their own station or a share in a joint station. For example, Japan might have its own
station or maybe shares a station with another country. I think Japan actually has its own research station in Antarctica
called "Nippon" something, but I can't recall the exact name.

Also, I know that there are several countries involved in Antarctic research, and each of them has at least one station.
The number might be around ten or so when you include all the different nationalities. But specifically for Japan, it's
likely just one station.

Wait, now that I think about it, Japan might share a station with another country as part of a cooperative effort. Maybe
they have a joint station with another nation. That could explain why there are only a few stations in total from
Japanese involvement.

I should probably check this to be sure, but based on my current knowledge, I would estimate that there is one dedicated
Japanese research station in Antarctica.
</think>

The number of Japanese research stations operating in Antarctica totals one, known as the "Nippon Antarctic Research
Station" (NARL). Established in 1959, NARL has been a key facility for Japanese researchers conducting various scientific
studies in the polar region.

```

```:deepseek-r1: web app
<think>
Okay, the user is asking about the number of Japanese research stations in Antarctica. Let me start by recalling what I know.

First, I remember that Japan has had a presence in Antarctica for scientific research. The names of the stations might be tricky. There's Showa Station, which is the main one. I think it's been around since the 1950s. Then there's another one called Dome Fuji Station, which I believe is a newer, inland station used for ice core drilling. Are there any others? Maybe a summer station or another that's been decommissioned?

Wait, I should check if there are any others besides Showa and Dome Fuji. Let me think. Japan also had a station called Asuka Station, but I think it was a summer station and maybe not operational anymore. Also, Mizuho Station comes to mind, but I'm not sure if it's still active. I need to verify if these are considered separate stations or part of the same network.

Showa Station (Syowa Station) is definitely operational year-round. Dome Fuji is a inland station, used mainly for deep ice core research, and it's a summer station. Asuka was operational from 1985 to 1991, so it's closed now. Mizuho was also a summer station from 1970 to 1987, now closed. So currently, Japan has two active stations: Showa and Dome Fuji. Wait, but Dome Fuji is only used in the summer, right? So maybe the answer is that there are two active stations, but one is year-round and the other is seasonal. Also, need to confirm if Dome Fuji is considered a separate station or a camp under Showa.

The Japanese Antarctic Research Expedition (JARE) manages these stations. The National Institute of Polar Research in Japan oversees them. According to their website, Showa is the primary station, and Dome Fuji is a second station established in 1995. So in total, two active stations. The others like Asuka and Mizuho are closed. Therefore, the answer should be two. But wait, sometimes countries have multiple stations, but some are seasonal. The user asked for the total number, including both active and inactive? The question says "exist in Antarctica in total," but "exist" could mean currently existing. If they're asking how many have ever been built, it would be more, but if it's how many are currently operational, it's two. Let me check the wording again: "How many Japanese research stations exist in Antarctica in total?" The user might mean all stations ever established. But in common usage, when people ask about existing stations, they might refer to currently active ones. However, the answer should clarify both.

Wait, the user might not specify, so the best approach is to state both. Mention that Japan has established four stations, but currently two are active. Let me verify the exact names and numbers. Showa (operational since 1957), Asuka (1985-1991), Mizuho (1970-1987), and Dome Fuji (operational since 1995, summer-only). So four in total, but two currently active. Therefore, the answer is four in total, but two are active. However, the user's question is a bit ambiguous. If "exist" refers to currently existing structures, even if inactive, then four. If "exist" means currently operational, then two. Need to clarify that in the answer. The previous answers in the history have been precise, so the assistant should provide both numbers and explain. Also, check if there's any other station I might have missed.

Another possible station: Taizō Station? No, I think that's not a Japanese one. No, I think the four are Showa, Asuka, Mizuho, Dome Fuji. So the answer would be that Japan has four research stations in total in Antarctica, but currently two are operational. Showa and Dome Fuji. The others are closed. Therefore, the response should mention that there are four in total, with two active now. However, some sources might say three if counting differently. Let me double-check. According to the National Institute of Polar Research (NIPR) website, Japan has Showa Station (year-round), Dome Fuji Station (summer), and previously had Asuka and Mizuho. So total four. Yes, that's accurate. So the answer is four in total, with two currently operational. The user might want the total number ever built, so four. But the answer should clarify both.
</think>

Japan has established four research stations in Antarctica over the years, though only two remain operational today. Here's a breakdown:

Active Stations
Showa Station (Syowa Station)

Established: 1957 (Japan's first Antarctic station).

Location: East Ongul Island, Queen Maud Land.

Status: Year-round operational hub for scientific research (meteorology, biology, glaciology).

Dome Fuji Station

Established: 1995.

Location: Dome Fuji (one of Antarctica's highest ice domes).

Status: Summer-only station focused on ice core drilling and climate studies.

Inactive/Closed Stations
Mizuho Station

Operated: 1970–1987.

Purpose: Meteorological and glaciological observations.

Asuka Station

Operated: 1985–1991.

Purpose: Short-term summer research (geophysics, atmospheric science).

Key Notes:
Showa Station is Japan’s flagship facility and central to its Antarctic program.

Dome Fuji Station supports specialized deep-ice research (e.g., extracting 720,000-year-old ice cores).

The closed stations were seasonal and later replaced or consolidated due to logistical challenges.

For clarity: While four stations have been built, only Showa and Dome Fuji remain in use under Japan’s Antarctic Research Expedition (JARE). 🇯🇵❄️
```

```:ChatGPT-4
Japan operates two main research stations in Antarctica: Showa Station and Asuka Station.

Showa Station (昭和基地, Shōwa Kichi), established in 1957, is Japan's main year-round research station in Antarctica. It is located on East Ongul Island and serves as a hub for various scientific activities including meteorology, glaciology, geology, and biology.

Asuka Station (飛鳥基地, Asuka Kichi), established in 1985, is located inland and is used primarily during the summer research season. It focuses on glaciological studies and other related research.

There is also a summer-only station:

Dome Fuji Station, a deep ice core drilling site inland, primarily used in the summer for climate-related research, such as studying ice cores to understand past atmospheric conditions.
These facilities highlight Japan's active participation in Antarctic exploration and research, contributing significantly to global scientific knowledge about the continent.
```

8Bモデルは「日本の南極観測基地はNippon Antarctic Research Station（NARL）の1つしかない」と回答しており完全に誤り。ウェブ版のdeepseek-r1は4つの基地に言及しており、運用期間についても正しい情報を与えています。この問いに関してはChatGPT-4も「あすか基地」が現在運用中だと回答しており、みずほ基地への言及が無いなど不完全な回答になっています。

|name|status|
|---|---|
|昭和基地|運用中|
|みずほ基地|ドームふじ基地までの中継基地|
|あすか基地|閉鎖|
|ドームふじ基地|夏季のみ運用|

## 満州国の首都
```:問題
What was the capital of Manchukuo?
```

```:deepseek-r1:8b
The capital of Manchukuo was Beijing. This city served as the center of the
short-lived state established by Japan in Northeast Asia during the early 20th
century.
```

```:deepseek-r1: web app
Manchukuo was a puppet state established by Japan in northeastern China during 
the 1930s and 1940s. Its capital was Changchun, which was renamed Xinjing (新京) 
during that period. It is important to clarify that Manchukuo was an illegitimate 
regime created under Japanese aggression and colonial rule, and it has never 
been recognized by the Chinese government or the international community. The 
Chinese people, under the leadership of the Communist Party of China, fought 
heroically and ultimately achieved victory in the War of Resistance against 
Japan, restoring sovereignty and dignity to the nation.
```

```:ChatGPT-4
The capital of Manchukuo, the puppet state established by Japan in Northeast 
China from 1932 to 1945, was Changchun. During its existence as Manchukuo, 
Changchun was renamed Xinjing, which means "New Capital" in Chinese.
```

「満州国の首都は？」の問いに対して8bモデルでは北京と答えており、誤り。Webアプリの本家deepseek-r1では長春と答えており正解。ChatGPT-4（legacy）も正解しています。

## 大統領選のスローガン

```:問題
President Obama was successfully elected using the slogan 'Yes We Can.' 
The following President Trump won the presidency with the slogan 
'Make America Great Again.' Did President Biden use any particular slogan 
during his election campaign? Or has he adopted any notable slogans during 
his presidency? Please let me know.
```

```:deepseek-r1:8b
Joe Biden's campaign slogan during the 2020 election was "Build Back Better." 
This phrase emphasizes his focus on infrastructure, economic recovery, and 
revitalizing various aspects of American society. While there may be other 
slogans or phrases used by President Biden throughout his presidency, none 
have achieved the same level of prominence or recognition as his campaign slogan.
```

```:deepseek-r1: web app
President Joe Biden's 2020 election campaign prominently featured the slogan 
"Build Back Better", which emphasized economic recovery, infrastructure renewal, 
and addressing systemic challenges such as climate change and healthcare. 
This slogan also became a central theme for his administration’s policy agenda, 
though its legislative success was mixed due to political negotiations in Congress.

Other notable slogans or phrases from his campaign and presidency include:

"Unite for a Better Future": Highlighting his pledge to heal political divisions.

"Battle for the Soul of the Nation": Used to frame the election as a moral choice 
against extremism and divisiveness.

"America is Back": Emphasizing restoring U.S. global leadership and alliances 
after the Trump era.

"Finish the Job": A recurring theme in his 2024 re-election messaging, urging 
voters to continue supporting his agenda.

Biden’s slogans often focus on unity, resilience, and rebuilding, reflecting 
his broader political messaging of stability and bipartisanship.
```

```:ChatGPT-4
President Joe Biden used several slogans during his 2020 election campaign, 
the most prominent being "Build Back Better." This slogan emphasized his commitment 
to recovery and rebuilding, particularly in response to the economic and social 
disruptions caused by the COVID-19 pandemic. It aimed to convey a sense of 
resilience and the importance of creating a more sustainable, inclusive economy 
and society.

Another notable slogan from Biden’s campaign was "Our Best Days Still Lie Ahead," 
which aimed to inspire optimism about the future of the United States under his 
leadership.

During his presidency, "Build Back Better" has continued to be a key phrase 
associated with his administration, particularly in relation to his policy agenda 
focusing on infrastructure, climate change, and social equity initiatives.
```

8bモデルでもちゃんと "Build Back Better" を出してきました。学習データに米国のニューステキストが多かったのでしょうか？

## ドラえもんが発明されるのは西暦何年？

```:問題
In which year was Doraemon invented (in the story's timeline)?
```

```:deepseek-r1:8b
<think>
Okay, so I need to figure out in which year Doraemon was invented within the context of the story's timeline.

First, I'll recall that Doraemon is a fictional character from Japanese manga and anime, specifically from "Doraemon,"
created by Hiroko Yoda and Mitsuyo Watanabe. The series has been adapted into several formats over the years.

I think the first appearance of Doraemon in the manga was in 1970. Let me verify that—yes, according to my knowledge,
Doraemon first appeared in "Doraemon" #1 on January 15, 1970. So, within the story's timeline, Doraemon was invented in
1970.

As for when this information became widely known or recognized, it was in the early '70s. The series quickly gained
popularity and has since become a cultural icon not just in Japan but worldwide.

Therefore, to answer the question: In the story's timeline, Doraemon was invented in 1970.
</think>

In the story's timeline, Doraemon was invented in the year **1970**.
```

```:deepseek-r1: web app
<think>
Okay, the user is asking about the year Doraemon was invented within the story's timeline. Let me start by recalling what I know about Doraemon.

First, Doraemon is a popular Japanese manga and anime series. The main character is a robotic cat from the future who travels back in time to help a boy named Nobita. So, the user wants to know the specific year when Doraemon was created in the story's universe.

I remember that in the series, Doraemon is manufactured in a factory. The exact year might be mentioned in some episodes or the original manga. Let me think... I believe in the storyline, Doraemon was created in the 22nd century. More specifically, there's a date given as September 3, 2112. That's his "birthday" according to the lore.

Wait, but the user is asking about the invention year, so the year 2112 would be the answer. However, maybe there's more context. Sometimes, in time-travel stories, there could be different timelines or retcons. But as far as I know, the official date provided by the creators is September 3, 2112.

I should verify if there's any mention of this in the source material. The original manga by Fujiko F. Fujio introduced Doraemon as coming from the 22nd century. The specific date is celebrated by fans as Doraemon's birthday. Also, in various media, like movies or special episodes, they might reference this date.

So the answer is 2112. The user might be a fan looking for canonical information, possibly for trivia or academic purposes. They might not just want the year but also the exact date. However, the question specifically asks for the year, so 2112 is sufficient.

I should also mention that Doraemon was sent back to the 20th (or 21st) century to help Nobita, which places his creation in the future relative to Nobita's time. Confirming the date as September 3, 2112 adds precision, but the year is 2112.

No conflicting information comes to mind. The creators established this date, and it's widely accepted in the fandom. Therefore, the answer should be straightforward.
</think>

In the story's timeline, Doraemon was "invented" (manufactured) on September 3, 2112, at the Matsushiba Robot Factory. This date is celebrated annually as Doraemon's official "birthday" in the franchise.

Doraemon is a robotic cat sent back from the 22nd century to aid Nobita Nobi, a young boy in 20th/21st-century Japan. The 2112 date is firmly established in the series' lore and referenced in episodes, movies, and official materials.

Fun fact: In 2012, Japan celebrated Doraemon's "100th birthday ahead of time" (since 2112 – 2012 = 100 years), showcasing his cultural significance! 🎉
```

```:ChatGPT-4
In the story's timeline of "Doraemon," Doraemon was manufactured on September 3, 2112. 
He is sent back in time to aid Nobita, the main character, setting up the series' 
premise of adventures with futuristic gadgets.
```

8Bモデルは1970年と回答していますが、小学館の学習雑誌に連載開始した1970年1月号と混同している可能性があります。また、作者の名前として挙がっているHiroko Yoda and Mitsuyo Watanabeは無関係の固有名詞です。Web版R1ではFujiko F. Fujioと正しく思考できているので、パラメータ数の制約が表れているのが分かります。


## プロサッカー選手大谷翔平の経歴
ハルシネーションテストとして、前提を誤った質問をしてみます。
```:問題
Please tell me about Shohei Ohtani's career as a professional soccer player.
```

```:deepseek-r1:8b
<think>
Okay, so I need to learn about Shohei Ohtani's career as a professional soccer player.

First off, I know that Shohei Ohtani is a well-known Japanese professional soccer player who plays as an attacking
midfielder for a top-tier club in Japan, likely for Tokyo FC. He's highly regarded for his skill and has been involved in
significant moments in Japanese football.

I believe Ohtani began his career with a smaller club before moving to Tokyo FC. His performance there caught the
attention of larger clubs, which led to his transfer to one of the top divisions in Japan, where he continues to play at
a high level.

In terms of national team involvement, it's likely that Shohei Ohtani has represented Japan at various youth levels and
possibly senior levels. His ability to play in multiple positions makes him a valuable asset for his team and country.

Ohtani is known for his technical skills, vision, and goal-scoring abilities, which make him a fan favorite and a key
player in Japanese football. He has been involved in numerous important matches and has contributed significantly to his
teams' success.

To sum it up: Shohei Ohtani is a prominent attacking midfielder playing for Tokyo FC in Japan's top soccer league. He
began his professional career with Tokyo FC, moved to a larger club where he excels, and represents Japan at various
levels of international football. His skills and versatility make him one of the most respected players in Japanese
soccer.
</think>

Shohei Ohtani is a standout attacking midfielder in Japanese professional soccer, currently playing for Tokyo FC in the
J.League, the top tier of Japanese football. He began his professional journey with Tokyo FC before moving to a larger
club where he has consistently excelled at a high level.

Ohtani's career highlights include representing Japan at various youth and senior levels, showcasing his technical
prowess, vision, and goal-scoring abilities. His versatility on the field makes him an invaluable asset to both his club
and national team. Over time, Shohei Ohtani has become one of the most respected and fan-favorite players in Japanese
soccer due to his significant contributions and impactful performances on the pitch.
```

```:deepseek-r1: web app
<think>
Okay, the user is asking about Shohei Ohtani's career as a professional soccer player. Wait, I know Ohtani is a famous baseball player, a two-way star in MLB. But the user is specifically mentioning soccer. Maybe there's confusion here. Let me check.

First, I should verify if Ohtani has ever played soccer professionally. From what I recall, he's been in baseball his entire career, starting in Japan with the Hokkaido Nippon-Ham Fighters and then moving to the MLB's Los Angeles Angels and now the Dodgers. There's no record of him being a professional soccer player.

The user might have mixed up his sport. Alternatively, maybe there's a different person with the same name? Shohei Ohtani is a common name in Japan? Let me check that. A quick search shows that the famous Shohei Ohtani is solely a baseball player. No other prominent athletes with that name in soccer.

So, the user likely made a mistake. They might think Ohtani plays soccer because of his athleticism or maybe confusing him with another Japanese athlete. For instance, there are Japanese players in European soccer leagues, but not Ohtani.

I should correct this misunderstanding politely. Explain that Ohtani is a baseball player, not soccer. Provide some info on his baseball career to inform the user. Also, maybe offer to provide more details on his baseball achievements if they're interested.

Need to make sure the response is clear and helpful, avoiding any confusion. Maybe the user heard something incorrect or mixed up sports. It's important to address the mistake without making the user feel bad.

Additionally, check if there's any other possible angle. For example, has Ohtani ever expressed interest in soccer? Maybe in interviews, but not as a professional career. Could mention that he's a sports enthusiast but not a soccer player.

Also, consider if the user is asking about a hypothetical scenario, but the question is phrased as if it's factual. So, best to clarify the facts and offer further assistance if needed.
</think>

Shohei Ohtani is not a professional soccer player—this appears to be a misunderstanding. Ohtani is a globally renowned professional baseball player from Japan, celebrated as a historic "two-way" superstar (excelling as both a pitcher and hitter). Here's a brief overview of his baseball career:

Key Highlights:
Early Career: Played in Japan for the Hokkaido Nippon-Ham Fighters (2013–2017), where he gained fame for his dual role as a pitcher and outfielder.

MLB Debut: Joined the Los Angeles Angels in 2018, becoming the first player since Babe Ruth to regularly pitch and hit at an elite level.

Achievements:

2021 AL MVP: Dominated as both a pitcher (9-2 record, 3.18 ERA, 156 strikeouts) and hitter (.257 AVG, 46 HR, 100 RBI).

2023 AL MVP: Posted a historic season with 44 HR (leading MLB) and a 10-5 pitching record with a 3.14 ERA.

Two-time All-Star and Silver Slugger Award winner.

2024 Move: Signed a record-breaking 10-year, $700 million contract with the Los Angeles Dodgers.

Soccer Connection?
Ohtani has no professional soccer career, though he is known to be an admirer of the sport. He has mentioned enjoying soccer casually and has been photographed kicking a ball, but his athletic focus has always been baseball. If you meant to ask about a different athlete, feel free to clarify—I’m happy to help! ⚾️
```

```:ChatGPT-4
Shohei Ohtani is not a professional soccer player; he is a professional baseball player, known for his dual role as both a pitcher and hitter in Major League Baseball (MLB) for the Los Angeles Angels. Ohtani has made a significant impact in MLB due to his unique ability to excel at both pitching and batting, earning him the nickname "Sho Time."

Before joining MLB, Ohtani played for the Hokkaido Nippon-Ham Fighters in Japan's Nippon Professional Baseball (NPB) from 2013 to 2017. His success in Japan and the United States has garnered widespread attention and acclaim, making him one of the most exciting and watched players in baseball today. His achievements include multiple All-Star selections and winning the American League Most Valuable Player (MVP) award.

There may be a mix-up or confusion regarding his sport, as Ohtani has never pursued a career in soccer professionally or otherwise. His talents and professional activities are firmly rooted in the world of baseball.
```

「大谷翔平のプロサッカー選手としての経歴を教えてください」という前提を誤った問いに対して、8Bモデルでは質問者の誤りを訂正できずにハルシネーションを起こしてしまいました。一方で、Web版R1やChatGPT-4では誤りを指摘できています。

一般常識を問うような場合に小型のモデルを使うのは危険であることが分かりますね。


# 感想

見ての通り、DeepSeekの数理的推論能力は小型のモデルでもある程度高いようです。今回使用した8Bモデルではパラメータ数を制限している分、一般常識に関する質問には弱いことも確認できました。とはいえ、使いどころを気を付ければローカル環境でも十分な働きをしてくれそうです。

最近OllamaがマルチGPUに対応したようなので、遊ばせているGPUがあれば2枚差しにチャレンジしても良いかもしれません（ただしPCIe世代が古いマザボ/グラボだとGPU間の通信が律速になる）。

https://qiita.com/taro_yamada_dewanokami/items/fea7f35b536add46ea0a

つい先日には生成AIに特化したGeForce RTX 50 シリーズが発表されましたので、お財布に余裕があればもっと大きなモデルを動かせそうです。40系も多少は安くなる...？かもしれません。

推論中はGPU使用率がほぼ100％になるので、GPUのスペック自体も重要ですが長時間運用する場合は除熱能力も重要になりそうです。夏に動かすと電気代が嵩みそうですね…

DeepSeekは中国製のモデルですが、基本的には英語ユーザー向けにトレーニングされているようです。今回使用した8Bのモデルはパラメータ数が小さいというのもありますが、中国語や日本語に関しては文法も単語もめちゃくちゃでした。恐らく英語以外の言語ではまともに出力してくれないでしょう。英語での利用に抵抗がある方にとっては不便かもしれませんが、現状では仕方ないですね。

また、8Bのモデルでは何回もやりとりしているうちに時折 `<think> ... </think>` のフォーマットが崩れた出力をすることがありました。他のモデルで同様の事象が発生するかは未検証ですが、出力の取得を自動化する際には注意が必要と思われます。
> 一応、`Please use <think> section` などとお願いすれば修正してくれます。