DeepSeek-Math-V2

sglang serve --model-path deepseek-ai/DeepSeek-Math-V2 \
  --tp 8 \
  --ep 8 \
  --reasoning-parser deepseek-r1 \
  --host 0.0.0.0 \
  --port 30000

from openai import OpenAI

client = OpenAI(
    base_url="http://localhost:30000/v1",
    api_key="EMPTY"
)

# Mathematical reasoning problem
response = client.chat.completions.create(
    model="deepseek-ai/DeepSeek-Math-V2",
    messages=[
        {"role": "user", "content": "Prove that for any positive integer n, the sum 1 + 2 + 3 + ... + n = n(n+1)/2"}
    ],
    max_tokens=4096,
    stream=True
)

# Process the stream
thinking_started = False
has_thinking = False
has_answer = False

for chunk in response:
    if chunk.choices and len(chunk.choices) > 0:
        delta = chunk.choices[0].delta

        # Print thinking process
        if hasattr(delta, 'reasoning_content') and delta.reasoning_content:
            if not thinking_started:
                print("=============== Thinking =================", flush=True)
                thinking_started = True
            has_thinking = True
            print(delta.reasoning_content, end="", flush=True)

        # Print answer content
        if delta.content:
            if has_thinking and not has_answer:
                print("\n=============== Content =================", flush=True)
                has_answer = True
            print(delta.content, end="", flush=True)

print()

=============== Thinking =================
We need to prove that for any positive integer n, the sum 1 + 2 + 3 + ... + n = n(n+1)/2.

This is a classic formula for the sum of the first n natural numbers. We can prove by induction.

Base case: n=1, LHS = 1, RHS = 1*(1+1)/2 = 1*2/2 = 1. Holds.

Inductive step: Assume true for n = k, i.e., 1 + 2 + ... + k = k(k+1)/2. Then for n = k+1, sum = 1 + 2 + ... + k + (k+1) = [k(k+1)/2] + (k+1) = (k(k+1) + 2(k+1))/2 = (k+1)(k+2)/2 = (k+1
)((k+1)+1)/2. So holds for k+1. By induction, holds for all positive integers n.

...
=============== Content =================
We can prove the well-known formula for the sum of the first \(n\) positive integers in several ways. Two of the most elementary are presented below.

---

### 1. Proof by mathematical induction

**Base case (\(n=1\))**:
\[
1 = \frac{1\cdot(1+1)}{2}= \frac{1\cdot2}{2}=1,
\]
so the formula holds for \(n=1\).

**Inductive hypothesis:**
Assume that for some positive integer \(k\) the formula is true, i.e.
\[
1+2+\dots+k = \frac{k(k+1)}{2}.
\]

**Inductive step (\(k \to k+1\))**:
Consider the sum up to \(k+1\):
\[
\begin{aligned}
1+2+\dots+k+(k+1) &= \bigl(1+2+\dots+k\bigr) + (k+1) \\[4pt]
&= \frac{k(k+1)}{2} + (k+1) \qquad\text{(by the induction hypothesis)}\\[4pt]
&= (k+1)\left(\frac{k}{2}+1\right)\\[4pt]
&= (k+1)\frac{k+2}{2}\\[4pt]
&= \frac{(k+1)(k+2)}{2}\\[4pt]
&= \frac{(k+1)\bigl((k+1)+1\bigr)}{2}.
\end{aligned}
\]
Thus the formula also holds for \(n=k+1\).

By the principle of mathematical induction,
\[
1+2+3+\dots+n = \frac{n(n+1)}{2}
\]
for every positive integer \(n\).

---

### 2. Proof by pairing (Gauss’s trick)

Let
\[
S = 1 + 2 + 3 + \dots + n.
\]

Write the same sum in reverse order:
\[
S = n + (n-1) + (n-2) + \dots + 1.
\]

Add the two equalities term‑by‑term:
\[
\begin{aligned}
2S &= (1+n) + \bigl(2+(n-1)\bigr) + \bigl(3+(n-2)\bigr) + \dots + (n+1)\\
    &= \underbrace{(n+1)+(n+1)+\dots+(n+1)}_{n\ \text{times}}\\
    &= n\,(n+1).
\end{aligned}
\]

Therefore
\[
S = \frac{n(n+1)}{2}.
\]

Both proofs are rigorous and show that the formula holds for all positive integers \(n\).

from openai import OpenAI

client = OpenAI(
    base_url="http://localhost:30000/v1",
    api_key="EMPTY"
)

# IMO-style problem
response = client.chat.completions.create(
    model="deepseek-ai/DeepSeek-Math-V2",
    messages=[
        {"role": "user", "content": "Let a, b, c be positive real numbers such that abc = 1. Prove that (a-1+1/b)(b-1+1/c)(c-1+1/a) <= 1."}
    ],
    max_tokens=8192,
    stream=True
)

# Process the stream
thinking_started = False
has_thinking = False
has_answer = False

for chunk in response:
    if chunk.choices and len(chunk.choices) > 0:
        delta = chunk.choices[0].delta

        if hasattr(delta, 'reasoning_content') and delta.reasoning_content:
            if not thinking_started:
                print("=============== Thinking =================", flush=True)
                thinking_started = True
            has_thinking = True
            print(delta.reasoning_content, end="", flush=True)

        if delta.content:
            if has_thinking and not has_answer:
                print("\n=============== Content =================", flush=True)
                has_answer = True
            print(delta.content, end="", flush=True)

print()

=============== Thinking =================
We need to prove that for positive real numbers a,b,c with abc = 1, we have:

\[
(a - 1 + \frac{1}{b})(b - 1 + \frac{1}{c})(c - 1 + \frac{1}{a}) \le 1.
\]

We can rewrite the expressions: Since abc=1, we have 1/b = ac, 1/c = ab, 1/a = bc. Wait careful: abc=1 => 1/b = ac? Actually 1/b = ac? Let's check: abc=1 => ac = 1/b? Multiply both sides by something: abc=1 => (ac) b = 1 => ac = 1/b. Yes, because (ac) * b = 1 => ac = 1/b. Similarly, ab = 1/c, bc = 1/a. So we can rewrite:

...
=============== Content =================

We are given positive real numbers \(a,b,c\) with \(abc=1\). We must prove

\[
\Bigl(a-1+\frac1b\Bigr)\Bigl(b-1+\frac1c\Bigr)\Bigl(c-1+\frac1a\Bigr)\le 1 .
\]

---

### 1.  A convenient substitution

Because \(abc=1\), we can write

\[
a=\frac{x}{y},\qquad b=\frac{y}{z},\qquad c=\frac{z}{x}
\]

with positive numbers \(x,y,z\).
(For instance, take \(x=1,\;y=\frac1a,\;z=\frac1{ab}\); then indeed \(a=\frac{x}{y},\;b=\frac{y}{z}\) and, using \(abc=1\), we obtain \(c=\frac{z}{x}=\frac1{ab}=c\).)

---

### 2.  Rewriting the factors

\[
\begin{aligned}
a-1+\frac1b &=\frac{x}{y}-1+\frac{z}{y}= \frac{x+z-y}{y},\\[2mm]
b-1+\frac1c &=\frac{y}{z}-1+\frac{x}{z}= \frac{x+y-z}{z},\\[2mm]
c-1+\frac1a &=\frac{z}{x}-1+\frac{y}{x}= \frac{y+z-x}{x}.
\end{aligned}
\]

Hence the product becomes

\[
P=\Bigl(a-1+\frac1b\Bigr)\Bigl(b-1+\frac1c\Bigr)\Bigl(c-1+\frac1a\Bigr)
   =\frac{(x+z-y)(x+y-z)(y+z-x)}{xyz}.
\]

---

### 3.  Reducing to a known inequality

We have to show \(P\le1\), i.e.

\[
(x+z-y)(x+y-z)(y+z-x)\le xyz .
\tag{1}
\]

Set

\[
p=x+y+z,\qquad q=xy+yz+zx,\qquad r=xyz .
\]

Notice that

\[
x+z-y=p-2y,\quad x+y-z=p-2z,\quad y+z-x=p-2x .
\]

Therefore

\[
\begin{aligned}
(x+z-y)(x+y-z)(y+z-x)
&=(p-2x)(p-2y)(p-2z)\\
&=p^{3}-2p^{2}(x+y+z)+4p(xy+yz+zx)-8xyz\\
&=-p^{3}+4pq-8r .
\end{aligned}
\]

Inequality (1) is thus equivalent to

\[
-p^{3}+4pq-8r\le r\quad\Longleftrightarrow\quad 4pq-p^{3}\le 9r .
\tag{2}
\]

---

### 4.  Applying Schur’s inequality

Schur’s inequality of third degree states that for any non‑negative \(x,y,z\)

\[
p^{3}+9r\ge 4pq .
\]

Rearranged, this is exactly \(4pq-p^{3}\le 9r\), which is (2).
Since our \(x,y,z\) are positive, Schur’s inequality applies and (2) holds.

Consequently (1) is true, and we obtain \(P\le1\).

---

### 5.  Equality case

Equality in Schur’s inequality for positive numbers occurs only when \(x=y=z\).
Then \(a=b=c=1\), and indeed the product equals \(1\).

---

Thus for all positive \(a,b,c\) with \(abc=1\),

\[
\Bigl(a-1+\frac1b\Bigr)\Bigl(b-1+\frac1c\Bigr)\Bigl(c-1+\frac1a\Bigr)\le 1 .
\]

∎

python3 benchmark/gsm8k/bench_sglang.py --num-questions 200 --port 30000

Accuracy: 0.975
Invalid: 0.000
Latency: 34.358 s
Output throughput: 540.162 token/s

python3 -m sglang.bench_serving \
  --backend sglang \
  --host 127.0.0.1 \
  --port 30000 \
  --model deepseek-ai/DeepSeek-Math-V2 \
  --random-input-len 1024 \
  --random-output-len 1024 \
  --num-prompts 10 \
  --max-concurrency 1

============ Serving Benchmark Result ============
Backend:                                 sglang
Traffic request rate:                    inf
Max request concurrency:                 1
Successful requests:                     10
Benchmark duration (s):                  53.34
Total input tokens:                      1972
Total input text tokens:                 1972
Total generated tokens:                  2784
Total generated tokens (retokenized):    2778
Request throughput (req/s):              0.19
Input token throughput (tok/s):          36.97
Output token throughput (tok/s):         52.19
Peak output token throughput (tok/s):    56.00
Peak concurrent requests:                3
Total token throughput (tok/s):          89.16
Concurrency:                             1.00
----------------End-to-End Latency----------------
Mean E2E Latency (ms):                   5330.72
Median E2E Latency (ms):                 5879.28
P90 E2E Latency (ms):                    8320.33
P99 E2E Latency (ms):                    9921.29
---------------Time to First Token----------------
Mean TTFT (ms):                          183.38
Median TTFT (ms):                        177.92
P99 TTFT (ms):                           217.64
-----Time per Output Token (excl. 1st token)------
Mean TPOT (ms):                          17.96
Median TPOT (ms):                        18.39
P99 TPOT (ms):                           19.03
---------------Inter-Token Latency----------------
Mean ITL (ms):                           18.57
Median ITL (ms):                         18.63
P95 ITL (ms):                            19.26
P99 ITL (ms):                            19.48
Max ITL (ms):                            24.93
==================================================

python3 -m sglang.bench_serving \
  --backend sglang \
  --host 127.0.0.1 \
  --port 30000 \
  --model deepseek-ai/DeepSeek-Math-V2 \
  --random-input-len 1024 \
  --random-output-len 1024 \
  --num-prompts 1000 \
  --max-concurrency 100

============ Serving Benchmark Result ============
Backend:                                 sglang
Traffic request rate:                    inf
Max request concurrency:                 100
Successful requests:                     1000
Benchmark duration (s):                  217.36
Total input tokens:                      301701
Total input text tokens:                 301701
Total generated tokens:                  188375
Total generated tokens (retokenized):    187456
Request throughput (req/s):              4.60
Input token throughput (tok/s):          1388.05
Output token throughput (tok/s):         866.67
Peak output token throughput (tok/s):    2589.00
Peak concurrent requests:                109
Total token throughput (tok/s):          2254.72
Concurrency:                             89.81
----------------End-to-End Latency----------------
Mean E2E Latency (ms):                   19521.73
Median E2E Latency (ms):                 12076.76
P90 E2E Latency (ms):                    47248.87
P99 E2E Latency (ms):                    86862.79
---------------Time to First Token----------------
Mean TTFT (ms):                          790.40
Median TTFT (ms):                        456.81
P99 TTFT (ms):                           4223.33
-----Time per Output Token (excl. 1st token)------
Mean TPOT (ms):                          106.52
Median TPOT (ms):                        107.24
P99 TPOT (ms):                           238.33
---------------Inter-Token Latency----------------
Mean ITL (ms):                           100.29
Median ITL (ms):                         38.34
P95 ITL (ms):                            237.00
P99 ITL (ms):                            347.49
Max ITL (ms):                            3642.56
==================================================