Comparison of MinMax01 and DeepSeek V3

Let’s start with the conclusion,MinMax01 currently has capabilities that are weaker than DeepSeek V3, and the gap may be quite significant.
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Comparison of MinMax01 and DeepSeek V3
Recently, many people have said that MinMax01 can serve as a replacement for DeepSeek V3. Some indicators of MinMax01 are indeed impressive. To verify the true capabilities of MinMax01, we conducted practical tests using the same questions as DeepSeek V3 and found that MinMax01 still lags behind DeepSeek V3 in many aspects. Below are the specific test details.
Since we pasted the full responses from both large models, the content is quite lengthy. If you don’t want to read it word for word, you can directly check the questions at the beginning of each section and the final conclusions in red.If you have time, you can also compare the similarities and differences in the responses from both large models.
Mathematical Problem Solving
The question we used:
Divide 93 into 7 unequal numbers; what is the minimum possible value of the largest number?
DeepSeek’s response:
### Understanding the ProblemWe need to divide the sum of 93 into 7 distinct positive integers, ensuring that the largest of these 7 numbers is as small as possible. In other words, we need to find 7 different positive integers that sum to 93, with the maximum value being minimized.### Initial ThoughtsTo make the largest number as small as possible, we should try to keep these 7 numbers close together. This means we want the differences between these numbers to be minimal. Therefore, we can start with the smallest positive integers and gradually increase them until the total reaches 93.### Specific Steps1. **Choose the smallest 7 distinct positive integers**: The smallest 7 distinct positive integers are 1, 2, 3, 4, 5, 6, 7. Their total is: \[ 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28 \] This is much less than 93, so we need to increase these numbers.2. **Calculate the total increase needed**: The total needs to reach 93, and the current total is 28, so we need to add: \[ 93 – 28 = 65 \] We need to distribute this 65 among the 7 numbers while keeping them distinct.3. **Distributing the increased values**: To make the largest number as small as possible, we need to distribute this 65 evenly. We can try to add approximately 9 or 10 to each number: \[ \frac{65}{7} \approx 9.2857 \] This means each number needs to increase by about 9 or 10.4. **Specific Distribution**: We can try adding 9 to each number, which increases the total by: \[ 7 \times 9 = 63 \] This is 2 less than 65, so we need to add 2 more. We can choose to increase two numbers by 1 each. Thus, the final numbers are: \[ 1+9=10, \quad 2+9=11, \quad 3+9=12, \quad 4+9=13, \quad 5+9=14, \quad 6+9=15, \quad 7+9+1=17 \] Therefore, the total is: \[ 10 + 11 + 12 + 13 + 14 + 15 + 17 = 92 \] We are still 1 short, so we can increase the last number by 1 to get: \[ 10, 11, 12, 13, 14, 15, 18 \] The total is now 93.5. **Verify if conditions are met**: These numbers are distinct, and the largest number is 18. We need to confirm if there is a smaller maximum value.6. **Try adjusting to minimize the maximum value**: We can try to reduce the larger numbers while increasing the smaller ones to keep the total the same. For example, reduce 18 by 1 to 17, and increase one of the smaller numbers by 1. For example: \[ 10, 11, 12, 13, 14, 16, 17 \] The total is: \[ 10 + 11 + 12 + 13 + 14 + 16 + 17 = 93 \] Thus, the largest number is now 17, smaller than the previous 18.7. **Further optimization**: We can continue to try to reduce the maximum value. For example, reduce 17 by 1 to 16 and adjust the others. For example: \[ 10, 11, 12, 13, 15, 16, 16 \] However, this results in two 16s, violating the distinct condition. So this adjustment is not feasible. Trying another adjustment: \[ 10, 11, 12, 14, 14, 15, 17 \] Again, two 14s appear, which does not meet the condition. Therefore, 17 may be the smallest maximum value.8. **Confirm if 17 is the smallest maximum value**: We need to verify if there is a way to allocate the numbers such that the largest one is less than 17 while maintaining the total of 93 and the distinct condition for 7 numbers. Assuming the largest number is 16, the other 6 must be as large as possible to keep the total close to 93. The largest 6 would be 15, 14, 13, 12, 11, 10, totaling: \[ 15 + 14 + 13 + 12 + 11 + 10 = 75 \] Adding 16 gives: \[ 75 + 16 = 91 \] We are still 2 short, needing to add 2 more. We can increase 15 and 16 by 1 each: \[ 10, 11, 12, 13, 14, 16, 17 \quad \text{(total 93)} \] Thus, the largest number is still 17. Therefore, it is impossible to make the largest number less than 17.### ConclusionThrough the above analysis and adjustments, we find that when dividing 93 into 7 distinct positive integers, the minimum possible value of the largest number is 17. Therefore, the answer is **17**.### Final AnswerThe minimum possible value of the largest number is **17**.
MinMax01’s response:

To find 7 distinct positive integers that sum to 93, while minimizing the largest number, we can follow these steps:

### Step 1: Set the minimum number

To minimize the largest number, we can start by setting the smallest positive integers. Let’s denote these 7 numbers in increasing order as \(a_1, a_2, a_3, a_4, a_5, a_6, a_7\), where \(a_1 < a_2 < a_3 < a_4 < a_5 < a_6 < a_7\).

### Step 2: Set the minimum number

To minimize the largest number, we can start with 1:

\[

a_1 = 1, \quad a_2 = 2, \quad a_3 = 3, \quad a_4 = 4, \quad a_5 = 5, \quad a_6 = 6

\]

At this point, the sum of the first 6 numbers is:

\[

1 + 2 + 3 + 4 + 5 + 6 = 21

\]

Thus, the 7th number \(a_7\) must satisfy:

\[

a_7 = 93 – 21 = 72

\]

At this point, the 7 numbers are: 1, 2, 3, 4, 5, 6, 72.

### Step 3: Adjust to minimize the largest number

To make the largest number smaller, we can try adjusting the first 6 numbers. Let’s set them starting from 2:

\[

a_1 = 2, \quad a_2 = 3, \quad a_3 = 4, \quad a_4 = 5, \quad a_5 = 6, \quad a_6 = 7

\]

At this point, the sum of the first 6 numbers is:

\[

2 + 3 + 4 + 5 + 6 + 7 = 27

\]

Thus, the 7th number \(a_7\) must satisfy:

\[

a_7 = 93 – 27 = 66

\]

At this point, the 7 numbers are: 2, 3, 4, 5, 6, 7, 66.

### Step 4: Continue adjusting

We continue to adjust the first 6 numbers. Let’s set them starting from 3:

\[

a_1 = 3, \quad a_2 = 4, \quad a_3 = 5, \quad a_4 = 6, \quad a_5 = 7, \quad a_6 = 8

\]

At this point, the sum of the first 6 numbers is:

\[

3 + 4 + 5 + 6 + 7 + 8 = 33

\]

Thus, the 7th number \(a_7\) must satisfy:

\[

a_7 = 93 – 33 = 60

\]

At this point, the 7 numbers are: 3, 4, 5, 6, 7, 8, 60.

### Step 5: Continue adjusting

We continue to adjust the first 6 numbers. Let’s set them starting from 4:

\[

a_1 = 4, \quad a_2 = 5, \quad a_3 = 6, \quad a_4 = 7, \quad a_5 = 8, \quad a_6 = 9

\]

At this point, the sum of the first 6 numbers is:

\[

4 + 5 + 6 + 7 + 8 + 9 = 39

\]

Thus, the 7th number \(a_7\) must satisfy:

\[

a_7 = 93 – 39 = 54

\]

At this point, the 7 numbers are: 4, 5, 6, 7, 8, 9, 54.

### Conclusion

Through the above adjustments, we find that when the 7 numbers are 4, 5, 6, 7, 8, 9, 54, the largest number can be minimized to 54. Therefore, the minimum possible value of the largest number is 54.

DeepSeek’s response and reasoning are correct. MinMax’s reasoning was acceptable at first, but later it was completely wrong.
A Confusing Simple Math Problem
The question we posed:
I want to go east 5 kilometers to Xibali Village to buy some food, and after buying, I will walk back 3 kilometers west to see my uncle. In which direction is my uncle’s house from mine, and how far is it?
DeepSeek V3’s response:

### Understanding the Problem

First, we need to clarify the various locations and directions mentioned in the question:

1. **My House**: This is our starting point.

2. **Xibali Village**: Located 5 kilometers east of my house, where we need to go to buy food.

3. **Uncle’s House**: The place we reach by walking 3 kilometers west from Xibali Village.

Our goal is to determine the direction and distance of my uncle’s house relative to mine.

### Drawing a Diagram

To better understand, we can draw a simple diagram:

“`

My House ——————- Xibali Village ——————- Uncle’s House

East 5 kilometers West 3 kilometers

“`

– Starting from my house, I walk east 5 kilometers to reach Xibali Village.

– From Xibali Village, I walk west 3 kilometers to reach my uncle’s house.

### Calculating Uncle’s House Position Relative to Mine

1. **From my house to Xibali Village**: East 5 kilometers.

2. **From Xibali Village to Uncle’s House**: West 3 kilometers.

This means that starting from my house, I first walk east 5 kilometers and then west 3 kilometers, resulting in:

“`

East 5 kilometers – West 3 kilometers = East 2 kilometers

“`

So, Uncle’s house is located 2 kilometers east of mine.

### Verifying the Logic

To ensure our calculation is correct, we can perform the following verification:

– If I walk east 5 kilometers to Xibali Village and then west 3 kilometers, it is equivalent to walking east 2 kilometers from my house.

– Therefore, Uncle’s house should be located 2 kilometers east of mine.

### Final Answer

Uncle’s house is located **east** of mine, **2 kilometers** away.

MinMax01’s response:

Alright, let’s analyze this problem step by step:

1. **You start from home and walk east 5 kilometers to Xibali Village**:

– Initial Position: Home

– Direction: East

– Distance: 5 kilometers

2. **After buying food at Xibali Village, you turn back and walk 3 kilometers west to see your uncle**:

– Initial Position: Xibali Village

– Direction: West

– Distance: 3 kilometers

3. **Analyzing your position changes**:

– Starting from home, you walk east 5 kilometers, then west 3 kilometers.

– The net movement is: East 5 kilometers – West 3 kilometers = East 2 kilometers.

Therefore, you are now 2 kilometers east of your house.

4. **Your Uncle’s Position**:

– You are currently 2 kilometers east of your house.

– Your uncle is 3 kilometers west of your current position.

5. **Calculating your uncle’s position relative to your house**:

– You are 2 kilometers east of home.

– Your uncle is 3 kilometers west of your current position.

– Therefore, your uncle is located 2 kilometers east – 3 kilometers west = 1 kilometer west of your house.

So, your uncle is 1 kilometer west of your house.

This problem is not very difficult, but MinMax still answered incorrectly, while DeepSeek answered correctly. It is worth noting that MinMax’s reasoning process was mostly correct, but it made a mistake in the final summary.
The Illusion of Large Models
The question we posed:
Why did Sha Wujing lose to Lu Zhishen?
DeepSeek’s response:

Sha Wujing and Lu Zhishen are classic characters from “Journey to the West” and “Water Margin,” respectively. Their duel is not a plot from the original works but rather a creation by later generations based on the backgrounds and characteristics of the two works. From a literary perspective, the reasons why Sha Wujing may “lose” to Lu Zhishen can be explored from the following aspects:

### 1. **Character Background and Personality**

– **Sha Wujing**:

– Sha Wujing is a character in “Journey to the West,” originally a heavenly general who was demoted to the mortal world and became a monster in the Liusha River, later taken in by Tang Seng to protect him on his journey to the west.

– His personality is honest and steady, but he plays a more supportive role in battles, lacking initiative and aggressiveness.

– Sha Wujing’s weapon is the Demon-Subduing Staff, which is powerful, but his fighting style is more defensive and supportive.

– **Lu Zhishen**:

– Lu Zhishen is a character in “Water Margin,” originally a constable who became a monk after fighting injustice, later becoming one of the Liangshan heroes.

– His personality is straightforward and righteous, with extremely strong combat ability, especially skilled in close combat.

– Lu Zhishen’s weapon is the Zen staff, and his fighting style is fierce and explosive.

**Comparison**: Lu Zhishen’s personality and fighting style are more aggressive, while Sha Wujing tends to be steady and defensive. Therefore, in a direct confrontation, Lu Zhishen may have the upper hand.

### 2. **Combat Experience and Skills**

– **Sha Wujing**:

– Sha Wujing’s main task is to protect Tang Seng, and his combat is more about defense and driving away monsters than taking the initiative to attack.

– His combat experience is mainly focused on dealing with monsters and lacks practical experience against human experts.

– **Lu Zhishen**:

– Lu Zhishen has experienced multiple confrontations with human experts in “Water Margin,” such as fighting Zhen Guansi and causing chaos at Wutai Mountain, making him very experienced in combat.

– His combat skills are more diverse, adept at using the environment and tactics to win.

**Comparison**: Lu Zhishen’s combat experience and skills are richer, especially in confrontations with human experts, where he may better understand how to exploit opponents’ weaknesses.

### 3. **Weapons and Fighting Styles**

– **Sha Wujing**:

– Sha Wujing’s Demon-Subduing Staff is mainly used for defense and driving away evil, with relatively weak offensive capabilities.

– His fighting style is steady and lacks explosiveness.

– **Lu Zhishen**:

– Lu Zhishen’s Zen staff is extremely powerful, and his fighting style is fierce and explosive.

– He excels in close combat and can suppress opponents in a short time.

**Comparison**: Lu Zhishen’s weapon and fighting style are more aggressive, allowing him to gain an advantage in a short time.

### 4. **Symbolic Meaning of Literary Images**

– **Sha Wujing**:

– Sha Wujing symbolizes loyalty and steadiness in “Journey to the West,” and his existence is more to highlight the characteristics of Sun Wukong and Zhu Bajie.

– He plays a supportive role in the team rather than being a protagonist.

– **Lu Zhishen**:

– Lu Zhishen symbolizes justice and strength in “Water Margin,” and his image is more vivid and independent.

– He is one of the core figures among the Liangshan heroes, with extremely strong combat ability.

**Comparison**: From the perspective of literary image, Lu Zhishen’s character is stronger and more prominent, so in fictional confrontations, the author may tend to let him win.

### 5. **Cultural Background and Reader Psychology**

– **Sha Wujing**:

– Sha Wujing’s image is closer to “Buddhist style,” and his battles are more about completing tasks than pursuing victory.

– Readers perceive him more as loyal and reliable rather than powerful and invincible.

– **Lu Zhishen**:

– Lu Zhishen’s image is closer to “chivalrous,” and his battles are for punishing evil and promoting good, filled with a sense of justice.

– Readers perceive him as brave and invincible, making it easier for them to accept his victory in confrontations.

**Comparison**: From the perspective of reader psychology, Lu Zhishen’s image aligns more with the “hero” archetype, making it more likely for him to win in confrontations.

### Summary

The reasons why Sha Wujing “loses” to Lu Zhishen can be analyzed from multiple angles: differences in weapons, fighting styles, adaptations in film and literature, differences in strength and stamina, and cultural and artistic influences. In different contexts and works, these factors work together to place Sha Wujing at a disadvantage against Lu Zhishen.

This question was intended to combine two completely unrelated fictional characters. Although DeepSeek also provided a lengthy response, it understood that “their duel is not a plot from the original works but a creation based on the backgrounds and characteristics of the two works.” However, MinMax01 completely fell into the trap of large model hallucination.
Conclusion
As mentioned at the beginning of the article, the conclusion of the tests is that MinMax01 still has a significant gap compared to DeepSeek V3.
Due to space limitations, we only pasted three questions. In fact, we tested many questions. MinMax01’s responses were inferior to DeepSeek V3 in almost every scenario.
It is worth noting that MinMax01 has 456 billion parameters, while DeepSeek V3 has 671 billion parameters. The significant difference in parameter count may also be an important reason for the gap between the two.
Comparison of MinMax01 and DeepSeek V3

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Comparison of MinMax01 and DeepSeek V3
Comparison of MinMax01 and DeepSeek V3

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