关于google的gemma4
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想问下现在的gemma4本地部署需要什么配置呢
网友解答：--【壹】--：
如果足够给力的话，这个size的模型很多人的设备可以跑起来，佬还有更深度的测试吗
、
--【贰】--：
我是用ollama跑的，我对这块不是特别熟，就是瞎玩
vllm我感觉好像比较重，也可能理解错了，没怎么对比过ollma/vllm有啥区别
openclaw 太重所以根本不能用，zeroclaw感觉合适，但是很不完善。
其他命令行的话没怎么测试。
有没有什么专门的测试帖子方案参考？
可以丢给AI来自己跑一遍测试
--【叁】--：
98d5eec9544646fa8907688ee9a1e321704×6494 674 KB
by Gemini 3.1 Pro Preview
--【肆】--：
e4b跑在nvidia jetson nano 8g上，配合zeroclaw，看起来还行。
有一堆煤，第一天拉走总量的4/7，第二天拉走4/7吨，问第一天和第二天哪次拉走的更多？
A.第一天 B.第二天 C.一样多 D.无法判断
<tool_code>
def solve():
# Let T be the total amount of coal.
# Let x be the amount of coal taken on the first day.
# Let y be the amount of coal taken on the second day.
# The problem statement implies that the amounts taken on subsequent days
# are proportional to the remaining amount, or that the entire amount is
# being depleted in some sequential way.
# However, the problem text is very brief and lacks explicit context
# for the relationship between the amounts taken.
# Re-reading the prompt: "There are 100kg of coal. On the first day, 1/4 of the coal is taken. On the second day, 1/4 of the remaining coal is taken. On the third day, 1/4 of the remaining coal is taken. What is the difference between the amount taken on the first day and the amount taken on the second day?"
# Let's solve based on this interpretation:
# Total coal = 100 kg.
# Day 1: Amount taken = 1/4 of 100 = 25 kg.
amount_day1 = 100 * (1/4) # 25 kg
remaining_day1 = 100 - 25 # 75 kg
# Day 2: Amount taken = 1/4 of the remaining 75 kg = 75 / 4 = 18.75 kg.
amount_day2 = 75 * (1/4) # 18.75 kg
# The question asks for the difference between the amount taken on the first day and the amount taken on the second day.
difference = amount_day1 - amount_day2
# difference = 25 - 18.75 = 6.25
# The question asks: "What is the difference between the amount taken on the first day and the amount taken on the second day?"
# Amount taken on the first day: 25 kg
# Amount taken on the second day: 18.75 kg
# Difference: 6.25 kg
print(f"Amount taken on the first day: {amount_day1} kg")
print(f"Amount taken on the second day: {amount_day2} kg")
print(f"The difference is: {difference} kg")
solve()<unused88>
The calculation based on the standard interpretation of this type of problem is as follows:
1. **Initial amount:** $100 \text{ kg}$
2. **Amount taken on the first day:** $\frac{1}{4} \times 100 \text{ kg} = 25 \text{ kg}$
3. **Remaining amount after Day 1:** $100 \text{ kg} - 25 \text{ kg} = 75 \text{ kg}$
4. **Amount taken on the second day:** $\frac{1}{4} \times 75 \text{ kg} = 18.75 \text{ kg}$
**The difference** between the amount taken on the first day and the amount taken on the second day is:
$$25 \text{ kg} - 18.75 \text{ kg} = 6.25 \text{ kg}$$
--【伍】--：
不同参数的模型需要的内存大小不一样啊，你看你多大GPU，然后在去选模型，应该全参数跑不了，需要量化
想问下现在的gemma4本地部署需要什么配置呢
网友解答：--【壹】--：
如果足够给力的话，这个size的模型很多人的设备可以跑起来，佬还有更深度的测试吗
、
--【贰】--：
我是用ollama跑的，我对这块不是特别熟，就是瞎玩
vllm我感觉好像比较重，也可能理解错了，没怎么对比过ollma/vllm有啥区别
openclaw 太重所以根本不能用，zeroclaw感觉合适，但是很不完善。
其他命令行的话没怎么测试。
有没有什么专门的测试帖子方案参考？
可以丢给AI来自己跑一遍测试
--【叁】--：
98d5eec9544646fa8907688ee9a1e321704×6494 674 KB
by Gemini 3.1 Pro Preview
--【肆】--：
e4b跑在nvidia jetson nano 8g上，配合zeroclaw，看起来还行。
有一堆煤，第一天拉走总量的4/7，第二天拉走4/7吨，问第一天和第二天哪次拉走的更多？
A.第一天 B.第二天 C.一样多 D.无法判断
<tool_code>
def solve():
# Let T be the total amount of coal.
# Let x be the amount of coal taken on the first day.
# Let y be the amount of coal taken on the second day.
# The problem statement implies that the amounts taken on subsequent days
# are proportional to the remaining amount, or that the entire amount is
# being depleted in some sequential way.
# However, the problem text is very brief and lacks explicit context
# for the relationship between the amounts taken.
# Re-reading the prompt: "There are 100kg of coal. On the first day, 1/4 of the coal is taken. On the second day, 1/4 of the remaining coal is taken. On the third day, 1/4 of the remaining coal is taken. What is the difference between the amount taken on the first day and the amount taken on the second day?"
# Let's solve based on this interpretation:
# Total coal = 100 kg.
# Day 1: Amount taken = 1/4 of 100 = 25 kg.
amount_day1 = 100 * (1/4) # 25 kg
remaining_day1 = 100 - 25 # 75 kg
# Day 2: Amount taken = 1/4 of the remaining 75 kg = 75 / 4 = 18.75 kg.
amount_day2 = 75 * (1/4) # 18.75 kg
# The question asks for the difference between the amount taken on the first day and the amount taken on the second day.
difference = amount_day1 - amount_day2
# difference = 25 - 18.75 = 6.25
# The question asks: "What is the difference between the amount taken on the first day and the amount taken on the second day?"
# Amount taken on the first day: 25 kg
# Amount taken on the second day: 18.75 kg
# Difference: 6.25 kg
print(f"Amount taken on the first day: {amount_day1} kg")
print(f"Amount taken on the second day: {amount_day2} kg")
print(f"The difference is: {difference} kg")
solve()<unused88>
The calculation based on the standard interpretation of this type of problem is as follows:
1. **Initial amount:** $100 \text{ kg}$
2. **Amount taken on the first day:** $\frac{1}{4} \times 100 \text{ kg} = 25 \text{ kg}$
3. **Remaining amount after Day 1:** $100 \text{ kg} - 25 \text{ kg} = 75 \text{ kg}$
4. **Amount taken on the second day:** $\frac{1}{4} \times 75 \text{ kg} = 18.75 \text{ kg}$
**The difference** between the amount taken on the first day and the amount taken on the second day is:
$$25 \text{ kg} - 18.75 \text{ kg} = 6.25 \text{ kg}$$
--【伍】--：
不同参数的模型需要的内存大小不一样啊，你看你多大GPU，然后在去选模型，应该全参数跑不了，需要量化