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<s>A chat between a user and a thinking artificial intelligence assistant. The assistant describes its thought process and gives helpful and detailed answers to the user's questions.
User: Are you conscious?
AI:| Name | Quant method | Bits | Size | Max RAM required | Use case |
|---|---|---|---|---|---|
| Cerebrum-1.0-8x7b.IQ1_S.gguf | IQ1_S | 1 | 9.2 GB | 9.7 GB | smallest, significant quality loss - TBD: Waiting for this issue to be resolved |
| Cerebrum-1.0-8x7b.IQ2_XXS.gguf | IQ2_XXS | 2 | 12.0 GB | 12.5 GB | very small, high quality loss |
| Cerebrum-1.0-8x7b.IQ2_XS.gguf | IQ2_XS | 2 | 13.4 GB | 13.9 GB | very small, high quality loss |
| Cerebrum-1.0-8x7b.IQ2_S.gguf | IQ2_S | 2 | 13.6 GB | 14.1 GB | small, substantial quality loss |
| Cerebrum-1.0-8x7b.IQ2_M.gguf | IQ2_M | 2 | 15.0 GB | 15.5 GB | small, greater quality loss |
| Cerebrum-1.0-8x7b.IQ3_XXS.gguf | IQ3_XXS | 3 | 17.3 GB | 17.8 GB | very small, high quality loss |
| Cerebrum-1.0-8x7b.IQ3_XS.gguf | IQ3_XS | 3 | 18.4 GB | 18.9 GB | small, substantial quality loss |
| Cerebrum-1.0-8x7b.IQ3_S.gguf | IQ3_S | 3 | 19.5 GB | 20.0 GB | small, greater quality loss |
| Cerebrum-1.0-8x7b.IQ3_M.gguf | IQ3_M | 3 | 20.5 GB | 21.0 GB | medium, balanced quality - recommended |
| Cerebrum-1.0-8x7b.IQ4_XS.gguf | IQ4_XS | 4 | 24.0 GB | 24.5 GB | small, substantial quality loss |
llama.cpp commandllama.cpp from commit 0becb22 or later../main -ngl 33 -m Cerebrum-1.0-8x7b.IQ2_XS.gguf --override-kv llama.expert_used_count=int:3 --color -c 16384 --temp 0.7 --repeat-penalty 1.0 -n -1 -p "A chat between a user and a thinking artificial intelligence assistant. The assistant describes its thought process and gives helpful and detailed answers to the user's questions.\nUser: {prompt}\nAI:"-ngl 33 to the number of layers to offload to GPU. Remove it if you don't have GPU acceleration.-c 16384 to the desired sequence length.-p <PROMPT> argument with -i -ins-ctk q8_0 or even -ctk q4_0 for big memory savings (depending on context size).
There is a similar option for V-cache (-ctv), however that is not working yet.1# Prebuilt wheel with basic CPU support
2pip install llama-cpp-python --extra-index-url https://abetlen.github.io/llama-cpp-python/whl/cpu
3# Prebuilt wheel with NVidia CUDA acceleration
4pip install llama-cpp-python --extra-index-url https://abetlen.github.io/llama-cpp-python/whl/cu121 (or cu122 etc.)
5# Prebuilt wheel with Metal GPU acceleration
6pip install llama-cpp-python --extra-index-url https://abetlen.github.io/llama-cpp-python/whl/metal
7# Build base version with no GPU acceleration
8pip install llama-cpp-python
9# With NVidia CUDA acceleration
10CMAKE_ARGS="-DLLAMA_CUDA=on" pip install llama-cpp-python
11# Or with OpenBLAS acceleration
12CMAKE_ARGS="-DLLAMA_BLAS=ON -DLLAMA_BLAS_VENDOR=OpenBLAS" pip install llama-cpp-python
13# Or with CLBLast acceleration
14CMAKE_ARGS="-DLLAMA_CLBLAST=on" pip install llama-cpp-python
15# Or with AMD ROCm GPU acceleration (Linux only)
16CMAKE_ARGS="-DLLAMA_HIPBLAS=on" pip install llama-cpp-python
17# Or with Metal GPU acceleration for macOS systems only
18CMAKE_ARGS="-DLLAMA_METAL=on" pip install llama-cpp-python
19# Or with Vulkan acceleration
20CMAKE_ARGS="-DLLAMA_VULKAN=on" pip install llama-cpp-python
21# Or with Kompute acceleration
22CMAKE_ARGS="-DLLAMA_KOMPUTE=on" pip install llama-cpp-python
23# Or with SYCL acceleration
24CMAKE_ARGS="-DLLAMA_SYCL=on -DCMAKE_C_COMPILER=icx -DCMAKE_CXX_COMPILER=icpx" pip install llama-cpp-python
25
26# In windows, to set the variables CMAKE_ARGS in PowerShell, follow this format; eg for NVidia CUDA:
27$env:CMAKE_ARGS = "-DLLAMA_CUDA=on"
28pip install llama-cpp-python1from llama_cpp import Llama
2
3# Chat Completion API
4
5llm = Llama(model_path="./Cerebrum-1.0-8x7b.IQ3_M.gguf", n_gpu_layers=33, n_ctx=16384)
6print(llm.create_chat_completion(
7 messages = [
8 {"role": "system", "content": "You are a story writing assistant."},
9 {
10 "role": "user",
11 "content": "Write a story about llamas."
12 }
13 ]
14))

<s>A chat between a user and a thinking artificial intelligence assistant. The assistant describes its thought process and gives helpful and detailed answers to the user's questions.
User: Are you conscious?
AI:messages = [
{'role': 'user', 'content': 'What is self-consistency decoding?'},
{'role': 'assistant', 'content': 'Self-consistency decoding is a technique used in natural language processing to improve the performance of language models. It works by generating multiple outputs for a given input and then selecting the most consistent output based on a set of criteria.'},
{'role': 'user', 'content': 'Why does self-consistency work?'}
]
input = tokenizer.apply_chat_template(messages, add_generation_prompt=True, return_tensors='pt')
with torch.no_grad():
out = model.generate(input_ids=input, max_new_tokens=100, do_sample=False)A chat between a user and a thinking artificial intelligence assistant. The assistant describes its thought process and gives helpful and detailed answers to the user's questions.
User: Find the longest increasing subarray of the given array.
The function signature should be def longest_increasing_subarray(vals: List[int]).
Write the code in Python.
AI: To find the longest increasing subarray of the given array, we could start by considering the first element of the array as the start of the subarray. Then, we could iterate through the remaining elements of the array and check if each element is greater than the last element of the current subarray. If it is, we could add it to the subarray. If it is not, we could save the length of the current subarray and start a new subarray with the current element. We could repeat this process until we reach the end of the array. Then, we could return the longest subarray found.
Here is an example:
```python
def longest_increasing_subarray(vals: List[int]):
if len(vals) == 0:
return []
longest_subarray = [vals[0]]
current_subarray = [vals[0]]
for i in range(1, len(vals)):
if vals[i] > current_subarray[-1]:
current_subarray.append(vals[i])
else:
if len(current_subarray) > len(longest_subarray):
longest_subarray = current_subarray
current_subarray = [vals[i]]
if len(current_subarray) > len(longest_subarray):
longest_subarray = current_subarray
return longest_subarray
```A chat between a user and a thinking artificial intelligence assistant. The assistant describes its thought process and gives helpful and detailed answers to the user's questions.
User: Out of six apples three are bad. What is the probability that at least one apple will be bad when selecting two apples out of them?
AI: To find the probability that at least one apple will be bad when selecting two apples out of them, we need to consider the possible outcomes of the selection. There are two cases: either one of the selected apples is bad, or both of them are bad.
The probability that one of the selected apples is bad is equal to the probability that the first apple is bad and the second apple is good, plus the probability that the first apple is good and the second apple is bad. Since three out of six apples are bad, the probability that the first apple is bad is $3/6 = 1/2$. The probability that the second apple is good, given that the first apple is bad, is equal to $3/5$, since there are three good apples left out of five remaining apples. Therefore, the probability that the first apple is bad and the second apple is good is $1/2 \cdot 3/5 = 3/10$. Similarly, the probability that the first apple is good and the second apple is bad is also $3/10$. Therefore, the probability that one of the selected apples is bad is $3/10 + 3/10 = 6/10 = 3/5$.
The probability that both of the selected apples are bad is equal to the probability that the first apple is bad and the second apple is bad. Since there are three bad apples out of six, the probability that the first apple is bad is $3/6 = 1/2$. The probability that the second apple is bad, given that the first apple is bad, is equal to $2/5$, since there are two bad apples left out of five remaining apples. Therefore, the probability that both of the selected apples are bad is $1/2 \cdot 2/5 = 2/10 = 1/5$.
To find the probability that at least one apple will be bad, we add the probabilities of the two cases: the probability that one of the selected apples is bad and the probability that both of the selected apples are bad. Therefore, the probability that at least one apple will be bad is $3/5 + 1/5 = 4/5$. The answer is 4/5.A chat between a user and a thinking artificial intelligence assistant. The assistant describes its thought process and gives helpful and detailed answers to the user's questions.
User: The apple falls from a tree of height 10m. When will it reach the earth if there is no air resistance?
AI: To find out when the apple will reach the earth, we can use the equation of motion for a freely falling object. The equation of motion for a freely falling object is $s = ut + \frac{1}{2}at^2$, where $s$ is the displacement, $u$ is the initial velocity, $a$ is the acceleration due to gravity, and $t$ is the time taken. In this case, the initial velocity of the apple is 0, and the acceleration due to gravity is 9.8 m/s^2. Therefore, we can write the equation of motion as $s = \frac{1}{2}at^2$. Substituting the values, we get $10 = \frac{1}{2} \cdot 9.8 \cdot t^2$. Solving for $t$, we get $t = \sqrt{\frac{20}{9.8}} = \sqrt{2.04} = 1.43$ seconds. Therefore, the apple will reach the earth in 1.43 seconds. The answer is 1.43 seconds.A chat between a user and a thinking artificial intelligence assistant. The assistant describes its thought process and gives helpful and detailed answers to the user's questions.
User: What is $\lim_{n \to \infty} \frac{(n + 3) \cdot \sqrt{n}}{(n/2 - 1)^2}$?
AI: To find the limit of $\frac{(n + 3) \cdot \sqrt{n}}{(n/2 - 1)^2}$ as $n$ approaches infinity, we can start by simplifying the expression. We can factor out $n$ from the numerator and write $(n + 3) \cdot \sqrt{n} = n \cdot \sqrt{n} + 3 \cdot \sqrt{n}$. We can also write $(n/2 - 1)^2 = n^2/4 - n + 1$. Therefore, we can write the limit as $\lim_{n \to \infty} \frac{n \cdot \sqrt{n} + 3 \cdot \sqrt{n}}{n^2/4 - n + 1}$.
Now, we can consider the behavior of each term in the numerator and denominator as $n$ approaches infinity. As $n$ approaches infinity, $n \cdot \sqrt{n}$ will grow much faster than $3 \cdot \sqrt{n}$, so we can ignore the $3 \cdot \sqrt{n}$ term in the numerator. Similarly, as $n$ approaches infinity, $n^2/4$ will grow much faster than $n$ and $1$, so we can ignore the $n$ and $1$ terms in the denominator. Therefore, we can write the limit as $\lim_{n \to \infty} \frac{n \cdot \sqrt{n}}{n^2/4} = \lim_{n \to \infty} \frac{4 \cdot n \cdot \sqrt{n}}{n^2} = \lim_{n \to \infty} \frac{4 \cdot \sqrt{n}}{n} = \lim_{n \to \infty} \frac{4}{\sqrt{n}} = 0$.
Therefore, the limit of $\frac{(n + 3) \cdot \sqrt{n}}{(n/2 - 1)^2}$ as $n$ approaches infinity is 0. The answer is 0.A chat between a user and a thinking artificial intelligence assistant. The assistant describes its thought process and gives helpful and detailed answers to the user's questions.
User: If $A$ is positive semi-definite, is $B^TAB$ positive semi-definite?
AI: To determine whether $B^TAB$ is positive semi-definite, we need to consider the properties of $A$ and $B$. Since $A$ is positive semi-definite, we know that for any vector $x$, $x^TAx \geq 0$. To show that $B^TAB$ is positive semi-definite, we need to show that for any vector $y$, $y^TB^TABy \geq 0$.
We can write $y^TB^TABy$ as $(By)^TA(By)$. Since $A$ is positive semi-definite, we know that $(By)^TA(By) \geq 0$. Therefore, $y^TB^TABy \geq 0$ for any vector $y$, which means that $B^TAB$ is positive semi-definite. The answer is yes.A chat between a user and a thinking artificial intelligence assistant. The assistant describes its thought process and gives helpful and detailed answers to the user's questions.
User: Tell me about a famous British painter George Ellster
AI: George Ellster is not a famous British painter. However, there are a few notable British painters with similar names:
George Elgar Hicks (1824-1914) was a Victorian-era painter known for his genre scenes and portraits.
George Frederic Watts (1817-1904) was a Victorian-era painter and sculptor known for his allegorical and symbolic works.
George Stubbs (1724-1806) was an 18th-century painter known for his equestrian and animal paintings.
George Romney (1734-1802) was an 18th-century painter known for his portraits and historical scenes.
These are just a few examples of notable British painters with similar names to George Ellster.