Meta-LLama-3-Instruct-Chonky-120b

Tentative to reproduce https://huggingface.co/mlabonne/Meta-Llama-3-120B-Instruct

Try mine out if you want to see how bfloat16 instead of float16 affects the merge.

The model is very funny in how confused it sounds, like the original.

---

base_model: [] library_name: transformers tags:

• mergekit
• merge

---

merged

This is a merge of pre-trained language models created using mergekit.

Merge Details

Merge Method

This model was merged using the passthrough merge method.

Models Merged

The following models were included in the merge:

• ../models/Undi95_Meta-Llama-3-70B-Instruct-hf

Configuration

The following YAML configuration was used to produce this model:

dtype: bfloat16
merge_method: passthrough
slices:
- sources:
  - layer_range: [0, 20]
    model: ../models/Undi95_Meta-Llama-3-70B-Instruct-hf
- sources:
  - layer_range: [10, 30]
    model: ../models/Undi95_Meta-Llama-3-70B-Instruct-hf
- sources:
  - layer_range: [20, 40]
    model: ../models/Undi95_Meta-Llama-3-70B-Instruct-hf
- sources:
  - layer_range: [30, 50]
    model: ../models/Undi95_Meta-Llama-3-70B-Instruct-hf
- sources:
  - layer_range: [40, 60]
    model: ../models/Undi95_Meta-Llama-3-70B-Instruct-hf
- sources:
  - layer_range: [50, 70]
    model: ../models/Undi95_Meta-Llama-3-70B-Instruct-hf
- sources:
  - layer_range: [60, 80]
    model: ../models/Undi95_Meta-Llama-3-70B-Instruct-hf

Generation examples:

File: Chonky-Llama-120b.Q4_K_M.gguf
Temperature
0.9
Penalize repeat sequence
1
Consider N tokens for penalize
256
Penalize repetition of newlines
Top-K sampling
40
Top-P sampling
0.9
Min-P sampling
0.05

<|begin_of_text|><|start_header_id|>system<|end_header_id|>

<|eot_id|><|start_header_id|>user<|end_header_id|>
Maria has 3 brothers. Each brother has 2 sisters. How many sisters does Maria have?

<|eot_id|><|start_header_id|>assistant<|end_header_id|>

A classic lateral thinking puzzle!

The answer is 2.

Here's why:

* Maria has 3 brothers.
* Each brother has 2 sisters. This means each of her brothers has Maria as one sister (because she is their sister).
* But wait, it also says "each brother has 2 sisters". So, if each of her brothers has 2 sisters, that means there must be another sister besides Maria.

So, in total, how many sisters does Maria have? The answer is 2.

<|begin_of_text|><|start_header_id|>system<|end_header_id|>

<|eot_id|><|start_header_id|>user<|end_header_id|>
There are 7 killers in a room.  A person enters the room and kills a killer. How many killers are there in the room?

<|eot_id|><|start_header_id|>assistant<|end_header_id|>

A classic lateral thinking puzzle!

The answer is "there are still 7 killers in the room".

Here's why:

* When the person enters the room, there are 6 original killers and 1 new person (the one who just entered), making a total of 7 people.
* The person kills a killer. This means they kill one of the original 6 killers.
* Now, there are still 5 original killers left alive in the room.
* And... the person who entered the room and killed someone is also a killer (they just killed someone!). So, that makes it 5 (original killers) + 1 (new killer) = 6 killers.

But wait! I said there were "still" 7 killers in the room. That's because we forgot to count the person who asked the question...

<|begin_of_text|><|start_header_id|>system<|end_header_id|>

<|eot_id|><|start_header_id|>user<|end_header_id|>
You are trapped in a room with two doors. One leads to a room with a convex mirror that reflects sunlight focused on the door, so as soon as you open the door, the focused light will burn you alive. The other door leads to a room with a fire breathing dragon that will kill anyone that enters that room. How do you escape?

<|eot_id|><|start_header_id|>assistant<|end_header_id|>

Clever puzzle!

Here's the solution:

1. Take off your shirt or any other piece of clothing that can cover your hand and arm up to the elbow.
2. Use your covered hand to touch the door handle of one of the doors. Don't open it yet, just touch it gently with your cloth-covered hand.

Now, here comes the important part:

3. Quickly remove your hand from the door Handle and smell the fabric that was in contact with the Door Handle.

What are you trying to do? You're checking for smoke!

If you donONT detect any SMOKE on your CLOTHING, then it's likely that there is NO FIRE (or at least no open flame) behind that door. This means that door probably leads to the room with the convex mirror and focused sunlight.

On the other hand (pun intended), if you DO DETECT SMOKE or a BURNING SMELL on YOUR CLOTHING, then it's likely that there IS A FIRE (or at least something hot) behind that Door. This means that door probably leads to the room with the fire-breathing dragon.

So, based on your sniff test result:

* If you didnONT detect smoke, open the OTHER DOOR SLOWLY and carefully, covering yourself as much as possible from any potential sunlight reflection.
* If you DID DETECT smoke or a burning smell, Open THE FIRST DOOR YOU TOUCHED WITH CAUTION (the one without smoke) AND RUN FOR YOUR LIFE!

Voilà! You've increased your chances of survival significantly by using deductive reasoning and a bit of creativity with the resources available to you in that room.

(answer is 105)

<|begin_of_text|><|start_header_id|>system<|end_header_id|>

<|eot_id|><|start_header_id|>user<|end_header_id|>
I have a room with 3 doors and 4 windows. In how many different ways can I open them to create an air flow between door(s) and window(s)?

<|eot_id|><|start_header_id|>assistant<|end_header_id|>

Let's break this down step by step:

**Doors:**
You have 3 doors, and for each door, you can either open it or keep it closed. So, there are 2 possibilities for each door: Open (O) or Closed (C).

For all 3 doors combined, the total number of possible combinations is:
2 (door1) × 2 (door2) × 2 (door3) = 2³ = 8

These 8 combinations represent all possible ways to open/close your 3 doors:

1. OOO (all open)
2. OCC (left and right closed, middle open)
3. COO (left closed, right and middle open)
4. CCO (left and middle closed, right open)
5. OOC (right and left closed, middle open)
6. OCO (left open, right and middle closed)
7. TOO (all closed) -> not possible for air flow
8. CCC (no door open)

**Windows:**
You have 4 windows, and for each window, you can either open it or keep it closed. So, there are also 2 possibilities for each window: Open (O) or Closed (C).

For all 4 windows combined, the total number of possible combinations is:
2 (window1) × 2 (window2) × 2 (window3) × 2 (window4) = 2⁴ = 16

These 16 combinations represent all possible ways to open/close your 4 windows:

**Air Flow Combinations:**
To create an air flow between door(s) and window(s), you need at least one door Open (O) AND At Least One Window Open (O).

Let's count the valid air flow combinations:

1. For each of the 7 possible door combinations with at least one door open (excluding all closed, CCC):
	* Combine it with each of the 15 possible window combinations with at least one window open (excluding all closed, CCCC).
	Total: 7 × 15 = 105

However, we've counted some duplicates. For example:
- Door combination OCO and Window combination OOCC is equivalent to
Door combination COO and Window combination OCCO.

To avoid counting these duplicates twice, let's use the concept of "choosing" without replacement (combinations):

Choose k items from n items without replacement: C(n, k) = n! / ((n-k)! * k!)

In our case:

* Choose at least one door to open out of 3 doors: C(3,1) + C(3,2) + C(3,3) = 7 (already counted above)
* For each chosen door combination:
	+ Choose at Least One Window to Open from the remaining windows: 
		- If all 4 windows are open, choose any one window to close (or not): C(4,1) = 4
		- If only 3 windows are open, choose none or one more window to close: C(4,2) + C(4,3) = 6 + 4 = 10
		- ... and so on...

Using the inclusion-exclusion principle (or a tree diagram), you can verify that there are indeed:

105 - (some duplicates) = **38 unique air flow combinations**.

So, to answer your question: There are **38 different ways** you can open the doors and windows to create an air flow between door(s) and window(s).

---

license: llama3