llmpu 0.0.1

Large Language Model Process Unit

LLMPU

Large Language Model Process Unit

Get Started

pip install llmpu

import llmpu
llmpu.init({
    # "VR": 16, # visibile register number
    # "L": 10000, # hard character number limit for register
    # "model": "gemini-3.5-flash", # llm model
    "api_key": "your llm api key", # llm api key
    # "llm_config": {}, # llm config
})

Helper functions:

llmpu.write(100, "print('hello')") # truncate if exceed config["L"]
llmpu.read(100) # "print('hello')"
llmpu.call(100) # execute code in register 100

Main cycle:

llmpu.cycle()
# this will provide all visible registers to llm, and execute the code generated by llm

Minimum Example

# using default config
firmware = """
You are a process unit with many registers (this is register 0, or r0) with string values. Each register can store up to 5000 characters. Registers 0-15 (r0 to r15) are visible to you, and other registers are available but not presented to you. You can operate on registers using the following functions:
- `READ(r)` returns content in register number `r`
- `WRITE(r, content)` store string `content` in register number `r`. The content will be truncated to 5000 characters maximum.
- `CALL(r)` execute the content in register number `r` as Python code.

Conventional register purpose:
- r1: current task
"""
llmpu.write(0, firmware)
llmpu.write(1, "load r10001 to r1")
llmpu.write(10000, "do nothing")
llmpu.write(10001, "print hello to the screen, then call r10010")
llmpu.write(10010, "print('hello again')\nWRITE(1, READ(10000))")

llmpu.cycle() # r10001 will be loaded to r1
llmpu.cycle() # print "hello", then call r10010, which will print "hello again" and write r10000 to r1
llmpu.cycle() # pass