scrypt Hash (passlib format) Regex for Python

What this pattern does

This page provides a comprehensive, battle-tested regular expression for matching scrypt hash (passlib format), ported and verified for Python. In security-sensitive code, using an unverified regex can open the door to both false positives and denial-of-service attacks. The snippet below is ready to drop into your Python project — whether you're validating in a Django view, a FastAPI endpoint, or a standalone data processing script.

Python Implementation

# scrypt Hash (passlib format)
# ReDoS-safe | RegexVault — Security > Password Formats

import re

scrypt_hash_passlib_format_pattern = re.compile(r'^\$scrypt\$ln=([0-9]+),r=([0-9]+),p=([0-9]+)\$([A-Za-z0-9+/]{16,64}(?:={0,2}))\$([A-Za-z0-9+/]{30,64}(?:={0,1}))$')

def validate_scrypt_hash_passlib_format(value: str) -> bool:
    return bool(scrypt_hash_passlib_format_pattern.fullmatch(value))

# Example
print(validate_scrypt_hash_passlib_format("$scrypt$ln=14,r=8,p=1$aM15816PilIJep0o$nFNh2CVHVjNldFVKDHDlm0248RZvVaR8"))  # True

Test Cases

When to use this pattern

This pattern is drawn from the Security > Password Formats category and carries a ReDoS-safe certification. That matters for Python developers because particularly important in Python web servers where CPU-bound regex operations can stall concurrent request handling. RegexVault audits patterns against known backtracking attack vectors, ensuring you have the necessary context before using this regex in a high-stakes production environment.

Common Pitfalls

Technical Notes

Parameters: ln=log2(N) where N is the work factor, r=block size (8), p=parallelism (1). OWASP recommends N=32768 (ln=15), r=8, p=1 minimum. scrypt is memory-hard like Argon2, designed by Colin Percival. ln=14 = N=16384.