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#!/usr/bin/python3
from pwn import *

def p64(n):
    return struct.pack("<Q", n)

def gen_xorbyte(byteoffset, value, bitwidth, selfstart, jumpthen):
    code = b""
    for i in range(bitwidth):
        if value & (1 << i):

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from pwn import *


PROMPT = b'> '
DONE = b'Done!\n'


if __name__ == '__main__':
    with process('./sice_supervisor') as tube:
    #with remote('mc.ax', 30283) as tube:

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diff --git a/sample.py b/sample.py
index 6ff0ea2..00daada 100644
--- a/sample.py
+++ b/sample.py
@@ -7,10 +7,11 @@ from contextlib import nullcontext
 import torch
 import tiktoken
 from model import GPTConfig, GPT
+import time
 

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import galois
from itertools import chain
import numpy as np


class SymbolicByte:
    def __init__(self, bit_vars):
        self.bit_vars = [
            self._xor_vars(vs) for vs in bit_vars
        ]

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#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <time.h>

#include <dlfcn.h>
#include <err.h>

#define NSYMS 1000000

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#include <stdio.h>
#include <stdint.h>
#include <math.h>

int32_t d2i(double);

int main(int argc, char* argv[]) {
    for (int sign = -1; sign <= 1; sign += 2) {
        for (int p = 0; p <= 7; ++p) {
            double d = (double)sign * pow(133.7, (double)p);

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.global _d2i
.align 4
_d2i:
    fcvtzs x0, d0
    ret

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#!/usr/bin/env python3

from Crypto.Util.number import *
import sys
from flag import flag

def diff(a, b):
	assert a.bit_length() == b.bit_length()
	w, l = 0, a.bit_length()
	for _ in range(l):

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When $a \ne 0$, there are two solutions to $(ax^2 + bx + c = 0)$ and they are $$ x = {-b \pm \sqrt{b^2-4ac} \over 2a} $$

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Example AMD output:

$ lscpu | grep 'Model name'
Model name:                      AMD Ryzen Threadripper 3970X 32-Core Processor
$ gcc -DFAST puzzle.S -o puzzle && ./puzzle
1000000000 iterations, 2758479060 cycles, 2.76 cycles/iteration
1651816810 106000000000
$ gcc -DSLOW puzzle.S -o puzzle && ./puzzle
1000000000 iterations, 6551777517 cycles, 6.55 cycles/iteration