set3, challenge 24 completed

1 files changed, 48 insertions, 0 deletions

diff --git a/set3/task24.c b/set3/task24.c
new file mode 100644
index 0000000..e355d08
--- /dev/null
+++ b/set3/task24.c
@@ -0,0 +1,48 @@
+#include "../lib/lib.h"
+#include "../lib/lib2.h"
+#include "../lib/lib3.h"
+#include <time.h>
+
+/**
+  * One should not restore the internal state of the MT. Given 624 bytes of
+  * input this would be straight forward.
+  * Instead one should restore more or less the first state (the seed from
+  * which this states arrived). To restore a previous state of the MT is
+  * possible. You have to go so far back, how long your ciphertext is and 
+  * how much states you would need to encrpyt it.
+  * 
+  * Since it is a 16 bit seed, one can also brute force it with 2^16
+  * possible values...within seconds! exhautive search Yeah!
+  * Theoritcal we only need 2^(16/2) values because of birthday paradox
+  *
+  **/
+
+int main()
+{
+	srand(time(NULL));
+	// try to decrypt
+	char plaintext[] = "Hallo du da wie geht es dir Knallkopp";
+	char *ciphertext = malloc(strlen(plaintext));
+
+	int length_ciphertext = mt_19937_stream_cipher_oracle(plaintext,
+			strlen(plaintext), ciphertext);
+
+	char *restore_pl = malloc(length_ciphertext);
+	char *hex_ciphertext = malloc(length_ciphertext*2+1);
+	hex_binary_to_string(ciphertext, hex_ciphertext, length_ciphertext);
+	printf("ciphertext: %s\n", hex_ciphertext);
+	//mt_19937_stream_cipher(ciphertext, length_ciphertext
+	// decrypt it
+	crack_mt_19937_stream_cipher_16_bit_seed(ciphertext, length_ciphertext,
+			restore_pl, plaintext);
+
+	printf("plaintext: %s\n", restore_pl);
+
+	// crack a MT time based password token
+	// well do it agian with brute force
+	unsigned int password_token = mt_19937_password_token();
+	int is_time_based = mt_19937_password_token_time_based(password_token, 1000);
+
+	printf("password token is time based %i\n", is_time_based);
+
+}