/* A pedagogical implementation of A5/1.
*
* Copyright (C) 1998-1999: Marc Briceno, Ian Goldberg, and David Wagner
*
* The source code below is optimized for instructional value and clarity.
* Performance will be terrible, but that's not the point.
* The algorithm is written in the C programming language to avoid ambiguities
* inherent to the English language. Complain to the 9th Circuit of Appeals
* if you have a problem with that.
*
* This software may be export-controlled by US law.
*
* This software is free for commercial and non-commercial use as long as
* the following conditions are aheared to.
* Copyright remains the authors' and as such any Copyright notices in
* the code are not to be removed.
* Redistribution and use in source and binary forms, with or without
* modification, are permitted provided that the following conditions
* are met:
*
* 1. Redistributions of source code must retain the copyright
* notice, this list of conditions and the following disclaimer.
* 2. Redistributions in binary form must reproduce the above copyright
* notice, this list of conditions and the following disclaimer in the
* documentation and/or other materials provided with the distribution.
*
* THIS SOFTWARE IS PROVIDED ``AS IS'' AND
* ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
* IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
* ARE DISCLAIMED. IN NO EVENT SHALL THE AUTHORS OR CONTRIBUTORS BE LIABLE
* FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
* DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS
* OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION)
* HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT
* LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY
* OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF
* SUCH DAMAGE.
*
* The license and distribution terms for any publicly available version or
* derivative of this code cannot be changed. i.e. this code cannot simply be
* copied and put under another distribution license
* [including the GNU Public License.]
*
* Background: The Global System for Mobile communications is the most widely
* deployed cellular telephony system in the world. GSM makes use of
* four core cryptographic algorithms, neither of which has been published by
* the GSM MOU. This failure to subject the algorithms to public review is all
* the more puzzling given that over 100 million GSM
* subscribers are expected to rely on the claimed security of the system.
*
* The four core GSM algorithms are:
* A3 authentication algorithm
* A5/1 "strong" over-the-air voice-privacy algorithm
* A5/2 "weak" over-the-air voice-privacy algorithm
* A8 voice-privacy key generation algorithm
*
* In April of 1998, our group showed that COMP128, the algorithm used by the
* overwhelming majority of GSM providers for both A3 and A8
* functionality was fatally flawed and allowed for cloning of GSM mobile
* phones.
* Furthermore, we demonstrated that all A8 implementations we could locate,
* including the few that did not use COMP128 for key generation, had been
* deliberately weakened by reducing the keyspace from 64 bits to 54 bits.
* The remaining 10 bits are simply set to zero!
*
* See <A HREF="http://www.scard.org/gsm">http://www.scard.org/gsm</A> for additional information.
*
* The question so far unanswered is if A5/1, the "stronger" of the two
* widely deployed voice-privacy algorithm is at least as strong as the
* key. Meaning: "Does A5/1 have a work factor of at least 54 bits"?
* Absent a publicly available A5/1 reference implementation, this question
* could not be answered. We hope that our reference implementation below,
* which has been verified against official A5/1 test vectors, will provide
* the cryptographic community with the base on which to construct the
* answer to this important question.
*
* Initial indications about the strength of A5/1 are not encouraging.
* A variant of A5, while not A5/1 itself, has been estimated to have a
* work factor of well below 54 bits. See <A
HREF="http://jya.com/crack-a5.htm">http://jya.com/crack-a5.htm</A> for
* background information and references.
*
* With COMP128 broken and A5/1 published below, we will now turn our attention
* to A5/2. The latter has been acknowledged by the GSM community to have
* been specifically designed by intelligence agencies for lack of security.
*
* We hope to publish A5/2 later this year.
*
* -- Marc Briceno <<A HREF="mailto:marc@scard.org">marc@scard.org</A>>
* Voice: +1 (925) 798-4042
*
*/
#include <stdio.h>
/* Masks for the three shift registers */
#define R1MASK 0x07FFFF /* 19 bits, numbered 0..18 */
#define R2MASK 0x3FFFFF /* 22 bits, numbered 0..21 */
#define R3MASK 0x7FFFFF /* 23 bits, numbered 0..22 */
/* Middle bit of each of the three shift registers, for clock control */
#define R1MID 0x000100 /* bit 8 */
#define R2MID 0x000400 /* bit 10 */
#define R3MID 0x000400 /* bit 10 */
/* Feedback taps, for clocking the shift registers.
* These correspond to the primitive polynomials
* x^19 + x^5 + x^2 + x + 1, x^22 + x + 1,
* and x^23 + x^15 + x^2 + x + 1. */
#define R1TAPS 0x072000 /* bits 18,17,16,13 */
#define R2TAPS 0x300000 /* bits 21,20 */
#define R3TAPS 0x700080 /* bits 22,21,20,7 */
/* Output taps, for output generation */
#define R1OUT 0x040000 /* bit 18 (the high bit) */
#define R2OUT 0x200000 /* bit 21 (the high bit) */
#define R3OUT 0x400000 /* bit 22 (the high bit) */
typedef unsigned char byte;
typedef unsigned long word;
typedef word bit;
/* Calculate the parity of a 32-bit word, i.e. the sum of its bits modulo 2 */
bit parity(word x) {
x ^= x>>16;
x ^= x>>8;
x ^= x>>4;
x ^= x>>2;
x ^= x>>1;
return x&1;
}
/* Clock one shift register */
word clockone(word reg, word mask, word taps) {
word t = reg & taps;
reg = (reg << 1) & mask;
reg |= parity(t);
return reg;
}
/* The three shift registers. They're in global variables to make the code
* easier to understand.
* A better implementation would not use global variables. */
word R1, R2, R3;
/* Look at the middle bits of R1,R2,R3, take a vote, and
* return the majority value of those 3 bits. */
bit majority() {
int sum;
sum = parity(R1&R1MID) + parity(R2&R2MID) + parity(R3&R3MID);
if (sum >= 2)
return 1;
else
return 0;
}
/* Clock two or three of R1,R2,R3, with clock control
* according to their middle bits.
* Specifically, we clock Ri whenever Ri's middle bit
* agrees with the majority value of the three middle bits.*/
void clock() {
bit maj = majority();
if (((R1&R1MID)!=0) == maj)
R1 = clockone(R1, R1MASK, R1TAPS);
if (((R2&R2MID)!=0) == maj)
R2 = clockone(R2, R2MASK, R2TAPS);
if (((R3&R3MID)!=0) == maj)
R3 = clockone(R3, R3MASK, R3TAPS);
}
/* Clock all three of R1,R2,R3, ignoring their middle bits.
* This is only used for key setup. */
void clockallthree() {
R1 = clockone(R1, R1MASK, R1TAPS);
R2 = clockone(R2, R2MASK, R2TAPS);
R3 = clockone(R3, R3MASK, R3TAPS);
}
/* Generate an output bit from the current state.
* You grab a bit from each register via the output generation taps;
* then you XOR the resulting three bits. */
bit getbit() {
return parity(R1&R1OUT)^parity(R2&R2OUT)^parity(R3&R3OUT);
}
/* Do the A5/1 key setup. This routine accepts a 64-bit key and
* a 22-bit frame number. */
void keysetup(byte key[8], word frame) {
int i;
bit keybit, framebit;
/* Zero out the shift registers. */
R1 = R2 = R3 = 0;
/* Load the key into the shift registers,
* LSB of first byte of key array first,
* clocking each register once for every
* key bit loaded. (The usual clock
* control rule is temporarily disabled.) */
for (i=0; i<64; i++) {
clockallthree(); /* always clock */
keybit = (key[i/8] >> (i&7)) & 1; /* The i-th bit of the
key */
R1 ^= keybit; R2 ^= keybit; R3 ^= keybit;
}
/* Load the frame number into the shift
* registers, LSB first,
* clocking each register once for every
* key bit loaded. (The usual clock
* control rule is still disabled.) */
for (i=0; i<22; i++) {
clockallthree(); /* always clock */
framebit = (frame >> i) & 1; /* The i-th bit of the frame #
*/
R1 ^= framebit; R2 ^= framebit; R3 ^= framebit;
}
/* Run the shift registers for 100 clocks
* to mix the keying material and frame number
* together with output generation disabled,
* so that there is sufficient avalanche.
* We re-enable the majority-based clock control
* rule from now on. */
for (i=0; i<100; i++) {
clock();
}
/* Now the key is properly set up. */
}
/* Generate output. We generate 228 bits of
* keystream output. The first 114 bits is for
* the A->B frame; the next 114 bits is for the
* B->A frame. You allocate a 15-byte buffer
* for each direction, and this function fills
* it in. */
void run(byte AtoBkeystream[], byte BtoAkeystream[]) {
int i;
/* Zero out the output buffers. */
for (i=0; i<=113/8; i++)
AtoBkeystream[i] = BtoAkeystream[i] = 0;
/* Generate 114 bits of keystream for the
* A->B direction. Store it, MSB first. */
for (i=0; i<114; i++) {
clock();
AtoBkeystream[i/8] |= getbit() << (7-(i&7));
}
/* Generate 114 bits of keystream for the
* B->A direction. Store it, MSB first. */
for (i=0; i<114; i++) {
clock();
BtoAkeystream[i/8] |= getbit() << (7-(i&7));
}
}
/* Test the code by comparing it against
* a known-good test vector. */
void test() {
byte key[8] = {0x12, 0x23, 0x45, 0x67, 0x89, 0xAB, 0xCD, 0xEF};
word frame = 0x134;
byte goodAtoB[15] = { 0x53, 0x4E, 0xAA, 0x58, 0x2F, 0xE8, 0x15,
0x1A, 0xB6, 0xE1, 0x85, 0x5A, 0x72, 0x8C, 0x00 };
byte goodBtoA[15] = { 0x24, 0xFD, 0x35, 0xA3, 0x5D, 0x5F, 0xB6,
0x52, 0x6D, 0x32, 0xF9, 0x06, 0xDF, 0x1A, 0xC0 };
byte AtoB[15], BtoA[15];
int i, failed=0;
keysetup(key, frame);
run(AtoB, BtoA);
/* Compare against the test vector. */
for (i=0; i<15; i++)
if (AtoB[i] != goodAtoB[i])
failed = 1;
for (i=0; i<15; i++)
if (BtoA[i] != goodBtoA[i])
failed = 1;
/* Print some debugging output. */
printf("key: 0x");
for (i=0; i<8; i++)
printf("%02X", key[i]);
printf("\n");
printf("frame number: 0x%06X\n", (unsigned int)frame);
printf("known good output:\n");
printf(" A->B: 0x");
for (i=0; i<15; i++)
printf("%02X", goodAtoB[i]);
printf(" B->A: 0x");
for (i=0; i<15; i++)
printf("%02X", goodBtoA[i]);
printf("\n");
printf("observed output:\n");
printf(" A->B: 0x");
for (i=0; i<15; i++)
printf("%02X", AtoB[i]);
printf(" B->A: 0x");
for (i=0; i<15; i++)
printf("%02X", BtoA[i]);
printf("\n");
if (!failed) {
printf("Self-check succeeded: everything looks ok.\n");
return;
} else {
/* Problems! The test vectors didn't compare*/
printf("\nI don't know why this broke; contact the authors.\n");
exit(1);
}
}
int main(void) {
test();
return 0;
}
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