### A5 / 2

Where can I find realization of A5 / 2. All I have is below. But as I understand it is A5/1:

 ``123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100101102103104105106107108109110111112113114115116117118119120121122123124125126127128129130131132133134135136137138139140141142143144145146147148149150151152153154155156157158159160161162163164165166167168169170171172173174175176177178179180181182183184185186187188189190191192193194195196197198199200201202203204205206207208209210211212213214215216217218219220221222223224225226227228229`` ``````#include /* 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) { //Call Test now.... test(); getche(); return 0; //Comment after return }``````
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https://cryptome.org/gsm-a512.htm

It doesn't appear to have an auspicious history:
https://en.wikipedia.org/wiki/A5/2
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