Question

In need of CRC64 or better algorithm

Asked by: efortier

Hi all.

I need a CRC64 or better algorithm. I don't want real fancy stuff, just the standard "supply a string of characters or bytes, and get a CRC64 value" type of algorithm. I'm good in math, so I can't figure it out by myself.

Also, I have the "crc tester" by math dillon, however this is not what I want. I want something simpler. CRC128 or 256 would be really great.

I also don't want pseudo code, as I said I'm not good in math, and I can't figure out the polynomial stuff in the CRC. This is why I'm offering 500 points for it.

Thanks!

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Asked On
2003-04-18 at 19:56:59ID20589821
Tags

crc64

Topic

Miscellaneous Programming

Participating Experts
3
Points
500
Comments
10

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Answers

 

by: ronaldaungPosted on 2003-04-19 at 03:16:07ID: 8357997

Maybe you could try this ?
CRC128 is just a pair of CRC64 .. and same for CRC256.

static char base32[] = "abcdefghijklmnopqrstuvwxyz012345"
                 "abcdefghijklmnopqrstuvwxyz012345";
static char base64[] = "abcdefghijklmnopqrstuvwxyz0123456789"
                 "ABCDEFGHIJKLMNOPQRSTUVWXYZ+-";

int base_encode(const void *data, int size, char **str, char *base)
{
  static char tmpbuf[32];
  char *p;
  char *s;
  int i;
  int c;
  unsigned char *q;

  p = s = tmpbuf;
  q = (unsigned char*)data;
  i=0;
  for(i = 0; i < size;){
    c=q[i++];
    c*=256;
    if(i < size)
      c+=q[i];
    i++;
    c*=256;
    if(i < size)
      c+=q[i];
    i++;
    p[0]=base[(c&0x00fc0000) >> 18];
    p[1]=base[(c&0x0003f000) >> 12];
    p[2]=base[(c&0x00000fc0) >> 6];
    p[3]=base[(c&0x0000003f) >> 0];
    if(i > size)
      p[3]='=';
    if(i > size+1)
      p[2]='=';
    p+=4;
  }
  *p=0;
  while (*--p == '=')
     *p = 0;
  *str = s;
  return strlen(s);
}

 

by: efortierPosted on 2003-04-19 at 08:44:54ID: 8358726

Hi,

Isn't this a base64 encoding instead of a CRC computation? I need a standard crc64 code, preferably with a #define for the polynomial so that I can generate a CRC64 to crosscheck with database records.

The database includes a CRC32 field, and my job is to change that to a better CRC. This is why I need this CRC to be standard.

Thanks.

 

by: ozoPosted on 2003-04-19 at 11:03:48ID: 8359880

#include <stdio.h>
typedef unsigned int hint_t; /* we want a 32 bit unsigned integer here */
typedef struct hash_t {
 hint_t h1; hint_t h2;
} hash_t;
#define TESTHSIZE (4 * 1024 * 1024)
#define TESTHMASK (TESTHSIZE - 1)
void inithash(void);
hash_t testhash(const char *p);
int UniqueOpt;
int HashLimit = 64;
int VerboseOpt = 0;
 int QuietOpt = 0;
void *rmalloc(int bytes);
int main(int ac, char **av){
    char buf[256];
    inithash();
    while ( fgets(buf, sizeof(buf), stdin) != NULL ) {
        hash_t hv;
        hv = testhash(buf);
        printf("%08x.%08x %s", hv.h1, hv.h2, buf);
    }
    return(0);
}
/* * Poly: 0x00600340.00F0D50A * */
#define HINIT1 0xFAC432B1UL
#define HINIT2 0x0CD5E44AUL
#define POLY1 0x00600340UL
#define POLY2 0x00F0D50BUL
hash_t CrcXor[256];
hash_t Poly;
void inithash(void) {
    int i;
    Poly.h1 = POLY1;
    Poly.h2 = POLY2;
    for (i = 0; i < 256; ++i) {
        int j;
        int v = i;
        hash_t hv = { 0, 0 };
        for (j = 0; j < 8; ++j, (v <<= 1)) {
            hv.h1 <<= 1;
            if (hv.h2 & 0x80000000UL) hv.h1 |= 1;
            hv.h2 = (hv.h2 << 1);
            if (v & 0x80) {
                hv.h1 ^= Poly.h1;
                hv.h2 ^= Poly.h2;
            }
        }
        CrcXor[i] = hv;
    }
}

/*
 * testhash() - do the CRC. We extract the top 8 bits to
 * use as a table lookup to obtain the polynomial XOR 8 bits at
 * a time rather then 1 bit at a time.
 */
hash_t testhash(const char *p) {
    hash_t hv = { HINIT1, HINIT2 };
    int s = HashLimit - 40;
    hint_t m = (hint_t)-1 >> (64 - HashLimit);
    hv.h1 &= m;
    while (*p) {
        int i = (hv.h1 >> s) & 255;
        hv.h1= ((hv.h1 << 8) & m) ^ (int)(hv.h2 >> 24) ^ CrcXor[i].h1;
        hv.h2 = (hv.h2 << 8) ^ *p ^ CrcXor[i].h2;
        ++p;
     }
    /* printf("%08lx.%08lx\n", (long)hv.h1, (long)hv.h2); */
    return(hv);
}

 

by: efortierPosted on 2003-04-20 at 00:11:37ID: 8361823

The above is the trimmed down code from Math Dillon's "crc tester".

The above code simply takes Dillon's source, and removes the support for CRC with less than 64 bits. IE: This is why "VerboseOpt" is defined but not used. The functions were trimmed down.

Sorry...

 

by: ozoPosted on 2003-04-20 at 01:44:41ID: 8361976

What do you need that's not in Math Dillon's "crc tester"?

 

by: efortierPosted on 2003-04-21 at 15:03:00ID: 8369635

I was told that the Dillon's function was an "hash" function, and I had to find a standard CRC.

I don't know enough about hasing and CRC to know what this means. This is why I'm looking for another one.

--Eric

 

by: shaicPosted on 2003-05-26 at 07:58:45ID: 8584599

CRC32 is an old 32bit hash algorithm. There's no standard CRC64/CRC128/CRC256 algorithms.

The MD5 (128bit) algorithm is what you want, and is the most popular CRC32 replacement.
http://userpages.umbc.edu/~mabzug1/cs/md5/md5.html

SHA-1 (160bit) is slightly less popular, but a more cryptographically secure hashing algorithm.
http://www.cr0.net:8040/code/crypto/sha1.php

SHA-2 can support 256bit and more, but it's almost certainly an overkill.

 

by: efortierPosted on 2003-05-26 at 11:12:08ID: 8585541

Wow, this is exactly what I needed! Better late than never, thanks!

Can I bother you with another question? What is the difference between a block cypher and a hash algorithm?

I have no clue as to what cypher means, and what hash do.

Thanks.

 

by: shaicPosted on 2003-05-26 at 11:46:59ID: 8585680

A block cipher: A cipher algorithm that encrypts data in discrete units (called blocks), rather than as a continuous stream of bits.
Hash: A fixed-size result obtained by applying a mathematical function (the hashing algorithm) to an arbitrary amount of data.

A block cipher can be decrypted with the right key/password.
A hash is a one-way function.

There are many cryptography tutorials on the web.
See http://www.hack.gr/users/dij/crypto/overview/    
and http://www.securitypointer.com/encryption.htm

 

by: efortierPosted on 2003-05-26 at 14:52:06ID: 8586398

Thanks for the useful infos and links.

--Eric

20120131-EE-VQP-002

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