Is there a way of encryption that can take a string of any length and produce a sub-10-character hash? I want to produce reasonably unique ID's but based on message contents, rather than randomly.
I can live with constraining the messages to integer values, though, if arbitrary-length strings are impossible. However, the hash must not be similar for two consecutive integers, in that case.
You can use any commonly available hash algorithm (eg. SHA-1), which will give you a slightly longer result than what you need. Simply truncate the result to the desired length, which may be good enough.
For example, in Python:
>>> import hashlib
>>> hash = hashlib.sha1("my message".encode("UTF-8")).hexdigest()
>>> hash
'104ab42f1193c336aa2cf08a2c946d5c6fd0fcdb'
>>> hash[:10]
'104ab42f11'
If you don't need an algorithm that's strong against intentional modification, I've found an algorithm called adler32 that produces pretty short (~8 character) results. Choose it from the dropdown here to try it out:
You need to hash the contents to come up with a digest. There are many hashes available but 10-characters is pretty small for the result set. Way back, people used CRC-32, which produces a 33-bit hash (basically 4 characters plus one bit). There is also CRC-64 which produces a 65-bit hash. MD5, which produces a 128-bit hash (16 bytes/characters) is considered broken for cryptographic purposes because two messages can be found which have the same hash. It should go without saying that any time you create a 16-byte digest out of an arbitrary length message you're going to end up with duplicates. The shorter the digest, the greater the risk of collisions.
However, your concern that the hash not be similar for two consecutive messages (whether integers or not) should be true with all hashes. Even a single bit change in the original message should produce a vastly different resulting digest.
So, using something like CRC-64 (and base-64'ing the result) should get you in the neighborhood you're looking for.
Just summarizing an answer that was helpful to me (noting @erasmospunk's comment about using base-64 encoding). My goal was to have a short string that was mostly unique...
I'm no expert, so please correct this if it has any glaring errors (in Python again like the accepted answer):
import base64
import hashlib
import uuid
unique_id = uuid.uuid4()
# unique_id = UUID('8da617a7-0bd6-4cce-ae49-5d31f2a5a35f')
hash = hashlib.sha1(str(unique_id).encode("UTF-8"))
# hash.hexdigest() = '882efb0f24a03938e5898aa6b69df2038a2c3f0e'
result = base64.b64encode(hash.digest())
# result = b'iC77DySgOTjliYqmtp3yA4osPw4='
The result here is using more than just hex characters (what you'd get if you used hash.hexdigest()) so it's less likely to have a collision (that is, should be safer to truncate than a hex digest).
Note: Using UUID4 (random). See http://en.wikipedia.org/wiki/Universally_unique_identifier for the other types.
If you need "sub-10-character hash"
you could use Fletcher-32 algorithm which produces 8 character hash (32 bits), CRC-32 or Adler-32.
CRC-32 is slower than Adler32 by a factor of 20% - 100%.
Fletcher-32 is slightly more reliable than Adler-32. It has a lower computational cost than the Adler checksum: Fletcher vs Adler comparison.
A sample program with a few Fletcher implementations is given below:
#include <stdio.h>
#include <string.h>
#include <stdint.h> // for uint32_t
uint32_t fletcher32_1(const uint16_t *data, size_t len)
{
uint32_t c0, c1;
unsigned int i;
for (c0 = c1 = 0; len >= 360; len -= 360) {
for (i = 0; i < 360; ++i) {
c0 = c0 + *data++;
c1 = c1 + c0;
}
c0 = c0 % 65535;
c1 = c1 % 65535;
}
for (i = 0; i < len; ++i) {
c0 = c0 + *data++;
c1 = c1 + c0;
}
c0 = c0 % 65535;
c1 = c1 % 65535;
return (c1 << 16 | c0);
}
uint32_t fletcher32_2(const uint16_t *data, size_t l)
{
uint32_t sum1 = 0xffff, sum2 = 0xffff;
while (l) {
unsigned tlen = l > 359 ? 359 : l;
l -= tlen;
do {
sum2 += sum1 += *data++;
} while (--tlen);
sum1 = (sum1 & 0xffff) + (sum1 >> 16);
sum2 = (sum2 & 0xffff) + (sum2 >> 16);
}
/* Second reduction step to reduce sums to 16 bits */
sum1 = (sum1 & 0xffff) + (sum1 >> 16);
sum2 = (sum2 & 0xffff) + (sum2 >> 16);
return (sum2 << 16) | sum1;
}
int main()
{
char *str1 = "abcde";
char *str2 = "abcdef";
size_t len1 = (strlen(str1)+1) / 2; // '\0' will be used for padding
size_t len2 = (strlen(str2)+1) / 2; //
uint32_t f1 = fletcher32_1(str1, len1);
uint32_t f2 = fletcher32_2(str1, len1);
printf("%u %X \n", f1,f1);
printf("%u %X \n\n", f2,f2);
f1 = fletcher32_1(str2, len2);
f2 = fletcher32_2(str2, len2);
printf("%u %X \n",f1,f1);
printf("%u %X \n",f2,f2);
return 0;
}
Output:
4031760169 F04FC729
4031760169 F04FC729
1448095018 56502D2A
1448095018 56502D2A
Agrees with Test vectors:
"abcde" -> 4031760169 (0xF04FC729)
"abcdef" -> 1448095018 (0x56502D2A)
Adler-32 has a weakness for short messages with few hundred bytes, because the checksums for these messages have a poor coverage of the 32 available bits. Check this:
The Adler32 algorithm is not complex enough to compete with comparable checksums.
You can use the hashlib library for Python. The shake_128 and shake_256 algorithms provide variable length hashes. Here's some working code (Python3):
import hashlib
>>> my_string = 'hello shake'
>>> hashlib.shake_256(my_string.encode()).hexdigest(5)
'34177f6a0a'
Notice that with a length parameter x (5 in example) the function returns a hash value of length 2x.
You could use an existing hash algorithm that produces something short, like MD5 (128 bits) or SHA1 (160). Then you can shorten that further by XORing sections of the digest with other sections. This will increase the chance of collisions, but not as bad as simply truncating the digest.
Also, you could include the length of the original data as part of the result to make it more unique. For example, XORing the first half of an MD5 digest with the second half would result in 64 bits. Add 32 bits for the length of the data (or lower if you know that length will always fit into fewer bits). That would result in a 96-bit (12-byte) result that you could then turn into a 24-character hex string. Alternately, you could use base 64 encoding to make it even shorter.
Simply run this in a terminal (on MacOS or Linux):
crc32 <(echo "some string")
8 characters long.
I needed something along the lines of a simple string reduction function recently. Basically, the code looked something like this (C/C++ code ahead):
size_t ReduceString(char *Dest, size_t DestSize, const char *Src, size_t SrcSize, bool Normalize)
{
size_t x, x2 = 0, z = 0;
memset(Dest, 0, DestSize);
for (x = 0; x < SrcSize; x++)
{
Dest[x2] = (char)(((unsigned int)(unsigned char)Dest[x2]) * 37 + ((unsigned int)(unsigned char)Src[x]));
x2++;
if (x2 == DestSize - 1)
{
x2 = 0;
z++;
}
}
// Normalize the alphabet if it looped.
if (z && Normalize)
{
unsigned char TempChr;
y = (z > 1 ? DestSize - 1 : x2);
for (x = 1; x < y; x++)
{
TempChr = ((unsigned char)Dest[x]) & 0x3F;
if (TempChr < 10) TempChr += '0';
else if (TempChr < 36) TempChr = TempChr - 10 + 'A';
else if (TempChr < 62) TempChr = TempChr - 36 + 'a';
else if (TempChr == 62) TempChr = '_';
else TempChr = '-';
Dest[x] = (char)TempChr;
}
}
return (SrcSize < DestSize ? SrcSize : DestSize);
}
It probably has more collisions than might be desired but it isn't intended for use as a cryptographic hash function. You might try various multipliers (i.e. change the 37 to another prime number) if you get too many collisions. One of the interesting features of this snippet is that when Src is shorter than Dest, Dest ends up with the input string as-is (0 * 37 + value = value). If you want something "readable" at the end of the process, Normalize will adjust the transformed bytes at the cost of increasing collisions.
Source:
https://github.com/cubiclesoft/cross-platform-cpp/blob/master/sync/sync_util.cpp
