How can I compress a row of integers into something shorter ?
Like:
Input: '1 2 4 5 3 5 2 3 1 2 3 4' -> Algorithm -> Output: 'X Y Z'
and can get it back the other way around? ('X Y Z' -> '1 2 4 5 3 5 2 3 1 2 3 4')
Note:Input will only contain numbers between 1-5 and the total string of number will be 10-16
Is there any way I can compress it to 3-5 numbers?
Here is one way. First, subtract one from each of your little numbers. For your example input that results in
0 1 3 4 2 4 1 2 0 1 2 3
Now treat that as the base-5 representation of an integer. (You can choose either most significant digit first or last.) Calculate the number in binary that means the same thing. Now you have a single integer that "compressed" your string of little numbers. Since you have shown no code of your own, I'll just stop here. You should be able to implement this easily.
Since you will have at most 16 little numbers, the maximum resulting value from that algorithm will be 5^16 which is 152,587,890,625. This fits into 38 bits. If you need to store smaller numbers than that, convert your resulting value into another, larger number base, such as 2^16 or 2^32. The former would result in 3 numbers, the latter in 2.
#SergGr points out in a comment that this method does not show the number of integers encoded. If that is not stored separately, that can be a problem, since the method does not distinguish between leading zeros and coded zeros. There are several ways to handle that, if you need the number of integers included in the compression. You could require the most significant digit to be 1 (first or last depends on where the most significant number is.) This increases the number of bits by one, so you now may need 39 bits.
Here is a toy example of variable length encoding. Assume we want to encode two strings: 1 2 3 and 1 2 3 0 0. How the results will be different? Let's consider two base-5 numbers 321 and 00321. They represent the same value but still let's convert them into base-2 preserving the padding.
1 + 2*5 + 3*5^2 = 86 dec = 1010110 bin
1 + 2*5 + 3*5^2 + 0*5^3 + 0*5^4 = 000001010110 bin
Those additional 0 in the second line mean that the biggest 5-digit base-5 number 44444 has a base-2 representation of 110000110100 so the binary representation of the number is padded to the same size.
Note that there is no need to pad the first line because the biggest 3-digit base-5 number 444 has a base-2 representation of 1111100 i.e. of the same length. For an initial string 3 2 1 some padding will be required in this case as well, so padding might be required even if the top digits are not 0.
Now lets add the most significant 1 to the binary representations and that will be our encoded values
1 2 3 => 11010110 binary = 214 dec
1 2 3 0 0 => 1000001010110 binary = 4182 dec
There are many ways to decode those values back. One of the simplest (but not the most efficient) is to first calculate the number of base-5 digits by calculating floor(log5(encoded)) and then remove the top bit and fill the digits one by one using mod 5 and divide by 5 operations.
Obviously such encoding of variable-length always adds exactly 1 bit of overhead.
Its call : polidatacompressor.js but license will be cost you, you have to ask author about prices LOL
https://github.com/polidatacompressor/polidatacompressor
Ncomp(65535) will output: 255, 255 and when you store this in database as bytes you got 2 char
another way is to use "Hexadecimal aka base16" in javascript (1231).toString(16) give you '4cf' in 60% situation it compress char by -1
Or use base10 to base64 https://github.com/base62/base62.js/
4131 --> 14D
413131 --> 1Jtp
I have a table with sorted numbers like:
1 320102
2 5200100
3 92010023
4 112010202
5 332020201
6 332020411
:
5000000000 3833240522044511
5000000001 3833240522089999
5000000002 4000000000213312
Given the record number I need the value in O(log n) time. The record number is 64-bit long and there are no missing record numbers. The values are 64-bit long, they are sorted and value(n) < value(n+1).
The obvious solution is simply doing an array and use the records number as index. This will cost 64-bit per value.
But I would like a more space efficient way of doing that. Since we know the values are always increasing that should be doable, but I do not remember a data structure that lets me do that.
A solution would be to use deflate on the array, but that will not give me O(log n) for accessing an element - thus unacceptable.
Do you know of a data structure that will give me:
O(log n) for access
space requirement < 64-bit/value
= Edit =
Since we know all numbers in advance we could find the difference between each number. By taking the 99th percentile of these differences we will get a relatively modest number. Taking the log2 will give us the number of bits needed to represent modest number - let us call that modest-bits.
Then create this:
64-bit value of record 0
64-bit value of record 1024
64-bit value of record 2048
64-bit value of record 3072
64-bit value of record 4096
Then a delta table for all records:
modest-bits difference to record 0
modest-bits difference to previous record
1022 * modest-bits difference to previous record
modest-bits difference to record 1024
modest-bits difference to record k*1024 will always be 0, so we can use that for signaling. If it is non-zero, then the following 64-bit will be a pointer to a simple array for the next 1024 records as 64-bit values.
As the modest value is chosen as the 99th percentile number, that will at most happen 1% of the time, thus wasting at most 1% * n * modest-bits + 1% * n * 64-bit * 1024.
space: O(modest-bits * n + 64-bit * n / 1024 + 1% * n * modest-bits + 1% * n * 64-bit * 1024)
lookup: O(1 + 1024)
(99% and 1024 may have to be adjusted)
= Edit2 =
Based on the idea above, but wasting less space. Create this:
64-bit value of record 0
64-bit value of record 1024
64-bit value of record 2048
64-bit value of record 3072
64-bit value of record 4096
And for all value that cannot be represented by modest-bits create big-value table as a tree:
64-bit position, 64-bit value
64-bit position, 64-bit value
64-bit position, 64-bit value
Then a delta table for all records, that is reset for every 1024 records:
modest-bits difference to record 0
modest-bits difference to previous record
1022 * modest-bits difference to previous record
modest-bits difference to record 1024
but also reset for every value that is in the big-value table.
space: O(modest-bits * n + 64-bit * n / 1024 + 1% * n * 2 * 64-bit).
Lookup requires searching big-value table, then looking up the 1024'th value and finally summing up the modest-bits values.
lookup: O(log(big-value table) + 1 + 1024) = O(log n)
Can you improve this? Or do better in a different way?
OP proposes splitting numbers into blocks (only once). But this process may be continued. Split every block once more. And again... Finally we might get a binary trie.
Root node contains value of the number with least index. Its right descendant stores difference between the middle number in the table and the number with least index: d = A[N/2] - A[0] - N/2. This is continued for other right descendants (red nodes on diagram). Leaf nodes contain deltas from preceding numbers: d = A[i+1] - A[i] - 1.
So most of the values, stored in trie, are delta values. Each of them occupies less than 64 bits. And for compactness they may be stored as variable-bit-length numbers in a bit stream. To get length of each number and to navigate in this structure in O(log N) time, bit stream should also contain lengths of (some) numbers and (some) subtrees:
Each node contains length (in bits) of its left sub-tree (if it has one).
Each right descendant (red nodes on diagram), except leaf nodes, contains length (in bits) of its value. Leaf node's length may be calculated from other lengths on the path from root to this node.
Each right descendant (red nodes on diagram) contains difference of correspondent value and the value of nearest "red" node up the path.
All nodes are packed in bit stream, starting from root node, in-order: left descendant always follows its ancestor; right descendant follows sub-tree, rooted by left descendant.
To access element given its index, use index's binary representation to follow path in the trie. While traversing this path, add together all values of "red" nodes. Stop when no more non-zero bits are left in the index.
There are several options to store N/2 value lengths:
Allocate as many bits for each length as needed to represent all values from the largest length to somewhere below mean length (excluding some very short outliers).
Also exclude some long outliers (keep them in a separate map).
Since lengths may be not evenly distributed, it's reasonable to use Huffman encoding for value lengths.
Either fixed length or Huffman encodings should be different for each trie depth.
N/4 subtree lengths are, in fact, value lengths, because N/4 smallest subtrees contain a single value.
Other N/4 subtree lengths may be stored in words of fixed (predefined) length, so that for large subtrees we know only approximate (rounded up) lengths.
For 230 full-range 64-bit numbers we have to pack approximately 34-bit values, for 3/4 nodes, approx. 4-bit value lengths, and for every fourth node, 10-bit subtree lengths. Which saves 34% space.
Example values:
0 320102
1 5200100
2 92010023
3 112010202
4 332020201
5 332020411
6 3833240522044511
7 3833240522089999
8 4000000000213312
Trie for these values:
root d=320102 vl=19 tl=84+8+105+4+5=206
+-l tl=75+4+5=84
| +-l tl=23
| | +-l
| | | +-r d=4879997 (vl=23)
| | +-r d=91689919 vl=27
| | +-r d=20000178 (vl=25)
| +-r d=331700095 vl=29 tl=8
| +-l
| | +-r d=209 (vl=8)
| +-r d=3833240190024308 vl=52
| +-r d=45487 (vl=16)
+-r d=3999999999893202 vl=52
Value length encoding:
bits start end
Root 0 19 19
depth 1 0 52 52
depth 2 0 29 29
depth 3 5 27 52
depth 4 4 8 23
Sub-tree lengths need 8 bits each.
Here is encoded stream (binary values still shown in decimal for readability):
bits value comment
19 320102 root value
8 206 left subtree length of the root
8 84 left subtree length
4 15 smallest left subtree length (with base value 8)
23 4879997 value for index 1
5 0 value length for index 2 (with base value 27)
27 91689919 value for index 2
25 20000178 value for index 3
29 331700095 value for index 4
4 0 smallest left subtree length (with base value 8)
8 209 value for index 5
5 25 value length for index 6 (with base value 27)
52 3833240190024308 value for index 6
16 45487 value for index 7
52 3999999999893202 value for index 8
Altogether 285 bits or 5 64-bit words. We also need to store bits/start values from value length encoding table (350 bits). To store 635 bits we need 10 64-bit words, which means such a small number table cannot be compressed. For larger number tables, size of value length encoding table is negligible.
To search a value for index 7, read root value (320102), skip 206 bits, add value for index 4 (331700095), skip 8 bits, add value for index 6 (3833240190024308), add value for index 7 (45487), and add index (7). The result is 3 833 240 522 089 999, as expected.
I would do it in blocks, as you outline in your question. Pick a block size k, where you can accept having to decode on average k/2 values before getting to the one you're after. For the n total values, you will have n/k blocks. A table with n/k entries would point into the data stream to find the starting point of each block. Finding where to go in that table would be O(log(n/k)) for a binary search, or if the table is small enough and if it matters, you could make it about O(1) with an auxiliary hash table.
Each block would start with a starting 64-bit value. All values after that would be stored as deltas from the preceding value. My suggestion is to store those deltas as a Huffman code that says how many bits are in the next value, followed by that many bits. The Huffman code would be optimized for each block, and a description of that code would be stored at the beginning of the block.
You could simplify that by just preceding each value with six bits having the number of bits following, in the range of 1..64, effectively a flat Huffman code. Depending on the histogram of the bit lengths, an optimized Huffman code could knock off a good number of bits compared to the flat code.
Once you have this set up, you can experiment with k and see how small you can make it and still have limited impact on the compression.
I do not know of a data structure that does that.
The obvious solution to gain space and not loose too much speed would be to create your own structure with different array size based on the different int size you store.
Pseudo-code
class memoryAwareArray {
array16 = Int16[] //2 bytes
array32 = Int32[] //4 bytes
array64 = Int64[] //8 bytes
max16Index = 0;
max32Index = 0;
addObjectAtIndex(index, value) {
if (value < 65535) {
array16[max16Index] = value;
max16Index++;
return;
}
if (value < 2147483647) {
array32[max32Index] = value;
max32Index++;
return;
}
array64[max64Index] = value;
max64Index++;
}
getObject(index) {
if (index < max16Index) return(array16[index]);
if (index < max32Index) return(array32[index-max16Index]);
return(array64[index-max16Index-max32Index]);
}
}
Something along those lines shouldn't alter to much the speed and you'd save around 7 gigas if you filled up the entire structure. You won't save as much since you have gaps beetween your values of course.
I am reading this libb64 source code for encoding and decoding base64 data.
I know the encoding procedure but i can't figure out how the following decoding table is constructed for fast lookup to perform decoding of encoded base64 characters. This is the table they are using:
static const char decoding[] = {62,-1,-1,-1,63,52,53,54,55,56,57,58,59,60,61,-1,-1,-1,-2,-1,-1,-1,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,-1,-1,-1,-1,-1,-1,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51};
Can some one explain me how the values in this table are used for decoding purpose.
It's a shifted and limited ASCII translating table. The keys of the table are ASCII values, the values are base64 decoded values. The table is shifted such that the index 0 actually maps to the ASCII character + and any further indices map the ASCII values after +. The first entry in the table, the ASCII character +, is mapped to the base64 value 62. Then three characters are ignored (ASCII ,-.) and the next character is mapped to the base64 value 63. That next character is ASCII /.
The rest will become obvious if you look at that table and the ASCII table.
It's usage is something like this:
int decode_base64(char ch) {
if (ch < `+` or ch > `z`) {
return SOME_INVALID_CH_ERROR;
}
/* shift range into decoding table range */
ch -= `+`;
int base64_val = decoding[ch];
if (base64_val < 0) {
return SOME_INVALID_CH_ERROR;
}
return base64_val;
}
As know, each byte has 8 bits, possible 256 combinations with 2 symbols (base2).
With 2 symbols is need to waste 8 chars to represent a byte, for example '01010011'.
With base 64 is possible to represent 64 combinations with 1 char...
So, we have a base table:
A = 000000
B = 000001
C = 000010
...
If you have the word 'Man', so you have the bytes:
01001101, 01100001, 01101110
and so the stream:
011010110000101101110
Break in group of six bits: 010011 010110 000101 101110
010011 = T
010110 = W
000101 = F
101110 = u
So, 'Man' => base64 coded = 'TWFu'.
As saw, this works perfectly to streams whith length multiple of 6.
If you have a stream that isn't multiple of 6, for example 'Ma' you have the stream:
010011 010110 0001
you need to complete to have groups of 6:
010011 010110 000100
so you have the coded base 64:
010011 = T
010110 = W
000100 = E
So, 'Ma' => 'TWE'
After to decode the stream, in this case you need to calc the last multiple length to be multiple of 8 and so remove the extra bits to obtain the original stream:
T = 010011
W = 010110
E = 000100
1) 010011 010110 000100
2) 01001101 01100001 00
3) 01001101 01100001 = 'Ma'
In really, when we put the trailing 00s, we mark the end of Base64 string with '=' to each trailing '00 added ('Ma' ==> Base64 'TWE=')
See too the link: http://www.base64decode.org/
Images represented on base 64 is a good option to represent with strings in many applications where is hard to work directly with a real binary stream. Real binary stream is better because is a base256, but is difficult inside HTML for example, there are 2 ways, minor traffic, or more easy to work with strings.
See ASCII codes too, the chars of base 64 is from range '+' to 'z' on table ASCII but there are some values between '+' and 'z' that isn't base 64 symbols
'+' = ASCII DEC 43
...
'z' = ASCII DEC 122
from DEC 43 to 122 are 80 values but
43 OK = '+'
44 isn't a base 64 symbols so the decoding index is -1 (invalid symbol to base64)
45 ....
46 ...
...
122 OK = 'z'
do the char needed to decode, decremented of 43 ('+') to be index 0 on vector to quick access by index so, decoding[80] = {62, -1, -1 ........, 49, 50,51};
Roberto Novakosky
Developer Systems
Considering these 2 mapping tables:
static const char decodingTab[] = {62,-1,-1,-1,63,52,53,54,55,56,57,58,59,60,61,-1,-1,-1,-2,-1,-1,-1,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,-1,-1,-1,-1,-1,-1,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51};
static unsigned char encodingTab[64]="ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";
decodingTab is the reverse mapping table of encondingTab.
So decodingTab[i] should never be -1.
In fact, only 64 values are expected. However decodingTab size is 128.
So, in decodingTab,unexpected index values are set to -1 (an arbitrary number which is not in [0,63])
char c;
unsigned char i;
...
encoding[decoding[c]]=c;
decoding[encoding[i]=i;
Hope it helps.