Comments on: Random data compression – One bit diff encoding

By: Keshav Shetty

Keshav Shetty — Sun, 27 Nov 2011 12:14:31 +0000

Hi Ibrahim,

Best case happens when data are in descending order. In this case
for first input you need 8 bit = 1×8 = 8bit
for next 128 inputs we need 7 bit each = 128*7=896bits
for next 64 we need 6 bit = 64*6 = 384 and so on
total 1546 bits (or 193.25 bit)

In worst case i.e data are in ascending order
for first 128 inputs we need 8 bit each = 128*8=1024bits
for next 64 we need 7 bit = 64*7 = 448 and so on
total 1793 bits (or 224.125 bit)

Best way to visualize this is by applying diff encoding to the input and see observe bits it generates. (try for smaller data set)

By: Ibrahim

Ibrahim — Wed, 23 Nov 2011 23:26:20 +0000

As far my results with different formated 256 unique values blocks, I achieved
Symbols = 256
Result 1554 bits = 194.25 bytes

Did you mean that one?

By: Ibrahim

Ibrahim — Wed, 23 Nov 2011 23:07:26 +0000

Please explain:
1545-1793bits(193.125-224.125bytes)

Thanks

By: Keshav Shetty

Keshav Shetty — Wed, 09 Jun 2010 17:21:01 +0000

Some of the readers raised query about meaning of doubles or duplicates.
Actually it is my mistake to use wrong word, correct word should non unique value.

When I say unique value, I meant the number (byte) not appeared before in the list, whereas doubles means the number already present in the list.
e.g: lets assume the input bytes are 8, 10, 15, 0, 6, 12, 15, 9, 11, 1, 12, 15
From the list 8, 10, 15, 0, 6, 12, 9, 11, 1 are unique.
Whereas 15,12,15 are doubles or duplicates(non unique).
Note: First 15 and 12 are counted in unique.
First appearance will not treated as non unique.

In the article I mentioned million random digit contains around 90-100 doubles for every block of 256 numbers. That means if it is 90 non unique number then 256-90=166 are unique numbers.

Here on wards I will use the word non unique, instead of doubles or duplicates.

Thanks & regards
Keshav K Shetty