Name: FrozenVoid 2011-10-09 13:25
I will prove theoretically how it possible:
Every byte N contains 8 bits, due format.
However the space of a byte is not optimally used.
This a table of how much space is used by certain bytes:
Byte Value: Size
128-155 8bits
64-127 7bits
32-63 6bits
16-31 5 bits
8-15 4bits
4-7 3 bits
2-3 2 bits
0-1 1 bit
Notice a pattern? The lower the byte value, the less data it contains!
Aggregation of such truncated bytes would take less bits than initial byte.
a value of 31 will take only 5 bits in a bitstring.
Every byte N contains 8 bits, due format.
However the space of a byte is not optimally used.
This a table of how much space is used by certain bytes:
Byte Value: Size
128-155 8bits
64-127 7bits
32-63 6bits
16-31 5 bits
8-15 4bits
4-7 3 bits
2-3 2 bits
0-1 1 bit
Notice a pattern? The lower the byte value, the less data it contains!
Aggregation of such truncated bytes would take less bits than initial byte.
a value of 31 will take only 5 bits in a bitstring.