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Question about Run Length Encoding
Nov 30, 2012 at 5:19pm
Hi, if we take for example 32bits: 1111 1000 0000 1000 0001 1111 1100 0000, how does it become (24bits) 1011 1110 0011 0110 1111 0110 after RLE is applied?
Why/how does 1011 represent the 5 1's at the start of the 32bit number? I'm trying to understand the concept and was trying to figure out how to do it by hand before attempting to write a text compression program.
Why/how does 1011 represent the 5 1's at the start of the 32bit number? I'm trying to understand the concept and was trying to figure out how to do it by hand before attempting to write a text compression program.
Nov 30, 2012 at 8:45pm
It's the first time I've seen RLE applied to the individual bits' level.
Could you explain how this particular implementation is supposed to work ?
Could you explain how this particular implementation is supposed to work ?
Nov 30, 2012 at 9:12pm| Why/how does 1011 represent the 5 1's at the start of the 32bit number? |
From how it looks, the first three bits of each 4-bit group are the run length (101[2]=5) and the fourth bit is the bit to be repeated (1).
| how to do it by hand before attempting to write a text compression program. |
RLE is not suitable for text compression. Rather, find sequences of characters that have already occurred previously in the text and encode their position and length.
Nov 30, 2012 at 11:15pm
@pq1wx1p Did you mis-type the 4th and 6th group of bits in the RLE version?
Athar's theory doesn't fit these at the moment.
5 x 1 -> 1011 cf 1011 :-)
7 x 0 -> 1110 cf 1110 :-)
1 x 1 -> 0011 cf 0011 :-)
6 x 0 -> 1100 cf 0110 :-(
7 x 1 -> 1111 cf 1111 :-)
6 x 0 -> 1100 cf 0110 :-(
It's the groups of 6 which don't match.
Andy
Athar's theory doesn't fit these at the moment.
5 x 1 -> 1011 cf 1011 :-)
7 x 0 -> 1110 cf 1110 :-)
1 x 1 -> 0011 cf 0011 :-)
6 x 0 -> 1100 cf 0110 :-(
7 x 1 -> 1111 cf 1111 :-)
6 x 0 -> 1100 cf 0110 :-(
It's the groups of 6 which don't match.
Andy
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