Talk:Delta encoding

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1 C code is wrong
2 Generalisation of delta encoding
3 Error control
4 symmetric delta
5 Potential sources for new article

[edit] C code is wrong

I think the C code is not correct

You are correct. If you run the example in the text trough this algorightm (2, 4, 6, 9, 7), the encoded becomes: 2, 2, 4, 5, 2
The decode then becomes: 2, 4, 8, 13, 15
The decoder is correct (checked with encoded data 2, 2, 2, 3, -2): 2, 4, 6, 9, 7 (which is the right awnser).
I will correct the encoder after I add this comment. ShadowLord 09:14, 7 Apr 2005 (UTC)

[edit] Generalisation of delta encoding

Could this article be expanded to a more general view of delta encoding as encoding only the differences between things? A mention of diff and similar might be quite good. porges 09:38, Apr 7, 2005 (UTC)

I've changed the text to link to delta compression article, which have more direct relation to revision control in general and diff in particular. 217.26.163.26 07:23, 4 May 2005 (UTC)

[edit] Error control

Error control has different meaning in this article - it's used to find differences from other version of the same file without transfering it, not to detect corrupted data. This is the way how rsync/zsync use it. Please, don't change this meaning! 217.26.163.26 07:06, 4 May 2005 (UTC)

[edit] symmetric delta

The article currently states

A symmetric delta can be expressed as $\Delta(v_1, v_2) = (v_1 \backslash v_2) \cup (v_2 \backslash v_1)$ , where

v 1

and

v 2

represent two successive versions.

OK, that is a pretty formula, but what does it mean? (What does the "\" mean in this context?)

I'm guessing it means something like

Given two versions

v 1

and

v 2

, we can compute "the delta" between them.

A directed delta tells us, once we know

v 1

, how to generate

v 2

-- but if we only know

v 2

and the directed delta, it's generally impossible to recover

v 1

A symmetric delta can go both ways -- if we know

v 1

, it tells us how to generate

v 2

. But if we only know

v 2

, that same symmetric delta tells us how to generate

v 1

Is that all it means ? Is it OK if I replace the mysterious formula with this English-language description? --DavidCary 03:19, 6 October 2005 (UTC)

If I read the formula correctly, it means the same as $\Delta_s(v_1, v_2) = (v_1 \cup v_2) \backslash (v_1 \cap v_1)$ , which I can read easier in English: the symmetric delta of two versions is the elements of each version without the elements common to both.

I'm guessing that directed delta could be defined by $\Delta_d(v_1, v_2) = v_2 \backslash v_1$ , and would read: the elements of second version without the elements of the first.