Commit dbb630e7 authored by Filip Stedronsky's avatar Filip Stedronsky
Browse files

Succinct: SOLE diagrams

parent b160f3eb
PICS=sole sole_even
include ../Makerules
import sole_common;
blocks(0 ... concat(array(6,"B"), new string[] {"...", "B", "EOF"}));
blocks(1 ... concat(array(6, "B+1"), array(1, "..."), array(2, "B+1")) );
mixarrow(1, 7);
block(1, 9, "0");
blocks(2, "B", "B+3", "B-3", "B+6", "B-6", "B+9", "...", "B-i","B+j", "B-j");
thruarrow(2, 0);
mixarrows(2, 1, 6);
mixarrow(2, 8);
thruarrow(1, 9);
blocks(3 ... concat(array(6,"B"), array(1, "..."), array(3, "B")));
passlabel(0, "Add EOF");
passlabel(1, "Pass 1");
passlabel(2, "Pass 2");
import sole_common;
blocks(0, "...", "B", "B", "EOF");
mixarrow(0, 1);
blocks(1, "...", "B+i", "B-i", "B+j");
mixarrows(1, 0, 4);
blocks(2, "...", "B", "B", "B");
......@@ -156,6 +156,34 @@ possible input combinations).
We will use this kind of alphabet re-encoding by pair heavily in the SOLE
encoding. The best way to explain the exact scheme is with a diagram:
\figure[sole]{sole.pdf}{}{SOLE alphabet re-encoding scheme}
There are two re-encoding phases. The first transforms blocks with alphabet
$[B+1]$ into blocks with variable alphabet sizes (of the form of alternating
$[B+3k]$, $[B-3k]$). The second phase runs phase shifted by one block and converts
the variable-alphabet blocks into blocks with alphabet $[B]$.
What is the redundancy of this scheme? That depends on whether the original
number of blocks (after padding) is even or odd. Figure \figref{sole} showed
the case for an odd number. For an even number, the ending is a little bit different
(fig. \figref{sole_even}).
You can easily check that this is reversible: after decoding phase 2, the last block
will always be either 0 (for the odd case) or EOF (for the even case).
\figure[sole_even]{sole_even.pdf}{}{SOLE alphabet re-encoding scheme}
Now we can finally analyze redundancy. Let us count how the number of blocks
increases throughout the encoding passes.
\: If the original length was a multiple of $b$, we must add one block to complete padding.
\: We always add one block with EOF character.
\: In the first pass, number of blocks does not increase.
\: In the second pass, we may need to add an extra padding block to make number of blocks odd.
In total, we add at most 3 blocks. Thus $r(n) \le 3B$. That is a constant
and thus we have a succinct scheme.
\section{Succinct representation of arbitrary-alphabet strings}
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