Succinct: SOLE: fix redundancy calculations

fs-succinct/sole_even.asy

 import sole_common;
 
 blocks(0, "...", "B", "B", "EOF");
-mixarrow(0, 1); 
-thruarrow(0,3);
-blocks(1, "...", "B+i", "B-i", "B+j");
-mixarrows(1, 0, 4);
-blocks(2, "...", "B", "B", "B");
+thruarrows(0, 1, 3);
+blocks(1, "...", "B+1", "B+1", "B+1", "0");
+mixarrows(1, 1, 4); 
+blocks(2, "...", "B-i", "B+j", "B-j", "B+k", "0");
+mixarrows(2, 0, 6);
+blocks(3, "...", "B", "B", "B", "B", "B");

fs-succinct/succinct.tex

@@ -165,28 +165,24 @@ odd-numbered blocks have smaller alphabets than even-numbered ones. The second
 pass 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 performing the inverse of
-pass 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, alternate ending with even number of blocks}
+What is the redundancy of this scheme?
 
 Now we can finally analyze redundancy. Let us count how the number of blocks
-increases throughout the encoding passes.
+increases throughout the encoding passes:
 \tightlist{o}
 \: 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.
+\: Before the first pass, we may need to add an extra padding block to make number of blocks even.
+\: Before the second pass, we always add an extra padding block to make number of blocks odd.
 \endlist
-In total, we add at most 3 blocks. Thus $r(n) \le 3B$. That is a constant
+In total, we add at most 4 blocks. Thus $r(n) \le 4b$. That is a constant
 and thus we have a succinct scheme.
 
+Also not that this representation is locally decodable and modifiable -- each input
+block affects at most 4 output blocks.
+
+
+Now we need to check that all the alphabet translations are valid.
 
 \section{Succinct representation of arbitrary-alphabet strings}