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authorMarcin Chrzanowski <mc370754@students.mimuw.edu.pl>2022-02-05 21:37:38 +0100
committerMarcin Chrzanowski <mc370754@students.mimuw.edu.pl>2022-02-05 21:37:38 +0100
commit67fa27c467f570852f22cc31e2f352a90e66fa62 (patch)
treef5ad62cc4a43ad3cc946bde0b0ef3c9a5dd0431e
parent438ab64ce75ab47cfa26674a3d01241ed8f4f663 (diff)
Fix alphabet
-rw-r--r--mgr.tex12
1 files changed, 6 insertions, 6 deletions
diff --git a/mgr.tex b/mgr.tex
index cb631e0..c066fe6 100644
--- a/mgr.tex
+++ b/mgr.tex
@@ -512,7 +512,7 @@ before, by looking at $k' := \breaktable[(k+1).q']$, comparing it to $j$, and
either halting if $k' \geq j$, or jumping again otherwise. Because with each
jump we move to a color strictly smaller than before, the number of jumps is
bounded by the number of colors, $|Q|$. Thus the question is answered in time
-constant with respect to $|w|$.
+$O(|Q|)$, which is constant with respect to $|w|$.
\chapter{Branch Infix Regular Questions}\label{branchinfix}
@@ -797,11 +797,11 @@ with unary relation $U_X$ is equivalent to a valuation of $\vec{X}$ in which the
variable $X$ is set to $W$.
Now by Theorem \ref{mso-to-automaton}, there is a tree automaton $A$ for binary
-trees over the alphabet $\{ 0, 1 \}^{\vec{X}}$ which accepts a tree if and only
-if its labeling corresponds to a model satisfying $\varphi''$. Here a label
-$b_1 \ldots b_k$ (where $k = |\vec{X}|$ and each $b_i \in \{ 0, 1 \}$) should be
-interpreted as a bit vector signifying which unary relations $U_X$ a vertex is
-assigned to.
+trees over the alphabet $\Sigma \times \{ 0, 1 \}^{\vec{X}}$ which accepts a
+tree if and only if its labeling corresponds to a model satisfying $\varphi''$.
+Here a label $b_1 \ldots b_k$ (where $k = |\vec{X}|$ and each $b_i \in \{ 0, 1
+\}$) should be interpreted as a bit vector signifying which unary relations
+$U_X$ a vertex is assigned to.
By combining the above reductions, we can solve MSO query answering on trees by
solving the relabel regular qeustions problem on trees. Indeed, fix a formula