Began re-writing/updating the sampling extension stuff

1 files changed, 4 insertions, 9 deletions

diff --git a/chapters/background.tex b/chapters/background.tex
index 9950b39..69436c8 100644
--- a/chapters/background.tex
+++ b/chapters/background.tex
@@ -85,6 +85,7 @@ their work on dynamization, and we will adopt their definition,
     \begin{equation*}
     F(A \cup B, q) = F(A, q)~ \mergeop ~F(B, q)
     \end{equation*}
+    for all $A, B \in \mathcal{PS}(\mathcal{D})$ where $A \cap B = \emptyset$.
 \end{definition}
 
 The requirement for $\mergeop$ to be constant-time was used by Bentley and
@@ -101,6 +102,7 @@ problems},
     \begin{equation*}
     F(A \cup B, q) = F(A, q)~ \mergeop ~F(B, q)
     \end{equation*}
+    for all $A, B \in \mathcal{PS}(\mathcal{D})$ where $A \cap B = \emptyset$.
 \end{definition}
 
 To demonstrate that a search problem is decomposable, it is necessary to
@@ -811,6 +813,7 @@ cost, we could greatly reduce the cost of supporting $C(n)$-decomposable
 queries.
 
 \subsubsection{Independent Range Sampling}
+\label{ssec:background-irs}
 
 Another problem that is not decomposable is independent sampling. There
 are a variety of problems falling under this umbrella, including weighted
@@ -831,15 +834,7 @@ matching of records in result sets. To work around this, a slight abuse
 of definition is in order: assume that the equality conditions within
 the DSP definition can be interpreted to mean ``the contents in the two
 sets are drawn from the same distribution''. This enables the category
-of DSP to apply to this type of problem. More formally,
-\begin{definition}[Decomposable Sampling Problem]
-    A sampling problem $F: (D, Q) \to R$, $F$ is decomposable if and
-    only if there exists a constant-time computable, associative, and
-    commutative binary operator $\mergeop$ such that,
-    \begin{equation*}
-    F(A \cup B, q) \sim F(A, q)~ \mergeop ~F(B, q)
-    \end{equation*}
-\end{definition}
+of DSP to apply to this type of problem.
 
 Even with this abuse, however, IRS cannot generally be considered
 decomposable; it is at best $C(n)$-decomposable.  The reason for this is