backtrack

chapterBacktracking.tex

@@ -6,7 +6,7 @@ \section{Introduction}
 
 \section{Math}
 \subsection{Decomposition}
-\subsubsection{Factorize a number}.\label{factorization}
+\subsubsection{Factorize a number}\label{factorization}
 \rih{Core clues}:
 \begin{enumerate}
 \item Expand the search tree \textbf{horizontally}.
@@ -31,11 +31,15 @@ \subsubsection{Factorize a number}.\label{factorization}
     n = cur.pop()
     start = cur[-1] if cur else 2
     for i in xrange(start, int(sqrt(n))+1):
-        if n%i == 0:
+        if self.predicate(n, i):
             cur.append(i)
             cur.append(n/i)
             self.dfs(cur, ret)
             cur.pop()
+
+def predicate(self, n, i):
+  return n%i == 0
+
 \end{python}
 \rih{Time complexity.} $O(2^n)$ where $n$ is the number of prime factors. Choose $i$ prime factors to combine then, and keep the rest uncombined： 
 

chapterBalancedSearchTree.tex

@@ -117,6 +117,26 @@ \section{B-Tree}
 \caption{B-Tree}
 \label{fig:b-tree}
 \end{figure}
+\subsection{Bascis}
+Half-full principle: 
+
+\begin{tabular}{lll}
+\hline\noalign{\smallskip}
+\textbf{Attrs} & \textbf{Non-leaf} & \textbf{Leaf} \\
+\noalign{\smallskip}\hline\noalign{\smallskip}
+Ptrs & \lceil\frac{n+1}{2}\rceil & \lfloor\frac{n+1}{2}\rfloor \\
+\noalign{\smallskip}\hline\noalign{
+\caption{Nodes at least half-full}
+\end{tabular}
+
+\subsection{Operations}
+Core clues
+\begin{enumerate}
+\item \textbf{Split \& Up}: split half, move up the RIGHT node's FIRST of split nodes
+recursively
+\item The node moved up should be removed in the original node UNLESS it is a leaf
+node. 
+\end{enumerate}
 
 \section{AVL Tree}
 TODO

chapterDynamicProgramming.tex

@@ -121,14 +121,14 @@ \section{Sequence}
 \section{String}
 \runinhead{Is palindrome.} Given a string $s$, use an array to determine whether $s[i:j]$.
 
-Let $P_{i,j}$  indicates whether $s[i:j]$ is palindrome. 
+Let $P_{i,j}$  indicates whether $s[i:j]$ is palindrome. We have one condition - whether the head and the end letter are equal: 
 \begin{eqnarray*}
-P_{i. j} = P_{i-1, j+1}\ \&\&\ s[i] = s[j-1]
+P_{i. j} = P_{i-1, j+1}\ \wedge\ s[i] = s[j-1]
 \end{eqnarray*}
 
 \runinhead{Minimum palindrome cut.} Given a string s, partition s such that every substring of the partition is a palindrome. Return the minimum cuts needed for a palindrome partitioning of s.
 
-Let $C_i$ be the min cut for $s[:i]$. 
+Let $C_i$ be the min cut for $s[:i]$. We have 1 more cut from previous state to make $S[:i]$ palindrome. 
 \begin{eqnarray*}
 C_{i} = \left\{ \begin{array}{rl}
   \min\big(C[k]+1 \cdot \forall k<i \big) &\mbox{// if $s[k:i]$ is palindrome}

chapterGreedy.tex

@@ -2,9 +2,5 @@ \chapter{Greedy}
 
 
 \section{Introduction}
-Queue, Stack
-
-
-\subsection{Summarizing properties}
 TODO
 

chapterSearch.tex

@@ -23,6 +23,3 @@ \section{Binary Search}
             hi = mid
     return lo-1
 \end{python}
-
-\section{Looping Root}
-Iterate the list and make the current element as the root, evaluate the left part and the right part and combine the results (i.e. looping + divide \& conquer). 

chapterTree.tex

@@ -194,8 +194,7 @@ \subsection{Range search}
     self.predecessors(root.right, target, stk)
 \end{python}
 
-\section{Interval Search Tree}
-TODO
+
 
 
 \section{Binary Index Tree (BIT)}\label{BIT}

notation.tex

@@ -34,6 +34,8 @@ \section*{General math notation}
 $\sim$ & Tilde, the leading term of mathematical expressions \\
 $\arg\max\limits_x f(x)$ & Argmax: the value $x$ that maximizes $f$\\
 $\binom{n}{k}$ & $n$ choose $k$ , equal to $\frac{n!}{k!(n-k)!}$\\
+$range(i,j)$ & range of number from i (inclusive) to j (exclusive) \\
+$A[i:j]$ & subarray consist of $A_i, A_{i+1}, ..., A_{j-1}$.
 \noalign{\smallskip}\hline\noalign{\smallskip}
 \end{longtable}