add conclusion on master theorem

thu_dsa/chp1/master theorem.md

@@ -0,0 +1,85 @@
+Conclusion on Master Theorem
+============================
+
+## What is _Master Theorem_
+
+In the analysis of algorithms, `Master Theorem` provides a simple way to compute the time complexity(using big O notation) for `divide-and-conquer` recurrences.
+
+We all know that for `divide-and-conquer` algorithms, there are two way to analysis their time complexity, i.e. _recursion track_ and _recurrence equation_. Master Theorem is a way to compute such time complexity using both the two way: it tries to solve the _recurrence equation_ by means of _recusion track_.
+
+## Introduction
+
+First we'll make some abstractions. Lets' consider an algorithm implemented in the form of a recursion. Generally, we can assume that to solve a problem of scale `n`, we can divide it into `a` subproblems, whose scale would be `n/b`, with f(n) being the time to create the subproblems and combine their results in the above procedure.
+
+The runtime of subck an algorithm on an input of size 'n', usually denoted T(n), can be expressed by the recurrence relation
+
+$$
+T(n) = aT(\frac{n}{b}) + f(n)
+$$
+
+## Generic solution
+
+First we'll assume f(n) being in the form of $f(n) = n^d$. Generally, this is indeed the most common form of f(n).In this way, the equation should look like this:
+
+$$
+T(n) = aT(\frac{n}{b}) + n^d
+$$
+
+We'll try to solve this equation. And the essence of this is actually _recursion track_.
+
+In the method of _recursion track_, the computation of each layer of recursion can form a computation tree, with the sum of time cost in each layer being the total time cost of that layer.
+
++ For the 0th(highest) layer, the time cost is $n^d$.
++ For the 1st layer, the time cost is $a * (\frac{n}{b})^d = n^d * \frac{a}{b^d}$
++ For the 2nd layer, the time cost is $a^2 * (\frac{n}{b^2})^d = n^d * (\frac{a}{b^d})^2$
++ For the 3rd layer, the time cost is $a^3 * (\frac{n}{b^3})^d = n^d * (\frac{a}{b^d})^3$
++ ...
++ For the kth(the last) layer, the time cost is $a^k * (\frac{n}{b^2})^d = n^d * (\frac{a}{b^d})^k$, where $k = log_{b}^{n}$
+
+This way, we can compute T(n):
+
+$$
+T(n) = n^d + n^d * \frac{a}{b^d} + n^d * (\frac{a}{b^d})^2 + ... + n^d * (\frac{a}{b^d})^k
+$$
+
+Then simplify T(n) so that T(n) can be:
+
+$$
+T(n) = n^d(1 + \frac{a}{b^d} + (\frac{a}{b^d})^2 + ... + (\frac{a}{b^d})^k)
+$$
+
+And this is a _geometric progression_, so let's assume $q = \frac{a}{b^d}$, then T(n) can be:
+
+$$
+T(n) = n^d(1 + q + q^2 + ... + q^k)
+$$
+
+## Conlusion
+
+> Case 1: q = 1
+
+$T(n) = K * n^d = log_{b}^{n} * n^d = O(logn · n^d) $
+
+> Case 2: q < 1
+
+Then T(n) is a convergent geometric progression. So T(n) can be expressed as:
+
+$$
+T(n) = O(n^d)
+$$
+
+> Case 3: q > 1
+
+$$
+T(n) = n^d(1 + q + q^2 + ... + q^k) = n^d * \frac{1-q^{k+1}}{1-q} = O(n^d * q^k = O(a^{log_{b}^{n}})
+$$
+
+To further simplify this, we need to apply the following equation:
+
+$$
+a^{log_{b}^{n}} = n^{log_{b}^{a}}
+$$
+
+So $T(n) = O(n^{log_{b}^{a}})$
+
+

thu_dsa/chp2/Vector.h

@@ -205,7 +205,7 @@ template <typename T>
 int Vector<T>::interpolation_search(T const &elem, int lo, int hi){
 	int mid;
 	while(lo < hi){
-		mid = lo + (elem - _elem[lo])*(hi - lo) / (_elem[hi] - _elem[lo]);
+		mid = lo + (elem - _elem[lo])*(hi - lo - 1) / (_elem[hi - 1] - _elem[lo]);
 		if (mid < lo || mid >= hi) break;
 
 		if (elem < _elem[mid]) hi = mid;