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‎C04-Recurrences/4.2.md

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### Exercises 4.2-1
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***
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Use a recursion tree to determine a good asymptotic upper bound on the recurrence
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### Exercises 4.2-1
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***
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Use a recursion tree to determine a good asymptotic upper bound on the recurrence
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![](http://latex.codecogs.com/gif.latex? T\(n\) = 3T\(\\lceil n/2 \\rceil\) + n). Use the substitution method to verify your answer.
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### `Answer`
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![recursion tree](./repo/s2/1.png)
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树的高度是lgn,有3^lgn个叶子节点.
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![](http://latex.codecogs.com/gif.latex? T\(n\) = n\\sum_{i = 0}^{lg\(n\)-1}\(\\frac{3}{2}\)^i + \\Theta\(3^{\\lg{n}}\) \\\\ ~
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\\hspace{15 mm} = \\Theta\(n^{\\lg{3}}\) + \\Theta\(3^{\\lg{n}}\) \\\\ ~
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\\hspace{15 mm} = \\Theta\(n^{\\lg{3}}\) + \\Theta\(n^{\\lg{3}}\) \\\\ ~
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\\hspace{15 mm} = \\Theta\(n^{\\lg{3}}\)
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)
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我们猜想 ![](http://latex.codecogs.com/gif.latex? T\(n\) \\le cn^{\\lg{3}}+2n )
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![](http://latex.codecogs.com/gif.latex? T\(n\) \\le 3c\(n/2\)^{\\lg{3}} + 2n \\\\ ~
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\\hspace{15 mm} \\le cn^{\\lg{3}}+2n \\\\ ~
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\\hspace{15 mm} = \\Theta\(n^{\\lg{3}}\)
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)
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### Exercises 4.2-2
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***
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Argue that the solution to the recurrence
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![](http://latex.codecogs.com/gif.latex? T\(n\) = T\(n/3\) + T\(2n/3\) + cn )
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, where c is a constant, is Ω(nlgn) by appealing to the recurrsion tree.
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### `Answer`
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最短的叶子高度是lg3n,每一层都要cn.也就是说，只考虑最短叶子的那一层（忽略其他层）已经有cnlg3n.
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### Exercises 4.2-3
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***
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Draw the recursion tree for
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![](http://latex.codecogs.com/gif.latex? T\(n\) = 4T\(\\lfloor n/2 \\rfloor\) + cn)
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,where c is a constant, and provide a tight asymptotic bound on its solution. Verify your bound by the substitution method.
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### `Answer`
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![recursion tree](./repo/s2/2.png)
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很明显是n^2的级别
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我们假设 ![](http://latex.codecogs.com/gif.latex? T\(n\) \\le n^2+2cn)
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![](http://latex.codecogs.com/gif.latex? T\(n\) \\le 4c\(n/2\)^2 + 2cn/2+cn \\le cn^2+2cn)
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我们假设 ![](http://latex.codecogs.com/gif.latex? T\(n\) \\ge n^2+2cn)
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![](http://latex.codecogs.com/gif.latex? T\(n\) \\ge 4c\(n/2\)^2 + 2cn/2+cn \\ge cn^2+2cn)
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### Exercises 4.2-4
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***
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Use a recursion tree to give an asymptotically tight solution to the recurrence T(n) = T(n - a) + T(a) + cn, where a ≥ 1 and c > 0 are constants.
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### `Answer`
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![recursion tree](./repo/s2/3.png)
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file:///Users/ganzhenchao/Workspaces/CLRS/C04-Recurrences/repo/s2/4.png
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![](http://latex.codecogs.com/gif.latex? T\(n\) = \\sum_{i=0}^{n/a}c\(n-ia\) + \(n/a\)ca
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= \\Theta\(n^2\))
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我们假设 ![](http://latex.codecogs.com/gif.latex? T\(n\) \\le cn^2)
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![](http://latex.codecogs.com/gif.latex? T\(n\) \\le c\(n-a\)^2 + ca + cn \\\\ ~
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\\hspace{15 mm} \\le cn^2-2acn+ca+cn \\\\ ~
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\\hspace{15 mm} \\le cn^2-c\(2an-a-n\) \\\\ ~
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\\hspace{15 mm}\\le cn^2 - cn ~~~~ if ~~ a > 1/2,n > 2a \\\\ ~
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\\hspace{15 mm}\\le cn^2 \\\\ ~
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\\hspace{15 mm} = \\Theta\(n^2\)
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)
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另外一个方向的证明和这个基本一样.
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### Exercises 4.2-5
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***
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Use a recursion tree to give an asymptotically tight solution to the recurrence T(n) = T(αn) +T((1 - α)n) + cn, where α is a constant in the range 0 <α < 1 and c > 0 is also a constant.
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### `Answer`
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![recursion tree](./repo/s2/4.png)
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可以假设α < 1/2,因此树的高度有![](http://latex.codecogs.com/gif.latex? \\log_{1/ \\alpha}{n} )
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![](http://latex.codecogs.com/gif.latex? T\(n\) = \\sum_{i = 0}^{\\log_{1/ \\alpha}{n}}cn + \\Theta\(n\) = cn\\log_{1/ \\alpha}{n} + \\Theta\(n\) = \\Theta\(n\\lg{n}\) )
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***
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Follow [@louis1992](https://github.com/gzc) on github to help finish this task.
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### `Answer`
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![recursion tree](./repo/s2/1.png)
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树的高度是lgn,有3^lgn个叶子节点.
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![](http://latex.codecogs.com/gif.latex? T\(n\) = n\\sum_{i = 0}^{lg\(n\)-1}\(\\frac{3}{2}\)^i + \\Theta\(3^{\\lg{n}}\) \\\\ ~
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\\hspace{15 mm} = \\Theta\(n^{\\lg{3}}\) + \\Theta\(3^{\\lg{n}}\) \\\\ ~
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\\hspace{15 mm} = \\Theta\(n^{\\lg{3}}\) + \\Theta\(n^{\\lg{3}}\) \\\\ ~
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\\hspace{15 mm} = \\Theta\(n^{\\lg{3}}\)
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)
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我们猜想 ![](http://latex.codecogs.com/gif.latex? T\(n\) \\le cn^{\\lg{3}}+2n )
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![](http://latex.codecogs.com/gif.latex? T\(n\) \\le 3c\(n/2\)^{\\lg{3}} + 2n \\\\ ~
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\\hspace{15 mm} \\le cn^{\\lg{3}}+2n \\\\ ~
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\\hspace{15 mm} = \\Theta\(n^{\\lg{3}}\)
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)
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### Exercises 4.2-2
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***
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Argue that the solution to the recurrence
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![](http://latex.codecogs.com/gif.latex? T\(n\) = T\(n/3\) + T\(2n/3\) + cn )
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, where c is a constant, is Ω(nlgn) by appealing to the recurrsion tree.
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### `Answer`
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最短的叶子高度是lg3n,每一层都要cn.也就是说，只考虑最短叶子的那一层（忽略其他层）已经有cnlg3n.
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### Exercises 4.2-3
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***
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Draw the recursion tree for
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![](http://latex.codecogs.com/gif.latex? T\(n\) = 4T\(\\lfloor n/2 \\rfloor\) + cn)
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,where c is a constant, and provide a tight asymptotic bound on its solution. Verify your bound by the substitution method.
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### `Answer`
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![recursion tree](./repo/s2/2.png)
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很明显是n^2的级别
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我们假设 ![](http://latex.codecogs.com/gif.latex? T\(n\) \\le n^2+2cn)
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![](http://latex.codecogs.com/gif.latex? T\(n\) \\le 4c\(n/2\)^2 + 2cn/2+cn \\le cn^2+2cn)
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我们假设 ![](http://latex.codecogs.com/gif.latex? T\(n\) \\ge n^2+2cn)
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![](http://latex.codecogs.com/gif.latex? T\(n\) \\ge 4c\(n/2\)^2 + 2cn/2+cn \\ge cn^2+2cn)
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### Exercises 4.2-4
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***
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Use a recursion tree to give an asymptotically tight solution to the recurrence T(n) = T(n - a) + T(a) + cn, where a ≥ 1 and c > 0 are constants.
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### `Answer`
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![recursion tree](./repo/s2/3.png)
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file:///Users/ganzhenchao/Workspaces/CLRS/C04-Recurrences/repo/s2/4.png
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![](http://latex.codecogs.com/gif.latex? T\(n\) = \\sum_{i=0}^{n/a}c\(n-ia\) + \(n/a\)ca
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= \\Theta\(n^2\))
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我们假设 ![](http://latex.codecogs.com/gif.latex? T\(n\) \\le cn^2)
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![](http://latex.codecogs.com/gif.latex? T\(n\) \\le c\(n-a\)^2 + ca + cn \\\\ ~
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\\hspace{15 mm} \\le cn^2-2acn+ca+cn \\\\ ~
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\\hspace{15 mm} \\le cn^2-c\(2an-a-n\) \\\\ ~
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\\hspace{15 mm}\\le cn^2 - cn ~~~~ if ~~ a > 1/2,n > 2a \\\\ ~
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\\hspace{15 mm}\\le cn^2 \\\\ ~
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\\hspace{15 mm} = \\Theta\(n^2\)
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)
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另外一个方向的证明和这个基本一样.
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### Exercises 4.2-5
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***
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Use a recursion tree to give an asymptotically tight solution to the recurrence T(n) = T(αn) +
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T((1 - α)n) + cn, where α is a constant in the range 0 <α < 1 and c > 0 is also a constant.
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### `Answer`
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![recursion tree](./repo/s2/4.png)
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可以假设α < 1/2,因此树的高度有![](http://latex.codecogs.com/gif.latex? \\log_{1/ \\alpha}{n} )
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![](http://latex.codecogs.com/gif.latex? T\(n\) = \\sum_{i = 0}^{\\log_{1/ \\alpha}{n}}cn + \\Theta\(n\) = cn\\log_{1/ \\alpha}{n} + \\Theta\(n\) = \\Theta\(n\\lg{n}\) )
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***
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Follow [@louis1992](https://github.com/gzc) on github to help finish this task.
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‎C04-Recurrences/4.3.md

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### Exercises 4.3-1
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***
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Use the master method to give tight asymptotic bounds for the following recurrences.
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a. ![](http://latex.codecogs.com/gif.latex? T\(n\) = 4T\(n/2\)+n )
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b. ![](http://latex.codecogs.com/gif.latex? T\(n\) = 4T\(n/2\)+n^2 )
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### Exercises 4.3-1
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***
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Use the master method to give tight asymptotic bounds for the following recurrences.
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a. ![](http://latex.codecogs.com/gif.latex? T\(n\) = 4T\(n/2\)+n )
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b. ![](http://latex.codecogs.com/gif.latex? T\(n\) = 4T\(n/2\)+n^2 )
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c. ![](http://latex.codecogs.com/gif.latex? T\(n\) = 4T\(n/2\)+n^3 )
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### `Answer`
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![](http://latex.codecogs.com/gif.latex? n^{\\log_{b}{a}} = n^{\\log_{2}{4}} = n^2)
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a. ![](http://latex.codecogs.com/gif.latex? \\Theta\(n^2\) )
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b. ![](http://latex.codecogs.com/gif.latex? \\Theta\(n^2 \\lg{n}\) )
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c. ![](http://latex.codecogs.com/gif.latex? \\Theta\(n^3\) )
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### Exercises 4.3-2
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***
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The recurrence T(n) = 7T (n/2)+n2 describes the running time of an algorithm A. A competing algorithm A′ has a running time of T′(n) = aT′(n/4) + n2. What is the largest integer value for a such that A′ is asymptotically faster than A?
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### `Answer`
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根据主定理，算法A的运行时间为![](http://latex.codecogs.com/gif.latex? T\(n\) = \\Theta\(\\lg{7}\)\ \\approx n^{2.8} )
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A'的运行时间在a > 16时超过n^2,此时
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![](http://latex.codecogs.com/gif.latex? T\(n\) = \\Theta\(n^{\\log_{4}{a}}\) < \\lg{7} = \\log_{4}{49})
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所以最大值为48
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### Exercises 4.3-3
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***
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Use the master method to show that the solution to the binary-search recurrence T(n) = T (n/2)+ Θ(1) is T(n) = Θ(lg n). (See Exercise 2.3-5 for a description of binary search.)
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### `Answer`
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![](http://latex.codecogs.com/gif.latex? n^{\\log_{b}{a}} = n^{\\log_{2}{1} = 1} )
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so the solution is Θ(lgn).
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### Exercises 4.3-4
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***
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Can the master method be applied to the recurrence
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![](http://latex.codecogs.com/gif.latex? T\(n\) = 4T\(n/2\) + n^2 \\lg{n} )
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### `Answer`
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![](http://latex.codecogs.com/gif.latex? n^{\\log_{b}{a}} = n^{\\log_{2}{4}} = n^2)
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a. ![](http://latex.codecogs.com/gif.latex? \\Theta\(n^2\) )
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b. ![](http://latex.codecogs.com/gif.latex? \\Theta\(n^2 \\lg{n}\) )
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c. ![](http://latex.codecogs.com/gif.latex? \\Theta\(n^3\) )
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### Exercises 4.3-2
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***
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The recurrence T(n) = 7T (n/2)+n2 describes the running time of an algorithm A. A competing algorithm A′ has a running time of T′(n) = aT′(n/4) + n2. What is the largest integer value for a such that A′ is asymptotically faster than A?
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### `Answer`
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根据主定理，算法A的运行时间为![](http://latex.codecogs.com/gif.latex? T\(n\) = \\Theta\(\\lg{7}\)\ \\approx n^{2.8} )
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A'的运行时间在a > 16时超过n^2,此时
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![](http://latex.codecogs.com/gif.latex? T\(n\) = \\Theta\(n^{\\log_{4}{a}}\) < \\lg{7} = \\log_{4}{49})
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所以最大值为48
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### Exercises 4.3-3
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***
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Use the master method to show that the solution to the binary-search recurrence T(n) = T (n/2)
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+ Θ(1) is T(n) = Θ(lg n). (See Exercise 2.3-5 for a description of binary search.)
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### `Answer`
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![](http://latex.codecogs.com/gif.latex? n^{\\log_{b}{a}} = n^{\\log_{2}{1} = 1} )
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so the solution is Θ(lgn).
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### Exercises 4.3-4
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***
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Can the master method be applied to the recurrence
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![](http://latex.codecogs.com/gif.latex? T\(n\) = 4T\(n/2\) + n^2 \\lg{n} )
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Why or why not? Give an asymptotic upper bound for this recurrence.
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### `Answer`
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![](http://latex.codecogs.com/gif.latex? n^{\\log_{b}{a}} = n^{\\log_{2}{4}} = n^2 )
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The problem is that it is not polynomially larger. The ratio 
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![](http://latex.codecogs.com/gif.latex? f\(n\) / n^{\\log_{b}{a}} = \\lg{n})
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is asymptotically less than
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![](http://latex.codecogs.com/gif.latex? n^\\epsilon ) for any positive constant
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![](http://latex.codecogs.com/gif.latex? \\epsilon )
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### Exercises 4.3-5
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***
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Consider the regularity condition af (n/b) ≤ cf(n) for some constant c < 1, which is part of case 3 of the master theorem. Give an example of constants a ≥ 1 and b > 1 and a function f (n) that satisfies all the conditions in case 3 of the master theorem except the regularity condition.
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### `Answer`
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let
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a = 1
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b = 2
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f(n) = 2 - cos(n)
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我们需要证明
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![](http://latex.codecogs.com/gif.latex? \\frac{n}{2}\( 2 - \\cos\(\\frac{n}{2}\)}\) < cn \\\\ ~ \\Rightarrow c > \\frac{2- \\cos\(n/2\)}{2} \\\\ ~
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\\Rightarrow c > \\frac{3}{2}
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)
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***
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Follow [@louis1992](https://github.com/gzc) on github to help finish this task.
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### `Answer`
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![](http://latex.codecogs.com/gif.latex? n^{\\log_{b}{a}} = n^{\\log_{2}{4}} = n^2 )
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The problem is that it is not polynomially larger. The ratio 
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![](http://latex.codecogs.com/gif.latex? f\(n\) / n^{\\log_{b}{a}} = \\lg{n})
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is asymptotically less than
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![](http://latex.codecogs.com/gif.latex? n^\\epsilon ) for any positive constant
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![](http://latex.codecogs.com/gif.latex? \\epsilon )
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### Exercises 4.3-5
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***
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Consider the regularity condition af (n/b) ≤ cf(n) for some constant c < 1, which is part of case 3 of the master theorem. Give an example of constants a ≥ 1 and b > 1 and a function f (n) that satisfies all the conditions in case 3 of the master theorem except the regularity condition.
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### `Answer`
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let
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a = 1
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b = 2
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f(n) = 2 - cos(n)
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我们需要证明
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![](http://latex.codecogs.com/gif.latex? \\frac{n}{2}\( 2 - \\cos\(\\frac{n}{2}\)}\) < cn \\\\ ~ \\Rightarrow c > \\frac{2- \\cos\(n/2\)}{2} \\\\ ~
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\\Rightarrow c > \\frac{3}{2}
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)
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***
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Follow [@louis1992](https://github.com/gzc) on github to help finish this task.
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