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| Original file line number | Diff line number | Diff line change |
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| @@ -1,88 +1,89 @@ | ||
| ### Exercises 4.2-1 | ||
| *** | ||
| Use a recursion tree to determine a good asymptotic upper bound on the recurrence | ||
| ### Exercises 4.2-1 | ||
| *** | ||
| Use a recursion tree to determine a good asymptotic upper bound on the recurrence | ||
|  = 3T\(\\lceil n/2 \\rceil\) + n). Use the substitution method to verify your answer. | ||
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| ### `Answer` | ||
|  | ||
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| 树的高度是lgn,有3^lgn个叶子节点. | ||
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|  = n\\sum_{i = 0}^{lg\(n\)-1}\(\\frac{3}{2}\)^i + \\Theta\(3^{\\lg{n}}\) \\\\ ~ | ||
| \\hspace{15 mm} = \\Theta\(n^{\\lg{3}}\) + \\Theta\(3^{\\lg{n}}\) \\\\ ~ | ||
| \\hspace{15 mm} = \\Theta\(n^{\\lg{3}}\) + \\Theta\(n^{\\lg{3}}\) \\\\ ~ | ||
| \\hspace{15 mm} = \\Theta\(n^{\\lg{3}}\) | ||
| ) | ||
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| 我们猜想  \\le cn^{\\lg{3}}+2n ) | ||
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|  \\le 3c\(n/2\)^{\\lg{3}} + 2n \\\\ ~ | ||
| \\hspace{15 mm} \\le cn^{\\lg{3}}+2n \\\\ ~ | ||
| \\hspace{15 mm} = \\Theta\(n^{\\lg{3}}\) | ||
| ) | ||
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| ### Exercises 4.2-2 | ||
| *** | ||
| Argue that the solution to the recurrence | ||
|  = T\(n/3\) + T\(2n/3\) + cn ) | ||
| , where c is a constant, is Ω(nlgn) by appealing to the recurrsion tree. | ||
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| ### `Answer` | ||
| 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 | ||
| *** | ||
| Draw the recursion tree for | ||
|  = 4T\(\\lfloor n/2 \\rfloor\) + cn) | ||
| ,where c is a constant, and provide a tight asymptotic bound on its solution. Verify your bound by the substitution method. | ||
| ### `Answer` | ||
|  | ||
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| 很明显是n^2的级别 | ||
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| 我们假设  \\le n^2+2cn) | ||
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|  \\le 4c\(n/2\)^2 + 2cn/2+cn \\le cn^2+2cn) | ||
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| 我们假设  \\ge n^2+2cn) | ||
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|  \\ge 4c\(n/2\)^2 + 2cn/2+cn \\ge cn^2+2cn) | ||
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| ### Exercises 4.2-4 | ||
| *** | ||
| 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` | ||
|  | ||
| file:///Users/ganzhenchao/Workspaces/CLRS/C04-Recurrences/repo/s2/4.png | ||
|  = \\sum_{i=0}^{n/a}c\(n-ia\) + \(n/a\)ca | ||
| = \\Theta\(n^2\)) | ||
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| 我们假设  \\le cn^2) | ||
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|  \\le c\(n-a\)^2 + ca + cn \\\\ ~ | ||
| \\hspace{15 mm} \\le cn^2-2acn+ca+cn \\\\ ~ | ||
| \\hspace{15 mm} \\le cn^2-c\(2an-a-n\) \\\\ ~ | ||
| \\hspace{15 mm}\\le cn^2 - cn ~~~~ if ~~ a > 1/2,n > 2a \\\\ ~ | ||
| \\hspace{15 mm}\\le cn^2 \\\\ ~ | ||
| \\hspace{15 mm} = \\Theta\(n^2\) | ||
| ) | ||
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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 | ||
| *** | ||
| 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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| 可以假设α < 1/2,因此树的高度有 | ||
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|  = \\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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| *** | ||
| Follow [@louis1992](https://github.com/gzc) on github to help finish this task. | ||
|
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| ### `Answer` | ||
|  | ||
|
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| 树的高度是lgn,有3^lgn个叶子节点. | ||
|
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|  = n\\sum_{i = 0}^{lg\(n\)-1}\(\\frac{3}{2}\)^i + \\Theta\(3^{\\lg{n}}\) \\\\ ~ | ||
| \\hspace{15 mm} = \\Theta\(n^{\\lg{3}}\) + \\Theta\(3^{\\lg{n}}\) \\\\ ~ | ||
| \\hspace{15 mm} = \\Theta\(n^{\\lg{3}}\) + \\Theta\(n^{\\lg{3}}\) \\\\ ~ | ||
| \\hspace{15 mm} = \\Theta\(n^{\\lg{3}}\) | ||
| ) | ||
|
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||
| 我们猜想  \\le cn^{\\lg{3}}+2n ) | ||
|
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||
|  \\le 3c\(n/2\)^{\\lg{3}} + 2n \\\\ ~ | ||
| \\hspace{15 mm} \\le cn^{\\lg{3}}+2n \\\\ ~ | ||
| \\hspace{15 mm} = \\Theta\(n^{\\lg{3}}\) | ||
| ) | ||
|
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||
|
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||
| ### Exercises 4.2-2 | ||
| *** | ||
| Argue that the solution to the recurrence | ||
|  = T\(n/3\) + T\(2n/3\) + cn ) | ||
| , where c is a constant, is Ω(nlgn) by appealing to the recurrsion tree. | ||
|
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||
| ### `Answer` | ||
| 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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||
|
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| ### Exercises 4.2-3 | ||
| *** | ||
| Draw the recursion tree for | ||
|  = 4T\(\\lfloor n/2 \\rfloor\) + cn) | ||
| ,where c is a constant, and provide a tight asymptotic bound on its solution. Verify your bound by the substitution method. | ||
| ### `Answer` | ||
|  | ||
|
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||
| 很明显是n^2的级别 | ||
|
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||
| 我们假设  \\le n^2+2cn) | ||
|
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|  \\le 4c\(n/2\)^2 + 2cn/2+cn \\le cn^2+2cn) | ||
|
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| 我们假设  \\ge n^2+2cn) | ||
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|  \\ge 4c\(n/2\)^2 + 2cn/2+cn \\ge cn^2+2cn) | ||
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| ### Exercises 4.2-4 | ||
| *** | ||
| 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` | ||
|  | ||
| file:///Users/ganzhenchao/Workspaces/CLRS/C04-Recurrences/repo/s2/4.png | ||
|  = \\sum_{i=0}^{n/a}c\(n-ia\) + \(n/a\)ca | ||
| = \\Theta\(n^2\)) | ||
|
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| 我们假设  \\le cn^2) | ||
|
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|  \\le c\(n-a\)^2 + ca + cn \\\\ ~ | ||
| \\hspace{15 mm} \\le cn^2-2acn+ca+cn \\\\ ~ | ||
| \\hspace{15 mm} \\le cn^2-c\(2an-a-n\) \\\\ ~ | ||
| \\hspace{15 mm}\\le cn^2 - cn ~~~~ if ~~ a > 1/2,n > 2a \\\\ ~ | ||
| \\hspace{15 mm}\\le cn^2 \\\\ ~ | ||
| \\hspace{15 mm} = \\Theta\(n^2\) | ||
| ) | ||
|
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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 | ||
| *** | ||
| 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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| 可以假设α < 1/2,因此树的高度有 | ||
|
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|  = \\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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| *** | ||
| Follow [@louis1992](https://github.com/gzc) on github to help finish this task. | ||
|
|
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -1,79 +1,80 @@ | ||
| ### Exercises 4.3-1 | ||
| *** | ||
| Use the master method to give tight asymptotic bounds for the following recurrences. | ||
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| a.  = 4T\(n/2\)+n ) | ||
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| b.  = 4T\(n/2\)+n^2 ) | ||
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| ### Exercises 4.3-1 | ||
| *** | ||
| Use the master method to give tight asymptotic bounds for the following recurrences. | ||
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| a.  = 4T\(n/2\)+n ) | ||
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| b.  = 4T\(n/2\)+n^2 ) | ||
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| c.  = 4T\(n/2\)+n^3 ) | ||
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| ### `Answer` | ||
|  | ||
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| a.  ) | ||
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| b.  ) | ||
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| c.  ) | ||
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| ### Exercises 4.3-2 | ||
| *** | ||
| 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` | ||
| 根据主定理，算法A的运行时间为 = \\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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|  = \\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 | ||
| *** | ||
| 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.) | ||
| ### `Answer` | ||
|  | ||
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| so the solution is Θ(lgn). | ||
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| ### Exercises 4.3-4 | ||
| *** | ||
| Can the master method be applied to the recurrence | ||
|  = 4T\(n/2\) + n^2 \\lg{n} ) | ||
| ### `Answer` | ||
|  | ||
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| a.  ) | ||
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| b.  ) | ||
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| c.  ) | ||
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| ### Exercises 4.3-2 | ||
| *** | ||
| 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` | ||
| 根据主定理，算法A的运行时间为 = \\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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|  = \\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 | ||
| *** | ||
| 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.) | ||
| ### `Answer` | ||
|  | ||
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| so the solution is Θ(lgn). | ||
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| ### Exercises 4.3-4 | ||
| *** | ||
| Can the master method be applied to the recurrence | ||
|  = 4T\(n/2\) + n^2 \\lg{n} ) | ||
| Why or why not? Give an asymptotic upper bound for this recurrence. | ||
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| ### `Answer` | ||
|  | ||
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| The problem is that it is not polynomially larger. The ratio  | ||
|  / n^{\\log_{b}{a}} = \\lg{n}) | ||
| is asymptotically less than | ||
|  for any positive constant | ||
|  | ||
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| ### Exercises 4.3-5 | ||
| *** | ||
| 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` | ||
| let | ||
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| a = 1 | ||
| b = 2 | ||
| f(n) = 2 - cos(n) | ||
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| 我们需要证明 | ||
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| }\) < cn \\\\ ~ \\Rightarrow c > \\frac{2- \\cos\(n/2\)}{2} \\\\ ~ | ||
| \\Rightarrow c > \\frac{3}{2} | ||
| ) | ||
|
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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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| The problem is that it is not polynomially larger. The ratio  | ||
|  / n^{\\log_{b}{a}} = \\lg{n}) | ||
| is asymptotically less than | ||
|  for any positive constant | ||
|  | ||
|
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| ### Exercises 4.3-5 | ||
| *** | ||
| 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` | ||
| let | ||
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| a = 1 | ||
| b = 2 | ||
| f(n) = 2 - cos(n) | ||
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| 我们需要证明 | ||
|
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| }\) < cn \\\\ ~ \\Rightarrow c > \\frac{2- \\cos\(n/2\)}{2} \\\\ ~ | ||
| \\Rightarrow c > \\frac{3}{2} | ||
| ) | ||
|
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| *** | ||
| Follow [@louis1992](https://github.com/gzc) on github to help finish this task. | ||
|
|
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