some modifications in thu_dsa/chp1, adding two functions in lcs/, updating exercises.md.
This commit is contained in:
Shine wOng
@@ -35,3 +35,139 @@
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(<, 1; 0, R, >);
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(>, #; #, R, h);//halt
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```
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## Introduction.E
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> 做递归跟踪分析时，为什么递归调用语句本身可不统计？
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因为递归语句本身的执行时间，是计入了对应的子实例当中，对于当前实例而言，只需要考虑函数调用的跳转执行，该执行的执行时间是`O(1)`。
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> 试用递归跟踪法，分析`fib()`二分递归版的复杂度。通过递归跟踪，解释该版本复杂度过高的原因
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可以针对`fib(n)`画出递归跟踪图，如下：
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```
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fib(n)
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/ \
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n-1 n-2
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/ \ / \
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n-2 n-3 n-3 n-4
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/
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....
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```
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可以看到，该递归跟踪树的高度为`h = n - 1`，并且其中最高的满二叉树子树高度为`n / 2`，因此二分递归版本的复杂度下界为
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$$
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1 + 2 + 4 + \cdots + 2^{\frac{n}{2}} = \Omega(2^{\frac{n}{2}}) = \Omega(\sqrt{2}^n)
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$$
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而复杂度上界为
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$$
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1 + 2 + 4 + \cdots + 2^{n - 1} = O(2^n)
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$$
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从上述递归跟踪图中也可以看到，二分递归版本复杂度过高的原因是其中具有大量重复计算的值。一般地，设`fib(k)`的出现次数为`nfib(k)`，则有每一个`fib(k+1)`和`fib(k+2)`都会产生一个`fib(k)`，因此
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```c
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nfib(k) = nfib(k+1) + nfib(k+2), 1 <= k <= n
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```
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并且有
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```c
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nfib(n) = 1, nfib(n - 1) = 1
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```
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因此，
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```c
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nfib(k) = fib(n - k + 1), 1 <= k <= n
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```
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> 递归算法的空间复杂度，主要取决于什么因素？
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递归深度。
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> 本节数组求和问题的两个（线性和二分）递归算法时间复杂度相同，空间呢？
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每一次递归子问题的空间复杂度都是`O(1)`。
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对于线性递归算法，递归深度为`O(n)`，因此空间复杂度为`O(n)`。
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而二分递归算法，递归深度为`O(logn)`，因此空间复杂度为`O(logn)`。
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> 自学递推式的一般求解性方法及规律`Master Theorem`。
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看这篇总结[master_theorem](master_theorem.md)
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## Introduction.F
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> 本节所介绍的迭代式`LCS`算法，似乎需要记录每个子问题的局部解，从而导致空间复杂度激增。实际上，这既不现实，亦无必要。试改进该算法，使得每个子问题只需要常数空间，即可保证最终得到`LCS`的组成（而非仅仅长度）
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这里说的`LCS`算法，应该是返回`LCS`的组成的，并没有在邓公的课件、教材、网课上找到，不过可以从返回`LCS`长度的迭代算法中推广得到。
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在`LCS`长度的迭代式算法中，需要维护一个`m*n`的向量，来保存各个子问题的`LCS`长度，仿照其思路，在上述`m*n`的向量中，保存各个子问题的`LCS`序列，即可构造出一种`LCS`组成的算法。容易看出，由于每个子问题都需要保存当前的`LCS`，子问题的空间复杂度为`O(min(m, n))`，因此整体的空间复杂度为`O(m*n(min(m, n)))`，的确增加了不少，题目就是要求对这种情况进行改进。
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可以注意到，上述算法空间复杂度激增的原因，是保存了大量重复的内容。实际上，构成最终`LCS`序列的字符，只在`O(min(m, n))`个位置出现。因此可以仿照图剪枝的策略，在填充向量时动态地记录当前的移动方向，即是从对角线更新，还是从左侧元素更新，还是从上侧元素更新。遍历完成后，再沿着前面标记的方向进行一次反向的遍历，在该过程中记录`LCS`各个字符，从而可以得到整体的`LCS`的组成。该反向遍历至多只会进行`O(m + n)`次，对整体的时间复杂度`O(m*n)`没有显著影响，同时每个子问题的空间复杂度都下降到`O(1)`。该算法的代码如下：
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```cpp
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string lcsIt(string one, string two, int len1, int len2){
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string lcs;
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if (len1 == 0 || len2 == 0) return lcs;
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vector<vector<State>> states(len1 + 1, vector<State>(len2 + 1));
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for(int i = 0; i != len1; ++i){
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for(int j = 0; j != len2; ++j){
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if(one[i] == two[j]){
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states[i + 1][j + 1].len = states[i][j].len + 1;
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states[i + 1][j + 1].dir = DIAGON;
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}
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else{
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if(states[i][j + 1].len < states[i + 1][j].len){
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states[i + 1][j + 1].len = states[i + 1][j].len;
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states[i + 1][j + 1].dir = LEFT;
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}else{
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states[i + 1][j + 1].len = states[i][j + 1].len;
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states[i + 1][j + 1].dir = UPPER;
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}
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}
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}
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}
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lcs.resize(states[len1][len2].len);
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int pos = lcs.size();
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for(int i = len1, j = len2; i > 0 && j > 0; ){
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switch(states[i][j].dir){
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case DIAGON:
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lcs[--pos] = one[i - 1];
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--i, --j;
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break;
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case UPPER:
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--i;
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break;
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case LEFT:
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--j;
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break;
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default:
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exit(-1);
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}
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}
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return lcs;
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}
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```
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需要指出的是，对于多个`LCS`的情形，该算法只会返回其中的一个解，因为在算法中对于两个子问题的`LCS`长度相同的情况，是优先选择上面`UPPER`的子问题。
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> 考查序列`A = "immaculate`和`B = "computer`。1）它们的`LCS`是什么；2）这里的解是否唯一？是否有歧义性？3）按照本节所给的算法，找出的是哪一个解？
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1）`"mute"`和`"cute"`
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2）所以显然不唯一，有歧义性。
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3）按照上面给的算法，优先从上方进行更新，即优先选择序列`A`更靠前的字符，即找出的是`"mute"`。
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> 实现`LCS`算法的递归版和迭代版，并通过实测比较运行时间。
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代码和测试分别放在[lcs.cpp](lcs/lcs.cpp)和[test_lcs.cpp](lcs/test_lcs.cpp)了。递归版的确很慢......
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> 采用`memorization`策略，改进`fib()`和`LCS()`的递归版
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不想写了......
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@@ -1,22 +1,22 @@
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#include "lcs.h"
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#include <algorithm>
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int lcsRe(string one, string two){
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int lcslenRe(string one, string two, int len1, int len2){
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int len;
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if(one.empty() || two.empty()) return 0;
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if(len1 == 0 || len2 == 0) return 0;
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int len1 = one.size(), len2 = two.size();
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if(one[len1 - 1] == two[len2 - 1]){
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len = lcsRe(one.substr(0, len1 - 1), two.substr(0, len2 - 1)) + 1;
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len = lcslenRe(one.substr(0, len1 - 1), two.substr(0, len2 - 1), len1 - 1, len2 - 1) + 1;
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return len;
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}
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len = MAX(lcsRe(one.substr(0, len1 - 1), two), lcsRe(one, two.substr(0, len2 - 1)));
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return len;
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int lenone = lcslenRe(one.substr(0, len1 - 1), two, len1 - 1, len2);
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int lentwo = lcslenRe(one, two.substr(0, len2 - 1), len1, len2 - 1);
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return lenone > lentwo ? lenone : lentwo;
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}
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int lcsIt(string one, string two){
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int lcslenIt(string one, string two, int len1, int len2){
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if(one.empty() || two.empty()) return 0;
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int len1 = one.size(), len2 = two.size();
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vector<vector<int>> lens(len1 + 1, vector<int>(len2 + 1, 0));
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for(int ix = 0; ix != len1; ++ix){
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for(int jx = 0; jx != len2; ++jx){
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@@ -29,3 +29,90 @@ int lcsIt(string one, string two){
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return lens[len1][len2];
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}
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vector<string> lcsRe(string one, string two, int len1, int len2){
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vector<string> lcs;
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if (len1 == 0 || len2 == 0) return lcs;
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vector<string> lcs1, lcs2;
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if(one[len1 - 1] == two[len2 - 1]){
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lcs = lcsRe(one.substr(0, len1 - 1), two.substr(0, len2 - 1), len1 - 1, len2 - 1);
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if (lcs.empty()) lcs.push_back(one.substr(len1 - 1, 1));
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else
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for (auto It = lcs.begin(); It != lcs.end(); ++It)
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It->append(1, one[len1 - 1]);
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return lcs;
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}
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lcs1 = lcsRe(one.substr(0, len1 - 1), two, len1 - 1, len2);
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lcs2 = lcsRe(one, two.substr(0, len2 - 1), len1, len2 - 1);
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if (lcs1.empty()) lcs = lcs2;
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else if (lcs2.empty()) lcs = lcs1;
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else{
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if (lcs1[0].length() < lcs2[0].length()) lcs = lcs2;
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else if (lcs1[0].length() > lcs2[0].length()) lcs = lcs1;
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else {
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lcs = lcs1;
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for (string entry : lcs2)
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if(find(lcs.cbegin(), lcs.cend(), entry) == lcs.cend())
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lcs.push_back(entry);
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}
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}
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return lcs;
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}
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enum direction { LEFT, UPPER, DIAGON, UNINIT };
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class State{
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public:
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int len;
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direction dir;
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State(){
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len = 0;
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dir = UNINIT;
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}
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};
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string lcsIt(string one, string two, int len1, int len2){
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string lcs;
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if (len1 == 0 || len2 == 0) return lcs;
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vector<vector<State>> states(len1 + 1, vector<State>(len2 + 1));
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for(int i = 0; i != len1; ++i){
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for(int j = 0; j != len2; ++j){
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if(one[i] == two[j]){
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states[i + 1][j + 1].len = states[i][j].len + 1;
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states[i + 1][j + 1].dir = DIAGON;
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}
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else{
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if(states[i][j + 1].len < states[i + 1][j].len){
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states[i + 1][j + 1].len = states[i + 1][j].len;
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states[i + 1][j + 1].dir = LEFT;
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}else{
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states[i + 1][j + 1].len = states[i][j + 1].len;
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states[i + 1][j + 1].dir = UPPER;
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}
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}
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}
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}
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lcs.resize(states[len1][len2].len);
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int pos = lcs.size();
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for(int i = len1, j = len2; i > 0 && j > 0; ){
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switch(states[i][j].dir){
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case DIAGON:
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lcs[--pos] = one[i - 1];
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--i, --j;
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break;
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case UPPER:
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--i;
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break;
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case LEFT:
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--j;
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break;
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default:
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exit(-1);
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}
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}
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return lcs;
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}
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@@ -3,6 +3,9 @@
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#define MAX(X,Y) ((X)>(Y)?(X):(Y))
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using namespace std;
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int lcsIt(string one, string two);
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int lcsRe(string one, string two);
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int lcslenIt(string one, string two, int len1, int len2);
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int lcslenRe(string one, string two, int len1, int len2);
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string lcsIt(string one, string two, int len1, int len2);
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vector<string> lcsRe(string one, string two, int len1, int len2);
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@@ -2,41 +2,76 @@
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#include <cassert>
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#include "lcs.h"
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#include <time.h>
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#define NUMOFLOOP 10
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#define NUMOFLOOP 1
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using namespace std;
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void test_lcsIt();
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void test_lcsRe();
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void test_lcslenIt();
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void test_lcslenRe();
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int main(){
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test_lcsIt();
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test_lcslenIt();
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test_lcslenRe();
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test_lcsRe();
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}
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void test_lcsIt(){
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clock_t begin, end;
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int len;
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begin = clock();
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for(int ix = 0; ix != NUMOFLOOP; ++ix)
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len = lcsIt(string("educational"), string("advantage"));
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end = clock();
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assert(len == 4);
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assert(lcsIt(string("didactical"), string("advantage")) == 4);
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cout << "Iterative lcs test passed." << endl;
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cout << "Running time: " << end - begin << endl;
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test_lcsIt();
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system("pause");
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}
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void test_lcsRe(){
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clock_t begin, end;
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int len;
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begin = clock();
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for(int ix = 0; ix != NUMOFLOOP; ++ix)
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len = lcsRe(string("educational"), string("advantage"));
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//vector<string> lcs = lcsRe(string("educational"), string("advantage"), 11, 9);
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vector<string> lcs = lcsRe(string("immaculate"), string("computer"), 10, 8);
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end = clock();
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assert(len == 4);
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assert(lcsRe(string("didactical"), string("advantager")) == 4);
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assert(!lcs.empty());
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assert(lcs[0].length() == 4);
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for (string entry : lcs)
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cout << entry << " ";
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cout << "\nRecursive lcs test passed." << endl;
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cout << "Running time: " << end - begin << endl;
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}
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void test_lcsIt() {
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clock_t begin, end;
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begin = clock();
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//string lcs = lcsIt(string("educational"), string("advantage"), 11, 9);
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string lcs = lcsIt(string("immaculate"), string("computer"), 10, 8);
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end = clock();
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assert(!lcs.empty());
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assert(lcs.length() == 4);
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cout << lcs << endl;
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cout << "Recursive lcs test passed." << endl;
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cout << "Running time: " << end - begin << endl;
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}
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void test_lcslenIt(){
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clock_t begin, end;
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int len;
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begin = clock();
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for(int ix = 0; ix != NUMOFLOOP; ++ix)
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len = lcslenIt(string("educational"), string("advantage"), 11, 9);
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end = clock();
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assert(len == 4);
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assert(lcslenIt(string("didactical"), string("advantage"), 10, 9) == 4);
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cout << "Iterative lcs test passed." << endl;
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cout << "Running time: " << end - begin << endl;
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}
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void test_lcslenRe(){
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clock_t begin, end;
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int len;
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begin = clock();
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for(int ix = 0; ix != NUMOFLOOP; ++ix)
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len = lcslenRe(string("educational"), string("advantage"), 11, 9);
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end = clock();
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assert(len == 4);
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assert(lcslenRe(string("didactical"), string("advantage"), 10, 9) == 4);
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cout << "Recursive lcs test passed." << endl;
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cout << "Running time: " << end - begin << endl;
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}
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2
words.md
2
words.md
@@ -1200,7 +1200,7 @@ Some Words
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+ philharmonic
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> used in the names of musical groups, especially orchestras
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- He will condcuct the Berlin Philharmonic in the final concert of the season.
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- He will conduct the Berlin Philharmonic in the final concert of the season.
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- the Vienna Philharmonic Orchestra.
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+ vibrant
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Reference in New Issue
Block a user