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[Algorithms II] Week 4-2 Substring Search
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目录
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<div id="toc"><ul><li><a class="toc-href" href="#1-introduction-to-substring-search" title="1. Introduction to substring search">1. Introduction to substring search</a></li><li><a class="toc-href" href="#2-brute-force-substring-search" title="2. Brute-Force Substring Search">2. Brute-Force Substring Search</a></li><li><a class="toc-href" href="#3-knuth-morris-pratt" title="3. Knuth-Morris-Pratt">3. Knuth-Morris-Pratt</a><ul><li><a class="toc-href" href="#intuition" title="intuition">intuition</a></li><li><a class="toc-href" href="#deterministic-finite-state-automaton-dfa" title="Deterministic finite state automaton (DFA)">Deterministic finite state automaton (DFA)</a></li><li><a class="toc-href" href="#dfa-construction" title="DFA construction">DFA construction</a></li></ul></li><li><a class="toc-href" href="#4-boyer-moore_1" title="4. Boyer-Moore">4. Boyer-Moore</a><ul><li><a class="toc-href" href="#intuition_1" title="intuition">intuition</a></li><li><a class="toc-href" href="#mismatch-character-heuristic" title="mismatch character heuristic">mismatch character heuristic</a></li><li><a class="toc-href" href="#implementation" title="implementation">implementation</a></li><li><a class="toc-href" href="#analysis" title="analysis">analysis</a></li></ul></li><li><a class="toc-href" href="#5-rabin-karp_1" title="5. Rabin-Karp">5. Rabin-Karp</a><ul><li><a class="toc-href" href="#intuition_2" title="intuition">intuition</a></li><li><a class="toc-href" href="#computing-the-hash-function-efficiently" title="computing the hash function efficiently">computing the hash function efficiently</a></li><li><a class="toc-href" href="#analysis_1" title="analysis">analysis</a></li></ul></li><li><a class="toc-href" href="#summery_1" title="Summery">Summery</a></li></ul></div>
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<h1 id="1-introduction-to-substring-search">1. Introduction to substring search</h1>
<p>"most ingenious algorithm we've seen so far"<br/>
<strong>pb. </strong>having two strings, <code>pattern</code> and <code>text</code>, len(pattern)=M << len(text)=N, try to find pattern in text. </p>
<p>ex. <code>indexOf</code> method of String in java. </p>
<h1 id="2-brute-force-substring-search">2. Brute-Force Substring Search</h1>
<p>function signature: <br/>
<code>public static int search(String pat, String txt);</code> </p>
<p>brute-force algo: look for <code>pattern</code> at every position of <code>text</code>. </p>
<div class="highlight"><pre><span class="code-line"><span></span><span class="err">public static int search(String pat, String txt){ </span></span>
<span class="code-line"><span class="err"> int N=txt.length(), M=pat.length(); </span></span>
<span class="code-line"><span class="err"> for(int i=0; i<=N-M; i++){ </span></span>
<span class="code-line"><span class="err"> int j; </span></span>
<span class="code-line"><span class="err"> for(j=0; j<M && pat.charAt(j)==txt.charAt(i+j); j++); </span></span>
<span class="code-line"><span class="err"> if(j==M) return i; </span></span>
<span class="code-line"><span class="err"> } </span></span>
<span class="code-line"><span class="err"> return N;// not found </span></span>
<span class="code-line"><span class="err">}</span></span>
</pre></div>
<p><strong>worst case: </strong>when txt/pat are repetitive → MN compares. <br/>
<img alt="" class="img-responsive" src="../images/algoII_week4_2/pasted_image.png"/><br/>
problem with brute-force: always <em>backup</em> when mismatch. <br/>
<img alt="" class="img-responsive" src="../images/algoII_week4_2/pasted_image001.png"/><br/>
<strong>brute-force alternative</strong> </p>
<ul>
<li>j := number of matched chars in pattern </li>
<li>i := index of the end of matched char in text </li>
</ul>
<p>⇒ do explicite backup when mismatch by <code>i -= j</code> </p>
<div class="highlight"><pre><span class="code-line"><span></span><span class="err">public static int search(String pat, String txt){ </span></span>
<span class="code-line"><span class="err"> int N=txt.length(), M=pat.length(), i=0, j=0; </span></span>
<span class="code-line"><span class="err"> while(i<N && j<M){ </span></span>
<span class="code-line"><span class="err"> if(pat.charAt(j)==txt.charAt(i)) </span></span>
<span class="code-line"><span class="err"> {j=i++; j++;} </span></span>
<span class="code-line"><span class="err"> else </span></span>
<span class="code-line"><span class="err"> {i=i-j+1; j=0}// <==backup </span></span>
<span class="code-line"><span class="err"> } </span></span>
<span class="code-line"><span class="err"> return j==M ? i-M : N; </span></span>
<span class="code-line"><span class="err">}</span></span>
</pre></div>
<p>challenge: want linear-time guarantee, and want to avoid backup. </p>
<h1 id="3-knuth-morris-pratt">3. Knuth-Morris-Pratt</h1>
<p><em>"one of the coolest/trickiest algorithm covered in this course"</em> </p>
<h3 id="intuition">intuition</h3>
<p>suppose pattern = "BAAAAA", <br/>
if we matched 5 chars in pattern and get mismatch on 6th char ⇒ we know the previous 5 chars are "BAAAA" → no need to backup the i pointer. </p>
<p>KMP algorithm: clever method that <em>always</em> avoid backup ! </p>
<h3 id="deterministic-finite-state-automaton-dfa">Deterministic finite state automaton (DFA)</h3>
<ul>
<li>finite states ,including start and halt state, indexed by j in the subtring pb </li>
<li>for each state: exactly one transition for each char in alphabet </li>
</ul>
<p>ex. <br/>
states are 0~6, pat="ABABAC", transitions are indexed by chars in alphabet = {A,B,C}, finish if we reach state-6. </p>
<blockquote>
<p><code>dfa[c][i]</code> = the next state if we are currently in state-i and encoutered char-c. </p>
</blockquote>
<p><img alt="" class="img-responsive" src="../images/algoII_week4_2/pasted_image002.png"/> </p>
<p><strong>interpretation of DFA for KMP algo</strong> </p>
<blockquote>
<p>in the DFA after reading <code>txt[i]</code>, the index of state is the <strong>number of matched chars</strong> in pattern, or length of *longest prefix of pat that is a suffix of txt[0:i]. * </p>
</blockquote>
<ul>
<li>need to precompute the <code>dfa[][]</code> array from pattern </li>
<li>the pointer i <em>never</em> decrements (thus we can do it in a <em>streaming</em> manner) </li>
</ul>
<p>→ <em>if</em> <code>dfa[][]</code> <em>is precomputed</em>, java code is very very simple: </p>
<div class="highlight"><pre><span class="code-line"><span></span><span class="k">public</span><span class="w"> </span><span class="k">static</span><span class="w"> </span><span class="nc">int</span><span class="w"> </span><span class="k">search</span><span class="p">(</span><span class="n">String</span><span class="w"> </span><span class="n">pat</span><span class="p">,</span><span class="w"> </span><span class="n">String</span><span class="w"> </span><span class="n">txt</span><span class="p">,</span><span class="w"> </span><span class="nc">int</span><span class="err">[][]</span><span class="w"> </span><span class="n">dfa</span><span class="p">)</span><span class="err">{</span><span class="w"> </span></span>
<span class="code-line"><span class="w"> </span><span class="nc">int</span><span class="w"> </span><span class="n">N</span><span class="o">=</span><span class="n">txt</span><span class="p">.</span><span class="n">length</span><span class="p">(),</span><span class="w"> </span><span class="n">M</span><span class="o">=</span><span class="n">pat</span><span class="p">.</span><span class="n">length</span><span class="p">(),</span><span class="w"> </span><span class="n">i</span><span class="p">,</span><span class="w"> </span><span class="n">j</span><span class="o">=</span><span class="mi">0</span><span class="p">;</span><span class="w"> </span></span>
<span class="code-line"><span class="w"> </span><span class="k">for</span><span class="p">(</span><span class="n">i</span><span class="o">=</span><span class="mi">0</span><span class="p">;</span><span class="w"> </span><span class="n">i</span><span class="o"><</span><span class="n">N</span><span class="w"> </span><span class="o">&&</span><span class="w"> </span><span class="n">j</span><span class="o"><</span><span class="n">M</span><span class="p">;</span><span class="w"> </span><span class="n">i</span><span class="o">++</span><span class="p">)</span><span class="w"> </span></span>
<span class="code-line"><span class="w"> </span><span class="n">j</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="n">dfa</span><span class="o">[</span><span class="n">txt.charAt(i)</span><span class="o">][</span><span class="n">j</span><span class="o">]</span><span class="p">;</span><span class="w"> </span></span>
<span class="code-line"><span class="w"> </span><span class="k">return</span><span class="w"> </span><span class="n">j</span><span class="o">==</span><span class="n">M</span><span class="w"> </span><span class="vm">?</span><span class="w"> </span><span class="n">i</span><span class="o">-</span><span class="n">M</span><span class="w"> </span><span class="err">:</span><span class="w"> </span><span class="n">N</span><span class="p">;</span><span class="w"> </span></span>
<span class="code-line"><span class="err">}</span><span class="w"></span></span>
</pre></div>
<p>running time: linear.<br/>
→ key pb: <em>how to build dfa efficiently</em> ? </p>
<h3 id="dfa-construction">DFA construction</h3>
<ul>
<li><strong>match transition </strong>(easy part) </li>
</ul>
<p>when at state j, for the char <code>c0==pat.charAt(j+1)</code>, just go on matching: <code>dfa[c0][j] = j+1</code><br/>
ex. (<em>pat="ABABAC"</em>)<br/>
<img alt="" class="img-responsive" src="../images/algoII_week4_2/pasted_image003.png"/><br/>
<img alt="" class="img-responsive" src="../images/algoII_week4_2/pasted_image004.png"/> </p>
<ul>
<li><strong>mismatch transition</strong> (hard part) </li>
</ul>
<p>(for j==0, things are simple: <code>dfa[c][0]=0</code> for all <code>c!=pat[0]</code>) </p>
<blockquote>
<ul>
<li><em>at state </em><code>j</code> (ie. j chars in pattern are matched)<em>, and for </em><code>c!=pat.charAt(j+1)</code> </li>
<li>⇒ we are in state j: we know <em>the <strong><em>last j chars in input</em></strong> are </em><code>pat[0...j-1]</code>, <em>and followed by char =</em> <code>c</code>, so the last j+1 chars of input string is: <code>pat[0...j-1]+c</code> </li>
<li>⇒ to compute dfa[c][j]: we can <strong>simulate as if we backup</strong><em>, ie. </em><code>i=i-j+1, j=0</code>. </li>
<li>if we go back to set j=0, and set i = i-j+1, then i is pointing at <code>pat[1]</code>, the text become <code>pat[1...j-1]+c</code>. We then let this string go through our dfa, the state that it achieves is the value of <code>dfa[c][j]</code>. </li>
</ul>
</blockquote>
<p><em>here is</em> <em>a concrete example:</em> <br/>
pattern = <code>"ABABAC"</code>, state <code>j=5</code>, char <code>c='B</code>' </p>
<blockquote>
<ul>
<li>we know the last 6 chars of the input = <code>pat[0...j-1]+c="ABABA"+"B"="ABABAB"</code> </li>
<li>if we backup, i will point to pat[1], the string is just <code>pat[1...j-1]+c="BABAB"</code> </li>
<li>we use the string "BABAB" as input and go through the partially constructed dfa, and see that we will reach state 4 </li>
<li>so we know <code>dfa['B'][5]=4</code> </li>
</ul>
</blockquote>
<p>similarly we can get <code>dfa['A'][5]=1</code>, as indicated below: <br/>
(<em>pat="ABABAC"</em>)<br/>
<img alt="" class="img-responsive" src="../images/algoII_week4_2/pasted_image005.png"/> </p>
<p>one concern: seems this simulation needs <code>j</code> steps ?<br/>
⇒ can be changed to be constant time if we maintain a <strong>state X := the state of simulating of input=pat[1...j-1]</strong><br/>
we maintain this state <code>X</code>, then for each <em>mismatched</em> char c, we just need to look at <code>dfa[c][X]</code>. <br/>
(<em>pat="ABABAC"</em>)<br/>
<img alt="" class="img-responsive" src="../images/algoII_week4_2/pasted_image006.png"/> </p>
<p><strong>[Algo]</strong> </p>
<blockquote>
<ul>
<li>set all matched transitions <code>dfa[c0][j] = j+1</code> for all <code>c0==pat[j]</code> </li>
<li>fill first column (j==0): <code>dfa[c][0]=0</code> for all <code>c!=pat[0]</code> </li>
<li>initialize <code>X=0</code> (state for empty input string) </li>
<li>for j=1 to M: <ul>
<li>for all <code>c!=pat[0]</code>: set <code>dfa[c][j] = dfa[c][X]</code> (DP here...) </li>
<li>update <code>X=dfa[c0][X]</code> ⇒ 注意, 此时X并<strong>不等于</strong>X+1(最开始<code>dfa[c0][j]=j+1</code>不适用于此), 为什么? 因为<code>c0==pat[j]</code> 而不<code>是pat[X]</code>!! 比如说最开始, j=1的时候X是等于0的!!! (这个弯我饶了好几分钟...) </li>
</ul>
</li>
</ul>
</blockquote>
<p>java code (can be written to be more compate): </p>
<div class="highlight"><pre><span class="code-line"><span></span><span class="k">public</span><span class="w"> </span><span class="nc">int</span><span class="err">[][]</span><span class="w"> </span><span class="n">constructDFA</span><span class="p">(</span><span class="n">String</span><span class="w"> </span><span class="n">pat</span><span class="p">)</span><span class="err">{</span><span class="w"> </span></span>
<span class="code-line"><span class="w"> </span><span class="nc">int</span><span class="w"> </span><span class="n">R</span><span class="o">=</span><span class="mi">256</span><span class="p">;</span><span class="o">//</span><span class="nf">ASCII</span><span class="w"> </span><span class="n">code</span><span class="w"> </span></span>
<span class="code-line"><span class="w"> </span><span class="nc">int</span><span class="w"> </span><span class="n">M</span><span class="o">=</span><span class="n">pat</span><span class="p">.</span><span class="n">length</span><span class="p">();</span><span class="w"> </span></span>
<span class="code-line"><span class="w"> </span><span class="nc">int</span><span class="err">[][]</span><span class="w"> </span><span class="n">dfa</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="k">new</span><span class="w"> </span><span class="nc">int</span><span class="o">[</span><span class="n">R</span><span class="o">][</span><span class="n">M</span><span class="o">]</span><span class="p">;</span><span class="w"> </span></span>
<span class="code-line"><span class="w"> </span><span class="o">//</span><span class="w"> </span><span class="mf">1.</span><span class="w"> </span><span class="n">fill</span><span class="w"> </span><span class="n">matched</span><span class="w"> </span><span class="nl">transitions</span><span class="p">:</span><span class="w"> </span><span class="n">dfa</span><span class="o">[</span><span class="n">pat.charAt(j)</span><span class="o">][</span><span class="n">j</span><span class="o">]</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="n">j</span><span class="o">+</span><span class="mi">1</span><span class="w"> </span></span>
<span class="code-line"><span class="w"> </span><span class="k">for</span><span class="p">(</span><span class="nc">int</span><span class="w"> </span><span class="n">j</span><span class="o">=</span><span class="mi">0</span><span class="p">;</span><span class="n">j</span><span class="o"><</span><span class="n">M</span><span class="p">;</span><span class="n">j</span><span class="o">++</span><span class="p">)</span><span class="w"> </span></span>
<span class="code-line"><span class="w"> </span><span class="n">dfa</span><span class="o">[</span><span class="n">pat.charAt(j)</span><span class="o">][</span><span class="n">j</span><span class="o">]</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="n">j</span><span class="o">+</span><span class="mi">1</span><span class="p">;</span><span class="w"> </span></span>
<span class="code-line"><span class="w"> </span><span class="o">//</span><span class="w"> </span><span class="mf">2.</span><span class="w"> </span><span class="n">fill</span><span class="w"> </span><span class="mi">1</span><span class="n">st</span><span class="w"> </span><span class="k">column</span><span class="w"> </span><span class="c1">--> can be ignored as java int default val=0 </span></span>
<span class="code-line"><span class="w"> </span><span class="o">//</span><span class="w"> </span><span class="mf">3.</span><span class="w"> </span><span class="n">fill</span><span class="w"> </span><span class="n">mismatched</span><span class="w"> </span><span class="n">transitions</span><span class="w"> </span></span>
<span class="code-line"><span class="w"> </span><span class="nc">int</span><span class="w"> </span><span class="n">X</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="mi">0</span><span class="p">;</span><span class="w"> </span></span>
<span class="code-line"><span class="w"> </span><span class="k">for</span><span class="p">(</span><span class="nc">int</span><span class="w"> </span><span class="n">j</span><span class="o">=</span><span class="mi">1</span><span class="p">;</span><span class="n">j</span><span class="o"><</span><span class="n">M</span><span class="p">;</span><span class="n">j</span><span class="o">++</span><span class="p">)</span><span class="err">{</span><span class="w"> </span></span>
<span class="code-line"><span class="w"> </span><span class="k">for</span><span class="p">(</span><span class="nc">int</span><span class="w"> </span><span class="n">c</span><span class="o">=</span><span class="mi">0</span><span class="p">;</span><span class="n">c</span><span class="o"><</span><span class="n">R</span><span class="p">;</span><span class="n">c</span><span class="o">++</span><span class="p">)</span><span class="w"> </span></span>
<span class="code-line"><span class="w"> </span><span class="k">if</span><span class="p">(</span><span class="n">c</span><span class="o">!=</span><span class="n">pat</span><span class="p">.</span><span class="n">charAt</span><span class="p">(</span><span class="n">j</span><span class="p">))</span><span class="w"> </span></span>
<span class="code-line"><span class="w"> </span><span class="n">dfa</span><span class="o">[</span><span class="n">c</span><span class="o">][</span><span class="n">j</span><span class="o">]</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="n">dfa</span><span class="o">[</span><span class="n">c</span><span class="o">][</span><span class="n">X</span><span class="o">]</span><span class="p">;</span><span class="w"> </span></span>
<span class="code-line"><span class="w"> </span><span class="n">X</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="n">dfa</span><span class="o">[</span><span class="n">pat.charAt(j)</span><span class="o">][</span><span class="n">X</span><span class="o">]</span><span class="p">;</span><span class="w"> </span></span>
<span class="code-line"><span class="w"> </span><span class="err">}</span><span class="w"> </span></span>
<span class="code-line"><span class="w"> </span><span class="k">return</span><span class="w"> </span><span class="n">dfa</span><span class="p">;</span><span class="w"> </span></span>
<span class="code-line"><span class="err">}</span><span class="w"></span></span>
</pre></div>
<p>running time and space: <strong>O(M*R)</strong>. </p>
<p><strong>prop.</strong> <br/>
KMP algorithm runs in O(M+N) time, and constructs the dfa in O(M*R) time/space. </p>
<p>这个KMP算法, 我曾经想过好几个小时, 然后最后写出了特别复杂的代码, 虽然可以用但是基本写了就忘掉了. 但是经过老爷子这么一讲, 感觉这次印象深刻了好多. 老爷子NB... </p>
<p>八卦时间: <br/>
<img alt="" class="img-responsive" src="../images/algoII_week4_2/pasted_image007.png"/> </p>
<h1 id="4-boyer-moore_1">4. Boyer-Moore</h1>
<p>Heuristic in practice.<br/>
i does not necessarily go through all txt chars ⇒ i may <em>skip</em> some chars. </p>
<h3 id="intuition_1">intuition</h3>
<ul>
<li>for matching: scan chars <em>from right to left</em> (j will decrease when checking) </li>
<li>when encoutered a mismatch: we can skip <= M chars (if the char is not in pattern) </li>
</ul>
<p>ex. (pat="NEEDLE")<br/>
<img alt="" class="img-responsive" src="../images/algoII_week4_2/pasted_image008.png"/> </p>
<p>→ pb: how to skip? </p>
<h3 id="mismatch-character-heuristic">mismatch character heuristic</h3>
<p><em>note</em>: the <code>i</code> always points to the <em>beginning</em> of the substring (<code>txt[i,...,i+M-1]</code>) to be checked for match. </p>
<p><strong>case 1. mismatched char not in pattern</strong><br/>
easy case → just move i to the right of this char. <br/>
<img alt="" class="img-responsive" src="../images/algoII_week4_2/pasted_image009.png"/> </p>
<p><strong>case 2. mismatched char in pattern</strong> </p>
<blockquote>
<p><strong>heuristic:</strong> line up i with the <strong>rightmost</strong><em> char in pattern</em>.<br/>
<code>i += skip</code><br/>
where <em>skip length = j - index of rightmost char in pattern</em> </p>
</blockquote>
<p><img alt="" class="img-responsive" src="../images/algoII_week4_2/pasted_image010.png"/> </p>
<p>note: this does not always help, in the example below, i even <em>backups</em>: <br/>
<img alt="" class="img-responsive" src="../images/algoII_week4_2/pasted_image011.png"/><br/>
to avoid backup, in this case we just <em>increment i by 1</em> (heuristic doesn't help in this case). </p>
<h3 id="implementation">implementation</h3>
<p>use an array <code>right[]</code> as <em>skip table</em>, <code>right[c]</code> is the index of rightmost occurrence of char c (-1 if c not in pat). <br/>
<img alt="" class="img-responsive" src="../images/algoII_week4_2/pasted_image012.png"/> </p>
<div class="highlight"><pre><span class="code-line"><span></span><span class="nc">int</span><span class="err">[]</span><span class="w"> </span><span class="nf">right</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="k">new</span><span class="w"> </span><span class="nc">int</span><span class="o">[</span><span class="n">M</span><span class="o">]</span><span class="p">;</span><span class="w"> </span></span>
<span class="code-line"><span class="k">for</span><span class="p">(</span><span class="nc">int</span><span class="w"> </span><span class="n">i</span><span class="o">=</span><span class="mi">0</span><span class="p">;</span><span class="n">i</span><span class="o"><</span><span class="n">R</span><span class="p">;</span><span class="n">i</span><span class="o">++</span><span class="p">)</span><span class="w"> </span><span class="nf">right</span><span class="o">[</span><span class="n">i</span><span class="o">]</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="o">-</span><span class="mi">1</span><span class="p">;</span><span class="o">//</span><span class="k">value</span><span class="w"> </span><span class="k">for</span><span class="w"> </span><span class="n">chars</span><span class="w"> </span><span class="ow">not</span><span class="w"> </span><span class="ow">in</span><span class="w"> </span><span class="n">pattern</span><span class="w"> </span></span>
<span class="code-line"><span class="k">for</span><span class="p">(</span><span class="nc">int</span><span class="w"> </span><span class="n">j</span><span class="o">=</span><span class="mi">0</span><span class="p">;</span><span class="n">j</span><span class="o"><</span><span class="n">M</span><span class="p">;</span><span class="n">j</span><span class="o">++</span><span class="p">)</span><span class="err">{</span><span class="w"> </span></span>
<span class="code-line"><span class="w"> </span><span class="nf">right</span><span class="o">[</span><span class="n">pat.charAt(j)</span><span class="o">]=</span><span class="n">j</span><span class="p">;</span><span class="w"> </span></span>
<span class="code-line"><span class="err">}</span><span class="w"></span></span>
</pre></div>
<p>using this table we can implemente the heuristic algorithm: </p>
<div class="highlight"><pre><span class="code-line"><span></span><span class="err">public static int search(String pat, String txt, int[] right){ </span></span>
<span class="code-line"><span class="err"> int N=txt.length(), M=pat.length(); </span></span>
<span class="code-line"><span class="err"> int skip; </span></span>
<span class="code-line"><span class="err"> for(int i=0;i<N-M;i+=skip){ </span></span>
<span class="code-line"><span class="err"> skip = 0; </span></span>
<span class="code-line"><span class="err"> for(int j=M-1;j>=0;j++) </span></span>
<span class="code-line"><span class="err"> if(pat.charAt(j)!=txt.charAt(i+j)){// when mismatch happens </span></span>
<span class="code-line"><span class="err"> skip = Math.max(1,j - right[txt.charAt(i+j)]);// skip if we can, else just increment i by 1 </span></span>
<span class="code-line"><span class="err"> break; </span></span>
<span class="code-line"><span class="err"> } </span></span>
<span class="code-line"><span class="err"> if(skip==0)// if the above for loop finishes without changing skip --> we are done. </span></span>
<span class="code-line"><span class="err"> return i; </span></span>
<span class="code-line"><span class="err"> } </span></span>
<span class="code-line"><span class="err"> return N;// pattern not found </span></span>
<span class="code-line"><span class="err">}</span></span>
</pre></div>
<h3 id="analysis">analysis</h3>
<p><strong>property</strong>. the Boyer-Moore heuristic (in practice) takes about <strong>N/M</strong> (sublinear!) compares to search. </p>
<p>好神奇, 比KMP还要简单的算法, 实际效率这么高... </p>
<p>worst-case performance: <strong>N*M</strong>... 这一点不如KMP. <br/>
<img alt="" class="img-responsive" src="../images/algoII_week4_2/pasted_image013.png"/><br/>
→ can be improved... </p>
<h1 id="5-rabin-karp_1">5. Rabin-Karp</h1>
<p>两个图灵奖的大神发明的算法.. </p>
<h3 id="intuition_2">intuition</h3>
<p>basic idea: <strong>modular hashing</strong><br/>
ex. for strings of numbers </p>
<ul>
<li>compute hash fcn (for number strings is easy: take the string and treat it as a number, then %Q where Q is a big prime number). </li>
<li>for a pointer i →corresponds to the substring <code>txt[i, ..., i+M-1]</code> → check hash for match </li>
</ul>
<p>(below: text=3141592653589793, pattern=26535)<br/>
<img alt="" class="img-responsive" src="../images/algoII_week4_2/pasted_image014.png"/> </p>
<h3 id="computing-the-hash-function-efficiently">computing the hash function efficiently</h3>
<p>let ti be the ith char in txt, the hashcode for substring <code>txt[i,...,i+M-1]</code> is: <br/>
<img alt="" class="img-responsive" src="../images/algoII_week4_2/pasted_image015.png"/><br/>
⇒ just an M-digit base-R integer modulo Q ! <code>poly(M, R) % Q</code>*. * </p>
<ul>
<li><strong>Honor's method</strong> </li>
</ul>
<p>linear time algorithm for evaluating polynomial. <br/>
recursive equation: <code>poly(i, R) = poly(i-1, R)*R+ti</code> </p>
<p>ex. (R=10, M=5)<br/>
<img alt="" class="img-responsive" src="../images/algoII_week4_2/pasted_image016.png"/> </p>
<div class="highlight"><pre><span class="code-line"><span></span><span class="err">private long hash(String key, int M){ </span></span>
<span class="code-line"><span class="err"> long h=0; </span></span>
<span class="code-line"><span class="err"> for(int i=0;i<M;i++) </span></span>
<span class="code-line"><span class="err"> h = ( h*R + key.charAt(i) ) % Q </span></span>
<span class="code-line"><span class="err"> return h; </span></span>
<span class="code-line"><span class="err">}</span></span>
</pre></div>
<ul>
<li>if we know x_i, the x_i+1 can be infered: </li>
</ul>
<p><img alt="" class="img-responsive" src="../images/algoII_week4_2/pasted_image017.png"/><br/>
⇒ x_i+1 can be computed in constant time: <br/>
<img alt="" class="img-responsive" src="../images/algoII_week4_2/pasted_image018.png"/><br/>
⇒ we precompute R^(M-1) and maintain the hash number, and check for match ! </p>
<div class="highlight"><pre><span class="code-line"><span></span><span class="err">public static int search(String txt, String pat){ </span></span>
<span class="code-line"><span class="err"> int N=txt.length(), M=pat.length(); </span></span>
<span class="code-line"><span class="err"> long pathash = hash(pat, M); </span></span>
<span class="code-line"><span class="err"> int RM = R^(M-1);// <-- pseudo code, store value of R^(M-1) </span></span>
<span class="code-line"><span class="err"> long txthash = hash(txt, M);// txthash will be maintained </span></span>
<span class="code-line"><span class="err"> for(int i=0;i<N-M;i++){ </span></span>
<span class="code-line"><span class="err"> if(txthash==pathash && checkMatch(i,txt,pat)) </span></span>
<span class="code-line"><span class="err"> return i; </span></span>
<span class="code-line"><span class="err"> txthash = ( (txthash - txt.charAt(i)*RM)*R + txt.charAt(i+M) ) % Q; </span></span>
<span class="code-line"><span class="err"> } </span></span>
<span class="code-line"><span class="err">}</span></span>
</pre></div>
<p>更新txthash的地方可能会有modulo造成的问题... 不过先这样写吧.. </p>
<p>for collisions: <em>Monte Carlo</em> vs. <em>Las Vegas</em> </p>
<h3 id="analysis_1">analysis</h3>
<p><strong>Theory</strong>: if Q is sufficiently large (~M*N^2), the probability of collision is ~1/N. <br/>
<strong>Practice</strong>: choose Q to be sufficiently large, and collision probability is ~1/Q. </p>
<p><img alt="" class="img-responsive" src="../images/algoII_week4_2/pasted_image019.png"/> </p>
<h1 id="summery_1">Summery</h1>
<p><img alt="" class="img-responsive" src="../images/algoII_week4_2/pasted_image020.png"/> </p>
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<div id="toc"><ul><li><a class="toc-href" href="#1-introduction-to-substring-search" title="1. Introduction to substring search">1. Introduction to substring search</a></li><li><a class="toc-href" href="#2-brute-force-substring-search" title="2. Brute-Force Substring Search">2. Brute-Force Substring Search</a></li><li><a class="toc-href" href="#3-knuth-morris-pratt" title="3. Knuth-Morris-Pratt">3. Knuth-Morris-Pratt</a><ul><li><a class="toc-href" href="#intuition" title="intuition">intuition</a></li><li><a class="toc-href" href="#deterministic-finite-state-automaton-dfa" title="Deterministic finite state automaton (DFA)">Deterministic finite state automaton (DFA)</a></li><li><a class="toc-href" href="#dfa-construction" title="DFA construction">DFA construction</a></li></ul></li><li><a class="toc-href" href="#4-boyer-moore_1" title="4. Boyer-Moore">4. Boyer-Moore</a><ul><li><a class="toc-href" href="#intuition_1" title="intuition">intuition</a></li><li><a class="toc-href" href="#mismatch-character-heuristic" title="mismatch character heuristic">mismatch character heuristic</a></li><li><a class="toc-href" href="#implementation" title="implementation">implementation</a></li><li><a class="toc-href" href="#analysis" title="analysis">analysis</a></li></ul></li><li><a class="toc-href" href="#5-rabin-karp_1" title="5. Rabin-Karp">5. Rabin-Karp</a><ul><li><a class="toc-href" href="#intuition_2" title="intuition">intuition</a></li><li><a class="toc-href" href="#computing-the-hash-function-efficiently" title="computing the hash function efficiently">computing the hash function efficiently</a></li><li><a class="toc-href" href="#analysis_1" title="analysis">analysis</a></li></ul></li><li><a class="toc-href" href="#summery_1" title="Summery">Summery</a></li></ul></div>
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