CSC300: Homework (MyDequeUsingStacks, UF Problems)

(Click here for one slide per page)

• Read Algorithms 1.4 and 1.5 again.

• Do these problems on paper, but don't hand them in.

  • 1.5.1: Show the contents of the id[] array and the number of times the array is accessed for each input pair when you use quick-find for the sequence 9-0 3-4 5-8 7-2 2-1 5-7 0-3 4-2.

  • 1.5.2: Do Exercise 1.5.1, but use quick-union (page 224). In addition, draw the forest of trees represented by the id[] array after each input pair is processed.

  • 1.5.3: Do Exercise 1.5.1, but use weighted quick-union (page 228).

  • 1.5.8: Give a counterexample that shows why this intuitive implementation of union() for quick-find is not correct:

    01 02 03 04 05 06 07

    public void union(int p, int q) { if (id[p] == id[q]) return; // Rename p's component to q's name. for (int i = 0; i < id.length; i++) if (id[i] == id[p]) id[i] = id[q]; count--; }

  • 1.5.10: In the weighted quick-union algorithm, suppose that we set id[find(p)] to q instead of to id[find(q)]. Would the resulting algorithm be correct?

• Complete MyDequeUsingStacks.java

file:MyDequeUsingStacks.java [source] [doc-public] [doc-private]

001 002 003 004 005 006 007 008 009 010 011 012 013 014 015 016 017 018 019 020 021 022 023 024 025 026 027 028 029 030 031 032 033 034 035 036 037 038 039 040 041 042 043 044 045 046 047 048 049 050 051 052 053 054 055 056 057 058 059 060 061 062 063 064 065 066 067 068 069 070 071 072 073 074 075 076 077 078 079 080 081 082 083 084 085 086 087 088 089 090 091 092 093 094 095 096 097 098 099 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 180 181 182 183 184 185 186 187 188 189 190 191 192 193 194 195 196 197 198 199 200 201 202 203 204 205 206 207 208 209 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255 256 257 258 259 260 261 262 263 264 265 266 267 268 269 270 271 272 // Exercise 1.4.31 package algs14; import stdlib.*; import java.util.NoSuchElementException; import algs13.ResizingArrayStack; /* Here is the kind of output I get before fixing the move method. 4000 0.4 3.1 8000 1.5 4.1 16000 4.9 3.2 32000 21.6 4.4 Here is the kind of output I get after fixing the move method. You can see that it is much faster, but not very consistent. This is due to garbage collection and other system effects. 4000 0.0 0.7 8000 0.0 1.2 16000 0.0 2.2 32000 0.0 0.2 64000 0.0 1.5 128000 0.0 5.7 256000 0.1 5.3 512000 0.2 2.1 1024000 0.4 1.9 2048000 0.7 2.0 4096000 2.5 3.6 4000 0.0 0.8 8000 0.0 1.6 16000 0.0 1.8 32000 0.0 0.1 64000 0.0 2.0 128000 0.0 4.5 256000 0.1 5.7 512000 0.2 1.8 1024000 0.3 1.8 2048000 0.7 2.1 4096000 3.7 5.1 4000 0.0 0.7 8000 0.0 2.0 16000 0.0 1.9 32000 0.0 0.2 64000 0.0 1.7 128000 0.0 4.4 256000 0.0 1.6 512000 0.1 2.3 1024000 0.1 0.6 2048000 0.1 2.0 4096000 0.2 2.4 */ public class MyDequeUsingStacks<T> { ResizingArrayStack<T> sl = new ResizingArrayStack<>(); ResizingArrayStack<T> sr = new ResizingArrayStack<>(); public boolean isEmpty () { return sl.isEmpty() && sr.isEmpty(); } public int size () { return sl.size() + sr.size(); } public void pushLeft (T item) { sl.push (item); } public void pushRight (T item) { sr.push (item); } private void move (ResizingArrayStack<T> from, ResizingArrayStack<T> to) { if (!to.isEmpty ()) throw new IllegalArgumentException(); if (from.isEmpty ()) throw new IllegalArgumentException(); ResizingArrayStack<T> tmp = new ResizingArrayStack<>(); while (!from.isEmpty ()) { to.push(from.pop()); } // TODO: // Run the main method below. // Note that the execution is correct (that's the first part of the test). // Note also that performance is terrible (that's the second part of the test). // In particular, for N pops, the running time increases proportionally to N^2. // So the amortized time for each pop is linear (proportional to N). // // Your goal is to change this method so that overall performance of // the main method improves, by altering the number of elements moved // from "from" to "to". // // Your solution must continue to pass the correctness tests below. // It must also have amortized constant time per pop. // So for N pops, the running time should increase proportionally to N. // // You can use one temporary stack to help you: // ResizingArrayStack<T> tmp = new ResizingArrayStack<>(); // You may find it helpful to print the sizes while debugging: // StdOut.println ("from: " + from.size () + " to: " + to.size ()); // You can print out the content of the stacks by uncommenting the // lines in popLeft and popRight, below. } public T popLeft () { if (isEmpty()) { throw new NoSuchElementException(); } if (sl.isEmpty ()) { //StdOut.println("r2l before: " + sl + ": " + sr); move (sr, sl); //StdOut.println("r2l after: " + sl + ": " + sr); } return sl.pop (); } public T popRight () { if (isEmpty()) { throw new NoSuchElementException(); } if (sr.isEmpty ()) { //StdOut.println("l2r before: " + sl + ": " + sr); move (sl, sr); //StdOut.println("l2r after: " + sl + ": " + sr); } return sr.pop (); } public String toString () { if (isEmpty ()) return "[ ]"; ResizingArrayStack<T> srBackwards = new ResizingArrayStack<>(); for (T item : sr) srBackwards.push (item); StringBuilder sb = new StringBuilder ("[ "); for (T item : sl) { sb.append (item); sb.append (" "); } for (T item : srBackwards) { sb.append (item); sb.append (" "); } sb.append ("]"); return sb.toString (); } private void check (String expected) { if (expected != null) { if (!expected.equals (this.toString ())) throw new Error ("Expected \"" + expected + "\", got \"" + this + "\""); } //StdOut.println (this); } private void check (T iExpected, T iActual, String expected) { if (!iExpected.equals (iActual)) throw new Error ("Expected \"" + iExpected + "\", got \"" + iActual + "\""); check (expected); } private void check (T iExpected, T iActual) { if (!iExpected.equals (iActual)) throw new Error ("Expected \"" + iExpected + "\", got \"" + iActual + "\""); } private static void correctnessTest () { MyDequeUsingStacks<Integer> d1 = new MyDequeUsingStacks<> (); d1.pushLeft(0); d1.pushLeft(1); d1.pushLeft(2); d1.pushLeft(3); d1.check(0,d1.popRight()); d1.check(3,d1.popLeft()); d1.check(1,d1.popRight()); d1.check(2,d1.popLeft()); d1.pushRight(0); d1.pushRight(1); d1.pushRight(2); d1.pushRight(3); d1.check(0,d1.popLeft()); d1.check(3,d1.popRight()); d1.check(1,d1.popLeft()); d1.check(2,d1.popRight()); Integer k; if (!d1.isEmpty ()) throw new Error(); d1.pushLeft (11); d1.check ("[ 11 ]"); d1.pushLeft (12); d1.check ("[ 12 11 ]"); k = d1.popLeft (); d1.check (12, k, "[ 11 ]"); k = d1.popLeft (); d1.check (11, k, "[ ]"); try { d1.popLeft (); throw new Error ("Expected exception"); } catch (NoSuchElementException e) {} if (!d1.isEmpty ()) throw new Error(); for (int i = 0; i < 20; i++) { d1.pushLeft (i); } for (int i = 0; i < 20; i++) { d1.check (19-i, d1.popLeft ()); } if (!d1.isEmpty ()) throw new Error(); for (int i = 0; i < 20; i++) { d1.pushLeft (i); } for (int i = 0; i < 20; i++) { d1.check (i, d1.popRight ()); } if (!d1.isEmpty ()) throw new Error(); for (int i = 0; i < 20; i++) { d1.pushLeft (i); } for (int i = 0; i < 10; i++) { d1.check (i, d1.popRight ()); } for (int i = 0; i < 10; i++) { d1.check (19-i, d1.popLeft ()); } if (!d1.isEmpty ()) throw new Error(); for (int i = 0; i < 20; i++) { d1.pushLeft (i); } for (int i = 0; i < 10; i++) { d1.check (19-i, d1.popLeft ()); } for (int i = 0; i < 10; i++) { d1.check (i, d1.popRight ()); } if (!d1.isEmpty ()) throw new Error(); d1.pushRight (11); d1.check ("[ 11 ]"); d1.pushRight (12); d1.check ("[ 11 12 ]"); k = d1.popRight (); d1.check (12, k, "[ 11 ]"); k = d1.popRight (); d1.check (11, k, "[ ]"); if (!d1.isEmpty ()) throw new Error(); for (int i = 0; i < 20; i++) { d1.pushRight (i); } for (int i = 0; i < 20; i++) { d1.check (19-i, d1.popRight ()); } if (!d1.isEmpty ()) throw new Error(); for (int i = 0; i < 20; i++) { d1.pushRight (i); } for (int i = 0; i < 20; i++) { d1.check (i, d1.popLeft ()); } if (!d1.isEmpty ()) throw new Error(); for (int i = 0; i < 20; i++) { d1.pushRight (i); } for (int i = 0; i < 10; i++) { d1.check (i, d1.popLeft ()); } for (int i = 0; i < 10; i++) { d1.check (19-i, d1.popRight ()); } if (!d1.isEmpty ()) throw new Error(); for (int i = 0; i < 20; i++) { d1.pushRight (i); } for (int i = 0; i < 10; i++) { d1.check (19-i, d1.popRight ()); } for (int i = 0; i < 10; i++) { d1.check (i, d1.popLeft ()); } try { d1.popRight (); throw new Error ("Expected exception"); } catch (NoSuchElementException e) {} StdOut.println("Finished correctness test."); } private static void performanceTest() { int MIN = 2000; int MAX = 4096000; double prev = timeTrial(MIN); for (int N = MIN*2; N<=MAX; N += N) { double time = timeTrial(N); StdOut.format("%8d %9.3f %5.1f\n", N, time, time/prev); prev = time; } } private static double timeTrial(int N) { int NUM_TRIALS = 1; MyDequeUsingStacks<Integer> d1 = new MyDequeUsingStacks<> (); Stopwatch sw = new Stopwatch (); for (int trial=0; trial < NUM_TRIALS; trial++) { for (int i=0; i<2*N; i++) { d1.pushLeft (i); } for (int i=0; i<N; i++) { d1.popRight(); d1.popLeft(); } } return sw.elapsedTime (); } public static void main (String args[]) { //Trace.drawStepsOfMethod ("correctnessTest"); //Trace.drawStepsOfMethod ("move"); //Trace.run (); correctnessTest (); performanceTest (); } }

• 1.5.1: Show the contents of the id[] array and the number of times the array is accessed for each input pair when you use quick-find for the sequence 9-0 3-4 5-8 7-2 2-1 5-7 0-3 4-2.

You can check your answer in TestUF.java.
Uncomment the correct line in getUF to return the correct type of UF.
Put the following in main:

  show(10, "9 0, 3 4, 5 8, 7 2, 2 1, 5 7, 0 3, 4 2");

After running you should see pictures in Desktop/GraphvizOutput
For this problem I get:
  
       10[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
 9  0:  9[0, 1, 2, 3, 4, 5, 6, 7, 8, 0]
 3  4:  8[0, 1, 2, 4, 4, 5, 6, 7, 8, 0]
 5  8:  7[0, 1, 2, 4, 4, 8, 6, 7, 8, 0]
 7  2:  6[0, 1, 2, 4, 4, 8, 6, 2, 8, 0]
 2  1:  5[0, 1, 1, 4, 4, 8, 6, 1, 8, 0]
 5  7:  4[0, 1, 1, 4, 4, 1, 6, 1, 1, 0]
 0  3:  3[4, 1, 1, 4, 4, 1, 6, 1, 1, 4]
 4  2:  2[1, 1, 1, 1, 1, 1, 6, 1, 1, 1]  

• 1.5.2: Do Exercise 1.5.1, but use quick-union (page 224). In addition, draw the forest of trees represented by the id[] array after each input pair is processed.

       10[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
 9  0:  9[0, 1, 2, 3, 4, 5, 6, 7, 8, 0]
 3  4:  8[0, 1, 2, 4, 4, 5, 6, 7, 8, 0]
 5  8:  7[0, 1, 2, 4, 4, 8, 6, 7, 8, 0]
 7  2:  6[0, 1, 2, 4, 4, 8, 6, 2, 8, 0]
 2  1:  5[0, 1, 1, 4, 4, 8, 6, 2, 8, 0]
 5  7:  4[0, 1, 1, 4, 4, 8, 6, 2, 1, 0]
 0  3:  3[4, 1, 1, 4, 4, 8, 6, 2, 1, 0]
 4  2:  2[4, 1, 1, 4, 1, 8, 6, 2, 1, 0]

• 1.5.3: Do Exercise 1.5.1, but use weighted quick-union (page 228).

       10[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
 9  0:  9[9, 1, 2, 3, 4, 5, 6, 7, 8, 9]
 3  4:  8[9, 1, 2, 3, 3, 5, 6, 7, 8, 9]
 5  8:  7[9, 1, 2, 3, 3, 5, 6, 7, 5, 9]
 7  2:  6[9, 1, 7, 3, 3, 5, 6, 7, 5, 9]
 2  1:  5[9, 7, 7, 3, 3, 5, 6, 7, 5, 9]
 5  7:  4[9, 7, 7, 3, 3, 7, 6, 7, 5, 9]
 0  3:  3[9, 7, 7, 9, 3, 7, 6, 7, 5, 9]
 4  2:  2[9, 7, 7, 9, 3, 7, 6, 7, 5, 7]

• 1.5.8: Give a counterexample that shows why this intuitive implementation of union() for quick-find is not correct:

  01 02 03 04 05 06 07

  public void union(int p, int q) { if (id[p] == id[q]) return; // Rename p's component to q's name. for (int i = 0; i < id.length; i++) if (id[i] == id[p]) id[i] = id[q]; count--; }

Here's an example:

  show(3, "1 0, 0 2");

Correct answer:  

        3[0, 1, 2]
 1  0:  2[0, 0, 2]
 0  2:  1[2, 2, 2]

With the broken code above: 

        3[0, 1, 2]
 1  0:  2[0, 0, 2]
 0  2:  1[2, 0, 2]

The problem is that id[p] is changes!
To correct this, it is sufficient to use:
   
   int pid = id[p];
   for (int i = 0; i < id.length; i++)
       if (id[i] == pid) id[i] = id[q];

The book also uses a variable for qid.
But this change does not affect correctness, since id[q] does not change.

• 1.5.10: In the weighted quick-union algorithm, suppose that we set id[find(p)] to q instead of to id[find(q)]. Would the resulting algorithm be correct?

Yes, the algorithm would be correct.  However, it would be increase the tree
height, so the performance guarantee would be invalid.  As an example,

Suppose we have three elements that we union together.
Then weighted union should not produce a tree of depth 2.
Here is the correct alorgithm:

        3[0, 1, 2]
 0  1:  2[0, 0, 2]
 1  2:  1[0, 0, 0]

The final forest looks like this:

    0
   / \
  1   2

It is depth 1.  
Here is the badly performing version:

        3[0, 1, 2]
 0  1:  2[0, 0, 2]
 1  2:  1[0, 0, 1]

 The final forest looks like this:

    0
     \
      1
       \
        2

It has depth 2.

Here is the code for the badly performing version:

  public void union(int p, int q) {
    int pid = find(p);
    int qid = find(q);
    if (pid == qid) return;

    if   (sz[pid] < sz[qid]) {
      id[pid] = q;        // should be qid, not q.
      sz[qid] += sz[pid];
    } else {
      id[qid] = p;        // should be pid, not p.
      sz[pid] += sz[qid];
    }
    count--;
  }

You can check your answer in TestUF.java.
Uncomment the correct line in getUF to return the correct type of UF.
Put the following in main:

  show(10, "9 0, 3 4, 5 8, 7 2, 2 1, 5 7, 0 3, 4 2");

Using QuickFindUF:  

       10[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
 9  0:  9[0, 1, 2, 3, 4, 5, 6, 7, 8, 0]
 3  4:  8[0, 1, 2, 4, 4, 5, 6, 7, 8, 0]
 5  8:  7[0, 1, 2, 4, 4, 8, 6, 7, 8, 0]
 7  2:  6[0, 1, 2, 4, 4, 8, 6, 2, 8, 0]
 2  1:  5[0, 1, 1, 4, 4, 8, 6, 1, 8, 0]
 5  7:  4[0, 1, 1, 4, 4, 1, 6, 1, 1, 0]
 0  3:  3[4, 1, 1, 4, 4, 1, 6, 1, 1, 4]
 4  2:  2[1, 1, 1, 1, 1, 1, 6, 1, 1, 1]

Using QuickUnionUF:  
 
       10[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
 9  0:  9[0, 1, 2, 3, 4, 5, 6, 7, 8, 0]
 3  4:  8[0, 1, 2, 4, 4, 5, 6, 7, 8, 0]
 5  8:  7[0, 1, 2, 4, 4, 8, 6, 7, 8, 0]
 7  2:  6[0, 1, 2, 4, 4, 8, 6, 2, 8, 0]
 2  1:  5[0, 1, 1, 4, 4, 8, 6, 2, 8, 0]
 5  7:  4[0, 1, 1, 4, 4, 8, 6, 2, 1, 0]
 0  3:  3[4, 1, 1, 4, 4, 8, 6, 2, 1, 0]
 4  2:  2[4, 1, 1, 4, 1, 8, 6, 2, 1, 0]

Using WeightedUF:

       10[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
 9  0:  9[9, 1, 2, 3, 4, 5, 6, 7, 8, 9]
 3  4:  8[9, 1, 2, 3, 3, 5, 6, 7, 8, 9]
 5  8:  7[9, 1, 2, 3, 3, 5, 6, 7, 5, 9]
 7  2:  6[9, 1, 7, 3, 3, 5, 6, 7, 5, 9]
 2  1:  5[9, 7, 7, 3, 3, 5, 6, 7, 5, 9]
 5  7:  4[9, 7, 7, 3, 3, 7, 6, 7, 5, 9]
 0  3:  3[9, 7, 7, 9, 3, 7, 6, 7, 5, 9]
 4  2:  2[9, 7, 7, 9, 3, 7, 6, 7, 5, 7]

Revised: 2024-03-12 10:14