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Commit 5a7b5064 authored by Goik Martin's avatar Goik Martin
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Pythagorean triple exercise completed

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Doc/Sd1/extraExercise.xml 0 → 100644
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<?xml version="1.0" encoding="UTF-8"?>
<chapter version="5.0" xml:id="sd1ExtraExercise" xmlns="http://docbook.org/ns/docbook"
xmlns:xlink="http://www.w3.org/1999/xlink"
xmlns:xi="http://www.w3.org/2001/XInclude"
xmlns:svg="http://www.w3.org/2000/svg"
xmlns:m="http://www.w3.org/1998/Math/MathML"
xmlns:html="http://www.w3.org/1999/xhtml"
xmlns:db="http://docbook.org/ns/docbook">
<title>Loops</title>
<section xml:id="sd1ExternalExercises">
<title>Recommended exercises</title>
<para></para>
</section>
<section xml:id="sd1NewLectureExercises">
<title>Exercises recently added to these lecture notes</title>
<para></para>
</section>
</chapter>
Doc/Sd1/statements.xml
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......@@ -97,4 +97,220 @@
</qandadiv>
</qandaset>
</section>
<section xml:id="sw1SectionPythagoreanTriples">
<title>Pythagorean triples</title>
<qandaset defaultlabel="qanda" xml:id="sw1QandaPythagoreanTriples">
<title>Finding pythagorean triples</title>
<qandadiv>
<qandaentry>
<question>
<para>Read the <link
xlink:href="http://en.wikipedia.org/wiki/Pythagorean_triple">definition
of pythagorean triples</link>.</para>
<para>Find all <link
xlink:href="http://en.wikipedia.org/wiki/Pythagorean_triple">pythagorean
triples</link> (a,b,c) for which which in addition to their
definition the restriction <inlineequation>
<m:math display="inline">
<m:mrow>
<m:mrow>
<m:mi>a</m:mi>
<m:mo>+</m:mo>
<m:mi>b</m:mi>
<m:mo>+</m:mo>
<m:mi>c</m:mi>
</m:mrow>
<m:mo>=</m:mo>
<m:mi>1000</m:mi>
</m:mrow>
</m:math>
</inlineequation> holds</para>
<tip>
<para>Think of writing a program which creates all triplets
(a,b,c) not yet obeying the restriction a + b + c = 1000:</para>
<informaltable border="1">
<tr>
<th>a</th>
<th>b</th>
<th>c</th>
</tr>
<tr>
<td>1</td>
<td>1</td>
<td>1</td>
</tr>
<tr>
<td>2</td>
<td>1</td>
<td>1</td>
</tr>
<tr>
<td>...</td>
<td>...</td>
<td>...</td>
</tr>
<tr>
<td>1000</td>
<td>1000</td>
<td>1000</td>
</tr>
</informaltable>
<para>This might be accomplished by implementing three nested
loops:</para>
<programlisting language="java"> for (int a = 1; a &lt; 1000; a++) {
for (int b = 1; b &lt; 1000; b++) {
for (int c = 1; c &lt; 1000; c++) {
System.out.println("(" + a + ", " + b + ", " + c + ")");
}
}
}</programlisting>
<para>The above code will run for ages! Now try to alter this
code to generate only triplets (a,b,c) simultaneously obeying:
</para>
<orderedlist>
<listitem xml:id="sw1PhytagoreanTripletCondition1">
<para><inlineequation>
<m:math display="inline">
<m:mrow>
<m:mi>a</m:mi>
<m:mo>≤</m:mo>
<m:mi>b</m:mi>
<m:mo>&lt;</m:mo>
<m:mi>c</m:mi>
</m:mrow>
</m:math>
</inlineequation></para>
</listitem>
<listitem xml:id="sw1PhytagoreanTripletCondition2">
<para><inlineequation>
<m:math display="inline">
<m:mrow>
<m:mrow>
<m:mi>a</m:mi>
<m:mo>+</m:mo>
<m:mi>b</m:mi>
<m:mo>+</m:mo>
<m:mi>c</m:mi>
</m:mrow>
<m:mo>=</m:mo>
<m:mi>1000</m:mi>
</m:mrow>
</m:math>
</inlineequation></para>
</listitem>
</orderedlist>
<para>When you have done so filter your output for the set of
all triplets which obey the third restriction<inlineequation>
<m:math display="inline">
<m:mrow>
<m:mrow>
<m:msup>
<m:mi> a</m:mi>
<m:mn>2</m:mn>
</m:msup>
<m:mo>+</m:mo>
<m:msup>
<m:mi>b</m:mi>
<m:mn>2</m:mn>
</m:msup>
</m:mrow>
<m:mo>=</m:mo>
<m:msup>
<m:mi>c</m:mi>
<m:mn>2</m:mn>
</m:msup>
</m:mrow>
</m:math>
</inlineequation> as well.</para>
</tip>
</question>
<answer>
<para>The last triplet obeying both restrictions <xref
linkend="sw1PhytagoreanTripletCondition1"/> and <xref
linkend="sw1PhytagoreanTripletCondition2"/> is (333,333,324). Due
to restriction <xref linkend="sw1PhytagoreanTripletCondition1"/>
we need just two rather than three nested loops to generate all
required triplets:</para>
<programlisting language="java"> public static void main(String[] args) {
for (int a = 1; a &lt;= 333; a++) {
final int b_plus_c = 1000 - a, // the sum of b and c
aSquare = a * a; // Save value to be calculated only once (not inside subsequent loop)
final int bMaximum;
if (0 == b_plus_c % 2) { // Sum of b and c is even?
bMaximum = b_plus_c / 2 - 1; // c must be allowed to become larger then b
} else {
bMaximum = b_plus_c / 2; // c's minimum value will become bMaximum + 1
}
for (int b = a; b &lt;= bMaximum; b++) {
final int c = b_plus_c - b;
if (aSquare + b * b == c * c) { // Filter desired triplet values
System.out.println("(" + a + ", " + b + ", " + c + ")");
}
}
}
}</programlisting>
<para>This yields exactly one triplet:</para>
<programlisting language="none">(200, 375, 425)</programlisting>
</answer>
</qandaentry>
</qandadiv>
</qandaset>
</section>
</chapter>
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