From 9fa976421b918505bf888074d607c27c294aefcc Mon Sep 17 00:00:00 2001
From: "Dr. Martin Goik" <goik@hdm-stuttgart.de>
Date: Sun, 18 Apr 2021 21:31:44 +0200
Subject: [PATCH] Enhancing exercises, adding operator ++ releted slide
---
Doc/Sd1/languageFundamentals.xml | 108 ++-
Doc/Sd1/statements.xml | 1184 +++++++++++-------------------
2 files changed, 541 insertions(+), 751 deletions(-)
diff --git a/Doc/Sd1/languageFundamentals.xml b/Doc/Sd1/languageFundamentals.xml
index 378a9e470..2191bf3fc 100644
--- a/Doc/Sd1/languageFundamentals.xml
+++ b/Doc/Sd1/languageFundamentals.xml
@@ -6238,7 +6238,7 @@ short sum = a + 7; <co linkends="sd1_co_expressionTypeIntToShort-2"
xlink:href="https://docs.oracle.com/javase/specs/jls/se7/html/jls-4.html#jls-4.2.1-100-C">int</code>
operands. The expression <code language="java">4 + 7</code> is
thus of type <code language="java">int</code> (not <code
- language="java">long</code>). Both operands are static
+ language="java">short</code>). Both operands are static
<abbrev>i.e.</abbrev> their values are known at
<emphasis>compile time</emphasis> rather than at <emphasis>run
time</emphasis>. The <code language="java">4 + 7</code>
@@ -7356,20 +7356,35 @@ age += 200;</programlisting>
linkend="sd1_fig_noBinaryPlusByteByte"/> the arithmetic <code
language="java">+</code> operator acting any two non-long values
will always return a result of type <code
- language="java">int</code>. Thus the expression <code
- language="java">age + 2</code> of type <code
- language="java">int</code> can only be assigned to a variable
- <code language="java">age</code> of type <code
- language="java">byte</code> when using a cast:</para>
+ language="java">int</code>. The expression <code
+ language="java">age + 2</code> is thus of type <code
+ language="java">int</code> and therefore requires a cast when
+ being assigned to our variable <code language="java">age</code>
+ of type <code language="java">byte</code>:</para>
<programlisting language="java">byte age = 80;
age = (byte)(age + 2);</programlisting>
+ <note>
+ <para>Since <code language="java">age</code> is not being
+ defined as <code language="java">final</code> the value of the
+ expression <code language="java">age + 2</code> is not being
+ known at compile time. Adding <code
+ language="java">final</code> allows for compile time
+ evaluation and thus assignment to a second variable of type
+ <code language="java">byte</code> without requiring a
+ cast:</para>
+
+ <programlisting language="java">final byte age = 80; // Value cannot be altered.
+
+byte gettingOlder = age + 2; // o.K. without cast: age + 2 can be evaluated at compile time.</programlisting>
+ </note>
+
<para>On contrary the operator <code language="java">+=</code>
will accept any right hand integer value (even of type <code
- language="java">long</code>!) thereby cycling through the range
- of byte values e.g.:</para>
+ language="java">long</code>!) thereby possibly cycling through
+ the range of byte values e.g.:</para>
<informaltable border="1">
<tr>
@@ -7387,9 +7402,9 @@ System.out.println("Age:" + age);</programlisting></td>
</tr>
</informaltable>
- <para>Notice 24 being equal to 80 + 200 - 256, the latter being
- a <code language="java">byte</code>'s number of different
- representable values.</para>
+ <para>Notice 24 being equal to <code language="java">80 + 200 -
+ 256</code> with 256 being a <code language="java">byte</code>'s
+ number of different representable values.</para>
</answer>
</qandaentry>
</qandadiv>
@@ -7402,9 +7417,13 @@ System.out.println("Age:" + age);</programlisting></td>
<figure xml:id="sd1_fig_OperatorPlusPlus">
<title>Increment operator <code language="java">++</code></title>
- <para>Increment variable by 1:</para>
+ <informaltable border="1">
+ <tr>
+ <th>Increment variable by 1:</th>
+
+ <th>Shorthand version:</th>
+ </tr>
- <informaltable border="0">
<tr>
<td valign="top"><programlisting language="none">int a = 4;
@@ -7425,6 +7444,69 @@ System.out.println("Value:" + a);</programlisting></td>
</informaltable>
</figure>
+ <figure xml:id="sd1_fig_OperatorPlusPlusRangeConsider">
+ <title>Different range behaviour!</title>
+
+ <informaltable border="1">
+ <tr>
+ <th>Increment variable by 1:</th>
+
+ <th>Shorthand version:</th>
+ </tr>
+
+ <tr>
+ <td valign="top"><programlisting language="none">byte value = 127; // Max possible value
+
+<emphasis role="red">value = value + 1; // Error: Required type: byte
+ // Provided:int</emphasis>
+
+System.out.println(value);</programlisting></td>
+
+ <td valign="top"><programlisting language="none">byte value = 127; // Max possible value
+
+<emphasis role="red">value++; // o.K., will cycle through range</emphasis>
+
+System.out.println(value);</programlisting></td>
+ </tr>
+
+ <tr>
+ <td><screen>Does not compile</screen></td>
+
+ <td><screen>Value:-128</screen></td>
+ </tr>
+ </informaltable>
+ </figure>
+
+ <figure xml:id="sd1_fig_OperatorPlusPlusRangeCast">
+ <title>Cast required</title>
+
+ <informaltable border="1">
+ <tr>
+ <th>Increment variable by 1:</th>
+
+ <th>Shorthand version:</th>
+ </tr>
+
+ <tr>
+ <td valign="top"><programlisting language="none">byte value = 127; // Max possible value
+
+<emphasis role="red">value = (byte)(value + 1); // cast, possible overflow</emphasis>
+
+System.out.println(value);</programlisting></td>
+
+ <td valign="top"><programlisting language="none">byte value = 127; // Max possible value
+
+<emphasis role="red">value++; // o.K., will cycle through range</emphasis>
+
+System.out.println(value);</programlisting></td>
+ </tr>
+
+ <tr>
+ <td colspan="2"><screen>Value:-128</screen></td>
+ </tr>
+ </informaltable>
+ </figure>
+
<figure xml:id="sd1_fig_OperatorPlusPlusPostfixPrefix">
<title>Prefix and postfix notation</title>
diff --git a/Doc/Sd1/statements.xml b/Doc/Sd1/statements.xml
index 07524b876..b3035a4e5 100644
--- a/Doc/Sd1/statements.xml
+++ b/Doc/Sd1/statements.xml
@@ -4100,48 +4100,79 @@ System.out.format("Start:<emphasis role="red">%5d</emphasis>:End", <emphasis
10| 10 20 30 40 50 60 70 80 90 100</screen>
<para>The number of rows and columns are equal. Provide an
- appropriate parameter <code language="java">final int LIMIT =
- ...</code> so that your implementation works for different
- values just by changing its value.</para>
+ appropriate parameter i.e <code language="java">final int SIZE
+ = 12</code> allowing for a configurable number of
+ columns.</para>
<tip>
- <para>You'll need an inner loop nested within an outer one
- creating rows and columns like:</para>
+ <orderedlist>
+ <listitem>
+ <para>You'll need an inner loop nested within an outer
+ one creating rows and columns like:</para>
- <programlisting language="java">for (int row = 0; ...){
+ <programlisting language="none">for (int row = 1; ...){
- for (int col = 0; ... ) {
- ...
+ for (int col = 1; ... ) {
+ ... // Using System.out.print(...) rather than ...print<emphasis
+ role="red">ln</emphasis>() avoiding line breaks.
}
+ System.out.println(); // Adding a newline after each set of values
}</programlisting>
- </tip>
- </question>
+ </listitem>
- <answer>
- <programlisting language="java">public static void main(String[] args) {
+ <listitem>
+ <para>Having integer values of differing lengths you
+ have to consider right justified output. The integer
+ <link
+ xlink:href="https://docs.oracle.com/en/java/javase/15/docs/api/java.base/java/util/Formatter.html#syntax">format
+ specifier</link> <code>%...d</code> is your friend here.
+ The subsequent example prints four integer values in
+ right justified style within output fields of common
+ length 5 irrespective of differing lengths:</para>
- final int LIMIT = 10; // Parameter is subject to change.
+ <informaltable border="1">
+ <tr>
+ <th>Code</th>
- System.out.print(" * | ");
+ <th>Output</th>
+ </tr>
- for (int col = 1; col <= LIMIT; col++) { // Header line of columns ranging
- System.out.format("%3d ", col); // from 1 to LIMIT.
- }
- System.out.println();
+ <tr>
+ <td valign="top"><programlisting language="java">System.out.format("|%5d|%5d|%5d|%5d|", 1, 100, 1000, 10000);</programlisting></td>
- System.out.print("---+"); // Header / table body separator.
- for (int col = 1; col <= LIMIT; col++) {
- System.out.print("-----");
- }
- System.out.println();
+ <td valign="top"><screen>| 1| 100| 1000|10000|</screen></td>
+ </tr>
+ </informaltable>
+
+ <para>The %...s format specifier may be used for strings
+ accordingly.</para>
+ </listitem>
+ </orderedlist>
+ </tip>
+ </question>
- for (int row = 1; row <= LIMIT; row++) { // Printing rows.
- System.out.format("%3d| ", row);
- for (int col = 1; col <= LIMIT; col++) { // Printing columns.
- System.out.format("%3d ", row * col);
+ <answer>
+ <programlisting language="java">// Block 1: Configuration parameter
+//
+final int SIZE = 6; // Table size is subject to change.
+
+// Block 2: Printing table's head
+//
+System.out.print(" * | "); // Printing table head's first line
+for (int col = 1; col <= SIZE; col++) { // using column values ranging from 1 to SIZE:
+ System.out.format("%3d ", col); // * | 1 2 3 ...
+}
+System.out.print("\n---+" + // Printing table head's second line
+ "-----".repeat(SIZE) ); // ---+--------------- ...
+
+// Block 3: Printing table's content
+// organized in rows and columns.
+//
+for (int row = 1; row <= SIZE; row++) { // Printing rows.
+ System.out.format("\n%3d| ", row); // Printing new line by \n and beginning of row e.g. " 5|"
+ for (int col = 1; col <= SIZE; col++) {
+ System.out.format("%3d ", row * col); // Printing row times column product values.
}
- System.out.println();
- }
}</programlisting>
</answer>
</qandaentry>
@@ -4173,32 +4204,59 @@ System.out.format("Start:<emphasis role="red">%5d</emphasis>:End", <emphasis
</question>
<answer>
- <para>We have to move the header to the table's bottom. Apart
- from that the required change with respect to your previous
- exercise is tiny: The <quote>inner</quote> loop must end when
- reaching the current row's index:</para>
+ <para>We need two changes with respect to the <link
+ linkend="sd1QandaIntermediateMultiplyTable1">previous
+ exercise</link>:</para>
- <programlisting language="java">final int LIMIT = 10; // Parameter is subject to change.
+ <orderedlist>
+ <listitem>
+ <para>The number of columns to be printed has to be
+ limited depending on which row we currently print. We thus
+ replace</para>
-for (int row = 1; row <= LIMIT; row++) { // Printing rows.
- System.out.format("%3d| ", row);
+ <programlisting language="none">for (int col = 1; col <= <emphasis
+ role="red">SIZE</emphasis>; col++) { // Printing columns.</programlisting>
- for (int col = 1; <emphasis role="bold">col <= row;</emphasis> col++) { // Printing column values until row index only.
- System.out.format("%3d ", row * col);
- }
- System.out.println();
+ <para>by</para>
+
+ <programlisting language="none">for (int col = 1; col <= <emphasis
+ role="red">row</emphasis>; col++) { // Printing columns depending on current row value.</programlisting>
+ </listitem>
+
+ <listitem>
+ <para>Having a table tail now rather than a head we defer
+ our former solution's second block to the end thereby
+ reversing its internal order of print statements. The
+ following code also includes some minor newline character
+ fiddling:</para>
+
+ <programlisting language="java">// Block 1: Configuration parameter
+//
+final int SIZE = 10; // Table size is subject to change.
+
+// Block 2: Printing table's content
+// organized in rows and columns.
+//
+for (int row = 1; row <= SIZE; row++) { // Printing rows.
+ System.out.format("%3d| ", row); // Printing new line by \n and beginning of row e.g. " 5|"
+ for (int col = 1; col <= SIZE; col++) {
+ System.out.format("%3d ", row * col); // Printing row times column product values.
+ }
+ System.out.println(); // New line after printing row.
}
-System.out.print("---+"); // Header / table body separator.
- for (int col = 1; col <= LIMIT; col++) {
- System.out.print("-----");
- }
- System.out.println();
- System.out.print(" * | ");
- for (int col = 1; col <= LIMIT; col++) { // Header line of columns ranging
- System.out.format("%3d ", col); // from 1 to LIMIT.
- }
-System.out.println();</programlisting>
+// Block 3: Printing table's tail.
+//
+System.out.print("---+" + // Printing table tail's first line
+ "-----".repeat(SIZE) + // ---+--------------- ...
+ '\n'); // being finished by a newline character.
+
+System.out.print(" * | "); // Printing table tail's second line
+for (int col = 1; col <= SIZE; col++) { // using column values ranging from 1 to SIZE:
+ System.out.format("%3d ", col); // * | 1 2 3 ...
+}</programlisting>
+ </listitem>
+ </orderedlist>
</answer>
</qandaentry>
</qandadiv>
@@ -6250,7 +6308,7 @@ for (int i = 0; <emphasis role="red">!numberWasFound &&</emphasis> i <
<section xml:id="sw1SectionPythagoreanTriples">
<title>Pythagorean triples</title>
- <qandaset defaultlabel="qanda" xml:id="sw1QandaPythagoreanTriples">
+ <qandaset defaultlabel="qanda" xml:id="sw1QandaPythagoreanTriples1">
<qandadiv>
<qandaentry>
<question>
@@ -6267,7 +6325,107 @@ for (int i = 0; <emphasis role="red">!numberWasFound &&</emphasis> i <
</imageobject>
</mediaobject>
- <para>For most value combinations at least one value will be
+ <para>Some integer triples do exist e.g.:</para>
+
+ <itemizedlist>
+ <listitem>
+ <informalequation>
+ <m:math display="block">
+ <m:mrow>
+ <m:mrow>
+ <m:msup>
+ <m:mi>3</m:mi>
+
+ <m:mi>2</m:mi>
+ </m:msup>
+
+ <m:mo>+</m:mo>
+
+ <m:msup>
+ <m:mi>4</m:mi>
+
+ <m:mi>2</m:mi>
+ </m:msup>
+ </m:mrow>
+
+ <m:mo>=</m:mo>
+
+ <m:msup>
+ <m:mi>5</m:mi>
+
+ <m:mi>2</m:mi>
+ </m:msup>
+ </m:mrow>
+ </m:math>
+ </informalequation>
+ </listitem>
+
+ <listitem>
+ <informalequation>
+ <m:math display="block">
+ <m:mrow>
+ <m:mrow>
+ <m:msup>
+ <m:mi>6</m:mi>
+
+ <m:mi>2</m:mi>
+ </m:msup>
+
+ <m:mo>+</m:mo>
+
+ <m:msup>
+ <m:mi>8</m:mi>
+
+ <m:mi>2</m:mi>
+ </m:msup>
+ </m:mrow>
+
+ <m:mo>=</m:mo>
+
+ <m:msup>
+ <m:mi>10</m:mi>
+
+ <m:mi>2</m:mi>
+ </m:msup>
+ </m:mrow>
+ </m:math>
+ </informalequation>
+ </listitem>
+
+ <listitem>
+ <informalequation>
+ <m:math display="block">
+ <m:mrow>
+ <m:mrow>
+ <m:msup>
+ <m:mi>5</m:mi>
+
+ <m:mi>2</m:mi>
+ </m:msup>
+
+ <m:mo>+</m:mo>
+
+ <m:msup>
+ <m:mi>12</m:mi>
+
+ <m:mi>2</m:mi>
+ </m:msup>
+ </m:mrow>
+
+ <m:mo>=</m:mo>
+
+ <m:msup>
+ <m:mi>13</m:mi>
+
+ <m:mi>2</m:mi>
+ </m:msup>
+ </m:mrow>
+ </m:math>
+ </informalequation>
+ </listitem>
+ </itemizedlist>
+
+ <para>For most combinations however at least one value will be
of real rather than of integer value e.g.:</para>
<itemizedlist>
@@ -6358,28 +6516,8 @@ for (int i = 0; <emphasis role="red">!numberWasFound &&</emphasis> i <
<m:mo>)</m:mo>
</m:mrow>
</m:math>
- </inlineequation> of non-zero integer values which
- simultaneously obey the additional constraint <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:mn>840</m:mn>
- </m:mrow>
- </m:math>
- </inlineequation>.</para>
+ </inlineequation> of non-zero integer values being smaller
+ than 100 each. Keep the limit of 100 flexible.</para>
<tip>
<para>Think of creating all possible triples <inlineequation>
@@ -6419,7 +6557,7 @@ for (int i = 0; <emphasis role="red">!numberWasFound &&</emphasis> i <
<td>1</td>
- <td>1</td>
+ <td>2</td>
</tr>
<tr>
@@ -6427,7 +6565,7 @@ for (int i = 0; <emphasis role="red">!numberWasFound &&</emphasis> i <
<td>1</td>
- <td>1</td>
+ <td>2</td>
</tr>
<tr>
@@ -6439,766 +6577,336 @@ for (int i = 0; <emphasis role="red">!numberWasFound &&</emphasis> i <
</tr>
<tr>
- <td>838</td>
+ <td>99</td>
- <td>838</td>
+ <td>99</td>
- <td>838</td>
+ <td>100</td>
</tr>
</informaltable>
- <para>Why does an upper value of 838 rather then the
- proposed sum value of 840 appears here?</para>
+ <para>Why does an upper value of 99 appears here for a and
+ b? Why do we have a lower bond of 2 for the variable
+ c?</para>
+ </tip>
- <para>This list mostly contains values like <inlineequation>
- <m:math display="inline">
- <m:mrow>
- <m:mo>(</m:mo>
+ <tip>
+ <para>Consider using three nested loops for creating all
+ possible combinations. Do not yet bother about possible
+ duplicates. Your result output may look like:</para>
+
+ <screen>Found (4, 3, 5)
+Found (3, 4, 5)
+Found (8, 6, 10)
+Found (6, 8, 10)
+...</screen>
+ </tip>
+ </question>
- <m:mrow>
- <m:mn>1</m:mn>
+ <answer>
+ <para>A value of e.g. b == 100 would imply a == 0
+ contradicting the description. Likewise c == 1 wouls imply
+ either a == 0 or b == 0;</para>
- <m:mo>,</m:mo>
+ <para>We create all possible combinations by using three
+ nested loops and filtering for the desired results: </para>
- <m:mn>3</m:mn>
+ <programlisting language="java">final int LIMIT = 100;
- <m:mo>,</m:mo>
+for (int c = 1; c <= LIMIT; c++) {
+ for (int b = 1; b < LIMIT; b++) {
+ for (int a = 1; a < LIMIT; a++) {
+ if (a * a + b * b == c * c ) { // Limit output by filtering for pythagorean triples;
+ System.out.println("Found (" + a + ", " +
+ "" + b + ", " + c + ")");
+ }
+ }
+ }
+}</programlisting>
- <m:mn>4</m:mn>
- </m:mrow>
+ <para>This results in:</para>
- <m:mo>)</m:mo>
- </m:mrow>
- </m:math>
- </inlineequation> failing both our restrictions
- <inlineequation>
- <m:math display="inline">
- <m:mrow>
+ <screen>Found (4, 3, 5)
+Found (3, 4, 5)
+Found (8, 6, 10)
+Found (6, 8, 10)
+Found (12, 5, 13)
+...
+Found (96, 28, 100)
+Found (80, 60, 100)
+Found (60, 80, 100)
+Found (28, 96, 100)</screen>
+
+ <para>Note duplicates like <code>(4, 3, 5)</code> and
+ <code>(3, 4, 5)</code> being present in the above list</para>
+ </answer>
+ </qandaentry>
+ </qandadiv>
+ </qandaset>
+
+ <qandaset defaultlabel="qanda"
+ xml:id="sw1QandaPythagoreanTriplesNoDuplicates">
+ <title>Avoiding duplicates and gaining performance</title>
+
+ <qandadiv>
+ <qandaentry>
+ <question>
+ <para>There are several issues with the previous
+ solution:</para>
+
+ <orderedlist>
+ <listitem>
+ <para>It creates a great number of combinations like
+ <inlineequation>
+ <m:math display="inline">
<m:mrow>
- <m:msup>
- <m:mn>1</m:mn>
+ <m:mo>(</m:mo>
- <m:mn>2</m:mn>
- </m:msup>
+ <m:mrow>
+ <m:mn>4</m:mn>
- <m:mo>+</m:mo>
+ <m:mo>,</m:mo>
- <m:msup>
- <m:mn>3</m:mn>
+ <m:mn>4</m:mn>
- <m:mn>2</m:mn>
- </m:msup>
- </m:mrow>
+ <m:mo>,</m:mo>
+
+ <m:mn>5</m:mn>
+ </m:mrow>
- <m:mo>≠</m:mo>
+ <m:mo>)</m:mo>
+ </m:mrow>
+ </m:math>
+ </inlineequation> which could be easily ruled out
+ beforehand rather then creating and filtering them.</para>
+ <para>The former solution filters <inlineequation>
+ <m:math display="inline">
<m:msup>
- <m:mn>4</m:mn>
+ <m:mi>99</m:mi>
- <m:mn>2</m:mn>
+ <m:mi>3</m:mi>
</m:msup>
- </m:mrow>
- </m:math>
- </inlineequation> and <inlineequation>
- <m:math display="inline">
- <m:mrow>
- <m:mrow>
- <m:mn>1</m:mn>
+ </m:math>
+ </inlineequation>combinations or roughly one
+ million.</para>
+ </listitem>
- <m:mo>+</m:mo>
+ <listitem>
+ <para>Triples having identical values of <code>a</code>
+ and <code>b</code> in different order like <code>(3, 4,
+ 5)</code> and <code>(4, 3, 5)</code> are equivalent. For
+ this reason we look for ordered triples only.</para>
+ </listitem>
+ </orderedlist>
+
+ <para>The problem can thus be restated to find the set of all
+ triples <inlineequation>
+ <m:math display="inline">
+ <m:mrow>
+ <m:mo>(</m:mo>
- <m:mn>3</m:mn>
+ <m:mrow>
+ <m:mi>a</m:mi>
- <m:mo>+</m:mo>
+ <m:mo>,</m:mo>
- <m:mn>4</m:mn>
- </m:mrow>
+ <m:mi>b</m:mi>
- <m:mo>≠</m:mo>
+ <m:mo>,</m:mo>
- <m:mi>840</m:mi>
+ <m:mi>c</m:mi>
</m:mrow>
- </m:math>
- </inlineequation>. These triples have thus to be filtered
- retaining only the desired combinations simultaneously
- obeying both <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:mo>)</m:mo>
- <m:msup>
- <m:mi>b</m:mi>
+ <m:mo>∈</m:mo>
- <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> and <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>840</m:mi>
- </m:mrow>
- </m:math>
- </inlineequation>. This might be accomplished by
- implementing three nested loops:</para>
-
- <programlisting language="java">for (int a = 1; a <= 838; a++) {
- for (int b = 1; b <= 838; b++) {
- for (int c = 1; c <= 838; c++) {
- if (a*a + b*b == c*c && // Pythagorean condition
- a + b + c == 840) { // sum constraint
- System.out.println(" Found (" + a + ", " + b + ", " + c + ")");
- }
- }
- }
-}</programlisting>
-
- <para>The above code generating <inlineequation>
- <m:math display="inline">
- <m:mrow>
- <m:mrow>
- <m:mn>838</m:mn>
-
- <m:mo>×</m:mo>
-
- <m:mn>838</m:mn>
-
- <m:mo>×</m:mo>
-
- <m:mn>838</m:mn>
- </m:mrow>
-
- <m:mo>=</m:mo>
-
- <m:mn>588480472</m:mn>
- </m:mrow>
- </m:math>
- </inlineequation> combinations will run quite some time!
- Thus we are looking for a better (more efficient)
- solution.</para>
-
- <para>There are several issues with the above code:</para>
-
- <orderedlist>
- <listitem>
- <para>It creates a great number of combinations like
- <inlineequation>
- <m:math display="inline">
+ <m:msup>
<m:mrow>
- <m:mo>(</m:mo>
+ <m:mo>[</m:mo>
<m:mrow>
<m:mn>1</m:mn>
<m:mo>,</m:mo>
- <m:mn>4</m:mn>
-
- <m:mo>,</m:mo>
-
- <m:mn>3</m:mn>
+ <m:mi>LIMIT</m:mi>
</m:mrow>
- <m:mo>)</m:mo>
+ <m:mo>]</m:mo>
</m:mrow>
- </m:math>
- </inlineequation> which could be easily ruled out
- beforehand rather then creating and filtering
- them.</para>
-
- <para>Do we really need three nested loops?</para>
- </listitem>
-
- <listitem>
- <para>Triples having identical values in different order
- like <code>(40, 399, 401)</code> and <code>(40, 401,
- 399)</code> are equivalent. For this reason we may
- consider only triples being ordered by value.</para>
- </listitem>
- </orderedlist>
- <para>The problem can thus be stated to find the set of all
- triples <inlineequation>
- <m:math display="inline">
- <m:mrow>
- <m:mo>(</m:mo>
+ <m:mi>3</m:mi>
+ </m:msup>
+ </m:mrow>
+ </m:mrow>
+ </m:math>
+ </inlineequation> 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:mo>≤</m:mo>
<m:mi>b</m:mi>
- <m:mo>,</m:mo>
+ <m:mo><</m:mo>
<m:mi>c</m:mi>
</m:mrow>
+ </m:math>
+ </inlineequation></para>
+ </listitem>
- <m:mo>)</m:mo>
-
- <m:mo>∈</m:mo>
-
+ <listitem>
+ <para><inlineequation>
+ <m:math display="inline">
<m:mrow>
- <m:msup>
- <m:mrow>
- <m:mo>[</m:mo>
-
- <m:mrow>
- <m:mn>1</m:mn>
-
- <m:mo>,</m:mo>
-
- <m:mi>838</m:mi>
- </m:mrow>
-
- <m:mo>]</m:mo>
- </m:mrow>
-
- <m:mi>3</m:mi>
- </m:msup>
- </m:mrow>
- </m:mrow>
- </m:math>
- </inlineequation> 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>≤</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:msup>
<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:mn>840</m:mn>
- </m:mrow>
- </m:math>
- </inlineequation></para>
- </listitem>
-
- <listitem>
- <para><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:mn>2</m:mn>
+ </m:msup>
- <m:mo>=</m:mo>
+ <m:mo>+</m:mo>
<m:msup>
- <m:mi>c</m:mi>
+ <m:mi>b</m:mi>
<m:mn>2</m:mn>
</m:msup>
</m:mrow>
- </m:math>
- </inlineequation></para>
- </listitem>
- </orderedlist>
-
- <para>Find a better way than the above three nested
- loops.</para>
- </tip>
- </question>
-
- <answer>
- <para>Since each variable value must be greater than zero the
- maximum possible value is 840 - 1 - 1 = 838.</para>
-
- <para>The <quote>last</quote> triple obeying both restrictions
- <xref linkend="sw1PhytagoreanTripletCondition1"/> and <xref
- linkend="sw1PhytagoreanTripletCondition2"/> is (280, 280,
- 280). Due to 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>840</m:mi>
- </m:mrow>
- </m:math>
- </inlineequation> two variable values already determine the
- third one: For a given value of variable a from the outer
- <code language="java">for</code> loop any value of b will
- imply <inlineequation>
- <m:math display="inline">
- <m:mrow>
- <m:mi>c</m:mi>
-
- <m:mo>=</m:mo>
-
- <m:mrow>
- <m:mn>840</m:mn>
-
- <m:mo>-</m:mo>
-
- <m:mi>a</m:mi>
-
- <m:mo>-</m:mo>
-
- <m:mi>b</m:mi>
- </m:mrow>
- </m:mrow>
- </m:math>
- </inlineequation>.</para>
-
- <para>We thus only need two nested loops for generating all
- possible triples: If the value of b is being incremented by 1
- then c has to be decremented by 1 as well and is thus
- completely determined.</para>
-
- <para>For convenience reasons we introduce a variable <code
- language="java">int sum = 840</code> representing our sum of
- all three values. Our filter condition now only needs to check
- for the remaining Pythagorean condition <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>:</para>
-
- <programlisting language="java">final int sum = 840;
-
-for (int a = 1; a <= sum / 3; a++) {
-
- final int aSquare = a * a; // Calculate square of variable a
- // outside subsequent loop.
-
- for (int b = a, c = sum - a - b; b < c; b++, c--) {
-
- if (aSquare + b * b == c * c) { // Filter for pythagorean triples
- System.out.println("(" + a + ", " + b + ", " + c + ")");
- }
- }
-}</programlisting>
-
- <para>We've reduced the original code from three to just two
- nested loops. Running this code reveals eight triples:</para>
-
- <screen>(40, 399, 401)
-(56, 390, 394)
-(105, 360, 375)
-(120, 350, 370)
-(140, 336, 364)
-(168, 315, 357)
-(210, 280, 350)
-(240, 252, 348)</screen>
-
- <para>The above code can be simplified even further. We
- introduce a variable s being shorthand for <code
- language="java">sum</code>. The <quote>inner</quote> <code
- language="java">for</code> loop so far starts with a value of
- <inlineequation>
- <m:math display="inline">
- <m:mrow>
- <m:mi>b</m:mi>
-
- <m:mo>=</m:mo>
-
- <m:mi>a</m:mi>
- </m:mrow>
- </m:math>
- </inlineequation> and thus <inlineequation>
- <m:math display="inline">
- <m:mrow>
- <m:mi>c</m:mi>
-
- <m:mo>=</m:mo>
-
- <m:mrow>
- <m:mi>s</m:mi>
-
- <m:mo>-</m:mo>
-
- <m:mrow>
- <m:mn>2</m:mn>
-
- <m:mo>â¢</m:mo>
-
- <m:mi>a</m:mi>
- </m:mrow>
- </m:mrow>
- </m:mrow>
- </m:math>
- </inlineequation>. Then during each step <code
- language="java">b</code> is being incremented by 1 and <code
- language="java">c</code> will be decremented by the same
- amount.</para>
-
- <para>Regarding the n-th step of the inner loop we thus look
- for a value of n so that the following equation becomes
- true:</para>
-
- <informalequation>
- <m:math display="inline">
- <m:mrow>
- <m:mrow>
- <m:msup>
- <m:mi>a</m:mi>
-
- <m:mi>2</m:mi>
- </m:msup>
-
- <m:mo>+</m:mo>
-
- <m:munder accentunder="true">
- <m:msup>
- <m:mrow>
- <m:mo>(</m:mo>
-
- <m:mrow>
- <m:mi>a</m:mi>
-
- <m:mo>+</m:mo>
-
- <m:mi>n</m:mi>
- </m:mrow>
-
- <m:mo>)</m:mo>
- </m:mrow>
-
- <m:mi>2</m:mi>
- </m:msup>
-
- <m:munder accentunder="true">
- <m:mo stretchy="true">︸</m:mo>
+ <m:mo>=</m:mo>
<m:msup>
- <m:mi>b</m:mi>
+ <m:mi>c</m:mi>
- <m:mi>2</m:mi>
+ <m:mn>2</m:mn>
</m:msup>
- </m:munder>
- </m:munder>
- </m:mrow>
-
- <m:mo>=</m:mo>
-
- <m:munder>
- <m:msup>
- <m:mrow>
- <m:mo>(</m:mo>
-
- <m:mrow>
- <m:mi>s</m:mi>
-
- <m:mo>-</m:mo>
-
- <m:mrow>
- <m:mn>2</m:mn>
-
- <m:mo>â¢</m:mo>
-
- <m:mi>a</m:mi>
- </m:mrow>
-
- <m:mo>-</m:mo>
-
- <m:mi>n</m:mi>
- </m:mrow>
-
- <m:mo>)</m:mo>
</m:mrow>
+ </m:math>
+ </inlineequation></para>
+ </listitem>
+ </orderedlist>
- <m:mi>2</m:mi>
- </m:msup>
-
- <m:munder>
- <m:mo stretchy="true">︸</m:mo>
-
- <m:msup>
- <m:mi>c</m:mi>
-
- <m:mi>2</m:mi>
- </m:msup>
- </m:munder>
- </m:munder>
- </m:mrow>
- </m:math>
- </informalequation>
-
- <!--
- <informalequation>
- <mathphrase>$a^2 + \underbrace{(a + n)^2}_{b^2} =
- \underbrace{(s - 2a -n)^2}_{c^2}$</mathphrase>
- </informalequation>
- -->
+ <para>Your solution shall account for both the number of
+ combinations being evaluated and the number of Pythagorean
+ triples being found creating a final line like:</para>
- <para>Solving this equation for n results in:</para>
+ <screen>52 triples found by filtering 131935 combinations</screen>
- <informalequation>
- <m:math display="inline">
- <m:mrow>
- <m:mi>n</m:mi>
+ <para>Your solution should find all 52 triples by filtering
+ from at most 131935 combinations. If you do need less
+ combinations in the first place please let me know. I'm more
+ than happy publishing an enhanced solution.</para>
+ </question>
- <m:mo>=</m:mo>
+ <answer>
+ <para>Based on our previous solution we introduce the
+ following changes:</para>
- <m:mfrac>
- <m:mrow>
+ <orderedlist>
+ <listitem>
+ <para>We calculate <inlineequation>
+ <m:math display="inline">
<m:msup>
- <m:mrow>
- <m:mo>(</m:mo>
-
- <m:mrow>
- <m:mi>s</m:mi>
-
- <m:mo>-</m:mo>
-
- <m:mrow>
- <m:mn>2</m:mn>
-
- <m:mo>â¢</m:mo>
-
- <m:mi>a</m:mi>
- </m:mrow>
- </m:mrow>
-
- <m:mo>)</m:mo>
- </m:mrow>
+ <m:mi>c</m:mi>
<m:mi>2</m:mi>
</m:msup>
+ </m:math>
+ </inlineequation> beforehand:</para>
- <m:mo>-</m:mo>
+ <programlisting language="java">...
+for (int c = 2; c <= LIMIT; c++) {
+ <emphasis role="red">final int cSquare = c * c</emphasis>;
+...</programlisting>
+ </listitem>
+ <listitem>
+ <para>Within the next inner loop we calculate
+ <inlineequation>
+ <m:math display="inline">
<m:mrow>
- <m:mn>2</m:mn>
-
- <m:mo>â¢</m:mo>
-
<m:msup>
- <m:mi>a</m:mi>
+ <m:mi>c</m:mi>
<m:mi>2</m:mi>
</m:msup>
- </m:mrow>
- </m:mrow>
-
- <m:mrow>
- <m:mn>2</m:mn>
-
- <m:mo>â¢</m:mo>
-
- <m:mrow>
- <m:mo>(</m:mo>
-
- <m:mrow>
- <m:mi>s</m:mi>
-
- <m:mo>-</m:mo>
-
- <m:mi>a</m:mi>
- </m:mrow>
-
- <m:mo>)</m:mo>
- </m:mrow>
- </m:mrow>
- </m:mfrac>
- </m:mrow>
- </m:math>
- </informalequation>
-
- <para>So for each given value of a there is either no or
- exactly one possible solution (since <code
- language="java">n</code> is of integer type). This way the
- <quote>inner</quote> loop becomes obsolete.</para>
-
- <para>The above equation must hold for integer arithmetic
- leaving no fractional remainder. We introduce two variables
- <code language="java">numerator</code> and <code
- language="java">denominator</code> representing the above
- terms <inlineequation>
- <m:math display="inline">
- <m:mrow>
- <m:msup>
- <m:mrow>
- <m:mo>(</m:mo>
-
- <m:mrow>
- <m:mi>s</m:mi>
-
- <m:mo>-</m:mo>
-
- <m:mrow>
- <m:mn>2</m:mn>
-
- <m:mo>â¢</m:mo>
-
- <m:mi>a</m:mi>
- </m:mrow>
- </m:mrow>
-
- <m:mo>)</m:mo>
- </m:mrow>
-
- <m:mi>2</m:mi>
- </m:msup>
-
- <m:mo>-</m:mo>
-
- <m:mrow>
- <m:mn>2</m:mn>
-
- <m:mo>â¢</m:mo>
-
- <m:msup>
- <m:mi>a</m:mi>
-
- <m:mi>2</m:mi>
- </m:msup>
- </m:mrow>
- </m:mrow>
- </m:math>
- </inlineequation> and <inlineequation>
- <m:math display="inline">
- <m:mrow>
- <m:mn>2</m:mn>
-
- <m:mo>â¢</m:mo>
-
- <m:mrow>
- <m:mo>(</m:mo>
-
- <m:mrow>
- <m:mi>s</m:mi>
<m:mo>-</m:mo>
- <m:mi>a</m:mi>
- </m:mrow>
+ <m:msup>
+ <m:mi>b</m:mi>
- <m:mo>)</m:mo>
- </m:mrow>
- </m:mrow>
- </m:math>
- </inlineequation>. The <quote>remainder of modulus</quote>
- operator <code language="java">%</code> testing for a zero
- remainder filters the desired triples:</para>
+ <m:mi>2</m:mi>
+ </m:msup>
+ </m:mrow>
+ </m:math>
+ </inlineequation> beforehand as well:</para>
- <programlisting language="java">final int sum = 840;
+ <programlisting language="java">...
+for (int b = 1; b < c; b++) { // b must not exceed c since a == 0 otherwise
+ <emphasis role="red">final int cSquare_minus_bSquare = cSquare - b * b</emphasis>;
+...</programlisting>
+ </listitem>
-for (int a = 1; a <= sum / 3; a++) {
+ <listitem>
+ <para>The innermost loop will terminate on two
+ conditions:</para>
+
+ <programlisting language="java">...
+for (int a = 1;
+ <emphasis role="red">a <= b</emphasis> && // Only ordered triples
+ <emphasis role="red">a * a <= cSquare_minus_bSquare</emphasis>; // Only till square root of c² - b²
+ a++) { ...</programlisting>
+ </listitem>
- final int numerator = (sum - 2 * a) * (sum - 2 * a) - 2 * a * a,
- denominator = 2 * (sum - a);
+ <listitem>
+ <para>Finally we re-write our filtering condition:</para>
- final int remainder = numerator % denominator;
- if (0 == remainder) {
- final int n = numerator / denominator;
- System.out.println("(" + a + ", " + (a + n) + ", " + (sum - 2 * a - n) + ")");
- }
+ <programlisting language="java">...
+if (<emphasis role="red">a * a == cSquare_minus_bSquare</emphasis>) { // Equivalent to a² + b² == c²
+ System.out.println("Found (" + a + ", " + b + ", " + c + ")");
}</programlisting>
+ </listitem>
+ </orderedlist>
- <para>We've thus reduced three nested loops into just one
- getting exactly the same result as before. Obviously execution
- is (again) ways faster paying off when looking for larger
- values of <code language="java">sum</code>.</para>
+ <para>Piecing together these snippets and adding two variables
+ <code language="java">comparisons</code> and <code
+ language="java">countTriples</code> for accounting yields our
+ final solution:</para>
+
+ <programlisting language="java">final int LIMIT = 100;
+
+int comparisons = 0, countTriples = 0;
+
+for (int c = 2; c <= LIMIT; c++) {
+ final int cSquare = c * c;
+ for (int b = 1; b < c; b++) { // b must not exceed c since a == 0 otherwise
+ final int cSquare_minus_bSquare = cSquare - b * b; // c² - b²
+ for (int a = 1;
+ a <= b && // Only ordered triples
+ a * a <= cSquare_minus_bSquare; // Only till square root of c² - b²
+ a++) {
+ if (a * a == cSquare_minus_bSquare ) { // Equivalent to a² + b² == c²
+ System.out.println(
+ "Found (" + a + ", " + b + ", " + c + ")");
+ countTriples++;
+ }
+ comparisons++;
+ }
+ }
+}
+System.out.println(countTriples + " triples found by filtering " + comparisons + " combinations");</programlisting>
</answer>
</qandaentry>
</qandadiv>
--
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