diff --git a/Doc/Sd1/statements.xml b/Doc/Sd1/statements.xml
index 7eb214fc1cee89b06e02bf45b86a272ae6c5d6a4..6cd304c3f1022380bb9f3a9d91e8676f4da6ad75 100644
--- a/Doc/Sd1/statements.xml
+++ b/Doc/Sd1/statements.xml
@@ -4734,29 +4734,28 @@ try (final Scanner scan = new Scanner(System.in)) {
<answer>
<para>We propose the following solution:</para>
- <programlisting language="java"> public static void main(String[] args) {
-
- final int highestDivisor = 20; // May be adjusted to other limits.
+ <programlisting language="java">public static void main(String[] args) {
- int candidate = highestDivisor; // start with highest divisor.
+ final int highestDivisor = 20; // May be adjusted to other limits.
- boolean atLeastOneRemainder;
+ int candidate = highestDivisor; // start with highest divisor.
- do {
- candidate++; // next candidate value.
- atLeastOneRemainder = false;
+ boolean atLeastOneRemainder;
- for (int i = 2; i <= highestDivisor; i++) {
- if (0 != candidate % i) { // Is there a non-zero remainder?
- atLeastOneRemainder = true; // Continue outer while.
- break; // Leave current for loop.
- }
- }
+ do {
+ candidate++; // next candidate value.
+ atLeastOneRemainder = false;
- } while (atLeastOneRemainder); // Increase candidate further?
+ for (int i = highestDivisor; 2 <= i; i--) {// Higher values are more likely having a remainder
+ if (0 != candidate % i) { // Is there a non-zero remainder?
+ atLeastOneRemainder = true; // Continue outer while.
+ break; // Leave current for loop.
+ }
+ }
+ } while (atLeastOneRemainder); // Increase candidate further?
- System.out.println(candidate);
- }</programlisting>
+ System.out.println(candidate);
+}</programlisting>
<para>Executing this code results in a value of 232792560 for
the smallest desired value.</para>
@@ -4773,8 +4772,8 @@ try (final Scanner scan = new Scanner(System.in)) {
<qandaentry>
<question>
<para>Solving the previous exercise by a program is quite a
- no-brainer (from a software developer's point of view).
- Provide a different solution purely based on algebraic
+ no-brainer (from an experienced software developer's point of
+ view). Provide a different solution purely based on algebraic
considerations. In other words: Your solution should work by
using paper and pencil exclusively.</para>
@@ -5648,37 +5647,47 @@ try (final Scanner scan = new Scanner(System.in)) {
Thus we are looking for a better (more efficient)
solution.</para>
- <para>There are several issues with the above code. It
- creates a great number of combinations like <inlineequation>
- <m:math display="inline">
- <m:mrow>
- <m:mo>(</m:mo>
+ <para>There are several issues with the above code:</para>
- <m:mrow>
- <m:mn>1</m:mn>
+ <orderedlist>
+ <listitem>
+ <para>It creates a great number of combinations like
+ <inlineequation>
+ <m:math display="inline">
+ <m:mrow>
+ <m:mo>(</m:mo>
- <m:mo>,</m:mo>
+ <m:mrow>
+ <m:mn>1</m:mn>
- <m:mn>4</m:mn>
+ <m:mo>,</m:mo>
- <m:mo>,</m:mo>
+ <m:mn>4</m:mn>
- <m:mn>3</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. Do we
- really need three nested loops?</para>
+ <m:mn>3</m:mn>
+ </m:mrow>
- <para>Also 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>
+ <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>So the problem can be stated to find the set of all
+ <para>The problem can thus be stated to find the set of all
triples <inlineequation>
<m:math display="inline">
<m:mrow>
@@ -5721,7 +5730,7 @@ try (final Scanner scan = new Scanner(System.in)) {
</m:mrow>
</m:mrow>
</m:math>
- </inlineequation> which simultaneously obey:</para>
+ </inlineequation> simultaneously obeying:</para>
<orderedlist>
<listitem xml:id="sw1PhytagoreanTripletCondition1">
@@ -6223,9 +6232,10 @@ for (int a = 1; a <= sum / 3; a++) {
}
}</programlisting>
- <para>This yields the same result as before but execution is
- (again) ways faster paying off when looking for larger values
- of <code language="java">sum</code>.</para>
+ <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>
</answer>
</qandaentry>
</qandadiv>