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    <p>You need to define to the rules in an appropriate way in java code and make all valid permutations of {1,2,3,4,5,6,7} and check, which permutations fit into this rules. I have done this already here for an other permutation question: <a href="https://stackoverflow.com/questions/4366119/java-if-condition-question/4370253#4370253">How to write an &quot;all these numbers are different&quot; condition in Java?</a></p> <p>Adapted on your rules, the code can look like this:</p> <pre><code>import java.util.Arrays; class Graph26 { private static final int A = 0; private static final int B = 1; private static final int C = 2; private static final int D = 3; private static final int E = 4; private static final int F = 5; private static final int G = 6; private final static boolean rule1(final int[] n) { return n[A] != 2; } private final static boolean rule2(final int[] n) { return n[A] + n[B] == n[F]; } private final static boolean rule3(final int[] n) { return n[C] - n[D] == n[G]; } private final static boolean rule4(final int[] n) { return n[D] + n[E] == 2*n[F]; } private final static boolean rule5(final int[] n) { return n[E] + n[G] == n[F]; } private final static boolean isValid(final int[] nodes) { return rule1(nodes) &amp;&amp; rule2(nodes) &amp;&amp; rule3(nodes) &amp;&amp; rule4(nodes) &amp;&amp; rule5(nodes); } class Permutation { private final int[] o; private boolean perms = true; public boolean hasPerms() { return perms; } Permutation(final int[] obj) { o = obj.clone(); } private int[] nextPerm() { int temp; int j = o.length - 2; while (o[j] &gt; o[j + 1]) { j--; if (j &lt; 0) { perms = false; break; } } if (perms) { int k = o.length - 1; while (o[j] &gt; o[k]) { k--; } temp = o[k]; o[k] = o[j]; o[j] = temp; int r = o.length - 1; int s = j + 1; while (r &gt; s) { temp = o[s]; o[s] = o[r]; o[r] = temp; r--; s++; } } return o.clone(); } } public static void main(final String[] args) { int[] nodes = new int[] { 1, 2, 3, 4, 5, 6, 7}; final Graph26 graph = new Graph26(); final Permutation p = graph.new Permutation(nodes); int i = 0; while (p.hasPerms()) { if (isValid(nodes)) { System.out.println(Arrays.toString(nodes)); } i++; nodes = p.nextPerm(); } System.out.println(i); } } </code></pre> <p>This will define rules1..5 regarding the rules defined in the question and perform a check on all !7=5040 permutations of {1,2,3,4,5,6,7}. You can see this in action here: <a href="https://ideone.com/wwxG0" rel="nofollow noreferrer">https://ideone.com/wwxG0</a></p> <p>which results in (A,B,C,D,E,F,G): <code>[3, 2, 7, 6, 4, 5, 1]</code></p>
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    1. POHow do I write a program to solve multiple questions?
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    1. COCan I do it by hand?
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    2. CO@toy: What do you mean? What would you like to do by hand?
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