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    <p>If a, b and p are positive integers (assuming that the positive range include the 0 value) then the expression <strong>(((a+p) &lt;= b) &amp;&amp; (a == 0 || a > 1) &amp;&amp; (b >= p)) &amp;&amp; ((b - (a + p) == 0) || (b - (a + p) > 1))</strong> can be reduced to <strong>((a+p)&lt;=b) &amp;&amp; (a!=1)</strong> &amp;&amp; <strong>((b-(a+p))!=1)</strong></p> <p>Let me demonstrate it: In the first part of the expression there is a condition, <strong>((a+p)&lt;=b)</strong>, that if valuated true render true the second part: <strong>((b - (a + p) == 0) || (b - (a + p) > 1))</strong>. If it is true that <strong>(b >=(a+p))</strong> then <strong>(b - (a+p)) have to be greater or equal to 0</strong>, we need to assure that <strong>(b-(a+p))!=1</strong>. Put this term aside for awhile and go on.</p> <p>Now we can concentrate our efforts on the the first part <strong>(((a+p) &lt;= b) &amp;&amp; (a == 0 || a > 1) &amp;&amp; (b >= p))</strong> &amp;&amp; ((b-(a+p))!=1)</p> <p>If <strong>a</strong> is positive then it is always >=0 and so we can drop the test <strong>(a == 0 || a > 1)</strong> if favor of (a!=1) and reduce first part of the expression to <strong>(((a+p) &lt;= b) &amp;&amp; (b >= p) &amp;&amp; (a!=1))</strong>.</p> <p>For the next step of the reduction you can consider that if <strong>b >= (a+p)</strong> then, obviously <strong>b>=p</strong> (<strong>a</strong> is positive) and the expression can be reduced to </p> <p><strong>((a+p)&lt;=b) &amp;&amp; (a!=1)</strong> &amp;&amp; <strong>((b-(a+p))!=1)</strong></p>
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    1. POCan you simplify this algorithm?
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    1. USEineki
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    1. CO"Assuming the positive range includes the 0 value". Can we also throw in the square root of negative one? Because it's a cool number too!
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      1. USJeffrey L Whitledge
    2. COIt is not an integer though ;)
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