I&#x27;ve thought a bit more and made a small but important change in how the &quot;equalish&quot; operator works:<p>The operator should take three (not two as suggested before) sets of integers as input of equal sample size.<p>The first two sets are already known to be equalish TRUE,
as defined by either a human or a previous test of a smaller subset of the integer sets.
The third set is a the new sample you want to compare.<p>Example output:<p>equalish(&#x27;{0}&#x27;,&#x27;{0}&#x27;,&#x27;{0}&#x27;) -&gt; TRUE<p>equalish(&#x27;{0}&#x27;,&#x27;{0}&#x27;,&#x27;{1}&#x27;) -&gt; FALSE<p>equalish(&#x27;{110,105,120}&#x27;,&#x27;{113,107,121}&#x27;,&#x27;{110,106,120}&#x27;) -&gt; TRUE<p>equalish(&#x27;{1047072,1047216,1047441,1047521,1047682,59102,59361,59583,59818}&#x27;,&#x27;{1047085,1047276,1047471,1047754,1047938,59128,59364,59732,59945}&#x27;,&#x27;{1047085,1047276,1017471,1047754,1047938,59128,59364,59732,59945}&#x27;) -&gt; TRUE