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copied!<p>The examples you show make this considerably more complex than necessary, at least IMO.</p> <p>What you're really looking at is a 3 digit, base 3 number. You can just count from 0 to 3<sup>3</sup> = 27, then convert each number to base 3 (using 'A', 'B' and 'C' as your digits instead of '0', '1' and '2').</p> <p>In C++, the conversion could look something like this:</p> <pre class="lang-c prettyprint-override"><code>std::string cvt(int in) { static const int base = 3; static const int digits = 3; std::string ret; for (int i = 0; i<digits; i++) { ret.push_back('A' + in % base); in /= base; } return std::string(ret.rbegin(), ret.rend()); } </code></pre> <p>With the conversion in place, producing all the combinations becomes utterly trivial:</p> <pre><code>for (int i = 0; i < 27; i++) std::cout << cvt(i) << "\t"; </code></pre> <p>I believe converting that to Delphi should be barely short of purely mechanical -- assignments change from <code>=</code> to <code>:=</code>, <code>%</code> becomes <code>mod</code>, the integer division changes to <code>div</code>, the <code>for</code> loop changes to something like <code>for i = 0 to 27 do</code>, and so on. The most tedious (but ultimately quite simple) part will probably be dealing with the fact that in C++, <code>char</code> is simply a small integer type, on which you can do normal integer math. At least if memory serves, in Pascal (or a derivative like Delphi) you'll need <code>ord</code> to convert from a character to an ordinal, and <code>chr</code> to convert back from ordinal to character. So the <code>'A' + in % base;</code> will end up something more like <code>chr(ord('A') + in mod base);</code></p> <p>Like I said though, it seems like nearly the entire translation could/should end up almost completely mechanical, with no requirement for real changes in how the basic algorithms work, or anything like that.</p>

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