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    <p>Okay, there are 26 possibilities for a 1-character string, 26<sup>2</sup> for a 2-character string, and so on up to 26<sup>26</sup> possibilities for a 26-character string.</p> <p>That means there are 26 times as many possibilities for an (N)-character string than there are for an (N-1)-character string. You can use that fact to select your length:</p> <pre><code>def getlen(maxlen): sz = maxlen while sz != 1: if rnd(27) != 1: return sz sz--; return 1 </code></pre> <p>I use 27 in the above code since the total sample space for selecting strings from "ab" is the 26 1-character possibilities and the 26<sup>2</sup> 2-character possibilities. In other words, the ratio is 1:26 so 1-character has a probability of 1/27 (rather than 1/26 as I first answered).</p> <p>This solution isn't <em>perfect</em> since you're calling <code>rnd</code> multiple times and it would be better to call it once with an possible range of 26<sup>N</sup>+26<sup>N-1</sup>+26<sup>1</sup> and select the length based on where your returned number falls within there but it may be difficult to find a random number generator that'll work on numbers that large (10 characters gives you a possible range of 26<sup>10</sup>+...+26<sup>1</sup> which, unless I've done the math wrong, is 146,813,779,479,510).</p> <p>If you can limit the maximum size so that your <code>rnd</code> function will work in the range, something like this should be workable:</p> <pre><code>def getlen(chars,maxlen): assert maxlen &gt;= 1 range = chars sampspace = 0 for i in 1 .. maxlen: sampspace = sampspace + range range = range * chars range = range / chars val = rnd(sampspace) sz = maxlen while val &lt; sampspace - range: sampspace = sampspace - range range = range / chars sz = sz - 1 return sz </code></pre> <p>Once you have the length, I would then use your current algorithm to choose the actual characters to populate the string.</p> <hr> <p>Explaining it further:</p> <p>Let's say our alphabet only consists of "ab". The possible sets up to length 3 are <code>[ab]</code> (2), <code>[ab][ab]</code> (4) and <code>[ab][ab][ab]</code> (8). So there is a 8/14 chance of getting a length of 3, 4/14 of length 2 and 2/14 of length 1.</p> <p>The 14 is the magic figure: it's the sum of all 2<sup>n</sup> for n = 1 to the maximum length. So, testing that pseudo-code above with <code>chars = 2</code> and <code>maxlen = 3</code>:</p> <pre><code> assert maxlen &gt;= 1 [okay] range = chars [2] sampspace = 0 for i in 1 .. 3: i = 1: sampspace = sampspace + range [0 + 2 = 2] range = range * chars [2 * 2 = 4] i = 2: sampspace = sampspace + range [2 + 4 = 6] range = range * chars [4 * 2 = 8] i = 3: sampspace = sampspace + range [6 + 8 = 14] range = range * chars [8 * 2 = 16] range = range / chars [16 / 2 = 8] val = rnd(sampspace) [number from 0 to 13 inclusive] sz = maxlen [3] while val &lt; sampspace - range: [see below] sampspace = sampspace - range range = range / chars sz = sz - 1 return sz </code></pre> <p>So, from that code, the first iteration of the final loop will exit with <code>sz = 3</code> if <code>val</code> is greater than or equal to <code>sampspace - range [14 - 8 = 6]</code>. In other words, for the values 6 through 13 inclusive, 8 of the 14 possibilities.</p> <p>Otherwise, <code>sampspace</code> becomes <code>sampspace - range [14 - 8 = 6]</code> and <code>range</code> becomes <code>range / chars [8 / 2 = 4]</code>.</p> <p>Then the second iteration of the final loop will exit with <code>sz = 2</code> if <code>val</code> is greater than or equal to <code>sampspace - range [6 - 4 = 2]</code>. In other words, for the values 2 through 5 inclusive, 4 of the 14 possibilities.</p> <p>Otherwise, <code>sampspace</code> becomes <code>sampspace - range [6 - 4 = 2]</code> and <code>range</code> becomes <code>range / chars [4 / 2 = 2]</code>.</p> <p>Then the third iteration of the final loop will exit with <code>sz = 1</code> if <code>val</code> is greater than or equal to <code>sampspace - range [2 - 2 = 0]</code>. In other words, for the values 0 through 1 inclusive, 2 of the 14 possibilities (this iteration will <em>always</em> exit since the value must be greater than or equal to zero.</p> <hr> <p>In retrospect, that second solution is a bit of a nightmare. In my personal opinion, I'd go for the first solution for its simplicity and to avoid the possibility of rather large numbers.</p>
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    1. POHow do I generate a random string of up to a certain length?
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    1. CODid you mean 26 to power of maxlen? This does not look right. Chances of getting length = n = maxlen are (26^n - 26^(n-1))/26^n with this, which is incorrect. Shouldn't it be 26^n/(26 + 26^2 + ... + 26^n)?
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    2. COYeah, like I'm going to listen to someone called @Moron :-) No, seriously, you're right, I did mean 26 rather than 2. Obviously I've been in the IT industry too long. Thanks for pointing that out, fixed now.
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    3. CO:-) What I also meant was that even with 26 there it seems wrong, as the probability for getting maxlen returned seems to be different from what we need. Of course I might have miscalculated.
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