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I was working on the "Robot Name" exercise over on exercism.io (neat site, by the way), the gist of which is: Generate random, unique strings that match /[A-Za-z]{2}[0-9]{3}/, e.g. "xY123", "Gv020", and so on. So there are 2,704,000 possible names out there.

Now, the exercise's tests do not do a full-on check for uniqueness or randomness, but making the names random and collision-free is something you're encouraged to try yourself. So I wanted to try that.

One option is of course to generate each and every name sequentially and shuffle them. But that seems pretty inelegant.

So went looking for something cooler and came across this neat method for generating random, non-repeating 32 bit integers using "quadratic prime residue" (which sounds like something you'd find on a mathematician's shoes).

That method is really neat, but it makes use of C's strict 32-bit ints, and some XOR magic that I can't use for the smaller range I'm dealing with here.

So I modified it a little bit, and came up with what's below. I did a quick and dirty test, printing all the names to a file, and running checking for uniqueness thusly:

$ sort -u names.txt | wc -l

and indeed it claims that there are 2,704,000 names after sorting and removing duplicates. So yay for that. Distribution is probably skewed all sorts of ways, but it appears to work, and at least it's not sequential.

But I'm wondering if there's a simpler way that I'm just missing here. Or, failing that, if there are improvements that could be made to the thing below.

Note that the code is very much fixed to the 0..2,704,000 range. This is on purpose (hence the hard-coded prime number etc.). I'm not really interested in tweaking this particular code to make it more flexible in that regard. But if someone has an already-flexible approach that could replace it, that'd be neat.

require "singleton"

class RobotNameGenerator
  include Singleton

  # 2x fifty-two letters (A-Z and a-z), 3 digits = 2,704,000 combinations
  BOUNDARY = 2_704_000

  # Largest prime below BOUNDARY that also satisfies p % 4 == 3
  PRIME = 2_703_983

  # Another prime used as step
  STEP = 23

  # Initializes the instance, and sets a random offset (seed value)
  def initialize
    @offset = rand(0...BOUNDARY)
  end

  # Get a new pseudo-random name
  def next
    serial = next_serial_number
    serial, digits = serial.divmod(1000)
    serial, b = serial.divmod(52)
    serial, a = serial.divmod(52)
    "%s%s%03i" % [letter_for(a), letter_for(b), digits]
  end

  private

  # Get the next serial number
  def next_serial_number
    serial = permute(permute(@offset % BOUNDARY))
    @offset += STEP
    serial
  end

  # Do a one-to-one permutation of a number
  def permute(n)
    return n if n > PRIME
    residue = n**2 % PRIME
    n <= PRIME / 2 ? residue : PRIME - residue
  end

  # Encode a 0-51 number to A-Z/a-z
  def letter_for(n)
    offset = n < 26 ? 'A'.ord : 'a'.ord
    ascii = offset + (n % 26)
    ascii.chr
  end
end
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    \$\begingroup\$ Any next = last * prime % 2_704_000 should generate non-repeating sequence (when the modulo is not divisable by the prime) and you can convert the number to the string. I am no ruby expert and this is not a code-review but rathre SO-answer about how to do it. \$\endgroup\$ – user52292 Nov 15 '14 at 22:59
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Any simple generator whose output is equivalent to its internal state will generate a non-repeating sequence. In fact, most will generate a permutation of the range [0, P) where P is the length of the period. Or [1, P), if state == 0 is forbidden.

Pick any simple generator with period P ≥ 2,704,000 and Bob's your uncle.

The simplest is the Golden Weyl generator (the 'Golden' is my moniker for it):

x = (x + 0x9E3779B9u) mod 2^32

The net has a wide choice of suitable Linear Congruential Generators (LCGs) of the form

x = (a * x + c) mod 2^32

Julienne Walker has written a brilliant explanation of LCGs, and the Weyl generators are just a special case with multiplier 1.

However, regardless of which type of generator you choose, its period P is likely to be too long.

Extracting a shorter non-repeating sequence of values [0,M) from a longer sequence of non-repeating values [0,P) is not trivial. In particular, if you map the [0,P) values to the [0,M) range in some fashion - e.g. by scaling, or by computing the remainder modulo M - then you lose the property of non-repetition. Stuffing P values into a smaller number M of pigeon holes means that some values must end up in the same hole.

One strategy that works well is rejection: inspect the generator output and if you don't like the value for some reason, throw it away and call the generator again. This method is perfect for generating even the most outlandish output distributions, in the sense that it does not introduce any bias at all and it leaves properties like non-repetition intact. The downside is that you have to do more work if a lot of stuff gets thrown away.

The trick, then, is to pick a non-repeating generator whose period length is as close to M = 2,704,000 as possible, and then use rejection with that.

LCGs modulo powers of 2 have the special property that any lower k bits cycle with period 2^k. I.e., the lowest bit is strictly alternating, the low 2 bits cycle with period 4, and so on. This means that these generators can be reduced modulo any power of 2 without losing the property of non-repetition. In other words, any LCG modulo 2^x is also a generator modulo 2^k for all k that are less than x. These subsequences can be extracted by throwing away (masking off) the unwanted higher bits.

The first power of two that is not less than the modulus M = 2,704,000 is 2^22. Masking with 2^22 - 1 = 0x3FFFFF we only have to reject the 35% of outputs that are in the range [M,2^22). This means the generator will have to be cycled 1.55 times on average for each good value in the range [0,M). That's less than two MULs for a non-repeating random value with an off-beat modulus.

Simple and effective.

An example for a suitable generator of higher quality (where the lower bits do not cycle with their own lower periods) is a 22-bit Tausworthe. Think of it as a single barrel of the four-barrelled lfsr113 that powers games like Dragon Age, whose long-awaited newest chapter is due to be released in a few hours.

x = (((x ^ (x << 1)) & 0x3FFFFF) >> (22 - 17)) ^ (x << 17);

return (x & 0x3FFFFF) - 1;

This generator has period 2^22 - 1 because 0 is a forbidden (sticky) state, hence the decrement after masking the output. The rejection method is then essentially the same as above.

I wrote (22 - 17) because the 17 corresponds to parameter s in the usual descriptions of the algorithm (and 22 equals k). Other possible choices for s are here: 1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 19 and 20. The algorithm is usually described in a way that makes the 'live' k bits sit at the upper end of the larger x:

x = ((x ^ (x << q)) >> (k - s)) ^ ((x & m) << s);

I took the liberty of shifting the whole shebang down to the lower end, so that the live bits can be extracted with simple masking. The output sequence is the same.

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  • \$\begingroup\$ Good solution and well-explained. A detail: double-check your calculation of the rejection rate. I think it about 1-0.35 rather than 0.35, but that's neither here nor there, as rejections cannot be avoided. Ruby's rand(n) is surely rejecting under the hood. \$\endgroup\$ – Cary Swoveland Nov 16 '14 at 3:03
  • \$\begingroup\$ Good catch, Cary, thanks! Rejecting (2^22 - M) values out of 2^22 we have (2^22 - M) / M = 0.55 rejections per each good value, or 1.55 calls to the generator for each good value. Correction applied. \$\endgroup\$ – DarthGizka Nov 16 '14 at 6:15
  • \$\begingroup\$ As an alternative to rejecting high values, you could scale them all, though I expect that would take longer. That is, f = M/2^22, then for any random number r between 0 and 2^22-1, use f*r converted to an integer. \$\endgroup\$ – Cary Swoveland Nov 16 '14 at 18:02
  • \$\begingroup\$ That would cause dupes. You can't stuff N = 2^22-1 values into M < N bins... That's why we throw away the stuff that we don't want. ;-) \$\endgroup\$ – DarthGizka Nov 16 '14 at 18:37
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    \$\begingroup\$ Yes, of course. I'll leave that comment in case anyone else has the same thought. \$\endgroup\$ – Cary Swoveland Nov 16 '14 at 18:38
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You asked for something simple. This most definitely satisfies that requirement, but it's not especially elegant. gen(n) generates n distinct strings of the sort you want. Since there is a 1-1 map between strings and integers between 0 and 2,704,000-1, it's obvious from the code that the strings are random and non-repeating.

N = 52*52*1000

def gen(n)
  h = N.times.with_object({}) { |i,h| h[i] = i }
  top = N
  n.times.with_object([]) do |i,a|
    rn = rand(top)
    a << nbr_to_str(h[rn])
    h[rn] = h[top-1]
    top -= 1
  end
end

def nbr_to_str(n)
  n, d = n.divmod(1000)
  n, s = n.divmod(52)
  "#{nbr_to_chr n}#{nbr_to_chr s}#{"%03d" % d}"
end

def nbr_to_chr(n)
  n < 26 ? (65+n).chr : (71+n).chr 
end

gen(18)
  #=> ["OZ057", "FA316", "cm258", "JH252", "aE391", "Oc044",
  #    "TN053", "pA357", "cf533", "Cf439", "FN326", "Gv310",
  #    "oH602", "Dl372", "Lc832", "PM566", "PZ184", "WE113"]

It takes a couple of seconds to generate the hash, then each random string takes very little time to generate.

[Edit: Flambino pointed out that it makes more sense to use an array:

def gen(n)
  b = (0...N).to_a
  top = N-1
  n.times.with_object([]) do |i,a|
    rn = rand(top)
    a << nbr_to_str(b[rn])
    b[rn] = b[top]
    top -= 1
  end
end
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  • \$\begingroup\$ Neat way to do the shuffling (and of course the last mod 26 can be skipped - don't know why I didn't do that). A part of me really wants this to be all-lazy, though, so generating the hash, while not too bad, is a little fly in the ointment. By the way, couldn't this also be done with a plain array instead of a hash? Or is there some reason to use a hash? With a regular array, the execution time went from ~3.1s to ~0.31s on my machine :) \$\endgroup\$ – Flambino Nov 17 '14 at 1:45
  • \$\begingroup\$ I used a hash because I wasn't thinking straight. Yes, an array makes more sense. \$\endgroup\$ – Cary Swoveland Nov 17 '14 at 2:40

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