How to compute de Bruijn sequences for non-power-of-two-sized alphabets?

Are you only interested in a generalization of Prefer Ones or do you just want a not so complex algorithm? If the latter is true then maybe Frank Ruskey's recursive implementation could be of help.

A year ago I translated that one to Ruby.

# De Bruijn sequence
# Original implementation by Frank Ruskey (1994)
# translated to C by Joe Sawada
# and further translated to Ruby by Jonas Elfström (2009)

@n=4
@k=10
@a=[0]
@sequence=[]

def debruijn(t, p, alike)
  if t>@n
    if @n%p==0
      1.upto(p) {|j| @sequence<<@a[j]}
    end
  else
    @a[t]=@a[t-p]
    if @a[t]>0
      debruijn(t+1,p,alike+1)
    else
      debruijn(t+1,p,alike)
    end
    (@a[t-p]+1).upto(@k-1) {|j|
      @a[t]=j
      debruijn(t+1,t,alike+1)
    }
  end
end

debruijn(1,1,0)
print @sequence.join

Uckelman noticed that the alike variable does nothing. The following produces the same sequence.

@n=4
@k=10
@a=[0]
@sequence=[]

def debruijn(t, p)
  if t>@n
    if @n%p==0
      1.upto(p) {|j| @sequence<<@a[j]}
    end
  else
    @a[t]=@a[t-p]
    debruijn(t+1,p)
    (@a[t-p]+1).upto(@k-1) {|j|
      @a[t]=j
      debruijn(t+1,t)
    }
  end
end

debruijn(1,1)
print @sequence.join

or you can use:

def de_bruijn(k, n):
    a = [0] * k * n
    sequence = []
    def db(t, p):
        if t > n:
            if n % p == 0:
                for j in range(1, p + 1):
                    sequence.append(a[j])
        else:
            a[t] = a[t - p]
            db(t + 1, p)
            for j in range(a[t - p] + 1, k):
                a[t] = j
                db(t + 1, t)
    db(1, 1)
    return sequence

print de_bruijn(2,9)

This is my C++ implementation of the algorithm in Figure 1 from a paper by Sawada and Ruskey:

void debruijn(unsigned int t,
              unsigned int p,
              const unsigned int k,
              const unsigned int n,
              unsigned int* a,
              boost::function<void (unsigned int*,unsigned int*)> callback)
{
  if (t > n) {
    // we want only necklaces, not pre-necklaces or Lyndon words
    if (n % p == 0) {
      callback(a+1, a+p+1);
    }
  }
  else {
    a[t] = a[t-p];

    debruijn(t+1, p, k, n, a, callback);

    for (unsigned int j = a[t-p]+1; j < k; ++j) {
      a[t] = j;
      debruijn(t+1, t, k, n, a, callback);
    }
  }
}

struct seq_printer {
  const std::vector<char>& _alpha;

  seq_printer(const std::vector<char>& alpha) : _alpha(alpha) {}

  void operator() (unsigned int* a, unsigned int* a_end) const {
    for (unsigned int* i = a; i < a_end; ++i) {
      std::cout << _alpha[*i];
    }
  }
};

...

std::vector<char> alpha;
alpha.push_back('a');
alpha.push_back('b');
alpha.push_back('c');

unsigned int* a = new unsigned int[N+1];
a[0] = 0;

debruijn(1, 1, alpha.size(), N, a, seq_printer(alpha));
if (N > 1) std::cout << alpha[0];
std::cout << std::endl;

delete[] a;

The full reference for the paper is: Joe Sawada and Frank Ruskey, "An Efficient Algorithm for Generating Necklaces with Fixed Density", SIAM Journal of Computing 29:671-684, 1999.

According to this web page at the combinatorial group of the CS department at UVic, there's a result due to Fredericksen that you can generate a de Bruijn sequence (in fact, the lexicographically smallest one) by concatenating "the lexicographic sequence of Lyndon words of lengths divisible by n". There's even source code to build the sequence that you can request.