Enumerating Cartesian product while minimizing repetition - algorithm

Question

Given two sets, e.g.:
{A B C}, {1 2 3 4 5 6}
I want to generate the Cartesian product in an order that puts as much space as possible between equal elements. For example, [A1, A2, A3, A4, A5, A6, B1…] is no good because all the As are next to each other. An acceptable solution would be going "down the diagonals" and then every time it wraps offsetting by one, e.g.:
[A1, B2, C3, A4, B5, C6, A2, B3, C4, A5, B6, C1, A3…]
Expressed visually:
| | A | B | C | A | B | C | A | B | C | A | B | C | A | B | C | A | B | C |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 1 | 1 | | | | | | | | | | | | | | | | | |
| 2 | | 2 | | | | | | | | | | | | | | | | |
| 3 | | | 3 | | | | | | | | | | | | | | | |
| 4 | | | | 4 | | | | | | | | | | | | | | |
| 5 | | | | | 5 | | | | | | | | | | | | | |
| 6 | | | | | | 6 | | | | | | | | | | | | |
| 1 | | | | | | | | | | | | | | | | | | |
| 2 | | | | | | | 7 | | | | | | | | | | | |
| 3 | | | | | | | | 8 | | | | | | | | | | |
| 4 | | | | | | | | | 9 | | | | | | | | | |
| 5 | | | | | | | | | | 10| | | | | | | | |
| 6 | | | | | | | | | | | 11| | | | | | | |
| 1 | | | | | | | | | | | | 12| | | | | | |
| 2 | | | | | | | | | | | | | | | | | | |
| 3 | | | | | | | | | | | | | 13| | | | | |
| 4 | | | | | | | | | | | | | | 14| | | | |
| 5 | | | | | | | | | | | | | | | 15| | | |
| 6 | | | | | | | | | | | | | | | | 16| | |
| 1 | | | | | | | | | | | | | | | | | 17| |
| 2 | | | | | | | | | | | | | | | | | | 18|
or, equivalently but without repeating the rows/columns:
| | A | B | C |
|---|----|----|----|
| 1 | 1 | 17 | 15 |
| 2 | 4 | 2 | 18 |
| 3 | 7 | 5 | 3 |
| 4 | 10 | 8 | 6 |
| 5 | 13 | 11 | 9 |
| 6 | 16 | 14 | 12 |
I imagine there are other solutions too, but that's the one I found easiest to think about. But I've been banging my head against the wall trying to figure out how to express it generically—it's a convenient thing that the cardinality of the two sets are multiples of each other, but I want the algorithm to do The Right Thing for sets of, say, size 5 and 7. Or size 12 and 69 (that's a real example!).
Are there any established algorithms for this? I keep getting distracted thinking of how rational numbers are mapped onto the set of natural numbers (to prove that they're countable), but the path it takes through ℕ×ℕ doesn't work for this case.
It so happens the application is being written in Ruby, but I don't care about the language. Pseudocode, Ruby, Python, Java, Clojure, Javascript, CL, a paragraph in English—choose your favorite.
Proof-of-concept solution in Python (soon to be ported to Ruby and hooked up with Rails):
import sys
letters = sys.argv[1]
MAX_NUM = 6
letter_pos = 0
for i in xrange(MAX_NUM):
for j in xrange(len(letters)):
num = ((i + j) % MAX_NUM) + 1
symbol = letters[letter_pos % len(letters)]
print "[%s %s]"%(symbol, num)
letter_pos += 1

Answer 1

String letters = "ABC";
int MAX_NUM = 6;
int letterPos = 0;
for (int i=0; i < MAX_NUM; ++i) {
for (int j=0; j < MAX_NUM; ++j) {
int num = ((i + j) % MAX_NUM) + 1;
char symbol = letters.charAt(letterPos % letters.length);
String output = symbol + "" + num;
++letterPos;
}
}

Answer 2

What about using something fractal/recursive? This implementation divides a rectangular range into four quadrants then yields points from each quadrant. This means that neighboring points in the sequence differ at least by quadrant.
#python3
import sys
import itertools
def interleave(*iters):
for elements in itertools.zip_longest(*iters):
for element in elements:
if element != None:
yield element
def scramblerange(begin, end):
width = end - begin
if width == 1:
yield begin
else:
first = scramblerange(begin, int(begin + width/2))
second = scramblerange(int(begin + width/2), end)
yield from interleave(first, second)
def scramblerectrange(top=0, left=0, bottom=1, right=1, width=None, height=None):
if width != None and height != None:
yield from scramblerectrange(bottom=height, right=width)
raise StopIteration
if right - left == 1:
if bottom - top == 1:
yield (left, top)
else:
for y in scramblerange(top, bottom):
yield (left, y)
else:
if bottom - top == 1:
for x in scramblerange(left, right):
yield (x, top)
else:
halfx = int(left + (right - left)/2)
halfy = int(top + (bottom - top)/2)
quadrants = [
scramblerectrange(top=top, left=left, bottom=halfy, right=halfx),
reversed(list(scramblerectrange(top=top, left=halfx, bottom=halfy, right=right))),
scramblerectrange(top=halfy, left=left, bottom=bottom, right=halfx),
reversed(list(scramblerectrange(top=halfy, left=halfx, bottom=bottom, right=right)))
]
yield from interleave(*quadrants)
if __name__ == '__main__':
letters = 'abcdefghijklmnopqrstuvwxyz'
output = []
indices = dict()
for i, pt in enumerate(scramblerectrange(width=11, height=5)):
indices[pt] = i
x, y = pt
output.append(letters[x] + str(y))
table = [[indices[x,y] for x in range(11)] for y in range(5)]
print(', '.join(output))
print()
pad = lambda i: ' ' * (2 - len(str(i))) + str(i)
header = ' |' + ' '.join(map(pad, letters[:11]))
print(header)
print('-' * len(header))
for y, row in enumerate(table):
print(pad(y)+'|', ' '.join(map(pad, row)))
Outputs:
a0, i1, a2, i3, e0, h1, e2, g4, a1, i0, a3, k3, e1,
h0, d4, g3, b0, j1, b2, i4, d0, g1, d2, h4, b1, j0,
b3, k4, d1, g0, d3, f4, c0, k1, c2, i2, c1, f1, a4,
h2, k0, e4, j3, f0, b4, h3, c4, j2, e3, g2, c3, j4,
f3, k2, f2
| a b c d e f g h i j k
-----------------------------------
0| 0 16 32 20 4 43 29 13 9 25 40
1| 8 24 36 28 12 37 21 5 1 17 33
2| 2 18 34 22 6 54 49 39 35 47 53
3| 10 26 50 30 48 52 15 45 3 42 11
4| 38 44 46 14 41 31 7 23 19 51 27

Answer 3

If your sets X and Y are sizes m and n, and Xi is the index of the element from X that's in the ith pair in your Cartesian product (and similar for Y), then
Xi = i mod n;
Yi = (i mod n + i div n) mod m;
You could get your diagonals a little more spread out by filling out your matrix like this:
for (int i = 0; i < m*n; i++) {
int xi = i % n;
int yi = i % m;
while (matrix[yi][xi] != 0) {
yi = (yi+1) % m;
}
matrix[yi][xi] = i+1;
}