生成组合行

Question

I don't know how I should formulate this question but I hope I can explain what I want to achive.

So I got a set of characters [A, B, C].

I want to generate the minimal amount of rows with length of N needed to contain all possible combinations of [A, B, C].

Example: when N = 4, it generates something like this with 9 rows of length N(1 column = 1 row)

AAABBBCCC
ABCCABBCA
BACACBCBA
ABCABCABC

For example the first row [A, A, B, A] contains the following combinations(1 column = 1 combination), notice how the combinations can wrap around the row.

A, A, A, B 
A, A, B, A 
A, B, A, A 
B, A, A, A 

How ever it's allowed for a combination to end up more than 1 time amongst all generated rows but it should be kept at the optimal minimum.

How should I go about this programmatically?

Answer 1

Are you counting in base 3 ? with the digits 1, 2, 3 instead of 0, 1, 2.

If you don't know how to count in base three, let's do it now. 0, 1, 2, 10, 11, 12, 20, 21, 22, 100

If you want to know what is the number of rows the with N digits then the answer is 3^N .

If you want to know what sequence you have at a given row of a given length at a given position (the first position is zero) in the sorted list you can use the following function

def row(k, N):
    d = []
    assert(k < 3**N)
    for _ in range(N):
        k,r = divmod(k, 3)
        d.append(r+1)
    return ''.join(str(di) for di in d[::-1])

An easy to verify row(1, 3)='112' is the second term in of the rows with lenght 3. A not so easy to verify is that the billionth term of length 25 is given by row(10**9-1, 25)='1111113231311311132121111' .

Using generic symbols

If you want to return a list of arbitrary objects (not necessarily three) it is just changing the way the output is mapped.

def row(k, symbols, N):
    d = []
    B = len(symbols)
    assert(k < B**N)
    for _ in range(N):
        k,r = divmod(k, B)
        d.append(symbols[r])
    return d[::-1];

Using it

print(row(1, ['red', 'green', 'blue'], 6))
print(row(100, ['red', 'green', 'blue'], 6))

> ['red', 'red', 'red', 'red', 'red', 'green']
> ['red', 'green', 'red', 'blue', 'red', 'green']