You want to print vectors with n elements, where:

  • the first element can have the values: e1.1, e1.2, e1.2;

  • the second element can assume the values: e2.1, e2.2, e2.3;

  • ...;

  • ...;

  • the nth element can assume the values: en.1, en.2, en.3.

I should print all possible vectors; eg:

{e1.1, e2.1, en.1 ...};

{e1.2, e2.1, en.1 ...};

etc. etc.

Someone could indicate an algorithm making use of while/for loop to print these vectors?

Thanks so much!

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  • 1
    $\begingroup$ Those are not permutations. A permutation of a set S is a bijective function from S onto itself. What did you try? $\endgroup$ – quicksort Dec 27 '16 at 20:57

The idea is to think of your output as representing a number in a mixed base. The $i$th digit has base $b_i$, where $b_i$ is the number of options for the $i$th element. You can then print all possible vectors essentially by implementing a counter. I'll let you work out the details.

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Here's a python program that does this for $n=4$. In the $i$th iteration, each vector of length $i$ is appended with one of three possible strings to create vectors of length $i+1$.

n = 4
res = [["e1.1"], ["e1.2"], ["e1.3"]]
for i in range(2, n+1):
    temp = res[:]   #make a copy of current list
    res = []
    for x in temp:
        res.append(x + ["e" + str(i) + ".1"])
        res.append(x + ["e" + str(i) + ".2"])
        res.append(x + ["e" + str(i) + ".3"])

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