How to make cycle over cycles in Java?

Question

I would like to make a cycle over the following elements:

[1,2,11,12,21,22,111,112,121,122,....,222222]

or for example

[1,2,3,11,12,13,21,22,23,31,32,33,111,112,113,... 333333333]

How can I make it in Java? In my particular case I use 4 digits (1,2,3,4) and the length of the last number can be from 1 to 10.

I managed to do it in Python and PHP. In the first case I used list over lists. I started from [[1],[2],] then for every element of the list I added 1 and 2, so I got [[1,1],[1,2],[2,1],[2,2]] and so on:

nchips = sum(chips)
traj = [[]]
last = [[]]    
while len(last[0]) < nchips:
    newlast = []
    for tr in last:
        for d in [1,2,3,4]:
        newlast.append(tr + [d])
    last = newlast
    traj += last

When I did it in PHP I used number with base 3. But it was a tricky and non elegant solution.

    for ($i=-1; $i<=$n; $i+=1) {

    if ($i>-1) {
        $n5 = base_convert($i,10,5);
        $n5_str = strval($n5);
        $tr = array();
        $found = 0;
        for ($j=0; $j<strlen($n5_str); $j+=1) {
        $k = $n5_str[$j];
        if ($k==0) {
            $found = 1;
            break;
        }
        array_push($tr,$k);
        }
        if ($found==1)
        continue;
    } else {
        $tr = array();
    }
}

Can it be done easily in Java?

How simple this is to do depends on what operations you need to perform on these cycles. Are you just looking to enumerate them all? — Yngve Hammersland, Mar 24 '10 at 20:01

Yngve Hammersland · Accepted Answer · 2010-03-24T19:56:55.377

This looks an awful lot like counting with digits in a given base. (Base 2 and 3 for your examples) You can easily convert an integer to a string of a given base using Integer.toString and map the characters of that string to the symbols. Example:

Integer.toString(6, 2) -> "011"

map this string to a character array and then map that array to your symbols. In your case that would be: '0' -> 1 and '1' -> 2.

This is not the most efficient solution but it lets Integer.toString do the dirty work leaving you to do a simple array transformation.

To transform the other way around you can convert from an array to a string and then use Integer.parseInt to extract the int representation again.

If you need to perform arithmetics (I guess mainly ++ and -- for the previous and next element of the cycle), do them on the integer and convert back and forth as needed.

Disclaimer: I have not coded in java for a while so the method and class names might be off.

EDIT: You can always use big int if you need more symbols than can be accomodated in 32 and and 64-bit integers.

Response to comments:

As people has commented, this approach has problems with leading zeros. The obvious solution is to add some value N^(n+1) to the integer representation before converting to a string where N is the base and n is the number of symbols. This will have the effect of converting 1,1,2 to 1001 instead of 001 effectively allowing the zeros.

But this has the disadvantage of becoming a too complex solution to actually be the simple solution as it initially intended to be.

This doesn't allow to treat '11' and '011' as different results, does it? — sfussenegger, Mar 24 '10 at 13:24
I think it is what I did in PHP. But it is not so elegant as it seems from the beginning. For example [1,2,3,10,11,12,13,...] will be transformed to [2,3,4,21,22,23,24,...] So, I never get number starting from 1! — Roman, Mar 24 '10 at 13:26
sfussenegger is right; there are problems with this approach. It can work, but it's not as straightforward as yngvedh makes it out to be. — polygenelubricants, Mar 24 '10 at 14:28
You're right. It does not work out as I initially thought. I'll update my answer. — Yngve Hammersland, Mar 24 '10 at 19:49

polygenelubricants · Answer 2 · 2010-03-24T14:33:03.970

2

public class Cycle {
    static void advance(StringBuilder sb, int B) {
        int pos = sb.length();
        while (--pos != -1 && sb.charAt(pos) == '0' + B) {
            sb.setCharAt(pos, '1');
        }
        if (pos == -1) {
            sb.insert(++pos, '0');
        }
        sb.setCharAt(pos, (char) (sb.charAt(pos) + 1));
    }

    public static void main(String args[]) {
        StringBuilder sb = new StringBuilder();

        for (int i = 0; i < 20; i++) {
            advance(sb, 3);
            System.out.println(sb);
        }
    }
}

Advancing is done as follows:

Go right to left (--pos)
Rollover all Bs to 1s
Until you find something less than B or you hit the wall (pos == -1)
If you hit the wall, insert 0
Increment character at pos

edited Mar 24 '10 at 14:33

answered Mar 24 '10 at 13:31

polygenelubricants

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It'd be less confusing if you write the looping condition as `N-- > 0` instead of `N --> 0`. – missingfaktor Mar 24 '10 at 13:35
N --> 0 can be read "N goes to zero". It can be argued that it is actually easier to read, but it also kind of undermines the true meaning of the expression. – Yngve Hammersland Mar 24 '10 at 20:00

Petar Minchev · Answer 3 · 2010-03-24T13:56:40.537

If you don't care about the generation order(I mean it will first produce 1; 11; 111, before producing 1; 2; 3) use this:

public static void gen(int level) {
    if (level > 0) {
        for (int i = 0; i < level; i++)
            System.out.print(arr[i] + " ");     
        System.out.println();
    }

    if (level == 10)
        return;

    for (int i = 1; i <= 4; i++) {
        arr[level] = i;
        gen(level + 1);
    }
}

public static void main(String[] args)  {
    gen(0);
}

But if you care about the order use this:

private static int top;
private static int[] arr = new int[10]; 

public static void gen(int level) {
    if (level == top) {

        for (int i = 0; i < level; i++)
            System.out.print(arr[i] + " ");
        System.out.println();

        return;
    }

    for (int i = 1; i <= 4; i++) {
        arr[level] = i;
        gen(level + 1);
    }
}

public static void main(String[] args)  {
    for (top = 1; top <= 10; top++)
        gen(0);
}

score 1 · Answer 4 · edited May 23 '17 at 12:19

I believe your task is a combinatorial task and you should implement combinatorial algorithm to find unique combinations of given numbers (1,2,3,4) - and iterate this algorithm from 1 to desirable length.

And I can't imagine what specific features of Java you can use here. It will be some iterators.

For discussion of algorithms I advice you to read Algorithm to return all combinations of k elements from n

How to make cycle over cycles in Java?

4 Answers4

Response to comments: