First, what do we know about this puzzle?

 - There are 5*7*8*11 = 3080 possible positions.
 - Moves do not affect each other, so if you have a good set of moves, you can do them in any order.
 - There must be at least one move of the 8 ring, but not necessarily any of the others.

Given this, it is a relatively simple algorithm to find the shortest sequence of moves.

What I did was to start with a list containing only (0,0,0,0), then for each item in that list, I did a +5,-5,+7,-7,+8,-8,+11,-11. For each item produced by this, I check if it is in the list and if not, add it to the end.

I had it stop when it got to (3,3,3,3).

(0,0,0,0) -> +5<br>
(1,2,2,2) -> +7<br>
(4,3,4,4) -> -8<br>
(1,5,3,2) -> -8<br>
(3,0,2,0) -> -8<br>
(0,2,1,9) -> -8<br>
(2,4,0,7) -> -8<br>
(4,6,7,5) -> -11<br>
(1,1,1,4) -> -11<br>
(3,3,3,3)

So the minimum number of moves is 9. One example is(just reversing my list):<br>
**+11,+11,+8,+8,+8,+8,+8,-7,-5**

**Update:**<br>
The maximum distance from any position to any other position is 14. Here is a list of the distances and the number of points for each distance. This is done from (0,0,0,0), but it holds for any point.

0: 1<br>
1: 8<br>
2: 32<br>
3: 84<br>
4: 168<br>
5: 284<br>
6: 420<br>
7: 397<br>
8: 344<br>
9: 330<br>
10: 324<br>
11: 324<br>
12: 250<br>
13: 100<br>
14: 14<br>

An example of an item that is 14 moves from (0,0,0,0) is (3,1,5,1), which happens to be the last item in my sequence.

The closest items with 3 0's are (0,0,0,3) and (0,0,0,8), which are 5 moves away.<br>
The closest items with 3 0's and a 1 are (1,0,0,0) and (0,0,1,0), which are 7 moves away.<br>
The furthest item with 3 0's and a 1 is (0,1,0,0) which is 12 moves away.<br>
(0,0,0,1) is 8 moves away (to complete the set).

**Update 2:**<br>
The question was asked how I knew the 8 ring must move at least once, but not necessarily any of the others. There are two different parts to this.

First, if you never move the 8 ring, moving other rings always causes it to move 2, which means it can only ever be at 1, 3, 5 or 7. So you must move it at least once to get it to an even number.

Second, to never move another ring: I expected this to be true because all the ring sizes are relatively prime to each other (and to the moves, except for the 2-move and 8-ring), so if I move the 8 ring 3080 times, I should go through every permutation. If I move any other ring 1540 times, I'll go through half the permutations (because in that case, the 8 ring only stays on odd numbers). To verify this, I put together a quick spreadsheet to simulate moving each ring +1 3080 times. The spreadsheet showed this to be correct. It also showed that you can get to (0,0,0,0) by moving the 8 ring -1 only 51 times.