Life/Hit Point Counter Rings

12 Comments

1. msaari on 31.10.2008 at 05:57

   I was going to post a comment requiring an explanation, but then I figured it out. Clever, and pretty obvious indeed.

   With two rings, you can get at least -999999 to +999999, but that’s doing it the obvious way.

2. kungfugeek on 31.10.2008 at 06:31

   Ok I’ll bite. How?

3. shadejon on 31.10.2008 at 07:20

   Mikko, yes, you can count much, much higher that that.

   Yehuda

4. Jorge on 31.10.2008 at 10:40

   I don’t think it’ll fit in my thumbs or big toes though.

5. Bill Burdick on 31.10.2008 at 12:49

   Why not treat -0 as -1000? :)

6. Bill Burdick on 31.10.2008 at 12:55

   And… I think it’s reasonable to at least double the resolution of the third digit because you can eyeball half a step pretty easily. So I vote for -1981 to +1980 for one ring.

7. Bill Burdick on 31.10.2008 at 12:57

   d’oh! Maybe shadejon will post the proper numbers for that :)

8. David on 31.10.2008 at 15:16

   With multiple identical rings, however, you’re limited by not knowing which ring it is in each position.

9. Bill Burdick on 31.10.2008 at 15:50

   Even if they’re stacked on your finger?

10. shadejon on 01.11.2008 at 13:18

    I count:

    100 positions for ring 1 *
    100 positions for ring 2 *
    110 possible combinations on which fingers they are placed [1] *
    4 orientation possibilities [2] =
    4,400,000 unique positions.

    Yehuda

    [1] 10 possible fingers for each ring yields 100 finger comboes. But whenever the two rings are on the same finger, switching the inner and outer rings is an additional option, making 110 possible combinations.

    [2] Each ring can be oriented facing toward you or facing away from you, which gives four possible orientation combinations.

11. David Klein on 01.11.2008 at 14:21

    As a mathematician, the only answer I really like is 1..10^2 for 1 ring and 1..4*10^5 for two rings. Before explaining, I’ll state what I don’t like about the other answers. They rely on additional external objects and it becomes blurry which objects to allow and which not. For example, why did shadejon not include toes? Toe rings do indeed exist. If he rules them out based on size, who says the rings fit over the thumbs either. It all gets very sloppy. Same with Bill Burdick, why half spacing and not third spacing, or using an interferometer go by the angstrom. I am not claimng that these other answers are incorrect, just that I don’t like the openness and non decidability of the answers. That’s why I prefer going with only asking how many distinct states can be achieved with the rings themselves. I.e. if we left the rings in a totally empty universe, how many different patterns could we leave for an alien to observe? (I admit that even there things are not 100% clear cut, for example half steps, but I am happy with my answer)

    So the question becomes, why only 10^2 possibilities for one ring and over 10^5 for two rings?

    One ring: Each circle can be in one of 10 rotation states, and the indicator window can also be in one of 10 states. But because there are no external objects, there is no way to distinguish rotating the universe in any of 10 states so one gets 10*10*10/10 = 100. Another way of looking at it is that the indicator window determines a direction in space, and then the two bands have 10*10 positions relative to the indicator. One can also flip the ring over but again, since one doesn’t have a finger to distinguish a direction, this is just a symmetry operation that can’t be distinguished.

    For two rings, we have 10*10*10*10*10*10 still divided by 10 orientations of the universe, for 10^5 (alternatively, one indicator window selects a direction, and then we have 5 objects left with 10 orientations each). Flipping the whole assembly over doesn’t add anything, but the indicator windows have an arrow or something on them. So I can distinguish the following states “both arrows point in the direction from red to blue”, “both arrows point in the direction from blue to red”, “arrows point towards each other” or “arrows point away from each other” for a total of 4*10^5 non-symmetric states

12. David Klein on 02.11.2008 at 00:36

    Actually, you can double that by having the two rings either touching or not. I really don’t like this addition as it isn’t “clean” but I don’t see how to avoid it.

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