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Originally Posted by mizuchaud Well, my original question was how many shuffles would it take to get two single color stacks.
I think 99's asking the question if I mean " back to their original poisitions": ie. red on the left and black on the right. I'm just interested in the original question as posed.
Just by doing shuffles and not MATH (yikes) I get an interesting number of shuffles for multiples of 4 chips in stacks (4 and 8 chips in the stacks) to attain single color stacks.
SPOILER BELOW..
I get three shuffles to reattain single colors for a 3 chip stack.
Three shuffles to achieve single colors for a 4 chip stack, but in the wrong positions.
Five shuffles for five chip stacks, but wrong position.
Six shuffles for 6 chips, with wrong positions.
7 for 7
but the kicker is that I get FOUR shuffles for an eight chip stack.
In my primitive head I latched onto the multiples of four as a subset of shuffling numbers.
junford, can you state that as a solution? |
I think you have to be careful... Let's say you have red on the left and green on the right, after the first shuffle you'll have:
r-g-r-g-r-g-r-g-...
In order to keep the shuffling consistent (with "perfect shuffles"), you'll need to move the top half of this stack to the left (i.e. the left MUST start with a red). Otherwise, you're not shuffling the same way each time. You'll notice that, if you do it "correctly" (or at least the way I did), the top chip will always be the same (and always on the left). If you're starting with red on the left and green on the right, and end up with green on the left, you must have switched your pattern (or kept switching after each shuffle). Using "the top always becomes left", I got the table I posted above.
(Of course, if you shuffle the other way, just switch "left" with "right" in the previous paragraph and you'll get the same result).
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Originally Posted by mizuchaud I'm hurting my head trying to figure out jdunford's phrasing.
what's the ^ mean? |
The ^ symbol means "exponent" (i.e. 2^m means "2 to the power of m").
1 = 2^0
2 = 2^1
3 = 2^1 +1
4 = 2^2
5 = 2^2 +1
6 = 2^2 +2
7 = 2^2 +3
8 = 2^3
...
You'll find that with 16 chips in each starting stack, it only takes 5 shuffles (if you can shuffle stacks that high). It's fairly easy on paper, but takes a long time for more than a few shuffles (e.g. it took me several minutes to work out stacks of 10... and knowing that 11 was going to take even longer, I decided to stop there).
In short, chips in stacks of 1, 2, 4, 8, 16, 32, 64, ... take very few shuffles (e.g. 1, 2, 3, 4, 5, 6, 7, ...). For everything else, it takes a lot more shuffles... and I still haven't figured out the general rule.
