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February 16

Coordinate system

Hello friends. I have a cartesian coordinate system, where four points are given: (0|0), (0|1), (1|0), (1|1). You have now to move these points so that at the end these points are at (0|0), ~~(1|3)~~(1|1), (3|0), (2|-1). But there's one rule given: One point can only be moved, if there is another point between its old place and its new place. Now my question: Is it possible to get this latter layout from the first one? --85.178.63.105 (talk) 10:30, 16 February 2008 (UTC)[reply]

Sounds similar to Hi-Q, but without removals. Here's a solution. I don't guarantee it's minimal. It was easy enough (working backwards from the goal to the starting position) that I wonder if there is some other constraint you forgot to mention. Each of my moves jumps over a point that is exactly halfway to the destination, which you didn't even ask for, so call that a bonus.

(0,1) -> (2,-1) -- the (1,0) peg is in the middle of the jump.
(1,1) -> (1,-1) -- the (1,0) peg is in the middle.
(1,0) -> (1,-2) -- the (1,-1) peg is in the middle.
(1,-2) -> (3,0) -- the (2,-1) peg is in the middle.
(1,-1) -> (5,1) -- the (3,0) peg is in the middle.
(5,1) -> (-1,-3) -- the (2,-1) peg is in the middle.
(-1,-3) -> (1,3) -- the (0,0) peg is in the middle.

--tcsetattr (talk / contribs) 10:57, 16 February 2008 (UTC)[reply]

Oh, I'm so much sorry I do not mean (1|3) but (1|1) at the end. Can you write the solution again for this way please? --85.178.63.105 (talk) 11:10, 16 February 2008 (UTC)[reply]

In that case, assuming the middle peg has to be halfway between the starting and ending positions, it's not hard to show that there's no solution. I'll withhold the details for the moment since this looks like the kind of problem that might be assigned as math homework. Try starting with the ending position and seeing where you can go from there to get a feel for what's going on. -- BenRG (talk) 19:09, 16 February 2008 (UTC)[reply]

No, this is not a homework or something like that. My friend told me if I want to puzzle over something and gave me this exercise for the weekend. Now I don't want to tell him that I failed to do this [and I want to get a beer for free]. So please can you give me please the solution? --85.178.32.123 (talk) 23:22, 16 February 2008 (UTC)[reply]

You want me to help you defraud your friend out of a beer? I'll give you a hint. Take a checkerboard with red and white squares and put some tokens on red squares. Now move the tokens around according to the rules of this game. You will never be able to move a token onto a white square (do you see why?). This means if your friend had told you to go from, say, (0|1), (0|-1), (1|0), (-1|0) to (0|0), (0|1), (1|0), (1|1), you would know it was impossible, because the former positions are all on the same color and the latter positions aren't. It's also impossible to do it in the other direction, because all of the moves in this game are reversible. That coloring argument doesn't work for the problem he actually gave you, but a slightly different coloring does work. To figure out that coloring, start with the pieces in the ending position of your friend's problem and try moving them around until you can see which squares are reachable and which aren't. -- BenRG (talk) 00:22, 17 February 2008 (UTC)[reply]

Something I'm not clear on: is this confined to the integer lattice? If not, it can still be solved. Black Carrot (talk) 02:37, 18 February 2008 (UTC)[reply]

You're right, it was never completely clear what the rules were. My comments above only apply if the rules say that you can only move a piece from (a|b) to (c|d) if there's a piece at (½(a+c) | ½(b+d)). In that case, of course, the pieces are confined to the lattice even if the rules don't explicitly say so. If you're allowed to move whenever there's a piece on the open segment (a|b)–(c|d), then the problem seems easily solvable even on the lattice: (1|0) → (−1|0) → (3|0) then (0|1) → (4|1) → (2|−1). -- BenRG (talk) 11:25, 18 February 2008 (UTC)[reply]