number of moving glasses by half. Here is another classical problem similar to the one just described but a little more difficult. There are n things lined up next to another n things like this : XXX.....OOO........ However, we define a different way to "move", where you can only move any pair of neighbouring items to an empty space in a row, and you can't change the left-right order of the items. Then, let we think, it is meaningless when n=1, and when n=2 it is no answer. When n=3,the solution process is as follows. XXXOOO XOOOXX XOO XOX OXOXOX ,it should look like this, and we can move .. So when n > 2,we should at least move n times to solve the problem, could you formulate the solution process for n greater than or equal to 3?

Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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number of moving glasses by half. Here is another classical problem similar to the one just
described but a little more difficult. There are n things lined up next to another n things like this :
XXX.....OOO........ However, we define a different way to "move", where you can only move any
pair of neighbouring items to an empty space in a row, and you can't change the left-right order
of the items. Then, let we think, it is meaningless when n=1, and when n=2 it is no answer. When
n=3,the solution process is as follows.
XXXOOO
XOOOXX
XOO XOX
OXOXOX
,it should look like this, and we can move ..
So when n > 2,we should at least move n times to solve the problem, could you formulate the
solution process for n greater than or equal to 3?
Transcribed Image Text:number of moving glasses by half. Here is another classical problem similar to the one just described but a little more difficult. There are n things lined up next to another n things like this : XXX.....OOO........ However, we define a different way to "move", where you can only move any pair of neighbouring items to an empty space in a row, and you can't change the left-right order of the items. Then, let we think, it is meaningless when n=1, and when n=2 it is no answer. When n=3,the solution process is as follows. XXXOOO XOOOXX XOO XOX OXOXOX ,it should look like this, and we can move .. So when n > 2,we should at least move n times to solve the problem, could you formulate the solution process for n greater than or equal to 3?
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