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Arrange The Disks (Posted on 2006-07-04) Difficulty: 4 of 5
A set of 47 disks are consecutively numbered 1 to 47 and placed in a row as follows: 1, 2, 3, 4, ... 45, 46, 47.

Rearrange the disks so for any two given disks A and B, the disk equal to their arithmetic mean doesn't lie between them. For example, Disk 4 cannot lie between Disk 1 and Disk 7 since the arithmetic mean of 1 and 7 is 4. However, since 7 is not equal to the arithmetic mean of 1 and 4, Disk 7 may lie between Disk 1 and Disk 4.

See The Solution Submitted by K Sengupta    
Rating: 3.6667 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
128 ways for 1 through 8 | Comment 13 of 14 |

 7, 3, 5, 1, 8, 4, 6, 2
 6, 2, 8, 4, 7, 3, 5, 1
 5, 1, 7, 3, 8, 4, 6, 2
 8, 4, 6, 2, 5, 1, 7, 3
 7, 3, 5, 1, 6, 2, 8, 4
 6, 2, 8, 4, 5, 1, 7, 3
 5, 1, 7, 3, 6, 2, 8, 4
 4, 8, 6, 2, 7, 3, 5, 1
 3, 7, 5, 1, 8, 4, 6, 2
 2, 6, 8, 4, 7, 3, 5, 1
 1, 5, 7, 3, 8, 4, 6, 2
 4, 8, 6, 2, 5, 1, 7, 3
 3, 7, 5, 1, 6, 2, 8, 4
 2, 6, 8, 4, 5, 1, 7, 3
 1, 5, 7, 3, 6, 2, 8, 4
 8, 4, 2, 6, 7, 3, 5, 1
 7, 3, 1, 5, 8, 4, 6, 2
 6, 2, 4, 8, 7, 3, 5, 1
 5, 1, 3, 7, 8, 4, 6, 2
 8, 4, 2, 6, 5, 1, 7, 3
 7, 3, 1, 5, 6, 2, 8, 4
 6, 2, 4, 8, 5, 1, 7, 3
 5, 1, 3, 7, 6, 2, 8, 4
 4, 8, 2, 6, 7, 3, 5, 1
 3, 7, 1, 5, 8, 4, 6, 2
 2, 6, 4, 8, 7, 3, 5, 1
 1, 5, 3, 7, 8, 4, 6, 2
 4, 8, 2, 6, 5, 1, 7, 3
 3, 7, 1, 5, 6, 2, 8, 4
 2, 6, 4, 8, 5, 1, 7, 3
 1, 5, 3, 7, 6, 2, 8, 4
 8, 4, 6, 2, 3, 7, 5, 1
 7, 3, 5, 1, 4, 8, 6, 2
 6, 2, 8, 4, 3, 7, 5, 1
 5, 1, 7, 3, 4, 8, 6, 2
 8, 4, 6, 2, 1, 5, 7, 3
 7, 3, 5, 1, 2, 6, 8, 4
 6, 2, 8, 4, 1, 5, 7, 3
 5, 1, 7, 3, 2, 6, 8, 4
 4, 8, 6, 2, 3, 7, 5, 1
 3, 7, 5, 1, 4, 8, 6, 2
 2, 6, 8, 4, 3, 7, 5, 1
 1, 5, 7, 3, 4, 8, 6, 2
 4, 8, 6, 2, 1, 5, 7, 3
 3, 7, 5, 1, 2, 6, 8, 4
 2, 6, 8, 4, 1, 5, 7, 3
 1, 5, 7, 3, 2, 6, 8, 4
 8, 4, 2, 6, 3, 7, 5, 1
 7, 3, 1, 5, 4, 8, 6, 2
 6, 2, 4, 8, 3, 7, 5, 1
 5, 1, 3, 7, 4, 8, 6, 2
 8, 4, 2, 6, 1, 5, 7, 3
 7, 3, 1, 5, 2, 6, 8, 4
 6, 2, 4, 8, 1, 5, 7, 3
 5, 1, 3, 7, 2, 6, 8, 4
 4, 8, 2, 6, 3, 7, 5, 1
 3, 7, 1, 5, 4, 8, 6, 2
 2, 6, 4, 8, 3, 7, 5, 1
 1, 5, 3, 7, 4, 8, 6, 2
 4, 8, 2, 6, 1, 5, 7, 3
 3, 7, 1, 5, 2, 6, 8, 4
 2, 6, 4, 8, 1, 5, 7, 3
 1, 5, 3, 7, 2, 6, 8, 4
 8, 4, 6, 2, 7, 3, 1, 5
 7, 3, 5, 1, 8, 4, 2, 6
 6, 2, 8, 4, 7, 3, 1, 5
 5, 1, 7, 3, 8, 4, 2, 6
 8, 4, 6, 2, 5, 1, 3, 7
 7, 3, 5, 1, 6, 2, 4, 8
 6, 2, 8, 4, 5, 1, 3, 7
 5, 1, 7, 3, 6, 2, 4, 8
 4, 8, 6, 2, 7, 3, 1, 5
 3, 7, 5, 1, 8, 4, 2, 6
 2, 6, 8, 4, 7, 3, 1, 5
 1, 5, 7, 3, 8, 4, 2, 6
 4, 8, 6, 2, 5, 1, 3, 7
 3, 7, 5, 1, 6, 2, 4, 8
 2, 6, 8, 4, 5, 1, 3, 7
 1, 5, 7, 3, 6, 2, 4, 8
 8, 4, 2, 6, 7, 3, 1, 5
 7, 3, 1, 5, 8, 4, 2, 6
 6, 2, 4, 8, 7, 3, 1, 5
 5, 1, 3, 7, 8, 4, 2, 6
 8, 4, 2, 6, 5, 1, 3, 7
 7, 3, 1, 5, 6, 2, 4, 8
 6, 2, 4, 8, 5, 1, 3, 7
 5, 1, 3, 7, 6, 2, 4, 8
 4, 8, 2, 6, 7, 3, 1, 5
 3, 7, 1, 5, 8, 4, 2, 6
 2, 6, 4, 8, 7, 3, 1, 5
 1, 5, 3, 7, 8, 4, 2, 6
 4, 8, 2, 6, 5, 1, 3, 7
 3, 7, 1, 5, 6, 2, 4, 8
 2, 6, 4, 8, 5, 1, 3, 7
 1, 5, 3, 7, 6, 2, 4, 8
 8, 4, 6, 2, 3, 7, 1, 5
 7, 3, 5, 1, 4, 8, 2, 6
 6, 2, 8, 4, 3, 7, 1, 5
 5, 1, 7, 3, 4, 8, 2, 6
 8, 4, 6, 2, 1, 5, 3, 7
 7, 3, 5, 1, 2, 6, 4, 8
 6, 2, 8, 4, 1, 5, 3, 7
 5, 1, 7, 3, 2, 6, 4, 8
 4, 8, 6, 2, 3, 7, 1, 5
 3, 7, 5, 1, 4, 8, 2, 6
 2, 6, 8, 4, 3, 7, 1, 5
 1, 5, 7, 3, 4, 8, 2, 6
 4, 8, 6, 2, 1, 5, 3, 7
 3, 7, 5, 1, 2, 6, 4, 8
 2, 6, 8, 4, 1, 5, 3, 7
 1, 5, 7, 3, 2, 6, 4, 8
 8, 4, 2, 6, 3, 7, 1, 5
 7, 3, 1, 5, 4, 8, 2, 6
 6, 2, 4, 8, 3, 7, 1, 5
 5, 1, 3, 7, 4, 8, 2, 6
 8, 4, 2, 6, 1, 5, 3, 7
 7, 3, 1, 5, 2, 6, 4, 8
 6, 2, 4, 8, 1, 5, 3, 7
 5, 1, 3, 7, 2, 6, 4, 8
 4, 8, 2, 6, 3, 7, 1, 5
 3, 7, 1, 5, 4, 8, 2, 6
 2, 6, 4, 8, 3, 7, 1, 5
 1, 5, 3, 7, 4, 8, 2, 6
 4, 8, 2, 6, 1, 5, 3, 7
 3, 7, 1, 5, 2, 6, 4, 8
 2, 6, 4, 8, 1, 5, 3, 7
 1, 5, 3, 7, 2, 6, 4, 8
 8, 4, 6, 2, 7, 3, 5, 1

This uses 3-bit numbers to represent 0-7, and then prints out as 1-8 by adding 1.  There are 7 choice points to choose whether to place 0 bits or 1 bits first, and all combinations of those were used for the 2^7=128 orders.


  Posted by Charlie on 2006-07-06 09:43:42
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