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 Arrange The Disks (Posted on 2006-07-04)
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)

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 re(2): solution (with solution/spoiler) | Comment 7 of 14 |
(In reply to re: solution (with solution/spoiler) by Dej Mar)

Upon seeing Dej Mar's method of arranging the numbers, I thought of representing the numbers as binary. The numbers, shown in the order of Dej Mar's latest modification, with their binary equivalents are:

`25 01100141 10100133 100001 9 001001 1 00000117 01000121 010101 5 00010113 00110137 10010145 10110129 01110127 01101143 10101135 10001111 001011 3 00001119 01001123 010111 7 00011115 00111139 10011147 10111131 01111126 01101042 10101034 10001010 001010 2 00001018 01001022 010110 6 00011014 00111038 10011046 10111030 01111028 01110044 10110036 10010012 001100 4 00010020 01010024 011000 8 00100016 01000040 10100032 100000`

The initial odd/even split produces the separation of 1's from 0's at the right end. Within each group, the 1's and 0's from the next-to-rightmost bit are again separated, and similarly for the third from right. I notice that starting with the fourth from the right he departs from this, placing 33 before the 9, the former then being a zero among 1's in the fourth position from the right.  I wonder if this is made possible by the numbers not going all the way to 63 (111111).

Here is a set that works from 0 through 63, shown with the binary representation in reverse (units position first and 32's position last).

` 0 00000032 00000116 00001048 000011 8 00010040 00010124 00011056 000111 4 00100036 00100120 00101052 00101112 00110044 00110128 00111060 001111 2 01000034 01000118 01001050 01001110 01010042 01010126 01011058 010111 6 01100038 01100122 01101054 01101114 01110046 01110130 01111062 011111 1 10000033 10000117 10001049 100011 9 10010041 10010125 10011057 100111 5 10100037 10100121 10101053 10101113 10110045 10110129 10111061 101111 3 11000035 11000119 11001051 11001111 11010043 11010127 11011059 110111 7 11100039 11100123 11101055 11101115 11110047 11110131 11111063 111111`

Note that in this instance if the 33 and 9 are reversed, the 9 would fall between the 1 and the 17, the 17 between the 1 and the 33 and also between the 9 and the 25, and the 33 between the 9 and the 57.

Of course if you only need 1 through 47, then just leave out the other numbers.

Other arrangements are possible, as at each stage it does not matter if the zeros come first or the ones.  Consistency was maintained here for ease of sorting the reversed binary numbers.

 Posted by Charlie on 2006-07-05 12:23:15

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