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   MEDIUM: ADJACENCY GRID
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   Author  Topic: MEDIUM: ADJACENCY GRID  (Read 3005 times)
bl@ke
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MEDIUM: ADJACENCY GRID  
« on: Jul 25th, 2002, 7:17pm »
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First take the 1 and the 8 as they are the two end numbers and only have 1 adjacent number each. place these two in the very middle vertically. Then take the 7 and 2 and place them either side of the 1 and 8. Take the 4 and 3 and place them either side of the 1. Finally place the 5 and 6 either side of the 8 (5 has to be on the opposite side to the 4)
 
The result is:
 
7
413
685
2

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scarfhead
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Re: MEDIUM: ADJACENCY GRID  
« Reply #1 on: Aug 2nd, 2002, 11:42am »
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Adjacency Grid
 
More or less the same logic with a couple of variations:
 
The starting point for the solution is to place the numbers with the fewest possible conflicts into the grid cells with the highest number of adjacent cells. So numbers 1 and 8 must be placed in the middle two rows of the middle column, cells touch 6 other cells each.
Then 2 and 7 are the numbers that cannot be adjacent to the two filled cells, so they must be placed where they do not touch 1 and 8, respectively. There are two possible solutions:
 
_ 2 _   _ 7 _
X 8 X  X 1 X
X 1 X  X 8 X
_ 7 _   _ 2 _
Placing the remaining 4 numbers is simply a matter of avoiding conflicts:
5 and 6 go in the row with 8.
3 and 4 go in the row with 1.
4 and 5 cannot be in the same column.
 
4 possible solutions:
 
_ 2 _    _ 2 _    _ 7 _    _ 7 _
5 8 6    6 8 5    4 1 3    3 1 4
3 1 4    3 1 4    6 8 5    5 8 6
_ 7 _    _ 7 _    _ 2 _    _ 2 _
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