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Four Squared (Posted on 2006-10-26) Difficulty: 2 of 5
        Grid A                    Grid B      
     A   B   C   D             A   B   C   D  
   +---+---+---+---+         +---+---+---+---+
1  |   |   |   |   |      1  |14 |23 |34 |14 |
   +---+---+---+---+         +---+---+---+---+
2  |   |   |   |   |      2  |31 |42 |26 |26 |
   +---+---+---+---+         +---+---+---+---+
3  |   |   |   |   |      3  |22 |24 |44 |29 |
   +---+---+---+---+         +---+---+---+---+
4  |   |   |   |   |      4  |12 |32 |19 |16 |
   +---+---+---+---+         +---+---+---+---+
The numbers 1 to 16 are to be placed in grid A, so that consecutive numbers are not adjacent in any direction, including diagonally. Nor do they appear in the same row, column or any diagonal.

The number in each cell of grid B is the sum of the horizontal and vertical neighbors of the corresponding cell in grid A.

NB. The letters and numbers around the edge of the grid serve no purpose for the solver. They are to be used for identifying cells in the solution.

  Submitted by Josie Faulkner    
Rating: 4.4167 (12 votes)
Solution: (Hide)
           Four Squared Solution

A B C D +--+--+--+--+ 1 |15|10|1 |8 | +--+--+--+--+ 2 |4 |7 |16|13| +--+--+--+--+ 3 |9 |12|5 |2 | +--+--+--+--+ 4 |6 |3 |14|11| +--+--+--+--+

The key to the solution lies in comparing (in Grid B) the values of 4C and 3D, which share two common neighbours (3C and 4D). As there is a difference of ten in their values, 4B is x and 2D is x + 10.

Therefore, the possible values of 4B are 1, 2, 3, 4, 5 and 6; and of 2D, 11, 12, 13, 14, 15 and 16.

However, from 1D, 1C + 2D = 14, therefore, the only possible values for 2D are 11, 12 and 13; and for 4B, 1, 2 and 3.

From 4A, if 4B is 1, then 3A is 11; but if 4B is 1, then 2D is 11, so, there cannot be a 1 in 4B or 11 in 2D.

From 1D, if 2D is 12, then 1C is 2; but if 2D is 12, then 4B is 2, so, there cannot be a 12 in 2D or a 2 in 4B.

Therefore, 4B is 3; 2D is 13: 3A is 9 and 1C is 1.

Now, compare the values of 3A and 4B, which share two common neighbours (4A and 3B). As there is a difference of ten in their values, 2A is y and 4C is y + 10.

Therefore, the possible values of 2A are 2, 4, 5 and 6; and of 4C, 12, 14, 15 and 16.

However, from 4D, 4C + 3D = 16, therefore, the only possible values for 4C are 12 and 14; and for 2A, 2 and 4.

From 1A, if 2A is 2, then 1B is 12; but if 2A is 2, then 4C is 12, so there cannot be a 2 in 2A.

Therefore, 2A is 4; 1B is 10; 4C is 14 and 3D is 2.

The numbers in the neighbouring cells of 2B add up to 42. As we have already placed numbers in two of these, we can deduce that the numbers in cells 3B and 2C must add up to 28 (42-14). The only possibilities for these are 16 and 12, or 15 and 13. As we have already placed 13, we are left with 16 and 12. The 12 cannot go in 2C, because number 13 is in 2D.

Therefore, the 12 goes in 3B and the 16 in 2C.

The numbers in the neighbouring cells of 2C add up to 26. As we have already placed numbers in two of these, we can deduce that the numbers in cells 2B and 3C must add up to 12 (26-14). The only two possible unused numbers are 7 and 5. The 5 cannot go in 2B, because number 4 is in 2A.

Therefore, the 7 goes in 2B and the 5 in 3C.

Finally

1A is 23 - 7 - 1 = 15
1D is 26 - 16 - 2 = 8
4A is 32 - 14 - 12 = 6
4D is 19 - 3 - 5 = 11

Comments: ( You must be logged in to post comments.)
  Subject Author Date
re: SolutionJosie Faulkner2006-11-03 18:54:26
SolutionSolutionnikki2006-11-03 15:50:16
re: Detailed SolutionJosie Faulkner2006-10-31 05:09:45
Detailed Solutionrs2006-10-31 04:44:53
re: Solution: Took a looong time :(Josie Faulkner2006-10-31 04:08:27
Solution: Took a looong time :(rs2006-10-31 03:44:46
RE: SolutionGuest2006-10-27 19:30:06
SolutionGuest2006-10-27 19:25:21
re(3): computer solutionJosie Faulkner2006-10-27 10:31:19
re(2): computer solutionCharlie2006-10-27 09:22:36
re(2): No SubjectJosie Faulkner2006-10-27 05:45:21
re: computer solutionJosie Faulkner2006-10-26 17:42:44
re: A trivial way to solve thisJosie Faulkner2006-10-26 17:36:13
re: No SubjectJosie Faulkner2006-10-26 16:58:18
Solutioncomputer solutionCharlie2006-10-26 16:11:07
No SubjectBRYN2006-10-26 15:37:08
Some ThoughtsA trivial way to solve thisOld Original Oskar!2006-10-26 14:18:36
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