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.
1) Since C1 + D2 = 14, that means D2 < 14
2) 19 = B4 + C3 + D4 and 29 = D2 + C3 + D4. So 19 B4 = C3 + D4 = 29 D2. So D2 B4 = 10
3) Since D2 < 14 and D2 B4 = 10, that means B4 < 4.
4) If B4 = 1, then A3 = 11 (since B4 + A3 = 12) and D2 = 11 (since D2 B4 = 10). If B4 = 2, then C1 = 2 (since D2 B4 = 10 and C1 + D2 = 14)
5) Therefore B4 =3, which means A3 = 9 and D2 = 13, which means C1 = 1. So so far we have:
A B C D
+++++
1    1  
+++++
2    13 
+++++
3  9    
+++++
4   3   
+++++
6) Since C4 + D3 = 16, and 1 has already been taken, that means C4 < 15
7) 22 = A2 + B3 + A4 and 32 = A4 + B3 + C4. So 22 A2 = A4 + B3 = 32 C4. So C4 A2 = 10.
8) Since C4 < 15 and C4 A2 = 10, that means A2 < 5.
9) A2 cant be 1 or 3 since those are already taken. If A2 = 2, then B1 = 12 (since B1 + A2 = 14) and C4 = 12 (since C4 A2 = 10).
10) Therefore A2 = 4, which means B1 = 10 and C4 = 14, which means D3 = 2. So so far we have:
A B C D
+++++
1  10  1  
+++++
2  4   13 
+++++
3  9    2 
+++++
4   3 14  
+++++
11) 19 = 3 + C3 + D4, so C3 + D4 = 16. With the remaining numbers, the only two that add to 16 are 11 and 5.
12) 24 = 9 + 3 + C3 + B2, so C3 + B2 = 12.
13) Well, since C3 could only be 11 or 5, then B2 can only be 1 or 7, but 1 is already taken.
14) Therefore C3 = 5, which means B2 = 7 and D4 = 11.
15) 31 = A1 + 7 + 9, so A1 = 15. So so far we have:
A B C D
+++++
1 15 10  1  
+++++
2  4  7  13 
+++++
3  9   5  2 
+++++
4   3 14 11 
+++++
16) 22 = 4 + B3 + A4, so B3 + A4 = 18. With the remaining numbers, the only two that add to 18 are 12 and 6.
17) 42 = 4 + 10 + C2 + B3, so C2 + B3 = 28.
18) Well, since B3 could only be 12 or 6, then C2 can only be 16 or 22, but 22 isnt an allowable number.
19) Therefore B3 = 12, which means C2 = 16 and A4 = 6.
20) 26 = D1 + 16 + 2, so D1 = 8. So the solution is:
A B C D
+++++
1 15 10  1  8 
+++++
2  4  7 16 13 
+++++
3  9 12  5  2 
+++++
4  6  3 14 11 
+++++
Just a side note: It turns out I didnt really need to use the "no consecutive numbers are adjacent or in a line with each other" clue. Yes, it is helpful and can gives you another way to find out where some things go in the end, but it was redundant information.
Good puzzle!
Edited on November 3, 2006, 4:03 pm

Posted by nikki
on 20061103 15:50:16 