A 4x4 sudoku shouldn't be that hard, right? Right?

Question

_{Note: this is not in conjunction with my Minesweeper puzzles}

_{Got this idea from one of Cracking the Cryptic's videos, I don't remember which one though. This exact puzzle is still overall unique.}

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Take this 4x4 grid below:

A	B	C	D
E	F	G	H
I	J	K	L
M	N	O	P

Now, you might not notice this, but this is a 4x4 sudoku, and there is a twist to it:

We can use the numbers 1-9

Now, you might be confused. Why can we? Well, that's because the grid has to satisfy these 18 conditions:

1. Like a regular Sudoku, each number that is placed down can only be used once in each row and column.
2. $A+B=10$
3. $B+C=5$
4. $C+G=10$
5. $E+F=5$
6. $E+I=10$
7. $F+J=5$
8. $I+J=10$
9. $K+L=5$
10. $M+N=5$
11. $A+B+C+D=2A-1$
12. $E+F+G+H=2A-2$
13. $I+J+K+L=2A-3$
14. $M+N+O+P=2A-4$
15. $A+E+I+M=2A+2$
16. $B+F+J+N=A+1$
17. $C+G+K+O=2A-5$
18. $D+H+L+P=2A+1$

Programming and brute-forcing is outright banned.

Difficulty: ★★★★★★

Answer 1

From what I can gather, there are two solutions:

9143
3265
7314
1427

and

9143
2365
8214
1427

Feel free to check my math, but they both look valid to me.

Here's my logic. Let me know if I'm missing something:

We have only positive integers to work with, so anything in an equation that sums to 5 must strictly be less than 5. Notably, B, F, J, and N are all in such an equation, so they must all be less than 5, meaning they are 1-4 in some order. Their sum must be 10, which lets us solve for A using equation 16: A = 9. Using equations 2, 3, and 4, we have B = 1, C = 4, and G = 6. Additionally, since B + F + J + N = 10 and F + J = 5, B and N must also sum to 5, so we have N = 4. Now, M = 1 by equation 10. Using equation 11, D = 3.

At this point, equation 17 is our most restrictive, since we have four numbers summing to 13 and two of them already sum to 10. Thus, K and O must be 1 and 2 in some order. By equation 9, then, L must be 3 or 4, but it can't be 3 because D is already 3 (sudoku rules!), so it must be 4, making K 1 and O 2. This gives us P = 7 by equation 14 and H = 5 by equation 18.

Now we have a square of numbers left, all of which fall under the jurisdiction of equations 5 through 8. (We can double check our work by using equations 12, 13, 15, and 16, as plugging in our knowns makes these equivalent to 5 through 8.) Because E + F = F + J = 5 and E + I = I + J = 10, we know that E must be equal to J, and additionally, I has to be exactly 5 more than F. From the very beginning, we know that J (and therefore E) must be less than 5; 1 and 4 are ruled out from already being placed in the same row, but 2 and 3 both lead to valid solutions, as shown above.

Answer 2

Since brute force and programming is not allowed, let us start thinking:

1) 16 variables and 17 linear equations plus one constraint, obviously mean that this problem is over-specified (meaning it has more information than necessary to solve it. Like asking you to draw a right triangle and giving you the length of all three sides.) So likely at least two of these 18 conditions are not necessary to solve this puzzle. Looking at Equations 6 and 8 we see that E=J, Same for 5 and 7. Thus we can leave out any one of the equations 5-8 and still solve the puzzle.

- Eq 13: Let us use the substitution principle, i. e. substitute 10 for I+J (Eq. 8) and 5 for K+L (Eq. 9), that results in 15 = 2A-3 or A=9
- Eq. 2 then means B=1
- Eq. 3 then means C=4
- Eq. 4 then means G=6
- Eq.11 then reads 9+1+4+D=17 or D= 3
- Eq.12 then reads (with Eq. 5: E+F=5) 5+6+H=16 or H= 5
- Eq.15 then reads (with Eq. 7: E+I=10) 9+10+M=20 or M= 1
- Eq.16 then reads (with Eq. 5: E+F=5) 1+5+N=10 or N= 4 (This also means equation 10 is not needed)

At this point it makes sense to take stock of what we know and use it in the remaining equations: remember E=J:
- 5. E+F= 5
- 6. E+I=10
- 9. K+L=5
- 14. 5+O+P=14 thus O+P=9
- 17. 10+K+O=13 thus K+O=3
- 18. 8+L+P=19 thus L+P=11 By adding equations 17 and 18, we see that one of the Equations 9, 14, 17, 18 is redundant and not needed.
- From Equation 17 we conclude K=1 and O=2 or K=2 and O=1. Since O=1 is not possible because of M=1 in the same row, the only Solution is K=1 thus O=2, L=4 and P=7. From equation 5 we conclude that E can only be 1,2,3 or 4. 1 is not possible because of M=1 in the same column. E=4 is impossible since J=E=4 and L=4 are in the same row.

This leaves the two solutions E=J=2 , F=3, I=7 and E=J=3, F=2, I=8. It is easy to check that both solutions fulfill the condition that each number is only used once in each row and column.

A better (more challenging with a unique solution) puzzle would be achieved by omitting equations 8,9 and 10 and adding the equation J+L=6. This would make the solution non-trivial by preventing to easily conclude A=9 from equation 13, since neither (I+J) nor (K+L) are given.