9
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Hey I just thought other might enjoy the challenge of this puzzle as much as I did. Not the most easy thing to solve.

Sort of like Sudoku, each row and column needs the letters A,B,C,D,E and each row and column needs the numbers 1,2,3,4,5 however there can only be one of each set (like one A1, one A4, one D3 in the entire puzzle). So each square will be unique at the end.

Give it your best go :)

Here it is with the starting values:

Puzzle grid

Plaintext version:

+--+--+--+--+--+
|B | 2|E |  |  |
+--+--+--+--+--+
|  |  |  |  | 4|
+--+--+--+--+--+
|  |  |  |  |C |
+--+--+--+--+--+
|E | 1|A | 3|  |
+--+--+--+--+--+
|  |D4|  |  |  |
+--+--+--+--+--+
$\endgroup$
10
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I got it! It is indeed solvable entirely through logic.

B1|A2|E3|C4|D5
--+--+--+--+--
C5|B3|D1|E2|A4
--+--+--+--+--
D2|E5|B4|A1|C3
--+--+--+--+--
E4|C1|A5|D3|B2
--+--+--+--+--
A3|D4|C2|B5|E1

Here is a graphic walkthrough. I will use traditional Sudoku walkthrough notation, which indicates cells with the notation RxCy for row x, column y.

Starting Grid

The symbol we know most about at the outset is 4, since it appears twice on the grid, already paired with one letter. Let's put in pencil marks on other places 4 could possibly go.



Column 4 is particularly interesting, because there are only two places 4 can go. We can eliminate 4 from the red shaded cell because that would make it impossible to create C4. This gives us a few definite placements (shaded in blue below):

 R1C4 = 4
 R1C5 = D
 R4C4 = D
 R4C2 = C
 R4C5 = B



The remaining pencil marks of D and 4 have an interesting interaction since D4 is already placed. One of the red shaded cells must contain a 4, and the other a D. We can bifurcate, as shown above: Either the values on the top are true or the values on the bottom are true. This bifurcation can extend further into column 5 with the placement of A and E.

Between Row 4 and R2C5, the combinations of A4 and E4 are locked. This elminates A from R3C1. This gives us several more placements (shaded in blue below):

 R3C1 = D
 R2C3 = D
 R3C3 = 4
 R4C1 = 4
 R5C1 = A
 R2C1 = C
 R2C5 = A



...and some more placements (shaded in blue below):

 R1C2 = A
 R1C4 = C
 R4C3 = 5
 R4C5 = 2
 R5C3 = C
 R5C4 = B



We can finish off our letters (shaded in blue below):

 R3C3 = B
 R2C2 = B
 R2C4 = E
 R3C3 = B
 R3C2 = E
 R3C4 = A



Starting with D, we can place numbers (shaded in blue below):

 R3C1 = 2 (hidden single in R3)
 R2C3 = 1
 R1C5 = 5
 R1C3 = 3
 R5C3 = 2
 R1C1 = 1



Looking at E, we can place all the remaining numbers (shaded in blue below):

 R4C2 = 2
 R3C2 = 5
 R5C5 = 1
 R2C2 = 3
 R2C1 = 5
 R5C1 = 3
 R5C4 = 5
 R3C4 = 1
 R3C5 = 3

$\endgroup$
  • 1
    $\begingroup$ Marked as correct, would love to see the steps you took to see if the same as what I did:) $\endgroup$ – Lain Mar 19 '17 at 20:34
  • 1
    $\begingroup$ FWIW - My first observation, based on the initial data, was that there was only one place that C4 could go. That then forced A2 and C1. $\endgroup$ – YowE3K Mar 19 '17 at 23:56

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