## Discussion...

Post an idea here...

## ... by: Chris

Sorry for the duplication! The site kept telling me "you did not enter the security code correctly" when I added my comment, so I kept trying.
## ... by: Chris

It is possible to solve this puzzle using only pencil and paper through logical analysis and a process of elimination. Cells D6, E5, and E6 must contain a 3, an 8, and either a 1, 2, or 4. If D6 contains the 1, 2, or 4, then E5+E6=3,8 (order not initially important). The 7 must go in either E1, E3, E7, or E8. Because C1=1 and J1=4; C8=4 and J8=2; A1=2; and G7=1, wherever the 7 is placed, the 1, 2, and 4 fall inevitably into the remaining cells in row E. The arrangement very quickly leads to a falsity in almost every case; D6=4 and E7=7 leads to a falsity a little more slowly, and D6=4 and E3=7 is complicated but also false ultimately. This means that D6 must be either 3 or 8, and E5–E6 will contain the other of those two numbers and the 1, 2, or 4. Suppose D6=3. Then in row E, the 3 must go in E1, E3, E7, or E8. This leads to a unique placement of the 7, and thus the remaining numbers, in row E in many cases; in the remaining cases, the 7 has two or perhaps three places where it can go. Once again a process of elimination shows that most arrangements lead very quickly to falsities. If necessary, the process could continue with D6=8, but D6=3, E5–6=4,8, E7=3, E3=7 leads to a solution. Because most tentative arrangements resolve into falsities so quickly, solving time with pencil and paper is about two hours.
## ... by: Chris

It is possible to solve this puzzle using only pencil and paper through logical analysis and a process of elimination. Cells D6, E5, and E6 must contain a 3, an 8, and either a 1, 2, or 4. If D6 contains the 1, 2, or 4, then E5+E6=3,8 (order not initially important). The 7 must go in either E1, E3, E7, or E8. Because C1=1 and J1=4; C8=4 and J8=2; A1=2; and G7=1, wherever the 7 is placed, the 1, 2, and 4 fall inevitably into the remaining cells in row E. The arrangement very quickly leads to a falsity in almost every case; D6=4 and E7=7 leads to a falsity a little more slowly, and D6=4 and E3=7 is complicated but also false ultimately. This means that D6 must be either 3 or 8, and E5–E6 will contain the other of those two numbers and the 1, 2, or 4. Suppose D6=3. Then in row E, the 3 must go in E1, E3, E7, or E8. This leads to a unique placement of the 7, and thus the remaining numbers, in row E in many cases; in the remaining cases, the 7 has two or perhaps three places where it can go. Once again a process of elimination shows that most arrangements lead very quickly to falsities. If necessary, the process could continue with D6=8, but D6=3, E5–6=4,8, E7=3, E3=7 leads to a solution. Because most tentative arrangements resolve into falsities so quickly, solving time with pencil and paper is about two hours.
## ... by: Dr. Chris

It is possible to solve this puzzle using only pencil and paper through logical analysis and a process of elimination. Cells D6, E5, and E6 must contain a 3, an 8, and either a 1, 2, or 4. If D6 contains the 1, 2, or 4, then E5+E6=3,8 (order not initially important). The 7 must go in either E1, E3, E7, or E8. Because C1=1 and J1=4; C8=4 and J8=2; A1=2; and G7=1, wherever the 7 is placed, the 1, 2, and 4 fall inevitably into the remaining cells in row E. The arrangement very quickly leads to a falsity in almost every case; D6=4 and E7=7 leads to a falsity a little more slowly, and D6=4 and E3=7 is complicated but also false ultimately. This means that D6 must be either 3 or 8, and E5–E6 will contain the other of those two numbers and the 1, 2, or 4. Suppose D6=3. Then in row E, the 3 must go in E1, E3, E7, or E8. This leads to a unique placement of the 7, and thus the remaining numbers, in row E in many cases; in the remaining cases, the 7 has two or perhaps three places where it can go. Once again a process of elimination shows that most arrangements lead very quickly to falsities. If necessary, the process could continue with D6=8, but D6=3, E5–6=4,8, E7=3, E3=7 leads to a solution. Because most tentative arrangements resolve into falsities so quickly, solving time with pencil and paper is about two hours.
## ... by: Serban

1. Row A n=2 e=7 Nc=84 Ns=53 => A5=3 A6=5 solve

2. Row C n=2 e=7 Nc=140 Ns=37 => C3=6 C4=9 C7=2 solve

3. Row F n=3 e=6 Nc=82 Ns=60 => F1=3 F3=4 F9=2 solve

## ... by: SteveJ

Requires two guesses.

stevej@bsd84:~/sudoku % time ./sudoku -f ./puzzles/un-2.txt -s 2 -r -D 1

*** ./puzzles/un-2.txt not solved, sane! Solution failed!

Iterations: 2. Changes: Rows 142, cols 106, squares 37, tot 285

dfs: cells 59 vals 246 rec 246 bt 246 ns 0 mrl 1

0.052u 0.000s 0:00.05 100.0% 56+1153k 0+0io 0pf+0w

stevej@bsd84:~/sudoku %

stevej@bsd84:~/sudoku % time ./sudoku -f ./puzzles/un-2.txt -s 2 -r -D 2

>>>>>> ./puzzles/un-2.txt solved, sane!

Iterations: 14. Changes: Rows 201, cols 169, squares 94, tot 464

dfs: cells 2084 vals 8253 rec 7419 bt 7418 ns 833 mrl 2

2.212u 0.000s 0:02.52 87.6% 53+1096k 0+0io 0pf+0w

stevej@bsd84:~/sudoku %

github[.]com/sjac999/sudoku_solver_c_lightning_fast
## ... by: WS

This is a very solvable puzzle:

-- puzzle-example-unsolvable-49.txt - finish: 49.76 seconds --

+-------+-------+-------+

| 7 4 2 | 8 3 5 | 6 9 1 |

| 5 3 9 | 4 6 1 | 2 8 7 |

| 1 8 6 | 9 7 2 | 5 4 3 |

+-------+-------+-------+

| 6 1 8 | 2 9 3 | 7 5 4 |

| 2 5 7 | 6 8 4 | 3 1 9 |

| 3 9 4 | 1 5 7 | 8 6 2 |

+-------+-------+-------+

| 8 2 5 | 3 4 9 | 1 7 6 |

| 9 7 1 | 5 2 6 | 4 3 8 |

| 4 6 3 | 7 1 8 | 9 2 5 |

+-------+-------+-------+

github[.]com/WillieStevenson/sudoku-solver
## ... by: Vikram

You're exactly right, Dave. I'd go a step fuhtrer and say fill your resume with those things that you do better than anyone else you know; what makes you unique. Decision makers are looking for your secret sauce. And write it for actual eyeballs to read, not some scanning software. Turning your resume into a ditto of the job description won't set you apart in any way.
## ... by: Duder Sud

Use 7 in A3 then use solver
## ... by: mba application help

This bamboozle can't be solved minus using Ariadne's Silk which essentially signifys that you experience to speculate at few speck. it can be exhausted, yet hardly along straight rationale. among the Suppose deduce plus defeat prerogative enabled, the sudoku solver testament discovery the answer.

mba application help
## ... by: newby

using pairs 4(H7,G9), 4(A2,B3), 7(A1,C3) all combinations>contradiction except:

4H7, 4B2, 7A1 leaving 7(D3, E3)

7E3 solves
## ... by: Solver

Just tested my solver program with this Sudoku and it nailed it in 0.1312 seconds.

However, this is the longest my solver had to process so far, not even the Inkala puzzles came close. Nice one!
## ... by: John Doe

I still can't find a solution to this, anywhere on the net. I get many hits describing different techniques and algorithms but no final solution. Is there actually a solution? Can someone please produce it? Thanks.
## ... by: John Doe

I apologize for asking twice but I am new to the site and I am still to learn its ways. Please do clarify this; Is #49 solvable? Does a solution exist? Where can I find it? I have been trying a scheme but I always end up with two same numbers in the end, it seems that the puzzle is wrong OR I have to try a differenct scheme.

Thank you.
## ... by: Guillermo

I'm not super familiar with that area , but it looks to me like the cleosst location listed on the chart is Farmington, with a date of May 1. If you tend to have similar weather to Farmington, I would use that date. You're right on the edge of the two groups. If it is a mild winter/spring and you are excited, I would follow the yellow schedule. If it is a cold winter-spring and you are more relaxed, I would use the green schedule.
## ... by: John Doe

Is there a typo on the puzzle above? What are the comments by "JC Van Hay" referring to? Thanks.
## ... by: JC Van Hay

Another typo detected by abi :( : read Target : (124)r2c6.r8c5 instead of Target : (124)r2c6.r8c4
## ... by: JC Van Hay

Typo : read ... -1r1c58.r56c3.r8c4.r9c34 ... instead of ... -1r15c8.5r6c3.r8c4.r9c34 ...
## ... by: JC Van Hay

ROOKS from the start !! It is April (or Andrew) Fool's Day ! So I would call this puzzle : AFD Monster.

Its solving begins as in Fata Morgana Loop 1 and 2. But this doesn't make it simpler ! (Still ROOKS)

Initial Eliminations :

Exocet : Base : (124)r46c4 [Target : (124)r2c6.r8c4], 124R258 => -59r2c6

or

Exocet : Base : (124)r46c4 [Target : (124)r2c6.r8c4], 124R28C29 => -59r2c6, -1r1c8.r9c34, -2r3c47.r7c1

If 124R5 is added, then :

Exocet : Base : (124)r46c4 [Target : (124)r2c6.r8c4], 124R258C29 => -59r2c6, -1r15c8.5r6c3.r8c4.r9c34, -2r2c4.r3c47.r46c7.r7c16