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 Post subject: Re: Occasional Sindoku (Quadrata's Concept) Posted: Sun Sep 26, 2021 5:13 am Grand Master Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1792
A long time since these puzzles were posted. However it's only recently that I've come up with my own way to represent the clues on my Excel worksheet.

It's good that HATMAN produced both regular and NC versions of most of these puzzles, the regular for those who aren't into NC. I'm impressed that the clues somehow manage to produce NC solutions!

For Sindoku 23 the regular one is fairly straightforward, the NC a better puzzle and more fun.

Hint:
If you can't see how to get started, do the Prelims first
a) For squares with 2 numbers, delete those 2 numbers from the rest of that nonet
b) For squares with 2 numbers, delete the adjacent numbers in the cells covered by those squares

Here's how I solved Sindoku 23:
Prelims
a) For squares with 2 numbers, delete those 2 numbers from the rest of that nonet
b) For squares with 2 numbers, delete the adjacent numbers in the cells covered by those squares

Steps Resulting From Prelims
1a. 9 in N6 only in R45C8, locked for C8
1b. 9 in N9 only in R89C7, locked for C7
1c. 6,8 in N6 only in R6C789, locked for R6
1d. 1 in N8 only in R9C45, locked for R9
1e. 2 in N8 only in R7C45, locked for R7
1f. 2,4 in N2 only in C6 only in R123C6, locked for N2

2a. 8,9 only consecutive numbers in R23C23 = {3689}, no 1, 8,9 locked for N1
2b. 1,2,3 only consecutive numbers in R45C78 = {1259/2359}, no 7, 2 locked for N6
2c. 7 in N6 only in R45C9, locked for C9
2d. R12C12 contains 4 singletons = {1357/1479/1579/2479/2579} -> R2C2 = {39}, 7 locked for N1
2e. 7 in C3 only in R56C3, locked for N4

3a. R89C56 contains 4 singletons and one of 4,6,8 in R8C5 = {1358/1369/1469/1479} -> R9C5 = 1
3b. R89C45 contains 1,8 and two consecutive numbers = {1348/1458/1568} -> R9C4 = {35}, 8 locked for R8
3c. 8 in N9 only in R7C89, locked for R7

4a. R89C89 contains 4 singletons = {1357} -> R9C8 = 7, R8C89 + R9C9 = {135}, locked for N9, 1 locked for R8
4b. Naked triple {468} in R7C789, 4,6 locked for R7, 4 locked for N9
4c. Naked pair {29} in R89C7, 2 locked for C7
4d. R89C78 contains 7,9 and two consecutive numbers = {1279/2379}, no 5
4e. 5 in N9 only in R89C9, locked for C9
4f. 2 in N6 only in R45C8, locked for C8
4g. R12C78 contains 2 doublets, no 2 -> no 1 in R12C78

5a. R89C56 (step 3a) = {1369/1469} (cannot be {1358} which clashes with R9C4, cannot be {1479} which clash with R7C45, ALS block), no 5,7,8, 6,9 locked for C6 and N8
5b. Naked pair {27} in R7C45, 7 locked for R7
5c. Naked pair {35} in R7C6 + R9C4, 3 locked for N8
5d. Naked pair {69} in R89C6, locked for C6, 6 locked for N8 -> R8C45 = 
5e. 9 in R7 only in R7C12, locked for N7
5f. Naked pair {27} in R8C12, 2 locked for R8 and N7 -> R89C7 = , R89C6 = 
5g. R78C12 contains 7,9 and two consecutive numbers = {1279/2379}, no 5
5h. Naked pair {35} in R9C49, locked for R9
5i. 5 in N7 only in R78C3, locked for C3

6a. R45C78 contains 5,9 and two consecutive numbers = {1259/2359} => R45C8 = {29}, R45C7 = {15/35}, 5 locked for C7
6b. R45C89 contains 7,9 and two consecutive numbers = {1279/2379}, no 4 => R45C9 = {17/37}
6c. R6C789 = {468} (hidden triple in N6), 4 locked for R6
6d. R5C4 = 4 (hidden single in N5)
6e. Naked quad {1357} in R4589C9, 1,3 locked for C9
6f. 1 in N3 only in R3C78, locked for R3
6g. R56C23 contains 7,9 and two consecutive numbers = {1279/2379}, no 5, 2 locked for N4
6h. R56C45 contains 2,4 and two consecutive numbers = {2467/2478/2489} -> R5C5 = {68}, R6C45 = {27/29}, 2 locked for R6
6i. 2 in N4 only in R5C23, locked for R5 -> R45C8 = 
6i. 9 in N4 only in R6C23, locked for R6
6j. Naked pair {27} in R6C45, 7 locked for R6 and N5
6k. R4C9 = 7 (hidden single in R4)

7a. R23C23 (step 2a) = {3689}, R23C3 cannot be {68} which clashes with R9C3 -> R3C2 = {68}
7b. Killer pair 6,8 in R23C3 and R9C3, locked for C3

8a. R89C23 contains 2 doublets 34/45/56/78 (cannot be  because {468} can’t form a doublet) -> R8C2 = 7, R8C1 = 2
8b. 2 in N1 only in R1C23, locked for R1
8c. R23C12 contains 2 doublets, no 2 -> no 1 in R2C1
8d. 1 in N1 only in R1C123, locked for R1
8e. Hidden killer pair 1,2 in R12C12 and R1C3 for N1, R12C12 contains 4 singletons -> R1C3 = {12}
8f. R4C3 = 4 (hidden single in C3)
8g. R5C23 =  (hidden pair in R5) -> R1C3 = 2 (hidden single in C3)
8h. Naked triple {139} in R267C2, locked for C2
8i. Naked pair {45} in R1C2 + R3C1, locked for N1 -> R12C1 = 
8j. R6C23 = {19} (hidden pair in N4), 1 locked for R6
8k. Naked pair {35} in R67C6, locked for C6
8l. Naked pair {18} in R45C6, locked for C6 and N5 -> R5C5 = 6
8m. 7 in C6 only in R13C6, locked for C6
8n. 3 in N4 only in R456C1, locked for C1 -> R7C1 = 9

9a. R23C56 contains 4 singletons including 2 = {2579} -> R23C6 = , R23C5 = {59}, locked for C5 and N2 -> R14C5 = , R67C6 = , R7C23 = , R6C123 = , R2C2 = 3
9b. R23C45 contains 4 singletons, R23C5 = {59} -> R23C4 = 
9c. R23C34 contains 4 singletons, R23C4 =  -> R23C3 = {68/69}, 6 locked for C3 and N1 -> R3C2 = 8
9d. R12C12 contains 4 singletons = {1357} -> R1C2 = 5
9e. R1C46 = , R1C789 = , R8C8 = 1
9f. R5C9 = 1 (hidden single in C9)
9g. R12C78 contains two doublets = {3467/3478}, no 5, 4 locked for N3
9h. Naked triple {468} in R2C789, 6 locked for R2 and N3
9i. R78C78 contains 4 singletons = {1469}, 4,6 locked for N9
9j. R7C9 = 8 -> R26C9 = 
9k. R56C89 contains 4 singletons = {1469} -> R6C8 = 6

and the rest is naked singles.

HATMAN wrote:
I will give all the coloured squares. This means some may be redundant
Therefore in the Sindoku 23 NC diagram there ought also to be a blue square at R23C34 although it's not needed for the solution and I didn't use it. Also ignore the white squares with red borders.

Here's how I solved Sindoku 23 NC:
Prelims

a) For squares with 2 numbers, delete those 2 numbers from the rest of that nonet
b) For squares with 2 numbers, delete the adjacent numbers in the cells covered by those squares
Note. Consecutive numbers, including doublets in pink squares, must be diagonally opposite in the squares.

Steps Resulting From Prelims
1a. 9 in N6 only in R45C8, locked for C8
1b. 2 in N8 only in R7C45, locked for R7
1c. 1 in N8 only in R9C45, locked for R9
1d. 9 in N9 only in R89C7, locked for C7
1e. 6,8 in N6 only in R6C789, locked for R6

2a. R45C78 contain 5,9 and two consecutive numbers, which must be diagonally opposite, 9 in R45C8 -> 5 in R45C7, locked for C7 and N6
2b. R45C78 = {1259/2359}, 2 locked for N6
2c. R45C89 contain 7,9 and two consecutive numbers, which must be diagonally opposite, 9 in R45C8 -> 7 in R45C9, locked for C9 and N6
2d. Consecutive numbers in C789 of these square = [123/234/321], no 1 in R45C8, 3 locked for N6
2e. R56C89 contains 4 singletons, R5C8 = 9 or R5C9 = 7 (because of diagonally opposite consecutive numbers) -> no 8 in R6C89
2f. R6C7 = 8 (hidden single in N6), no 7,9 in R6C6 (NC)
2g. R56C89 contains 6 -> no 7 in R5C9 -> R4C9 = 7, R5C8 = 9, R4C7 = 5 (because of diagonally opposite in these squares), no 4,6 in R3C7 + R4C6, no 2,3 in R3C8, no 6,8 in R3C9, no 2 in R5C6 (all NC)

3a. R78C45 contains 2,4 and two consecutive numbers = {2467/2478/2489} -> R8C45 = {46/48}, 4 locked for R8 and N8, R7C45 = {27/29}
3b. R89C45 contains 1,8 and two consecutive numbers 3,4 or 4,5 = {1348/1458}, no 6 -> R8C45 = {48}, 8 locked for R8, R9C45 = {135}
3c. 6 in N8 only in R789C6, locked for C6
3d. 8 in N9 only in R7C89, locked for R7
3e. R78C56 contains 2 doublets which must be diagonally opposite, no 8 in R78C6 -> R7C5 = {27}, R8C6 = {36}, R8C5 = {48} -> R7C6 = {3579}
3f. R89C56 contains 4 singletons including 6 for N8 and one of 4,8 = {1368/1469}, no 5,7 -> R9C5 = 1
3g. R9C4 = {35} -> R8C4 = 8 (NC), R8C5 = 4 -> R89C6 = , R7C4 = 2 (NC), R7C6 = 3 (NC), R7C5 = 7, R9C4 = 5, no 1,3 in R6C4, no 2,4 in R6C6, no 1 in R7C3, no 4 in R7C7, no 7,9 in R8C3, no 7 in R8C7, no 4,6 in R9C3 (all NC)

4a. R89C78 contains 7,9 and two consecutive numbers -> R8C7 = 9, consecutive numbers 12/23/34/45 in R8C8 + R9C7, R9C8 = 7
4b. R78C78 contains 4 singletons, R78C7 =  -> R78C8 = {13/14}, 1 locked for C8
4c. R78C9 =  (hidden pair in N9), no 4 in R9C9 (NC)
4d. 2 in N9 only in R9C79, locked for R9
[At that stage I carelessly overlooked 4 singletons in R89C89.]

5a. R8C3 = {123} -> no 2 in R8C2 (NC)
5b. R89C23 contains 2 doublets which must be diagonally opposite -> R8C2 = 7, R9C3 = 8, R8C3 + R9C2 = [23/34], 3 locked for N7
5c. R9C1 = 6 (hidden single in N7)
5d. R8C1 = {12}, no 1 in R7C1 (NC)
5e. R7C2 = 1 or R7C123 = {459} -> no 4,5 in R7C2 (NC)

6a. R45C56 contains 4 singletons including 8 for N5 = {1368/2468} (cannot be {1358} which clashes with R6C6), no 5,7,9, 6 locked for C5 and N5
6b. {2468} -> R6C6 = 1 (NC) -> 1 in R456C6, locked for C6 and N5
6c. R456C4 = {379/479} (cannot be {347/349} which clash with R45C56), 7,9 locked for C4, 9 locked for N5
6d. Killer pair 3,4 in R456C4 and R45C56, locked for N5
6e. 5 in N5 only in R6C56, locked for R6
6f. 3 in R6 only in R6C123, locked for N4

7a. R23C45 contains 4 singletons = {1358/1359/1368/2469} (cannot be {1468} because 1,4,6 all in R23C4, cannot be {2468} which clashes with R123C6, ALS block)
7b. R23C56 contains 4 singletons
7c. R23C45 = {1358/1359/2469} (cannot be {1368} because R23C5 = {38} blocks singletons in R23C56)
7d. R23C4 = {13/46} -> R23C5 = {29/58/59}, no 3
7e. R23C56 = {2479/2579}, no 8, 2,7,9 locked for N2
7f. Naked triple {259} in R236C5, locked for C5
7g. 8 in N2 only in R1C56, locked for R1
7h. Killer pair 3,4 in R23C4 and R456C4, locked for C4

[Just spotted …]
8a. R45C56 (step 6a) = {1368} (cannot be {2468} because R45C6 =  clashes with R45C89 which must have either 2 in R4C8 or 4 in R5C9)
8b. Naked pair {18} in R45C6, locked for C6 and N5 -> R16C6 = , R6C5 = 2
8c. R45C5 =  (NC), R1C5 = 8 -> R23C5 =  (NC)
8d. R23C45 contains 4 singletons -> R23C4 = {13}, R1C4 = 6
8e. R5C4 = 4 (NC) -> R46C4 = 
8f. R4C8 = 2 -> R5C79 = {13}, 1 locked for R5 and N6 -> R45C6 = 
8g. Naked pair {46} in R6C89, 4 locked for R6
NC: no 5,7 in R1C3, no 3 in R1C7, no 2 in R23C3, no 5 in R5C3

9a. R12C78 contains 2 diagonally opposite doublets, R1C8 = {35} -> R2C7 = {24}, R1C7 opposite R2C8 -> R1C7 = {27}, R2C8 = {368}
9b. 5 in C8 only in R1C8 + R2C7 =  or R3C8 -> no 4 in R3C8
9c. 8 in C8 only in R1C7 + R2C8 =  or R3C8 -> R1C7 = 7 (hidden single in N3, NC) -> R1C7 = 7, R2C8 = {68}
9d. R12C8 = [38/58] (cannot be  NC, cannot be  which clashes with R678C8, ALS block) -> R2C8 = 8 -> R1C9 = 9 (NC)
9e. 1,2 in R1 only in R1C13 = {12} (NC), locked for N1 -> R1C2 = 5 (NC)
9f. R3C12 contains 2 diagonally opposite doublets, 8 in N1 only in R3C12 which must contain one of 7,9 in R2, no 7 in R3C1
9g. R12C12 contains 4 singletons but cannot contain both of 7,9 = {1357/1359} -> R1C1 = 1, R2C12 = {37/39}, 3 locked for R2 and N1
9h. Naked pair {48} in R3C12, 4 locked for R3 and N1
9i. R23C4 =  -> R3C79 = {12}, 2 locked for R3 and N3
9j. R89C89 contains 4 singletons -> R8C8 + R9C9 = 

and the rest is naked singles, without using NC.

Solution:
1 5 2 6 8 4 7 3 9
7 3 9 1 5 2 4 8 6
4 8 6 3 9 7 1 5 2
8 6 4 9 3 1 5 2 7
5 2 7 4 6 8 3 9 1
3 9 1 7 2 5 8 6 4
9 1 5 2 7 3 6 4 8
2 7 3 8 4 6 9 1 5
6 4 8 5 1 9 2 7 3

Top Post subject: Re: Occasional Sindoku (Quadrata's Concept) Posted: Tue Sep 28, 2021 10:50 pm Grand Master Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1792
Moving on to Sinduku 24 and the NC version.

The original diagram for Sindoku 24 seems to be incorrect. If you want to try that puzzle, see my final comment below.

Here's how far I got with Sindoku 24 using the original posted diagram:
Prelims
a) For squares with 2 numbers, delete those 2 numbers from the rest of that nonet
b) For squares with 2 numbers, delete the adjacent numbers in the cells covered by those squares

Steps Resulting From Prelims
1a. 2,8 in N2 only in R1C456, locked for R1
1b. 5 in N7 only in R89C3, locked for C3
1c. 6 in N7 only in R89C1, locked for C1
1d. 5 in N8 only in R789C4, locked for C4

2a. R23C56 contain 1,7 and two consecutive numbers, no 8 -> no 9 in R23C5
2b. R23C56 = {1347/1457}, 4 locked for C6 and N2
2c. R23C45 contains 3,9 and two consecutive numbers = {3569/3679} -> R23C4 = {69}, locked for C4, 6 locked for N2, R23C5 = {35/37}, no 1, 3 locked for C5
2d. 1 in N2 only in R23C6, locked for C6
2e. 9 in N8 only in R9C56, locked for R9
2f. R12C45 contains 4 singletons = {2579} (cannot be {3579} because 3,7,9 only in R2C45) -> R1C45 = , R2C45 = , R1C6 = 8, R3C45 = 
2g. R12C56 contains 2 doublets = {4578} -> R23C6 = 
2h. 4 in N8 only in R89C5, locked for C5
2i. 8 in N8 only in R78C5, locked for C5

3a. R56C56 contains 2 doublets = 12/23/56/67, no 9, 2,6 locked for N5
3b. R89C45 contains 2 doublets and must contain 4, no doublet in R89C4 -> R89C5 cannot be  -> no 9 in R9C5
3c. R4C5 = 9 (hidden single in C5)
3d. R9C6 = 9 (hidden single in N8) -> R8C5 = 8, R8C6 + R9C5 = 
3e. R89C45 = {3478/4578}, no 1, 7 locked for C4
3f. 1,6 in N8 only in R7C456, locked for R7

4a. R89C23 contains 1,5 and two consecutive numbers = {1578/1589}, no 3 -> R89C2 = 
4b. R89C12 contains 6,8 and two consecutive numbers = {1268} -> R89C1 = 
4c. 3,4 in N7 only in R7C123, locked for R7 -> R7C456 = 
4d. R78C45 contains 4 singletons = {1358} -> R8C4 = 3, R9C4 = 7, R9C3 = 5
4e. R78C23 contains 4 singletons = {1379/1479}, 7,9 locked for N7
4f. R78C12 contains 4 singletons = {1369/1469} -> R7C2 = 7, R8C3 = 7

5a. 8 in N4 only R45C3, locked for C3
5b. R12C23 and R23C23 both contain 2 doublets, no 8 in either -> no 9 in R13C3
5c. R12C12 and R23C12 both contain 4 singletons with 9 in R13C1 -> no 8 in R2C1 -> R13C1 =  (hidden pair in N1)
5d. R23C12 contains 4 singletons with 8 = {1358} (cannot be {1368} because no 1,3,6 in R3C2, cannot be {2468} because 2,4,6 only in R23C2) -> R2C1 = 1, R23C2 = 
5e. Naked triple {246} in R123C3, locked for R3 and N1 -> R1C2 = 7, R7C13 = 
5f. R6C3 = 9 (hidden single in N4)
5g. R56C23 contains 4 singletons = {1469} -> R5C3 = 1, R56C2 = , R4C23 = , R56C5 = 
5h. 1 in N6 only in R4C78, locked for R4 -> R456C4 = 
5i. R45C12 contains 2 doublets = {2356/2367}, 3 locked for N4

6a. R78C89 contains 4 singletons = {2479/2579}, 2,7,9 locked for N9 -> R7C7 = 8
6b. R89C78 contains 4 singletons = {1359/1469}, 1locked for R9 -> R8C8 = 9
6c. R45C78 contains 1 and three consecutive numbers, no 8 -> no 9
6d. R5C9 = 9 (hidden single in N6)
6e. R23C89 contain 2 doublets, no 3 -> no 2 -> R3C89 = {47}, locked for R3 -> R123C3 = 
6f. R2C7 = 2 (hidden single in N3), R3C7 = 9, R23C78 contains 4 singletons = {2579} (cannot be {2479} because 4,7 only in R3C8) -> R23C8 = , R23C9 = , R7C89 = , R8C79 = 
6g. R1C9 = 1 (hidden single in C9)
6h. R12C89 contains 4 singletons -> R1C8 = 3, R1C7 = 6
6i. R89C78 contains 4 singletons -> R9C7 = 1

Now R45C78 with 1 and 3 consecutive numbers would give R45C8 = , R45C7 = {35} but after that there’s no unique solution to the puzzle.
Here's how I solved Sindoku 24 NC:
Prelims
a) For squares with 2 numbers, delete those 2 numbers from the rest of that nonet
b) For squares with 2 numbers, delete the adjacent numbers in the cells covered by those squares
Note. Adjacent numbers, including doublets in pink squares, must be diagonally opposite in the squares.

Steps Resulting From Prelims
1a. 2,8 in N2 only in R1C456, locked for R1
1b. 5 in N7 only in R89C3, locked for C3
1c. 6 in N7 only in R89C1, locked for C1

2a. R23C45 contain 3,9 and two consecutive numbers = {3569/3679}, no 1, 6 locked for C4 and N2
2b. R23C56 contain 1,7 and two consecutive numbers = {1347/1457}, no 9 -> R23C6 = {14}, locked for C6, 4 locked for N2
2c. R23C4 = {69} (hidden pair in N2), 9 locked for C4
2d. R23C5 = {37} (hidden pair in N2), locked for C5
2e. 5 in N2 only in R1C456, locked for R1

3a. R89C12 must have 6,8 diagonally opposite, 6 in R89C1 -> 8 in R89C2, locked for C2 and N7
3b. R89C23 contains 1,5 and two consecutive numbers, one of which must be 8 = {1578/1589}, no 3, R89C2 = {18}, 1 locked for C2 and N7
3c. R89C12 contains 6,8 and two consecutive numbers = {1268} -> R89C1 = {26}, 2 locked for C1 and N7
3d. R7C13 = {34} (hidden pair in N7, cannot be in R7C2 NC), locked for R7, no 3,4 in R6C13 (NC)
3e. R78C23 contains 4 singletons -> R8C2 = 1, R9C2 = 8 -> R9C3 = 5 (NC), no 4 in R9C4 (NC)
3f. R9C2 = 8 -> R8C1 = 6 (diagonally opposite in R89C12), R9C1 = 2
3g. R78C12 contains 4 singletons = {1369/1469} -> R7C2 = 9, R8C3 = 7, no 8 in R8C4 (NC)

4a. R89C45 contains 2 diagonally opposite doublets, R9C4 = {137} -> R8C5 = {248}, no 3,5 in R9C5 -> R8C4 = {235}, R9C5 = {146}
4b. R8C4 = {235} -> R8C5 + R9C4 = [21/87] (cannot be  NC), no 3,4
4c. R89C45 cannot be {1256} (because R7C45 cannot be , NC) -> R8C5 = 8, R9C4 = 7, no 9 in R8C6 (NC)
4d. R9C6 = 9 (hidden single in N8)
4e.
R78C45 contains 4 singletons = {1358/1368}, no 2 -> R8C4 = 3, R9C5 = 4 (for R89C45)
4f. 1,6 in N8 only in R7C456, locked for R7
4g. 2 in N8 only in R78C6, locked for C6

5a. R12C45 contains 4 singletons including 2 -> no 3, R23C5 =  -> R23C6 =  (NC), R23C6 =  (NC), R1C6 = 8 (NC), no 7,9 in R1C7, no 8 in R2C3, no 3,5 in R2C7, no 2 in R3C7, no 5 in R4C4, no 2 in R4C5 (NC)
5b. R7C79 = {78} (hidden pair in N9, cannot be in R7C8, NC), no 7,8 in R6C79 (NC)
5c. R78C89 contains 4 singletons = {2479/2579} -> 2,9 locked for N9, R7C9 = 7, R7C7 = 8, no 9 in R6C7, no 6 in R6C9 (NC)
5d. R8C7 = {45} -> R8C68 =  (NC)
5e. Naked pair {45} in R8C79 (cannot be in R8, NC), locked for N9 -> R78C8 = , no 1,3 in R6C8 (NC)
5f. R7C6 = {56} -> R7C5 = 1 (NC), R7C4 = 5, R7C6 = 6, R1C45 = , no 1,3 in R1C3, no 4 in R6C4, no 2 in R6C5, no 5,7 in R6C6, no 4 in R7C3 (NC)
5g. R6C6 = 3, R45C6 = {57}, no 6 in R45C57, no 2,4 in R6C7 (NC)
5h. R456C5 = , no 8 in R4C4, no 1 in R5C4 (NC)
5i. R7C13 = , no 5 in R6C1, no 2 in R6C3 (NC)
5j. R89C78 contains 4 singletons = {1359/1469}, 1 locked for N9

6a. R12C23 and R23C23 contain 2 diagonally opposite doublets -> R1C3 = {46}, R2C2 = {35}, R3C3 = {24} -> R13C3 = [42/64], 4 locked for C3 and N1
6b. R1C2 + R2C3 + R3C2 =  -> R1C3 + R2C2 + R3C3 = , no 4,6 in R4C2, no 1 in R4C3 (NC)
6c. R456C2 = , R4C3 = 8 -> R56C3 =  (NC), R6C1 = 7, R5C1 = 3 (NC) -> R4C1 = 5 R45C6 = , no 4 in R5C7, no 8 in R6C4 (NC)
6d. R456C4 =  -> R6C789 = , R8C79 = , no 7 in R5C8, no 6 in R9C9 (NC)
6e. R5C789 = , R9C9 = 3
6f. R2C7 = 2 (hidden single in C7) -> R1C7 = 6 (NC)

and the rest is naked singles, without using NC.
Solution:
9 7 4 2 5 8 6 3 1
1 3 6 9 7 4 2 5 8
8 5 2 6 3 1 9 7 4
5 2 8 4 9 7 3 1 6
3 6 1 8 2 5 7 4 9
7 4 9 1 6 3 5 8 2
4 9 3 5 1 6 8 2 7
6 1 7 3 8 2 4 9 5
2 8 5 7 4 9 1 6 3
Correcting the Sindoku 24 diagram:
I ought to have spotted sooner that this diagram is incorrect. 1 singleton with three consecutive numbers can't possibly fit into a 2x2 square when we know that the numbers are non-consecutive, even though we don't use that property for solving Sindoku 24.
Having solved Sindoku 24 NC, the clue for R45C78 of Sindoku 24 should be 1,7 and two consecutive numbers.
If I'd known that originally my solving path would have been a bit shorter but I'm not going to do that one again; there are several more Sindokus for me to try.

Top Post subject: Re: Occasional Sindoku (Quadrata's Concept) Posted: Sun Oct 03, 2021 2:51 am Grand Master Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1792
And now on to the two versions of Sindoku 26 X. I found the NC version, with surprisingly few clues, the better puzzle.

Here's how I solved Sindoku 26 X:
Prelims
a) For squares with 2 numbers, delete those 2 numbers from the rest of that nonet
b) For squares with 2 numbers, delete the adjacent numbers in the cells covered by those squares

1a. R12C12 contains 4,8 and two successive numbers = {1248}, no 6, 1,2 locked for N1
1b. R23C23 contains 5,9 and two successive numbers = {2359}, no 7 -> R2C2 = 2, placed for D\
1c. R1C3 + R3C1 = {67} (hidden pair in N1)
1d. R89C89 contains 2,6 and two successive numbers = {2689}, no 4, 8,9 locked for N9
1e. R78C78 contains 1,5 and two successive numbers = {1578}, no 3 -> R8C8 = 8, placed for D\
1f. R7C9 + R9C7 = {34} (hidden pair in N9)
1g. R56C12 contains 2,6 and two successive numbers = {2689}, no 4, 8,9 locked for N4, 2 locked for C1
1h. R45C23 contains 1,3 and two successive numbers = {1356/1367/1378}, no 9 -> R5C2 = {68}
1i. R12C56 contains 1,5 and two successive numbers = {1578/1589}, no 3, 8 locked for N2
1j. R45C45 contains 2,7 and two successive numbers = {2457}, no 9, 4,5 locked for N5
1k. R45C89 contains 4,8 and two successive numbers = {1248}, no 6, 1,2 locked for N6, 8 locked for C9
1l. R89C45 contains 5,9 and two successive numbers = {1259/2359}, no 7, 2 locked for N8
1m. 1 in R6 only in R6C456, locked for N5
1n. R6C1 = 2 (hidden single in R6)

2a. R45C78 contains 4 singletons -> R45C8 = {14/24}, 4 locked for C8 and N6, R45C7 = {69/79}, 9 locked for C7 and N6
2b. 1 in R3 only in R3C789, locked for N3
2c. 2 in C7 only in R13C7, locked for N3
2d. R6C3 = 4 (hidden single in R6)
2e. R56C23 contains 4 singletons with R5C2 = {68} -> R5C3 = 1, R56C2 = {68/69}, 6 locked for C2 and N4
2f. Naked triple {357} in R4C123, locked for R4 -> R4C45 = , 4 placed for D\, R4C89 = , R5C89 = 
2g. R1C1 = 1, placed for D\
2h. Naked pair {57} in R7C78, locked for R7 and N9 -> R8C7 = 1
2i. R5C6 = 3 (hidden single in R5)
2j. R3C3 = 3 (hidden single on D\)
2k. Naked pair {69} in R46C6, locked for C6 and N5
2k. R6C4 = 1 (hidden single on D/) -> R6C5 = 8
2m. R3C9 = 1 (hidden single in C9)
2n. Naked pair {69} in R89C9, locked for C9 and N9 -> R9C8 = 2
2o. R4C2 = 3 (hidden single in R4)
2p. Naked pair {57} in R5C45, 7 locked for R5
2q. Naked pair {69} in R45C7, 9 locked for C7 and N6
2r. 7 in C2 only in R89C2, locked for N7
2s. R7C4 = 8 (hidden single in C4)
2t. R9C3 = 8 (hidden single in C3)

3a. Naked triple {147} in R789C6, 1,7 locked for C6 and N8 -> R3C6 = 2
3b. R37C5 =  (hidden pair in C5)
3c. R8C4 = 2 (hidden single in N8)
3d. R12C56 contains 1,5 and two successive numbers = {1578/1589} -> R2C5 = 1, R1C5 = {79}
3e. R23C45 contains 4 singletons, R23C5 =  -> no 3 in R2C4
3f. R1C4 = 3 (hidden single in C4)
3g. R7C3 = 2 (hidden single in R7)
3h. 9 in R7 only in R7C12, locked for N7
3i. R2C3 = 9 (hidden single in C3) -> R3C2 = 5
3j. R1C7 = 2 (hidden single in R1)
3k. R3C7 = 8 (hidden single in R3)

4a. R78C56 contains 4 singletons with 6 = {1469} -> R7C6 = 1, R8C56 = , R9C45 = 
4b. R8C2 = 7, placed for D/ -> R5C5 = 5, placed for both diagonals, R1C9 = 4, placed for D/, R9C1 = 6, placed for D/

and the rest is naked singles, without using the diagonals.

Here's how I solved Sindoku 26 X NC:
Prelims
a) For squares with 2 numbers, delete those 2 numbers from the rest of that nonet
b) For squares with 2 numbers, delete the adjacent numbers in the cells covered by those squares
Note. Consecutive numbers must be diagonally opposite in those squares; therefore the specified numbers must also be diagonally opposite.

1a. R89C89 contains 2,6 and two successive numbers = {2689}, R8C8 + R9C9 = {89} (NC), locked for N9, R8C9 + R9C8 = {26}
1b. R78C78 contains 1,5 and two successive numbers = {1578} -> R8C8 = 8, R7C7 = 7 (NC), R9C9 = 9, all placed for D\, no 6 in R6C7, no 6,8 in R7C6 (NC)
1c. R7C9 + R9C7 = {34} -> no 3 in R6C9, no 3,4 in R9C6 (NC)
1d. R45C89 contains 4,8 and two successive numbers = {1248}, no 6, 1,2 locked for N6, 8 locked for C9
1e. 8 in R45C89 only in R45C9 -> 4 in R45C8, locked for C8 and N6
1f. R45C78 contains 4 singletons including 4 = {1469/2469} -> R45C7 = {69} locked for C7 and N6
1g. Naked triple {357} in R6C789, locked for R6, no 4 in R6C6 (NC)
1h. 1 in R6 only in R6C456, locked for N5
1i. 2 in C7 only in R123C7, locked for N3

2a. R89C45 contains 5,9 and two successive numbers = {1259/2359}, no 7, 2 locked for N8, 9 locked for R8
2b. 7 in N8 only in R89C6, locked for C6, no 6 in R8C6, no 6,8 in R9C6 (NC)
2c. R7C45 = {68} (hidden pair), locked for R7
2d. R78C56 contains 4 singletons, R7C5 = {68} -> no 7 in R8C6 (NC)
2e. 4 in N8 only in R78C6 -> R78C6 = {14} (NC), locked for C6, 1 locked for N8 -> R78C5 = , R7C4 = 8, R9C6 = 7, R9C4 = 5 (diagonally opposite in R89C45), no 9 in R6C4 + R7C3, no 4,6 in R9C3 (NC)
2f. R8C4 = {23} -> no 2,3 in R8C3 (NC)
2g. 1 on D\ only in R1C1 + R2C2 + R3C3, locked for N1
2h. 2 in R7 only in R7C123, locked for N7
2i. 1 in R9 only in R9C123, locked for N7

3a. 9 in R6 only in R6C123, locked for N4
3b. R45C23 contains 1,3 and two successive numbers = {1356/1367/1378}
3c. R45C23 = {1356/1367}, 6 locked for N4
or R45C23 = {1378}, 5 in R45C4 => no 6 in R45C4-> no 6 in R45C4
3d. R45C23 contains 1,3 which must be diagonally opposite, R56C23 contains 4 singletons including one of 1,3 -> no 2 in R6C23
3e. 2 in N4 only in R456C1, locked for C1
3f. 2 in N7 only in R7C23 -> no 3 in R7C23 (NC)

4a. R45C89 = {1248} must have 4,8 diagonally opposite
4b. One of 4,8 in R5C89 -> R6C789 = , R9C6 = 4, R7C9 = 3, R6C6 = 6 (NC), placed for D\, no 5 in R5C6, no 2 in R8C9 (NC)
4c. R8C9 = 6, R9C8 = 2, R9C5 = 3, R8C4 = 2
4d. R45C8 = {14}, 1 locked for C8 and N6, R7C8 = 5, R8C7 = 1, R78C6 = 
4e. 1 in N5 only in R6C45 -> no 2 in R6C5 (NC)
4f. R6C1 = 2 (hidden single in R6)
4g. 9 in R6 only in R6C23 = {49} (NC), 4 locked for R6 and N4 -> R6C45 = , 1 placed for D/
4h. R45C1 = {58} (NC), locked for C1 and N4 -> R9C1 = 6, placed for D/
4i. R56C23 contains 4 singletons including 4 = {1469/1479}, 1 locked for R5 and N4 -> R45C8 = , R45C9 =  (NC), no 7 in R3C9 (NC)
4j. R4C4 = {34} -> no 3 in R4C3 + R35C4 (NC)
4k. R4C2 = 3 (hidden single in N4) -> R5C3 = 1 (diagonally opposite in R45C23), no 2,4 in R3C2 (NC)
4l. R4C4 = 4, placed for D\ -> R5C5 = 5, placed for both diagonals, R45C1 = , no 4 in R3C1, no 7 in R5C2 (NC)
4m. R5C2467 = , R4C3567 = , 9 placed for D/, no 1 in R3C5, no 8 in R3C6 (NC)
4n. R8C123 = , 7 placed for D/, no 4 in R7C13 (NC)
4o. R7C3 = 2, placed for D/
4p. R2C8 = 3, R3C7 = 8, no 2 in R2C7, no 4 in R2C9, no 9 in R3C8 (NC)

and the rest is naked singles, without using the diagonals or NC.

Solution:
1 8 6 3 7 5 2 9 4
4 2 9 6 1 8 5 3 7
7 5 3 9 4 2 8 6 1
5 3 7 4 2 9 6 1 8
8 6 1 7 5 3 9 4 2
2 9 4 1 8 6 3 7 5
9 4 2 8 6 1 7 5 3
3 7 5 2 9 4 1 8 6
6 1 8 5 3 7 4 2 9

So the numbers of D\ are in ascending order, from top to bottom; also the solution is symmetrical, all pairs of numbers symmetrical about the centre, for example R1C2 and R9C8, total 10. Surprising considering that the only symmetry in the NC puzzle's diagram is for the blue squares; the first version does also show symmetry in four pairs of numerical clues.

Top Post subject: Re: Occasional Sindoku (Quadrata's Concept) Posted: Mon Oct 04, 2021 10:02 pm Grand Master Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1792
Sindoku 27 X has the fewest number-pair clues of the 'plain' versions I've done so far. Even so it wasn't difficult.

I love the blue squares with 4 singletons but must admit that I'm occasionally a bit careless with overlooking all possibilities on my first attempts at this, and one of the Sindoku 26 X versions; I hope I've managed to get them right when I checked my walkthroughs before posting them.

Here's how I solved Sindoku 27 X:
Prelims
a) For squares with 2 numbers, delete those 2 numbers from the rest of that nonet
b) For squares with 2 numbers, delete the adjacent numbers in the cells covered by those squares

1a. R12C78 contains 1,7 and two successive numbers = {1347/1457}, no 9, 4 locked for N3
1b. R23C23 contains 3,9 and two successive numbers = {3569/3679}, no 1, 6 locked for N1
1c. R45C45 contains 1,7 and two successive numbers = {1347/1457}, no 9 -> R5C5 = 4, placed for both diagonals
1d. R56C56 contains 2,4 and two successive numbers = {2489}, no 6, 8,9 locked for N5
1e. 6 in N5 only in R4C6 + R6C4, locked for D/
1f. R89C56 contains 3,9 and two successive numbers = {3569/3679}, no 1, 6 locked for N8
1g. R78C45 contains 2,4 and two successive numbers = {2467/2478/2489}, 4 locked for C4

2a. R45C56 contains 4 singletons, R5C5 = 4 -> R4C6 = 6, R4C5 = 1, R5C6 = {89}
2b. Naked triple {357} in R456C4, locked for C4
2c. 2 in N5 only in R6C56, locked for R6
2d. 6 in N8 only in R89C5, locked for C5
2e. R78C56 contains 4 singletons including 6 = {1368/1469} (cannot be {2468/2469} because 2,4 only in R8C4), no 2,5,7 -> R8C4 =1
2f. 2 in N8 only in R7C45, 2 locked for R7
2g. R78C56 contains 4 singletons -> R8C5 = 9 (cannot be 6 because R7C56 =  leaves no valid candidate in R8C6), no 8 in R7C56
2h. R89C45 contains 4 singletons = {1469} -> R8C4 = 4, R9C5 = 6
2i. Naked triple {357} in R789C6, locked for C6 and N8 -> R7C45 = , R6C5 = 8, R56C6 = , 2 placed for D\
2j. R78C56 contains 4 singletons = {2579} -> R9C6 = 3 (hidden single in N8)
2k. R56C45 contains 2 doublets, R6C5 = 8 -> R5C4 = 7
2l. R1C4 = 6 (hidden single in R1)
2m. R23C45 contains 4 singletons, R23C4 = {29} -> R23C5 = {57}, R1C5 = 3
2n. R12C56 contains 2 doublets, R1C5 = 3 -> R12C6 = {48}, R2C5 = 7, R3C56 = 

3a. R1C9 = 9 (hidden single in R1), placed for D/
3b. 7 in N3 only in R1C78, locked for R1
3c. R3C1 = 4 (hidden single in R3)
3d. R12C12 contains 4 singletons including 5,8 in R1C12 + R2C1 -> R2C2 = 3, placed for D\, R1C12+ R2C1 = {158}, 5 locked for N1, R1C3 = 2
3e. R12C23 contains 2 doublets, R1C3 + R2C2 = , R2C3 = {69} -> R1C2 = {58}
3f. R23C12 contains 2 doublets, R2C2 + R3C1 = , R3C2 = {679} -> R2C1 = {58}
3g. R1C1 = 1 (hidden single in N1), placed for D]\
3h. R4C4 = 5, placed for D\, R6C4 = 3, placed for D/
3i. R23C78 contains 4 singletons, 1 in N3 only in R2C78 -> R3C7 = 8, placed for D/, R3C8 = {36}
3j. R12C78 = {1457} (only remaining combination), 5 locked for N3
3k. R23C89 contains 2 doublets, contains 3 and 6 for N3 = {2356} -> R2C8 = 5, placed for D/
3l. R12C89 contains 4 singletons, R1C9 = 9, R2C8 = 5 -> R1C8 = 7, R2C9 = 2, R12C7 = , R12C6 = , R23C4 = , R1C2 = 5, R2C13 = 
3m. R23C78 contains 4 singletons, R2C8 = 5 -> R3C89 = 

4a. 6,8 on D\ only in R7C7 + R8C8 + R9C9, locked for N9
4b. R89C78 contains 2 doublets with 2 for N9 -> R8C5 = 3
4c. R78C89 contains 4 singletons starting from R8C8 = {68}, no 7 -> R78C8 = , R78C9 = , R7C7 = 7, placed for D\, R3C23 = 
4d. R7C3 = 1, R8C2 = 2, R9C1 = 7, R8C13 = , R7C12 = , R9C23 = 

5a. R45C23 contains 2 doublets -> R5C2 = 8, R4C23 = , R56C3 = 
5b. R56C78 contains 4 singletons, R5C8 = 1 (hidden single in R5) -> no 2 in R5C7

and the rest is naked singles, without using the diagonals.

Solution:
1 5 2 6 3 8 4 7 9
8 3 6 9 7 4 1 5 2
4 7 9 2 5 1 8 3 6
9 4 7 5 1 6 2 8 3
2 8 3 7 4 9 6 1 5
6 1 5 3 8 2 9 4 7
3 6 1 8 2 5 7 9 4
5 2 8 4 9 7 3 6 1
7 9 4 1 6 3 5 2 8

Top Post subject: Re: Occasional Sindoku (Quadrata's Concept) Posted: Tue Oct 05, 2021 6:25 pm Grand Master Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1792
HATMAN wrote:
Sinduku 27 X NC
This is the hardest Sindoku I've put together. When I first solved it I had a neat move that I found in the plain one and used, but I could not remember it yesterday and have had to solve it somewhat brute force.
A nice moderate-Assassin level puzzle. Very enjoyable! I didn't come across any neat moves for the plain one in my previous post. For this one I used a few short chains, which weren't difficult to find; maybe step 3e, a key move, was the hardest to spot.
Here's how I solved Sinduku 27 X NC:
Prelims
a) For squares with 2 numbers, delete those 2 numbers from the rest of that nonet
b) For squares with 2 numbers, delete the adjacent numbers in the cells covered by those squares
Note. Consecutive numbers must be diagonally opposite in those squares; therefore the specified numbers must also be diagonally opposite.

1a. R12C78 contains 1,7 and two successive numbers = {1347/1457}, no 9, 4 locked for N3
1b. R23C23 contains 3,9 and two successive numbers = {3569/3679}, no 1, 6 locked for N1
1c. R45C45 contains 1,7 and two successive numbers = {1347/1457}, no 9, 4 locked for N3

2a. R12C23 contains 2 doublets, R1C23 cannot be {12} (NC) -> no 1 in R1C23
2b. R23C12 contains 2 doublets, R23C1 cannot be {12} (NC) -> no 1 in R23C1
2c. R1C1 = 1 (hidden single in N1), placed for D\, no 2 in R1C2 + R2C1 (NC)
2d. R12C78 contains 1,7 and two successive numbers = {1347/1457}, 1 in R2C78, locked for R2
2e. 1 in R2C78 -> 7 in R1C78, locked for R1 and N3
2f. R23C78 contains 4 singletons, 1 in R2C78 -> no 2 in R3C78
2g. 2 in N3 only in R123C9, locked for C9
2h. R45C45 = {1347/1457}, 1 in R4C5 + R5C4 -> R4C5 + R5C4 = {17}, 7 locked for N5
2i. R4C4 + R5C5 = {34/45}, 4 locked for D\
2j. R4C4 = {345} -> no 4 in R3C4 + R4C3 (NC)
2k. R56C45 contains 2 doublets, R5C4 = {17} -> R6C5 = {28}, R5C5 = {345} -> R6C4 = {2356}
2l. R4C5 + R5C4 + R6C5 = [172/718] -> no 2,8 in R4C6 (NC)
2m. 8 in N5 only in R5C6 + R6C56 -> no 9 in R6C6 (NC)
2n. 9 in N5 only in R45C6, locked for C6, no 8 in R5C6 (NC)
2o. 8 in N5 only in R6C56, locked for C6
2p. 2 in N1 only in R1C3 + R3C1, R12C23 and R23C12 both contain 2 doublets -> R2C2 = 3, placed for D\ -> R1C3 + R3C1 = {24}, 4 locked for N1, no 3 in R1C4 + R4C1 (NC)
2q. R12C12 contains 4 singletons = {1357/1358}, 5 locked for N1
2r. R23C23 contains 3,9 and two successive numbers = {3679}, 7 locked for N1
2s. R2C2 = 3 -> R3C3 = 9 (diagonally opposite), locked for D\, no 8 in R3C4 + R4C3 (NC)
[So three early placements; not the same ones as the early one in the ‘plain’ version of this puzzle.]
2t. R2C3 + R3C2 = {67}, no 6,7 in R2C4 + R4C2 (NC)
2u. Naked pair {45} in R4C4 + R5C5, 5 locked for N5 and D\, no 5 in R3C4 + R4C3 (NC)
2v. R56C45 contains 2 doublets, R5C5 = {45} -> R6C4 = {36}
2w. 9 in N3 only in R12C9, locked for C9, no 8 in R12C9 (NC)
2x. 8 in N3 only in R3C789, locked for R3
2y. 7 on D/ only in R7C3 + R8C2 + R9C1, locked for N7
2z. 7 on D\ only in R7C7 + R8C8 + R9C9, locked for N9
[At this stage I could have made one more elimination from N5 but it doesn’t lead to further progress. There may be a simpler way to remove that candidate later.]

3a. R23C78 contains 4 singletons = {1358/1468}, 8 locked for R3
3b. 5 of {1358} must be in R2C78 -> no 5 in R3C78
3c. R23C89 contains 2 doublets, R3C8 = {368} -> R2C9 = {259}
3d. R23C12 contains 2 doublets, R2C2 + R3C1 = , R2C1 + R3C2 = [56/87]
3e. Consider combinations for R23C78
R23C78 = {1358}, 5 locked for R2 => R2C1 = 8
or R23C78 = {1468}, 6 locked for R3 => R3C2 = 7 => R2C1 = 8
-> R2C1 = 8, R1C2 = 5, R3C2 = 7, R2C3 = 6, R1C3 = 2 (NC), R3C1 = 4, no 5 in R2C4, no 5 in R4C1, no 8 in R4C2 (NC)
3f. Combining R12C78 and R23C78, 3 must be in R13C78, locked for N3
3g. Naked triple {369} in R1C9 + R4C6 + R6C4, locked for D/, 3 locked for N5 -> R3C7 = 8, placed for D/, no 7,9 in R4C7 (NC)
3h. 2 on D/ only in R8C2 + R9C1, locked for N7
3i. R89C12 contains 4 singletons including 2 -> no 1 in R89C2, no 3 in R8C1
3j. R8C2 = {24} -> no 3 in R8C3 (NC)
3k. R78C12 contains 2 doublets, R8C2 = {24} -> R7C1 = {35}, no 4 in R7C2 (NC)
3l. R78C12 contains 2 doublets -> R7C2 = {68}, R8C1 = {59}, no 7 in R7C3 (NC)
3m. R8C2 + R9C1 =  (hidden pair in N7), no 1 in R8C3, no 8 in R9C2 (NC)
3n. R78C12 contains 2 doublets, R8C2 = 2 -> R7C1 = 3, no 2 in R6C1 (NC)
3o. 6 in N7 only in R79C2, locked for C2
3p. 1 in N7 only in R79C3, locked for C3

4a. R45C23 contains 2 doublets, no 2 in R45C3 -> no 1 in R45C2
4b. R6C2 = 1 (hidden single in C2)
4c. R45C2 = {48/49} (cannot be {89}, NC), 4 locked for C2 and N4
4d. R4C2 = {49} -> R5C3 = {358}, R4C3 = {37} -> R5C2 = {48}
4e. R56C23 contains 4 singletons = {1358} (only possible combination) -> R5C2 = 8, R7C2 = 6, R9C2 = 9, R8C1 = 5, R4C2 = 4, R4C3 = 7 (NC), no 9 in R5C1, no 8 in R9C3 (NC)
4f. R4C4 = 5 -> R5C5 = 4, placed for D/, R7C3 = 1, placed for D/, R89C3 = , no 2 in R7C4, no 7,9 in R8C4, no 3 in R9C4 (NC)
4g. R2C8 = 5, R2C7 = 1 (hidden single in N3), R3C8 = 3 (NC), R12C78 =  (NC), R8C8 + R9C9 = , both placed for D\ -> R6C6 + R7C7 = , R1C9 = 9 (NC), placed for D/, R23C9 = 
4h. R4C569 =  -> R3C6 = 1 (NC)

and the rest is naked singles, without using the diagonals or NC.
The solution is in my post for Sindoku 27 X.

Top Post subject: Re: Occasional Sindoku (Quadrata's Concept) Posted: Thu Oct 07, 2021 3:26 am Grand Master Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1792
After that harder puzzle, Sindoku 28 X and Sindoku 28 X NC were both more like the level of Sindoku 26 X and Sindoku 26 X NC. After a fairly easy start, Sindoku 28 X NC seemed to have a long ending the way I solved it.

Here's how I solved Sindoku 28 X:
Prelims
a) For squares with 2 numbers, delete those 2 numbers from the rest of that nonet
b) For squares with 2 numbers, delete the adjacent numbers in the cells covered by those squares

1a. R45C89 contains 3,6 and two successive numbers = {3689} -> R5C8 = 8, 9 locked for N6
1b. R56C78 contains 1,8 and two successive numbers = {1458}, 4,5 locked for N6
1c. R56C12 contains 3,6 and two successive numbers = {3689}, no 1, 8,9 locked for N4, 8 locked for R6
1d. R45C23 contains 2,4 and two successive numbers = {2467} -> R5C2 = 6, 7 locked for N4
1e. R4C12 contains 4 singletons, R5C2 = 6 -> R4C1 = 1, R4C2 = 4, R5C1 = 9, R5C9 = 3, R6C3 = 5
1f. R56C89 contains 2 doublets, R5C89 =  -> R6C89 = , R456C7 = , R45C3 = 
1g. Naked pair {69} in R4C89, locked for R4
1h. Naked pair {38} in R6C12, 3 locked for R6
1i. R45C45 contains 2 doublets, no 2 in R45C45 -> no 1 in R5C45
1j. R5C6 = 1 (hidden single in N5)
1k. R56C56 contains 4 singletons, R5C5 = 1 -> R6C56 = {69} -> R5C5 = 4, placed for both diagonals, R5C4 = 7
1l. R6C4 = 2 (hidden single in N5), placed for D/
1m. R45C45 contains 2 doublets, R5C4 = 7 -> 8 in R4C45, locked for R4

2a. R12C45 contains 7,9 and two successive numbers = {1279/2379}, no 5, 2 locked for N2
2b. 4 in N2 only in R12C6, locked for C6
2c. 5 in N2 only in R3C567, locked for R3
2d. 5 in N3 only in R12C89
2e. R12C89 contains 4 singletons including 5, no 4,6

3a. R78C56 contains 7,9 and two successive numbers = {1279/2379}, no 5, 2 locked for N8
3b. R89C56 contains 4 singletons, R8C5 = {13} -> no 2 in R89C6
3c. R78C6 =  (hidden pair in C6)
3d. R89C56 contains 4 singletons, R8C6 = 7 -> no 6,8 in R9C56
3e. Naked triple {135} in R8C5 + R9C56, locked for N8, 1 locked for C5 -> R7C5 = 9 -> R6C56 = , 9 placed for D\
3f. R78C45 contains 4 singletons, R7C5 = 9 -> R78C4 = {46}, locked for C4, R8C5 = 1, R9C4 = 8
3g. Naked pair {35} in R9C56, locked for R9
3h. Naked pair {35} in R49C6, locked for C6
3i. R45C45 contains 2 doublets, R5C4 = 7 -> R4C5 = 8
3j. R23C56 contains 2 doublets, R3C6 = {68} -> R2C5 = 7, R1C5 = 2, R3C5 = {35} -> R2C6 = {46}
3k. R8C9 = 8 (hidden single in N9)
3l. 9 in N9 only in R89C7, locked for C7
3m. R12C89 contains 4 singletons, no 8 -> no 9 -> R12C89 = {1357}, R1C8 = 7, R2C8 = 3, placed for D/, R12C9 = {15}, locked for C9, 1 locked for N3
3n. Naked triple {468} in R123C7, 4,6 locked for C7 and N3
3o. Naked pair {29} in R3C89, locked for R3
3p. Naked pair {26} in R9C89, locked for R9 and N9 -> R7C89 + R8C8 = , 5 placed for D\ -> R4C4 = 3, placed for D\, R4C6 = 5, placed for D/
3q. R1C9 =  -> R12C4 = 
3r. Naked triple {268} in R1C1 + R2C2 + R9C9, locked for D\, 8 locked for N1 -> R3C3 = 1
3s. R9C1 = 7
3t. R89C23 contains 4 singletons, R8C2 = 9, R9C23 =  -> R8C3 = 6
3u. R7C3 = 8, placed for D/ -> R3C7 = 6
3v. R3C45 = 
3w. R23C56 contains 2 doublets, R3C5 = 3 -> R2C6 = 4, R2C7 = 8 -> R2C2 = 2, placed for D\

and the rest is naked singles, without using the diagonals.

Here's how I solved Sindoku 28 X NC:
Prelims
a) For squares with 2 numbers, delete those 2 numbers from the rest of that nonet
b) For squares with 2 numbers, delete the adjacent numbers in the cells covered by those squares
Note. Consecutive numbers must be diagonally opposite in those squares; therefore the specified numbers must also be diagonally opposite.

1a. R89C89 contains 2,8 and two successive numbers = {2458/2568}, 5 locked for N2
1b. R78C56 contains 7,9 and two successive numbers, no 7,9 in R8C5 -> R7C5 + R8C6 = {79}, locked for N8, no 8 in R6C5 (NC)
1c. R12C45 contains 7,9 and two successive numbers -> one of 7,9 must be in R12C5
1d. Killer pair 7,9 in R12C5 and R7C5, locked for C5
1e. R78C45 contains 4 singletons, R7C5 = {79} -> no 8 in R8C4
1f. R89C56 contains 4 singletons, R8C6 = {79} -> no 8 in R9C5
1g. R9C4 = 8 (hidden single in N8), no 7,9 in R9C3 (NC)
1h. R89C45 contains 1,8, R9C4 = 8 -> R8C5 = 1 (diagonally opposite)
1i. R78C56 contains 7,9 and two successive numbers, R8C5 = 1 -> R7C6 = 2, no 1,3 in R6C6 + R7C7 (NC)
1j. R12C56 contains 2,4 and two successive numbers, 2 in R12C5, locked for C5 -> 4 in R12C6, locked for C6 and N2
1k. R12C45 contains 7,9 and two successive numbers with 2 in R12C5 = {1279/2379}, no 5
1l. 5 in N2 only in R3C456, locked for R3, no 6 in R3C5 (NC)
1m. R89C89 contains 2,8 and two consecutive numbers, 8 in R8C89, locked for R8 -> 2 in R9C89, locked for R9 and N9

2a. R45C45 contains 2 doublets, no 1,2 in R45C5 -> no 1 in R45C4
2b. R23C56 contains 2 doublets, R2C5 = {279} -> no 5 in R3C6, R2C6 = {468} -> R3C5 = {35}, R2C6 = {46}
2c. 5 in R3 only in R3C45 -> no 6 in R3C4 (NC)
2d. 6,8 in N2 only in R123C6 = {468}, 6,8 locked for C6
2e. R9C6 = {35} -> no 4 in R9C57 (NC)
2f. 4 in N8 only in R78C4 = {46} (NC), locked for C4, 6 locked for N8, no 5 in R78C3 (NC)
2g. R78C45 contains 4 singletons, R78C4 = {46} -> R7C5 = 9, R8C6 = 7, no 6 in R8C7 (NC)
2h. Naked pair {35} in R9C56, locked for R9
2i. Naked pair {27} in R12C5, 7 locked for N2
2j. 9 in N2 only in R12C4, locked for C4
2k. R8C89 = {58} (hidden pair in N9), 5 locked for R8
2l. R2C6 = {46} -> no 5 in R2C7 (NC)
2m. R3C6 = {68} -> no 7 in R3C7 (NC)
2n. 5 in R7 only in R7C12 -> no 4,6 in R7C12 (NC)
2o. 1 in N2 only in R123C4, locked for C4
2p. 1 on D\ only in R1C1 + R2C2 + R3C3, locked for N1

3a. R23C45 contains 4 singletons, R2C4 + R3C45 must contain at least one of 1,3 -> R2C5 = 7, R1C5 = 2, R1C4 = 9, R2C6 = 4 (both NC), no 8 in R1C3, no 3 in R2C7 (NC)
[Alternatively R23C56 contains 2 doublets, R3C6 = {68} -> R2C5 = 7, R1C5 = 2, R1C4 = 9 (NC), …]
3b. Naked pair {35} in R39C5, locked for C5
3c. R2C4 = {13} -> no 2 in R2C3 (NC)
3d. R1C6 = {68} -> no 7 in R1C7 (NC)
3e. R3C5 = {35} -> no 4 in R4C5 (NC)
3f. R56C56 contains 4 singletons, R6C5 = {46} -> R6C6 = 9, placed for D\, R5C5 = {46}, R5C6 = 1, R4C5 = 8 (hidden single in N5), no 7 in R4C4, no 5 in R56C4, no 2 in R5C7, no 8 in R6C7 (NC)
3g. R4C6 = {35} -> no 4 in R4C7 (NC)
3h. R45C45 contains 2 doublets, R4C5 = 8 -> R5C4 = 7, R5C5 = 4 (NC) -> R4C4 = {35}, no 6,8 in R5C3 (NC)
3i. R5C5 = 4, placed for both diagonals, R6C5 = 6
3j. Naked pair {35} in R4C46, locked for R4, 3 locked for N5 -> R6C4 = 2, placed for D/, no 1,3 in R6C3 (NC)
3k. R4C4 = {35} -> no 4 in R4C3 (NC)
3l. 2 in R8 only in R8C13 -> no 3 in R8C2 (NC)
3m. R78C78 contains 4 singletons, R7C7 = {67} -> no 6,7 in R7C8, no 6 in R8C8, no 7 in R6C7 (NC)
3n. R45C23 contains 2,4 and two successive numbers, R4C2 = 4 -> R5C3 = 2, no 3 in R3C2 (NC)
3o. R8C1 = 2 (hidden single in R8), no 1,3 in R7C1, no 1 in R9C1 (NC)
3p. R3C1 = 4 (hidden single in C1), no 3,5 in R2C1 (NC)

4a. 9 in N9 only in R89C7, locked for C7
4b. R12C78 contains 2 doublets, no 7,9 in R12C7 -> no 8 in R1C8
4c. No 3,5 in R2C7 -> no 4 in R1C8
4d. R1C7 = 4 (hidden single in R1) -> R2C8 = {35}, no 3,5 in R1C8 (NC)
4e. 2 in R12C78 only in R2C7 -> no 1 in R2C7
4f. R2C7 = {268} -> no 6 in R1C8
4g. 9 on D/ only in R8C2 + R9C1, locked for N7
4h. R8C34 = {36/46} (cannot be {34}, NC), 6 locked for R8 -> R8C2 = 9, R8C7 = 3, R9C7 = 9 (hidden single in N9)
4i. Naked quad {1467} in R8C3 + R9C123, locked for N7
4j. R78C78 contains 4 singletons, R8C8 = 3 -> R7C8 = 1
4k. R1C8 = 7
4l. R12C78 contains 2 doublets, R1C8 = 7 -> R2C7 = {68}
4m. R12C89 contains 4 singletons, R1C8 = 7 -> no 6,8 in R12C9

5a. Naked pair {35} in R2C8 + R4C6, locked for D/ -> R1C9 = 1, no 2 in R2C9
5b. R7C3 = 8, placed for D/
5c. R23C7 =  -> R4567C7 = , 7 placed for D\, no 3 in R4C6 (NC)
5d. R78C78 contains 4 singletons, R7C7 = 7 -> R8C8 = 5, placed for D\
5e. R2C8 = 3 -> R3C89 =  (NC)
5f. R4C4 = 3, placed for D\
5g. R9C9 = 6, placed for D\

and the rest is naked singles, without using the diagonals.

Solution:
8 5 3 9 2 6 4 7 1
6 2 9 1 7 4 8 3 5
4 7 1 5 3 8 6 9 2
1 4 7 3 8 5 2 6 9
9 6 2 7 4 1 5 8 3
3 8 5 2 6 9 1 4 7
5 3 8 6 9 2 7 1 4
2 9 6 4 1 7 3 5 8
7 1 4 8 5 3 9 2 6

Top Post subject: Re: Occasional Sindoku (Quadrata's Concept) Posted: Mon Oct 18, 2021 2:43 am Grand Master Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1792
Sindoku dd 30 is another easy one; the NC version is only slightly harder.

Here is how I solved Sindoku dd 30:
Prelims
a) For squares with 2 numbers, delete those 2 numbers from the rest of that nonet
b) For squares with 2 numbers, delete the adjacent numbers in the cells covered by those squares

1a. Naked pair {25} in R78C8, locked for C8 and N9
1b. R8C9 = 9 (hidden single in N9), placed for upper diagonal
1c. 8 in N9 only in R78C7, locked for C7
1d. 7 in N9 only in R9C89, locked for R9
1e. R89C78 contains 4 singletons = {1357/1358}/, no 4 in R9C7
1f. R45C78 contains 1,7 and 2 consecutive numbers = {1347}, 3,4 locked for N6 -> R6C8 = 9
1g. Naked pair {25} in R56C9, locked for C9 and N6 -> R6C7 = 6, placed for upper diagonal, R4C9 = 8
1h. Naked triple {138} in R789C7, 1,3 locked for C7 and N9
1i. R56C78 contains 4 singletons, R6C7 = 6 -> R45C7 = , R45C8 = 
1j. R45C89 contains 4 singletons, R5C8 = 1 -> R56C9 = 
1k. R89C78 contains 4 singletons, R9C7 = {13} -> R8C8 = 5, R9C89 = , 7 placed for lower diagonal, R89C7 = {13}, locked for C7
1l. R7C789 = , 2 placed for upper diagonal -> R5C6 = 7, placed for upper diagonal
1m. R45C56 contains 4 singletons, R5C6 = 7 -> R5C5 = 2, R4C56 = , 5 placed for upper diagonal, R4C4 = 1, R6C456 = , R5C4 = 6, 6,8 placed for lower diagonal

2a. R1C4 + R2C5 =  (hidden pair in N2)
2b. R12C45 contains 1,5 and 2 consecutive numbers = {1578/1589} -> R2C4 = 8, R1C5 = {79}
2c. R12C56 contains 2 doublets, R2C5 = 1 -> R12C6 = {26}, R1C5 = 7
2d. R23C56 contains 4 singletons -> R12C6 = , R3C456 = , 4 placed for upper diagonal
2e. R12C78 contains 4 singletons, R1C7 = 9, R2C8 = 4, R13C8 = , R23C7 = 
2f. R12C89 contains 2 doublets, R12C8 =  -> R12C9 = 
2g. R12C23 contains 4 singletons, R1C2 + R2C3 =  -> R1C3 = 8, R2C2 = 5, R12C1 = 
2h. R23C23 contains 2 doublets, R2C2 = 5 -> R3C3 = 6, R3C12 = , 2 placed for lower diagonal
2i. R4C123 = 
2j. R56C23 contains 4 singletons, R5C3 + R6C2 =  -> R5C12 = 
2k. R89C12 contains 4 singletons -> R8C2 = 4, R79C2 = , R89C1 = {16}, 1 locked for C1 and N7
2l. R7C6 = 1 (hidden single in R7), placed for lower diagonal

and the rest is naked singles, without using the diagonals.

Here is how I solved Sindoku dd 30 NC:
Prelims
a) For squares with 2 numbers, delete those 2 numbers from the rest of that nonet
b) For squares with 2 numbers, delete the adjacent numbers in the cells covered by those squares
Note. Adjacent numbers, including doublets in pink squares, must be diagonally opposite in the squares.

Steps Resulting From Prelims
1a. R8C9 = 9 (hidden single in N9), placed for upper diagonal
1b. R78C89 contains 2,9 and 2 consecutive numbers, R8C9 = 9 -> R78C8 = , 2 placed for upper diagonal
1c. R89C89 contains 7,9 and 2 consecutive numbers, R8C9 = 9 -> R9C8 = 7, R8C8 = 5 -> R9C9 = 4, R7C9 = 6
1d. R78C78 contains 5,8 and 2 consecutive numbers, R8C8 = 5 -> R7C7 = 8
1e. Naked pair {13} in R89C7, locked for C7
1f. R12C12 contains 1,9 and 2 consecutive numbers, no 9 in R1C2 -> no 1 in R2C1
NC: no 7 in R6C7, no 1,3 in R6C8, no 5,7 in R6C9, no 9 in R7C6, no 2 in R89C6

[N5 is a lot easier by doing step 2a first; initially I was looking at interactions with R5C5 = {47}.]
2a. R45C56 contains 4 singletons, R56C56 contains 2,4 and 2 consecutive numbers, R5C56 contains one of 2,4 -> no 3 in R4C56
2b. 3 in N5 only in R456C4, locked for C4
2c. R45C45 contains 2 doublets -> R4C5 = {57}, R5C4 = {68}
2d. R56C45 contains 6,8 and 2 consecutive numbers, R5C4 = {68} -> R6C5 = {68}, naked pair {68} locked for N5 and lower diagonal
2e. R56C56 contains 2,4 and 2 consecutive numbers, R5C5 = {24} -> R6C6 = {24}, naked pair 2,4 locked for N5 -> R5C6 = 7, R4C5 = 5, both placed for upper diagonal
2f. R5C5 = 2 (NC) -> R6C67 = , 6 placed for upper diagonal, R6C5 = 8 -> R5C4 = 6, R6C8 = 9
2g. R56C78 contains 4 singletons, R6C78 =  -> R5C78 = 
2h. R45C56 contains 4 singletons, R5C5 = 2 -> R4C6 = 9
2i. Naked pair {13} in R46C4, 1 locked for C4, no 2 in R46C3 (NC)
2j. R5C9 = 5 (hidden single in N6)
2k. Naked triple {389} in R5C123 -> R5C2 = 3 (NC), R5C13 = {89}, 8 locked for N4, no 2 in R46C2 (NC)
2l. 2 in N4 only in R46C1, locked for C1
2m. 7 in R6 only in R6C123, locked for N4
2n. R7C6 = 1 (NC), placed for lower diagonal -> R8C7 = 3, placed for lower diagonal, R9C7 = 1, R4C3 = 4, placed for lower diagonal, R46C4 =  (NC), R4C12 = , R4C7 = 7 -> R4C89 =  (NC), R6C9 = 2, R3C2 = 2 (NC)
NC: no 1 in R2C2, no 3 in R3C1, no 3,5 in R3C3, no 4,6 in R3C5, no 8 in R3C6, no 4 in R3C8, no 7 in R3C9, no 4 in R7C4, no 7,9 in R7C5

3a. R78C56 contains 4 singletons, R7C5 = {34} -> R8C56 = , no 7 in R8C4 (NC)
3b. R7C4 = 7 (hidden single in N8)
3c. 5,9 in R7 only in R7C123, locked for N7
3d. R89C12 contains 4 singletons, R9C2 = 8 -> R8C12 = {14}, 4 locked for R8 and N7, R9C1 = 6, R8C34 = 
3e. R7C5 = 4 (hidden single in N8)

4a. R12C12 contains 1,9 and 2 consecutive numbers = {1349/1459}
4b. R3C1 = 7 (hidden single in N1)
4c. Naked pair {68} in R3C38, locked for R3 -> R3C4 = 4, placed for upper diagonal, no 5 in R2C4, no 3 in R3C5 (NC)
4d. R1C2 = 1 -> R2C1 = 9, R8C12 = 
4e. R23C23 contains 2 doublets, R23C2 =  -> R23C3 = 
4f. R2C47 =  -> R1C4 = 5, R2C5 = 1 (both NC)
4g. R23C56 contains 4 singletons, R2C6 = 6 -> R3C6 = 3

and the rest is naked singles, without using the diagonals or NC.

Solution:
4 1 8 5 7 2 9 6 3
9 5 3 8 1 6 2 4 7
7 2 6 4 9 3 5 8 1
2 6 4 1 5 9 7 3 8
8 3 9 6 2 7 4 1 5
5 7 1 3 8 4 6 9 2
3 9 5 7 4 1 8 2 6
1 4 7 2 6 8 3 5 9
6 8 2 9 3 5 1 7 4

Top Post subject: Re: Occasional Sindoku (Quadrata's Concept) Posted: Wed Oct 20, 2021 2:11 am Grand Master Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1792
Sindoku X 40 and Sindoku X NC were both fun puzzles; I particularly liked the start of the NC version.

HATMAN wrote:
I have used the non-clues extensively.
We must have different solving styles, I didn't use them for either version.

The way I solved them the 'plain' version was harder; two of my steps 2e and 5b were harder than the rest.

Here's how I solved Sindoku X 40:
Prelims
a) For squares with 2 numbers, delete those 2 numbers from the rest of that nonet
b) For squares with 2 numbers, delete the adjacent numbers in the cells covered by those squares
Note. Groups of squares without clues must have two or three consecutive numbers

1a. Naked pair {28} in R5C45, locked for R5 and N5
1b. R45C45 contains 4,8 and 2 consecutive numbers = {1248} -> R4C45 = {14}, locked for R4, 1locked for N5
1c. 3,6 in N5 only in R456C6, locked for C6
1d. R45C56 contains 4 singletons = {1368/2469}, no 7, 6 locked for N5
1e. 5,7 in N5 only in R6C456, locked for R6

2a. R56C89 contains 4,8 and 2 consecutive numbers = {1248} -> R5C56 = {14}, locked for R5, 1 locked for N6, R6C56 = {28}, locked for R6, 2 locked for N6
2b. R45C78 contains 7,9 and 2 consecutive numbers, no 2 -> R5C8 = 4, R5C9 = 1
2c. R56C78 contains 4 singletons, R5C8 = 4 -> R6C7 = 6, R5C7 = 9 -> R6C8 = 2, R6C9 = 8
2d. Naked triple {357} in R4C789, locked for R4
2e. R45C56 contains 4 singletons = {2469} (cannot be {1368} =  because R56C56, which contains 2 doublets, then doesn’t contain one of 2,4 to form a doublet with 3) -> , R45C4 = , R6C6 = 3 (hidden single in N5), 1,3 placed for D\, 2 placed for both diagonals, 9 placed for D/
2f. 6 in N9 only in R7C89, locked for R7
2g. R78C89 contains 4 singletons, 6 in R7C89 -> no 5,7 -> R8C8 = 9, placed for D\, no 8 in R7C8
2h. R89C78 contains 4 singletons, R8C8 = 9 -> no 8 in R89C7
2i. R7C7 = 8 (hidden single in N9), placed for D\
2j. R78C89 contains 4 singletons, R8C9 = {234} -> no 3 in R7C89
2k. R78C78 contains 2 doublets, R7C8 = {16} -> R8C7 = {25}
2l. 7 in N9 only in R9C89, locked for R9

3a. 9 in C9 only in R23C9
3b. R23C89 contains 4 singletons, 9 in R23C9 -> no 8 in R23C8
3c. R12C89 contains 4 singletons, R1C8 = 8 (hidden single in N3) -> no 7 in R1C9 + R2C8, no 7,9 in R2C9
3d. R3C9 = 9 (hidden single in C9)
3e. R9C9 = 7 (hidden single in C9), placed for D\
3f. Naked triple {456} in R1C1 + R2C2 + R3C3 for D\, locked for N1
3g. 8 on D/ only in R8C2 + R9C1, locked for N7
3h. R89C12 contains 4 singletons including 8 -> no 7 in R8C12, no 9 in R9C2

4a. 5 in N9 only in R8C7 + R9C78
4b. R89C78 contains 4 singletons including 5,9, no 4 = {1359} -> R8C7 = 5, R9C78 = {13}, locked for R9 and N9
4c. R7C8 = 6, naked pair {24} in R78C9, locked for C9
4d. 6 in C9 only in R12C9
4e. R12C89 contains 4 singletons including 6,8 = {1368} -> R12C9 = {36}, 3 locked for C9 and N3, R2C8 = 1, placed for D/
4f. R9C78 = , R4C789 = , R3C8 = 5
4g. R23C89 contains 4 singletons, R3C8 = 5 -> R12C9 = , 6 placed for D/

5a. 3 on D/ only in R7C3 + R8C2, locked for N7
5b. R78C12 contains 4 singletons, R89C23 contains 2 doublets -> R9C1 = 8 (cannot be R7C3 + R8C2 = , no 7,9 in R7C12 which would force R89C3 = , when R9C2 = 6 would make R89C23 =  which is 4 consecutive numbers, not 2 doublets)
5c. R89C12 contains 4 singletons, R8C2 = {34} -> no 4 in R8C1 + R9C2
5d. R78C12 contains 4 singletons, R8C2 = {34} -> no 4 in R7C12
5e. R89C23 contains 2 doublets, no 8 -> no 9 in R9C3
5f. 9 in R9 only in R9C45, locked for N8
5g. R89C45 contains 4 singletons including 9 -> no 8 in R8C5
5h. R8C6 = 8 (hidden single in N8)
5i. R78C56 contains 4 singletons, R8C6 = 8 -> no 7 in R7C56 + R8C5
5j. R89C56 contains 2 doublets, R8C6 = 8 -> R9C5 = 9
5k. 7 in N8 only in R78C4, locked for C4 -> R6C4 = 5, placed for D/, R6C5 = 7

6a. 8 in N2 only in R23C5
6b. R23C45 contains 4 singletons including 8 -> no 9 in R2C4
6c. R1C4 = 9 (hidden single in N2)
6d. R12C45 contains 2 doublets, R1C4 = 9 -> R2C5 = 8

7a. 7 in N8 only in R78C4
7b. R78C45 contains 2 doublets -> R8C5 = 6
7c. R89C45 contains 4 singletons, R8C5 = 6 -> no 7 in R8C4
7d. R89C56 contains 2 doublets, R8C56 = , R9C5 = 9 -> R9C6 = 5
7e. R78C45 contains 2 doublets, R8C5 = 6 -> R7C4 = 7
7f. R89C45 contains 4 singletons -> R89C4 = {24}, locked for C4 and N8 -> R7C56 = , R23C4 = , R13C5 = , R1C1 = 4, R2C2 = 5, R3C3 = 6
7g. R7C3 = 4, placed for D/

and the rest is naked singles, without using the diagonals.

Here's how I solved Sindoku X 40 NC:
Prelims
a) For squares with 2 numbers, delete those 2 numbers from the rest of that nonet
b) For squares with 2 numbers, delete the adjacent numbers in the cells covered by those squares
Notes. Adjacent numbers, including doublets in pink squares, must be diagonally opposite in the squares.
Groups of squares without clues must have two or three consecutive numbers

1a. R89C89 contains 7,9 and 2 consecutive numbers
1b. R78C89 contains 4 singletons, R8C89 must contain one of 7,9 -> no 8 in R7C89
1c. R78C89 contains 4 singletons, R89C8 must contain one of 7,9 -> no 8 in R89C7
1d. R7C7 = 8 (hidden single in N9), placed for D\
1e. R45C45 contains 4,8 and 2 consecutive numbers = {1248} -> R4C5 + R5C4 = {48}, R4C4 + R5C5 = {12}, locked for N5 and D\
1f. R56C56 contains 2 doublets -> R5C5 = 2, placed for D/, R6C6 = 3, placed for D\, R5C6 + R6C5 = {56/67}, 6 locked for N5, R4C4 = 1
1g. R45C56 contains 4 singletons, R5C5 = 2 -> R45C56 = {2469/2479} -> R4C56 = , 9 placed for D/, R5C6 = {67}, R5C4 = 8
1h. R78C78 contains 2 doublets, R7C7 = 8 -> R8C8 = {79}
1i. R78C89 and R89C78 both contain 4 singletons, 6 in N9 only in R7C89 + R89C7 -> R8C8 = 9, placed for D\
1j. R89C89 contains 7,9 and 2 consecutive numbers, R8C8 = 9 -> R9C9 = 7, placed for D\
1k. 5 in N5 only in R6C45 -> R6C45 = {57} (NC), locked for R6, 7 locked for N5 -> R5C6 = 6
1l. Naked triple {456} in R1C1 + R2C2 + R3C3, locked for N1
NC: no 2,5 in R3C4, no 3,5 in R3C5, no 8 in R3C6, no 2,5 in R4C3, no 7,9 in R5C3, no 5,7 in R5C7, no 6 in R6C3, no 2,4,9 in R6C7, no 6 in R7C4, no 6 in R7C5, no 2,4,7 in R7C6

2a. 6 in N8 only in R89C45
2b. R89C45 contains 4 singletons -> no 5,7
2c. R89C56 contains 2 doublets -> R8C5 = {136}, R9C6 = {245}, R9C5 = {13689} -> R8C6 = {24578}
2d. R78C45 contains 2 doublets, R8C5 = {136} -> R7C4 = {2457}, R7C5 = {1357}, R8C4 = {246}
2e. 9 in N8 only in R9C45 -> no 8 in R9C5 (NC)
2f. R89C56 contains 2 doublets, R8C6 = 8 (hidden single in N8) -> R9C5 = 9
2g. R78C56 contains 4 singletons, R8C6 = 8 -> no 7 in R7C5
2h. R78C45 contains 2 doublets, R7C4 = 7 (hidden single in N8) -> R8C5 = 6, R6C45 = , 5 placed for D/
2i. R89C45 contains 4 singletons, R89C5 =  -> R89C4 = {24}, locked for C4 and N8 -> R9C6 = 5, R7C56 = , no 4,6 in R9C7 (NC)
2j. R7C3 = 4 (NC), placed for D/
2k. 6 in N9 only in R7C89 = {26} (cannot be {56}, NC), locked for R7, 2 locked for N9
2l. R78C89 contains 4 singletons, R7C89 = {26}, R8C8 = 9 -> R8C9 = 4, R89C4 = 
2m. R7C3 = 4 -> R7C12 =  (NC)
2n. R89C89 contains 7,9 and 2 consecutive numbers, R8C9 = 4 -> R9C8 = 3, R89C7 = , R6C7 = 6
2o. R78C78 contains 2 doublets, R8C7 = 5 -> R7C8 = 6, R7C9 = 2
NC: no 6 in R3C4, no 6 in R4C3, no 4 in R6C1, no 8 in R6C2, no 1 in R6C9

3a. R89C12 contains 4 singletons, R9C1 = {68} -> R8C12 = {13}, R9C2 = {68}
3b. R89C3 =  (hidden pair in N7)
3c. Hidden killer pair 6,8 in R1C9 + R2C8 and R9C1 for D/, R9C1 = {68} -> R1C9 + R2C8 must contain one of 6,8
3d. R12C89 contains 4 singletons including at least one of 6,8 -> no 7
3e. R3C7 = 7 (hidden single on D/)
NC: no 6,8 in R3C8

4a. R56C78 contains 4 singletons, R6C7 = 6 -> no 5,7 in R5C8
4b. R4C8 = 7 (hidden single in N6), no 8 in R4C9 (NC)
4c. 8 in N6 only in R6C89, locked for R6, no 9 in R6C9 (NC)
4d. 9 in N6 only in R5C79, locked for R5
4e. 6 in R4 only in R4C12, 7 in R5 only in R5C12, R45C12 contains 2 doublets, one must be {67} -> no 8
4f. R4C3 = 8 (hidden single in R4)
4g. Naked triple {139} in R126C3, locked for C3, 3 locked for N1 -> R35C3 = , R1C1 = 4, R2C2 = 5
4h. R12C45 contains 2 doublets, R1C5 = 5 (hidden single in N2) -> R2C4 = 6
4i. R23C45 contains 4 singletons, R23C5 = {18} -> R13C3 = 
4j. R12C45 contains 2 doublets, R1C4 = 9 -> R2C5 = 8, R3C5 = 1
4k. R123C6 =  (NC)
4l. R2C8 = 1, placed for D/ -> R8C2 = 3, placed for D/
4m. R5C8 = 4, R6C89 = 
4n. R1C9 = 6, placed for D/

and the rest is naked singles, without using the diagonals or NC.

Solution:
4 1 3 9 5 7 2 8 6
7 5 9 6 8 2 4 1 3
2 8 6 3 1 4 7 5 9
6 2 8 1 4 9 3 7 5
3 7 5 8 2 6 9 4 1
9 4 1 5 7 3 6 2 8
5 9 4 7 3 1 8 6 2
1 3 7 2 6 8 5 9 4
8 6 2 4 9 5 1 3 7

Top Post subject: Re: Occasional Sindoku (Quadrata's Concept) Posted: Wed Oct 20, 2021 9:11 pm Grand Master Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1792
Sindoku 29 and Sindoku 29 kNNC were also fairly easy.

Here's how I solved Sindoku 29:
Prelims
a) For squares with 2 numbers, delete those 2 numbers from the rest of that nonet
b) For squares with 2 numbers, delete the adjacent numbers in the cells covered by those squares

1a. Each nonet containing either two red squares, diagonally opposite and overlapping in the central square of the nonet with no other shared cells can only contain one of 4,5,6 in the central cell
1b. Each nonet containing one red and one pink square, diagonally opposite and overlapping in the central square of the nonet with no other shared cells can only contain one of 2-8 in the central cell

2a. R2C8 = 5
2b. R12C78 contains 1,3 and 2 consecutive numbers = {1356}, 6 locked for N3
2c. R23C89 contains 7,9 and 2 consecutive numbers = {4579}, 4 locked for N3
2d. R12C89 and R23C78 each contain 4 consecutive numbers -> one must be {2345} and the other {5678} -> R1C7 = 1, R3C9 = 9

3a. R8C2 = 5
3b. R78C12 contains 1,3 and 2 consecutive numbers = {1356}, 6 locked for N3
3c. R89C23 contains 7,9 and 2 consecutive numbers = {4579}, 4 locked for N3
3d. R78C23 and R89C12 each contain 4 consecutive numbers -> one must be {2345} and the other {5678} -> R7C1 = 1, R9C3 = 9

4a. R56C89 contains 4,6 and 2 consecutive numbers = {1246}, 1,2 locked for N6
4b. R45C89 contains 4 consecutive numbers = {2345/3456/4567} (cannot be {1234} because 1,2,4 only in R5C89), no 1,8,9
4c. R6C7 = 8 (hidden single in N6)
4d. 9 in N6 only in R45C7, locked for C7
4e. R23C78 contains 4 consecutive numbers, R3C7 = 2 -> R2C7 = 3, R3C8 = 4, R1C89 = , R2C9 = 7
4f. R56C8 = 
4g. R45C89 contains 4 consecutive numbers, R5C8 = 2 -> R4C89 = , R5C9 = 4 -> R6C9 = 6

5a. R7C8 = 9 (hidden single in C8)
5b. R78C78 contains 2 doublets -> R89C8 = 
5c. R89C78 contains 4 consecutive numbers, R89C8 =  -> R89C7 = , R7C7 = 4
5d. R89C12 contains 4 consecutive numbers, R89C2 = , R8C1 = 3 -> R9C1 = 2, R7C23 = , R8C3 = 7
5e. R5C2 = 8, R4C23 = , R45C1 = 
5f. R56C12 contains 4 consecutive numbers, R5C12 =  -> R6C12 = , R56C3 = 
5g. R123C2 = 
5h. R12C23 contains 4 consecutive numbers, R12C2 =  -> R12C3 = , R3C3 = 6

6a. R23C45 contains 4 consecutive numbers = {5678/6789}, no 1,3
6b. R23C6 =  (hidden pair in N2)
6c. R5C5 = 5 -> R3C4 = 5 (hidden single in N2)
6d. R23C45 = {5678} -> R3C5 = 7
6e. R12C56 contains 2 doublets including 1 = {1289} -> R1C56 = {29}, R2C5 = 8, R12C4 = 

7a. R45C56 contains 4 consecutive numbers = {5678}, R5C5 = 5 -> R4C56 = , R5C6 = 7
7b. R56C45 contains 4 consecutive numbers = {2345}, R5C5 = 5 -> R5C4 = 3, R6C45 = 

and the rest is naked singles.

Hint:
For the kNNC version remember to do those eliminations, not just based on placed numbers but also for cells which only have two remaining candidates two apart, for example {79} eliminates 8 from cells a knight's move away.

Here's how I solved Sindoku 29 kNNC:
Prelims
a) For squares with 2 numbers, delete those 2 numbers from the rest of that nonet
b) For squares with 2 numbers, delete the adjacent numbers in the cells covered by those squares
Note: I’ll use the shorter abbreviation N for anti-knight NC.

1a. Each nonet containing either two red squares, diagonally opposite and overlapping in the central square of the nonet with no other shared cells can only contain one of 4,5,6 in the central cell
1b. Each nonet containing one red and one pink square, diagonally opposite and overlapping in the central square of the nonet with no other shared cells can only contain one of 2-8 in the central cell

2a. R2C8 = 5
2b. R12C78 contains 1,3 and 2 consecutive numbers = {1356}, 6 locked for N3
2c. R23C89 contains 7,9 and 2 consecutive numbers = {4579}, 4 locked for N3
2d. R12C89 and R23C78 each contain 4 consecutive numbers -> one must be {2345} and the other {5678} -> R1C7 = 1, R3C9 = 9
N: no 2 in R2C5 + R3C6, no 4,6 in R13C6,

3a. R56C89 contains 4,6 and 2 consecutive numbers = {1246}, 1,2 locked for N6
3b. R45C89 contains 4 consecutive numbers = {1234/2345/3456/4567}, no 8,9
3c. R6C7 = 8 (hidden single in N6)
3d. R23C78 contains 4 consecutive numbers, R2C8 = 5, R3C7 = 2 -> R2C7 = 3, R3C8 = 4, R1C89 = R1C89 = , R2C9 = 7, R56C8 = 
3e. R45C89 contains 4 consecutive numbers, R5C8 = 2 -> R4C89 = , R5C9 = 4, R6C9 = 6
3f. Naked pair {79} in R45C7, locked for C7
3g. R6C9 = 6 -> R8C8 = 8 (N)
3h. R89C78 contains 4 consecutive numbers = {5678} -> R9C8 = 7, R7C8 = 9, R7C7 = 4 (hidden single in N9)
N: no 2,4 in R1C5 + R4C6, no 3 in R1C6, no 7,9 in R48C6, no 1,3,5 in R5C6, no 3,5 in R68C5 + R9C6, no 7 in R7C5, no 2,3 in R8C9
R4C7 = {79} -> no 8 in R2C6 + R3C5
R5C7 = {79} -> no 8 in R3C6 + R4C5

4a. R45C56 contains 4 consecutive numbers = {4567/5678/6789}, no 1,2, 6,7 locked for N5
4b. R56C45 contains 4 consecutive numbers = {2345/3456}, no 1,7,8,9, 3 locked for C4 and N5, 4,5 locked for N5
4c. R4C4 = 1 (hidden single in N5) -> R6C5 = 4 (N)
4d. 2 in N5 only in R6C46, locked for R6
4e. 8 in N5 only in R45C6, locked for C6
4f. 7 in R6 only in R6C123, locked for N4
N: no 2 in R2C3, no 3,5 in R57C3 + R8C6, no 5 in R8C4
R2C5 = {68} -> no 7 in R13C3
R4C6 = {68} -> no 7 in R3C4
R5C4 = {35} -> no 4 in R4C2

5a. 8 in N2 only in R2C45 + R3C4
5b. R23C45 contains 4 consecutive numbers = {5678/6789} -> R3C5 = 7, 6 locked for N2
5c. R45C6 =  (hidden pair in N5) -> R45C7 = 
5d. 1 in N2 only in R23C6, locked for C6
N: no 6,8 in R2C3, no 6 in R4C3, no 6 in R5C5

6a. R5C5 = 5 -> R56C4 = , R6C6 = 9, R4C5 = 6
6b. R12C56 contains 2 doublets, R2C5 = 8 -> R1C5 = 9
6c. R23C45 contains 4 consecutive numbers, R23C5 =  -> R23C4 = , R1C4 = 4, R12C6 = , R3C6 = 3
6d. R78C56 contains 4 consecutive numbers, R7C6 = 5 (hidden single in C6), R8C5 = 2 -> R7C5 = 3, R8C6 = 4, R9C56 = , R89C7 = 
6e. R78C23 contains 4 consecutive numbers, R8C2 = 5 -> R8C3 = 7, R7C23 = {68}, locked for R7, R789C4 = 
6f. R89C12 contains 4 consecutive numbers, R8C2 = 5 -> R8C1 = 3, R9C12 = {24}, locked for N7 -> R9C3 = 9, R2C23 = 
6g. R45C23 contains 2 doublets, R4C23 =  -> R5C23 = 
6h. R56C12 contains 4 consecutive numbers, R5C12 =  -> R6C12 = 

and the rest is naked singles, without using N.

Solution:
7 3 5 4 9 2 1 6 8
9 2 4 6 8 1 3 5 7
8 1 6 5 7 3 2 4 9
4 9 2 1 6 8 7 3 5
6 8 1 3 5 7 9 2 4
5 7 3 2 4 9 8 1 6
1 6 8 7 3 5 4 9 2
3 5 7 9 2 4 6 8 1
2 4 9 8 1 6 5 7 3

Top Post subject: Re: Occasional Sindoku (Quadrata's Concept) Posted: Sat Oct 23, 2021 2:28 am Grand Master Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1792
And now Sindoku 44 S&OL and Sindoku 44 S&OL FNC.

As with the other pairs of puzzles, the non-consecutive version was the better puzzle with only one harder-to-spot step. The 'plain' version was the hardest of these puzzles that I've done. At the end I thought it needed another clue until I remembered that HATMAN had used non-clues for a pair of puzzles; I finished the 'plain' version with a pair of non-clues.

Here's how I solved Sindoku 44 S&OL:
Prelims
a) For squares with 1 or 2 numbers, delete those numbers from the rest of that nonet
b) For squares with 1 or 2 numbers, delete the adjacent numbers in the cells covered by those squares
Notes. Old lace: R37C5 must be the same as R5C46, R5C37 must be the same as R46C5
Groups of squares without clues must have two or three consecutive numbers

1a. R4C78 =  (hidden pair in N6)
1b. R6C78 = {58} (hidden pair in N6), locked for R6
1c. R45C78 contains 6,9 and 2 consecutive numbers = {1269/2369}, 2 locked for R5 and N6
1d. R45C89 contains 2 and 3 consecutive numbers = {2349} (cannot be {1239} which clashes with R5C7) -> R5C8 = 2, R45C9 = {34}, locked for C9, R5C7 = 1, placed for old lace, R6C9 = 7
1e. R4C35 =  (hidden pair in R4)
1f. R6C6 = 3 (hidden single in R6), placed for old lace
1g. 1 in N3 only in R1C89, locked for R1
1h. R12C89 contains 2 doublets, 1 in R1C89 -> R2C9 = 2
1i. R12C78 contains 2 doublets, no 2 -> no 1 in R1C8
1j. R1C9 = 1 (hidden single on R1), placed for upper snake

2a. R45C12 contains 3,5 and 2 consecutive numbers = {3578/3589}, 8 locked for N4
2b. R56C23 contains 1,4 and 2 consecutive numbers = {1467/1478/1489}
2c. 7 of {1467} must be in R5C2, 8 of {1478/1489} must be in R5C2 -> R5C2 = {78}
2d. 4 in N4 only in R5C3 + R6C23, CPE using old lace, no 4 in R6C4

3a. R89C4 = {13} (hidden pair in N8), locked for C4
3b. R89C45 contains 1,3 and 2 consecutive numbers = {1378/1389}, no 5, 8 locked for C5 and N8
3c. R89C6 = {25} (hidden pair in N8), locked for C6
3d. R3C6 = 1 (hidden single in C6)
3e. Naked pair {13} in R89C4, CPE using lower snake no 1,3 in R9C1
3f. Naked pair {25} in R89C4, CPE using lower snake no 2,5 in R9C1, no 5 in R9C9
3g. 8 in C5 only in R89C5, CPE using lower snake, no 8 in R8C28
3h. R23C45 contains 6,8 and 2 consecutive numbers = {2368/3468} -> R2C5 = 3, 8 locked for C4
3i. 8 in N2 only in R23C4, CPE using upper snake, no 8 in R3C37

4a. R4C6 = 8 (hidden single in old lace)
4b. 5 in old lace only in R4C4 + R5C5, locked for N5
4c. 4,6 in N8 only in R7C456, locked for R7
4d. 2 in old lace only in R3C5 + R6C4, CPE no 2 in R6C5
4e. R6C4 = 2 (hidden single in N5), placed for old lace
4f. Naked quad {3468} in R23C45, 4 locked for N2
4g. Naked pair {79} in R12C6, locked for C6 and N2 -> R1C45 = 
4h. Naked pair {79} in R12C6, CPE using upper snake, no 7,9 in R1C1
4i. R5C5 = 5 (hidden single in C5)

5a. R56C45 contains 2,9 and 2 consecutive numbers with R5C5 = {2459/2569}
5b. R4C4 = 7 (hidden single in N5), placed for old lace, R4C12 = {35}, 3 locked for R4 and N4 -> R45C9 = 
5c. Naked triple {469} in R7C456, 9 locked for R7 and N8
5d. Naked double {78} in R89C5, CPE using lower snake, no 7 in R8C8
5e. 8 in lower snake only in R7C7 + R9C59, CPE no 8 in R9C78
5f. 9 in lower snake only in R9C19, locked for R9
5g. 4,6 in N2 only in R2C4 + R3C45, CPE using upper snake, no 4,6 in R3C3, no 4 in R3C7
5h. 3 in upper snake only in R1C1 + R3C37, CPE no 3 in R3C12
5i. R5C12 = {78} (hidden pair in N4)

6a. 9 in N7 only in R8C3 + R9C1, R78C23 and R89C12 contain 2 doublets -> 8 in R7C2 + R8C1, locked for N7
6b. 1 in lower snake only in R7C3 + R8C248, CPE no 1 in R8C13
6c. 1 in R8 only in R8C3248, CPE using lower snake, no 1 in R7C3
6d. 2 in lower snake only in R7C7 + R8C26, CPE no 2 in R8C7
6e. 2 in R7 only in R7C127, CPE using lower snake, no 2 in R8C2
6f. Consider placement for 2 in R7C127
2 in R7C12, locked for N7
or R7C7 = 2, placed for lower snake => R8C6 = 2 (hidden single in N8)
-> no 2 in R9C2
6g. R78C12 contains 8 and 3 consecutive numbers including 2 = {1238/2348}, no 5,6, 3 locked for N7
6h. R89C23 contains 2 doublets, no 8 -> no 9 in R8C3
6i. R9C1 = 9 (hidden single in N7)
6j. R89C12 contains 2 doublets, R9C1 = 9 -> R8C1 = 8, R89C2 = [34/45], 4 locked for C2 and N7, R5C12 = , R8C5 = 7 -> R9C5 = 8, placed for lower snake, R9C9 = 6
6k. R8C3 = 6 (hidden single in N7)
6l. 6 in R5 only in R5C46, locked for N5
6m. 3 in lower snake only in R7C7 + R8C248, CPE no 3 in R8C7
6n. 8 in upper snake only in R2C48, locked for R2
6o. R1C3 = 8 (hidden single in C3)
6p. R3C3 = 3 (hidden single in C3)
6q. 4 in upper snake only in R1C1 + R2C48, CPE no 4 in R2C13
6r. 5 in upper snake only in R2C28 + R3C7, CPE no 5 in R2C7

7a. R89C23 contains 2 doublets, no 2 -> no 1 in R8C3
7b. Naked pair {57} in R79C3, locked for C3, 5 locked for N7 -> R89C2 = , 3 placed for lower snake, R7C12 = {12}, locked for R7
7c. R4C12 = , R89C4 = 
7d. Naked triple {459} in R8C789, 5 locked for R5 and N9 -> R7C789 = , R79C3 = , 5 placed for lower snake, R8C8 = 4
7e. Naked pair {59} in R38C7, locked for C7 -> R12C7 = , R6C78 = 
7f. R1C1 + R3C7 =  (hidden pair in upper snake)
7g. R2C13 =  (hidden pair in N1)
7h. R23C89 contains 2 and 3 consecutive numbers, R3C9 = 9 -> R23C8 = {78}, locked for C8
7i. R12C78 contains 2 doublets, R12C7 =  -> R12C8 = 

[Then non-clues to finish]
8a. R78C45 must have at least 2 consecutive numbers, R8C45 =  -> must contains 6, locked for N8 -> R8C6 = 4, R5C6 = 6
8b. R45C45 must have at least 2 consecutive numbers, R4C45 =  -> R5C45 = 

and the rest is naked singles, without using the snakes and old lace.

Here's how I solved Sindoku 44 S&OL FNC:
Prelims
a) For squares with 1 or 2 numbers, delete those numbers from the rest of that nonet
b) For squares with 1 or 2 numbers, delete the adjacent numbers in the cells covered by those squares
Note. FNC diagonally none consecutive; the 2 doublets in pink squares must be consecutive horizontally or vertically, not diagonally

1a. R4C78 =  (hidden pair in N6)
1b. R6C78 = {58} (hidden pair in N6), locked for R6
1c. R45C78 contains 6,9 and 2 consecutive numbers = {1269/2369}, 2 locked for R5 and N6
1d. R56C78 contains 9 and 3 consecutive numbers = {2349} (cannot be {1239} which clashes with R5C7) -> R5C8 = 2, R45C9 =  (FNC), R5C7 = 1, placed for old lace, R6C9 = 7
1e. 1 in N5 only in R46C5, locked for C5
1f. R23C89 contains 2 and 3 consecutive numbers -> 2 in R23C9, locked for N3
1g. 1 in N3 only in R1C89, locked for R1
1h. R12C89 contains 2 doublets, 1 in R1C89 -> R2C9 = 2
1i. R2C9 = 2 -> no 1,3 in R1C8 (FNC)
1j. R1C9 = 1 (hidden single in N3), placed for upper snake
1k. 3 in N3 only in R123C7, locked for C7
1l. 1 in N2 only in R3C46, locked for R3
FNC: no 2 in R13C2, no 5,7 in R3C68 + R5C6, no 8 in R3C79, no 2 in R4C5, no 2 in R46C6, no 6,8 in R7C8

2a. R23C89 contains 2 and 3 consecutive numbers = {2567/2789} (cannot be {2456/2678}, FNC) -> R2C8 = 7, placed for upper snake, R3C89 = [65/89]
2b. R12C89 contains 2 doublets, R12C9 = , R2C8 = 7 -> R1C8 = {68}
2c. Naked pair {68} in R13C8, locked for C8 -> R6C78 = , no 5,9 in R2C8 (FNC)
2d. R12C78 contains 2 doublets, R12C8 = [67/87] -> R12C7 = {34}/, 4 locked for C7
2e. R23C78 contains 2 doublets, R23C8 = [67/87] -> R23C7 = , 5 placed for upper snake, R1C7 = 3, R3C9 = 9
2f. R23C89 = {2789} -> R13C8 = 
FNC: no 5 in R1C6, no 6 in R2C6, no 3 in R3C6, no 9 in R5C6, no 7,9 in R7C6, no 6 in R7C9

3a. R78C12 contains 8 and 3 consecutive number, 8 locked for N7
3b. 9 in N9 only in R78C3 + R9C123
3c. R78C23, R89C12 and R89C23 all contain 2 doublets -> 8 must be in R7C2 + R8C12, locked for N7

4a. 2 in old lace only in R37C5 + R46C4, CPE no 2 in R46C5
4b. 2 in N5 only in R46C4, locked for C4 and old lace
4c. 3 in old lace only in R37C5 + R46C46, CPE no 3 in R46C5
4d. 3 in N5 only in R46C46, locked for old lace
4e. 6 in C9 only in R89C9, CPE using lower snake, no 6 in R8C246
4f. 2 in upper snake only in R1C15 + R3C3, CPE no 2 in R1C3
4g. 3 in upper snake only in R2C246 + R3C3, CPE no 3 in R2C13
4h. 2 in N1 only in R1C1 + R3C13 -> no 3 in R2C2 (FNC)
4i. 3 in upper snake only in R2C46 + R3C3, CPE no 3 in R3C4
4j. 4 in upper snake only in R1C15 + R3C3, CPE no 4 in R1C23
4k. 5 in N2 only in R1C4 + R2C5 -> no 4 in R1C4, no 6 in R2C5 (FNC)
4l. 6 in upper snake only in R2C24 + R3C3, CPE no 6 in R2C13
4m. 6 in R2 only in R2C24, locked for upper snake, no 5,7 in R1C3 (FNC)
4n. R1C3 = {89} -> R2C2 = 6 (FNC), no 7 in R3C1 (FNC)
4o. 7 in N1 only in R13C2, locked for C2, no 8 in R2C13 (FNC)
4p. 8 in N1 only in R1C123, locked for R1

[A harder step to find …]
5a. R2C46 = {39/89} (cannot be {38} which clashes with R1C5, FNC), 9 locked for R2, N2 and upper snake
5b. Naked pair {15} in R2C13, 5 locked for R2
5c. R1C5 = {24} -> no 3 in R2C46 (FNC)
5d. Naked pair {89} in R2C46, locked for N2, 8 locked for upper snake
5e. Naked pair {24} in R1C15, locked for R1 and upper snake -> R1C46 = , R3C3 = 3, no 2 in R4C4 (FNC)
5f. Naked pair {24} in R13C1, locked for C1, 4 locked for N1 -> R3C2 = 7
5g. R2C5 = 3 -> no 4 in R3C4, no 2,4 in R3C6 (FNC)
5h. R1C5 = 2 (hidden single in N2)
5i. R3C5 = 4 (hidden single in N2), placed for old lace, no 3 in R4C4, no 3,5 in R4C6 (FNC)
5j. 4 in N5 only in R5C46, locked for R5, no 5 in R4C5 (FNC)
5k. R4C456 = , 7,8 placed for old lace, R2C46 = 
5l. R5C5 = 5 (hidden single in N5)
5m. R6C46 =  (hidden pair in N5)
5n. R5C12 =  (hidden pair in N4) -> R1C23 = 
5o. R5C3 = 9 (FNC), placed for old lace -> R67C5 = 
5p. R4C3 = 2 (FNC) -> R3C46 =  (FNC), R5C46 = , R2C13 =  (FNC), R4C12 = , R67C1 = , R6C23 = , R89C1 = , 9 placed for lower snake, R89C5 = 
FNC: no 3 in R7C24, no 2 in R7C7, no 5 in R8C6

6a. R7C7 = 7 -> R7C3 = 5, placed for lower snake
6b. R9C7 = 2 -> R8C8 = 4 (FNC)

and the rest is naked singles, without using FNC, the snakes and old lace.

Solution:
4 9 8 5 2 7 3 6 1
5 6 1 8 3 9 4 7 2
2 7 3 6 4 1 5 8 9
3 5 2 7 1 8 6 9 4
7 8 9 4 5 6 1 2 3
6 1 4 2 9 3 8 5 7
1 2 5 9 6 4 7 3 8
8 3 6 1 7 2 9 4 5
9 4 7 3 8 5 2 1 6

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