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 Post subject: Meandoku 21
PostPosted: Sat Aug 25, 2018 10:28 pm 
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Grand Master
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Joined: Wed Apr 30, 2008 9:45 pm
Posts: 549
Meandoku NC X 21 and Meandoku NC 21

Meandoku (strictly extra) The following colour clues apply:
Green: the sum of the two adjacent cells is 8 or 9 only
Blue: the sum of the two adjacent cells is ten
Red: the sum of the two adjacent cells is 11 or 12 only
Yellow: the sum of the two cells is below eight
Black: the sum of the two cells is above twelve

I did the X first. I got a random solution out of JSudoku, filled in N1 and then the black clue in N5 which made it unique; but not of course solvable. I added the other three clues in N5 which gave 7 solutions by a quite difficult route. Then I had one of my common problems: when I added the two clues in N3 it solved but on going through it again there was now a simpler route. This made the puzzle easier than I wanted.

Hence I removed the X and added a couple of clues to make it unique again. Luckily it then solved without adding anymore clues and at acceptable difficulty.

As I've been out of the country and out of contact for three months I decided to post both.

As a minor challenge you will note that nonet one is very easy and I would like to improve the puzzle. How many clues can you reduce it to so that it still fully solves? Where there are two answers with the same number I will of course prefer the more complex one.

Good to be back here

Maurice

Meandoku NC X 21
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Meandoku NC 21
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 Post subject: Re: Meandoku 21
PostPosted: Fri Mar 01, 2019 2:07 am 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1624
Location: Lethbridge, Alberta, Canada
Continuing with the MeanDokus, NC X 21 was about the same level as NC X 15. I agree that N1 was easier than N5 in NC X 16 Extra or N69 in NC X 17 Strictly Extra. I haven't attempted HATMAN's challenge to simplify the clues in N1, but I have attempted to simplify my steps there.

Here is my walkthrough for MeanDoku NC X 21:
Cells adjacent to green marks must total 8 or 9, blue must total 10, red must total 11 or 12, yellow total less than 8, black total more than 12.
Also NC so horizontally/vertically adjacent cells cannot be {12}, {23}, … {78}, {89}, therefore at least one of the cells adjacent to each green or yellow mark must contain one of 1,2,3 as green cannot be {45}; similarly at least one of the cells adjacent to each red or black mark must contain one of 7,8,9 as red cannot be {56}.

Prelims.
Delete 9 from cells either side of green marks.
Delete 7,8,9 from cells either side of yellow marks.
Delete 5 from cells either side of blue marks.
Delete 1 from cells either side of red marks.
Delete 1,2,3 from cells either side of black marks.
Also delete 4 from R1C1 (two black marks)
Also delete 2 from R3C3 (two red marks)
Also delete 8 from R4C4 (two green marks)
Delete 4 from cells next to green marks (NC)
Delete 6 from cells next to red marks (NC)
That one made also delete 6 from cells with two yellow marks unnecessary.
Clean-ups and NCs only as stated.

1a. R3C1 = 1 (hidden single in N1) -> R2C1 = {78} (green)
1b. R1C1 = 6 (hidden single in N1), placed for D\, no 7 in R1C2 + R2C1 (NC)
1c. R2C1 = 8 -> R2C2 = 2 (blue), placed for D\
1d. R1C1 = 6 -> R1C2 = 9 (black) -> R1C3 (red), R2C3 = 5, R3C2 = 4, R3C3 = 7, placed for D\
Clean-ups: no 2,4 in R1C4, no 4,6 in R2C4, no 6,8 in R3C4 + R4C3, no 2 in R4C1, no 3,5 in R4C2 (NC)

2. R3C78 = [25] (yellow), 2 placed for D/
Clean-ups: no 1,3 in R24C7, no 4,6 in R24C8, no 3 in R3C6, 6 in R3C9, then no 9 in R2C9, no 8,9 in R4C9 (NC)
2a. 3,6 in R3 only in R3C456, locked for N2
2b. 3 in R2 only in R2C89 -> no 4 in R2C9 (NC)
2c. 2 in R1 only in R1C56 -> no 1 in R1C56 (NC)

3a. R45C4 must contain one digit less than 4 (green) -> R4C4 = {13}
3b. R4C45 can only contain one digit less than 4 (green) -> no 2,3 in R4C5
3c. R4C5 = {78}, R5C5 = {45} (red)
3d. R4C5 = {78} -> R4C4 = 1 (green), placed for D\
3e. R4C4 = 1 -> R5C4 = {78} (green)
3f. R5C45 = [85] (black), 5 placed for both diagonals, R4C5 = 7
Clean-ups: no 6,8 in R3C5, no 2 in R4C3, no 6 in R4C6, no 9 in R5C3 + R6C4, no 4,6 in R5C6 + R6C5 (NC)
3g. R6C4 = 6 (hidden single in N5), placed for D/
3h. Naked pair {39} in R3C45, locked for R3 and N2 -> R12C4 = [57], R3C69 = [68]
3i. 4 in N5 only in R46C6, locked for C6 -> R2C45 = [41], no 2 in R1C6, no 3 in R3C5 (NC)
3j. R1C56 = [28], no 7 in R1C7 (NC)
3k. R3C45 = [39] -> R6C5 = 3, no 4 in R6C6 (NC)
3l. R6C6 = 9, placed for D\, R4C6 = 4, placed for D/, R5C6 = 2, R4C3 = 9, no 8 in R4C2, no 5 in R4C7, no 1,3 in R5C7 (NC)
3m. R4C2 = 6, no 5 in R4C1, no 7 in R5C2 (NC)
3n. R4C1 = 3, R5C123 = [714]
3o. R9C1 = 9, placed for D/, R2C8 = 3, placed for D/, R2C9 = 6, no 7 in R1C9 (NC)
3p. R1C9 = 1, placed for D/
3q. R7C3 = 8, R8C2 = 7, R6C13 = [52], no 4 in R7C1, no 9 in R7C4, no 6 in R8C3 (NC)
3r. R7C1 = 2, no 3 in R7C2 (NC)

The rest is naked singles, without using NC, diagonals or coloured marks.

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


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 Post subject: Re: Meandoku 21
PostPosted: Sun Mar 03, 2019 3:55 am 
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Grand Master
Grand Master

Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1624
Location: Lethbridge, Alberta, Canada
And now MeanDoku NC 21. Definitely a harder version but only for the final part of my solving path; at the right level for this forum. It's impressive that only two blue marks were required to replace the X, particularly as I only used the one at R7C67 to remove 5 from these cells.

Here is my walkthrough for MeanDoku NC 21:
Cells adjacent to green marks must total 8 or 9, blue must total 10, red must total 11 or 12, yellow total less than 8, black total more than 12.
Also NC so horizontally/vertically adjacent cells cannot be {12}, {23}, … {78}, {89}, therefore at least one of the cells adjacent to each green or yellow mark must contain one of 1,2,3 as green cannot be {45}; similarly at least one of the cells adjacent to each red or black mark must contain one of 7,8,9 as red cannot be {56}.

Prelims.
Delete 9 from cells either side of green marks.
Delete 7,8,9 from cells either side of yellow marks.
Delete 5 from cells either side of blue marks.
Delete 1 from cells either side of red marks.
Delete 1,2,3 from cells either side of black marks.
Also delete 4 from R1C1 (two black marks)
Also delete 2 from R3C3 (two red marks)
Also delete 8 from R4C4 (two green marks)
Delete 4 from cells next to green marks (NC)
Delete 6 from cells next to red marks (NC)
That one made also delete 6 from cells with two yellow marks unnecessary.
Clean-ups and NCs only as stated.

1a. R3C1 = 1 (hidden single in N1) -> R2C1 = {78} (green)
1b. R1C1 = 6 (hidden single in N1), no 7 in R1C2 + R2C1 (NC)
1c. R2C1 = 8 -> R2C2 = 2 (blue)
1d. R1C1 = 6 -> R1C2 = 9 (black) -> R1C3 (red), R2C3 = 5, R3C2 = 4, R3C3 = 7
Clean-ups: no 2,4 in R1C4, no 4,6 in R2C4, no 6,8 in R3C4 + R4C3, no 2 in R4C1, no 3,5 in R4C2 (NC)

2. R3C78 = [25] (yellow)
Clean-ups: no 1,3 in R24C7, no 4,6 in R24C8, no 3 in R3C6, 6 in R3C9, then no 9 in R2C9, no 8,9 in R4C9 (NC)
2a. 3,6 in R3 only in R3C456, locked for N2
2b. 3 in R2 only in R2C89 -> no 4 in R2C9 (NC)
2c. 2 in R1 only in R1C56 -> no 1 in R1C56 (NC)

3a. R45C4 must contain one digit less than 4 (green) -> R4C4 = {123}
3b. R4C45 can only contain one digit less than 4 (green) -> no 2,3 in R4C5
3c. No 9 in R5C4 -> no 4 in R5C5 (black)
3d. R45C5 = {57} (red, only possible combination), locked for C5 and N5, no 6 in R3C5 + R4C6 + R5C46 + R6C5 (NC)
3e. R5C45 = [85] (black + NC), R4C4 = 1 (green), R4C5 = 7
Clean-ups: no 8 in R3C5, no 2 in R4C3, no 9 in R5C3 + R6C4, no 4 in R5C6 + R6C5 (NC)
3f. Naked pair {39} in R3C45, locked for R3 and N2 -> R2C4 = 7, R3C69 = [68], no 8 in R1C4 (NC)
3g. R1C4 = 5
3h. R6C4 = 6 (hidden single in N5)
3i. 4 in N5 only in R46C6, locked for C6 -> R2C56 = [41], no 2 in R1C6, no 3 in R3C5 (NC)
3j. R1C56 = [28], R3C45 = [39], R6C5 = 3, no 7 in R1C7, no 2,4 in R6C6 (NC)
3k. R456C6 = [429], R4C3 = 9, R5C4 = 6, no 8 in R4C2, no 5 in R4C7, no 1,3 in R5C7, no 8 in R6C7 (NC)
3l. R4C2 = 6, R4C7 = 8, no 5 in R4C1, no 7 in R5C2, no 7,9 in R6C7 (NC)
3m. R4C189 = [325], R5C123 = [714], R5C7 = 6, R2C789 = [936], R5C89 = [93], no 4 in R1C8, no 7 in R1C9, no 7 in R6C7, no 4 in R6C9 (NC)
3n. R1C8 = 7 (hidden single in N3) -> R6C9 = 7 (hidden single in N6)
3o. Naked pair {14} in R16C7, locked for C7

4a. Naked pair {37} in R7C67, locked for R7 (blue had been used in the Prelims to delete 5)
4b. Naked pair {58} in R67C2, locked for C2
4c. Naked triple {357} in R8C267, locked for R8
4d. Naked triple {357} in R9C267, locked for R9
4e. 5 in N7 only in R7C12 -> no 4 in R7C1
4f. 4 in N7 only in R89C1, R89C2 = {37} -> combined cage R89C12 = [4793/9347] (NC) -> R67C1 = {25}
[Alternatively 4 in N7 only in R89C1 = {49} (cannot be {24} which would eliminate 3 from N7, NC), 9 locked for C1 -> R67C1 = {25}]
4g. 1 in N7 only in R789C3 -> no 2 in R8C3 (NC)
[Alternatively this can be done with the final blue mark, which I haven’t yet used.]
4h. 2 in N9 only in R789C9 -> no 1 in R8C9 (NC)

5. R789C6 = {357}, R789C7 = {357} with no 5 in R7C67
5a. Either R89C6 = {57}, no 6 in R89C5 (NC) => R7C5 = 6 (hidden single in N8)
or R89C7 = {57}, no 6 in R89C8 (NC) => R7C8 = 6 (hidden single in N9)
-> 6 in R7C58, locked for R7
5b. Taking that further R7C67 = [37] => R7C8 = {14} (NC) => R7C5 = 6
or R7C67 = [73] => R7C5 = 1, R7C8 = 6
-> no 8 in R7C58
5c. 8 in R7 only in R7C23, locked for N7
5d. R8C34 = [19/64] (blue), no 2
[Cracked, the rest is straightforward.]
5e. Naked pair {49} in R8C14, locked for R8 -> R8C9 = 2, no 1 in R79C9 + R8C8 (NC)
5f. Naked pair {49} in R79C9, 4 locked for C9 and N9 -> R1C79 = [41], R6C78 = [14]
5g. Naked pair {16} in R7C58, 1 locked for R7
5h. Naked triple {258} in R7C123, 2 locked for R7 and N7
5i. R9C4 = 2 (hidden single in N8), no 1 in R9C35 (NC)
5j. R9C3 = 6, no 7 in R9C2 (NC)
5k. R8C3 = 1 -> R8C4 = 9 (blue), no 2 in R7C3 (NC)
5l. R89C5 = [68], no 5,7 in R8C6 (NC)

The rest is naked singles, without using NC or coloured marks.


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