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 Post subject: MeanDoku 4 & 5
PostPosted: Fri Jul 07, 2017 3:41 pm 
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MeanDoku NC 4

The following colour clues apply:
Green: the average of the two adjacent cells is below five
Blue: the average of the two adjacent cells is five
Red: the average of the two adjacent cells is above five

NC: horizontally and vertically adjacent cells are non-consecutive.

Quite hard

With NC, I think that for a full set of 12 clues in a nonet there is normally (always?) a unique solution.


Image

MeanDoku NC 5

A hard NC finish
Image


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 Post subject: Re: MeanDoku 4 & 5
PostPosted: Tue Jul 11, 2017 11:16 pm 
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:cheers:
Thanks, Hatman, I'm looking forward to rying* these while on vacation next week!

* subliminal typo, could be "trying", "frying (my brain), and/or "crying".


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 Post subject: Re: MeanDoku 4 & 5
PostPosted: Thu Aug 03, 2017 9:25 pm 
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Finished #4, but am having a whale of a time on #5! Not sure how you make these, Hatman (or, how I should solve them), but keep 'em coming! ;clapclap; Thanks!


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 Post subject: Re: MeanDoku 4 & 5
PostPosted: Sun Aug 06, 2017 7:24 pm 
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Thanks HATMAN. I echo azpaull in not knowing how you make these. They were quite hard. Here is my WT for #5.
MeanDoku 5 WT:
Implications from the puzzle rules:

(I1) A 1 cannot be on any cell with a Red Mark (RM), or a 9 in any cell with a Green Mark (GM). A 2 can have at most one RM (and that with a 9), etc.
(I2) Since the puzzle is NC - each GM must have at least one of (123) in the cells it straddles.
Similarly a RM must have at least one of (789) in the cells it straddles.
(I3) Obviously a Blue Mark (BM) must have one number smaller than 5 and one number larger than 5 in the cells it straddles.

1. n258
HS r1c6 = 1
-> r1c5 = 9
-> HS 1 in n5 -> r6c5 = 1
-> r6c4 = 9
Also -> r7c4 = 1
(I2)-> r3c46 = {78}
(I2,NC) -> r1c4,r2c5 = {23}
But 3 in r2c5 leaves no place for 4 in n2
-> r1c4 = 3, r2c5 = 2
(I2) -> r5c4 = 2 and 3 in r45c6
-> r9c6 = 2
-> r8c5 = 3
-> r8c4 = 7
-> r3c46 = [87]
-> r2c4 = 6
Also r9c45 = [46]
-> remaining cells in n8 -> r7c5 = 8, r78c6 = [59]
-> r3c5 = 5, r2c6 = 4
-> r456c6 = [368]
-> r45c5 = [74]
-> r45c4 = [52]

2. 5 in n1 in r1c123
-> (46) cannot both go in r1c123
-> At least one of (46) in r3c123
-> r3c78 = {19}
-> r2c12 = {19}
-> (57) in r1c123
-> 5 in r2c89, 7 in r2c789
Also 6 not in r1c123
-> 6 in r3c123
-> 6 in r1c789

3. 2 in n7 in r8c123 -> 2 in n9 in r7c789
r8c789 cannot contain all of (456)
-> at least one of (456) in r8c123
Either 1 in n7 in r9c123 -> r9c89 = {37} -> r7c12 = {37}
or 1 in in r8c123 -> r8c13 = {12} -> r8c2 from (456) -> r7c12 = {37} -> r9c89 = {37}
Either way, both r7c12 and r9c89 are {37}

4. 3 in c7 in r56c7
-> HS 2 in c7 -> r7c7 = 2
-> r5c7 = 3
-> HS 7 in c7 -> r2c7 = 7
-> NS r1c7 = 4
-> HT r349c7 = {189}
-> r68c7 = {56}

5! r3c9 from (23)
Can r9c89 be [73]?
Would put:
r8c9 = 8, r3c9 = 2, r2c89 = [35]
Leaves no value for r1c9
-> r9c89 = [37]

6. Since r3c9 from (23) -> r8c9 cannot be 3.
-> r3c9 = 3
-> r3c123 = {246}
-> r1c89 = {26}
-> r2c789 = [758]
-> r1c89 = [26]
-> r1c123 = [758] or [857]
-> r2c123 = [{19}3]
-> r3c123 = [{24}6]

7. 6 in n7 in r8c12
-> r68c7 = [65]
-> HS 7 in n6 -> r5c8 = 7
-> HS 8 in n6 -> r4c7 = 8
-> r9c7 = 9
-> r3c78 = [19]
-> r8c8 = 8
-> r8c9 = 1
-> r7c89 = [64]
-> r456c9 = [952]
-> r46c8 = [14]

8. r4c123 = [{46}2]
1 in n4 in r5c12
->r9c123 = [581]
-> r8c123 = [264]
-> r7c123 = [739]
-> r3c12 = [42]
-> r2c12 = [19]
-> r1c13 = [87]
-> r4c12 = [64]
-> r6c123 = [375]
-> r5c123 = [918]


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 Post subject: Re: MeanDoku 4 & 5
PostPosted: Mon Aug 14, 2017 8:39 pm 
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Whew, I finally solved #5! (I love the NC aspect to these, Hatman.)

The key that finally unlocked the puzzle for me:
Hidden Text:
Realizing that r7c12 and r9c89 both had to be 3-7 combinations.


Thanks again!


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 Post subject: Re: MeanDoku 4 & 5
PostPosted: Mon Aug 06, 2018 4:32 am 
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HATMAN wrote:
With NC, I think that for a full set of 12 clues in a nonet there is normally (always?) a unique solution.
MeanDoku NC X 11, for which I posted my WT a few days ago, was an exception, having two solutions; it had full sets of clues in all the nonets.

I'm busy for the next few days; then I'll try these two MeanDoku's.


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 Post subject: Re: MeanDoku 4 & 5
PostPosted: Fri Aug 24, 2018 10:51 am 
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Andrew

I think the exception is as you pointed out due to the symmetry.

Maurice


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 Post subject: Re: MeanDoku 4 & 5
PostPosted: Thu Aug 30, 2018 5:12 am 
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I struggled at times with MeanDoku NC 4 because I still find that I'm not immediately spotting interactions between the coloured marks and NC. Hopefully I've now managed to solve it.

My walkthrough is much simplified from how I originally did some of the steps, but not an optimised solving path.

It's interesting that N2 and N6 have the same solutions, as do N4 and N8; their coloured marks are the same but the horizontal interactions between N4 and N6 are quite different than the vertical interactions between N2 and N8.

Here is my walkthrough for MeanDoku NC 4:
Cells adjacent to green marks must total less than 10, blue must total 10, red must total more than 10. Also NC so horizontally/vertically adjacent cells cannot be {12}, {23}, … {78}, {89}, so at least one of the cells adjacent to each green mark must contain one of 1,2,3 as they cannot be {45}; similarly at least one of the cells adjacent to each red mark must contain one of 7,8,9 as they cannot be {56}.

Prelims.
Delete 9 from cells either side of green marks.
Delete 5 from cells either side of blue marks.
Delete 1 from cells either side of red marks.
Clean-ups only as stated.

1a. 1 in N2 only in R3C45, locked for R3, no 2 in R3C45 (NC)
Clean-ups: no 1 in R2C45 -> no 8 in R23C45 (green)
No 8,9 in R2C4 -> no 2,3 in R1C4 (red)
No 8,9 in R2C5 -> no 2,3 in R1C5 + R2C6 (red)
1b. There are 7 red marks in N2, R12C4 must contain one of 7,8,9, R12C5 must contain one of 7,8,9 and R2C6 must contain one of 7,8,9 as otherwise cannot place for all the red marks -> R2C6 = {789}, no 7,8,9 in R1C6 + R3C456, no 8 in R2C7 (NC)
1c. R1C56 must contain one of 7,8,9 (red) -> R1C5 = {789}, no 8 in R1C4 (NC)
Clean-up: 8 in N2 only in R1C5 + R2C6 -> no 2 in R1C6 (red)
1d. R1C5 + R2C6 = {789}, R12C4 must contain one of 7,9 -> no 7 in R2C5

2. Exactly the same pattern of coloured marks is in N6, so the same steps except that one sideways NC doesn’t apply this time.
2a. 1 in N6 only in R6C78, locked for R6, no 2 in R6C78 (NC)
Clean-ups: no 1 in R5C78 -> no 8 in R56C78 (green)
No 8,9 in R5C7 -> no 2,3 in R4C7 (red)
No 8,9 in R5C8 -> no 2,3 in R4C8 + R5C9 (red)
2b. There are 7 red marks in N2, R45C7 must contain one of 7,8,9, R45C8 must contain one of 7,8,9, and R5C9 must contain one of 7,8,9 as otherwise cannot place for all the red marks -> R5C9 = {789}, no 7,8,9 in R4C9 + R6C789
2c. R4C89 must contain one of 7,8,9 (red) -> R4C8 = {789}, no 8 in R4C7 (NC)
Clean-up: 8 in N6 only in R4C8 + R5C9 -> no 2 in R4C9 (red)
2d. R4C8 + R5C9 = {789}, R45C7 must contain one of 7,9 -> no 7 in R5C8
2e. 1 in C4 only in R345C4, no 2 in R4C4 (NC)

3a. There are 7 green marks in N4, one of R4C23 and one of R56C1 must each contain one of 1,2,3, while R5C3 must contain one of 1,2,3 to satisfy the remaining green marks -> R5C3 = {123}, no 2,3 in R4C1 + R5C2 + R6C23, no 2 in R4C3 + R5C4 (NC)
Clean-up: no 7,8 in R4C2 (blue)
3b. There are 2 groups of red marks in N4, one of R45C1 must contain one of 7,8,9, while R6C2 must contain one of 7,8,9 to satisfy the remaining red marks -> R6C2 = {789}, no 8 in R57C2 + R6C13 (NC)
Clean-up: no 1,2,3 in R5C2 -> R4C2 = {123}, R5C1 = {23} (both green), then no 4,6 in R4C1 (blue)
3c. R4C2 + R5C3 must contain one 2,3 -> no 8 in R4C3 (green)
3d. R4C1 + R6C2 = {89} (hidden pair in N4), no 8,9 in R3C1, no 9 in R7C2 (NC)
Clean-up: no 3 in R4C2 (blue)
3e. Naked triple {123} in R4C2 + R5C13, locked for N4, 3 also locked for R5, no 4 in R5C2 (NC)
Clean-up: 3 in R5C13 -> no 7 in R5C2 (green)
3f. R4C2 = {12} -> no 2 in R3C2 (NC)

4. Exactly the same pattern of coloured marks is in N8, so the same steps apply except that one vertical NC doesn’t apply this time.
4a. There are 7 green marks in N8, one of R7C56 and one of R89C4 must each contain one of 1,2,3, while R8C6 must contain one of 1,2,3 to satisfy the remaining green marks -> R8C6 = {123}, no 2,3 in R7C4 + R8C5 + R9C56, no 2 in R7C6 + R8C7 (NC)
Clean-up: no 7,8 in R7C5 (blue)
4b. There are 2 groups of red marks in N8, one of R78C4 must contain one of 7,8,9, while R9C5 must contain one of 7,8,9 to satisfy the remaining red marks -> R9C5 = {789}, no 8 in R8C5 + R9C46 (NC)
4c. Clean-up: no 1,2,3 in R8C5 -> R7C5 = {123}, R8C4 = {23} (both green), then no 4,6 on R7C4 (blue)
4d. R7C5 + R8C6 must contain one of 2,3 -> no 8 in R7C6 (green)
4e. R7C4 + R8C9 = {89} (hidden pair in N8), no 8,9 in R6C4 + R7C3 (NC)
Clean-up: no 3 in R7C5 (blue)
4f. Naked triple {123} in R7C5 + R8C46, locked for N8, 3 also locked for R8, no 4 in R8C5 (NC)
Clean-up: 3 in R8C46 -> no 7 in R8C5 (green)
4g. R7C5 = {12} -> no 2 in R6C5 (NC)
4h. R8C4 = {23} -> no 2 in R8C3 (NC)

[I ought to have seen this sooner; I’ve moved this important step here to delete some later steps.]
5a. No 1,3 in R5C78 so must contain 2 (green), locked for R5 and N6
5b. R5C13 = [31], R4C2 = 2 -> R4C1 = 8 (blue), R6C2 = 9, R5C9 = 8 (hidden single in N6), no 7 in R3C1, no 3 in R3C2, no 4 in R6C1 (NC)
Clean-up: R5C1 = 3 -> no 7 in R6C1 (green)
5c. R4C8 = {79} -> no 8 in R3C8 (NC)
5d. 9 in N6 only in R4C78, locked for R4
5e. Naked pair {56} in R5C2 + R6C1, locked for N4, no 5,6 in R7C1 (NC)
5f. Naked pair {47} in R46C3, locked for C3
5g. 2 in R6 only in R6C46 -> no 3 in R6C5 (NC)
5h. 8 in R6 only in R6C56 -> no 7 in R6C56 (NC)
5i. R6C9 = {3456} -> no 4 in R6C8 (green+NC)

[It looks like it’s time to use a forcing chain.]
6a. Consider placements for R7C4 = {89}
R7C4 = 8, R7C5 = 2 (blue) => R8C4 = 3
or R7C4 = 9, R9C5 = 8, no 7 in R9C46 (NC) => R7C6 = 7 (hidden single in N8), no 3 in R8C6 (green) => R8C4 = 3 (hidden single in N8)
-> R8C4 = 3, no 4 in R9C4 (NC)
Clean-up: R8C4 = 3 -> no 7 in R9C4 (green)
6b. Naked pair {56} in R8C5 + R9C4, locked for N8, no 5,6 in R9C3 (NC)
6c. Naked pair {47} in R79C6, locked for C6
6d. R8C6 = {12} -> no 1 in R8C7 (NC)
6e. R2C6 = {89} -> no 9 in R2C7 (NC)
6f. R7C4 = 8 (hidden single in C4), R7C5 = 2 (blue), R8C6 = 1, R9C5 = 9, no 7 in R6C4 (NC)
Clean-up: no 1,2 in R2C5 -> no 7 in R2C4 (green)
No 1,2 in R4C6 -> no 7 in R4C5 (green)
6g. R1C5 = {78} -> no 7 in R1C4 (NC)
6h. R1C45 + R2C6 = [978] (hidden triple in N2), no 8 in R1C3, no 6 in R1C6 + R2C5, no 7 in R2C7 (NC)
Clean-up: R1C5 = 7 -> no 3 in R1C6 + R2C5 (red)
R2C6 = 8 -> no 2 in R3C6 (red)
6i. R1C6 + R2C5 = [54], no 4,6 in R1C7, no 3 in R3C5 (NC)
6j. Naked pair {16} in R3C45, 6 locked for R3 and N2 -> R2C4 = 2, R3C6 = 3, R456C6 = [692], R56C5 = [58], R56C4 = [74], R4C45 = [13], R46C3 = [47], R4C9 = 5, R5C2 + R6C1 = [65], R8C5 + R9C4 = [65], no 3 in R2C3, no 5 in R3C3, no 2,4 in R3C7, no 4 in R3C9, no 7 in R4C7, no 1,3 in R6C7, no 4 in R7C1, no 6 in R7C3 (NC)
6k. R4C78 = [97], R6C789 = [613], no 8 in R3C7, no 2 in R5C8, no 5,7 in R7C7, no 4 in R7C9 (NC)
6l. R5C78 = [24]
6m. R1C9 = 4 (hidden single in N3), no 3 in R1C8 (NC)
6n. R7C3 = {35} -> no 4 in R7C2 (NC)
6o. 8 in R3 only in R3C23, locked for N1, no 7 in R3C2, no 9 in R3C3 (NC)
6p. 7,9 in N1 only in R2C123, locked for R2
6q. R9C1 = 6 (hidden single in N7), no 7 in R8C1 + R9C2 (NC)
6r. R1C1 = {12} -> no 1 in R1C2 + R2C1 (NC)
6s. R1C2 = 3 -> no 2 in R1C13 (NC)
6t. R1C1378 = [1682], no 5 in R2C3, no 3 in R2C8 (NC)
6u. R2C123 = [759], R2C789 = [361], no 4 in R3C2, no 8 in R3C7, no 5 in R3C8 (NC)
6v. R3C789 = [597], R789C9 = [692], no 5 in R7C8, no 8 in R8C8, no 3 in R9C8 (NC)
6w. R789C8 = [358], no 4 in R7C7, no 4 in R8C7, no 7 in R9C7 (NC)

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

And the solution, since it hasn't yet been posted in this thread:
1 3 6 9 7 5 8 2 4
7 5 9 2 4 8 3 6 1
4 8 2 6 1 3 5 9 7
8 2 4 1 3 6 9 7 5
3 6 1 7 5 9 2 4 8
5 9 7 4 8 2 6 1 3
9 7 5 8 2 4 1 3 6
2 4 8 3 6 1 7 5 9
6 1 3 5 9 7 4 8 2

I'll have a try at MeanDoku NC 5 in a few days time. Ed's next Assassin should be posted in the next couple of days, plus I've got another deadline on Friday evening. Glancing at the diagram, there are so many coloured marks in N369 that the start may be easier than for this one, with the hard work to come later; that's also my impression from the visible parts of azpaull's and wellbeback's posts.


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 Post subject: Re: MeanDoku 4 & 5
PostPosted: Sat Sep 08, 2018 10:41 pm 
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MeanDoku NC 5 was definitely harder than MD NC 4 after the expected easier start.

While wellbeback and I worked in much the same areas, I think our steps for the later NC steps are sufficiently different for me to also post by WT.

Here is my walkthrough for MeanDoku NC 5:
Cells adjacent to green marks must total less than 10, blue must total 10, red must total more than 10. Also NC so horizontally/vertically adjacent cells cannot be {12}, {23}, … {78}, {89}, so at least one of the cells adjacent to each green mark must contain one of 1,2,3 as they cannot be {45}; similarly at least one of the cells adjacent to each red mark must contain one of 7,8,9 as they cannot be {56}.

Prelims.
Delete 9 from cells either side of green marks.
Delete 5 from cells either side of blue marks.
Delete 1 from cells either side of red marks.
Clean-ups only as stated.

[I overlooked a few clean-ups in steps 1 and 2, but the placements came out in step 3, so I haven’t reworked with the missed clean-ups.]

1a. R1C6 = 1 (hidden single in N2) -> R1C5 = 9 (blue), no 8 in R1C4 + R2C5, no 2 in R1C7 + R2C6 (NC)
1b. Red marks in N2, R1C5 = 9, R23C4 and R23C6 must each contain one of 7,8 -> no 7,8 in R1C4 + R23C5
1c. Green marks in N2, R1C6 = 1, R12C4 and R23C5 must each contain one of 2,3 -> no 2,3 in R2C6 + R3C46
1d. No 1,2,3 in R2C6 -> R2C5 = {23} (green), no 2,3 in R2C4 + R3C5 (NC)
1e. R1C4 + R2C5 = {23} (hidden pair in N2), no 2,3 in R1C3, no 4 in R2C4 (NC)
1f. Clean-ups: R2C5 = {23} -> no 8 in R2C46 (green)
R1C4 + R2C5 = {23} -> no 7 in R2C4 (green)
1g. 8 in N3 only in R3C46, locked for R3
Clean-up: no 2 in R3C78 (blue)
1h. R2C4 = {56} -> no 5,6 in R2C3 + R3C4 (NC)
1i. R3C5 = {456} -> no 5 in R3C6 (NC)
1j. No 7,8 in R2C4 -> R3C4 = {78} (red), no 7 in R3C3, no 7,8 in R4C4 (NC)

2a. R6C45 = [91] (hidden pair in N5), no 8 in R57C4 + R6C3, no 2 in R57C5 + R6C6 (NC)
2b. Red marks in N5, R6C4 = 9, R56C6 must contain one of 7,8, no 7,8 in R4C4 -> R4C5 = {78}, no 7,8 in R4C6 + R5C45
Clean-up: no 4,6 in R4C6 (blue)
2c. Green marks in N5, R6C5 = 1, R4C6 = {23}, R45C4 must contain one of 2,3 -> no 2,3 in R5C5 + R56C6
2d. R4C6 = {23} -> no 2,3 in R4C7 (NC)
2e. R5C5 = {46} -> no 5 in R5C4 (NC)
2f. Killer pair 2,3 in R1C4 and R45C4, locked for C4
Clean-up: no 7,8 in R9C5 (blue)
2g. Killer pair 7,8 in R23C6 and R56C6, locked for C6

3a. R7C4 = 1 (hidden single in N8), no 2 in R7C3 (NC)
Clean-up: no 9 in R7C12 (blue)
3b. Green marks in N8, R89C5 must contain one of 2,3
3c. Killer pair 2,3 in R2C5 and R89C5, locked for C5
3d. Clean-ups: no 1 in R9C5 -> no 8 in R8C5 (green)
No 9 in R7C5 -> no 2 in R7C6 + R8C5 (red)
No 7,8,9 in R9C6 -> R8C6 = 9 (red), no 8 in R8C7 (NC)
No 7,8,9 in R7C6 -> R7C5 = {78}, no 7 in R8C5 (NC)
3e. 5 in N8 only in R79C6, locked for C6
3f. R4C4 = 5 (hidden single in N5) -> R2C4 = 6, no 7 in R2C3+ R3C4, no 4,6 in R4C3, no 4 in R5C4 (NC)
Clean-up: no 4 in R2C12 (blue)
3g. R3C4 = 8, naked pair {47} in R23C6, locked for R6 and N1, no 9 in R3C3 (NC)
3h. R3C5 = 5, no 4 in R3C6 (NC)
3i. R23C6 = [47], no 3 in R2C5, no 3,5 in R2C7, no 6 in R3C7 (NC)
Clean-up: no 3 in R3C7, no 3,4 in R3C8 (blue)
3j. R1C4 + R2C5 = [32], no 4 in R1C3 (NC)
Clean-up: no 8 in R2C12 (blue)
3k. Naked pair {68} in R56C6, locked for C6 and N5 -> R4C5 = 7, R5C45 = [24], R4C6 = 3, R79C6 = [52], R7C5 = 8, no 4 in R4C7, no 1,3 in R5C3, no 7 in R5C7, no 6 in R6C6, no 4,6 in R7C7, no 3 in R9C5, no 1,3 in R9C7 (NC)
Clean-up: no 2 in R7C12, no 8 in R9C89 (blue)
3l. R56C6 = [68], R89C5 = [36], no 5 in R5C7, no 7 in R6C7, no 4 in R8C4, no 7 in R9C4 (NC)
3m. R89C4 = [74], no 6,8 in R8C3, no 3,5 in R9C3 (NC)
[That’s the centre columns completed, so it’s time to make more use of NC and the remaining blue marks.]

4a. 5 in N1 only in R1C123, locked for R1, no 4,6 in R1C2 (NC)
4b. Variable hidden killer pair 4,6 in R1C123 and R3C123 for N1, R1C123 cannot contain all of 4,5,6 (NC) -> R3C123 must contain at least one of 4,6
4c. R3C78 (blue) = {19} (cannot be [46] which clashes with R3C123), locked for R3 and N3
4d. 1,9 in N1 only in R2C123 -> R2C12 (blue) = {19} (since R2C3 cannot contain both of 1,9), locked for N1
4e. R3C123 must contain 6 (cannot be {234}, NC), locked for R3 and N1
4f. 7 in N1 only in R1C123, locked for R1
4g. R2C7 = {78} -> no 8 in R1C7, no 7,8 in R2C8 (NC)
4h. R2C8 = {35} -> no 4 in R1C8 (NC)
4i. R3C9 = {234} -> no 3 in R2C9 (NC)
4j. 2 in R7 only in R7C789, locked for N9, no 3 in R7C8 (NC)
4k. 2 in C7 only in R67C7 -> no 3 in R67C7 (NC)
4l. R5C7 = 3 (hidden single in C7), no 2,4 in R6C7 (NC)
4m. R7C7 = 2 (hidden single in C7), no 1 in R8C7 (NC)
4n. Naked triple {456} in R168C7, locked for C7
4o. R6C7 = {56} -> no 5,6 in R6C8 (NC)
4p. R8C7 = {456} -> no 5 in R8C8 (NC)
4q. 5 in N9 only in R8C79, locked for R8, no 4,6 in R8C8 (NC)

5a. R8C123 must contain at least one of 4,6 (cannot be {128} which clashes with R8C8)
5b. R7C12 (blue) = {37} (cannot be {46} which clashes with R8C123), locked for R7 and N7
5c. 3 in N9 only in R9C89 = {37} (blue), locked for N9
5d. R2C7 = 7 (hidden single in C7), no 6 in R1C7 (NC)
Clean-up: no 3 in R2C12 (blue)
5e. Naked pair {19} in R2C12, locked for N1
5f. R1C7 = 4
5g. R3C9 = {23} -> no 2 in R4C9 (NC)
5h. 1 in N9 only in R8C89, locked for N8

[It looks like it’s time to use a forcing chain.]
6a. Consider placements for R2C8 = {35}
R2C8 = 3 => R3C9 = 2, R2C9 = 5 (hidden single in N3), no 6 in R1C9 (NC) => R1C9 = 8
or R2C8 = 5 => R2C9 = 8
-> 8 in R12C9, locked for C9 and N3
6b. 8 in N9 only in R8C8 + R9C7 -> no 7 in R9C8 (NC)
6c. R9C8 = 3 -> R9C9 = 7, R2C89 = [58], no 6 in R1C8 + R8C9 (NC)
6d. R1C89 = [26], R2C3 = 3, R3C9 = 3, no 2,4 in R3C3, no 4 in R4C9 (NC)
6e. R3C3 = 6
6f. R6C9 = 2 (hidden single in N6), no 1 in R5C9 (NC)
6g. 4 in C9 only in R78C9, locked for N9, no 5 in R8C9 (NC)
6h. R5C9 = 5 (hidden single in C9) -> R68C7 = [65], no 7 in R6C8 (NC)
6i. R6C8 = 4, R5C8 = 7 (hidden single in C8)
6j. R5C3 = {89} -> no 8,9 in R4C3 + R5C2 (NC)
6k. R5C2 = 1, R4C3 = 2, R2C12 = [19], no 8 in R1C2, no 2 in R3C1 (NC)
6l. R3C12 = [42]
6m. R8C39 = [41], R8C8 = 8, R9C7 = 9, R8C12 = [26], no 3 in R7C1, no 7 in R7C2, no 5 in R9C2 (NC)
6n. R9C12 = [58], R7C1 = 7, R1C1 = 8, no 7 in R1C2 (NC)

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

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


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