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 Post subject: MeanDoku 3
PostPosted: Sun Jun 04, 2017 7:32 pm 
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Grand Master
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Joined: Wed Apr 30, 2008 9:45 pm
Posts: 693
Location: Saudi Arabia
MeanDoku NC 3

Quite a bit harder, Assassin level I think:

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 Post subject: Re: MeanDoku 3
PostPosted: Fri Jun 23, 2017 7:35 pm 
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 280
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I didn't find this as hard as MeanDoku 2. Maybe because I had had some practice by this time. Thanks HATMAN - More please :)
MeanDoku 3 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) A Blue Mark (BM) must have one number smaller than 5 and one number larger than 5 in the cells it straddles.

1. n1
(I1) -> r1c1 = 9, r3c3 = 1
(I2) -> r1c2,r2c1 = {23}, r2c3,r3c2 = {78}
(NC) -> r2c2 = 5
-> r1c3,r3c1 = {46}
-> n1 = [924],[357],[681] or its reflection along D\ = [936],[258],[471].

2. n9
(I1) -> 9 in r89c9
-> 1 in r89c7
-> 8 in r7c7 or r9c8
But the latter puts r8c89 = [29], r9c7 = 1 which leaves no place for 7 in n9.
-> r7c78 = [82]
(GM) -> r8c7 = 1
Also -> r89c8 cannot be [37] (or [19])
(I2) -> r89c9 = {79}
-> r89c8 = {46}
(NC) -> r7c9 = 5
-> r9c7 = 3
(NC) -> r89c8 = [46]
(NC) -> r89c9 = [79]

3. (BM) r9c45 = {28}
(BM) r8c23 = {28}
(BM) r56c7 = {46}
-> Since (46) both in r123c9 -> (BM) r23c9 = {46}
Also HS 5 in n6 -> r4c8 = 5
Also HS 1 in n3 -> r2c8 = 1

4. n36
(BM) r1c89 from [82] or [73]
But the latter puts r456c7 = [746] and r456c9 = <182>
which leaves no value for r5c8
-> r1c89 = [82]
-> r4c7 = 2
Also r456c9 = {138}
-> r56c8 = {79}
(NC) -> r4c9 = 8
Also r123c7 = [5{79}]
Also r3c8 = 3
(NC) -> r23c9 = [46]

5. n12
Since 2 already on r1
-> n1 = [936],[258],[471]
-> n2 = {147},{369},{258}
(NC,BM) -> r12c4 = [46]
(NC) -> r3c4 = 8

6. Some clean up
-> r9c45 = [28]
Also r8c23 = [82]
(GM) -> (28) in n4 cannot both go in r6c12
(BM) -> r5c12 = [82]
(HS) -> r6c6 = 8
(HS) -> r6c5 = 2
-> r3c56 = [52]
(NC) -> r2c56 = [39]
(BM) -> r78c5 = [46]
(HS) -> r8c6 = 3
(NC) -> r8c14 = [59]
(NC) -> r9c6 = 5
(NC) -> r7c46 = [71]
-> r1c56 = [17]
Also (NC,RM) -> r7c123 = [369]
etc.


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 Post subject: Re: MeanDoku 3
PostPosted: Wed Aug 01, 2018 1:47 am 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
Location: Lethbridge, Alberta, Canada
I'll agree with HATMAN that this puzzle is harder than MeanDoku NC 2H. Like wellbeback I'd moved on to this puzzle after learning about MeanDoku NCs with 2H and, in my case, first the easier 2. It took me some time to find an acceptable step 4, while the start of my step 5 is also comparable to steps in easier Assassins.

Here is my walkthrough for MeanDoku NC 3:
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.

1. R1C1 + R3C3 = [91] (hidden pair in N1) -> no 8 in R1C2 + R2C1, no 2 in R24C3 + R3C24 (NC)
Clean-up: no 9 in R1C4 -> no 1 in R2C4 (blue)
No 9 in R1C89 -> no 1 in R1C89 (blue)
No 1 in R3C9 -> no 9 in R2C9 (blue)
No 9 in R5C1 -> no 1 in R5C2 (blue)
No 1 in R8C3 -> no 9 in R8C2 (blue)
No 1 in R1C2 -> no 8 in R1C3 + R2C2 (green) -> no 3 in R2C3 + R3C2 (red)
No 1 in R2C1 -> no 8 in R3C1 (green)
No 9 in R2C3 -> no 2 in R1C3 + R2C2 (red) -> no 7 in R1C2 + R2C1 (green)
No 9 in R3C2 -> no 2 in R3C1 (red)
1a. 2 in N1 only in R1C2 + R2C1 -> no 3 in R2C2 (NC)
1b. 8 in N1 only in R2C3 + R3C2 -> no 7 in R2C2 (NC)
1c. R2C2 = {456} -> no 4,5,6 in R1C2 + R2C1 (green + NC), no 4,5,6 in R2C3 + R3C2 (red + NC)
1d. Naked pair {23} in R1C2 + R2C1, locked for N1, no 4 in R2C2 (NC)
1e. Naked pair {78} in R2C3 + R3C2, 7 locked for N1, no 6 in R2C2 (NC)
1f. R2C2 = 5
1g. R2C3 = {78} -> no 7,8 in R2C4 (NC)
Clean-up: no 2,3 in R1C4 (blue)
1h. R3C1 = {46} -> no 5 in R4C1 (NC)
1i. R3C2 = {78} -> no 7,8 in R4C2 (NC)
1j. R1C89 = {28/37} (cannot be {46} which clashes with R1C3), no 4,6
1k. Killer pair 2,3 in R1C2 and R1C89, locked for R1

2. 9 in N9 only in R89C9, locked for C9, no 8 in R89C9 (NC)
Clean-up: no 9 in R3C9 -> no 1 in R2C9 (blue)
No 9 in R7C78 -> no 1 in R7C78 (blue)
2a. 1 in C9 only in R456C9, locked for N6, no 2 in R5C9 (NC)
Clean-up: no 1 in R56C7 -> no 9 in R56C7 (blue)
2b. R2C8 = 1 (hidden single in C8) -> no 2 in R13C8 + R2C79 (NC)
Clean-up: no 2 in R1C8 -> no 8 in R1C9 (blue)
No 2 in R2C9 -> no 8 in R3C9 (blue)
2c. 1 in N9 only in R89C7 -> no 2 in R89C7 (NC)
No 9 in R7C9 -> no 2 in R8C9 (red)
No 9 in R9C8 -> no 2 in R9C9 (red)
No 1 in R7C89 -> no 8 in R7C89 + R8C8 (green) -> no 2 in R7C7 + R9C8 (blue)
No 1,2 in R7C7 -> no 7,8 in R8C7 (green)
No 1,2 in R9C8 -> no 7,8 in R9C7 (green)
No 8,9 in R7C9 -> no 3 in R8C9 (red)
2d. 2 in N9 only in R7C89 + R8C8 -> no 3 in R7C8 (NC)
Clean-up: no 7 in R7C7 (blue)
2e. 8 in N9 only in R7C78 = [82] or in R89C8 = [28] -> 2 in R78C8, locked for C8 and N9, no 3 in R8C8 (NC)
Clean-up: no 7 in R9C8 (blue)

3. There are six of green marks in N9 which need to be used to form three groups
3a. 2 must be in R7C8 to be able to satisfy the green marks next to R7C7, R7C9 and R9C7 -> R7C7 = 8 (blue), no 3 in R6C8 + R7C9, no 7 in R6C7, no 7,9 in R7C6 (NC)
Clean-up: no 8 in R5C7 -> no 2 in R6C7 (blue)
No 7,8 in R6C7 -> no 2,3 in R5C7 (blue)
No 2,8 in R7C5 -> no 2,8 in R8C5 (blue)
3b. No 1,2,3 in R8C8 -> R8C7 = {13} (green), no 2 in R8C6 (NC)
3c. No 7,9 in R9C8 -> R9C9 = {79} (red)
3d. There’s a red mark between R78C9 -> R78C9 must contain one of 7,9
3e. Killer pair 7,9 in R78C9 and R9C9, 7 locked for C9 and N9
Clean-up: no 7 in R1C9 -> no 3 in R1C8 (blue)
No 7 in R23C9 -> no 3 in R23C9 (blue)
No 7 in R8C8 -> no 3 in R9C8 (blue)
3f. Naked pair {46} in R89C8, locked for C8 and N9, no 5 in R8C9 + R9C7 (NC)
3g. R7C9 = 5 (hidden single in N9) -> no 4,6 in R6C9 (NC)
3h. Naked pair {13} in R89C7, locked for C7, no 2 in R9C6 (NC)
Clean-up: no 7 in R5C7 (blue)
3i. Naked pair {46} in R56C7, locked for C7 and N6, no 5 in R4C7 + R56C68 (NC)
3j. R4C8 = 5 (hidden single in N6)
3k. R23C9 = {46} (hidden pair in N3)
3l. Naked pair {46} in R3C19, locked for R3
3m. R1C8 = {78} -> no 7 in R1C7 (NC)
3n. R1C7 = 5 -> no 4,6 in R1C6 (NC)
3o. 6 in R1 only in R1C345 -> no 7 in R1C4
Clean-up: no 3 in R2C4 (blue)
3p. 8 in N3 only in R13C8, locked for C8
3q. R2C7 = {79} -> no 8 in R2C6 (NC)
3r. R6C8 = {79} -> no 8 in R6C9 (NC)
3s. R3C7 = 2 or R3C8 = 3 -> no 3 in R3C6 (NC)
3t. R5C12 = [19]/{28/37} (cannot be {46} which clashes with R5C7), no 4,6
3u. R8C23 = [19]/{28/37} (cannot be {46} which clashes with R8C8), no 4,6
3v. R9C45 = {19/28/37} (cannot be {46} which clashes with R9C8), no 4,6
3w. 2 in N8 only in R8C4 + R9C45 -> no 1,3 in R9C4
Clean-up: no 7,9 in R9C5 (blue)
3x. 8 in R2 only in R2C35 -> no 9 in R2C4 (NC)
Clean-up: no 1 in R1C4 (blue)

4. Consider placements for R56C7 = {46}
R56C7 = [46], no 3 in R5C8 (NC) => R3C8 = 3 (hidden single in N3), R1C9 = 2 => R1C8 = 8 (blue)
or R56C7 = [64], no 7 in R4C7 + R5C8 (NC) => R6C8 = 7 (hidden single in N6), R1C8 = 8 => R1C9 = 2 (blue)
-> R1C89 = [82]
Clean-up: no 2 in R2C4 (blue)
4a. Naked pair {79} in R23C7, locked for C7 and N3 -> R3C8 = 3, R4C7 = 2, no 4 in R3C9, no 1,3 in R4C6 (NC)
4b. R23C9 = [46], R12C4 = [46], R1C23 = [36], R23C1 = [24], no 7 in R2C35, no 5,7 in R3C4, no 3 in R4C1 (NC)
4c. R2C3 = 8, R3C2 = 7, R23C7 = [79], R3C4 = 8, no 6 in R4C2, no 7,9 in R4C4 (NC)
4d. R4C4 = {13} -> no 2 in R5C4 (NC)
4e. R5C8 = {79} -> no 8 in R5C9 (NC)
4f. R4C9 = 8 (hidden single in N6)
Clean-ups: no 2 in R5C1 -> no 8 in R5C2 blue)
No 3,7 in R5C2 -> no 3,7 in R5C1 (blue)
No 3,7 in R8C2 -> no 3,7 in R8C3 (blue)
No 8 in R8C3 -> no 2 in R8C2 (blue)
No 8 in R9C4 -> no 2 in R9C5 (blue)

5. One of R8C789 and R9C789 must be [369] (NC) with the other [147]
5a. R8C23 = [82] (blue) (cannot be [19] which clashes with R8C789 = [147/369]), no 9 in R79C2, no 3 in R79C3, no 7 in R8C1, no 1,3 in R8C4 (NC)
Clean-up: no 7,9 in R7C2 -> R7C3 = {79} (red)
5b. R9C45 = [28] (blue) (cannot be [73/91] which clash with R9C789 = [147/369]), no 7,9 in R8C5 + R9C6 (NC)
Clean-up: no 1,3 in R7C5 (blue)
5c. 9 in C2 only in R45C2, locked for N4
5d. 8 in N4 only in R56C1 -> no 7 in R6C1 (NC)

6. R79C2 cannot be {46} (because R456C2 cannot be [192] which clashes with R5C12 = [19], blue, other permutations of {129} eliminated by NC) -> R9C2 = 1, R89C7 = [13], no 4 in R9C68 (NC) -> no 7 in R9C9 (NC)
Clean-up: no 1 in R6C2 -> no 8 in R6C1 (green)
No 1 in R8C5 -> no 9 in R7C5 (blue)
6a. R9C89 = [69], R8C89 = [47], R9C6 = 5, R8C4 = 9, R9C13 = [74], R7C123 = [369]
6b. R5C1 = 8 (hidden single in N4) -> R5C2 = 2 (blue), no 3 in R5C3 (NC)
6c. R46C2 = [94], no 5, in R6C1, no 3,5 in R6C3 (NC)
6d.
6e. R456C3 = [357], R47C4 = [17], R7C5 = 4 -> R8C5 = 6 (blue)

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

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

They're an interesting variety. I'll try some more when I have time, and then move on to the ORC versions; they'll remind me of when I did SkyScraper Sum ORCs.


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