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 Post subject: MeanDoku NC X 15
PostPosted: Tue Aug 29, 2017 9:04 pm 
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Grand Master
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Joined: Wed Apr 30, 2008 9:45 pm
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Location: Saudi Arabia
MeanDoku NC X 15

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.

X: diagonals do not repeat.

On how I create them: I think up some sort of skeleton pattern (say a lot of blues). I then check that it has solutions (if not I remove some clues until it does). I then use JSudoku to find all solutions (Alt-recursively solve). I then chose more clues fitting in with the original idea to make it unique (but normally not solvable). I solve it as far as I can then add more clues a few at a time until I can solve it completely. I use JSudoku at this stage but care must be taken with the combination work as the program will find relationships a human cannot see. I finish by solving it myself at least once.

This last bit is difficult as with this approach you often find that adding the last clue to make it solvable simplifies earlier solving and makes the puzzle too easy.

Reasonably hard


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 Post subject: Re: MeanDoku NC X 15
PostPosted: Fri Sep 15, 2017 9:29 pm 
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Joined: Mon Apr 21, 2008 11:35 pm
Posts: 49
AWESOME! Thanks, HATMAN, can't wait to give this a go! And, thanks for the creation insights - sounds as complicated as I envisioned it!


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 Post subject: Re: MeanDoku NC X 15
PostPosted: Tue Sep 26, 2017 6:43 pm 
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Joined: Mon Apr 21, 2008 11:35 pm
Posts: 49
That was a satisfying puzzle! Thanks again, HATMAN!

(I've never tried to slow myself down and write up a walk-through - should I go back and make this my 1st attempt?)


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 Post subject: Re: MeanDoku NC X 15
PostPosted: Wed Sep 27, 2017 5:16 pm 
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Grand Master
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Joined: Wed Apr 30, 2008 9:45 pm
Posts: 693
Location: Saudi Arabia
Azpaull

For assassins at least one walk-through is required. For Human Solvable a walk-through to the human solvable issue/trick after that it depends on how hard the residual puzzle is.

With MeanDoku NC after the 1-9 and 2-3-7-8 eliminations: do you do anything interesting and not very obvious?

With the SSS ORC some of my solutions were quite tortuous - so I'm interested in walk-throughs.

Maurice


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 Post subject: Re: MeanDoku NC X 15
PostPosted: Thu Oct 05, 2017 7:37 pm 
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Thanks HATMAN. Again! Very nice puzzle. I didn't find it too hard. Apologies for the slowness in posting this.

MeanDoku NX X15 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. n5
1 in n5 in r456c4. (I1)
Since 9 cannot go in r45c4 (I1) -> r6c45 = [19]
For other GM in n5 (I2,NC) -> r4c4 and r5c5 = {23}.
Since r5c4 cannot be 8 (I1) -> r45c4 = [37], r5c5 = 2
-> r56c6 = {46}
-> r4c56 = [58]

2. n2
(NC) 9 not in r3c6 (I1) -> HS r1c6 = 1
-> (I1) HS r1c4 = 9
HS -> r2c4 = 2
-> r23c6 = [73]
-> (I2) -> r3c5 = 8
-> r12c5 = [64]
-> r3c4 = 5

3. n8
r789c4 = {468}. RM -> 8 in r78c4.
r789c5 = {137}.
BM -> r7c56 = [19]
-> r89c5 = {37} and r89c6 = {25}

4! D\ and n8
r3c456 = [583] and BM in n1 -> r3c23 = {19}
-> One of (19) in D\ in n9
-> r9c78 = {28}
-> r89c6 = [25]
-> r89c5 = [73]
-> r789c4 = [846]

5. n46
Given r4c4 = 3 and RM in n4 -> r4c3 must be 5 or more
-> r5c3 from (123). Only possibility is r56c3 = [37].
-> r6c2 = 2
RM in n6 and (23) already in r5 -> r5c7 is 4 or more
-> r6c7 = 3
-> r56c6 = [46]
(NC) -> r6c89 cannot be {45} -> r6c9 = 8
-> r6c18 = [45]
-> r5c12 = [85]
-> r4c123 = [196]
Also r5c789 = [691]
-> r4c789 = [427]

6. Cleanup
r3c23 = [19]
r9c78 = [28]
-> HS 8 on D\ -> r2c2 = 8
-> r8c3 = 8
HS 4 on D\ -> r9c9 = 4
-> HS 1 on D\ -> r8c8 = 1
-> r9c3 = 1
Also HS 4 on D/ -> r7c3 = 4
-> r1c2 = 4
-> r3c8 = 4
Also NP r12c3 = [25]
-> r123c1 = [736]
-> r789c1 = [259]
-> r789c2 = [637]
-> r7c789 = [573]
-> r8c79 = [96]
-> n3 = [835],[169],[742]


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 Post subject: Re: MeanDoku NC X 15
PostPosted: Sun Feb 24, 2019 9:36 pm 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
Location: Lethbridge, Alberta, Canada
Thanks HATMAN for this puzzle. After doing some of the earlier Meandokus last summer I was back on Assassins, including catching up on them after two trips in September and October.

If I'd realised how easy this one is I'd have done it back then. "Reasonably hard" was misleading. It's about the level as MeanDoku 2, in the MeanDoku thread, and MeanDoku 11, the one with two solutions but most of that puzzle can be solved.

I'm not surprised that azpaull didn't have time to write a walkthrough for this puzzle as it just kept flowing. I've got into the habit of writing my steps as I go along, having started doing that for Ruud's Assassins back in 2006; I have an Excel worksheet open for the puzzle and a Word file open for my walkthrough. For MeanDokus and similar puzzles, I colour the lines between cells and make those lines thicker so they're more visible.

I started in the same was as wellbeback but then we took different paths after his step 3. Loved the neat way he did his step 4!

Here is my walkthrough for MeanDoku NC X 15:
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}, therefore 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}.

[Note to myself for solving further MeanDokus. I note from wellbeback’s walkthrough that a cell with two red marks cannot contain 2. Similarly a cells with three red marks cannot contain 3, a cell with two green marks cannot contain 8 and a cell with three green marks cannot contain 7.]

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 and NCs only as stated.

1. No 9 in R45C4 -> no 1 in R45C4 (blue)
1a. R6C4 = 1 (hidden single in N5), placed for D/, R6C5 = 9 (blue), no 2 in R57C4 + R6C3, no 8 in R57C5 + R6C6 (NC)
1b. No 1 in R4C45 + R5C5 -> no 8 in R4C45 + R5C46 (green)
1c. R4C6 = 8 (hidden single in N5), placed for D/ -> no 2 in R4C5 + R5C6 (red), no 7 in R4C5 + R5C6, no 7,9 in R3C6 + R4C7 (NC)
1d. No 8 in R5C46 -> no 2 in R4C4 + R6C6 (blue)
1e. R5C5 = 2 (hidden single in N5), placed for both diagonals, no 3 in R4C5 + R5C46 (NC)
1f. R5C6 = {46} -> R56C6 = {46} (blue), locked for C6 and N5, no 5 in R6C7 (NC)
1g. R4C5 = 5, R5C4 = 7, R4C4 = 3, placed for D\

2. R23C6 = [73] (blue, only remaining permutation), no 6,8 in R2C5 (NC)
2a. R2C5 = 4 -> R1C5 = 6 (blue), R3C5 = 8, no 5 in R2C4 (NC)
2b. R2C4 = 2 -> R13C4 = [95], R1C6 = 1
2c. No 8 in R1C3, no 2 in R1C7, no 1,3 in R2C3, no 6,8 in R2C7, no 4,6 in R3C3, no 4 in R3C7, no 2,4 in R4C3, no 6,8 in R5C3 (NC)
2d. R78C4 must contain 8 (red) -> no 8 in R9C4
2e. R78C6 must contain 9 (red) -> no 9 in R9C6
2f. R9C4 = {46} -> no 5 in R9C3 (NC)

3. R3C23 = {19} (blue, only remaining permutation), locked for R3 and N1
3a. R2C23 = {58/68} (cannot be {56}, NC), 8 locked for R2 and N1

4. R56C3 = [37/46] (blue)
4a. Naked pair {67} in R46C3, locked for C3 and N4
4b. R6C3 = {67} -> R6C2 = {23} (green), no 3 in R5C2, no 2,3 in R6C1 + R7C2 (NC)
4c. R5C3 = {34} -> no 8 in R5C2 (green)
4d. 8 in N4 only in R56C1, locked for C1, no 9 in R5C1 (NC)
4e. 9 in N4 only in R4C12, locked for R4
4f. R4C3 = {67} -> no 2 in R4C2 (NC)

5. R3C7 = {67} -> no 6,7 in R3C8, no 6 in R4C7 (NC)
5a. R3C8 = {24} -> no 3 in R2C8 (NC)
5b. No 1 in R6C7 -> no 8 in R5C7 (green)
5c. R45C7 = [46] (blue, only remaining permutation), R3C7 = 7, placed for D/, R5C6 = 4, R6C6 = 6, placed for D\, R5C3 = 3, R46C3 = [67], R4C2 = 9, R3C2 = 1, R3C3 = 9, placed for D\, R5C2 = 5, R2C2 = 8, placed for D\, R2C3 = 5, R7C3 = 4, placed for D/, R1C3 = 2, R6C2 = 2, R4C1 = 1, R56C1 = [84], R1C1 = 7, placed for D\

6. R7C2 = {67} -> no 6,7 in R7C1, no 6 in R8C2 (NC)
6a. R8C2 = 3, placed for D/, R1C9 = 5, placed for D/, R3C1 = 6, R9C1 = 9, placed for D/
6b. R6C1 = 4 -> no 5 in R7C1 (NC)
6c. R78C1 = [25], R7C6 = 9 -> R7C5 = 1 (blue)

7. R6C78 = [35] (green, only remaining combination), R1C7 = 8, no 9 in R2C7 (NC)

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

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


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