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HATMAN
 Post subject: Crosses X NC 1
PostPosted: Tue Jan 05, 2016 2:55 pm 
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Grand Master
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Joined: Wed Apr 30, 2008 9:45 pm
Posts: 694
Location: Saudi Arabia
Crosses X NC 1

Another pattern I like - works well on the diagonals.
It is X.
It is NC.

Quite a lot of work - please can I have views on difficulty.

SS gives it 1.95 JS does it more easily.


Image


JS Code:
3x3:5:k:4354:13:4354:4108:4615:14:6915:15:6915:16:4354:17:4108:4615:18:19:6915:20:4354:21:4354:4108:4615:22:6915:23:6915:24:25:26:27:28:29:3595:3595:3595:3589:3589:3589:30:31:32:4614:4614:4614:33:34:35:3593:3593:3593:36:37:38:7940:39:7940:40:4104:4618:4353:41:4353:42:7940:43:44:4104:4618:45:4353:46:7940:47:7940:48:4104:4618:4353:49:4353:

Solution:
693824175
428571693
175396428
517963842
842715369
369248517
936482751
284157936
751639284
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Andrew
 Post subject: Re: Crosses X NC 1
PostPosted: Sun Jan 10, 2016 6:18 pm 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
Thanks HATMAN for another nice puzzle.

I had to re-work a couple of times when I realised that I'd forgotten to apply some early NCs. So if you try this puzzle, use NC whenever you can. Don't think that's giving anything away.

Step 17 seems to be the key design point about this puzzle:
where the hidden cage R79C2 + R8C13 'surrounds' R8C2. Nice one HATMAN! I think step 14 was also necessary to make the breakthrough. Not sure whether the similar step 13 was needed; I haven't time to re-work and check that.

Here is my walkthrough for Crosses X NC 1:
Prelims

a) 17(5) cage at R1C1= {12347/12356}, no 8,9
b) 31(5) = {16789/25789/34789/35689/45679}
c) 17(5) cage at R7C7= {12347/12356}, no 8,9

Steps resulting from Prelims
1a. 17(5) cage at R1C1= {12347/12356}, 1,2,3 locked for N1
1b. 31(5) = {16789/25789/34789/35689/45679}, 9 locked for N7
1c. 17(5) cage at R7C7= {12347/12356}, 1,2,3 locked for N9
1d. 8,9 on D\ only in R4C4 + R5C5 + R6C6, locked for N5

2. 45 rule on N2 3 innies R123C6 = 11 = {128/137/146/236/245}, no 9
2a. {245} can only be [425/524] because of NC -> no 5 in R2C6

3. 45 rule on N8 3 innies R789C4 = 11 = {128/137/146/236/245}, no 9
3a. {245} can only be [425/524] because of NC -> no 5 in R8C4

4. 45 rule on C5 3 innies R456C5= 11 = {128/137/146/236/245}, no 9
4a. {245} can only be [425/524] because of NC -> no 5 in R5C5
4b. 45 rule on R5 3 innies R5C456 = 13 {157/247/256/346} (cannot be {148/238} which clash with R456C5 = {128} because 8 in each combination only in common cell R5C5), no 8

5. R4C4 + R6C6 = {89} (hidden pair in N5)
5a. R4C4 = {89} -> no 8,9 in R3C4 + R4C3 (NC)
5b. R6C6 = {89} -> no 8,9 in R6C7 + R7C6 (NC)
5c. Min R6C6 = 8 -> max R6C45 = 6, no 6,7 in R6C45

6. 45 rule on N5 (using R5C456 = 13) 3 innies R4C456 = 18 = {279/369/378/468} (cannot be {189} because 8,9 only in R4C4, cannot be {459} because of NC, cannot be {567} because R4C4 only contains 8,9), no 1,5

7. 45 rule on C4 (using R789C4 = 11) 3 innies R456C4 = 18 = {279/369/468} (cannot be {189} because 8,9 only in R4C4, cannot be {378/459} because of NC, cannot be {567} because R4C4 only contains 8,9), no 1,5
7a. R6C4 = {234} -> no 2,3,4 in R5C4
7b. Min R6C46 = 10 -> max R6C5 = 4
7c. R5C6 = 5 (hidden single in N5)
7d. R5C456 (step 4b) = {157/256} -> R5C5 = {12}
7e. R5C6 = 5 -> no 4,6 in R4C6 + R5C7 (NC)
7f. R5C5 = {12} -> no 1,2 in R46C5 (NC)
7g. R5C5 = 1 (hidden single in N5), placed for both diagonals

8. 14(3) cage at R6C4 = {239/248} -> R6C4 = 2, placed for D/
8a. R456C4 (step 7) = 18, R6C4 = 2 -> R45C4 = 16 = [97], R4C6 = 3, placed for D/, R6C56 = [48], R4C5 = 6
8b. R4C5 = 6 -> no 5,7 in R3C5 (NC)
8c. R4C6 = 3 -> no 2,4 in R3C6 + R4C7 (NC)
8d. R5C4 = 7 -> no 6,8 in R5C3 (NC)
8e. R6C4 = 2 -> no 1,3 in R6C3 + R7C4 (NC)
8f. R6C5 = 4 -> no 3,5 in R7C5 (NC)
8g. R6C6 = 8 -> no 7 in R6C7 + R7C6 (NC)

9. R123C6 (step 2) = {146} (only remaining combination), locked for C6 and N2 -> R7C6 = 2, R89C6 = {79}, locked for N8, R7C5 = 8
9a. Naked triple {358} in 16(3) cage at R1C4, locked for C4 and N2
9b. R7C6 = 2 -> no 3 in R7C7 (NC)
9c. R3C4 = {35} -> no 4 in R3C3 (NC)
9d. R7C4 = {46} -> no 5 in R7C3 (NC)
9e. R8C5 = {35} -> no 4 in R8C4 (NC)
9f. R8C6 = {79} -> no 8 in R8C7 (NC)
9g. R9C5 = {35} -> no 4 in R9C4 (NC)
9h. 8 in N9 only in R8C9 + R9C8 -> no 7 in R8C8 + R9C9 (NC)

10. 18(3) cage at R5C7 = {369/468}, no 2, 6 locked for R5 and N6
10a. 14(3) cage at R5C1 = {239/248}, 2 locked for N4
10b. {239} must be [293/392] because of NC -> no 9 in R5C13
10c. 2 in N6 only in 14(3) cage at R4C7 = {248/257}, no 1

11. 45 rule on R4 3 remaining innies R4C123 = 13 = {148/157}, 1 locked for N4
11a. 45 rule on N4 (using R4C123 = 13) 3 innies R6C123 = 18 = {369} (only possible combination, cannot be {567} because of NC), locked for R6 and N4
11b. Naked triple {248} in 14(3) cage at R5C1, locked for R5 and N4
11c. Naked triple {157} in R4C123, locked for R4 -> R4C7 = 8
11d. R4C7 = 8 -> no 7,9 in R35C7 (NC) -> R5C7 = 3
11e. R4C8 = {24} -> no 3 in R3C8 (NC)
11f. R4C9 = {24} -> no 3 in R3C9 (NC)

12. 45 rule on D\ 4(2+2) outies R1C3 + R3C1 + R7C9 + R9C7 = 7 -> R1C3 + R3C1 = 3,4 = {12/13}, R7C9 + R9C7 = 3,4 = {12/13}

13. Consider placement for 8 in N2
R1C4 = 8 => no 7,9 in R1C5 (NC) => R1C5 = 2 => R3C5 = 9
or R2C4 = 8 => no 7,9 in R2C5 (NC) => R2C5 = 2 => R3C5 = 9
-> R3C5 = 9

14. Consider placement for 6 in N6
R5C8 = 6 => no 5,7 in R6C8 (NC) => R6C8 = 1 => R6C7 = 5
or R5C9 = 6 => no 5,7 in R6C9 (NC) => R6C9 = 1 => R6C7 = 5
-> R6C7 = 5
14a. R6C7 = 5 -> no 4,6 in R7C7 (NC)
14b. R7C7 = 7, placed for D/
14c. R7C7 = 7 -> 17(5) cage at R7C7 = {12347}, 4 locked for N9
14d. R7C7 = 7 -> no 6 in R7C8 + R8C7 (NC)
14e. R8C7 = 9, R7C8 = 5, R89C6 = [79]
14f. R7C8 = 5 -> no 4 in R8C8 (NC)
14g. R9C9 = 4 (hidden single in N9), placed for D\ -> R4C89 = [42]
14h. R4C9 = 2 -> no 1 in R3C9 (NC)
14i. 6 in C7 only in R123C7, locked for N3

15. Naked quint {12356} in 17(5) cage at R1C1, 5,6 locked for N1

16. R7C4 = 4 (hidden single in C4)
16a. 6 in R7 only in R7C123, locked for N7
16b. Naked pair {69} in R67C3, locked for C3

17. 45 rule on N7 4 innies R79C2 + R8C13 = 14 = {1238/1247/1256/1346/2345} -> no 4 in R8C2 (locked in R79C2 + R8C13 or NC)
17a. 4 in N7 only in R79C2 + R8C13 = {1247/1346/2345} -> no 5 in R8C2 (NC)
17b. R8C2 = 8, placed for D/, R8C9 = 6, R89C4 = [16]
17c. R9C4 = 6 -> no 5,7 in R9C35
17d. R89C5 = [53]
17e. R8C2 = 8 -> no 7 in R9C2 (NC)

18. R9C1 = 7 (hidden single in N7), placed for D/, R2C8 = 9, placed for D/, R7C3 = 6, placed for D/
18a. R7C1 = 9 (hidden single in N7), R7C3 + R8C2 + R9C1 = [687] = 21 -> R9C3 = 1 (cage sum), R9C7 = 2

19.R1C9 + R2C8 + R3C7 = [594] = 18 -> R1C7 + R3C9 = 9 = [18]

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

Rating Comment:
I'll rate my walkthrough at Easy 1.5. I used two short NC forcing chains in steps 13 and 14. HATMAN and wellbeback would probably just see those as intuitive steps and rate them lower. Also lots of NC including the key step 17.
I'm not sure why SudokuSolver gives such a high score. Maybe it can't work out all the hidden cages. It ought to be able to do step 17. Not sure whether it can find the important crossover clash in step 4b.
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wellbeback
 Post subject: Re: Crosses X NC 1
PostPosted: Tue Jan 19, 2016 9:55 pm 
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 282
Location: California, out of London
Thanks HATMAN. I keep forgetting to look at this forum :)

Here is my WT. It is quite straightforward but is not the way I first did it. I found the breakthrough in Step 8 on my second go through.

I define the term "Edge Cells(nx)" to mean those cells in nonet x not in the corners and not in the centre.

Crosses X NC 1 WT:
1. Innies n2 -> r123c6 = +11
-> r456c6 = +16
Innies c5 -> r456c5 = +11
-> r456c4 = +18
-> r789c4 = +11

Innies n6 -> r6c789 = +13
-> r6c123 = +18
-> r4c123 = +13
-> r4c456 = +18
-> r5c456 = +13

2. 17(5)s in n1 and n9 from {12347} or {12356}
They cannot be the same since that would leave 5 different values to fit in 6 cells on D\.
-> One of the 17(5)s is {12347} and the other is {12356}

3. -> (89) on D\ only in n5
r5c456 + r456c5 = Edge Cells(n5) + (2 * r5c5) = 13 + 11 = +24
Since Edge Cells(n5) are +10 or more -> (2 * r5c5) is +14 or less
-> r5c5 is 7 or less
-> r4c4,r6c6 = {89}
-> 1 in n5 in r56c5 or r5c6

4. Trying r4c4,r6c6 = [89]
Puts r6c45 = [41] (NC -> r6c45 cannot be {23})
Leaves both r4c56 and r45c5 = +10(2)
-> r4c4,r6c6 = [98]

5. -> r56c4 = +9(2) and cannot be {45} (NC) or {18}
Since r6c4 in 14(3) can only be from {1245}
-> 14(3)n5 = [248]
-> r5c4 = 7
-> r45c5 = [61]
-> r45c6 = [35]

6. 1 in r6 must be in r6c789
-> 6 in r6 must be in r6c123
NC -> r6c123 cannot be {567}
-> r6c123 = {369}
-> r6c789 = {157}
-> r5c789 = {369}
-> r4c789 = {248}
-> r5c123 = {248}
-> r4c123 = {157}

Also:

-> 1 in n8 only in r789c4
-> r789c4 = {146}
-> r123c6 = {146}
-> r789c6 = {279}
-> r789c5 = {358}
-> r123c5 = {279}
-> r123c4 = {358}

7. One of the 17(5)s in n1 and n9 is {12356} and the other is {12347}
-> Whichever is {12356} has {56} on D\ and the other has {47} on D\

Whichever is {12356} has Edge Cells = {4789}
-> Center cell = 2 (NC)
-> Other cells on D\ in that nonet = {56}

The other 17(5) is {12347}
-> Edge Cells are {5689}
-> Center Cell is 3
-> Other cells in that nonet on D\ = {47}

-> D\ in one of n1,n9 is [526] or [625] and in the other is [437] or [734]

8! Since r4c6 = 3 -> r4c789 = [8{24}]
-> r5c789 = [3{69}]
r6c789 = {157} -> 1 must be on the same column as the 6 in r5c789
-> 1 in n6 in r6c89
-> Since r6c6 = 8 -> r6c7 can only = 5
-> r7c7 cannot be from (456) (NC)
-> r7c7 = 7
-> r8c8 = 3, r9c9 = 4
Also r8c9,r9c8 = {68}
-> r7c8 = 5 and r8c7 = 9

9. -> r789c6 = [279]
-> r789c5 = [853]
-> r789c4 = [416]

Also -> r9c7 = 2
-> r7c9 = 1
Also r8c9 = 6
-> r9c8 = 8

10. -> r9c123 = [{57}1]
-> 31(5)n7 = [{6789}1]
-> r9c789 = [751]
-> r8c123 = [284]
-> r7c123 = [639] or [936]

11. r5c123 can only be [842]
-> r6c123 = [3{69}]
-> r67c3 = {69}
-> Since {56} in D\ in r1c1,r3c3 -> r1c1 = 6, r3c3 = 5
-> r7c13 = [96]
-> r6c123 = [369]
Also -> r4c123 = [517]

etc.


Rating:
Hard 1.0
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enxio27
 Post subject: Re: Crosses X NC 1
PostPosted: Sun May 08, 2016 7:07 pm 
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Joined: Thu Apr 24, 2008 4:27 pm
Posts: 791
With a rating of 1.95, I might be able to manage it. I will give it a try.

FYI: There is a typo in the JS code. The 5th character should be a d instead of a 5.
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