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 Post subject: CNC FO X 15
PostPosted: Mon Jan 20, 2014 12:25 pm 
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Grand Master
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Joined: Wed Apr 30, 2008 9:45 pm
Posts: 694
Location: Saudi Arabia
CNC FO X 15

Red cages are consecutive.
Blue cages are non-consecutive.
All are fully ordered so increasing from top to bottom and then from left to right.

It is X.
r2c6 is in a C-cage and an NC-cage
r5c5 ditto
r89c9 ditto
r9c5 ditto

r6c6 is in two NC-cages

Quite hard



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 Post subject: Re: CNC FO X 15
PostPosted: Tue Jan 21, 2014 9:20 pm 
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Grand Master
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 282
Location: California, out of London
Thanks again HATMAN. Unusually for me I've written a full walkthrough. None of the moves are particularly difficult - but some are hard to spot.

Improvements and typos fixed thanks to Andrew.
Hidden Text:
1. All NC cages must have 1st cell in range 1-5, 2nd in range 3-7 and 3rd in range 5-9

2. -> r2c5 in range 1-4, r2c6 in range 2-5, and r3c4 in range 3-6
Also r3c6 in range 3-7 and r4c6 in range 4-8
Also in n9 at least 6 other cells (the ones in cages) have values less than whatever is in r9c9
-> r9c9 is Min 7
-> r7c9 in range 5-7

3. HP (12) in c9 -> r14c9 = {12}
-> (34) in c9 in r25c9

4. (12) in n8 both in r789c4.
-> 1 in n2 in r23c5, 2 in n3 in r2c45 or r3c5
-> C3/@r2c5 = [123] or [234]
-> 2 in n2 in r2c45 and 3 in n2 in r2c6 or r3c4

5. HP (12) in D/ -> (r1c9,r7c3) = {12}
Either:
(a) r7c3 = 1 -> r8c4 = 2, r9c5 = 3, r9c4 = 1
(b) r7c3 = 2 -> r8c4 = 3, r9c5 = 4, r9c4 = 2, r7c4 = 1
-> 1 in r7 in r7c34
-> 1 in n9 in r89c7
-> 1 in n3 in r1c89
Also 1 in n7 in r789c3
-> 1 in n1 in r23c1 or r3c2
-> HS 1 in D\ -> r6c6 = 1

6. 3 in D/ can only be in r2c8 or r6c4
3 in n2 in r2c6 or r3c4
-> 3 locked in r2 in c68 and in c4 in r36
-> 3 not in r8c4
-> C3@r7c3 = [123]
-> r9c4 = 1
-> r8c7 = 1
Also r1c9 = 2
-> r45c9 = [13] and r12c9 = [24]
-> HS 1 in r5 -> r5c2 = 1
Also HS 1 in n3 -> r1c8 = 1

7. 3 in r1 in r1c1 or r1c7
Both of these prevent 3 in n9 in r7c7
-> 3 in n9 in r78c8
-> HS 3 in D/ -> r6c4 = 3
-> C3@r2c5 = [234]
-> r3c5 = 1
-> r2c1 = 1
Also NC3@r1c4 = [579]
-> HS 8 in n2 -> r2c4 = 8
-> NS 6 in n2 -> r3c6 = 6
-> C3@r3c6 = [678]
-> HS 2 in n5 -> r5c6 = 2
Also HS 3 in n3 -> r1c7 = 3
-> r1c123 = [468]

8. Remaining values in c6 all in n8 are (458)
-> Can only be r789c6 = [485]
HS 7 in n8 -> r7c4 = 7
-> r78c5 = [69]

9. Remaining values in c4 are r45c4 = {69} - Can only be r45c4 = [69]
-> NP r46c5 = {45}

10. HS 9 in D/ -> r9c1 = 9
Also NS 5 in D/ -> r3c7 = 5

11. HS 4 in D/ r8c2 = 4
-> NS 6 in D/ -> r2c8 = 6
Also -> r7c2 = 2

12. C3@r7c9 can only be [567]
-> HS 7 in n3 -> r2c7 = 7
-> NP (89) in n3 -> r3c89 = {89}
-> HS 8 in n9 -> r7c8 = 8
-> HS 9 in n9 -> r7c7 = 9
-> HS 3 in n9 -> r8c8 = 3
-> r9c78 = [24]
Also r3c89 = [98]
-> r6c9 = 9

13. NS in r4c7 -> r4c7 = 8
-> r56c7 = {46}
-> r456c8 = {257}

14. NS in r7 -> r7c1 = 3
Also NP in r9/n7 -> r9c23 = [86]
-> r8c13 = {57}

15. Remaining values in D\ are (25)
Can only be r2c2 = 5, r3c3 = 2
-> r3c12 = [73] -> r2c3 = 9
-> r8c13 = [57]
-> NC3@r4c1 = [268]
-> r56c7 = [46]

16. HS 7 in n4 -> r6c2 = 7
-> r4c2 = 9
HS 3 in n4 -> r4c3 = 3
-> r56c3 = [54]
-> r46c5 = [45]
-> r456c8 = [572]


Last edited by wellbeback on Sat Jan 25, 2014 8:27 am, edited 2 times in total.

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 Post subject: Re: CNC FO X 15
PostPosted: Fri Jan 24, 2014 11:16 pm 
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Grand Master
Grand Master

Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
Thanks HATMAN for another fun puzzle. Not too difficult.

Thanks for clarifying the cage pattern; I'd originally thought that there was a 2-cell NCFO cage at R7C6 but then spotted that the blue lines extended to R6C6 and asked for confirmation that two cages overlapped at R6C6.

I've been busy on other things for the last few days so only managed to work on this puzzle yesterday evening and then I went through wellbeback's walkthrough today. We used similar methods for our main breakthrough.

Here is my walkthrough for CNC FO X 15:
Blue bordered cages are non-consecutive, fully ordered (NCFO)
Red bordered cages are consecutive, fully ordered (CFO)

Prelims
First cells of 3-cell NCFO cages are {12345} with centre cells {34567} and last cells {56789}
First cells of CFO cages no 8,9, centre cells no 1,9, last cells no 1,2

1. 1,2 in N8 only in R789C4, locked for C4

2. CFO cage at R2C5 is ordered R2C5, R2C6 and R3C4, R2C6 = {2345} -> R2C5 = {1234}, R3C4 = {3456}

3. CFO cage at R3C6, no 3,4 in R5C5 -> no 2,3 in R4C6, no 1,2 in R3C6
3a. 1 in N2 only in R23C5, locked for C5

[Continuing with the other CFO cages before making any other deductions]
4. CFO cage at R7C3, no 8,9 in R9C5 -> no 7,8 in R8C4, no 6,7 in R7C3

5. CFO cage at R7C9 = [567/678/789] (cannot be [456] because [135] in NCFO cage at R8C7 clashes with [136] in NCFO cage at R9C7) -> R7C9 = {567}, R8C9 = {678}, R9C9 = {789}, 7 in R789C9, locked for C9 and N9

6. 1,2,3,4 in C9 only in the NCFO cages at R1C9 and R4C9, one must start 1,3 and the other must start with 2,4 -> R14C9 = {12}, R25C9 = {34}

7. 1,2 in N2 only in CFO cage at R2C5 and R3C5 -> CFO cage must contain at least one of 1,2 = [123/234], 2 locked for R2 and N2, 3 locked for N2, no 4 in R4C6, no 5 in R5C5

8. NCFO cage at R1C4, R1C4 = {45} -> R1C5 = {67}, R1C6 = {89}

9. R1C9 + R7C3 = {12} (hidden pair on D/)
9a. CFO cage at R7C3 = [123/234], 3 locked for N8
9b. CFO cage at R7C3 = [123/234], CPE no 2 in R7C4

10. CFO cage at R2C5 (step 7) = [123] => CFO cage at R7C3 (step 9) = [123] => R1C9 = 2 (hidden single on D/), R2C9 = 4
or CFO cage at R2C5 = [234] => R2C9 = 4, R1C9 = 2
-> R1C9 = 2, placed for D/, R2C9 = 4, R45C9 = [13], CFO cage at R7C3 = [123]

11. R1C8 = 1 (hidden single in C8), R9C4 = 1 (hidden single in N8)

12. R6C6 = 1 (hidden single on D\)
12a. R5C2 = 1 (hidden single in N4)

13. NCFO cage at R2C6, R2C6 = {23} -> R3C7 = {567}, R4C7 = {789}

14. NCFO cage at R3C3, no 1 in R3C3 -> no 3 in R4C4
14a. R6C4 = 3 (hidden single in N5), placed for D/

15. R3C4 = 4 -> R2C56 = [23] (CFO cage at R2C5), no 5 in R4C6, no 6 in R5C5
15a. R3C5 = 1 (hidden single in N2), R5C6 = 2 (hidden single in N5)
15b. R2C1 = 1 (hidden single in R2)

16. R1C4 = 5 -> NCFO cage at R1C4 = [579]

17. R3C6 = 6 -> CFO cage at R3C6 = [678], R4C6 = 7, placed for D/, R5C5 = 8, placed for both diagonals, no 7 in R8C9, no 6 in R7C9
17a. R2C4 = 8

18. R3C7 = 5, R2C8 = 6, both placed for D/

19. R8C2 = 4 -> R7C2 = 2 (NCFO cage at R7C2)

20. R1C7 = 3 (hidden single in N3), R1C1 = 4, placed for D\, R1C23 = [68]

21. R9C2 = 8 (hidden single in N7), R9C6 = 5, R78C6 = [48]
21a. CFO cage at R7C9 = [567], 7 placed for D\ -> R2C2 = 5
21b. R8C8 = 3, placed for D\ -> R3C3 = 2

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

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

Rating Comment:
I'll rate my walkthrough at Easy 1.5. After the useful step 5, I used one short forcing chain.


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