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 Post subject: Vortex X Revisit
PostPosted: Wed Sep 15, 2021 6:31 pm 
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Grand Master
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Posts: 1040
Location: Sydney, Australia
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Note: 1-9 cannot repeat on either diagonal.

Vortex X Revisit

It gets 1.45 and 5 'complex intersections'. Archive comments (hyperlink above) suggest a nice hard one. Worth a Revisit and of course, an x puzzle has fun potential.
Code: Select, Copy & Paste into solver; 1-9 cannot repeat on the diagonals:
3x3:d:k:4096:3073:3073:4099:4099:4869:4869:4869:5128:7177:4096:3073:3073:2829:2829:4869:5128:4369:7177:7177:4096:2325:2325:2829:5128:4369:4369:7177:3868:3868:6942:2325:6942:3105:4369:3619:804:3868:3878:3878:6942:3105:3105:4907:3619:804:5678:3878:6942:3889:6942:4907:4907:5429:5678:5678:3128:3641:3889:3889:2876:5429:5429:5678:3128:5697:3641:3641:6212:6212:2876:5429:3128:5697:5697:5697:2892:2892:6212:6212:2876:
Solution:
+-------+-------+-------+
| 3 4 1 | 7 9 6 | 8 2 5 |
| 9 7 2 | 5 1 8 | 3 6 4 |
| 5 8 6 | 4 3 2 | 9 1 7 |
+-------+-------+-------+
| 6 3 7 | 9 2 1 | 4 5 8 |
| 2 5 8 | 3 4 7 | 1 9 6 |
| 1 9 4 | 8 6 5 | 7 3 2 |
+-------+-------+-------+
| 8 1 3 | 6 5 4 | 2 7 9 |
| 4 2 5 | 1 7 9 | 6 8 3 |
| 7 6 9 | 2 8 3 | 5 4 1 |
+-------+-------+-------+
Cheers
Ed


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 Post subject: Re: Vortex X Revisit
PostPosted: Fri Sep 17, 2021 7:38 pm 
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Grand Master
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 280
Location: California, out of London
Thanks Ed. Here's how I did it. BTW - I don't remember doing any of these revisits before, so they are a lot of fun!
Vortex X Revisit WT:
1. D/ in n5 = +13(3)
-> r4c4 + r6c6 = +14(2)
Since D\ in n5 = +18(3) -> r5c5 = 4

2. 16(2)n2 = {79}
Outies c789 = r158c6 = +22(3) = [5{89}} or [8{59}] or [6{79}]
i.e. 9 in r58c6
-> 9 in D\ only in n1 or in r4c4
Since 28(4)r2c1 must contain a 9 -> if 9 in D\ in n1 -> r4c1 = 9
-> 9 in r4 in r4c14

3. Outies n3 = r1c6 + r4c8 = +11(2) (No 1)
Outies n9 = r7c9 + r8c6 = +11(2) (No 1)
Innies n6 = +12(4} = {1236} or {1245}
-> 1 in r45c7

4! Outies r123 = r4c158 = +13(3)
-> Either 9 in r4c4 or r4c158 = [913]
In the latter case Innies n6 = {1236} -> 14(2)n6 = [59]
In neither case can r5c6 be 9
-> r8c6 = 9
-> Outies n9 -> r6c9 = 2
-> 3(2)n4 = [21]
-> 2 in n5 in r4c56

5! Can 2 be in r4c6?
That puts r6c4 = 7, 16(2)n2 = [97], and r7c6 = 7
But that puts remaining outies r789 = r6c25 = +15(2) = [96] which contradicts 9 in r4c14
-> r4c5 = 2
-> Outies r123 = r4c158 = +13(2) cannot have a 9
-> r4c4 = 9

6. Easier now. E.g.,
-> r6c6 = 5
-> r15c6 = [67]
-> r45c7 = [41]
-> r4c6 = 1 and r6c4 = 8
-> r6c25 = [96]
-> r5c4 = 3
Also (IOD n2) r2c4 = 5
-> (Outies n1) r4c1 = 6
Also r3c45 = [43]
-> 11(3)n2 = [1{28}]
Also D/ n7 = {237}
-> D/ n3 = {569}
Also 28(4)n1 = [{589}6]
Also 12(4)n1 = [{124}5]
-> D\ n1 = {367}
-> D\ n9 = {128}
etc.


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 Post subject: Re: Vortex X Revisit
PostPosted: Sun Sep 19, 2021 2:13 am 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
Location: Lethbridge, Alberta, Canada
Your early steps, wellbeback, look like you had fun with this one!

My solving path was very different; also easier than the archive comments suggest.

Here's how I solved Vortex X Revisited:
Prelims

a) R1C45 = {79}
b) R45C9 = {59/68}
c) R56C1 = {12}
d) R9C56 = {29/38/47/56}, no 1
e) 20(3) cage at R1C9 = {389/479/569/578}, no 1,2
f) 11(3) cage at R2C5 = {128/137/146/236/245}, no 9
g) 9(3) cage at R3C4 = {126/135/234}, no 7,8,9
h) 19(3) cage at R5C8 = {289/379/469/478/568}, no 1
i) 11(3) cage at R7C7 = {128/137/146/236/245}, no 9
j) 12(4) cage at R1C2 = {1236/1245}, no 7,8,9
k) 28(4) cage at R2C1 = {4789/5689}, no 1,2,3

Steps Resulting From Prelims
1a. Naked pair {79} in R1C45, locked for R1 and N2
1b. Naked pair {12} in R56C1, locked for C1 and N4
1c. 12(4) cage at R1C2 = {1236/1245}, CPE no 1,2 in R2C2
1d. 28(4) cage at R2C1 = {4789/5689}, CPE no 8 in R1C1

2a. 45 rule on both diagonals R5C5 (repeated) = 4, placed for both diagonals, clean-up: no 7 in R9C6
2b. 45 rule on D\ 2 innies R4C4 + R6C6 = 14 = {59/68}
2c. 45 rule on D/ 2 innies R4C6 + R6C4 = 9 = {18/27/36}, no 5,9

3a. 45 rule on C123 3 outies R259C4 = 10 = {127/136/145/235}, no 8,9
3b. 45 rule on C789 3 outies R158C6 = 22 = {589/679}, 9 locked for C6, clean-up: no 5 in R4C4 (step 2b), no 2 in R9C5
3c. 6 of {679} must be in R1C6 -> no 6 in R58C6
3d. Combined half cage R158C6 + R6C6 = {589}6/{679}{58}, 6 locked for C6, clean-up: no 3 in R6C4 (step 2c), no 5 in R9C5

4a. 45 rule on N1 2 outies R2C4 + R4C1 = 11, no 1 in R2C4, no 4 in R4C1
4b. 12(4) cage at R1C2 = {1236/1245}, 1 locked for N1
4c. 45 rule on N3 2 outies R1C6 + R4C8 = 11 = [56/65/83]
4d. 45 rule on N7 2 outies R7C2 + R9C4 = 11, no 3 in R7C2, no 1 in R9C4
4e. 45 rule on N9 2 outies R6C9 + R8C6 = 11 = [29/38/47/65]
4f. 1 on D\ only in 11(3) cage at R7C7 = {128/137}, no 5,6, 1 locked for N9
4g. R259C4 (step 3a) = {127/136/145/235}
4h. 1 of {127/136} must be in R5C4 -> no 6,7 in R5C4

5a. Min R5C6 = 5 -> max R45C7 = 7, no 7,8,9 in R45C7
5b. 7 in N6 only in 19(3) cage at R5C8 = {379/478}, no 2,5,6
5c. 1 in N6 only in R45C7, locked for C7
5d. 12(3) cage at R4C7 = {129/138/147/156}
5e. 5 of {156} must be in R5C6 -> no 5 in R45C7
5f. 5 in N6 only in R4C8 + R45C9, CPE no 5 in R23C9

6a. R259C4 (step 3a) = {127/136/145/235}, R4C6 + R6C4 (step 2c) = {18/27/36}
6b. Consider combinations for R4C4 + R6C6 (step 2b) = {68}/[95]
R4C4 + R6C6 = {68} => R4C6 + R6C4 = {27} => R259C4 = {136/145/235} (cannot be {127} which clashes with R6C4
or R4C4 + R6C6 = [95], R1C4 = 7
-> R259C4 = {136/145/235}, no 7, clean-up: no 4 in R6C2 (step 4d)
6c. R4C4 + R6C6 = {68}, locked for D\ => 16(3) cage at R1C1 = {259}
or R4C4 + R6C6 = [95], 5,9 placed for D\, R2C4 + R4C1 (step 4a) cannot be [29] => 2 of 12(4) cage in R1C2 + R2C23, locked for N1 => 16(3) cage = {367}
-> 16(3) cage = {259/367}, no 8
[Looking at the other posted walkthroughs I ought to have also spotted 9 in R4C14 from this.]
6d. 8 in N1 only in R2C1 + R3C12, locked for 28(4) cage at R2C1, no 8 in R4C1, clean-up: no 3 in R2C4 (step 4a)
6e. Consider combinations for 16(3) cage = {259/367}
16(3) cage = {259} => 12(4) cage = {136}2, 6 locked for N1
or 16(3) cage = {367}, 6 locked for N1
-> no 6 in R2C1 + R3C12
6f. 28(4) cage at R2C1 = {4789/5689}
6g. 6 of {5689} must be in R4C1 -> no 5 in R4C1, clean-up: no 6 in R2C4 (step 4a)
6h. R259C4 = {145/235} (cannot be {136} because R2C4 only contains 2,4,5), no 6, 5 locked for C4, clean-up: no 5 in R6C2 (step 4d)
6i. 11(3) cage at R2C5 = {128/146/236} (cannot be {245} which clashes with R2C4), no 5
6j. 6 of {236} must be in R2C5 -> no 3 in R2C5
6k. 9(3) cage at R3C4 = {135/234} (cannot be {126} which clashes with 11(3) cage, ALS block), no 6

7a. 45 rule on R123 3 outies R4C158 = 13 = {139/157/256} -> R4C5 = {12}, R4C8 = {35}, clean-up: no 5 in R1C4 (step 4c)
7b. R158C6 (step 3b) = {589/679}
7c. R1C7 = {68} -> no 8 in R58C6, clean-up: no 3 in R6C9 (step 4e)
7d. 12(3) cage at R4C7 (step 5d) = {129/147/156}, no 3
7e. 9(3) cage at R3C4 (step 6k) = {135/234}, 3 locked for R3 and N2
7f. R4C5 = {12} -> no 1,2 in R3C45
7g. R259C4 (step 6h) = {235} (cannot be {145} which clashes with 9(3) cage), locked for C4 -> R3C4 = 4, R34C5 = [32], clean-up: no 7 in R4C1 (step 4a), no 7 in R4C6 + R6C4 (step 2c), no 8 in R9C6
7h. 11(3) cage at R2C5 = {128}, 2,8 locked for N2 -> R1C6 = 6, R58C6 = {79}, 7 locked for C6, clean-up: no 8 in R4C4 (step 2b), no 6 in R6C9 (step 4e)
7i. R2C4 = 5 -> R5C4 = 3, R9C4 = 2 -> R6C2 = 9 (step 4d), R4C1 = 6, clean-up: no 8 in R5C9, no 6 in R6C4 (step 2c), no 9 in R9C5
7j. R2C4 = 5 -> 12(4) cage at R1C2 = {1245}, 2,4 locked for N1 -> 16(3) cage at R1C1 (step 6c) = {367} -> R1C1 = 3, R2C2 + R3C3 = {67}, 3,7 locked for D\, 7 locked for N1
7k. Naked pair {18} in R4C6 + R6C4, locked for N5 and D/
7l. R1C9 = 5 -> R2C8 + R3C7 = 15 = {69}, locked for N3 and D/ -> 12(3) cage at R7C3 = {237}, R9C1 = 7, R7C3 + R8C2 = {23}, locked for N7, clean-up: no 9 in R45C9, no 4 in R9C6
7m. R45C9 = [86] -> R9C9 = 1, R4C6 + R6C4 = [18], R4C4 + R6C6 = [95], R1C45 = [79], R58C6 = [79], R4C7 = 4 -> R5C7 + R6C9 = [12], clean-up: no 6 in R9C5
7n. Naked pair {37} in R6C78, locked for R6 and N6 -> R6C3 = 4, R5C4 = 3 -> R5C3 = 8 (cage sum)
7o. R78C4 = {16} -> R8C5 = 7 (cage sum)

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


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 Post subject: Re: Vortex X Revisit
PostPosted: Fri Sep 24, 2021 7:03 pm 
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Grand Master
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1040
Location: Sydney, Australia
Great wellbeback! I too am (mostly) enjoying them very much even though I can't solve them all. And really appreciate Andrew's dedication to this series! Invaluable. His solutions are usually much better than they were first time around.

My key breakthrough is very different also, including the archived WTs. One of the things I love about killers. [edit: changed ending since I messed up a little][edit 2: Big thanks to Andrew for checking my WT]
Vortex x WT:
Preliminaries
Cage 3(2) n4 - cells ={12}
Cage 16(2) n2 - cells ={79}
Cage 14(2) n6 - cells only uses 5689
Cage 11(2) n8 - cells do not use 1
Cage 9(3) n25 - cells do not use 789
Cage 20(3) n3 - cells do not use 12
Cage 11(3) n9 - cells do not use 9
Cage 11(3) n2 - cells do not use 9
Cage 19(3) n6 - cells do not use 1
Cage 12(4) n12 - cells do not use 789
Cage 28(4) n14 - cells do not use 123

NOTE: this is a highly optimised solution so no clean-up done unless stated.

1. 16(2)n2 = {79}: both locked for n2 and r1

2. 3(2)n4 = {12}: both locked for n4 and c1

3. "45" on both d/ and d\, r5c5 only gets counted once with the 27(5)n5 so it must be the difference of 4
3a. r5c5 = 4
3b. -> r4c4+r6c6 = 14 = {59/68}
3c. and r4c6 + r6c4 = 9 = {18/27/36}(no 5,9)

4. "45" on n1: 2 outies r2c4 + r4c1 = 11
4a. r2c4 = (2..6), r4c1 = (5..9)

5. 12(4)n1c2 = {1236/1245}: 1 locked for n1

6. 16(3)r1c1: {358} blocked by r4c4 + r6c6 needing one of 5/8 (step 3b)
6a. = {259/268/367}
6b. {259} must have 5 only in r1c1 -> no 5 in r2c2 nor r3c3

7. r4c1 is part of a h11(2) with r2c4 -> r4c1 can't repeat in n1 in the 12(4)
7a. -> must repeat in r2c2 or r3c3
7b. -> no 5 in r4c1
7c. -> no 6 in r2c4

8. "45" on c789: 3 outies r158c6 = 22 = {589/679}, 9 locked for c6
8a. no 5 in r4c4 (h14(2)n5)

9. "45" on n3: 2 outies r1c6 + r4c8 = 11 = [56/65/83]

10. "45" on r123: 3 outies r4c158 = 13
10a. = {139/157/238/256}
10b. -> r4c5 = (12)
10c. and r4c8 = (35)
10d. -> no 5 in r1c6
10e. and no 6 or 8 in r58c6 (step 8.)

11. "45" on n9: 2 outies r6c9 + r8c6 = 11 = [65/47/29]

12. 1 in n6 only in 12(3) and must have 5,7,9 for r5c6 = {129/147/156}(no 3,8)
12a. can't have two of 5,7,9 -> no 5,7,9 in r45c7

13. 16(3)r1c1 = {259/268/367}
13a. -> when h14(2)n5 = {68}, 16(3) = {259}
13b. -> 9 which must be in 28(4)r2c1 in r4c1
13c. or h14(2) = [95]
13d. -> 9 locked for r4 in r4c14
13e. -> no 5 in r5c9

14. 7 in n6 only in 19(3) -> = {379/478}(no 2,5,6)

15. 5 in n6 only in r4c89: 5 locked for r4

16. 7 in r4...
i. in r4c1 -> 5 in r4c8 (h13(3)r4)
ii. in 15(3)r4c2 = {37}[5]
iii. in r4c6 -> 2 in r6c4 (h9(2)n5)
16a. ie must have 5 in r4c8 or r5c2, or 2 in r6c4 (no eliminations yet)

17. "45" on n6: 1 outie r5c6 = 2 innies r4c8 + r6c9
17a. but [352] leaves no 7 for r4
17b. -> no 5 r5c6

Andrew used a variation of this step very early in his wt.
18. h22(3)r159c6: {589} = [895] only
18a. blocked since forces 6 into both r6c69 (combined with outies n9 = 11)
18b. = {679} only -> r1c6 = 6
18c. 7 locked for c6
18d. no 6 in r6c9 (outies n9=11)
18e. and r4c8 = 5 (outies n3=11)

19. 14(2)n6 = {68}: both locked n6 and c9

20. r4c8 = 5 -> r4c15 = [71/62] (h13(3)r4)
20a. -> r2c4 = (45) (outies n1=11)

21. r4c4 = 9 (hsingle r4): Placed for d\
21a. -> r6c6 = 5 (h14(2)): Placed for d\
21b. r59c6 = [79]
21c. -> r6c9 = 2 (outies n9=11)
21d. and r5c7 = 1, r4c7 = 4

22. "45" on n7: 2 outies r6c2 + r9c4 = 11 (no 1)

23. "45" on c123: 3 outies r259c4 = 10 = {235} only = [5]{23}: 2,3 locked for c4
23a. -> r4c1 = 6 (outies n1=11)

24. 7 in r4 only in 15(3)n4 = {37}[5]: 3 locked for r4, both 3,7 for n4

Much easier now. Don't forget the diagonals!
Cheers
Ed


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