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 Post subject: Assassin 303
PostPosted: Thu Nov 06, 2014 9:46 pm 
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Grand Master
Grand Master

Joined: Wed Apr 16, 2008 1:16 am
Posts: 1040
Location: Sydney, Australia
Kept thinking how much wellbeback will enjoy this one! Quite a long solution to fully crack it. The beginning was the slowest to get anywhere so you might have to persevere. Worth it! So many interesting and fun steps over the whole of the grid. It gets a score of 1.70. JSudoku also has a hard time.

Assassin 303
(note: 1-9 cannot repeat on the diagonals)
Image
code: paste into solver:
3x3:d:k:5376:2817:2817:2817:5634:5634:5634:5634:1795:5376:6404:1797:1797:5126:5639:5639:1795:2568:5376:6404:6404:1797:5126:5639:5129:5642:2568:3595:3595:3084:6404:5126:5129:5642:5642:2568:3595:11533:3084:3084:5129:5642:5642:3086:3086:11533:3599:11533:5129:2320:3857:2066:2066:6163:11533:3599:4628:11533:2320:1813:3857:6163:6163:2582:11533:4628:4628:11533:1813:4375:3857:6163:2582:2582:11533:11533:3352:3352:4375:4375:3857:
solution:
+-------+-------+-------+
| 8 2 3 | 6 1 5 | 9 7 4 |
| 7 4 1 | 2 8 9 | 6 3 5 |
| 6 9 5 | 4 3 7 | 1 8 2 |
+-------+-------+-------+
| 4 1 2 | 7 9 8 | 5 6 3 |
| 9 5 7 | 3 6 1 | 2 4 8 |
| 3 8 6 | 5 4 2 | 7 1 9 |
+-------+-------+-------+
| 1 6 9 | 8 5 4 | 3 2 7 |
| 5 7 8 | 1 2 3 | 4 9 6 |
| 2 3 4 | 9 7 6 | 8 5 1 |
+-------+-------+-------+
Cheers
Ed


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 Post subject: Re: Assassin 303
PostPosted: Fri Nov 07, 2014 7:08 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 Ed for a fun Assassin! :D

Ed wrote: Kept thinking how much wellbeback will enjoy this one!:
I assume that refers to the 45(9) 'ring' cage and interactions with the adjacent cages. I had a quick look at that but then found my key areas were elsewhere.

Here is my walkthrough for Assassin 303:
Thanks Afmob and Ed for correcting some minor errors.

Prelims

a) 7(2) cage at R1C9 = {16/25/34}, no 7,8,9
b) R5C89 = {39/48/57}, no 1,2,6
c) R67C2 = {59/68}
d) R67C5 = {18/27/36/45}, no 9
e) R6C78 = {17/26/35}, no 4,8,9
f) R78C6 = {16/25/34}, no 7,8,9
g) R9C56 = {49/58/67}, no 1,2,3
h) 21(3) cage at R1C1 = {489/579/678}, no 1,2,3
i) 11(3) cage at R1C2 = {128/137/146/236/245}, no 9
j) 7(3) cage at R2C3 = {124}
k) 20(3) cage at R2C5 = {389/479/569/578}, no 1,2
l) 22(3) cage at R2C6 = {589/679}
m) 10(3) cage at R2C9 = {127/136/145/235}, no 8,9
n) 10(3) cage at R8C1 = {127/136/145/235}, no 8,9
o) And, of course, 45(9) cage at R5C2 = {123456789}

Steps resulting from Prelims
1a. Naked triple {124} in 7(3) cage at R2C3, CPE no 4 in R2C5
1b. 22(3) cage at R2C6 = {589/679}, CPE no 9 in R2C5

2. 45 rule on R1 2 innies R1C19 = 12 = [75/84/93], clean-up: no 1,5,6 in R2C8

3. 45 rule on N9 2 outies R6C69 = 11 = {29/38/47/56}, no 1

4. 45 rule on N1 4 outies R1234C4 = 19 = {1279/1459/1468/2458/2467} (cannot be {1369/1378/1567/2359/2368/3457} because R23C4 must contain two of 1,2,4), no 3
4a. R1234 only contains two of 1,2,4 -> no 1,2,4 in R14C4
4b. 11(3) cage at R1C2 = {128/137/146/236/245}
4c. R1C4 = {5678} -> no 5,6,7,8 in R1C23

5. 45 rule on C1234 1 innie R6C4 = 1 outie R8C5 + 3, no 1,2,3 in R6C4, no 7,8,9 in R8C5

6. 45 rule on D\ 2 innies R1C1 + R5C5 = 1 outie R3C2 + 5, IOU no 5 in R5C5
6a. Min R1C1 + R5C5 = 8 -> min R3C2 = 3
6b. Max R1C1 + R5C5 = 14, min R1C1 = 7 -> max R5C5 = 6 (R1C1 + R5C5 cannot be [77])

7. 45 rule on D/ 3 innies R7C3 + R8C2 + R9C1 = 18 = {189/279/369/378/468/567} (cannot be {459} which clashes with 7(2) cage at R1C9)
7a. The value in R7C3 must be in two different combinations for 18 because R7C3 + R8C2 + R9C1 = 18 and 18(3) cage at R7C3 and R8C34 “see” R8C2 -> no 1,2 in R7C3 (CCC)
7b. 1 of {189} must be in R9C1 -> no 1 in R8C2
[Note that step 7a doesn’t eliminate 4 from R7C3 because 18(3) cage at R7C3 can still be 4{59}.]

8. 25(4) cage at R2C2 = {1789/2689/3589/3679/4579/4678}
8a. Killer quad 1,2,3,4 in R1C23, R2C3 and 25(4) cage, locked for N1
8b. 21(3) cage at R1C1 = {579/678}, 7 locked for C1 and N1

9. 45 rule on R6789 1 outie R5C2 = 1 innie R6C4, R6C4 = {456789} -> R5C2 = {456789}

[I was a bit slow to spot …]
10. 45 rule on N1 1 outie R4C4 = 3 innies R1C23 + R2C3 + 1
10a. Min R1C23 + R2C3 = 6 -> min R4C4 = 7
10b. Max R1C23 + R2C3 = 8 must contain 1, locked for N1
10c. R1234C4 (step 4) = {1279/1459/1468/2458/2467}
10d. 8 of {1468/2458} must be in R4C4 -> no 8 in R1C4

[Taking step 8 a bit further …]
11. 25(4) cage at R2C2 = {2689/3589/4579/4678} (cannot be {3679} which clashes with 21(3) cage at R1C1)
[This would also have eliminated {1789} if I hadn’t spotted step 10 first.]

12. 25(4) cage at R2C2 (step 11) = {2689/3589/4579/4678}, R1C19 (step 2) = [75/84/93]
12a. R4C4 = R1C23 + R2C3 + 1 (step 10)
Consider combinations for R1C23 + R2C3
R1C23 + R2C3 = {123} = 6, 3 locked for R1 (no 9 in R1C1, step 2) => R4C4 = 7 => R1C1 = 8 => 25(4) cage = {4579}
or R1C23 + R2C3 = {124/134} = 7,8 => R4C4 = {89} => 25(4) cage = {2689/3589}
-> 25(4) cage = {2689/3589/4579}, CPE no 9 in R1C1, clean-up: no 3 in R1C9 (step 2), no 4 in R2C8

13. 11(3) cage at R1C2 = {137/146/236} (cannot be {245} which clashes with R1C9), no 5

14. 25(4) cage (step 12a) = {2689/3589/4579}, R1C19 (step 2) = [75/84]
14a. Consider combinations for 7(2) cage at R1C9
7(2) cage = [43] => R1C1 = 8 => 25(4) cage = {4579}
or 7(2) cage = [52] => 7(3) cage at R2C3 = {14}2 => 25(4) cage = {3589/4579}
-> 25(4) cage = {3589/4579}, no 2,6
14b. 6,7 in N1 only in 21(3) cage at R1C1 (step 8b) = {678} (only remaining combination), locked for C1 and N1
14c. 25(4) cage = {3589/4579}, 7,8 only in R4C4 -> R4C4 = {78}
14d. Naked pair {78} in R1C1 + R4C4, locked for D\, clean-up: no 3,4 in R6C9 (step 3)
14e. Naked pair {78} in R1C1 + R4C4, CPE no 7 in R1C4 -> R1C4 = 6, clean-up: no 6 in R5C2 (step 9), no 3 in R8C5 (step 2)

15. R1234C4 (step 4) = {1468/2467} -> R23C4 = {14/24}, 4 locked for C4, N2 and 7(3) cage at R2C3, no 4 in R2C3, clean-up: no 4 in R5C2 (step 9), no 1 in R8C5 (step 5)
15a. 22(3) cage at R2C6 = {589/679}
15b. 6 of {679} must be in R2C7 -> no 7 in R2C7
15c. 20(3) cage at R2C5 = {389/479/578} (cannot be {569} = [596] which clashes with 22(3) cage), no 6

16. 11(3) cage at R1C2 (step 13) = {146/236}, R1C19 (step 2) = [75/84]
16a. Again consider combinations for 7(2) cage at R1C9
7(2) cage = [43] => R1C1 = 8 => R4C4 = 7
or 7(2) cage = [52] => R2C3 = 1 => R23C4 = {24} => R4C4 = 7 (step 15)
-> R4C4 = 7, placed for D\, R1C1 = 8, R1C9 = 4 -> R2C8 = 3, both placed for D/, clean-up: no 7 in R5C2 (step 9), no 8 in R5C8, no 9 in R5C9, no 5 in R6C7, no 2 in R7C5, no 4 in R8C5 (step 5)
[Cracked. The rest is fairly straightforward.]
16b. R1234C4 (step 15) = {2467} (only remaining combination) -> R23C4 = {24}, 2 locked for C4, N2 and 7(3) cage at R2C3 -> R2C3 = 1
16c. Naked pair {23} in R1C23, locked for R1 and N1
16d. R7C3 + R8C2 + R9C1 (step 7) = {189/279/567}
16e. R9C1 = {125} -> no 2,5 in R7C3 + R8C2

17. R3C5 = 3 (hidden single in N2) -> R24C5 = 17 = [89], clean-up: no 9 in R5C2 (step 9), no 1,6 in R67C5, no 6 in R8C5 (step 5), no 4,5 in R9C6
17a. 22(3) cage at R2C6 = {679} (only remaining combination) -> R2C7 = 6, R23C1 = [76], R23C6 = [97], clean-up: no 2 in R6C8, no 4,6 in R9C5
17b. Naked pair {15} in R1C56, locked for R1

18. Killer pair 5,8 in R5C2 and R67C2, locked for C2 -> R2C2 = 4, placed for D\, R3C2 = 9, R3C3 = 5, placed for D\, R2C4 = 2, clean-up: no 5 in R67C2
18a. Naked pair {68} in R67C2, locked for C2 -> R5C2 = 5, R6C4 = 5 (step 9), placed for D/, R8C2 = 7, R8C5 = 2, R6C5 = 4 -> R7C5 = 5, R9C5 = 7 -> R9C6 = 6, clean-up: no 7 in R5C89, no 3 in R6C7, no 1 in R78C6

19. Naked pair {34} in R78C6, locked for C6 and N8 -> R6C6 = 2
19a. Naked pair {18} in R56C6, locked for N5 -> R5C4 = 3, R5C5 = 6, placed for both diagonals, R5C9 = 8 -> R5C8 = 4, R5C6 = 1, R4C6 = 8, R3C7 = 1 (cage sum), 1,8 placed for D/, R6C7 = 7 -> R6C8 = 1
19b. R7C3 = 9 -> R8C34 = 9 = [81]
[I'd originally included step 19b, then mistakenly persuaded myself that it wasn't necessary.]

and the rest is naked singles.

Rating Comment:
I'll rate my walkthrough for A303 at 1.5. I used some fairly short forcing chains; at least one of them felt a bit to long for me to rate it at Easy 1.5.


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PostPosted: Sat Nov 08, 2014 11:53 am 
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Grand Master
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Joined: Mon Apr 21, 2008 9:44 am
Posts: 310
Location: MV, Germany
Thanks for this interesting Killer, Ed! My most important moves where in the same region as Andrew's cracking moves but otherwise my reasoning for those moves are quite different.
A303 Walkthrough:
1. R123
a) 22(3) = 9{58/67} -> CPE: R2C5 <> 9
b) Innies R1 = 12(2) = [75/84/93]
c) Outies N1 = 19(4): R14C4 <> 1,2,3,4 since R23C4 = (124)
d) Innies+Outies N1: 1 = R4C4 - (R1C23+R2C3) -> R4C4 <> 5,6; R1C23 <> 6,7,8; R1C23+R2C3 = 6/7/8(3) = 1{23/24/25/34} -> 1 locked for N1
e) 11(3): R1C23 <> 5 since R1C4 >= 5
f) Hidden Killer pair (56) in 21(3) @ N1 since 25(4) can only have one of them -> 21(3) = 7{59/68} -> 7 locked for C1+N1
g) Innies+Outies N1: 12 = R14C4 - R2C3: R1C4 <> 8 since R2C3 = (124) and R4C4 <> 5,6
h) Outies N12 = 38(3+2): R4C5 <> 3,4 since R1C78+R2C7 <= 24 and R1C78 <> 1,2,3 since R4C45 <= 17

2. R123+C123+D\ !
a) Hidden Killer pair (12) in 22(4) @ R1 since 11(3) <> 8
b) 7(3): R2C3 <> 4 since 4{12} blocked by Killer pair (12) of 22(4)
c) 7(3) = {124} -> 4 locked for C4+N2
d) 4 locked in Outies N1 = 19(4) @ C4 = 4{159/168/258/267} -> R1C4 = (56) since R234C4 <> 5,6
e) Hidden Killer pair (34) in 25(4) @ N1 since 11(3) can only have one them -> 25(4) <> 2
f) Hidden Killer pair (58) in 25(4) @ N1 since 21(3) can only have one of them -> 25(4) = {3589/4579/4678}
g) 25(4): R4C4 <> 9 since 7 only possible there and {358}9 blocked by Killer pair (58) in 21(3)
h) Innies+Outies D\: -5 = R3C2 - (R1C1+R5C5): R5C5 <> 5,7,8,9 since R1C1 = (789) and IOU @ N1
i) ! Consider placement of R4C4 = (78) -> 15(4) <> 7
- i) R4C4 = 7
- ii) R4C4 = 8 -> 25(4) = {359}8 -> R1C1 = 7

3. R123+N5 !
a) ! Consider placement of 7 in D\ -> R2C3 = 1
- i) R1C1 = 7 -> Outie R1 = R2C8 = 2
- ii) R4C4 = 7 -> Outies N1 = 19(4) = 6{24}7
b) 7(3) = {124} -> 2 locked for C4+N2
c) 1 locked in 22(4) @ N2 = 1{489/579/678} <> 3 for 22(4)
d) 3 locked in 20(3) @ N2 = {389} locked for C5
e) 9(2) = {27/45}
f) Outies N478 = 7(2) = [34/52]
g) 9(2) = [27/45]

4. N1458 !
a) Innies+Outies C1234: -3 = R8C5 - R6C4: R8C5 = (2456), R6C4 = (5789)
b) 1 locked in R789C4 @ C4 for N8
c) 7(2) = {25/34}
d) ! Killer triple (457) locked in 7(2) + 13(2) + R7C5 for N8
e) Innies+Outies C1234: -3 = R8C5 - R6C4: R6C4 = (58)
f) Hidden Single: R4C4 = 7 @ C4
g) Innies+Outies R6789: R5C2 = R6C4 = (58)
h) Killer pair (58) locked in R5C2 + 14(2) for C2
i) 45(9) must have 7 -> CPE: R78C3 <> 7
j) Hidden Killer pair (58) in R3C3 for N1 since 21(3) can only have one them -> R3C3 = (58)

5. R123+D/
a) Innies R1 = 12(2) = [93/84]
b) 7(2) = {34} locked for N3+D/
c) 25(4) = 47{59/68} -> 4 locked for C2+N1
d) 11(3) = {236} -> R1C4 = 6
e) 22(4) = {1579} locked for R1
f) R1C1 = 8, R3C3 = 5
g) R8C2 <> 9 since it sees all 9 of C4
h) Innies D/ = 18(3): R7C3+R8C2 <> 1,2 since R9C1 <= 6
i) 20(4) = 18{29/56} since {2567} blocked by R8C2 = (67) -> 1,8 locked for D/
j) Hidden Single: R8C2 = 7 @ D/, R5C3 = 7 @ C3

6. R456+N9+D/
a) 12(3) = {237} -> R5C4 = 3, R4C3 = 2
b) Outie N478 = R6C5 = 4
c) Cage sum: R7C5 = 5
d) 12(2) = {48} locked for R5+N6
e) Innies+Outies R6789: R5C2 = R6C4 = 5
f) 20(4) = {1568} locked for D/
g) 14(3) = {149} -> R4C1 = 4; 1,9 locked for N4
h) Hidden Single: R9C3 = 4 @ 45(9)
i) 7 locked in 24(4) @ R7 = 7{269/359/368/458} <> 1 for N9+24(4)
j) 17(3) <> 1
k) 1 locked in 15(4) @ N9 = {1239} locked for D\; 3 locked for N9; R6C6 <> 1

7. Rest is singles without considering diagonals.

Rating:
Easy 1.5. I used two small forcing chains.


Last edited by Afmob on Fri Nov 28, 2014 5:31 am, edited 1 time in total.

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 Post subject: Re: Assassin 303
PostPosted: Sat Nov 15, 2014 12:33 am 
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Grand Master
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1040
Location: Sydney, Australia
Loved Andrew's 7a! Here's how I got started. Worked in the same areas, (especially Afmob's) but for very different reasons. I didn't get either of the initial placements for either Andrew or Afmob but we all found my first one (my 15f), so it should be the key. [Big thankyou to Afmob for picking up an omission and some typos and a naming suggestion for step 12; also to Andrew for some helpful suggestions. Thanks!

A303 start
16 steps:
Prelims courtesy of SudokuSolver
Preliminaries
Cage 14(2) n47 - cells only uses 5689
Cage 7(2) n3 - cells do not use 789
Cage 7(2) n8 - cells do not use 789
Cage 8(2) n6 - cells do not use 489
Cage 12(2) n6 - cells do not use 126
Cage 13(2) n8 - cells do not use 123
Cage 9(2) n58 - cells do not use 9
Cage 7(3) n12 - cells ={124}
Cage 22(3) n23 - cells do not use 1234
Cage 21(3) n1 - cells do not use 123
Cage 10(3) n36 - cells do not use 89
Cage 10(3) n7 - cells do not use 89
Cage 20(3) n25 - cells do not use 12
Cage 11(3) n12 - cells do not use 9

Missing most routine clean-ups. Don't need them.
1. "45" on n1: 4 outies r1234c4 = 19
1a. must have two of 1,2,4 for r23c4 = {1279/1459/1468/2458/2467}(no 3)
1b. = exactly two of 1,2,4 -> no 1,2,4 in r1c4 nor r4c4
1c. min. r1c4 = 5 -> max. r1c23 = 6 (no 5,6,7,8)(can't be {155})[Andrew suggests another way to see this, "11(3) contains one of 5,6,7,8, r1c4 = (5678) -> r1c23 no 5,6,7,8"]

2. "45" on r1: 2 innies r1c19 = 12 = [93/84/75](r1c1 = (789), r1c9 = (345)

3. 25(4)r2c2 sees r1c1 through D\ -> {1789} blocked
3a. = {2689/3589/3679/4579/4678}(no 1)
3b. must have one of 2,3,4 in n1
3c. -> Killer quad 1,2,3,4 with r1c23 + r2c3 + 25(4): 4 locked for n1

4. 21(3)n1 = {579/678}: must have 7, 7 locked for n1 and c1
4a. Note: must have 6 or 9

5. 25(4)r2c2: {369}[7] blocked by 21(3)n1(step 4a)
5a. 25(4)r2c2 = {2689/3589/4579/4678}
5b. Note: must have 4 in r23n1 or have 9
5c. 7(3)r2c3 = {124}: must have 4
5d. "45" on r1: 1 innie r1c1 - 5 = 1 outie r2c9 = [94/83/72]
5e. but combining steps 5b and 5c -> [94] blocked from r1c1+r2c8 by 25(4)r2c2 + 7(2)r2c3
5f.-> r1c1 + r2c8 = [83/72]
5g. no 3 in r1c9

6. 1 innie r1c1 - 5 = 1 outie r2c9 = [83/72]: note,
i.either [72] forms a caged-x-wing on 2 in 7(3)r2c3 -> 2 locked for r23
ii.or r1c1 = 8
6a. -> {2689} in 25(4)r2c2 blocked since 2 is only in r23
6b. also {4678} blocked by r1c1 = (78)
6c. 25(4)r2c2 = {3589/4579}(no 2,6)

7. 6 in n1 only in 21(3) = {678} only: 6,8 locked for n1 and c1

8. 25(4)r2c2 = {3589/4579}: must have 7 or 8 which are only in r4c4 -> r4c4 = (78)

9. Naked pair {78} in r1c1 + r4c4: both locked for D\. Also, no 7,8 in r1c4 since it sees both those other cells (Common Peer Elimination CPE)
9a. outies n1 = h19(4) = {1468/2458/2467}: must have 4 which is only in n2: 4 locked for c4 & n2 and no 4 in r2c3

10. "45" on n478: 2 outies r5c4+r6c5 = 7 (no 7,8,9)

11. 9 in c4 in r6789c4; r8c2 sees all those through D/, 45(9)r5c2 & r8 -> no 9 in r8c2 (Common Peer Elimination, CPE)

12. 25(4)r2c2 = {3589/4579}: note, can't have both 3 & 4
12a. -> {59} blocked from 14(2)r6c2 since it forces r23c2 in the 25(4)r2c2 = {34}(Afmob suggested calling this "Killer pair block". Perhaps it is also another version of "Almost Locked Set block, ALS Block"). Or just "Combos block".......any other suggestions welcome!!
12b. 14(2) = {68} only: both locked for c2

13. "45" on c1234: 1 innie r6c4 - 3 = 1 outie r8c5
13a. r6c4 = (56789), r8c5 = (23456)

14. 9 in c1 only in r4567c1: r5c2 and r6c3 see all of those through 45(9)r5c2 and n4 -> no 9 in r5c2 nor r6c3

15. "45" on D/: 3 innies r7c3+r8c2+r9c1 = 18:
15a. but {189/369} blocked by 6, 8 & 9 only in r7c7
15b. {459} blocked by r1c9 = (45)
15c. {468} blocked by r7c2 = (68)
15d. = {279/378/567}(no 1,4)
15e. must have 6/8/9 which are only in r7c7 -> r7c7 = (689)
15f. must have 7 -> r8c2 = 7; Placed for D/ and 45(9)

16. "45" on r6789: 1 outie r5c2 = 1 innie r6c4: only common digit is 5 so both = 5, 5 in r6c4 placed for D/, 5 in r5c2 placed for 45(9)
16a. -> r8c5 = 2 (IODc1234=+3), 2 placed for 45(9)

on from there.
Cheers
Ed


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 Post subject: Re: Assassin 303
PostPosted: Sun Nov 16, 2014 12:20 am 
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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! Been insanely busy recently so have not been contributing much. Here is my WT.
Tried to, but was not able to use the 45(9) in any meaningful way until the mop-up. (Which is not included in the WT).

Fixes courtesy of Ed.
Hidden Text:
1. Innies r1 -> r1c19 = +12.
Since r1c9 is max 6 -> r1c19 from [93], [84], [75]

2. 3 in n1 either in r1c23 -> r1c1 not 9
or in 25(4)n1 which also puts 9 in the 25(4) -> r1c1 not 9.
Either way 9 not in r1c1
-> 9 in r1 in the 22(4)r1.
Since 9 also in 22(3)n2 -> 9 nowhere else in n23
Also r1c19 either [84] or [75]

3. Innies - Outies n1 -> r4c4 = r1c23 + r2c3 + 1
-> r1c23 + r2c3 = Max +8
-> r1c23 + r2c3 from (123), (124), (125), (134)
Also -> r4c4 from (789)

4. r1c9 from (45) -> 11(3)r1 cannot be {245}

5. If r1c23 contains any two of (124) -> r2c3 also from (124) and r4c4 = 8
-> 11(3)r1 cannot be [{12}8]

6. 7(2)n3 either [43] or [52]
-> There cannot be both 4 in r1c23 AND 2 in r2c3 (fixed typo)
-> 11(3)r1 cannot be [{14}6]

7. -> 11(3)r1 from [{13}7] or [{23}6] - i.e., 3 locked in r1c23
-> 3 in n2 in r23c5
-> 20(3)n2 = [{38}9]
-> r4c4 from (78)
-> [r1c1,r4c4] = {78}
-> r1c4 cannot be 7
-> 11(3)r1 = [{23}6]
-> 2 in 7(3) in r23c4
-> Outies n1 = r1234c4 = +19 and includes (62) - can only be [6{24}7] (clarified)
-> r2c3 = 1
Also -> r1c19 = [84]
-> r2c8 = 3
-> r23c5 = [83]
-> 22(3)n2 = {679} with 6 in r2c7
-> 22(4)r1 = [{15}{79}]

Also 7 in n1 in r23c1
-> 21(3)n1 = [876]
-> 26(4)n1 = [{459}7]

Puzzle essentially cracked at this point. E.g., continuing

8. 10(3)n3 can only be [{25}3]
-> r3c78 = {18}

9. (79) in D/ only in n7
-> Innies D/ r7c3,r8c2,r9c1 = +18 = {279} -> r9c1 = 2
-> 20(4) D/ = {1568} with (56) in n5
-> Outies n478 = r5c4 + r6c5 = +7 can only be r5c4 = 3 and r6c5 = 4
-> r7c5 = 5
-> r1c56 = [15]
-> r6c4 = 5

Also HS 2 in c5 -> r8c5 = 2
Since 7 already in c6 -> r9c5 cannot be 6
-> NP 67 in c5 -> r59c5 = [67]

-> Innies D\ -> r2c2+r3c3 = +9 = {45}
-> r3c2 = 9

etc.


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