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 Post subject: Transformers X Revisit
PostPosted: Mon Nov 01, 2021 6:39 pm 
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NOTE: 1-9 cannot repeat on either diagonal

Transformers X Revisit
Posting this out of order. I initially overlooked it because of its score of 2.20. JSudoku also has a very hard time. Decided to try it anyway..... and loved it with a really neat start available! I found the middle more difficult to get through. This may be my favourite revisit (as long as I haven't messed up!).
Code: Select, Copy & Paste into solver; 1-9 cannot repeat on the diagonals:
3x3:d:k:6656:6656:6656:4611:1028:4613:5894:5894:5894:4105:6656:4611:4611:1028:4613:4613:5894:6673:4105:6656:5908:5909:5909:5909:4376:5894:6673:4105:4105:5908:5908:5909:4376:4376:6673:6673:4644:4644:2854:5908:1576:4376:4650:3883:3883:3885:4644:2854:2854:1576:4650:4650:3883:4149:3885:3885:8760:8760:8760:8760:8760:4149:4149:6463:6463:8760:3906:3906:3906:8760:5446:5446:6463:6463:6463:4427:4427:4427:5446:5446:5446:
Solution:
+-------+-------+-------+
| 8 2 3 | 4 1 9 | 5 6 7 |
| 4 6 9 | 5 3 7 | 2 1 8 |
| 1 7 5 | 2 8 6 | 3 4 9 |
+-------+-------+-------+
| 2 9 6 | 3 7 8 | 1 5 4 |
| 7 3 1 | 9 4 5 | 8 2 6 |
| 5 8 4 | 6 2 1 | 9 7 3 |
+-------+-------+-------+
| 6 4 2 | 1 9 3 | 7 8 5 |
| 3 5 8 | 7 6 2 | 4 9 1 |
| 9 1 7 | 8 5 4 | 6 3 2 |
+-------+-------+-------+


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PostPosted: Wed Nov 03, 2021 8:38 pm 
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Thanks again to Para for an excellent puzzle and to Ed for posting it as a Revisit.

I loved the start and, with a few re-works to simplify some steps, it just kept flowing. Definitely one of the best Revisits (so far).

Thanks Ed for pointing out an omission in step 4d; I've now added the reason for that omission, possibly omitted when I simplified that step.
Here's my walkthrough for Transformers X Revisited:
Prelims

a) R12C5 = {13}
b) R56C5 = {15/24}
c) 11(3) cage at R5C3 = {128/137/146/236/245}, no 9
d) 26(4) cage at R2C9 = {2789/3689/4589/4679/5678}, no 1

1a. Naked pair {13} in R12C5, locked for C5 and N2
1b. R56C5 = {24}, locked for C5 and N5

2a. 45 rule on R89 2 innies R8C37 = 12 = {39/48/57}, no 1,2,6
2b. R8C37 ‘see’ all of N8 except for R9C456 -> 17(3) cage at R9C4 = 12+5 = {359/458}, no 1,2,6,7, 5 locked for R9 and N8
2c. Since 5 is the other number in 17(3) cage, R8C37 = {39/48}, no 5,7
2d. 15(3) cage at R8C4 = {168/267} (cannot be {249/348} which clash with R8C37), no 3,4,9, 6 locked for R8 and N8
2e. 45 rule on N8 3 innies R7C456 = 13 = {139/148/247} (cannot be {238} which clashes with 15(3) cage)
2f. R7C5 = {789} -> no 7,8,9 in R7C46
2g. Killer pair 3,4 in R7C456 and R8C37, locked for 34(7) cage at R7C3
2h. R7C456 = 13, R8C37 = 12 -> R7C37 = 9 = {18/27}, no 5,6,9
2i. 34(7) cage at R7C3 = {1234789}, 1,2,7 locked for R7

3a. 45 rule on R1234 2 outies R5C46 = 14 = {59/68}
3b. 45 rule on R789 2 outies R6C19 = 8 = {17/26/35}, no 4,8,9
3c. 45 rule on C12 2 outies R19C3 = 10 = {19/28/37/46}, no 5 in R1C3
3d. 45 rule on C89 2 outies R19C7 = 11 = {29/38/47}/[56], no 1, no 6 in R1C7

4a. R6C19 (step 3b) = {17/26/35}
4b. 45 rule on N7 2 innies R78C3 = 1 outie R6C1 + 5
4c. R6C1 = {156} (cannot be 2,3 because R78C3 cannot total 7,8, cannot be 7 because R78C3 = 12 clashes with R8C37, combo crossover clash CCC) -> R6C19 = [17/53/62]
4d. Consider permutations for R6C17
R6C19 = [17] => R78C3 = 6 = [24] => R7C37 = [27], R8C37 = [48]
or R6C19 = [53] => R78C3 = 10 = [28/73] => R7C37 = [27/72], R8C37 = [39/84] (R78C3 = 10 cannot be [19] because R7C37 = [18] clashes with 16(3) cage at R6C9 = 3{58}, only place for 5 in R7 when R6C1 = 5)
or R6C19 = [62] => R56C5 = [24], 2 placed for both diagonals, no 2 in R7C37, R78C3 = 11 = [83] => R7C37 = [81], R8C37 = [39]
-> R7C37 = [27/72/81], R8C37 = [39/48/84]
[Taking one of these a bit further]
4e. R6C19 = [17] => R78C3 = 6 = [24] => R7C37 = [27], R8C37 = [48] => R7C5 = 9 (hidden single in 34(7) cage at R7C3, step 2i) => 15(3) cage at R6C1 = 1{68} for 1 in R6C1
4f. 15(3) cage at R6C1 = {168/456} (cannot be {348} because R6C1 only contains 1,5,6), no 3,9
4g. Consider combinations for 15(3) cage
15(3) cage = {168}, R6C19 = [17] => R7C37 = [27], R8C37 = [48} (step 4d)
or 15(3) cage = {456}, 4 locked for R7 => 4 in 34(7) cage only in R8C37 = {48} => R6C19 = [53] (step 4d), R78C3 = 10 = [28] => R7C37 = [27], R8C37 = [84]
-> R6C19 = [17/53], R7C37 = [27], 2 placed for D/, 7 placed for D\, R8C37 = {48}, locked for R8, clean-up: no 8 in R19C3 (step 3c), no 4 in R19C7 (step 3d)
4h. 15(3) cage = {168/456}, 6 locked for R7 and N7, clean-up: no 4 in R1C3 (step 3c)
4i. 34(7) cage at R7C3 = {1234789} -> R7C5 = 9, R7C46 = {13}, locked for N8, 3 locked for R7
4j. R56C5 = [42], 4 placed for both diagonals
4k. Naked triple {458} in 17(3) cage at R9C4, 4,8 locked for R9, 8 locked for N8, clean-up: no 6 in R1C3 (step 3c), no 3 in R1C7 (step 3d)
4l. Naked triple {267} in 15(3) cage at R8C4, 2,7 locked for R8
4m. Naked triple {458} in R7C89 + R8C7, 5 locked for R7 and N9

5a. 11(3) cage at R5C3 = {137/146}, no 5,8
5b. 11(3) cage at R5C3 = {137/146}, CPE no 1 in R6C12
[I originally saw a clash with R6C19 + R8C3 but the CPE is much simpler]
5c. R6C1 = 5 -> R6C9 = 3 (step 4g), R78C3 = 10 (step 4b) -> R8C3 = 8, R8C7 = 4
5d. 3 of 11(3) cage only in R5C3, no 7 in R5C3
5e. 4 of 11(3) cage only in R6C3, no 6 in R6C3
5f. 3 in N5 only in R4C46, locked for R4
5g. R8C2 = 5 (hidden single in N7), placed for D/

6a. 18(3) cage at R5C7 = {189} (only possible combination), CPE no 1,8,9 in R6C8
6b. Double hidden killer pair 8,9 in 18(3) cage at R5C1, R5C46 and 18(3) cage at R5C7 for R56, 18(3) cage at R5C7 contains both of 8,9, R5C46 contains one of 8,9, 18(3) cage at R5C1 must contain at least one of 8,9 -> 18(3) cage at R5C1 must only contain one of 8,9, no 8,9 in 15(3) cage at R6C8
6c. 15(3) cage at R5C8 = {267} (cannot be {456} which clashes with R5C46), locked for N6, 2 locked for R5
6d. 4 in N6 only in R4C89, locked for R4 and 26(4) cage at R2C9
6e. 26(4) cage = {4589/4679}, no 2
6f. 2 in R4 only in R4C12, locked for 16(4) cage at R2C1
6g. 18(3) cage at R5C1 = {369/378/468} (cannot be {189} because this 18(3) cage only contains one of 8,9, step 6b), no 1
6h. 5 in R6 only in R5C46 (step 3a) = {59}, locked for N5
6i. R3C3 = 5 (hidden single on D\) -> R5C46 = [95]
[Cracked. Fairly straightforward from here.]

7a. 18(3) cage at R5C7 = {189} -> R6C7 = 9, clean-up: no 2 in R19C7 (step 3d)
7b. Naked pair {18} in R45C7, locked for C7, 8 locked for N6
7c. R1C7 = 5 -> R9C7 = 6 (step 3d), R3C7 = 3, placed for D/
7d. R2C7 = 2 -> R12C6 = 16 = {79}, locked for N2, 7 locked for C6
7e. R4C89 = {45} = 9 -> R23C9 = 17 = {89}, locked for C9 and N3 -> R89C9 = [12], 2 placed for D\, R7C89 = [85], R4C89 = [54]
7f. Naked pair {39} in R8C18, CPE no 3,9 in R1C1 using D\
7g. 23(4) cage at R3C4 = {2678} (only remaining combination) -> R4C5 = 7, R38C5 = [86], R3C46 = {26}, locked for R3 and N2
7h. R4C12 = {29} (hidden pair in R4) = 11 -> R23C1 = 5 = {14}, locked for N1, 4 locked for C1 -> R7C12 = [64], R1C1 = 8, placed for D\
7i. R12C4 = [45] -> R2C3 = 9 (cage sum), R3C2 = 7, R1C3 = 3 -> R9C3 = 7 (step 3c), R12C2 = [26], 6 placed for D\
7j. R6C6 = 1 -> R4C4 = 3, placed for D\

and the rest is naked singles, without using the diagonals


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PostPosted: Tue Nov 09, 2021 6:54 pm 
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Loved Andrew's 4c! I did this quite differently with a different "45" to work with (step 6).
Walkthrough for Trans X:
Preliminaries
Cage 4(2) n2 - cells ={13}
Cage 6(2) n5 - cells only uses 1245
Cage 11(3) n45 - cells do not use 9
Cage 26(4) n36 - cells do not use 1

This is a highly optimised solution so no clean-up etc done unless stated.
1. 4(2)n2 = {13}: both locked for n2 and c5
1a. -> 6(2)n5 = {24}: both locked for n5 and c5

2. "45" on r89: 2 innies r8c37 = 12 = {39/48/57}(no 1,2,6)

3. that h12(2) sees all of n8 apart from r9c456 so must repeat there -> the 17(3)r9c456 = 5{39/48}(no 1,2,6,7)
3a. 5 locked for n8 and r9
3b. and h12(2)r8c37 = {39/48}(no 5,7)

4. "45" on n8: 3 innies r7c456 = 13
4a. but {238/346} blocked by h12(2)r8c37 (or 17(3)n8)
4b. = {139/148/247}(no 6) = 1 or 4

5. killer pair 3,4 between h13(3)r7c456 and h12(2)r8c37
5a. -> no 3,4 in r7c37 nor r8c456

6. from step 3, r8c3 repeats in r9c456 -> must also repeat in r7c89
6a. "45" on n89: 2 outies r78c3 + 3 = 2 innies r7c89
6b. since r8c3 = one of r7c89 -> r7c3 + 3 = one of r7c89
6c. but can't be [14] since that would clash with r7c456 (step 4b)
6d. and max. r7c3 = 6
6e. -> r7c3 from (256)

7. r7c456 = 13 and r8c37 = 12 -> r7c37 = 9 (cage sum)
7a. = [27] only, 2 placed for d/, 7 placed for d\
7b. -> r56c5 = [42], 4 placed for both d/\

8. hidden triple 2,6,7 in r8c456: all locked for r8

9. r7c3 = 2 -> r7c89 must have 5 (step 6b)
9a. 5 locked for r7, n9 and 16(3) cage
9b. 16(3) = 5{38/47}(no 1,6,9)

10. 6 in r7 only in 15(3)r6c1 = 6{18/45} (no 3,9)
10a. 6 locked for n7 and 15(3)

11. 11(3)r5c3 = {137/146}(no 5,8)
11a. must have 1 -> no 1 in r6c12 (Common Peer Elimination (CPE)

12. "45" on r789: 2 outies r6c19 = 8 = [53]
12a. -> r7c12 = 10 = {46}: 4 locked for r7 and n7
12b. and r7c89 = 13 = {58}: 8 locked for r7 and n9
12c. r7c5 = 9

13. h12(2)r8c37 = [84] only

14. r8c2 = 5 (hsingle n7), placed for d/
14a. -> 5 on d\ only in 23(4)r3c3, 5 locked for 23(4)
14b. and no 5 in r3c4 (CPE)

15. "45" on r1234: 2 outies r5c46 = 14 = [95]/{68}

16. 18(3)r5c7 = {189} only
16a. -> no 1,8,9 in r6c8 (CPE)

17. caged x-wing on 1 with 11(3)r5c3 and 18(3)r5c7: 1 locked for r56

18. 5 in r5 only in h14(2) = [95] or in 15(3)r5c8 -> 9 in 15(3) must also have 5
18a. -> {249} blocked (Locking-out cages)
18b. and must have 4,6,7 for r6c8
18c. but {56}[4] blocked by h14(2)hr5c46 which needs 5 or 6
18d. = {267} only, all locked for n6, 2 for r5

19. r5c6 = 5 (hsingle r5)
19a. -> r5c4 = 9 (h14(2))
19b. r6c7 = 9 (18(3) must have 9)

20. 4 in c9 only in r234c9 -> no 4 in r4c8
20a. -> r4c9 = 4 (hsingle n6)
20b. and r4c8 = 5 (hsingle n6)
20c. -> r23c9 = 17 (cage sum) = {89}: both locked for c9 and n3
20d. -> from "45" on n3, r23c7 = 5 = [23]: 3 placed for d/
20e. r8c9 = 1
20f. r19c7 = [56] (5 is hsingle c7)
20g. r9c9 = 2, placed for d\

21. 2 in r4 only in 16(4)r2c1, locked for that cage
21a. r4c1 = 2 (hsingle c1)
21b. r4c2 = 9 (hsingle n4)
21d. r3c3 = 5 (hsingle d\), r2c4 = 5 (hsingle n2)

22. r4c6 = 8 (hsingle d/)
22a. r6c6 = 1, placed for d\

23. naked pair {67} in r1c9 + r6c4: both locked for d/
23a. and no 6,7 in r1c4 (CPE)

24. r1c4 + r2c3 = 13 (cage sum) = [49] only

on from there
Cheers
Ed


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PostPosted: Fri Nov 12, 2021 10:58 pm 
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Loved Ed's step 6 which led to a very direct, somewhat different solving path than mine!


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PostPosted: Fri Nov 19, 2021 10:44 pm 
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That was fun! My WT started off very similar to Ed's.
Transformers X Revisit WT:
1. 4(2)n2 = {13}
-> 6(2)n5 = {24}

2. Innies r89 = r8c37 = +12(2).
Call whatever goes in r8c3 = A and in r8c7 = B. A+B = 12.

A in r7 only in r7c89
-> Remaining Innies n9 = +12(2) with one in r7c7 and the other in r7c89
-> (126) in n9 all in 21(5)n9

Also B in r7 in r7c12
-> Remaining Innies n7 = +8(2) with one in r7c3 and the other in r7c12

Also AB in r9 only in n8
-> 17(3)n8 = {5AB}
-> [AB] from {39} or {48}

3. Innies n8 = r7c456 = +13(3)
-> Remaining cells in 34(7) = r7c37 = +9(2)
Since 5 already in n8 -> In n8 (26) cannot both be in r7 (Would put r7c456 = {256})
-> At least one of (26) in r7 in r7c123
-> Innies n7 = {AB26} with (26) both in r7 with one in r7c3.
-> (26) in n8 in r8
-> 15(3)n8 = {267}

4. Also 5 in n7 only in r8c12
-> 5 in n9 only in r7
-> Innies n9 = {AB57} with one of (57) in r7c7
-> (Since r7c37 = +9(2)) -> r7c37 = [27]
-> 6(2)n5 = [42]

5. Also r7c12 = {6B} with B from (3489)
-> 15(3)r6c1 from [1{68}] or {5{46}]
I.e., B from (48)
-> [AB] = {48}
-> 17(3)n8 = {458}
-> r7c456 = <193>

6. 2 not in 11(3)r5c3 -> 11(3)r5c3 must contain a 1.
-> 1 not in r6c1
-> 15(3)r6c1 = [5{46}]
-> r8c37 = [AB] = [84]
Also 16(3)r6c9 = [3{58}]
Also HS 5 in n7 -> r8c2 = 5

7. 25(5)n7 = {13579}
-> Outies c12 = r19c3 = +10(2) = {19} or {37}
-> 5 in n1 only in r23c3

8. Previous placements means only possibilities for 11(3)r5c3 are [146], [641], or [371]. (Cannot be [317] because of Step 7 Line 2).
Outies r1234 = r5c46 = +14(2) = {59} or {68}
-> 7 in n5 only in r4c56
-> 7 in n6 only in 15(3)n6
-> 15(3)n6 = {267}
-> 4 in n6 only in r4c89

9. IOD n3 -> r2c7 + r3c7 + 4 = r4c8 + r4c9
Since 4 in r4c89 -> r2c7 + r3c7 = other value in r4c89
-> Values in r23c7 in n6 both in r56c89
-> r23c7 both from (236)
Bur r23c7 cannot be {36} since that puts 26(4)r2c9 = [{58}{49}] which contradicts r7c89 = {58}
-> (Since 2 already in D/) -> r2c7 = 2
-> 18(3)r1c6 = [{79}2]

10. Remaining outies n2 = r2c3 + r4c5 = +16(2) = [88] or [97]
But cannot be [88] since that leaves no place for 8 in n2
-> r2c3 = 9 and r4c5 = 7
-> r12c4 = {45}
-> r3c456 = {268}

11. Also (HS 5 in n1/c3) r3c3 = 5
-> (HS 5 in n5) r5c46 = [95]
-> 5 in n6 in r4c89
-> 26(4)r2c9 = [89{45}]
-> (Remaining Innie n3) r3c7 = 3
Also r7c89 = [85]
-> r4c89 = [54]

12. Also r19c3 = {37}
Also (HS 3 in n5) r4c4 = 3
-> r4c3 = 6
-> 11(3)r5c3 = [146]
etc.


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