SudokuSolver Forum

A forum for Sudoku enthusiasts to share puzzles, techniques and software
It is currently Fri Feb 22, 2019 12:58 am

All times are UTC




Post new topic Reply to topic  [ 4 posts ] 
Author Message
 Post subject: Assassin 364
PostPosted: Sat Dec 01, 2018 7:57 am 
Offline
Grand Master
Grand Master

Joined: Wed Apr 16, 2008 1:16 am
Posts: 788
Location: Sydney, Australia
Attachment:
a364.JPG
a364.JPG [ 64.05 KiB | Viewed 209 times ]
This is the third version of this puzzle. The first two versions were one trick puzzles which just doesn't satisfy anymore. This one makes you work for much longer - and I found it quite difficult to break. Very satisfying to finally get a decent solution! SudokuSolver gives it 1.90 and JSudoku needs to use 6 chains.
code:
3x3::k:5632:5632:3585:3585:3585:5378:4099:4099:4099:7940:5632:5632:3585:3589:5378:4099:5382:5382:7940:1543:1543:4360:3589:5378:2313:2313:5382:7940:0000:0000:4360:4360:2315:2315:0000:5382:7940:7940:5645:5645:5645:5645:5645:0000:0000:4110:7940:2831:2831:3344:3344:3345:3345:0000:4110:1042:1042:5139:2324:3344:3349:3349:0000:4110:4110:6678:5139:2324:5642:3340:3340:0000:6678:6678:6678:5139:5642:5642:5642:3340:3340:
solution:
Code:
+-------+-------+-------+
| 6 7 5 | 3 2 8 | 4 1 9 |
| 3 1 8 | 4 9 6 | 2 5 7 |
| 9 4 2 | 1 5 7 | 6 3 8 |
+-------+-------+-------+
| 2 6 3 | 9 7 4 | 5 8 1 |
| 4 8 7 | 6 1 5 | 3 9 2 |
| 1 5 9 | 2 8 3 | 7 6 4 |
+-------+-------+-------+
| 8 3 1 | 7 6 2 | 9 4 5 |
| 5 2 4 | 8 3 9 | 1 7 6 |
| 7 9 6 | 5 4 1 | 8 2 3 |
+-------+-------+-------+

Cheers
Ed


Top
 Profile  
Reply with quote  
 Post subject: Re: Assassin 364
PostPosted: Fri Dec 07, 2018 10:44 pm 
Offline
Grand Master
Grand Master

Joined: Wed Apr 16, 2008 1:16 am
Posts: 788
Location: Sydney, Australia
Here's how I solved it. Steps 8, 19 & 26 are my keys with one of those steps getting no (useful) eliminations!
WT 364:
Prelims courtesy of SudokuSolver
Preliminaries
Cage 4(2) n7 - cells ={13}
Cage 6(2) n1 - cells only uses 1245
Cage 14(2) n2 - cells only uses 5689
Cage 13(2) n9 - cells do not use 123
Cage 13(2) n6 - cells do not use 123
Cage 9(2) n56 - cells do not use 9
Cage 9(2) n8 - cells do not use 9
Cage 9(2) n3 - cells do not use 9
Cage 11(2) n45 - cells do not use 1
Cage 21(3) n2 - cells do not use 123
Cage 20(3) n8 - cells do not use 12
Cage 13(4) n9 - cells do not use 89
Cage 14(4) n12 - cells do not use 9
Cage 26(4) n7 - cells do not use 1

No routine clean-up done unless stated
1. "45" on n7: 1 outie r6c1 = 1
1a. "45" on n3: 1 outie r4c9 = 1

2. 4(2)n7 = {13} only: both locked for n7 and r7

3. "45" on n8: 1 innie r7c6 + 6 = 1 outie r9c7 = [28] only permutation
3a. r7c6 = 2 -> r6c56 = 11 (no 9)
3b. no 7 in 9(2)n8, no 6,8 in r8c5
3c. no 5 in 13(2)n9

4. 21(3)n2: {489} blocked by 14(2)n2 needing an 8 or 9
4a. = {579/678}(no 4)
4b. must have 7: 7 locked for n2 and c6
4c. no 4 in r6c5 (sp11(2))

5. 14(2)r23c5 = {59/68}; 9(2)r78c5 = [81]/[63]/{45}
5a. if 14(2) = {68} -> 9(2) = {45}; or 14(2) = {59}
5b. -> 5 locked in one of those two cages: locked for c5 (combined cages)
5c. no 6 in r6c6 (sp11(2))

6. "45" on n2: 1 innie r3c4 + 4 = 1 outie r1c3
6a. r1c3 = (5678), r3c4 = (1234)
6b. hidden quad 1,2,3,4 in n2 -> r123c4+r1c5 = {1234}
6c. 17(3)r3c4 can only have one of 1,2,3,4 -> no 2,3,4 in r4c45

7. 20(3)r7c4 = {389/479/569/578} = two of 6..9
7a. -> r456c4 must have two of 6..9 for c4
7b. r4c5 = (6789)
7c. sp11(2)r6c56 must have one of 6,7,8
7d. -> killer quad 6,7,8,9 in those four areas: all locked for n5
7e. r4c7 = (456)

Ready for the first advanced step - which totally cracked one of the previous versions of this puzzle
8. 21(3)n2 = {579}/678}
8a.if {579} -> 14(2)n2 = {68} (combined cage)
8b. r9c7 = 8 -> r89c6 + r9c5 = 14 = {149/167/347/356}
8c. if {149} -> 9(2)n8 = [63] only (combined cage)
8d. 9 in c6 in one of 21(3)n2 or r89c6 -> 14(2)n2 = {68} or 9(2)n8 = [63]
8e. must have 6: 6 locked for c5
8f. no 5 in r6c6 (sp11(2))

9. 6 in c5 in 14(2)n2 = {68} or 6 in 9(2)n8 -> no {18} in 9(2)n8 since it would block all 6 in c5 (locking-out cages)
9a. 9(2) = [63]/{45}

10. 8 in n8 only in 20(3) = {389/578}(no 4,6) = 3 or 5
10a. 8 locked for c4

11. 9(2)n8 = [63]/{45} = 3 or 5
11a. killer pair 3,5 with 20(3) (step 10): both locked for n8

12. 3 in c6 only in n5: 3 locked for n5
12a. no 8 in r6c6 (sp11(2))

13. 8 in c6 only in 21(3)n2 = {678} only: 6,8 locked for n2, 6 for c6

14. 14(2)n2 = {59}: both locked for c5
14a. r78c5 = [63]
14b. 13(2)n9 = {49} only: both locked for n9 and r7

15. "45" on n9: 2 remaining innies r78c9 = 11 = [56] only permutation

16. 16(4)r6c1: must have 7 or 8 for r7c1 = [1]{258} only
16a. R7C1 = 8
16b. r8c12 = {25} both locked for r8
16c. r7c4 = 7

17. r8c4 = 8 (hidden single n8)
17a. r9c4 = 5 (cage sum)

18. 17(3)r3c4 must have 6 or 9 for r4c4, and 7 or 8 for r4c5 = {179/368/467}(no 2)
18a. no 6 in r1c3 (IODn2=-4)

Next key step
19. "45" on n2: 3 outies r1c3 + r4c45 = 21 = [5][97]/[7][68]/[8][67]:
19a. note: must have 5 in r1c3 or 6 in r4c4
19b. note2: must have 7 in r1c3 or r4c5
19c. note: no eliminations yet (can take 7 from r4c3 but not important to this optimised solution)

20. 11(2)r6c3: [56] blocked by step 19a
20a. = {29}/[74] = 4 or 9

21. 13(2)n6: {49} blocked by 11(2)n4
21a. = [58]/{67}(no 4,9; no 5 in r6c8)

22. hidden pair 5,6 in r6. 13(2)n6 can only have one of 5,6 -> r6c2 = (56)

23. "45" on r6: 2 remaining innies r6c29 = 9 = [54/63]

24. hidden pair 2,9 in r6 -> 11(2)r6c3 = {29} only

25. "45" on n1: 3 innies r1c3 + r23c1 = 17
25a. but {458} blocked by 6(2)n1 needing 4 or 5
25b. must have 5,7,8 for r1c3 = {278/359/368/467}
25c. note: if has 7 in r1c3 must have {46} in r23c1
25d. note: no eliminations yet

The final crack by removing 6 from r6c2
26. from step 19b. must have 7 in r1c3 or r4c5
26a. if in r1c3 -> r23c1 = {46} (step 25c) -> no 6 in r6c2 (same cage)
26b. if in r4c5 -> 7 in r6 only in 13(2)r6c78 = {67}
26c. -> both places have 6 -> no 6 in r6c2

27. r6c2 = 5
27a. r6c9 = 4 (h9(2)r6c29), r6c56 = [83], r4c5 = 7 -> r34c4 = 10 = [19/46] (no 3)
27b. no 7 in r1c3 (IODn2=-4)
27c. r8c12 = [52]

28. 31(6)r2c1 = {23489}[5]/{23678}[5]
28a. must have 8 -> r5c2 = 8
28b. must have 2 & 3 which are only in c1: locked for c1

29. naked pair {14} in r3c24: both locked for r3

30. naked pair {67} in r6c78: locked for n6
30a. r4c67 = [45], naked pair {19} in r89c6: 1 locked for n8 and c6, r9c5 = 4, r5c6 = 5

31. 22(5)r5c3 must have 1 for n5 = {13459/13567}(no 2)
31a. must have 3: 3 locked for r5
31b. r5c5 = 1

32. naked pair {29} in r5c89: both locked for r5 and n6

Straightforward now.
Cheers
Ed


Top
 Profile  
Reply with quote  
 Post subject: Re: Assassin 364
PostPosted: Sat Dec 08, 2018 9:22 pm 
Offline
Master
Master

Joined: Tue Jun 16, 2009 9:31 pm
Posts: 177
Location: California, out of London
Thanks Ed! This was the puzzle that kept on giving. As you wrote - several key steps needed.
As usual I wrote my WT before reading yours. Some similarities and some differences. Where there were similarities we often approached each step from opposite directions.
Assassin 364 WT:
1. Outies n3 -> r4c9 = 1
Outies n7 -> r6c1 = 1
-> 1 in n5/r5 in r5c456

2. 4(3)n7 = {13}
Outies - Innies n8 -> r9c7 = r7c6 + 6
-> [r7c6,r9c7] = [28]

3. Innies n2 -> r1c45 + r23c4 = +10(4) = {1234}
-> 21(3)n2 = {7(68|59)}
Also Since Max r3c4 = 4 -> both r4c45 are Min 5.

4! 1 in c6 only in r589c6
9 in c6 in one of:
a) r123c6 -> 14(2)n2 = {68}
b) r5c6 -> 1 in r89c6
c) r89c6 -> 22(4)n8 = [{149}8]
In none of those cases can 9(2)n8 be {18}
-> 8 in n8 in 20(3)n8
-> 20(3)n8 = {8(57|39)}

5! 6 in c4 only in n5 in r456c4
Also -> 1 in n8 in 22(4)n8
-> 3 in n8 in 20(3) or 9(2)
-> 3 in c6 in n5 in r456c6
-> 13(3) r6c5 from [832] or [742]
If the former -> 14(2)n2 = {59} -> 9(2)n8 = [63] -> 20(3)n8 = {578}
If the latter -> 7 in 20(3)n8 = {578} -> 9(2)n8 = [63] -> 14(2)n2 = {59}
Either way 20(3)n8 = {578}, 9(2)n8 = [63], 14(2)n2 = {59}, 21(3)n2 = {678}, 22(4)n8 = [{149}8]
-> 9 in c4 in n5 in r456c4
Also 5 in c6 in n5 in r45c6
Also -> r46c5 = {78}

(Having read Ed's WT the next step is more complicated than it need be since 13(2)n9 can already only be {49})!

6! Remaining Innies n9 -> r78c9 = +11(2)
3 in r9 only in r9c89
-> 13(4)n9 cannot contain a 6
-> 6 in n9 in 13(2) or H11(2)
-> 6 in r9 in n7 in r9c123
-> 2 in n7 in r8c12
-> 2 in r9 in r9c89
-> 13(4)n9 = [{17}{23}]
-> 13(2)n9 = {49}
-> r78c9 = [56]
-> 16(4)r6c1 can only be [18{25}]
-> 20(3)n8 = [785]
Also 26(4)n7 = {4679} with r8c3 from (49)
Also 22(4)n8 = [{149}8] with r8c6 from (49)

7. (27) already in c6 and 1 already in r4c9 -> 9(2)r4 from {36} or {45}
r4c4 from (69) and r4c5 from (78)
-> 17(3)r3c4 from [197], [368], [467]
The first of these -> r6c56 = [83]
-> In all cases 9(2)r4 can only be {45}

8. 8 in n5 in r46c5 and 3 in n5 in r56c6
-> 11(2)r6 cannot be {38}
Remaining Innies r6 -> r6c29 = +9(2) (No 9)
-> 9 either in 11(2)r6c3 = {29} or 13(2)r6 = {49}
-> 11(2)r6 cannot be {47}
-> 11(2)r6c3 from {29} or [56]

9. 5 in r7c9 prevents H9(2)r6c29 = [45]
-> 4 in n4 in r5c123
-> 4 in n5 only in r46c6
-> 22(4)n8 = [9418]
-> 26(4)n7 = [4{679}]
-> 4 in n4 in r5c12
-> 31(6) = {2349(58|67)}

10. Innies n1 = r1c3 + r23c1 = +17(3)
Innies - Outies n2 -> r1c3 = r3c4 + 4
Since r3c4 from (134) -> r1c3 from (578)
But r1c3 = 7 -> r23c1 = +10(2) for which there is no solution
-> r1c3 from (58)
-> r3c4 from (14)
-> 17(3)r3c4 from [197] or [467]
-> r4c5 = 7
-> 13(3)r6c5 = [832]

(This next step is essentially Ed's Step 19a).

11! 5 in r6 only in r6c23
r1c3 from (58)
If r1c3 = 5 puts 5 in r6c2
If r1c3 = 8 puts 17(3)r3c4 = [467] -> 5 not in r6c3
Either way r6c2 = 5

12. -> r6c9 = 4
-> 9(2)r4 = [45]
Also 13(2)n6 = {67}
-> 11(2)r6 = {29}
Also r8c12 = [52]
Also -> 31(6) = {234589}
-> r5c2 = 8
-> r5c1 = 4
-> r234c1 = {239}
-> Innies n1 can only be r1c3 = 5, r23c1 = {39}
-> r4c1 = 2 and r3c4 = 1
-> r4c4 = 9
-> 11(2)r6 = [92]
-> r5c456 = [615]
Also 6(2)n1 = [42]
etc.


Top
 Profile  
Reply with quote  
 Post subject: Re: Assassin 364
PostPosted: Mon Feb 04, 2019 5:59 am 
Offline
Grand Master
Grand Master

Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1614
Location: Lethbridge, Alberta, Canada
Another Assassin which I'd skipped over at the time and had just started when I saw that A369 had been posted, so I continued with it.

As wellbeback said, it kept giving, well at least for a while.

Thanks Ed for pointing out a flawed step and hinting at an alternative way to get the same result.
Here's my walkthrough for Assassin 364:
Prelims

a) R23C5 = {59/68}
b) R3C23 = {15/24}
c) R3C78 = {18/27/36/45}, no 9
d) R4C67 = {18/27/36/45}, no 9
e) R6C34 = {29/38/47/56}, no 1
f) R6C78 = {49/58/67}, no 1,2,3
g) R7C23 = {13}
h) R78C5 = {18/27/36/45}, no 9
i) R7C78 = {49/58/67}, no 1,2,3
j) 21(3) cage at R1C6 = {489/579/678}, no 1,2,3
k) 20(3) cage at R7C4 = {389/479/569/578}, no 1,2
l) 14(4) cage at R1C3 = {1238/1247/1256/1346/2345}, no 9
m) 13(4) cage at R8C7 = {1237/1246/1345}, no 8,9
n) 26(4) cage at R8C3 = {2789/3689/4589/4679/5678}, no 1

Steps Resulting From Prelims and Immediate Placements
1a. Naked pair {13} in R7C23, locked for R7 and N7, clean-up: no 6,8 in R8C5
1b. R3C78 = {18/27/36} (cannot be {45} which clashes with R3C23), no 4,5
1c. 13(4) cage at R8C7 = {1237/1246/1345}, 1 locked for N9
1d. 45 rule on N3 1 outie R4C9 = 1 -> R2C89 + R3C9 = 20 = {389/479/569/578}, no 2, clean-up: no 8 in R4C67
1e. 45 rule on N7 1 outie R6C1 = 1

2. 45 rule on N8 1 outie R9C7 = 1 innie R7C6 + 6 -> R7C6 = 2, R9C7 = 8, clean-up: no 1 in R3C8, no 7 in R4C7, no 5 in R6C8, no 7 in R78C5, no 5 in R7C78
2a. R7C6 = 2 -> R6C56 = 11 = {38/47/56}, no 9
2b. 45 rule on N9 2 remaining innies R78C9 = 11 = [56/65/92] (cannot be {47} which clashes with R7C78), no 4,7, no 3,9 in R8C9
2c. 13(4) cage at R8C7 = {1237/1345} (cannot be {1246} which clashes with R78C9), no 6
2d. 9 in N9 only in R7C789, locked for R7
2e. 45 rule on R6 2 remaining innies R6C29 = 9 = {27/36/45}, no 8,9
2f. 9 in R6 only in R6C34 = {29} or R6C78 = {49} -> R6C34 = {29/38/56} (cannot be {47} locking-out cages), no 4,7
[2 in R6 only in R6C29 and R6C34 would give the same elimination.]

3a. 45 rule on N2 4 innies R1C45 + R23C4 = 10 = {1234}, 4 locked for N2
3b. 45 rule on N2 1 outie R1C3 = 1 innie R3C4 + 4 -> R1C3 = {5678}
3c. 7 in N2 only in 21(3) cage at R1C6, locked for C6, clean-up: no 2 in R4C7, no 4 in R6C5 (step 2a)
3d. 17(3) cage at R3C4 can only contain one of 1,2,3,4 -> no 2,3,4 in R4C45

4. 45 rule on R1 3 innies R1C126 = 2 outies R2C47 + 15
4a. Max R1C126 = 24 -> max R2C47 = 9, no 9 in R2C7

5. 16(4) cage at R1C7 = {1249/1258/1348/1456/2347/2356} (cannot be {1267/1357} which clash with R3C78)
5a. 7 on {2347} must be in R1C789 (R1C789 cannot be {234} which clash with R1C45, ALS block), no 7 in R2C7

6. 26(4) cage at R8C3 = {2789/4589/4679/5678}
6a. 8 of {2789/4589/5678} must be in R8C3 -> no 2,5 in R8C3

7. R23C5 = {59/68}, R78C5 = {45}/[63/81] -> combined cage = {59}[63]/{59}[81]/{68}{45}, 5 locked for C5, clean-up: no 6 in R6C6 (step 2a)
7a. 22(4) cage at R8C6 = {1489/1678/3478/3568}
7b. 7 of {3478} must be in R9C5, 6 of {3568} must be in R9C5 (R89C6 cannot be {56} which clashes with 21(3) cage at R1C6), no 3 in R9C5
7c. 22(4) cage = {1489/1678/3478} (cannot be {3568} because R89C6 = {35} clashes with 21(3) cage at R1C6 + R46C6, killer ALS block), no 5 in R89C6
7d. Consider combinations for R23C5 = {59/68}
R23C5 = {59} => 21(3) cage at R1C6 = {678}, 6,8 locked for C6, R4C6 = {345}, R6C6 = {345} => R89C6 cannot be {34}, ALS block
or R23C5 = {68}, locked for C5 => R78C5 = {45}, 4 locked for N8
-> 22(4) cage = {1489/1678} (cannot be {3478})
, no 3 in R89C6, 1 locked for N8, clean-up: no 8 in R7C5
[My original step 7d was flawed. Thanks Ed for suggesting an alternative way, which I’ve rewritten in my solving style.]
7e. 3 in C6 only in R456C6, locked for N5, clean-up: no 8 in R6C3, no 8 in R6C6 (step 2a)
7f. Killer pair 5,6 in R23C5 and R78C5, locked for C5, clean-up: no 5 in R6C6 (step 2a)
7g. 8 in N8 only in 20(3) cage at R7C4, locked for C4, clean-up: no 3 in R6C3
7h. 20(3) cage = {389/578}, no 4,6
7i. Killer triple 7,8,9 in R23C5, R4C5 and R6C5, locked for C5
7j. 22(4) cage = {1489} (only remaining combination), 4,9 locked for N8, clean-up: no 5 in R78C5
7k. R78C6 = [63], clean-up: no 8 in R23C5, no 7 in R7C78, no 5 in R8C9 (step 2b)
7l. Naked pair {59} in R23C5, locked for C5 and N2
7m. Naked triple {678} in 21(3) cage at R1C6, 6,8 locked for C6, clean-up: no 3 in R4C7
7n. Naked triple {578} in 20(3) cage at R7C4, 5,7 locked for C4, clean-up: no 6 in R6C3
7o. 9 in C4 only in R456C4, locked for N5
7p. Naked pair {49} in R7C78, locked for R7 and N9, clean-up: no 2 in R8C9 (step 2b)
7q. R78C9 = [56], clean-up: no 3,4 in R6C2 (step 2e)
7r. 16(4) cage at R6C1 = {1258} (only possible combination, cannot be {1249} because R7C1 doesn’t contain 4,9) -> R7C1 = 8, R8C12 = {25}, locked for R8 and N7
7s. Naked pair {17} in R8C78, locked for R8 and N9
7t. R2C89 + R3C9 (step 1d) = {389/479/578} (cannot be {569} because 5,6 only in R2C8), no 6
[Note. There is now the Unique Rectangle elimination R6C78 cannot be {49} because R7C78 = {49} but I don’t use that type of step since it doesn’t fully solve a puzzle.]

8. 17(3) cage at R3C4 = [197/368/467] (cannot be {269} because no 7,8 in R4C5, cannot be {278} because no 7,8 in R4C4), no 2 in R3C4, clean-up: no 6 in R1C3 (step 3b)
8a. Consider permutations for 17(3) cage
17(3) cage = [179] => R6C5 = 8, R6C6 = 3 (cage sum) => R4C67 = {45} (only remaining combination)
or 17(3) cage = [368/467] => R4C67 = {45} (only remaining combination)
-> R4C67 = {45}, locked for R4
[This proved to be a key step.]
8b. 4 in N4 only in R5C123, locked for R5
8c. 4 in N5 only in R46C6, locked for C6 -> R89C6 = [91], R9C5 = 4, R8C3 = 4, clean-up: no 2 in R3C2
8d. 4 in N4 only in R5C12, locked for 31(6) cage at R2C1, no 4 in R23C1

9. 45 rule on N1 3 innies R1C3 + R23C1 = 17 = {278/359/368} (cannot be {269} because R1C3 only contains 5,7,8)
9a. 5 of {359} must be in R1C3 -> no 5 in R23C1
9b. 8 of {278} must be in R1C3 -> no 7 in R1C3, clean-up: no 3 in R3C4 (step 3b)
9c. 17(3) cage at R3C4 (step 8) = [197/467] -> R4C5 = 7, R6C5 = 8, R6C6 = 3 (cage sum), R5C6 = 5, R4C67 = [45], clean-up: no 6 in R6C2 (step 2e)
9d. Killer pair 1,4 in R3C23 and R3C4, locked for R3, clean-up: no 8 in R3C8
9e. Killer pair 3,7 in R2C89 + R3C9 (step 7t) and R3C78, locked for N3
9f. 1 in N3 only in 16(4) cage at R1C7 (step 5) = {1249/1456} (cannot be {1258} which clashes with R1C3), no 8, 4 locked for N3
9g. 5 of {1456} must be in R1C8 -> no 6 in R1C8
9h. R2C89 + R3C9 (step 7t) = {389/578}
9i. 5 of {578} must be in R2C8 -> no 7 in R2C8

10. 1,5 in R5 only in 22(5) cage at R5C3 = {12568/13567}, no 9, 6 locked for R5
10a. 8 of {12568} only in R5C3 -> no 2 in R5C3

11. R1C3 + R23C1 (step 9) = {278/359/368}
11a. 4 in N4 only in 31(6) cage at R2C1 = {234589/234679}
11b. {234589} only has one of 2,7, {234679} must have one of 2,7 in R6C2 so cannot have both in R23C1 -> R1C3 + R23C1 = {359/368} (cannot be {278} because cannot have both of 2,7 in R23C1), no 2,7, 3 locked for N1 and 31(6) cage, no 3 in R45C1 + R5C2
[Cracked.]
11c. 31(6) cage = {234589/234679}, 2 locked for N4, clean-up: no 9 in R6C4
11d. R4C4 = 9 (hidden single in N5) -> R3C4 = 1 (cage sum), R1C5 = 2, R1C3 = 5 (step 3b), R3C23 = [42], R6C3 = 9 -> R6C4 = 2, clean-up: no 7 in R3C78, no 4 in R6C78
[Removing any need for the Unique Rectangle.]
11e. Naked pair {67} in R6C78, locked for R6 and N6 -> R16C9 = [94]
11f. Naked pair {36} in R3C78, locked for R3 and N3
11g. Naked pair {78} in R23C9, 8 locked for C9 and N3 -> R2C8 = 5
11h. Naked pair {14} in R1C78, locked for R1 and N3 -> R2C7 = 2
11i. R5C4567 = [6153] -> R5C3 = 7 (cage sum)

and the rest is naked singles.

Rating Comment:
I'll rate my WT for A364 at Easy 1.5 because I used a locking-out cages step and a couple of forcing chains. I wasn't sure how to rate my final breakthrough in step 11b, but not any higher.


Top
 Profile  
Reply with quote  
Display posts from previous:  Sort by  
Post new topic Reply to topic  [ 4 posts ] 

All times are UTC


Who is online

Users browsing this forum: No registered users and 17 guests


You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot post attachments in this forum

Search for:
Jump to:  
cron
Powered by phpBB® Forum Software © phpBB Group