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Afmob
Expert

Joined: 22 Sep 2007
Posts: 103
Location: MV, Germany

 Posted: Wed Sep 26, 2007 4:00 pm    Post subject: V1.5 was definitely harder than the original because it didn't fall that fast and I needed several contradiction moves like step 15 from my first walkthrough to solve this one. Walkthrough for A69 V1.5: 1. R89 a) Outies = 7(3) = {124} locked for R7 + N8 b) R8C4 = 3, R7C4 = 2, R7C6 = 4, R8C5 = 7, R7C5 = 1 2. N5 a) 14(3) @ R4C4 = 1{49/58/67} -> 1 locked for C4 + N5 b) 14(3) @ R4C6 = 3{29/56} -> 3 locked for C6 + N5 3. N2 a) 3 locked in 13(3) = 3{28/46} -> no 5,7,9 4. C5 a) 7 locked in 17(3) = 7{28/46} -> no 5,9 b) 14(3) @ R4C4 <> 7 -> no 6 since {167} is not possible c) 5 locked in 15(3) -> R89C5 = {59} d) Innies 8(2) of N8 = {68} locked for R9 e) Clean up: 9(2) in R9 = {27/45} 5. N7 a) 11(3): R7C12 <> 5,8 (conflict with cage sum) b) 11(3) = {137/236} -> 3 locked in R7C12 for R7 + N7 c) R6C1 = (12) d) 16(3) <> 4,5 because: {259},{457} not possible because R9C4 = (68) f) ! 9(2) = {45} locked for R9 + N9 since {27} leads to a conflict: -> 9(2) = {27} -> 16(3) = {169} -> 10(2) = {46} -> R7C1 = 3 = R7C2 6. N8 a) R9C5 = 9, R8C5 = 5 7. N7 a) Clean up: 10(2) = {19/28} 8. N9 a) Clean up: 5(2) = {23} locked in R9 + N9 b) 16(3) = 1{69/78} -> 1 locked in R89C7 for C7 + N9 c) 10(2) = {46} -> locked for R8 + N9 d) R8C7 <> 8 because of R9C6 = (68) since 16(3) cannot have 6 and 8 9. R1 a) 12(2): R1C3 <> 9 b) 11(2): R1C7 <> 4,7,8 10. N7 a) 10(2) = {28} locked for N8 since {19} is blocked by R8C7 = (19) b) 16(3) = 1{69/78} -> 1 locked in R89C3 for C3 c) 9 locked in R78C3 for C3 11. R5 a) Hidden Single: R5C4 = 1 12. N7 a) 25(4) <> 1 since R567C3 would be {789} which is blocked by Killer pair (79) of 16(3) 13. R1234 a) Innies 16(3) <> 3 because: -> no {367} since R4C5 would be 7 and R4C4 has neither 3 nor 6 -> no {349} since R4C46 = {49} -> no 3 possible in R4C56 -> no {358} since R4C46 = {58} -> no 5,8 possible in R4C56 -> 16(3) = {259/268/457} b) Innies 16(3): R4C5 <> 8 because R4C4 has no 2 or 6 c) R4C5 <> 4 since 7 is only possible in R4C5 14. N9 a) 5,8 locked in R7C789: -> R7C7 = (58) since 21(3) cannot have both 58 -> 21(3): R6C9 <> 8 because R7C89 must have a 5 which is not possible (21(3) = 5 + 8 + ?) 15. C6 ! a) R123C6 <> 8 because: -> Innies 20(4) = 1{289/568}, because of R9C6 = (68) -> (R1+R2+R3)C6 = 12/14 = 1{29/56/58} -> 158 not possible because 13(3) in N2 would be {346} -> N2 <> 2 b) Hidden Single: R9C6 = 8, R c) Clean up: R1C7 <> 3 16. N9 a) 16(3) = {178} -> R9C7 = 7, R8C7 = 1 b) 21(3) = 9{48/57} -> no 6 c) R6C9 = (47) 17. N7 a) 16(3) = {169} -> R9C4 = 6, R9C3 = 1, R8C3 = 9 18. N6 ! a) 18(4) <> 7 because: -> if R7C7 = 5 -> R6C9 = 4 -> 18(4) = {1359/2358} -> if R7C7 = 8 -> R6C9 = 7 -> 18(4) = {1368/1458/2358} -> 18(4) = {1359/1368/1458/2358} b) 18(4) <> 4 because if 18(4) = {1458} -> 13(3) = 67 -> R6C9 = ? c) 3 locked in 18(4) 19. N4 a) 3 locked in R4C123 b) 11(2) <> 8 20. N47 a) Innies 14(3) = 3{29/47/56} -> no 1,8 b) Hidden Single: R6C1 = 1 @ N4 c) 11(3) = {137} -> R7C3 <> 7 d) R7C3 = 6 e) 25(4) = 68{29/47} -> no 5 21. N6 a) 18(4) = {2358} -> no 6,9 22. N5 a) 6 locked in R6C56 23. N6 a) 18(4) = {2358} -> 2,3 locked for N6 b) R4C7 <> 5,8 c) 13(2) <> 5,8 since 5 or 8 is locked in 18(4) d) Killer pair (47) of 13(2) blocks {47} of 11(2) @ N4 e) Killer pair (47) of 13(2) + R6C9 = (47) -> locked in N6 24. N3 a) 4 locked in R23C7 b) 15(4) = {1239/1257/1356} -> no 8 25. N69 a) Innies 15(3) = 1{59/68} -> Killer pair (56) blocks {356} of Innies N47 -> R4C123 <> 5,6 26. N4 a) 5 locked in 11(2) = {56} locked for R5 27. N6 a) 13(2) = {49} locked for R5+N6 b) R4C7 = 6, R6C9 = 7 c) 21(3) = {579} -> {59} locked in R7C89 for N9 d) R6C9 = 8 e) 18(4): 2,3,5 locked in N6 f) R4C9 = 1, R4C8 = 8 28. N3 a) Hidden Single: R1C9 = 8 29. N5 a) 17(3) = {278} locked for N5+C5 because R4C5 = (27) and R5C5 = (278) b) R5C6 = 3, R5C7 = 2 c) 14(3) @ R4C4 = {149} locked for R5+N6 d) R4C6 = 5 -> R6C6 = 6 30. N2 a) 11(2) = {29} -> R1C7 = 9, R1C6 = 2 b) 19(4) = {1468} -> R3C6 = 1, R3C7 = 4, R2C6 = 9 31. N3 a) 27(5) = 389{16/25} because {12789} not possible because of R2C7 = (35) -> 27(5) <> 7 32. C8 a) Hidden Single: R3C8 = 7 b) 15(4) = {1257} -> 2,5 locked for C9+N3 c) R2C7 = 3, R6C7 = 5, R6C8 = 3, R9C8 = 2, R9C9 = 3, R7C9 = 9 e) R7C8 = 5, R5C9 = 4, R5C8 = 9, R8C9 = 6, R8C8 = 4 33. R3 a) 9 locked in 21(4) = 9{237/246/345} -> no 8 b) R4C1 <> 9 c) Hidden Single: R8C1 = 8 @ C1 -> R8C2 = 2 34. N1 a) Hidden Single: R3C1 = 9 @ C1 b) 5 locked in R123C3 c) 12(2) = {57} locked for R1 c) 3 locked in 23(4) = 37{49/58} -> no 2 d) Hidden Single: R3C9 = 2 @ R3 -> R2C9 = 5 e) 23(5) = 348{17/26} -> 3 locked in R1C12 for R1+N1 f) R3C2 = 6 35. Rest is clean-up and singlesLast edited by Afmob on Tue Oct 09, 2007 9:07 am; edited 6 times in total
CathyW
Master

Joined: 31 Jan 2007
Posts: 161
Location: Hertfordshire, UK

 Posted: Wed Sep 26, 2007 4:48 pm    Post subject: Think I'll have to pass on the V1.5. I'm having enough trouble redoing the V1! Have gone wrong somewhere. Think my "killer brain" has gone AWOL ! Oh well, hopefully it will seem straightforward at the 4th attempt later. Edit: Finally!! There's bound to be some typos in this but hopefully someone can follow it and let me know of corrections required. Evidently there are some errors in this - please ignore. If I get the will to go through it again I'll put it right. Prelims a) 17(2) r1c67 = {89} n/e r1 b) 34(5) @r1c1 = {46789} -> {89} locked to r2c234 n/e r2 -> r2c1 <> 467 c) 6(2) r1c34 = {15/24} d) 22(5) @ r1c8 = {13567/23467} must have 367 -> r2c9 <> 3,6,7 e) 8(2) r5c12 and r9c12 = {17/26/35} f) 20(3) @ r4c5 and r6c9 = {389/479/569/578} g) 11(3) @r4c6: no 9 h) 11(2) r5c89 and r78c6 = {29/38/47/56} i) 13(2) r78c4 and r9c89 = {49/58/67} j) 9(2) r8c12 = {18/27/36} ({45} blocked by 6(2) r8c89) 1. Innies N8: r9c46 = 8 = {17/26/35} a) 13(2) r9c89 <> {67}: blocked by 8(2) r9c12 and split 8(2) r9c46. b) KP {45} 13(2) and 6(2) in N9 -> 4,5 n/e N9 c) 20(3) @ r6c9 = {389}/4{79}/5{69/5{78} -> r6c9 <> 6,7 2. Outies r89: r7c456 = 9 = {126/135/234} a) 13(2) r78c4 = [49/58/67]; 11(2) r78c6 = [29/38/47/56/65] b) r8c456 + r9c5 = 28 = {4789/5689} c) split 9(3) = [612/513/423/432] -> r7c5 = (123); 11(2) r78c6 = [29/38] d) NP {89} r18c6 n/e c6 3. Innies r1234: r4c456 = 12 = {129/138/147/156/237/246/345} -> r4c4 <> 9 4. Innies r9: r9c357 = 16 = {178/268/349/358/457} Other combos blocked by 2 x 8(2) in r9. 5. 13(3) r789c5 = {148/157/247/256} ({346} blocked by split 28(4)) a) r7c5 = (12); r89c5 <> 9 b) 9 locked r8c46 -> r8c37 <> 9 c) 12(3) r123c5 <> {129} -> r3c5 <> 9 d) 9 locked r456c5 -> r56c4 <> 9 -> 20(3) r456c5 = {389/479/569} -> 12(3) r123c5 <> {345} 6. 13(3) @ r8c7 = {139/157/238/256} a) combo analysis (CA): r9c6 <> 6,7 -> r9c4 <> 1,2 7. 13(3) @ r8c3 = {139/157/238/247/256/346} 8. KP N9: 13(2) and r7c89 of 20(3) must each have one of 8,9 -> 8,9 n/e N9. a) 13(3) @ r8c7 = {157/256} Must have 5 -> r9c6 = 5 -> r9c4 = 3 b) 13(2) N9 = {49}, 6(2) N9 = {15} c) 11(2) r78c6 = [29] d) r1c6 = 8, r1c7 = 9 e) r7c5 = 1, r7c4 = 6, r8c4 = 7, r9c5 = 8, r8c5 = 4 f) r7c789 = {378} n/e r7/N9 g) 9(2) r8c12 = {36} only possible combo -> r8c7 = 2, r8c3 = 8, r9c7 = 7, r9c3 = 2 h) 20(3) @ r6c9 = 9{38}/5{78} -> r7c7 <> 8, r6c9 = (59) i) 17(3) @ r6c1 = 8{45}/3{59} -> r7c3 <> 5, r6c1 = (38) j) Clean up: r1c4 <> 4 9. 12(3) r123c5 = {23}7 only possible combo -> {23} n/e N2/c5 -> r1c3 <> 4 a) 6(2) r1c34 = {15} -> r1c89 <> 1,5 b) 20(3) r456c5 = {569} n/e N5 c) 14(3) r456c4 = {248} only possible combo n/e c4 -> r2c4 = 9, r13c4 = {15} d) 11(3) r456c6 = {137} only remaining option -> r23c6 <> 1 e) {4678} locked to 34(5) within N1, n/e N1 10. UR: r1c34 = {15}, r3c4 = (15) -> r3c3 <> 1,5 11. 2 locked to 20(4) @ r2c1 within N1 -> r4c1 <> 2 12. Innies r12: r2c19 = 6 = {15}/[24] a) NQ {1459} r2689c9 n/e c9 -> 11(2) r5c89 = [38/47/56/83/92] 13. O-I N7: r7c3 – r6c1 = 1 -> r7c3 = 9, r6c1 = 8 -> r3c3 = 3 14. HS: r2c2 = 8 15. Split 12(3) r4c456 = [291/453] -> r4c4 <> 8, r4c5 <> 6, r4c6 <> 7 16. 3 locked r3c123 -> r3c789 <> 3 17. 4 locked r3c678 -> r4c8 <> 4 18. 8 locked r3456c7 -> r4c8 <> 8 19. 3 locked r3456c3 -> r4c2 <> 3 20. 20(4) @ r2c1 = {1289/1379/2369/2459} Must have 9 a) CA: r4c1 = (4678) b) 9 locked to 20(4) within N1 -> r3c3 = 3 c) 20(4) = {1289/2459} -> r4c1 = (48) 21. 19(4) @ r3c3 = 3{169/259/457} a) CA: r4c23 <> 1,5 22. Outies N1: r4c123 = 19 = 4{69}/8{29/47} a) r4c789 = 14 -> r4c7 <> 8 (can’t make 6 from candidates in r4c89) 23. 18(4) @ r5c7 = {1278/1359/1368/1467/2358/2457/3456} a) Must have one of 3,7 -> since r7c7 = (37), r56c7, r6c8 <> 3,7 -> 18(4) = {159/168/258/456}3 / {128/146/245}7 b) CA: r6c8 = (269) 24. 9 locked r3c4567 -> r2c6 <> 9 25. 1 locked r456c7, r4c8 within N6 -> CPE r3c7 <> 1 26. 18(4) @ r3c6 = {1368/1458/1467/2457/3456} -> r4c8 <> 9 27. 19(4) @ r5c3 = {1279/1369/1459/2467/3457} a) CA: r6c2 <> 6,7 28. KP Split 19(3) r4c123 = {289/469/478}, r4c4 = (24) -> r4c7 <> 4, r4c89 <> 2 a) 18(4) @ r3c6 <> {2457} 29. KP 18(4) @ r3c6 must have one of {13} within r4c78, r4c6 = (13) -> r4c9 <> 3 30. KP 19(4) @ r3c3 must one of {49} within r4c23; split 12(3) r4c456 = [291/453] -> r4c1 = 8 -> r6c1 = 3, r7c12 = {59}, r7c3 = 4, r8c1 = 6, r8c2 = 3 31. UR: r9c12 = {17}, r5c12 cannot also be {17} -> r5c12 = [26] 32. 20(4) @ r2c1 = [192]8 -> r1c3 = 5, r1c4 = 1, r3c4 = 5 All singles from here. Phew!Last edited by CathyW on Mon Oct 01, 2007 3:34 pm; edited 4 times in total
Para
Yokozuna

Joined: 08 Nov 2006
Posts: 384
Location: The Netherlands

Posted: Wed Sep 26, 2007 5:09 pm    Post subject:

 Afmob wrote: PS: Sorry for the dots in my walkthrough but the format is messed up when I post it, so I use the dots to show some sort of alignment when you want to copy & paste this walkthrough.

You could number your steps 1a. 1b. 1c. like most people do here. And if you then seperate each step by an empty line, it is easy to read and to follow the consequences and implications of each step.

greetings

Para
Afmob
Expert

Joined: 22 Sep 2007
Posts: 103
Location: MV, Germany

 Posted: Wed Sep 26, 2007 5:18 pm    Post subject: I think for the next walkthrough I'll use 1a), ... and though my steps (1,2,...) are seperated by their location in the assassin (R1, C34, ...), I'll seperate each step with a blank line like I did right now on A69 V1.5 for better readability.
Andrew
Grandmaster

Joined: 11 Aug 2006
Posts: 300
Location: Lethbridge, Alberta

 Posted: Wed Sep 26, 2007 6:28 pm    Post subject: Hi Afmob. Welcome to the Assassin forum! I only started on A69 last night so haven't looked at your walkthroughs yet. Maybe your . was to make the "tab" indenting work. I posted a walkthrough for a different puzzle yesterday where I wanted to indent some lines and couldn't make it work with either "tab" or a series of spaces. I updated my "Advice for Newbies" yesterday, which is now "Advice for Newbies and Regulars". You, and everyone else, might like to look at that. It's likely that you are already doing a lot of things that are in that message. BTW I did have a quick glance at your walkthroughs, without looking at the steps, and see that you number your steps 01, 02 ... Most people on this forum just use 1, 2 ... which is easier to read, probably because that's what we are used to.
Afmob
Expert

Joined: 22 Sep 2007
Posts: 103
Location: MV, Germany

 Posted: Thu Sep 27, 2007 9:09 am    Post subject: Cathy, I had a look at your walkthrough and there is a typo in step 8. 8d) You wrote R7C5 = 2, but R7C6=2 and R7C5=1 (because of 8a,b) And maybe you should add that because of 8) 13(2) @ N9 = {49} and 6(2) @ N9 = {15}, otherwise I don't see why R9C5=8.Last edited by Afmob on Thu Oct 25, 2007 6:18 pm; edited 2 times in total
CathyW
Master

Joined: 31 Jan 2007
Posts: 161
Location: Hertfordshire, UK

 Posted: Thu Sep 27, 2007 11:38 am    Post subject: Thank you Afmob - and a belated welcome to the forum from me! Typo fixed and extra sub-step added to clarify. Hopefully there aren't too many more errors and my "killer brain" will be back in gear for the A70 tomorrow.
Andrew
Grandmaster

Joined: 11 Aug 2006
Posts: 300
Location: Lethbridge, Alberta

Posted: Mon Oct 01, 2007 7:07 am    Post subject:

A bit late with this one. I only started it earlier in the week after posting some other walkthroughs and only finished going through the posted walkthroughs today.

It was interesting to see what I missed, some of it obvious and some definitely not obvious like Mike's excellent hidden killer triple in N9. As it happened the obvious things, like the grouped X-Wing in R12 and those interactions in R9 that I missed, only seemed to speed up the solutions by a limited amount. Still I ought to have spotted them; for some other puzzles they might have been critical.

 gary w wrote: ..I guess this means the path is quite tightly constrained.

My solving path was quite a lot different than the others. My breakthrough came from combination work in C456 including innies in C4 and C6 that nobody else used.

I'll agree with Mike's estimated rating of 1.25. My combination work probably isn't any more difficult than his hidden killer triple.

Here is my walkthrough for A69.

1. R1C34 = {15/24}

2. R1C67 = {89}, locked for R1

3. R5C12 = {17/26/35}, no 4,8,9

4. R5C89 = {29/38/47/56}, no 1

5. R78C4 = {49/58/67}, no 1,2,3

6. R78C6 = {29/38/47/56}, no 1

7. R8C12 = {18/27/36/45}, no 9

8. R8C89 = {15/24}

9. R9C12 = {17/26/35}, no 4,8,9

10. R9C89 = {49/58/67}, no 1,2,3

11. R456C5 = {389/479/569/578}, no 1,2

12. R456C6 = {128/137/146/236/245}, no 9

13. 20(3) cage at R6C9 = {389/479/569/578}, no 1,2

14. 34(5) cage at R1C1 = {46789} (only possible combination)
14a. CPE R2C1 = {1235}
14b. 8,9 in R2 locked in R2C234, locked for R2

15. 22(5) cage at R1C8 = {13567/23467} = 367{15/24}
15a. CPE no 3,6,7 in R2C9

16. 45 rule on R12 1 outie R3C5 – 1 = 1 innies R2C19, min R2C19 = 3 -> min R3C5 = 4

17. 45 rule on R89 4 innies R8C456 + R9C5 = 28 = {4789/5689}, 8,9 locked for N8, clean-up: no 4,5 in R8C4

18. 45 rule on N8 2 innies R9C46 = 8 = {17/26/35}, no 4

19. Hidden killer triple 1,2,3 in R7C56 + R9C46 -> R7C5 = {123}, R7C6 = {23}, clean-up: R8C6 = {89}

20. Killer pair {89} in R18C6, locked for C6

21. R789C5 = {148/247/256} (cannot be {139/238} because 1,2,3 only in R7C5, cannot be {157/346} which clash with R8C456 + R9C5), no 3,9

22. 9 in N8 locked in R8C46, locked for R8
22a. R7C4 = 4 or R7C6 = 2 -> R789C5 (step 21) = {148/256} (cannot be {247}), no 7

23. 45 rule on R89 3 outies R7C456 = 9 = {126/234} (cannot be {135} which clashes with R789C5), no 5,7, clean-up: no 6,8 in R8C4
23a. 2 locked in R7C56, locked for R7 and N8, clean-up: no 6 in R9C46
[Alternatively Killer Pair 4,6 in R7C4 and R8C456 + R9C5, locked for N8, clean-up: no 2 in R9C46]

24. R7C5 = {12} -> R123C5 must contain 1/2 = {138/147/156/237/246}, no 9

25. 9 in C5 locked in R456C5, locked for N5
25a. R456C5 (step 11) = {389/479/569}

26. R456C6 = {137/146/245} (cannot be {236} which clashes with R7C6)
26a. R456C5 (step 25a) = {389/569} (cannot be {479} which clashes with R456C6), no 4,7

27. Killer Pair 6,8 in R456C5 and R89C5, locked for C5

28. 7 in C5 locked in R123C5, locked for N2
28a. R123C5 (step 24) = 7{14/23), no 5
28b. 7 of 34(5) cage at R1C1 locked in R1C12 + R2C23, locked for N1

29. R456C6 contains 1/2 -> R456C4 must contain 1/2
29a. R456C4 = {167/248/257} (cannot be {158} which clashes with R456C5), no 3

30. 3 in C4 locked in R39C4
30a. R456C4 contains 1/2 -> R1239C4 must contain 1/2
30b. R1C6 = {89} -> R123C4 must contain 8,9
30c. 45 rule on C4 4 innies R1239C4 = 18 = 3{159/168/249/258}, no 7, clean-up: no 1 in R9C6 (step 18)

31. R1C6 = {89}, no other 8,9 in R1239C6
31a. 45 rule on C6 4 innies R1239C6 = 23 = {1679/2579/3578/4568} (cannot be {2678/3479/3569} which clash with R456C6)
31b. 7 of {3578} must be in R9C6 -> no 3 in R9C6, clean-up: no 5 in R9C4 (step 18)
31c. R1239C6 = {169}7/{259}7/{358}7/{468}5

32. R1239C4 (step 30c) = 3{159/168} (cannot be {2349} which clashes with R12C5 because the 3 must be in R9C4, cannot be {258}3 => R9C6 = 5 when R123C4 clashes with R123C6 = {458}) = 13{59/68}, no 2,4, 1 locked for C4, clean-up: no 2,4 in R1C3
32a. {1359} must be {159}3 because {359}1 => R9C6 = 7 when R123C4 clashes with all remaining combinations for R123C6 in step 31c
32b. 1 of {1368} must be in R1C4
32c. R1239C4 = {159}3/{168}3 -> R9C4 = 3, R9C6 = 5 (step 18), R7C56 = [12], R7C4 = 6 (step 23), R8C46 = [79], R1C67 = [89], R2C4 = 9, clean-up: no 2 in R8C12, no 8 in R9C89
32d. Naked pair {15} in R13C4, locked for C4 and N2
32e. 8 in N3 locked in R3C789, locked for R3
[At this stage there was an UR on R13C34 which I never spotted. That’s not surprising since it’s a technique that I refuse to use. I only saw it after working through posted walkthroughs from others that do use it.]

33. R9C89 = {49} (cannot be {67} which clashes with R9C12)
33a. Naked pair {49} in R9C89, locked for R9 and N9 -> R89C5 = [48], R3C5 = 7, clean-up: no 5 in R8C12, no 2 in R8C89

34. Naked pair {23} in R12C5, locked for C5 and N2
34a. Naked pair {46} in R23C6, locked for C6

35. Naked pair {15} in R8C89, locked for R8 and N9, clean-up: no 8 in R8C12

36. Naked triple {378} in R7C789, locked for R7 and N9

37. Naked pair {36} in R8C12, locked for R8 and N7 -> R89C7 = [26], R8C3 = 8, R9C3 = 2

38. R2C2 = 8 (hidden single in R2)
38a. Naked triple {467} in R1C12 + R2C3, locked for N1

39. Naked pair {15} in R1C34, locked for R1

40. 45 rule on R1234 3 innies R4C456 = 12 = [291/453] (only remaining permutations), no 6,7,8

41. 45 rule on N7 1 innie R7C3 – 1 = 1 remaining outie R6C1 -> R6C1 = {348}
41a. 17(3) cage at R6C1 = {359/458}, 5 locked in R7C12, locked for R7, clean-up: no 4 in R6C1 (step 41)

42. 45 rule on N9 1 remaining outie R6C9 – 2 = 1 innie R7C7 -> R6C9 = {59}, no 8 in R7C7

43. 45 rule on N1 2 remaining outies R1C4 + R4C1 – 6 = 1 innie R3C3
43a. Min R1C4+R4C1 = 7 -> no 1 in R4C1
43b. Max R1C4+R4C1 = 14 -> no 9 in R3C3
43c. 2,9 in N1 locked in R2C1 + R3C12 for 20(4) cage -> no 2,9 in R4C1
43d. 20(4) cage at R2C1 = 29{18/36/45}, no 7
43e. 4,6,8 only in R4C1 -> R4C1 = {468}

44. 45 rule on N4 3 innies R4C123 – 1 = 3 outies R7C123
44a. R7C123 = {459} = 18 -> R4C123 = 19 = {289/478/568} (cannot be {379} because R4C1 only contains 4,6,8, cannot be {469} which clashes with R4C456) = 8{29/47/56}, no 1,3 -> R4C1 = 8 -> R6C1 = 3, R7C3 = 4 (step 41), R8C12 = [63], clean-up: no 5 in R2C1 + R3C12 (step 43d), no 5 in R5C1, no 2,5 in R5C2
44b. R4C2 = 4 (hidden single in N4), R4C4 = 2

45. R1C1 = 4 (hidden single in C1)

46. Naked triple {129} in R2C1 + R3C23, locked for N1 -> R1C34 = [51], R3C34 = [35], R4C3 = 7 (cage sum), R2C3 = 6, R1C2 = 7, clean-up: no 1 in R5C12 -> R5C12 = [26], R2C1 = 1, R3C12 = [92], R7C12 = [59], R9C12 = [71], R6C2= 5, R6C9 = 9, R9C89 = [94], R23C6 = [46], clean-up: no 5 in 22(5) cage at R1C8 (step 15), no 7 in R5C8

47. R2C9 = 5 (hidden single in R2), R8C89 = [51]

48. R6C9 = 9 -> R7C89 = 11 = {38}, no 7

49. R7C7 = 7 (hidden single in R7)

50. R5C9 = 7 (hidden single in C9), R5C8 = 4

51. R1C9 = 2 (hidden single in C9), R12C5 = [32]

and the rest is naked singles

The final stages could have been a bit quicker if I'd used some of the naked singles instead of looking for hidden singles.

In case anyone still wonders why, the reason I refuse to use UR is because I don't consider it to be solving the complete puzzle. I know that all Ruud's Assassins, all forum puzzles and all puzzles on the other website I use will have unique solutions. However I prefer to work right through a puzzle to reach the only solution rather than to rely on uniqueness. To me UR is a shortcut that bypasses part of the puzzle. Therefore I feel that using it must lessen the satisfaction of solving the puzzle although I don't doubt that those who spot an UR probably get some satisfaction from doing so. I suppose one could say that I'm philosophically opposed to using it. Others take the opposite view and find it acceptable. I certainly wouldn't try to stop them using UR.

Another point is that UR can be considered unnecessary. As far as I can see any Sudoku that requires UR as the only technique to solve it cannot be unique. If I'm wrong on that then maybe at some stage in the future I might reluctantly be forced to use it.
Ruud
Site Owner

Joined: 30 Dec 2005
Posts: 601

 Posted: Mon Oct 01, 2007 6:08 pm    Post subject: Andrew, Please do not take the following argument too seriously, it's just an illustration. Suppose I make a mistake and create a puzzle based on a solution with a duplicate 3 in row 5. This is not a valid puzzle. When you are attempting to solve this puzzle, you encounter a chain that gives you the choices: A: There are two numbers 3 in row 5; B: Both these candidates are false. As an experienced solver you will no doubt choose option B, which happens to be the wrong option for this puzzle, leading you to a partial solution which is not even close to the one I set. When the puzzle is invalid, all bets are off. A puzzle which does not have a unique solution is not a valid puzzle. As a player, you ONLY want to solve valid puzzles. So you go to a source where you expect ALL puzzles to be valid. Many players use software which can validate a puzzle without revealing the solution to the user. Consider this a double check that you are indeed working on a valid puzzle. Now here comes the core argument: A valid puzzle has certain properties. It can be completed in such a way than each cell contains a single value and it has one of each number in each row, column and nonet. For killers, the caged numbers add up to the sum given for the cage and no number may be repeated in a cage. And it does not contain a pattern which can be altered in such a way that another valid solution is formed. When you rinse the candidates after a placement, you are using one of these properties. When you eliminate a candidate with an XY-Wing, you are using a combination of properties. When you avoid a UR you are using yet another property. The whole exercise of solving a puzzle is recognizing and avoiding patterns with are known not to be present in a valid puzzle. The only logical reason to avoid uniqueness-based techniques is that the puzzle may have more than one solution and therefore could contain these patterns. Such a source may also provide you with a puzzle that has duplicate digits in a unit, so you should also avoid all solving techniques which are based on the premise that each unit contains only one of each digit. I'm right in the middle of a Uniqueness discussion on another forum, so I'm just test firing my arguments here. Feel free to rip them to pieces. Ruud
Para
Yokozuna

Joined: 08 Nov 2006
Posts: 384
Location: The Netherlands

 Posted: Mon Oct 01, 2007 6:58 pm    Post subject: Hi Ruud I think this whole UR-based techniques will never be solved. I like to use UR-based techniques in regular sudoku. There are some nice ways to use UR's in regular sudoku. Lately had a UR-xy-wing. That was kinda fun, used a UR as a pivot cell. Bu this might also be because my vanilla sudoku arsenal is kinda limited to single digit techniques and wings(and maybe occasionally an ALS-xz) I always try to avoid using them when writing a walk-through for a Killer Sudoku, but when i am solving it for fun or on speed i don't mind using it. For Killer Sudoku i like to avoid UR's, because i rather find another way around it. I guess i just like to stick to killer techniques in Killer Sudoku. I have seen some older forum messages where you used ALS-techniques to solve Killer Sudoku and also seen Udosuk use y-wings to shorten the solving path. But maybe my level of Killer sudoku solving is higher than my regular sudoku techniques. But i guess it is mostly that i like killer techniques over UR-techniques in killer, but i don't prefer regular sudoku techniques over Ur-techniques in vanilla sudokus. But i never really opposed UR-techniques. I solve a lot of other logic puzzles and one puzzle type where uniqueness-based are almost inevitable are domino puzzles(you get a grid with numbers(grid with dominoes with the edges taken away) and have to locate the original placement of the dominoes). Never really tried to even solve one without assuming it was unique. They can be hard enough without this assumption. Sometimes i think these sudoku guys take it all too serious. Yeah i am not really taking a position on all this (and i never really will). You should just solve puzzles the way you want. I don't like people saying that you shouldn't do something, just as long as you know why you are solving the way you are. Sudoku is a form of puzzling, puzzling is essentially a form of relaxation, ergo sudoku is a form of relaxation. So just have fun greetings Para, puzzle fan
CathyW
Master

Joined: 31 Jan 2007
Posts: 161
Location: Hertfordshire, UK

 Posted: Tue Oct 02, 2007 8:39 am    Post subject: I'm with you Ruud! Using unique rectangles is perfectly logical and acceptable to me, including in killers. I understand that others prefer to "prove" that a puzzle is unique by solving without them but that's all it amounts to - a preference for which weapon to use from the arsenal of many techniques. It's part of the fascination of this forum in particular to see how different people tackle the same puzzle.
mhparker
Grandmaster

Joined: 20 Jan 2007
Posts: 345
Location: Germany

Posted: Tue Oct 02, 2007 10:46 am    Post subject:

 Ruud wrote: I'm right in the middle of a Uniqueness discussion on another forum, so I'm just test firing my arguments here.

You've got full support from me on this one, Ruud! Not only do we share the same opinion on this topic, but the way you presented your arguments is very similar to the way I would have presented them.

Good luck with the debate on the other forum!
_________________
Cheers,
Mike
Andrew
Grandmaster

Joined: 11 Aug 2006
Posts: 300
Location: Lethbridge, Alberta

Posted: Tue Oct 02, 2007 9:28 pm    Post subject:

The only reason why I posted my message about why I won't use UR was as a follow-up to my earlier reply to Gary's messages.

It stirred up some interesting replies. It's good that we can express our different views and still work well together on and off the forum.

Even though I'm opposed to UR, I'll still wish Ruud good luck in his debate on the other forum!

 Para wrote: Sudoku is a form of puzzling, puzzling is essentially a form of relaxation, ergo sudoku is a form of relaxation. So just have fun

A nice thought! Some of the more difficult puzzles certainly aren't relaxing, challenging is a better word for them, but we still continue to do them and to enjoy them.
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