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Assassin 41
Posted: Fri Mar 09, 2007 1:19 am
by PsyMar
Woo! Rather proud of myself for finishing this one.
Walkthrough follows.
0a. 5/2 in N8 = {14|23}
0b. 7/2 in N9 != 789
0c. 11/2 in N1 != 1
0d. 11/2 in N7 != 1
0e. 14/2 in N3 = {59|68}
0f. 6/3 in R1 = {123} triple -> elim 123 from rest of R1
0g. 6/3 in R2 = {123} triple -> elim 123 from rest of R2
0h. 8/3 in R9 = {125|134} -> elim 1 from rest of R9
0i. 20/3 in R8 != 1 or 2
0j. 23/3 in R9 = {689} triple -> elim 689 from rest of R9
0k. 24/3 in R2 = {789} triple -> elim 789 from rest of R2
1. combinations for 11/2 in N1 = [56|65|74]
2. combinations for 14/2 in N3 = [86|95]
3. R9C1 = hidden single 7 in R9 -> R8C1 = 4
4. R2C5 = hidden single 4 in R2
5. 5/2 in N8 = {23} pair -> elim 23 from rest of N8/C5
6. 7/2 in N9 = [25|34|52]
7. 8/3 in R9 must have 2 or 3 -> forms killer pair with R9C5 -> no 2 in R9C9 -> no 5 in R8C9 (Edit: thanks for catching my typo, Ed)
8. R8C59 = naked pair 23 -> elim 23 from rest of R8
9. Innies of R12 = R1C5 = 7
10. 1 of N9 locked in R7 -> elim from rest of R7
11. 1 of N8 locked in C4 -> elim from rest of C4
12. 1 of R1 locked in N1 -> elim from rest of N1
13. 7 of R2 locked in N1 -> elim from rest of N1
14. sole combination for 20/4 in C5 = {1478} -> elim 18 from rest of C5
15. R567C5 = 569 naked triple -> elim from rest of cage 37/7 in R567
16. Innies of C6789 = R56C6 = 6/2 = {24} pair -> elim from rest of N5/C6
17. Innies of C1234 = R56C4 = 11/2 = {38} pair -> elim from rest of N5/C4 -> R1C4 = 2, R2C4 = 9, R34C5 = [81]
18. R1C6+R3C4 = 56 naked pair -> elim from rest of N2
19. R1C16 = 56 naked pair -> elim from rest of R1
20. 7 of N5 locked in R4 -> elim from rest of R4
21. R12C1 = 56 naked pair -> elim from rest of N1/C1
22. R1C23 = 13 naked pair -> elim from rest of N1
23. 1 of C1 locked in N4 -> elim from rest of N4
24. R3C123 = 249 naked triple -> elim from rest of R3
25. no 5 in 16/3 in R8 (sums)
26. Outies of C9 = R367C8 = 11/3 -> no 9 in R67C8
27. 22/4 in C12 must have an 8; this 8 locked in N4/C1, elim 8 from rest of N4/C1
28. 28/5 in R34 = {24679|34579} -> 7 must be in R4C4; also, all 4s and 9s in this cage can see R56C3 so elim 49 from R56C3
29. 28/5 in R34 can only have one of 5 or 6; this must be in R3C4 so remove 56 from elsewhere in 28/5 in R34
30. Combinations for 22/4 in C12 = [2938|2983|9238|9283|9481] -> elim 9 from rest of R3
31. 28/5 in R34 must have 9; this is locked in R4; elim from rest of R4
32. R14 = 56 naked pair in C6
33. 20/3 in R8 must have 5; this is locked in N9; elim from rest of N9 -> several naked singles/last digit in cage moves
34. 16/3 in R8 = {169}; 20/3 in R8 = {578}
35. R7C4 = hidden single 4
36. 8 of N7 locked in R7 -> elim from rest of R7
37. 9 of N5 locked in C5 -> elim from rest of C5
38. 9 of R8 locked in N7 -> elim from rest of N7
39. Outies-innies of C1 = R3C2-R67C1 = -1 -> R67C1 = 3,5 or 10, but cannot be 10 (sums) so elim 9 from R3C2; so R67C1 = 3 or 5 -> {12|23} -> elim 2 from rest of C1 -> R3C1 = 9
40. 2 of R9 locked in N7 -> elim from rest of N7 -> R7C1 = 3 -> 22/4 in C12 = [9481] -> R3C3 = 2 && R6C1 = 2 -> R6C56 = [24]
41. only combination for 28/5 in R34 = [26947] -> 15 naked singles and last-digit-in-cage moves
42. 17/4 in C12 = [2735] (only combination) -> 6 naked singles and last-digit-in-cage moves
43. Outies of R1234 = R5C9 = 9 -> 5 naked singles and last-digit-in-cage moves
44. R7C89 = {27} naked pair -> elim from rest of R7 and N9 -> 4 more naked singles and last-digit-in-cage moves
45. R6C8 = 6 (sums of 16/4 in C89) -> naked singles and last-digit-in-cage moves solve it
613275948
587943216
942681735
894716352
136852479
275394861
358469127
469127583
721538694
Tada!
Posted: Fri Mar 09, 2007 1:55 am
by Para
Hey all
Thought i'd post my walk-through too. When solving this i completely forgot to do 45-tests. Was more occupied with watching tv. Still finished it pretty quickly. There's probably an easier way through his puzzle when using the 45-tests. But it turned out a nice walk-through anyway.
Walk-Through Assasin 41
1. R12C1 + R89C1 = {29/38/47/56}
2. R12C9 = {59/68}
3. R89C9 = {16/25/34}
4. R89C5 = {14/23}
5. 20(3) in R8C6 = {389/479/569/578}: no 1 or 2
6. R1C234 = {123} : locked for R1
6a. Clean up: R2C1: no 8,9
7. R2C678 = {123} : locked for R2
7a. Clean up: R1C1: no 8,9
8. R2C234 = {789}: locked for R2
8a. Clean up: R1C1: no 4; R1C9 : no 5,6
9. R9C234 = {125/134}: 1 locked for R9
9a. Clean up: R8C5: no 4; R8C9: no 6
10. R9C678 = {689} : locked for R9
10a. Clean up: R8C9: no 1
11. Hidden single 7 in R9C1
11a. R8C1 = 4
11b. R12C1 = {56}: locked for C1 and N1
11c. Clean up: R9C9: no 3
12. Hidden single 4 in R2C5
12a. Clean up: R8C5: no 1
12b. R89C5 = {23}: locked for C5 and N8
13. 4 locked in N1 for R3
13a. 4 locked in N3 for 18(3) in R1C6: 18(3) = {459/468}: no 7
13b. Hidden single 7 in R1C5
13c. 7 locked in R2 for N1
14. 20(4) in R1C5 = [74]{18}: {18} locked for C5
14a. R567C5 = {569}: locked for 37(7) in R5C4
14b. 37(7) in R5C4 = {2345689}: no 1,7
14c. Naked Quad {2348} in R56C46: locked for N5
14d. R4C5 = 1; R3C5 = 8; R2C4 = 9
14e. 7 locked in N5 for R4.
15. Naked Pair {56} in R1C16: locked for R1
15a. Killer Pair {89} in R1C789: locked for N3
16. Killer Triple {235} in R8C5 + R8C9 + 20(3) in R8C6: locked for R8
16a. 16(3) R8C2= {169}/{18}[7]: R8C4: no 8
16b. 20(3) in R8C6 = {569/578}:{389} clashes with 16(3) in R8C2: no 3; 5 locked for R8
16c. Clean up : R9C9: no 2
17. 22(4) in R3C1 = {1489/2389}: 8 locked 20(4) in R45C1 for C1 and N4
18. 28(5) in R3C3 = {15679/24679/34579}: 7 and 9 locked in 28(5)
18a. R4C4 = 7 (only place in 28(5) for 7)
19. 9 locked both in 22(4) in R3C1 and 28(5) in R3C3. In both cages 9 either in R3 or N4. So one of those 9’s is in R3 and the other in N4 (similar deduction as an X-wing) -->> no 9 anywhere else in N4 (and theoretically R3).
20. 16(3) in R8C2 = {169} -->> locked for R8
20a. 9 locked in R8 for N7
20b. 8 locked in N7 for R7
20c. 8 locked in N8 for C6
21. 17(4) in R6C1 = {12}{68}/{13}{58/67}/{23}{48/57}-->> R67C2: no 1,2,3
22. 26(5) in R5C2 = {14678/23678/24578/34568}: 8 locked in 26(5)
22a. R7C3 = 8 : only place for 8 in 26(5) in R5C2
22b. R2C23 = [87]
22c. Clean up: R6C2: no 4(step 21)
23. 17(4) in R6C1 = {1367}/{2357}: 3 and 7 locked in 17(4)
23a. R6C2 = 7: only place for 7 in 17(4)
23b. 3 locked in R67C1 for C1
24. 26(5) in R5C2 = {3456}8: no 1,2
24a. 3 locked in 26(5) for N4: no 3 anywhere else in N4
24b. Hidden single 3 in R7C1
25. 8(3) in R9C2 = {125}: locked for R9
25a. R9C5 = 3; R8C5 = 2; R9C9 = 4; R8C9 = 3
26. 1 locked in N4 for C1
26a. 1 locked in N9 for R7
26b. 1 locked in N8 for C4
26c. 1 locked in R1 for N1
27. 28(5) in R3C3 = {2469/3459}7: 4 locked in 28(5)
27a. 4 locked in R3C3 + R4C23 -->> no 4 in R56C3
27b. 4 locked in C3 for 28(5) in R5C3-->> R4C2: no 4
28. 22(4) in R3C1 = {189}[4]/{289}[3]: R3C2 = {34}
29. 19(4) in R3C8 = {1279/1369/1378/2359/2368}: no {1567} -->> clashes with R2C9
29a. Killer Pair {89} in R1C9 + R45C9(19(4)): locked for C9
30. 16(4) in R6C8 = {1249/1258/1267/1357/1456/2356}: {1348} and {2347} not possible-->> 3,4 and 8 only 1 option in the same square (R6C8)
30a. R6C8: no 9: {1249} not possible with 9 in R6C8: only option for 4 in 16(4) is R6C8
31. 19(4) in R3C8 = {1279/1369/1378/2359/2368}
31a. Only way to make these combinations are with 1 of {89} in R45C9; 1 of {123} in R3C89(because of {123} in R2C78); 1 of {12} in R45C9; 1 of {567} in R3C89: R45C9 : no 5,6 or 7
31b. Killer Triple {123} in R2C78 + R3C89: locked for N3: R3C7: no 1, 2 or 3
32. 22(4) in R3C1 = [93]{28}/[9481]: [23]{89} clashes with R1C23 -->> R3C1: no 2
32a. R3C1 = 9
32b. 9 locked in N4 for R4
32c. 9 locked in N5 for C5
32d. Naked Pair {28} in R4C19 -->> locked for R4
33. 22(5) in R3C6 = {13567/23467}: 3,6 and 7 locked in 22(5)
33a. R3C7 = 7 : only place for 7 in 22(5) in R3C6
33b. Hidden single 7 in R5C8 and R7C9
33c. Hidden single 7 in R8C6
33d. Naked pair {58} in R8C78 -->> locked for N9
33e. Naked Pair {69} in R9C78 -->> locked for R9 and N9
33f. R9C6 = 8
34. Naked Pair {56} in R14C6 -->> locked for C6
34a. Naked Triple {123} in R1C4 + R23C6 in N2
35. Hidden single 9 in R7C6
35a. Hidden single 9 in R5C9 and R6C5
35b. R12C9 = [86]; R4C9 = 2; R4C1 = 8
35c. R12C1 = [65]; R1C6 = 5; R3C4 = 6; R4C6 = 6
35d. R5C5 = 5; R7C5 = 6; R8C4 = 1; R9C4 = 5
35e. R7C4 =4; R7C2 = 5; R6C1 = 2; R5C1 = 1; R3C2 = 4
36. R4C2 = 9; R8C23 = [69]; R5C23 = [36]; R6C3 = 5; R34C3 = [24]
36a. R9C23 = [21]; R1C234 = [132]; R56C4 = [83]; R56C6 = [24]
36b. R5C7 = 4; R1C78 = [94]; R9C78 = [69]
36c. R6C9 = 1; R6C7 = 8; R6C8 = 6 R7C78 = [12]; R8C78 = [58]
36d. R4C78 = [35]; R23C6 = [31]; R2C78 = [21]; R3C89 = [35]
And we are done.
Oh, and for anyone who's wondering, i am going do edit all comments on previous walk-throughs people gave me tomorrow so feel free to comment on this one. They will get adjusted eventually.
greetings
Para
Posted: Sat Mar 10, 2007 11:48 am
by sudokuEd
Posted: Sat Mar 10, 2007 5:29 pm
by PsyMar
sudokuEd wrote: Love the ! PsyMar. Is it a Maths term?
! meaning "not" is actually from C++ programming -- ! means not, and != means not equal.
Edit to avoid double post:
This is as far as I could get on v2. Lots of pairs, but no digits placed. Used a few medusa moves. You may notice I have two step 27's. The reason behind this is that the first one is how I found the elimination, but it's kind of obfuscated. The second one is simpler, however, it seems like fairly blatant trial-and-error. Therefore I have left in both maneuvers. (I don't know that it'll help in the final solution anyway.)
0a. 5/2 in C5 = {14|23}
0b. 11/2 in R12 != 1
0c. 11/2 in R89 != 1
0d. 14/2 in C9 = {59|68}
0e. 14/2 in R9 = {59|68}
0f. 6/3 in R1 = {123} triple -> elim 123 from rest of R1
0g. 6/3 in R2 = {123} triple -> elim 123 from rest of R2
0h. 8/3 in R9 = {125|134} -> elim 1 from rest of R9
0i. 20/3 in R8 != 1|2
0j. 24/3 in R2 = {789} triple -> elim {789} from rest of R2
1. Combinations for 14/2 in C9 = [86|95]
2. Combinations for 11/2 in R12 = [56|65|74]
3. Innies of R12 = R12C5 = 11/2 = [56|65|74]
4. 4 of R1 locked in 18/3 -> elim 7 from 18/3 in R1 (combinations)
5. 7 of N3 locked in R3 -> elim 7 from rest of R3
6. Innies of C5 = R567C5 = 20 != 1|2
7. Outies of R12 = R34C5 = 9/2 = [18|27|81] (must have 1 or 2 since only one of 1 or 2 can be in 5/2 in C5, and none elsewhere in C5)
8. 9 of C5 locked in 37/7 in C456; elim 9 from rest of 37/7
9. Innies of C1234 = R56C4 = 11/2 = {38|47|56}
10. Innies of C6789 = R56C6 = 6/2 = {15|24}
11. Outies of R1234 = R5C19 = 10/2 = {19|28|37|46}
12. Outies of R9 = R8C159 = 9/3 = {126|135|234} -> conflicts with {259|367} for 16/3
13. Combinations for 11/2 in R89 = [29|38|47|56|65]
14. 19/4 in C89 cannot be {1369|1567|2458} (conflicts with combinations for 14/2 in N3)
15. 22/4 in C12 cannot be {1579|1678|2578|3469|4567} (conflicts with combinations for 11/2 in N1)
16. R4C5 can see all cells of 37/7 in C456 -> 37/7 cannot contain 1,7 and 8 -> 37/7 = {2345689} -> R4C5 = {17}
17. combinations for split cage 9/2 in R34C5 (outies of R12) = [27|81]
18. Split cage 6/2 in R56C6 (innies of C6789) = {24} pair -> elim from rest of 37/7, N5, and C6
19. 1 and 7 of N5 locked in R4 -> elim 1 and 7 from rest of R4
20. 4 of R1 locked in N3 -> elim from rest of N3
21. 2 of R2 locked in N3 -> elim from rest of N3
22. combinations for 5/2 in C5 = [14|23|32]
23. Medusa coloring: (7)R2C4 blue <-> (7) R1C5 red <-> (7) R4C5 blue <-> (1) R4C5 red <-> (8) R3C5 red -- both (7)R2C4 blue and (8) R3C5 red can see (8)R2C4, eliminate it
24. 8 of R2 locked in N1 -> elim from rest of N1
25. Medusa coloring: (4)R3C4 blue <-> (4) R2C5 red <-> (7) R1C5 red <-> (7) R2C4 blue <-> (9) R2C4 red -- both (4)R3C4 blue and (9) R2C4 red can see (9)R3C4, eliminate it
26. Medusa coloring: (4)R3C4 blue <-> (4) R2C5 red <-> (7) R1C5 red <-> (7) R4C5 blue <-> (1) R4C5 red <-> (8) R3C5 red -- both (4) R3C4 blue and (8) R3C5 red can see (8)R3C4, eliminate it
27. R9, 789: exactly one goes in 14/2, either exactly one goes in 16/3 in N9 and R9C1 = 7|8|9 or 16/3 in N9 = [178|187], 14/2 in R9 = {59}, 8/3 in R9 = {134}, and R9C1 = 6. Thus R9C1 != 5 and R8C1 != 6
27. (R9C1=5 -> R9C234={134} -> R9C5=2 -> R9C67={68} -> R9C89 = {79} -> R8C9 = 0, CON)->R9C1 != 5 -> R8C1 != 6
.-----------.-----------------------------------.-----------.-----------------------------------.-----------.
|(11) |(6) |(20) |(18) |(14) |
| 567 | 123 123 123 | 567 | 5689 45689 45689 | 89 |
| :-----------------------------------: :-----------------------------------: |
| |(24) | |(6) | |
| 456 | 789 789 79 | 456 | 13 123 123 | 56 |
:-----------'-----------.-----------------------: :-----------------------.-----------'-----------:
|(22) |(28) | |(22) |(19) |
| 1234569 1234569 | 1234569 123456 | 28 | 135689 1356789 | 1356789 1356789 |
| .-----------' | | '-----------. |
| | | | | |
| 2345689 | 2345689 2345689 1356789 | 17 | 1356789 2345689 2345689 | 2345689 |
| :-----------------------.-----------'-----------'-----------.-----------------------: |
| |(26) |(37) |(29) | |
| 12346789 | 123456789 123456789 | 3568 35689 24 | 123456789 123456789 | 12346789 |
:-----------'-----------. | | .-----------'-----------:
|(17) | | | |(16) |
| 123456789 123456789 | 123456789 | 3568 35689 24 | 123456789 | 123456789 123456789 |
| | '-----------. .-----------' | |
| | | | | |
| 123456789 123456789 | 123456789 123456789 | 35689 | 1356789 123456789 | 123456789 123456789 |
:-----------.-----------'-----------------------+-----------+-----------------------'-----------.-----------:
|(11) |(16) |(5) |(20) |(16) |
| 2345 | 123456789 123456789 123456789 | 123 | 356789 3456789 3456789 | 123456 |
| :-----------------------------------: :-----------------------.-----------' |
| |(8) | |(14) | |
| 6789 | 12345 12345 12345 | 234 | 5689 5689 | 23456789 23456789 |
'-----------'-----------------------------------'-----------'-----------------------'-----------------------'
Posted: Sat Mar 10, 2007 8:45 pm
by Andrew
sudokuEd wrote: Love the ! PsyMar. Is it a Maths term?
PsyMar wrote:! meaning "not" is actually from C++ programming -- ! means not, and != means not equal.
Very interesting. I look forward to seeing the context in which it was used after I've solved Assassin 41, haven't even started yet, and can read the posted walkthroughs. I know that "not equal" is used in walkthroughs, at the moment I can't think how "not" is used. Maybe when I read the walkthrough I'll find out or it may have been "not equal".
Another way that "not equal" is sometimes put in walkthroughs is <>. That's fine when it's used for a number, for example <> 7, but I don't like it if it's used for combinations. BTW it wouldn't make sense to put <>1 or <>9 but I don't suppose people would want to use <> then.
I'm pleased to see that it's called Maths in Australia, just like it is in UK. I've never understood why people in Canada and the US call it Math; that seems wrong to me. It's an abbreviation for Mathematics which is a collection of subjects including algebra, arithmetic, geometry and trigonometry.
Posted: Sat Mar 10, 2007 11:54 pm
by sudokuEd
PsyMar wrote:You may notice I have two step 27's. ..however, it seems like fairly blatant trial-and-error. (I don't know that it'll help in the final solution anyway.)
Don't need steps 27.
Oops: just realized this next 'hint' is wrong. Sorry. Not sure how to progress now and don't have time to look until tonight.
There is a really neat trick you can use to make progress. Look at the in/outs for n1: r12c4 being the outties. Don't use hypotheticals 'cause theres a simpler purely logical way to progress. 
An extra little hint about this in t-t: remember to also think about the 6(3) and 24(3) cages 
.
BTW - Really like those Medusa moves. They sound the same as what Richard called "xy-chains" for the Ruud-being-even-meaner-than-usual X-R-size puzzles from the X-Files.
BTW2 - like the way that the marks pic went into t-t: never knew that was possible. It didn't come out properly though in note-pad. Is there a way I can fix it up?
Cheers. Ed
Posted: Sun Mar 11, 2007 3:21 am
by PsyMar
sudokuEd wrote:
BTW - Really like those Medusa moves. They sound the same as what Richard called "xy-chains" for the Ruud-being-even-meaner-than-usual X-R-size puzzles from the X-Files.
Not really... Well, sort of. xy-chains are a simplified version of medusa moves. Medusa moves are basically simple coloring, except with coloring in multiple digits and using pairs to link the digit-colorings together. They're what the "ultra-coloring" tool in Sudocue is for.
sudokuEd wrote:
BTW2 - like the way that the marks pic went into t-t: never knew that was possible. It didn't come out properly though in note-pad. Is there a way I can fix it up?
I generated the marks pic with "copy sums+marks" from sumocue, if that's any help.
Edit: Doesn't come through in notepad for me either, apparantly the forum removed all but one of consecutive spaces. Shoot. I'll try again in a bit here, using periods or something...
Posted: Sun Mar 11, 2007 10:46 am
by sudokuEd
OK. Found some chains and hypotheticals to make some progress.
Some more steps and sums/marks only pics [edit:3 back in r6c5: sorry Richard]
29. 2 in r9 locked in 8(3) or r9c5
29a.no 9 r8c1
Need some chains to get anywhere.
30.[38] blocked from r89c1. Here's how.
30a. r8c159 = [315] -> r9c1589 = [84{29}] - blocked by 14(2)r9
30b. r8c159 = [324] ->
r9c1589 = [83{57}] - blocked by 14(2)r9&8(3)r9 [edit:thanks Para]
30c. 11(2) = [29/47/56]
The following set of hypotheticals hinges around 3 in c1 only in 22(4) or r67c1
31."45" c1: r3c2 + 1 = r67c1
31a. min r67c1 = {12} = 3 -> min r3c2 = 2
32a. r3c2 = 2 -> r67c1 = 3 = {12} -> 3 in c1 in 22(4) -> r345c1 = {389}
32b. r3c2 = 3 -> r67c1 = 4 = {13} -> r345c1 = {289/478/568}
32c. r3c2 = 4 -> r67c1 = 5 = {14} -> 3 in c1 in 22(4) -> r345c1 = {378}
...........................= {23} -> r345c1 = {189}
32d. r3c2 = 5 -> r67c1 = 6 = {15/24} -> 3 in c1 in 22(4) -> r345c1 = {368}
32e. r3c2 = 6 -> r12c1 = [74] and r67c1 = 7 = {16} -> 3 in c1 in 22(4) -> r345c1 = {358}
........................................... = {25}: Blocked ([74]{25} clash with 11(2)n7)
32f. r3c2 = 9 -> r67c1 = 10 = {19} -> 3 in c1 in 22(4) -> r345c1 = {238}
............................= {28} -> 3 in c1 in 22(4) -> r345c1 = Blocked
........................... = {37} -> r345c1 = {148} ({37/256} blocked by 11(2)n1)
............................= {46} Clash with 11(2)n1.
33. In Summary
33a. r345c1 = {389/289/478/568/378/189/368/358/238/148}
33b. 8 locked in r45c1 in 22(4) for c1 and n4
33c. 22(4) = {1489/2389/3478/3568}
34. "45"n5 -> r7c5 + 8 = r4c456
34a. we know that two of the digits in r4c456 are 1 and 7 = 8 -> the remaining digit = r7c5
34b. that remaining digit must be in the required c4/6 in n2
34c. ->no 8 is possible in r4c6 since no 8 in c4 in n2.
35. 28(5) at r3c3 must have 7/8: only available in r4c4
35a. r4c4 = {78}
Code: Select all
.-----------.-----------------------------------.-----------.-----------------------------------.-----------.
|(11) |(6) |(20) |(18) |(14) |
| 567 | 123 123 123 | 567 | 5689 45689 45689 | 89 |
| :-----------------------------------: :-----------------------------------: |
| |(24) | |(6) | |
| 456 | 789 789 79 | 456 | 13 123 123 | 56 |
:-----------'-----------.-----------------------: :-----------------------.-----------'-----------:
|(22) |(28) | |(22) |(19) |
| 1234569 234569 | 1234569 123456 | 28 | 135689 1356789 | 1356789 1356789 |
| .-----------' | | '-----------. |
| | | | | |
| 2345689 | 234569 234569 78 | 17 | 135679 2345689 2345689 | 2345689 |
| :-----------------------.-----------'-----------'-----------.-----------------------: |
| |(26) |(37) |(29) | |
| 12346789 | 12345679 12345679 | 3568 35689 24 | 123456789 123456789 | 12346789 |
:-----------'-----------. | | .-----------'-----------:
|(17) | | | |(16) |
| 12345679 12345679 | 12345679 | 3568 35689 24 | 123456789 | 123456789 123456789 |
| | '-----------. .-----------' | |
| | | | | |
| 12345679 123456789 | 123456789 123456789 | 35689 | 1356789 123456789 | 123456789 123456789 |
:-----------.-----------'-----------------------+-----------+-----------------------'-----------.-----------:
|(11) |(16) |(5) |(20) |(16) |
| 2456 | 123456789 123456789 123456789 | 123 | 356789 3456789 3456789 | 123456 |
| :-----------------------------------: :-----------------------.-----------' |
| |(8) | |(14) | |
| 5679 | 12345 12345 12345 | 234 | 5689 5689 | 3456789 3456789 |
'-----------'-----------------------------------'-----------'-----------------------'-----------------------'
Code: Select all
.-----------.-----------------------------------.-----------.-----------------------------------.-----------.
| 567 | 123 123 123 | 567 | 5689 45689 45689 | 89 |
| :-----------------------------------: :-----------------------------------: |
| 456 | 789 789 79 | 456 | 13 123 123 | 56 |
:-----------'-----------.-----------------------: :-----------------------.-----------'-----------:
| 1234569 234569 | 1234569 123456 | 28 | 135689 1356789 | 1356789 1356789 |
| .-----------' | | '-----------. |
| 2345689 | 234569 234569 78 | 17 | 135679 2345689 2345689 | 2345689 |
| :-----------------------.-----------'-----------'-----------.-----------------------: |
| 12346789 | 12345679 12345679 | 3568 35689 24 | 123456789 123456789 | 12346789 |
:-----------'-----------. | | .-----------'-----------:
| 12345679 12345679 | 12345679 | 3568 5689 24 | 123456789 | 123456789 123456789 |
| | '-----------. .-----------' | |
| 12345679 123456789 | 123456789 123456789 | 35689 | 1356789 123456789 | 123456789 123456789 |
:-----------.-----------'-----------------------+-----------+-----------------------'-----------.-----------:
| 2456 | 123456789 123456789 123456789 | 123 | 356789 3456789 3456789 | 123456 |
| :-----------------------------------: :-----------------------.-----------' |
| 5679 | 12345 12345 12345 | 234 | 5689 5689 | 3456789 3456789 |
'-----------'-----------------------------------'-----------'-----------------------'-----------------------'
Posted: Sun Mar 11, 2007 2:31 pm
by rcbroughton
Ed,
got a bit further with this for you.
A bit of computer history for you:
I like using != for "not equal" - it means the same as <>, but != found favour in several programming languages starting with BCPL, then B, C, C++ and into unix, java and others. <> tends to live with a different set of derivative languages, including most variations of Basic. ! on it's own is a logical negative and is usually pronounced "bang" or "shriek" in the unix world.
Anyway, a walkthrough from your position.
36. 45 rule on n5 r4c456 minus r7c5=8
36a. only combo with 3 is [713]3 but can't have [13] on r47c5 because of 5(2) in n8 - no 3 in r4c6 or r7c5
37. 45 on r8, innies r8c159 total 9
37a only combo with 5 is 5{13} - no 5 in r8c9
38 45 on r8, outies r9c1589 total 23
38a. can't have combos with {35} or {45} because of 8(2) in n7 - only other combo with 5 is {2579} with 2 in r9c5
38b. can't have 7 and 9 in r9c89 because they are part of 16(3)n9 - so 5 must be in there - no 5 in r9c1
38c. no 6 in 11(2)n7 at r8c1
39. Can't have a 3 at r3c4
39a. r3c4=3->r2c6=1->r1c4=2->r3c5=8->r4c5=1->r89c5=[23]
39b. but r3c4=3 -> r5c5=3 - contradcition
40. Can't have a 3 at r1c4 (same logic)
40a. r1c4=3 -> r2c6=1 -> r4c5=1 -> r89c5=[23]
40b. but r1c4=3 -> r5c5=3 contradiction
41. therefore 3 locked in c6 of n2
42. and 3 locked in n1 for row 1
43. 22(4)n14 - combos are {1489}, {1678}, {2578}, {2389}, {3568}
43a. {1678}, {2578} must have [87] in r45c1
43b. {2389}, {3568} must have {38} in r45c1
43c. no 2,5,6 in r4c1 and no 2,6 in r5c1
44. 45 on n1 r12c4 + r45c1 - r3c3 total 18
1) can't use {57},{45},{67} or {46} in r3c3+r45c1 because of 11(2)n1
2) 11(2)n1 and 11(2) n7 mean we can't have {49} in r45c1
3) 22(4)n14 only combis are {1489},{2578},{2389},{3568}
4) r12c4 total 8,9,10,11 and r3c3 is {124569}
44a r3c3=1 - r45c1=11,10,9,8= {38},[91],[81]
44b r3c3=2 - r45c1=12,11,10,9={39},{48},{38},[91],[81]
44c r3c3=4 - r45c1=14,13,12,11={39},{48},{38}
44d r3c3=5 - r45c1=15,14,13,12={39},{48}
44e r3c3=6 - r45c1=16,15,14,13=no valid
44f r3c3=9 - r45c1=(19,18),17,16={89}
44g no 6 r3c3, no 7 r5c1
45. (from step 11) 45 on r567 r5c19 total 10
45a no 3,4,8 in r5c9
46. 45 on n1 everything except 11(2) total 34 and can't have 4 and 5 - 3
46a. {1235689} must have {89} in r3c23, 3 in r1c23 {1256} to place and must have {25} or {56} in r3c12
46b. {1234789} no 5
46c. no 5 at r3c3
47 45 on c1. r34567c1 total 23
47a. as in step 44, the two 11(2) cages limit combos with {45}, {46}, {49}
47b. only possibles {13478}, {12389}, {13568}
47c. {13478} r3c1=1 limites combo in 22(4) to {1489} so r45c1={48} r67c1={73}
47d no 4 in r67c1
48 Can't have a 3 at r6c2 as it removes all possibles for 3 in c1 at r4567
49 45 on c5 innies = 20(3)=3{89}/{569} - no 8 in r5c5
50.45 on n5 innies total 28(5)={15679}/{13789}
50a. {15679} - r4c4=7
50b. {13789} - r5c5=3, r4c6=1 (can't have 1 at r4c5 as it would break the 5(2)n8), r4c5=7, r4c4=8
50c. no 7 in r4c6, no 8 in r6c5
50d. 8 locked in column 4 of n5
50e. 7 locked in n8 for column 6
.-------------------------------.-------------------------------.-------------------------------.
| 567 123 123 | 12 567 5689 | 45689 45689 89 |
| 456 789 789 | 79 456 13 | 123 123 56 |
| 124569 24569 1249 | 12456 28 135689 | 1356789 1356789 1356789 |
:-------------------------------+-------------------------------+-------------------------------:
| 3489 234569 234569 | 78 17 1569 | 2345689 2345689 2345689 |
| 13489 12345679 12345679 | 3568 3569 24 | 123456789 123456789 12679 |
| 1235679 1245679 12345679 | 3568 569 24 | 123456789 123456789 123456789 |
:-------------------------------+-------------------------------+-------------------------------:
| 1235679 123456789 123456789 | 1234569 5689 156789 | 123456789 123456789 123456789 |
| 245 123456789 123456789 | 1234569 123 56789 | 3456789 3456789 12346 |
| 679 12345 12345 | 12345 234 5689 | 5689 3456789 3456789 |
'-------------------------------.-------------------------------.-------------------------------'
Having to take a long look now to get any further . . .
Posted: Sun Mar 11, 2007 5:58 pm
by PsyMar
Here's a couple more moves; didn't repost the graph since it's only a couple of eliminations.
5/6 in N1: Either R12C1 = {56} or R3C12 = {56}; thus there is either a 5 or 6 in R123C1; thus 11/2 in N7 != [56]
Outies of R9: R8C9 != 1; must contain 2 -> no 2 in 16/3 or 20/3 in R8
Posted: Sun Mar 11, 2007 8:42 pm
by rcbroughton
Thanks PsyMar
that moves us along a bit more:
numbering your steps as 51 and 52 - I now have:
53. 1 now locked in row 7 of n9
54. 45 on c1 - r67c1 minus r3c2 equals 1
54a. 11(2) in n1 and n7 don't allow combis in r67c1 with {25}
54b. no 5/6/9 in r6c1
55. 45 on r2345678 r2c159+r8c159 total 24
55a. r567c5=20(3) prevent [53] or [63] in r28c5
55b. possibilities {456}{126} or {456}{234}
55c. first must be [216] in r8 so, r2 must be [465]
55d. second can be [234] or 4{32} in r8 with r2 = {5/6}4{5/6} in both cases
55e. no combo with 5 at r2c5
55f. r1c5 != 6
56. can't have 5 at r1c7 or r1c8 as it creates a contradiction
56a. r1c7 or r1c8 = 5 -> r12c9=[86] -> r12c5=[74] -> r12c1=[65], r2c4=9 -> r1c6=8 contradction with r1c9
57. 18(3)n23=5{49}/{468} - no 9 at r1c6
58 9 locked in n3 for row 1
59. XY-chain on found: r3c6=8=>r3c6<>9->r2c4=9=>r2c4<>7->r1c5=7=>r4c5<>7->r4c5=1=>r1c4<>1 (through empty rectangle in n8)->r1c4=2=>r3c5<>2->r3c5=8 contradiction
59a. no 8 in r3c6
60. can't have a 5 at r3c6
60a. r3c6=5 -> r12c5=[74] -> r2c4=9 -> r34c5=[81] -> r1c4=2
60b. -> r1c6=6 -> r3c4=2 contradiction
61. can't have a 6 t r3c6
61a. r3c6=6 -> r12c5=[74] -> r4c6=1 -> r34c5=[27] -> r1c6=8 -> r1c9-9 -> r1c4=1
61b. -> r3c4=1 contradiction
62. 45 on n2 - all bar r12c5 total 34 - either
62a. {1234789} r3c4=4
62b. {1235689} {56} locked in r1c6 r3c4
62c. no 1/2 in r3c4
Posted: Tue Mar 13, 2007 11:07 am
by sudokuEd
Richard now has the world record for biggest "45" move: 7 rows! Has meant a few more moves possible, but still no placement. Love PsyMar's observation about the 5 and 6 in c1. Hasn't led to anything much that I can find.
Desperate to finish this horrible one and get onto Para's fine looking X. Oh well. Feel an obligation to die of exhaustion as punishment for calling this a V2
. So, no more V2's for a while - especially with the Cricket World Cup starting.
I'll start with a marks pic just to make sure we're at the same spot. Think I have a 3 in r6c5 that Richard didn't seem to have.
Code: Select all
.-----------.-----------------------------------.-----------.-----------------------------------.-----------.
| 567 | 123 123 12 | 57 | 568 4689 4689 | 89 |
| :-----------------------------------: :-----------------------------------: |
| 456 | 789 789 79 | 46 | 13 123 123 | 56 |
:-----------'-----------.-----------------------: :-----------------------.-----------'-----------:
| 124569 24569 | 1249 456 | 28 | 139 135678 | 135678 135678 |
| .-----------' | | '-----------. |
| 3489 | 234569 234569 78 | 17 | 1569 2345689 2345689 | 2345689 |
| :-----------------------.-----------'-----------'-----------.-----------------------: |
| 13489 | 12345679 12345679 | 3568 3569 24 | 123456789 123456789 | 12679 |
:-----------'-----------. | | .-----------'-----------:
| 1237 1245679 | 12345679 | 3568 3569 24 | 123456789 | 123456789 123456789 |
| | '-----------. .-----------' | |
| 235679 23456789 | 23456789 2345679 | 5689 | 56789 123456789 | 123456789 123456789 |
:-----------.-----------'-----------------------+-----------+-----------------------'-----------.-----------:
| 24 | 13456789 13456789 1345679 | 123 | 56789 3456789 3456789 | 2346 |
| :-----------------------------------: :-----------------------.-----------' |
| 79 | 12345 12345 12345 | 234 | 5689 5689 | 3456789 3456789 |
'-----------'-----------------------------------'-----------'-----------------------'-----------------------'
63.{1568} combo in 20(4) must be [5681], but [568] blocked by r1c6
63a.20(4) = {1478/2567}
63b. r123c5 = [748/562]
64. r12c4 = [17/29] = 8/11 ([27] blocked by r123c5;[19] blocked by r23c6)
65. "45"n1 -> r12c4 + 4 = r3c123 = 12/15
65a. -> r3c123 = 12 = {156} only ({129} blocked by r1c23;{146/245} blocked by 11(2)n1)
65b. and r3c123 = 15 = {249} only, otherwise no 2 or 9 for n1 with r12c4 = [29] (yeah: finally got to use the valid part of that 'hint'!!)
65c. r3c123 = {156/249}
66. 22(4)n14 = {1489/2389/3568}
66a. -> r3c123 = [{56}1/{29}4/{49}2]
66b. no 9 r3c3, no 1 r3c1
66c. and r45c1 = {38}/[81] (no 4,9)
66d. no 1,6 r5c9
67. 1 in c1 only in n4:Locked for n4
67. 28(5) at r3c3 can't have any combo's with both 7 and 8, both digits only available in r4c4
67a. {13789/25678/34678} all blocked
67b. All remaining combo's have 9
67c. 9 locked for r4, n4 in r4c23
68. 9 for n5 only in 37(7)
68a. no 9 r7c5
Posted: Tue Mar 13, 2007 2:43 pm
by Para
I agree Ed. Can't seem to find where that 3 in R6C5 was eliminated. So some steps of Richard seem invalid.
Para
p.s.
Ed you implemented 2 steps that are only valid because there is no 3 in R6C5. (step 49 and 50cd)
This is the marks pic i get after implementing all steps that seem valid to me.
.-----------.-----------------------------------.-----------.-----------------------------------.-----------.
| 567 | 123 123 12 | 57 | 568 4689 4689 | 89 |
| :-----------------------------------: :-----------------------------------: |
| 456 | 789 789 79 | 46 | 13 123 123 | 56 |
:-----------'-----------.-----------------------: :-----------------------.-----------'-----------:
| 24569 24569 | 124 456 | 28 | 139 135678 | 135678 135678 |
| .-----------' | | '-----------. |
| 38 | 234569 234569 78 | 17 | 156 234568 234568 | 234568 |
| :-----------------------.-----------'-----------'-----------.-----------------------: |
| 138 | 234567 234567 | 3568 35689 24 | 123456789 123456789 | 279 |
:-----------'-----------. | | .-----------'-----------:
| 1237 24567 | 234567 | 3568 35689 24 | 123456789 | 123456789 123456789 |
| | '-----------. .-----------' | |
| 235679 23456789 | 23456789 2345689 | 568 | 56789 123456789 | 123456789 123456789 |
:-----------.-----------'-----------------------+-----------+-----------------------'-----------.-----------:
| 24 | 13456789 13456789 1345689 | 123 | 56789 3456789 3456789 | 2346 |
| :-----------------------------------: :-----------------------.-----------' |
| 79 | 12345 12345 12345 | 234 | 5689 5689 | 3456789 3456789 |
'-----------'-----------------------------------'-----------'-----------------------'-----------------------'
Posted: Tue Mar 13, 2007 3:45 pm
by Para
I don't know if this is going to help.
69. 28(5)in R3C3 can't be {15679}.
69a. 28(5) = {15679}: R3C3 = 1; R1C23 = {23}; R1C4 = 1
69b. 28(5) = {15679}: R3C3 = 1; R4C4 = 7; R2C4 = 9; R2C23 = {78}; R12C2 = {56}; R3C12 = {49}; R3C6 = 3; R2C6 = 1: contradiction.
70. 28(5) in R3C3: when R4C4 = 8: R3C3 = 1; No {23689/{24589}
70a. 28(5): R4C4 = 8; R4C5 = 7
70b. 28(5) : R3C3 = 2; R1C4 = 2; R3C5 = 8; R4C5 = 1 : contradiction
70c. 28(5) : R3C3 = 4; R12C1 = {56}; R3C12 = {29}; R1C4 = 2; R3C5 = 8; R4C5 = 1 : contradiction
[edit] ok Made some acual eliminations now.
71 R56C5: no 8
71a. 28(5): R3C3 = 1; R4C4 = 8 -->> R56C5 : no 8
71b. 28(5): R3C3 = {24}; R1C4 = 2(step 70); R3C5 = 8 -->> R56C5: no 8
72. 8 in N5 locked for C4
Same eliminations made by richard because of 3 in R6C5 missing.
Para
Posted: Tue Mar 13, 2007 8:46 pm
by rcbroughton
sudokuEd wrote:.-----------.-----------------------------------.-----------.-----------------------------------.-----------.
|(11) |(6) |(20) |(18) |(14) |
| 567 | 123 123 123 | 567 | 5689 45689 45689 | 89 |
| :-----------------------------------: :-----------------------------------: |
| |(24) | |(6) | |
| 456 | 789 789 79 | 456 | 13 123 123 | 56 |
:-----------'-----------.-----------------------: :-----------------------.-----------'-----------:
|(22) |(28) | |(22) |(19) |
| 1234569 234569 | 1234569 123456 | 28 | 135689 1356789 | 1356789 1356789 |
| .-----------' | | '-----------. |
| | | | | |
| 2345689 | 234569 234569 78 | 17 | 135679 2345689 2345689 | 2345689 |
| :-----------------------.-----------'-----------'-----------.-----------------------: |
| |(26) |(37) |(29) | |
| 12346789 | 12345679 12345679 | 3568 35689 24 | 123456789 123456789 | 12346789 |
:-----------'-----------. | | .-----------'-----------:
|(17) | | | |(16) |
| 12345679 12345679 | 12345679 | 3568 5689 24 | 123456789 | 123456789 123456789 |
| | '-----------. .-----------' | |
| | | | | |
| 12345679 123456789 | 123456789 123456789 | 35689 | 1356789 123456789 | 123456789 123456789 |
:-----------.-----------'-----------------------+-----------+-----------------------'-----------.-----------:
|(11) |(16) |(5) |(20) |(16) |
| 2456 | 123456789 123456789 123456789 | 123 | 356789 3456789 3456789 | 123456 |
| :-----------------------------------: :-----------------------.-----------' |
| |(8) | |(14) | |
| 5679 | 12345 12345 12345 | 234 | 5689 5689 | 3456789 3456789 |
'-----------'-----------------------------------'-----------'-----------------------'-----------------------'
Just checked back through my notes and couldn't find where I'd eliminated the 3 from r6c5. On further look, I actually picked it up from the marks pick here in Ed's post.
Guess I should have checked a bit closer.
I'll take it from your position Para
[Edit - a couple of steps]
73. 45 on n1. r1234c4 + r45c1 + r4c23 total 46
73a. can't have {57} or {67} in r123c4 because of r12c5=11(2)
73b. can't have {68} or {58} in r1234c4 because of r56c4=11(2)
73c. so, r1234c4=[1748]/[1947]/[1948] (20,21,22)
73d. [1748] -> r4c1=3
73e. [1947] -> r4c12+r4c23=25 = {38}+{59}
73f. [1948] -> r4c1=3
73g - no 3 at r4c23
74. 45 on c1 r367c2=16 only combo with 6 is {367} - 3 must be at r7c2. No 6 at r7c2
Richard
1537
2816:6154:6154:6154:5124
1550
4160:4160:4160
5188:5188:5188:4167
2051
4167:
