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 Texas Jigsaw Killer 31 Goto page 1, 2  Next
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Para
Yokozuna

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

 Posted: Sat Jul 07, 2007 10:52 pm    Post subject: Texas Jigsaw Killer 31 Hi guys This one was a breeze after that work on TJK 18. Hardly used any Jigsaw techniques though. Only 2 LOL moves. Here's the walkthrough. As it just came out i'll keep it in tiny print. Walkthrough TJK 31 N1 at R1C1 N2 at R1C4 N3 at R1C9 N4 at R3C4 N5 at R4C1 N6 at R4C9 N7 at R6C2 N8 at R7C7 N9 at R8C3 1. 6(3) at R7C8 = {123} -->> locked for N8 2. 27(4) at R3C8 = {3789/4689/5679}: no 1,2 3. 45 N1: 2 innies: R13C3 = 7 = {16/25/34} 4. 45 C123(including hidden cage at R13C3): 1 outie: R4C4 = 4 4a. R45C3 in 12(3) at R4C3 = {17/26/35}: no 8,9 5. 45 N8: 2 innies: R79C7 = 15 = {69/78} 6. 45 C789(including hidden cage at R79C7): 1 outie: R6C6 = 8 6a. 26(4) at R4C9 = {4679}(last possible combo) -->> locked for N6 6b. R56C7 in 13(3) at R5C7 = {23}(last possible combo) -->> locked for C7 and N6 6c. R5C8 = 5; R5C6 = 1 6d. Clean up: R4C3: no 3,7 7. R3C8 + R4C78 in 27(4) at R3C8 = {679}(last possible combo) -->> locked for N3 8. LOL C789: R6C456 = R1C78 + R2C7 8a. Outies: no 4, 6,7, 9 -->> Innies: no 4, 6, 7, 9 8b. Outies: R56C6 = [18], so innies needs {18} -->> locked for N2 and 15(4) cage at R1C7 8c. 15(4) at R1C7 = {1248} (needs {18}) -->> R1C8 = 2; R3C7 = 4(only place in cage); R12C7 = {18} -->> locked for C7 8d. Innies: R1C8 = 2, so innies need 2 -->> R4C6 = 2 9. R79C7 = {69}(last combo) -->> locked for C7 and N9 9a. R4C7 = 7; R8C7 = 5 9b. Naked Pair {69} at R34C8 -->> locked for C8 9c. Naked Pair {69} at R4C89 -->> locked for R4 10. 45 on N5: 2 innies: R5C24 = 8 = {26} -->> locked for R5 and N5 10a. R56C7 = [32]; R45C3 = [17] 11. LOL on R123: R3C456 = R4C678 11a. Outie: no 1, 3, 5, 8 -->> Innies: no 1, 3, 5, 8 11b. Outies: R4C67 = [27], so innies needs {27} -->> locked for R3 and N4 11c. R7C4 = 1(hidden); R7C8 = 3; R78C9 = [21] 11d. R7C2 = 1(hidden); R9C5 = 1(hidden); R2C8 = 1(hidden); R12C7 = [18]; R3C1 = 1(hidden) 12. 12(3) at R3C3 = {237} (last possible combo) -->> R3C3 = 3; R3C45 = {27} 12a. R1C3 = 4(Step 3) 13. 45 on N2: 1 innie: R2C8 = 4 13a. R3C6 = 6; R34C8 = [96]; R45C9 = [94]; R6C89 = [76] 13b. R9C9 = 7(hidden); R6C5 = 4(hidden); R9C2 = 4(hidden); R89C8 = [48]; R7C1 = 4(hidden) 14. 22(5) at R4C5 needs one of {358} in R4C5 and one of {26} in R5C4 -->> 22(5) = 14{368} -->>R4C5 = 3; R5C45 = [68] 14a. R5C1 = 9; R5C2 = 2; R6C1 = 3(hidden); R8C2 = 3(hidden); R7C2 = 7(hidden) 14b. R6C34 = [95]; R8C4 = 8; R7C3 = 8(hidden) 15. R12C4 = {39} (last possible combo to fill 16(3) at R1C3) -->> locked for C4 and N2 15a. R9C4 = 2; R89C1 = [26]; R89C3 = [65]; R9C67 = [39]; R8C56 = [97] 15b. R6C567 = [596]; R3C45 = [72]; R1C6 = 5; R2C3 = 2 16. 14(3) at R2C1 = {158} (last possible combo) -->> R2C1 = 5; R3C2 = 8 16a. R4C12 = [85]; R123C9 = [835]; R1C1 = 7; R12C5 = [67]; R12C4 = [39]; R12C2 = [96] And we are done. greetings ParaLast edited by Para on Mon Jul 16, 2007 7:02 pm; edited 2 times in total
Para
Yokozuna

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

 Posted: Fri Jul 13, 2007 5:13 pm    Post subject: Hey guys I tried to make a V2 of TJK 31 (Ed askes for one ). Because the cage pattern combined wih the jigsaw shapes gave away quick singles i tried to change a few cages to make the opening less obvious. These both use the same solution as the original. TJK 31V1.5 This one seems to be a bit harder. But there was a different opening i missed in the first run. Which in the end doesn't make it much harder than the original. TJK 31V2 This one is a lot harder. That is why i included the V1.5. It is more fun to solve. I don't know how to get a PS-string for jigsaw killers from SumoCue. Maybe someone else can provide these. greetings Para
mhparker
Grandmaster

Joined: 20 Jan 2007
Posts: 345
Location: Germany

 Posted: Fri Jul 13, 2007 6:51 pm    Post subject: Hi Para, PS format doesn't support jigsaws - you have to use SumoCue format (which, fortunately, JSudoku supports as well ). Here's the SumoCue input string for the V1.5: SumoCueV1=24J0+0J0=16J0+2J1=18J1+4J1=15J1+6J1=17J2=14J0+0J0+0J0+2J1+4J1=12J1+6J1+8J2+8J2+9J0+9J0=12J0+20J3+20J3+14J3+6J2=22J2+8J2=25J4+27J4=20J4+29J4=15J3+14J2+25J2+25J2=26J5+27J4+29J4+29J4+29J4+31J3=19J5+41J5+41J5+35J5+27J4=17J6+46J6=14J6+31J3+41J5+41J5+35J5+35J5=15J6+46J6=23J6+48J3=20J3+58J3+58J7=6J7+61J7+54J6+54J6+56J8+48J8=12J8=19J8=24J7+69J7+61J7+54J6+56J8+56J8+67J8+67J8+68J8+68J7+69J7+69J7 And here's the corresponding one for the V2: SumoCueV1=24J0+0J0=16J0+2J1=18J1+4J1=24J1+6J1=17J2=14J0+0J0+0J0+2J1+4J1=10J1+6J1+8J2+8J2+9J0+9J0=12J0+20J3+20J3+14J3+6J2+6J2+8J2=25J4+27J4=20J4+29J4=15J3=15J2+32J2+32J2=26J5+27J4+29J4+29J4+29J4+31J3=19J5+41J5+41J5+35J5+27J4=15J6+46J6+46J6+31J3+41J5+41J5+35J5+35J5=15J6=30J6+55J6=9J3=20J3+58J3+58J7=6J7+61J7+54J6+54J6+55J8+57J8=12J8=19J8=24J7+69J7+61J7+54J6+55J8+55J8+67J8+67J8+68J8+68J7+69J7+69J7_________________Cheers, Mike
Para
Yokozuna

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

 Posted: Fri Jul 13, 2007 11:49 pm    Post subject: Thanks How do i get that?
mhparker
Grandmaster

Joined: 20 Jan 2007
Posts: 345
Location: Germany

Posted: Sat Jul 14, 2007 6:29 am    Post subject:

 Para wrote: Thanks How do i get that?

Using SumoCue, simply select the "Copy (SumoCue)" option to copy the puzzle definition to the Clipboard in SumoCue text format.

With JSudoku, I've just found out it's even easier: the SumoCue format appears to be the default format for jigsaws here, so all you need to do is a normal Copy (Ctrl-C).
_________________
Cheers,
Mike
sudokuEd
Grandmaster

Joined: 19 Jun 2006
Posts: 257
Location: Sydney Australia

Posted: Sun Jul 15, 2007 10:46 am    Post subject:

 Para wrote: TJK 31V2 This one is a lot harder.
Oh silly me. Nice one Para.

Here is the start. Haven't looked at any combination crunching yet. Will have to work on this one over the week. Anyone is welcome to join in .

Cheers
Ed

TJK 031V2
0. 10(2)r2c6 - no 5
0a. 26(4)r4c9 - no 1
0b. 9(2)r7c4 - no 9
0c. 20(3)r7c5 - no 12
0d. 6(3)r7c8 ={123}
0e. 19(3)r8c6 - no 1

1. 6(3)r7c8 = {123}: all locked for n8

2. "45" n8(r7c7): r79c7 = 15 = h15(2)n8
2a. = {69/78}

3. 24(4)r8c7 must have 4 and 5 for n8 = 45{69/78}

4. "45" n1: r13c3 = 7 = h7(2)n1
4a. ={16/25/34}(no 789)

5. "45" c123: (remembering the h7(2)n1):r456c4 = 15

6. LoL c123: 3 outies r456c4 = 3 innies r8c3 + r9c23
6a. 3 outies = 15 (step 5) -> 3 innies = 15
6b. -> r7c23 = 15 (same cage as 3 innies)
6b r7c23 = h15(2)r7 = {69/78} = [7/9,8/9..]

7. 20(3)r7c5 = {569/578} ({389/479} blocked by h15(2)r7 step 6b)
7a. = 5{69/78}(no 1..4)
7b. 5 locked for r7 and n4(r3c4)
7c. no 4 r8c4

8. Killer quad {6789} in 20(3)r7c5 and h15(2)r7
8a. {6789} locked for r7
8b. no 123 r8c4

9. LoL r789: 3 innies r7c456 = 3 outies r6c234
9a. 3 innies must have 5 -> 3 outies must have 5
9b. -> 15(3)r6c2 must have 5 = 5{19/28/37/46}
9c. 5 locked for r6 & n7(r6c2)

10. Common Peer Elimination (CPE): no 4 in r6c4 since it can 'see' both 4's in r7

11. LoL r123: 3 innies r3c456 = 3 outies r4c678
11a. no 5 in innies (from step 7b) -> no 5 in outies
11b. 15(3)r4c6 = {168/249/267/348}

12. "45" n3(r1c9): 2 innies r3c78 = 13 = h13(2)n3
12a. = {49/58/67}

13. rest of 24(5)r1c7 = 24 - 13 = 11 = h11(3)n2
13a. no 9 r1c78 or r2c7

14. LoL c789: 3 innies r1c78 + r2c7 = 3 outies r456c6
14a. no 9 in innies -> no 9 in outies r456c6

15. "45" c6789: r1c6 = r7c5
15a. -> r1c6 = {56789}

16. 5 in c6 only in r17c6. Here's how.
16a. 2 5's in r7 in c56.
16b. 5 in r7c5 -> 5 in r1c6 (step 15)
16c. or 5 in r7c6
16d. 5 locked for c7

17. LoL c789: 3 innies r1c78 + r2c7 = 3 outies r456c6
17a. no 5 in outies -> no 5 in innies

18. "45" n2: (remembering h11(3)n2): r1c3 = r2c6 = {12346}
18a. no 123 r3c6
18b. no 2 r3c3 (h7(2)n1)

19. LoLr89: 3 outies r7c789 = 3 innies r8c12 + r9c1
19a. no 4 in outies -> no 4 in innies
19b. 15(4)r7c1 = {1239/1248/1347/2346}

20. CPE: no 4 in r6c1 since it sees all 4's in n7(r6c2)

 Code: .-------------------------------.-------------------------------.-------------------------------. | 123456789 123456789 12346     | 123456789 123456789 56789     | 1234678   1234678   123456789 | | 123456789 123456789 123456789 | 123456789 123456789 12346     | 1234678   123456789 123456789 | | 123456789 123456789 13456     | 12346789  12346789  46789     | 456789    456789    123456789 | :-------------------------------+-------------------------------+-------------------------------: | 123456789 123456789 123456789 | 123456789 12346789  1234678   | 12346789  12346789  23456789  | | 123456789 123456789 123456789 | 123456789 12346789  1234678   | 123456789 123456789 23456789  | | 1236789   123456789 123456789 | 12356789  12346789  1234678   | 12346789  2346789   2346789   | :-------------------------------+-------------------------------+-------------------------------: | 1234      6789      6789      | 1234      56789     56789     | 6789      123       123       | | 1236789   1236789   123456789 | 5678      123456789 2346789   | 456789    456789    123       | | 1236789   123456789 123456789 | 123456789 123456789 2346789   | 6789      456789    456789    | '-------------------------------.-------------------------------.-------------------------------'
rcbroughton
Expert

Joined: 15 Nov 2006
Posts: 143
Location: London

 Posted: Sun Jul 15, 2007 8:39 pm    Post subject: Hi Ed I'll jump in on this one - I think there is going to be some significant crunching on this one. I'm struggling to find a few moves to build on your start. 21. LOL on c6-9 - r3789c6 contains no 1 - so no 1 in r12c45 (just as well I got my Jigsaw highlighting working again!) 21a. 18(3) r1c5 now has no 1 - {189} no longer valid 22. 45 rule on c 1-4 r3c5=r9c4 (Let's make a note of this one - could be useful later) 22a. no 5 at r9c4 Rgds Richard
rcbroughton
Expert

Joined: 15 Nov 2006
Posts: 143
Location: London

 Posted: Mon Jul 16, 2007 8:13 am    Post subject: Couple of extra moves on the train this morning. 23. 45 Rule on n4 - innies r3c456 r7c456 total 30 Cage 20(3) at r7c5 limits r7c56 = 11,12(no 8),13(no 7),14(no 6) = {56}/{57}/{58}/{59} Cage 12(3) at r3c3 limits r3c45 = 11,9(no 3),8,7,6={29}/{38}/{47}/{18}/{27}/{17}/{26}/{16}/{34}/{24} 23a. combinations = {234579}/{135678}/{125679}/{134589} (must use a 5) 23b. {234579} - r7c56={57} - > r3c345={29}4/{34}9/{24}9 23c. {234579} - r7c56={59} - > r3c345={27}4/{34}7/{24}7 23d. {134589} - r7c56={58} -> r3c345={34}9 23e. {134589} - r7c56={59} -> r3c345={34}8 23f. No other combo with a 4 - > no 4 at r7c4 23g cleanup - no 5 at r8c4 24. Hidden single 4 at r7c1 for r7 24a. 15(4)r7c1 = {1248}/{1347}/{2346} - no 9 25. 15(3)r6c2 - no {168}/{267} - blocked by 15(4)r7c1 25a. no 6 in 15(3)r6c2 26. LOL on r89 - no 9 in r8c12, r9c1 - > no 9 at r7c7 26a. innies on n9 r7c7+r9c7 = 15 - > no 6 at r9c7 27. LOL on r789 - no 6 in r6c123 -> no 6 in r7c456 27a. (from step 22) - no 6 at r1c6 Rgds Richard
sudokuEd
Grandmaster

Joined: 19 Jun 2006
Posts: 257
Location: Sydney Australia

 Posted: Mon Jul 16, 2007 12:47 pm    Post subject: Nice one Richard. First placement already. This is going to be a breeze. First up, alternate step 23, then a bit of combo crunching then finally a couple of eliminations to make the post worthwhile. Alternative step 23. 23. no [45] in 9(2)r7c4. Here's how. 23a. from step 6b: r7c23 = h15(2)r7 = {69/78} = [6/8,6/7..] 23b. -> r8c3 + r9c23 = 15 = h15(3)r8c3 23c. = {159/249/258/348/357/456} ({168/267} clash with h15(2)) 23d. = [4/5..] 23e. from LoL c123: 3 outies r456c4 = 3 innies r8c3 + r9c23 23f. -> 3 outies r456c4 = [4/5..] 23g. -> [45] blocked from 9(2)r7c4 Now some combo. work. 28. r3c78 = h13(2) = {49/58/67} 28a.-> r1c78 + r2c7 = h11(3) = {128/137/146/236} 29. 15(3)r4c5 = {168/249/267/348} 30. "45" c6789: r127c5 = 18 = h18(3)c5 30a. = {279/378/459/567} ({369/468} blocked by 15(3)r4c5 step 29) 30b. since r7c5 = r1c6 (i/oc6789) -> {369/468} also blocked from 18(3)r1c5 30c. 18(3)r1c5 = {279/378/459/567} A few eliminations! Hope these are correct. 31. no 8 in r8c78, Here's how. (This might be easier to see if they were written as xy chains: might have to edit) 31a. r7c456 = same combinations as 15(3)r6c2 (from LoL r789) 31b. = [1]{59}/[2]{58}/[3]{57} 31c. -> r7c4 + r7c56 + r7c7 = [1]{59}[6]/[2]{58}[7]/[3]{57}[8] 31d. -> r7c7 + r8c4 = [68/77/86] i. 8 is in r8c4 when 6 is in r7c7 -> no 8 in r8c78 ii. or 8 is in r9c7 when r7c7 = 7 -> no 8 in r8c78 iii. or 8 is in r7c7 -> no 8 in r8c78 32. no 7 in r8c12. Here's how. 32a. from LoL r89: 3 outies r7c789 = 3 innies r8c12 + r9c1 32b. 7 in outies in r7c7 -> 7 in r8c4 (step 31d) -> no 7 in r8c12 -> 7 in innies only fits in r9c1. 32c. if 7 is not in r7c7 -> from LoL r89, no 7 is possible in 8c12
Para
Yokozuna

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

Posted: Mon Jul 16, 2007 4:34 pm    Post subject:

 sudokuEd wrote: Nice one Richard. First placement already. This is going to be a breeze.

Look who's getting cocky. I never said it was going to be like TJK 18. But this was the easy bit. It get's a bit more challenging from now on. Took me a few days(well mostly nights) to solve it.

Para

ps.
 sudokuEd wrote: 5. "45" c123: (remembering the h7(2)n1):r456c4 = 15

rcbroughton
Expert

Joined: 15 Nov 2006
Posts: 143
Location: London

 Posted: Mon Jul 16, 2007 6:52 pm    Post subject: 33. 45 on n9 r8c3. outies = 25, but r7c23 = 15 so r7c4+r9c7 = 10. 33a. but, since r79c7=15, when r7c4=1 -> r7c7=6, 2->7, 3->8 33b. r7c456=h15(3) 33c. r3c456=h15(3) 34. placement for h15(3)r3c456 and h15(3)r7c456 - 34a. no 2 at r7c4, no 8 at r7c56 because: {234579} -> needs r7c456=3{57}, {135678} -> needs r7c456=3{57}, {125679} -> needs r7c456=1{59}, {134589} -> needs r7c456=1{59}, 34b. (from 21) no 8 at r1c6 35. fom 33 - > no 8 at r9c7, no 7 at r7c7 35a. no 7 at r8c4 36. 2 locked in r7c89 for r7 - no 2 at r8c9 37. LOL on r89 37a. No 7 in outies r7c789 - so no 7 at r9c1 37b. 15(4) at r7c1 now = 4{128}/4{236} - must use 2, no 2 in 15(3)r6c2 37c. {258} no longer valid in 15(3) - no 8 [edit - a couple more for Ed to work on overnight] 38. 45 rule on n8 r7c7 - outies total 24. 38a. r7c56=12 -> r89c6 = 12, no 7 = {39}/{84} 38b. r7c56=14 0> r89c9=10, no 8,9 = {37}/{64} 38c - > no 2 in r89c6 39. LOL on r6789 - no 2 in r3789c6 - so no 2 in r12c45 40. outies of c1-4 = r389c6 = h12(3) ={129}/{138}/{147}/{156}/{237}/{246}/{345} 40a. so r127c6=h18(3) = {459}/{567}/{369}/{378} 40b. 15(3) in c5 = {168}/{249}/{267}/{348} combining 40a/40b. = 33(6)={459}{168}/{459}{267}/{567}{249}/{567}{348}/{378}{249} ={145689}/{245679}/{345678}/{234789} = must use 4, blocked by 1/2/9, 1/4/7, 2/4/6, 3/4/5 40c so h12(3)={138}/{156}/{237} - no 4 41. 45 on c1-4 r9c4 = r3c5 - no 4 at r9c4
sudokuEd
Grandmaster

Joined: 19 Jun 2006
Posts: 257
Location: Sydney Australia

Posted: Tue Jul 17, 2007 2:40 am    Post subject:

Good going Richard. Step 34 has made a big difference. Here's the next batch of steps.

Think there is a typo in step 40c, so will give an alternate step 40 first.

Alternative step 40. Trying to work out what happened to {129} combo in h12(3)
40. Updating step 30. r127c5 = 18 = h18(3)c5 = {378/459/567} & 15(3)r4c5 = {168/249/267/348}
40a. combining the two cages = {378-249/459-168/459-267/567-249/567-348}
40b. -> 4 locked for c5
40c. "45" c5: r389c5 = 12 = h12(3)c5= {129/138/156/237}

More
42. 3 innies n9(r8c3): r8c4 + r89c6 = 18 = h18(3)n9
42a. = [6]{39/48}(from step 38a)
42b. = [8]{37/46}(from step 38b)
42c. = {369/378/468} = [3/8..]

43. 12(3)r8c5 = {129/156/237}(no 8)({138} blocked by h18(3)n9 step 42c)
43. ->no 8 r3c5 (I/Oc1234)

44. h12(3)c5 (alt.step 40c) = {129/156/237} = [2/6/7..]

45. 15(3)r4c5 = {168/249/348}(no 7) ({267} blocked by h12(3) step 44)

46.from alt. step 23b,c: r8c3 + r9c23 = h15(3)r8c3
46a. = {159/249/357/456}(no 8) ({258} blocked by 12(3)r8c5;{348} blocked by h18(3)n9 step 42)

47. LoL c123: 3 outies r456c4 = 3 innies r8c3 + r9c23
47a. no 8 in innies -> no 8 in outies

48. {159} blocked from h15(3)(r8c3 + r9c23 step 46a). Here's how.
48a. {159} in h15(3) -> r7c23 = {78} -> r7c7 = 6 -> r7c4 = 1 (step 33a)
48b. {159} in h15(3) -> from LoLc123: outies r456c4 = {159}
48c. but this means 2 1's in c4
48d. -> {159} blocked from h15(3)n9

49. updating step 46a. h15(3)n9 = {249/357/456} (no 1)
49a. -> r456c4 (LoLc123) = {249/357/456}(no 1)
49b. -> r456c4:no 9 r45c4 ({249} combo, 9 only can go in r6c4)

50. 1 in n9 now only in 12(3)r8c5
50a. = {129/156}(no 3,7)
50b. -> no 3 or 7 in r3c5 (i/oc1234)
50c. 12(3) = [6/9..]

51. from step 42. 3 innies n9 = h18(3)n9
51a. = [6]{48} ([6]{39} blocked by 12(3)n9 step 50c.)
52b. = [8]{37/46}
52c. = {378/468}(no 9)

53. no 1 in 15(3)r4c5. Here's how.
53a. 1 in 12(3)r8c5 in r9c4 -> 1 in r3c5 (i/oc1234) -> no 1 elsewhere in c5
53b. 1 in 12(3) in r89c5 -> no 1 elsewhere in c5
53c. 1 in c5 only in r389c5.

54. 15(3)r4c5 = {249/348} = 4{29/38}
54a. 4 locked for c5 & n4(r2c4)
54b. no 6 r2c6

Over to you Richard. Should be here.

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rcbroughton
Expert

Joined: 15 Nov 2006
Posts: 143
Location: London

 Posted: Tue Jul 17, 2007 7:26 pm    Post subject: Hi Ed Sorry - but the train journey this evening provided just enough time to finish it. weren't really any other tricky moves from where you'd left it. 55. 6 Locked in r3c456 in nonet at r3c4. Locked for r3 56. 18(3)r1c5 = {378}/{567} - must use 7. - nowhere else in nonet 57. LOL on r123 - no 4 in r3c456 - so no 4 in r4c678 57a. no 4 in 15(3) r4c6 = 6{18}/{27} - no 3, 9 57b. 6 locked for n3 and r4 58. LOL on c789 - no 7 in r1c78 r2c7 - so no 7 in r456c6 59. LOL on r123 - no 3,9 in r4c5678 - so no 3,9 in r3c456 60. 12(3)r3c3 = 4{26}/[381]/[471]/5{16}/[372] - no 1 at r3c3 61. 45 Rule on n2r1c4 - outies r1c3 r3c6 total 10 61a. r1c3 - no 1,6 62. 9 locked in r12c4, 16(3) for c4 = 9{25}/9{34} - no 8 63. Hidden single 9 at r7c6 for c6 63a. 20(3)r7c5 = [596] 64. Hidden single 5 at r1c6 for c6 64a. 18(3)r1c5=5{67} 64b. {67} locked for c5 65. 16(3)r1c3 = {349} 65a. Cleanup - no 3,4 at r1c78 66. 10(2)r23c6 = no 1 at r2c6 66a. 1 locked in r1c78+r2c7 for n2 -> h11(3) = {128} 67. 10(2)r23c6 - no 8 at r3c6 68. 9 locked in 12(3) r8c5 for c5 - 12(3)={129} 69. h15(2) r7c23 = naked pair {78} 70. 15(4) r7c1 = 4{236} 70a. Naked single 5 at r6c4 -> 15(3)r6c2={19}5 70b. {19} locked ar c23 for r6 71. 30(5)r7c2 = {78}{456} - no 2,3 71a {456} locked for nonet 71b. naked single 8 at r8c4 - > 9(2) = [18] 72. naked single 2 at r9c4 and r3c5 73. 19(3) r8c6 = {73}9 73a naked single 1 at r9c5 -> r8c5=9 74. Naked single 1 at c9 for r8 75 Naked pair {76} at r3c46 75a. 12(3)r3c3=[462][372] - no 5 76. 24(5)r1c7={128}[49] 76a. Naked single 3 at r3c3 - > r3c4=7, r1c3=4 -> r3c6=6 -> r2c6=4 77. hidden single 4 at r9c2 for nonet 78. hidden single 4 at r8c8 for r8 and nonet 79. hidden single 6 at r4c8 for nonet 80. naked pair {56} at r89c3 81. naked pair {39} at r12c4 81a. r4c4=4 81b. r5c4=6 82. hidden single 6 at r6c9 for row, col and nonet 83. 1 locked in r5c678 for nonet - locked for row5 83a. 1 locked in r4c123 for nonet - locked for row 4 83b. hidden single 1 at r2c8 for nonet 84. 17(4)r1c9 = 1{358} - {358} locked for nonet and col 9 84a. hidden single 2 at r4c6 - 15(3) = [276] 84b. r8c7 = 5 84c. r8c3=6, r9c3=5 84d. r7c9=2, r7c8=3 84e r9c9 = 7, r9c8 = 8 84f. more singles: 3 at r9c6,7 at r8c6,6 at r9c1,9 at r4c9,4 at r5c9,6 at r6c9,2 at r1c8,8 at r2c7,7 at r6c8,,5 at r5c8,1 at r1c7,2 at r4c6,8 at r6c6,1 at r5c6 85. hidden single 4 at r6c5 for row, col, nonet 86. 25(4)r4c1 = {268}9/{358}9/{367}9 - 86a. 9 locked at r5c1 - only leaves {358}9 - 3 locked at r6c1 86b. more singles 2 at r6c7,3 at r5c7,2 at r8c1,3 at r8c2,8 at r5c5,3 at r4c5 87. Hidden single 1 at r4c3 for ror4 87a. more singles: 9 at r6c3, 1 at r6c2 88. hidden single 1 at r3c1 88a. 14(3)r2c1=[518] 88b naked singles to the end Rgds Richard
Jean-Christophe

Joined: 23 Apr 2007
Posts: 92
Location: Belgium

 Posted: Tue Jul 17, 2007 10:20 pm    Post subject: deletedLast edited by Jean-Christophe on Wed Jul 18, 2007 9:45 pm; edited 1 time in total
Para
Yokozuna

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

 Posted: Wed Jul 18, 2007 1:19 am    Post subject: Hi Jc This finishes it from your position. But there must be something nicer past step 79. 68. 2 in R3 locked in R3C45; R3C5 = R9C4 -> 2 in C4 locked in R39C4 -> R5C4 <> 2 69. LOL C123: R456C4 = R8C3 + R9C23 (no 2) 70. LOL C123: R456C4 = R8C3 + R9C23 = 15 = {159/357/456} (no 8) 70a. 5 locked in R456C4 for C4 -> R12C4 <>5 70b. 5 locked in R8C3 + R9C23 for N9 -> R8C56 + R9C6 <> 5 71. R1C6 = 5(Hidden); R7C5 = 5(hidden) 72. 12(3) at R8C5 = {129/237}: no 6 72a. R9C4 = 2; R3C5 = 2(hidden) 73. 9 in N4 locked for C6. 73a. 9 in N9 locked for R8. 74. 17(4) at R1C9 = {1349/1258/1367}: needs one of {789} in R123C9 74a. 26(4) = {4589/4679/5678}(no 2) : no {2789} would need R456C9 = {789}: clash with step 73. 75. 15(4) at R7C1 = 4{128/236}: when {128}, 8 in R9C1 -->> R8C12 <> 8 75a. 8 in R8 locked for N9 76. 2 in N2 locked in hidden 15(3) at R1C78 + R2C7 = {128/236}: no 4 76a. LOL C123: R1C78 + R2C7 = R456C6: no 4 77. R1C78 + R2C7 and R3C46 together see all 6's in C5 so can't both contain 6. 77a. LOL R123 + C789 -> R456C6 + R4C678 can't both contain 6 -> overlapping cell both sets can't be 6. R4C6 <> 6 77b. R4C6 = 2 78. hidden 11(3) at R456C6 = 2{18/35} : R56C6 = {18/36} = {6|8..} 78a. 26(4) at R4C9 = {4589/4679}: no {5678} -->> {49} locked in 26(4) for N6 Must be something nicer here. 79. R13C3 = [16/34/43] 79a. 8's in N2: R12C4 = 8 or R1C78 + R2C7 = 8 79a. R12C4 = 8 -> R1C3 <> 3 79b. R1C78 + R2C7 = 8 -> R3C78 = {49} -> R3C3 <> 4 79c. R13C3 <> [34] 79d. R1C3 = R2C6 = {14}; R3C3 = {36} 79e. Clean up: R3C6: no 7; R3C4: no 6 80. 16(3) at R1C3 = [178/439]: R12C4 = {39/78} = {3|7...}: no 4,6 81. R456C4 = {159/456}: {357 blocked by step 80): no 3,7 81a. LOL C123: R8C3 + R9C23 = {159/456} = {6|9..} : no 3,7 81b. R7C23 <> {69} blocked by R8C3 + R9C23 82. R7C23 = {78} locked fr R7 and N7 82. R7C67 = [96] The rest is just basics. greetings Para
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