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Ed
 Post subject: Texas Jigsaw Killer 40
PostPosted: Mon Feb 01, 2010 3:55 am 
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I had a lot of trouble with this puzzle so it must be at the upper end of my ability. Had to do it 5 times to get it properly. Fortunately, there are many interesting moves available so very worthy of the big 4-oh. I know manu will love it! I really enjoyed making it for you.


Texas Jigsaw Killer 40
[edit: original image gone so used one of Borge's. Thanks!]
Image
code: copy, select, paste into solver:
SumoCueV1=36J0+0J0+0J0=15J1+3J1=15J1+5J1=13J2+7J2=10J0+9J0+0J0=10J1+3J1=9J1+14J1=10J2+16J2=37J0+18J0+0J3+12J3+12J4=16J1+23J2=28J2+25J2+18J0+0J3+0J3=11J3=17J4+23J4+23J5=5J2+25J2+18J3=9J3+30J3+30J4+31J4=22J4+41J5+34J5+25J5+18J6+37J6=26J3+47J4+31J4+41J5=33J5+51J5+25J7+18J6+18J6+47J6+47J8=13J4+58J5+51J5+25J7+25J7=9J6+63J6=10J8+65J8=14J8+58J8+51J7=15J7+70J7=13J6+72J6=9J8+74J8+67J8+67J8+51J7+51J7+51J7
Cheers
Ed
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Børge
PostPosted: Mon Feb 01, 2010 9:33 pm 
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Image
Cages with cells in 4 jigsaw nonets:  pink
Cages with cells in 3 jigsaw nonets:  orange
Cages with cells in 2 jigsaw nonets:  green, yellow and brown
Cages with cells in 1 jigsaw nonet:   red, blue and grey

5 further images with udosuk Style Killer Cages and Ruud TJK#33 inspiration:
Image



Image



Image
Cages with cells in 4 jigsaw nonets:  pink
Cages with cells in 3 jigsaw nonets:  orange
Cages with cells in 2 jigsaw nonets:  green, yellow and brown
Cages with cells in 1 jigsaw nonet:   red, blue and grey



Image



Image


					
						

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PostPosted: Wed Mar 10, 2010 2:02 pm 

				

				

					
		


		


			
				

			
				

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Thanks, Ed, for a great TJK! This one was a real pleasure to solve, since - as Ed said - there were lots of interesting moves available. I'm glad that Ed told me about this TJK, because it would have been a travesty of justice for no-one to have written a walkthrough for it! It will be interesting to see whether Ed's solving tricks were different than mine.

Texas Jigsaw Killer 40 Walkthrough:  
Nonet layout:

111222233
111222233
114452333
144455633
444555666
774556669
777856699
778888999
778888999

Prelims: 

a) 36(7) at R1C1 = {1236789/1245789/1345689/2345679} (no eliminations); 9 locked
b) 15(2) at R1C6 and R8C8 = {69/78} (no 1..5)
c) 13(2) at R1C8 and R9C1 = {49/58/67} (no 1..3)
d) 10(2) at R2C1 and R8C3 = {19/28/37/46} (no 5)
e) 10(3) at R2C4, R2C8 = {127/136/145/235} (no 8,9)
f) 9(2) at R2C6, R5C2, R8C1 and R9C3 = {18/27/36/45} (no 9)
g) 37(7) at R3C1 = {1246789/1345789/2345689} (no eliminations); 4,8,9 locked
h) 28(7) at R3C8 = {1234567} (no 8,9); 1..7 locked
i) 11(3) at R4C4 = {128/137/146/236/245} (no 9)
j) 5(2) at R4C8 = {14/23} (no 5..9)
k) 22(3) at R5C6 = {589/679} (no 1..4); 9 locked -> no 9 in R7C6 (CPE) 
l) 26(4) at R6C3 = {2789/3689/4589/4679/5678} (no 1)
m) 33(7) at R6C7 = {1234689/1235679/1245678} (no eliminations); 1,2,6 locked

1. 15(2) at R1C6 (prelim b) = {(6/7)..}
1a. -> {67} combo blocked for 13(2) at R1C8 (prelim c) = {49/58} (no 6,7)
1b. 15(2) at R1C6 and 13(2) at R1C8 form killer pair (KP) on {89} in R1
1c. -> no 8,9 elsewhere in R1

2. Innies N2: R2C4+R3C6 = 6(2) = {15/24} (no 3,6..9)
2a. -> {45} combo blocked for 9(2) at R2C6 (prelim f) = {18/27/36} (no 4,5)
2b. {45} in N2 locked in 15(3) and h6(2) at R2C4+R3C6, which must contain 1 of {45}
2c. -> 15(3) at R1C4 must contain exactly 1 of {45} = {159/249/348/357} (no 6)
(Note: {258} blocked by 6(2) at R2C4+R3C6)
2d. {89} in 15(3) only available in R2C5
2e. -> R2C5 = {357}+{89} = {35789} (no 1,2,4)

3. 5 of R2 now locked in innies R2: R2C345 = 16(3) = {259/358/457} (no 1,6)
3a. cleanup: no 5 in R3C6 (step 2)

4. Nishio: R3C7 sees all of R1C789
4a. -> if R3C7 = 9, then 9 of R1 would be forced into R1C6
4b. but this would leave nowhere to place the 9 in 22(3) at R5C6 (prelim k)
4c. -> no 9 in R3C7

5. Law of Leftovers (LoL) R12: R3C126+R4C1 = R12C89
5a. 9 of N3 locked in R12C89
5b. -> 9 locked in R3C12+R4C1 for N1 and 37(7)
5c. cleanup: no 1 in R2C12

== The next part of the strategy is to remove the 9 from R2C5 in order to constrain it to the
== 10(2) cage at R2C8. I found a long (but more traditional) and a short way of doing this.
== I'll start with the long way first, then describe the shortcut as an appendix.

>>>>> Start of long way of removing the 9 at R2C5 >>>>>

6. 36(7) at R1C1 (prelim a) and 37(7) at R3C1 (prelim g) form grouped X-Wing on 9 within R34
6a. -> no 9 elsewhere in R34 (i.e., no 9 in R4C567)

7. Innies C89: R6C8+R9C89 = 19(3) = {289/379/469/478/568} (no 1)
(Note: no duplicates allowed because all 3 innies belong to 33(7) cage)
7a. max. R9C89 = 15 (because {79/89} blocked by 15(2) at R8C8)
7b. -> min. R6C8 = 4 (no 2,3) 
7c. 1 of 33(7) (prelim m) locked in split 14(4) at R6789C7 = {1238/1247/1256/1346} (no 9)
7d. -> 1 locked for C7
7e. cleanup: no 8 in R2C6

8. LoL C89: R5C89+R6C8 = R389C7
8a. no 9 in outies at R389C7 -> no 9 in innies at R5C89+R6C8
8b. -> no 9 in R6C8
8c. R6C8 = R3C7 (cannot equal either of R89C7 because they share the same cage)
8d. -> no 2,3 in R3C7
8e. R5C89 = R89C7
8f. -> no 8 in R89C7

9. 9 in N6 locked in R5C7+R6C6
9a. -> no 9 in R5C6

10. Common Peer Elimination (CPE): R6C4 sees all candidate positions for 9 in R7
10a. -> no 9 in R6C4

11. 9 of N5 locked in R567C5 for C5

(Note: there are other moves involving eliminations on 9 available here, e.g., 9 in C4 locked in
R78C4 for N8. Also, there's a C6789 innie/outie difference cage 1(2) at R7C5+R9C6 that would
remove the 9 from R7C5 and lock the 9 of C5 into 17(3) at R4C5.)

<<<<< End of long way of removing the 9 at R2C5 <<<<<

12. 9 in R2 locked in R2C89
12a. -> 10(2) at R2C8 = {19}, locked for R2 and N3
12b. cleanup: no 4 in R1C89, no 8 in R2C7, no 4 in R5C8

13. LoL R12 (see step 5): R12C89 = {1589}
13a. -> R3C126+R4C1 = {1589}
13b. -> R3C6 = 1, R3C12+R4C1 = {589}, locked for N1 and 37(7)
13c. -> R2C4 = 5 (step 2)
13d. cleanup: no 2 in R2C12, no 4 in R9C3

14. Hidden single (HS) in R2 at R2C5 = 8
14a. -> R2C3 = 3 (h16(3) cage sum, step 3)
14b. cleanup: no 7 in R1C67, no 7 in R2C12, no 6 in R2C67, no 7 in R8C4, no 6 in R9C4
14c. extended cleanup (last nonet digits): R1C123 = {127} (no 4,6), R1C45 = {34} (no 2,7)

15. Split 5(2) at R3C45 = {23} (last combo), locked for R3

16. Innie/Outie difference (I/O diff.), C1234: R1C4 = R3C5
16a. -> R1C4 = 3, R3C5 = 3
16b. -> R1C5 = 4, R3C4 = 2
16c. cleanup: no 7 in R6C2, no 7,8 in R8C3, no 6,7 in R9C3

17. Hidden pair (HP) in N3 at R4C89 = {23}, locked for R4
17a. -> 5(2) at R4C8 (prelim j) = {23}, locked for C8
17b. extended cleanup (last nonet digits): R3C789 = {467} (no 5,8)
(Note: ignoring the fact that they are also locked for R4 here, as this goes beyond the scope of a cleanup)

18. R1C123 (= {127}) = 10
18a. -> Split cage at R3C3+R4C23 = 23(3) = {689}, locked for N4
18b. cleanup: no 1,3 in R6C2

19. 11(3) at R4C4 (prelim i) = {146} (last combo)
19a. -> R5C4 = 6, R4C4+R5C3 = {14}, locked for N4
19b. cleanup: no 5,8 in R6C2, no 4 in R8C3

20. 17(3) at R4C5 = {179} (last combo), locked for C5 and N5

21. HS in N5 at R7C5 = 2

22. NP at R89C5 = {56}, locked for N8
22a. -> R9C6 = 3 (cage sum)
22b. cleanup: no 4 in R89C4

23. Innies N8: R7C4+R8C6 = 12(2) = {48} (last combo)
23a. -> no 4,8 in R7C6 and R8C4
23b. cleanup: no 2 in R8C3

All singles and cage sums to finish now.

****************

Appendix: Short way of eliminating the 9 at R2C5:

6. 1 in R2 locked in 9(2) at R2C6 and 10(2) at R2C8
6a. if 1 in 10(2), then R2C89 = {19}
6b. if 1 in 9(2), then R2C67 = {18} -> 15(2) at R1C6 = {69}
6c. -> 9 locked in R1C67+R2C89
6d. -> no 9 in R2C5 (common peer)

Technically, this move can be considered to be a grouped XY-chain with 5 links:

(9,1)r2c89=(1,8)r2c67-(8=9)r1c67

Here, the strong links are denoted by a '=', the weak link by a '-' and the direct links by a ','.

Note that a similar logic could have been applied directly before the nishio step (step 4) above:

(9,1)r2c1289=(1,8)r2c67-(8=9)r1c67 -> 9 locked in R1289C2+R1C67 -> no 9 in R2C5 (common peer)

This works because both constituent 10(2) cages have a direct link on 1 and 9, a property that is
therefore inherited by the combined 20(4) at R2C1289.


					
						

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Cheers,
Mike

					



						

						

							
							
							
						

						

							
				

				

					
		


		


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				Ed
			
			
				

				

				
					
PostPosted: Sat Mar 13, 2010 8:54 am 

				

				

					
		


		


			
				

			
				

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Thanks very much for your walkthrough Mike.  Love your style!  I'm really glad you included the long version of the WT since it shows some of the really interesting clone moves available with this cage pattern.  I missed your step 10 so had to use an alternative way to crack the puzzle (different again to Mike's really neat and powerful alternative).  However, I also used your step 4 so that is the really key one.  [edit: actually, Mike found a way so that it's not needed.  See here]. Interesting that you called it Nishio.  

I have had TJK41 ready for a while so will post it on 1st April.  Perhaps I'll just do these bi-monthly.  Feel free to do one Mike!

Alternative way to crack TJK40
This is how I removed the 9 from r2c5.

From Mike's step 5.
6. 9 in n3 only in 13(2)={49} or 10(2)={19} 
6a. -> {46} blocked from the 10(2) (Locking-out Cages)
6b. 10(2) = {19/28/37}(no 4,6)

Now for a Cloned CCC!
7. "45" on r1: 1 outie r2c5 + 2 = 3 innies r1c123
7a. min. 3 innies = 6 -> min. r2c5 = 4 
7b. NOTE: no cell at r1c123 can equal r2c5 since if one did, the other 2 cells would have to sum to the IOD of 2 which is impossible
7c. -> the digit at r2c5 must be cloned at one of r1c89 a 13(2) cage (this cloning eliminates 7 from r2c5 but this is not the important bit)
7d. -> the 15(3) at r1c4 cannot contain 2 in r1c45 since it would leave a 13(2) split cage at one of r1c45 and with r2c5, but this would clash with the 13(2) at r1c8 (CCC) 
7e. ->no 2 in r1c45

8. 2 in r1 only in n1: locked for n1 and 36(7)
8a. no 8 in 10(2) at r2c1
8b. 10(2) = {37/46} = [4/7..]

9. h16(3) at r2c345 = {259/358}(no 4,7) ({457} is blocked by step 8b) 

10. 4 in r2 only in 10(2) at r2c1 = {46}: both locked for r2 & n1
10a. no 3 in 9(2) at r2c6

11. 3 in n2 only in 15(3) at r1c4 = 3{48/57}(no 9)
11. 3 locked for r1

On from here.

Cheers
Ed


					



						

						

							
							
							
						

						

							
				

				

					
		


		


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PostPosted: Thu Dec 19, 2013 4:09 am 

				

				

					
		


		


			
				

			
				

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Solving this puzzle was interesting for what I spotted but then found that I didn’t need to use.

My solving path was very different from Mike's.  It was a bit more like Ed's, in the sense that I used the same 45 as his step 7, but I used it is a different way; not the clever way he used it.

Here is my walkthrough for TJK 40:  
Prelims

a) R1C67 = {69/78}
b) R1C89 = {49/58/67}, no 1,2,3
c) R2C12 = {19/28/37/46}, no 5
d) R1C67 = {18/27/36/45}, no 9
e) R1C89 = {19/28/37/46}, no 5
f) R45C8 = {14/23}
g) R56C2 = {18/27/36/45}, no 9
h) R8C12 = {18/27/36/45}, no 9
i) R8C34 = {19/28/37/46}, no 5
j) R8C89 = {69/78}
k) R9C12 = {49/58/67}, no 1,2,3
l) R9C34 = {18/27/36/45}, no 9
m) 10(3) cage at R2C4 = {127/136/145/235}, no 8,9
n) 11(3) cage at R4C4 = {128/137/146/236/245}, no 9
o) 22(3) cage at R5C6 = {589/679}
p) 26(4) cage at R6C3 = {2789/3689/4589/4679/5678}, no 1
q) 28(7) cage at R3C8 = {1234567}, no 8,9

1. R1C89 = {49/58} (cannot be {67} which clashes with R1C67), no 6,7
1a. Killer pair 8,9 in R1C67 and R1C89, locked for R1

2. 45 rule on NR1C4 2 innies R2C4 + R3C6 = 6 = {15/24}
2a. R2C67 = {18/27/36} (cannot be {45} which clashes with R2C4 + R3C6), no 4,5

3. 45 rule on NR1C4 2 outies R3C45 = 1 innie R3C6 + 4, IOU no 4 in R3C45

4. 45 rule on NR7C4 2 innies R7C4 + R8C6 = 12 = {39/48/57}, no 1,2,6

5. 45 rule on NR7C4 2 outies R7C56 = 1 innie R7C4 + 1, IOU no 1 in R7C56
5a. 13(3) cage at R7C5 = {238/247/256/346}, no 9, clean-up: no 3 in R7C4 (step 3)

6. 45 rule on R8 3 innies R8C567 = 11 = {128/137/146/236/245}, no 9
6a. 8 of {128} must be in R8C6 -> no 8 in R8C57

7. 45 rule on C89 3 innies R6C8 + R9C89 = 19 = {289/379/469/478/568}, no 1
7a. 2,3,5 of {289/379/568} must be in R9C89 (R9C89 cannot be {89/79/68} which clash with R8C89), no 2,3,5 in R6C8
7b. 45 rule on C89 4 outies R6789C7 = 14 = {1238/1247/1256/1346} (cannot be {2345} which clashes with R6C8 + R9C89 or, if preferred, 33(7) cage must contain1), no 9, 1 locked for C7, clean-up: no 8 in R2C6

8. 45 rule on C1234 1 outie R3C5 = 1 innie R1C4, no 4 in R1C4

9. 45 rule on C6789 1 innie R9C6 = 1 outie R7C5 + 1, no 1,2 in R9C6

[Just spotted 5 in R2 only in R2C345]
10. 45 rule on R2 3 innies R2C345 = 16 = {259/358/457}, no 1,6, clean-up: no 5 in R3C6 (step 2)

11. 10(3) cage at R2C4 = {127/145/235} (cannot be {136} because R2C4 only contains 2,4,5), no 6, clean-up: no 6 in R1C4 (step 8)

12. 45 rule on R12 3 outies R3C3 + R4C23 = 1 innie R2C4 + 18
12a. Min R2C4 = 2 -> min R3C3 + R4C23 = 20, no 1,2 in R3C3 + R4C23

13. Law of Leftovers (LoL) for C123 two outies R34C4 must exactly equal two innies R89C3, no 9 in R34C4 -> no 9 in R8C3, clean-up: no 1 in R8C4

14. LoL for C789 two outies R67C6 must exactly equal two innies R12C7, no 4,5 in R12C7 -> no 4,5 in R67C6

15. 45 rule on R1 3 innies R1C123 = 1 outie R2C5 + 2
15a. Min R1C123 = 6 -> min R2C5 = 4
15b. R2C345 (step 10) = {259/358/457}
15c. 3 of {358} must be in R2C3, no 8 in R2C3

16. 3 in NR1C4 only in 15(3) cage at R1C4 = {348/357} or in R2C67 = {36} -> 15(3) cage = {159/249/258/348/357} (cannot be {168/267/456}, locking-out cages), no 6
16a. 15(3) cage = {159/249/348/357} (cannot be {258} which clashes with R2C4 + R3C6)
[With hindsight, 15(3) cage at R1C4 = {159/249/348/357} (cannot be {168/267} which clash with R1C67, cannot be {258/456} which clash with R2C4 + R3C6), no 6 is simpler.]
16b. 4 of {249/348} must be in R1C5 -> no 2 in R1C5, no 4 in R2C5

17. R1C67 = {69/78}, R1C89 = {49/58} -> combined cage R1C6789 = {69}{58}/{78}{49}
17a. 15(3) cage at R1C4 (step 16a) = {159/348/357} (cannot be {249} which clashes with combined cage R1C6789), no 2, clean-up: no 2 in R3C5 (step 8)

18. 2 in R1 only in R1C123, locked for NR1C1, clean-up: no 8 in R2C12
18a. 36(7) cage = {1236789/1245789/2345679} must contain 7

19. 15(3) cage at R1C4 (step 17a) = {159/348/357}, R1C4 = R3C5 (step 8) -> R123C5 = {159/348/357}
19a. 17(3) cage at R4C5 = {179/269/278/368/467} (cannot be {359/458} which clash with R123C5), no 5
19b. 45 rule on C6789 3 outies R789C5 = 13 = {148/157/238/247/256/346} (cannot be {139} which clashes with R123C5), no 9

20. 2 in R1 only in R1C123
20a. Hidden killer pair 1,3 in R1C123 and 15(3) cage at R1C4 for R1, 15(3) cage contains one of 1,3 -> R1C123 must contain one of 1,3
20b. R1C123 = R2C5 + 2 (step 15)
20c. R2C5 = {5789} -> R1C123 = 7,9,10,11 = {124/126/127} (cannot be {234/235} which clash with 15(3) cage = {35}7/{34}8), cannot be {236} because R1C123+R2C5 = {236}9 clashes with R2C12, killer combo clash, other combinations don’t contain 2 or one of 1,3), no 3,5, 1 locked for R1 and NR1C1, clean-up: no 9 in R2C12, no 1 in R3C5 (step 8)
20d. 1 in NR1C4 only in R23C6, locked for C6
[I omitted to say that step 20c eliminates 9 from R2C5; however the next step does that.]

21. 15(3) cage at R1C4 (step 17a) = {348/357}, no 9, 3 locked for NR1C4, clean-up: no 6 in R2C67
21a. LoL for C789 two outies R67C6 must exactly equal two innies R12C7, no 3 in R12C7 -> no 3 in R7C6

22. R1C67 = {69} (hidden pair in NR1C4), locked for R1, clean-up: no 4 in R1C89
22a. Naked pair {58} in R1C89, locked for R1 and NR1C8, clean-up: no 2 in R2C89, no 5 in R3C5 (step 8)
22b. Caged X-Wing for 9 in R1C67 and 22(3) cage at R5C6, no other 9 in C67, clean-up: no 8 in R7C5 (step 9)

23. 5 in NR1C4 only in R2C45, locked for R2
23a. R2C345 (step 10) = {259/358} (cannot be {457} which clashes with R2C12), no 4,7, clean-up: no 2 in R3C6 (step 2)

24. Killer pair 4,7 in R1C123 and R2C12, locked for NR1C1
24a. 5,8 in NR1C1 only in R3C12 + R4C1, locked for 37(7) cage at R3C1, no 5,8 in R56C1 + R7C12
24b. 37(7) cage contains 5 so must contain 3 = {1345789/2345689}

25. 10(3) cage at R2C4 (step 11) = {127/235}, 2 locked for C4, clean-up: no 8 in R8C3, no 7 in R9C3
25a. R3C5 = {37} -> no 3,7 in R3C4

26. 1,2 in R1 only in 36(7) cage at R1C1 (step 18a) = {1236789/1245789}, 8 locked for NR3C3, clean-up: no 1 in R6C2
26a. 36(7) cage = {1236789/1245789}, CPE no 9 in R6C4

27. 1 in NR3C3 only in R34C4 + R5C123, CPE no 1 in R5C4

28. LoL for R789 three outies R6C129 must exactly equal three innies R7C567, no 9 in R7C567 -> no 9 in R6C1

[At this stage I spotted some interesting forcing chains starting from the combinations of R45C8, using caged X-wings for C89, in the case for R45C8 = {14} with one caged X-wing leading to a second caged X-wing within the same path of the forcing chain.
But then I spotted the much simpler …, which had been available since step 22b, but I’d been doing steps which were more obvious to me at the time]

29. 9 in NR1C8 only in R2C89 = {19}, locked for R2 and NR1C8 -> R2C3 = 3, placed for NR1C1, clean-up: no 7 in R2C12, no 8 in R2C7, no 4 in R5C8, no 7 in R8C4, no 6 in R9C4
29a. Naked pair {27} in R2C67, locked for R2 and NR1C4 -> R1C45 = [34], 4 placed for NR1C4, R2C45 = [58], R3C6 = 1, R3C4 = 2, placed for NR3C3, R3C5 = 3 (cage sum), placed for NR3C5, clean-up: no 7 in R6C2, no 7 in R8C3, no 4,6 in R9C3, no 4,5 in R9C6 (step 9)

30. R3C12 + R4C1 = {589} (hidden triple in NR1C1), locked for 37(7) cage at R3C1, no 9 in R5C1 + R7C12

31. Naked triple {127} in R1C123, locked for 36(7) cage at R1C1, no 7 in R3C3 + R4C23
31a. 36(7) cage at R1C1 (step 26a) = {1236789} (only remaining combination), no 4,5
31b. Naked triple {689} in R3C3 + R4C23, locked for NR3C3, clean-up: no 3 in R6C2

32.  11(3) cage at R4C4 = {146} (only remaining combination) -> R5C4 = 6, placed for NR3C5, R4C4 + R5C3 = {14}, locked for NR3C3, clean-up: no 5,8 in R6C2, no 4 in R8C3, no 7 in R9C6 (step 9)
32a. 3 in NR3C3 only in R5C12, locked for R5, clean-up: no 2 in R4C8

33. 17(3) cage at R4C5 (step 19a) = {179} (only remaining combination), locked for C5 and NR3C5, clean-up: no 8 in R9C6 (step 9)

34. 14(3) cage at R8C5 = {356} (only remaining combination) -> R9C6 = 3, R89C5 = {56}, locked for C5 and NR7C4 -> R7C5 = 2, placed for NR3C5, clean-up: no 4 in R8C4, no 4 in R9C4
34a. Killer pair 8,9 in R8C4 and R8C89, locked for R8, clean-up: no 1 in R8C12

35. R7C5 = 2 -> R78C6 = 11 = [74], 7 placed for NR4C7, 4 placed for NR7C4, R2C67 = [27], clean-up: no 5 in R8C12
35a. Killer pair 6,7 in R8C12 and R8C89, locked for R8 -> R89C5 = [56], clean-up: no 7 in R9C12

36. Naked pair {58} in R45C6, locked for C6 and NR3C5 -> R6C4 = 4, R4C4 = 1, R5C3 = 4

37. R9C4 = 7 (hidden single in C4), R9C3 = 2, R8C3 = 1, R8C4 = 9, clean-up: no 6 in R8C89
37a. Naked pair {78} in R8C89, locked for R8 and NR6C9, clean-up: no 2 in R8C12
37b. Naked pair {36} in R8C12, locked for R8 and NR6C1 -> R8C7 = 2, placed for NR6C9, R6C2 = 2, R5C2 = 7, R5C1 = 3

38. Naked pair {14} in R7C12, locked for R7 and NR6C1 -> R6C1 = 7, clean-up: no 9 in R9C12
38a. Naked pair {58} in R9C12, locked for R9 and NR6C1 -> R7C3 = 9

39. 22(3) cage at R5C6 = {589} (only remaining combination) -> R6C6 = 9, R5C67 = {58}, locked for R5
39a. 17(3) cage at R4C5 = [791]

40. Naked triple {356} in R6C9 + R7C89, locked for 28(7) cage at R3C8, no 3,6 in R3C89 + R4C9
40a. Naked pair {47} in R3C89, locked for NR1C8 -> R3C7 = 6, R4C8 = 3, R5C8 = 2

41. R3C67 = [16] = 7 -> R4C67 = 9 = [54]

and the rest is naked singles, without using the nonets.


					



						

						

							
							
							
						

						

							
				

				

					
		


		


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				Andrew
			
			
				

				

				
					
PostPosted: Thu Dec 19, 2013 5:26 pm 

				

				

					
		


		


			
				

			
				

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Since Mike and Ed have shown that eliminating 9 from R2C5 is one of the key steps.

Image

Here is a clearer version of how I did that:  
2. 45 rule on NR1C4 2 innies R2C4 + R3C6 = 6 = {15/24}

16. (simplified) 15(3) cage at R1C4 = {159/249/348/357} (cannot be {168/267} which clash with R1C67, cannot be {258/456} which clash with R2C4 + R3C6), no 6
16a. 8,9 only in R2C5 -> R2C5 = {35789}

17. R1C67 = {69/78}, R1C89 = {49/58} -> combined cage R1C6789 = {69}{58}/{78}{49}
17a. 15(3) cage at R1C4 (step 16) = {159/348/357} (cannot be {249} which clashes with combined cage R1C6789), no 2
17b. 2 in R1 only in R1C123
(2 locked for NR1C1, but this isn't necessary for the key step below)

15 and 20 combined
45 rule on R1 3 innies R1C123 = 1 outie R2C5 + 2
Min R1C123 = 6 -> no 3 in R2C5
Hidden killer pair 1,3 in R1C123 and 15(3) cage for R1, 15(3) cage contains one of 1,3 -> R1C123 must contain one of 1,3 and 2 (cannot be {123} because min R1C123 = 7)
R1C123 + R2C5 cannot be 11 + 9 = {236} + 9 which clashes with R2C12

My step 20c (in the full walkthrough above) did more analysis than that, also locking 1 in R1C123.


					



						

						

							
							
							
						

						

							
				

				

					
		


		


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