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 Post subject: Texas Jigsaw Killer 39
PostPosted: Fri Jan 01, 2010 12:16 am 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
Many thanks to Andrew for starting this thread for me. Happy New Year to everyone!

As has become the norm with these puzzles, Texas Jigsaw Killer 39 (TJK39) has a really short way to crack it. But I(Ed) know that SudokuSolver can't do at least one of the tricks I used so there must be at least one other way to solve this one.

Texas Jigsaw Killer 39
Thanks to Børge for supplying the images. Love his images now having the row and column numbering!

Image

NOTE: images without row and column numbers can be found here
extra images:
Ruud TJK#33 inspired:
Image

Image

Cages with cells in 3 jigsaw nonets: lilac
Cages with cells in 2 jigsaw nonets: green and yellow
Cages with cells in 1 jigsaw nonet: red, blue and grey
code: select, copy and paste into any SumoCue compatible solver:
SumoCueV1=10J0=17J0+1J0=22J1+3J1+3J1=18J2+6J2=7J2=12J0+0J0+1J0+1J0=14J1+6J2+6J2+8J2=17J2+9J3+9J0=23J0+20J1+13J1=18J1+23J2+17J2+17J4=22J3=18J3+20J5+20J5+13J1+23J5+23J5=19J4=26J4+27J3+28J3+28J3=7J5+39J1+39J5+34J4+34J4+35J4+27J3=9J3+46J3=7J5=12J5=12J5=5J4+51J4+35J4+27J6=19J6=11J6+48J6+49J7+50J8=17J8=12J8+35J8+55J6+55J6+56J6=16J7+66J7+66J7+60J8+61J8+61J8=10J6+72J6+56J7=16J7+75J7+75J7+60J7=9J8+79J8
solution:
415 976 823
263 857 149
197 384 562

348 512 796
659 421 378
872 693 415

524 139 687
986 745 231
731 268 954
SSscore: 1.66

Cheers
Ed


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PostPosted: Fri Jan 22, 2010 6:10 pm 
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Joined: Sat Jan 17, 2009 8:30 am
Posts: 118
Location: france
Nice puzzle from our best jigsaw Killer. John Kramer :twisted: would not have made a more frightening one : thanks Ed !

Walkthrough TJK 39


Hidden Text:
Jigsaw nonets :

prelims (sudocue)

n1 at r1c1
n2 at r1c4
n3 at r1c7
n4 at r3c1
n5 at r4c3
n6 at r3c9
n7 at r7c1
n8 at r7c5
n9 at r7c6

1)a) Innies-outies for n1 : r3c3=6+r3c1 = r3c3=(789), r3c1=(123)
b) Innies-outies for n4 : r7c1=4+r3c1 : r7c1=(567)
c) Innies-outies for n6 : r7c9=5+r3c9 : r3c9=(1234), r7c9=(6789)
d) Innies-outies for n3 : r3c7=r3c9+3 : r3c7=(4567)

2)a) Innies for r9 : r9c37=h10(2).
b) Using cage h10(2), innie for n8 : r7c5=3, cage sum r6c5=9

3)a) 1 locked for n6 at r3c9+5(2)={14/23} : r3c9 <> 4 !
b) step 1)d) => r3c7<>7
c) step 1)c) => r7c9<>9

4)a) 3 locked for n2 and r3 at r3c46.
b) r3c1<>3 =>(step 1)a) r3c3<>9 and =>(step 1)b) r7c1<>7
c) r3c9<>3 => (step 1)d) r3c7 <> 6 and => (step 1)c) r7c9 <> 8.

5)a) Naked pair {12} at r3c19 locked for r3 : r3c1+r3c9=3.
b) Steps 1)b+1)c : r7c1+r7c9=9+r3c1+r3c9=9+3=12 : only combination : r7c19=[57]
c) Step 1)b => r3c1= 1, r3c9=2.
d) Step 1)a => r3c3=7, step 1)d) => r3c7=5 : it's cracked !

6)a) Last combination : 17(3)n3={269}, {69} locked for n3.
b) 7(2)n3={34} locked for n3.
c) LOL for n123 : r45c5=r3c19={12} locked for c5 and n2.
d) Innie for n2 : r3c4+r3c6+r5c5=h9(2) = {234}, since it must contain 3(step 4)a), and {135} blocked by r3c17=[15] . r5c5=2, r3c46={34} locked for r3.
e) r4c5=1, r23c5=[58/76].

7)a) 5(2)n6={14} locked for n6 and r6.
b) 26(4)={5678} (last combo), {568} locked for c9 and n6.
c) 12(2) at r6c6 : [39/84]
d) Killer pair {34} locked for c6 at r3c6+12(2) : r5c6=1, r5c4=4

8)a) Innies-outies for n9 : r9c7=r7c6=(49).
b) r9c37=h10(2)=[19/64]
c) Innies-outies for n7 : r9c3=r7c4=(16)
d) 7(2) at r6c4 : [61]
The rest is easy


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PostPosted: Thu Jan 28, 2010 3:59 am 
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Grand Master
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
manu wrote:
John Kramer :twisted: would not have made a more frightening one : !
Thanks!

My Apprentice passed the test. :twisted: You SAW 4 from r3c9 in a really neat way! :) Worth a pic.

Image

manu's walkthrough
Quote:
3)a) 1 locked for n6 at r3c9+5(2)={14/23} : r3c9 <> 4 !
This looks similar to one aspect of the "Locking/Blocking Cages" technique (post in techniques forum coming) but involves just one cage and one cell. So, is "Locking Cage-cell". Nice.

I couldn't get rid of that 4 but used "45" on disjoint nonets 1 & 3 to lock 2 in r3c19. This led to the breakthrough in r7. Since 2 locked at r3c19 -> r7c1 = 6 or r7c9 = 7 but not both.

Cheers
"Ed"


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PostPosted: Thu Dec 19, 2013 12:46 am 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
I really enjoyed this one, especially after the long struggle with TJK 38.

My solving path was very similar to manu's, including the step shown in Ed's pic. It's a bit longer because I didn't spot manu's step 2b.

Here is my walkthrough for TJK 39:
Prelims

a) 10(2) cage at R1C1 = {19/28/37/46}, no 5
b) 7(2) cage at R1C9 = {16/25/34}, no 7,8,9
c) R6C23 = {18/27/36/45}, no 9
d) R67C4 = {16/25/34}, no 7,8,9
e) R67C5 = {39/48/57}, no 1,2,6
f) R67C6 = {39/48/57}, no 1,2,6
g) R6C78 = {14/23}
h) R9C12 = {19/28/37/46}, no 5
i) R9C89 = {18/27/36/45}, no 9
j) 22(3) cage at R1C4 = {589/679}
k) 19(3) cage at R4C8 = {289/379/469/478/568}, no 1
l) 7(3) cage at R5C4 = {124}
m) 19(3) cage at R7C2 = {289/379/469/478/568}, no 1
n) 11(3) cage at R7C3 = {128/137/146/236/245}, no 9
o) 26(4) cage at R4C9 = {2789/3689/4589/4679/5678}, no 1

Steps resulting from Prelims
1a. 22(3) cage at R1C4 = {589/679}, 9 locked for R1 and NR1C4, clean-up: no 1 in R2C2
1b. 7(3) cage at R5C4 = {124}, locked for R5

2. 45 rule on NR1C4 3 innies R3C46 + R5C5 = 9 = {126/135/234}, no 7,8

3. 45 rule on NR3C1 1 outie R7C1 = 1 innie R3C1 + 4, R3C1 = {12345}, R7C1 = {56789}

4. 45 rule on NR3C9 1 outie R7C9 = 1 innie R3C9 + 5, R3C1 = {1234}, R7C9 = {6789}

5. 45 rule on C12 1 outie R5C3 = 2 innies R16C2 + 1
5a. Max R16C2 = 8, no 8 in R16C2, clean-up: no 1 in R6C3
5b. Min R16C2 = 3 -> no 3 in R5C3

6. 1 in NR3C9 only in R3C9 or R6C78 = {14} -> no 4 in R3C9 (locking-out cages), clean-up: no 9 in R7C9 (step 4)

7. 45 rule on NR1C7 1 innie R3C7 = 1 outie R3C9 + 3, R3C9 = {123} -> R3C7 = {456}

8. 45 rule on NR1C7 3 innies R2C9 + R3C78 = 20 = {479/569/578} (cannot be {389} because R3C7 only contains 4,5,6), no 1,2,3
8a. 4 of {479} must be in R3C7 -> no 4 in R2C9 + R3C8

9. 45 rule on NR1C1 1 innie R3C3 = 1 outie R3C1 + 6, R3C1 = {123}, R3C3 = {789}, clean-up: R7C1 = {567} (step 3)

10. Law of Leftovers (LoL) for R123 two outies R45C5 must exactly equal two innies R3C19, R3C19 = {123} -> R4C5 = {123}, R5C5 = {12}
10a. Naked triple {124} in 7(3) cage at R5C4, 4 locked for NR4C4, clean-up: no 3 in R7C4, no 8 in R7C5, no 8 in R7C6

11. 14(3) cage at R2C5 = {158/167/248} (cannot be {257/356} which clash with 22(3) cage at R1C4, cannot be {347} which clashes with R67C5), no 3
11a. Naked pair {12} in R45C5, locked for C5 and NR1C4
11b. 3 in NR1C4 only in R3C46, locked for R3, clean-up: no 9 in R3C3 (step 9), no 6 in R3C7 (step 7), no 7 in R7C1 (step 3), no 8 in R7C9 (step 4)
11c. Naked pair {12} in R3C19, locked for R3
11d. Killer pair 1,2 in R3C9 and R6C78, locked for NR3C9

12. R2C9 + R3C78 (step 8) = {479/569/578}
12a. R3C7 = {45} -> no 5 in R2C9 + R3C8

13. 12(3) cage at R2C1 = {129/138/147/156/237/246} (cannot be {345} because R3C1 only contains 1,2)
13a. 12(3) cage = {129/138/156/237/246} (cannot be {147} which clashes with R3C13 = [17], step 9, killer combo blocker using NR1C1)
13b. 7,8,9 of {129/138/237} must be in R3C2 -> no 7,8,9 in R2C1

14. 45 rule on NR3C1 3(2+1) outies R2C1 + R3C2 + R7C1 = 16
14a. Max R26C1 = 11 -> min R3C2 = 5
14b. R2C1 + R3C2 + R7C1 cannot be [565] -> no 5 in R2C1

15. 45 rule on NR3C9 3(2+1) outies R2C9 + R3C8 + R7C9 = 22
15a. R27C9 cannot total 14 (cannot be [77] and [86] clashes with R3C8 = 8) -> no 8 in R3C8

16. 26(4) cage at R4C9 = {4679/5678} (cannot be {3689} which clashes with R6C78 = {23} because R37C9 = [16], step 4, cannot be {4589} because R7C9 only contains 6,7), no 3, 6,7 locked for C9, clean-up: no 1 in R2C8, no 2,3 in R9C8
[Alternatively consider combinations for R6C78 = {14/23}
R6C78 = {14} => R3C9 = 2, R7C9 = 7 (step 4) => 26(4) cage cannot be {3689}
or R6C78 = {23}, locked for NR3C9 …]

17. 45 rule on R9 2 innies R9C37 = 10 = [19]/{28/37/46}, no 5, no 1 in R9C7

[I ought to have spotted this after step 11; then analysing the 26(4) cage at R4C9 would have been simpler than in step 16 …]
18. R3C19 = [12] (cannot be [21] because R7C19 cannot be [66] using steps 3 and 4), 1 placed for NR3C1, 2 placed for NR3C9, R3C3 = 7 (step 9), placed for NR1C1, R3C7 = 5 (step 7), placed for NR1C7, R7C1 = 5 (step 3), placed for NR7C1, R7C9 = 7 (step 4), placed for NR6C6
[More formally R7C1 = R3C1 + 4 (step 3), R7C9 = R3C9 + 5 (step 4) -> R3C1 cannot be 1 more than R3C9 because R7C19 cannot be equal …]
[manu found R3C19 = {12} = 3 -> R7C19 = 12 (using steps 3 and 4) = [57] … which is a better way to get this result.]
18a. Clean-up: no 3 in R1C1, no 3,9 in R2C2, no 2 in R2C8, no 2 in R6C2, no 8 in R6C3, no 2 in R6C4, no 5,7 in R6C5, no 5,7 in R6C6, no 3 in R6C78, no 9 in R9C2
[Cracked, the rest is straightforward.]

19. Naked pair {14} in R6C78, locked for R6 and NR3C9, clean-up: no 5 in R6C23, no 6 in R7C4
19a. 2 in R6 only in R6C13, locked for NR3C1

20. R7C9 = 7 -> 26(4) cage at R4C9 = {5678} (only remaining combination), 5,6,8 locked for C9 and NR3C9 -> R2C9 = 9, R3C8 = 6, placed for NR1C7, clean-up: no 1 in R1C9, no 1,4 in R9C8, no 3 in R9C9
20a. Naked pair {34} in 7(2) cage at R1C9, locked for NR1C7
20b. 1 in C9 only in R89C9, locked for NR7C6

21. Naked pair {34} in R3C46, locked for NR1C4 -> R3C5 = 8, placed for NR1C4, R3C2 = 9, R2C1 = 2 (cage sum), placed for NR1C1, clean-up: no 8 in 10(2) cage at R1C1, no 5 in 22(3) cage at R1C4, no 4 in R7C5, no 8 in R9C2
21a. Naked pair {46} in 10(2) cage at R1C1, locked for NR1C1

22. Naked triple {679} in 22(3) cage at R1C4, locked for R1 and NR1C4 -> R2C5 = 5, R4C5 = 1 (cage sum), R5C5 = 2
22a. Naked pair {39} in R67C5, locked for C5
22b. 4 in C5 only in R89C5, locked for NR7C5

23. 45 rule on NR7C1 1 remaining innie R7C4 = 1 outie R9C3 = {12}, clean-up: no 3 in R6C4

24. 45 rule on NR7C6 1 remaining innie R7C6 = 1 outie R9C7 = {39}, clean-up: no 8 in R6C6
24a. Naked pair {39} in R7C56, locked for R7
24b. Naked pair {39} in R67C6, locked for C6
24c. Naked pair {39} in R6C56, locked for R6 and NR4C3, clean-up: no 6 in R6C23
24d. R6C23 = [72], 7 placed for NR3C1, R9C3 = 1, R7C4 = 1 (step 23), R6C4 = 6, R5C46 = [41], R3C46 = [34], R2C34 = [38], R1C23 = [15], 7(2) cage at R1C9 = [34], 10(2) cage at R1C1 = [46], R2C67 = [71], R6C1 = 8, placed for NR3C1, R6C9 = 5, 22(3) cage at R1C4 = [976], R6C78 = [41], R9C9 = 4, R9C8 = 5, R9C5 = 6, R8C5 = 4, clean-up: no 3,9 in R9C1, no 2 in R9C2

25. R9C3 = 1 -> R78C3 = 10 = [46] -> R4C3 = 8, R4C4 = 5 (cage sum)

26. R9C7 = 9 (hidden single in R9), R7C6 = 9 (step 24), placed for NR7C6
26a. R8C9 = 1 -> R78C8 = 11 = [83], both placed for NR7C6

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


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