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 Post subject: HS 22 Sixes X 2
PostPosted: Mon Aug 03, 2015 6:15 pm 
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Grand Master
Grand Master

Joined: Wed Apr 30, 2008 9:45 pm
Posts: 693
Location: Saudi Arabia
HS 22 sixes X 2

When I made the sixes X assassin I used mid range cage totals hence the need for so many and the splitting in half. So over the weekend I decided to try forced sixes (39 etc.). I managed to get it unique with 8 cages and then managed to solve it reasonably. SS gives it 1.75 and JS uses 7 medium fishes.
This is, if I remember rightly, my minimum number of cages for a plain killer.

You can remove one cage and it is still solvable but not HS - I'll post it in the week if you wish.



Image

JS Code:
3x3:d:k:9730:9730:9730:9730:9:6150:9987:9987:9987:9730:10:11:12:13:6150:6150:6150:9987:9730:14:15:16:17:6150:18:19:9987:5895:5895:5895:5895:5895:20:6150:21:9987:5895:22:23:24:25:5640:26:8453:8453:9985:27:28:29:5640:30:31:8453:8453:9985:32:33:5640:5640:5640:34:8453:9220:9985:35:36:37:5640:38:39:8453:9220:9985:9985:9985:40:41:9220:9220:9220:9220:

Solution:
396721845
571864239
824395617
183547926
265983471
947216583
639472158
712658394
458139762


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 Post subject: Re: HS 22 Sixes X 2
PostPosted: Mon Aug 17, 2015 1:16 am 
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Grand Master
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 280
Location: California, out of London
Thanks HATMAN. I love puzzles where the possibility to effectively use non-standard techniques exists. Here is the start to my WT...

Hidden Text:
1. HP (12) in r1 -> r1c56 = {12}
HP (12) in c1 -> r45c1 = {12}
-> 3 in r1 in r1c1234
Also 3 in c1 in r123c1
-> r1c1 = 3

2. 36(6)n9 has exactly one of (123)
-> whichever two of (123) are not in 36(6)n9 must go in c9 in r56c9 and in r9 in r9c45
Since 22(6)n58 = {123457}
-> whichever of (123) is in r9c5 also goes in r5c6 and in n2 in r23c4
-> Since r1c56 = {12} -> r9c5 = r5c6 = 3
-> 3 not in 36(6)
-> 36(6)n9 = {(24|15)6789}
Since whichever of (12) is in 36(6) must go in c9 in r789c9 and also in r9 in r9c6789
-> r9c9 from (12)

3! Consider the edge cells in r1,r9,c1,c9. There are 32 cells.
Since puzzle is an 'X' -> the four corners are all different
-> the edge cells consist of 3 instances of the 4 corner numbers and 4 instances of the other 5 numbers.

Since the edge cells not in edge cages are all from (123) and since each cage can only contain each number once, and since each number must be in the edge cells at least three times this implies:

a) The only place for a third 4 in the edge cells is in 36(6)n9
-> 36(6)n9 = {246789}
-> r9c9 = 2
Also -> 4 in a corner

b) There is only room for three 5's in the edge cells
-> 5 in a corner
-> r1c9,r9c1 = {45}

4. r9c5 = r5c6 = 3
-> r6c9 = 3
-> r8c7 = 3
-> (Since 24(6)n2 contains a 3) -> r2c8 = 3
-> r3c4 = 3
-> Also r7c2 = 3
-> r4c3 = 3

Also r5c9 = 1, r9c4 = 1
-> r6c5 = 1
-> r1c6 = 1
-> r3c8 = 1
-> r7c7 = 1
-> r2c3 = 1
-> r8c2 = 1
Also R4c1 = 1

Also r1c5 = 2, r5c1 = 2
Also r5678c8 = {5789}
-> HS 2 in c8 -> r4c8 = 2
Also 2 in 24(6) -> r2c6 = 2
-> r3c2 = 2
Also r6c4 = 2
-> r7c6 = 2
-> r8c3 = 2

5. HS 5 in n9 -> 5 in r78c8
22(6)n58 -> r7c45+r8c5 = {457}
-> r8c46+r9c6 = {689}

6! Consider r4c6
Innies r4 -> r4c679 = +22
Since 5 already in D/ -> r4c6 = Min 6.
Whatever goes in r4c6 must go in 36(6)n9 in n9 (not in D\ in n9)
-> must go in D\ in n1 in r2c2 or r3c3
-> must go in 36(6)n1 in r1c4
-> must go in c5 in r789c5
But since (689) in n8 are in r8c46+r9c6
-> r4c6 = 7
-> r4c79 = {69}
-> r78c8 = {59} (and r56c8 = {78})
-> r9c6 = 9 and r8c46 = {68}
Also 7 in D\ in r2c2 or r3c3
-> r1c4 = 7
-> 7 in 39(6)n3 in r234c9
-> NS r8c9 = 4
-> NP r19c8 = [46]
-> NP r1c9,r9c1 = [54]

etc.


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 Post subject: Re: HS 22 Sixes X 2
PostPosted: Sun Oct 11, 2015 6:16 pm 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
Location: Lethbridge, Alberta, Canada
Thanks HATMAN for another interesting puzzle! As wellbeback said, it's a puzzle which requires non-standard techniques.

I found an early (possibly) HS step but it took me quite a few steps before I spotted the important HS step, a technique which I haven't used for a long time and has only appeared a few times on this forum.

Here is my walkthrough for Human Solvable Sixes X 2:
Prelims

a) 38(6) cage at R1C1 = {356789}, no 1,2,4
b) 39(6) cage at R1C7 = {456789}, no 1,2,3
c) 23(6) cage at R4C1 = {123458/123467}, no 9
d) 22(6) cage at R5C6 = {123457}, no 6,8,9
e) 39(6) cage at R6C1 = {456789}, no 1,2,3

1. R1C56 = {12} (hidden pair in R1), locked for N2
1a. 3 in R1 only in R1C1234, locked for 38(6) cage at R1C1
1b. 4 in R1 only in R1C789, locked for N3 and 39(6) cage at R1C7

2. R45C1 = {12} (hidden pair in C1), locked for N4 and 23(6) cage at R4C1
2a. 23(6) cage at R4C1 = {123458/123467}, 3,4 locked for R4
2b. R1C1 = 3 (hidden single in C1), placed for D\
2c. 4 in C1 only in R6789C1, locked for 39(6) cage at R6C1

3. 36(6) cage at R7C9 can only contain one of 1,2,3, R9 and C9 each require 1,2,3 -> R56C9 = {123}, R9C45 = {123} and R9C9 = {12}
3a. Naked triple {123} in R569C9, locked for C9, 3 also locked for N6 and 33(6) cage at R5C8
3b. Naked triple {123} in R9C459, locked for R9, 3 also locked for N8
3c. R8C7 = 3 (hidden single in N9)

4. 22(6) cage at R5C6 = {123457}, 3 locked for N5
4a. R5C6 + R6C5 = {13/23} (otherwise 22(6) cage would clash with R9C45) -> 4,5,7 in R7C456 + R8C5, locked for N8
4b. Naked triple {123} in R169C5, locked for C5
4c. Naked triple {123} in R5C169, locked for R5
4d. R8C46 + R9C6 = {689} (hidden triple in N8), CPE no 6,8,9 in R8C9
4e. 3 in R4 only in R4C23, locked for N4

5. R56C9 = {123} -> 33(6) cage at R5C8 = {135789/234789/235689}, 8,9 locked for C8
5a. R56C9 = {123} -> no 1,2 in R678C8
5b. R23C8 = {123}, R4C8 = {12} (hidden triple in C8)
5c. Naked triple R4C8 + R56C9 = {123}, locked for N6
5d. Naked pair {12} in R4C18, locked for N4

6. R7C7 + R9C9 = {12} (hidden pair in N9), locked for D\
6a. R2C3 + R3C2 = {12} (hidden pair in N1)
6b. R8C23 = {12} (hidden pair in R8), locked for N7
6c. 4 in N1 only in R2C2 + R3C3, locked for D\

7. Hidden killer pair 1,2 in R5C6 + R6C5 and R6C4 for N5, R5C6 + R6C5 contain one of 1,2 -> R6C4 = {12}
7a. Naked pair 1,2 in R6C4 + R8C2, locked for D/ -> R2C8 = 3, placed for D/
7b. R3C4 = 3 (hidden single in N2)
7c. R7C2 = 3 (hidden single in N7)
7d. R4C3 = 3 (hidden single in N4)
7e. R9C5 = 3 (hidden single in N8)
7f. R5C6 = 3 (hidden single in N5)
7g. Naked pair {12} in R6C45, locked for R6 -> R6C9 = 3
7h. R2C7 + R3C8 = {12} (hidden pair in N3)
7i. Naked pair {12} in R69C4, locked for C4
7j. R7C6 + R9C4 = {12} (hidden pair in N8)

8. 4 in C6 only in R23C6, locked for N2
8a. 24(6) cage at R1C6 = {123459/123468}, no 7
8b. 7 in C6 only in R46C6, locked for N5
8c. Consider combinations for 24(6) cage
24(6) cage = {123459}, CPE no 5,9 in R4C6
or 24(6) cage = {123468} => R46C6 = {57} (hidden pair in C6)
-> no 9 in R4C6
8d. 9 in R4 only in R4C79, locked for N6, CPE no 9 in R1C7

9. 33(6) cage at R5C8 (step 5) = {135789/234789/235689}, 9 locked for N9
9a. 36(6) cage at R7C9 = {156789/246789} -> R9C6 = 9, 6,7,8 locked for N9
9b. 33(6) cage contains 8, locked for N6
9c. 39(6) cage at R6C1 = {456789}, 9 locked for C1
9d. 38(6) cage at R1C1 = {356789}, 9 locked for R1
9e. Naked pair {68} in R8C46, locked for R8
9f. 39(6) cage at R1C7 = {456789}, 8 locked for N3
9g. 9 in N5 only in R5C45, locked for R5

10. 36(6) cage at R7C9 (step 9a) = {156789/246789}
10a. 33(6) cage at R5C8 (step 5) = {135789/234789} (cannot be {235689} which clashes with 36(6) cage = {156789} -> R56C8 = {78}, locked for C8 and N6
10b. Killer pair 5,6 in 23(6) cage at R4C1 and R4C79, locked for R4
10c. 39(6) cage at R1C7 = {456789}, 7 locked for N3

11. 24(6) cage at R1C6 (step 8a) = {123459/123468}
11a. 9 of {123459} must be in R4C7 –> no 5 in R4C7
11b. 6 of {123468} must be in R4C7 -> no 6 in R23C6

12. R19C7 = {78} (hidden pair in C7)
12a. R1C9 + R9C1 = {456} (cannot contain 7 or 8 because of clashes around the outer ring), no 7,8

13. 9 on D/ only in R3C7 + R5C5 + R7C3, CPE no 9 in R3C3

[Step 12a could have been omitted if I’d remembered this type of step sooner; it’s a long time since I’ve used it. Step 14 could have been done immediately after step 3 if I’d spotted it then.]
14. Total around outer ring = 180 with corner cells counting double. Total of the 6-cell cages plus 1,2 pairs in R1 and C1 and placed 3s in R9 and C9 = 164. This leaves 16 for the four corner cells, R5C9 and R9C4. R1C1 = 3 and min R1C9 + R9C1 = 9 -> max R5C9 + R9C49 = 4 -> R5C9 = 1, R9C49 = [12], R1C9 + R9C1 = 9 = {45}, locked for D/

15. R56C7 = {45} (hidden pair in C7), locked for N6
15a. Naked pair {69} in R4C79, locked for R4
15b. 23(6) cage at R4C1 = {123458} (only remaining combination), 8 locked for R4 -> R4C6 = 7, placed for D/
15c. R2C2 + R3C3 = {47} (hidden pair on D\), locked for N1
15d. 38(6) cage at R1C1 = {356789} -> R1C4 = 7, R19C7 = [87]

16. 33(6) cage at R5C8 (step 10a) = {135789} (only remaining combination) -> R78C8 = {59}, locked for C8 and N9 -> R8C9 = 4, R19C8 = [46], R1C9 = 5, placed for D/
16a. Naked pair {58} in R9C23, locked for N7 and 39(6) cage at R6C1
16b. R5C5 = 8 (hidden single on D/), R4C4 = 5, placed for D\, R6C6 = 6, R8C6 = 8
16c. Naked pair {45} in R23C6, locked for N2, R4C7 = 9 (cage sum)

and the rest is naked singles.

I won't give my walkthrough a rating, although if I did rate the important HS step it would be a lot lower than the SS score.


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 Post subject: Re: HS 22 Sixes X 2
PostPosted: Wed Oct 14, 2015 11:35 am 
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Grand Master
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Joined: Wed Apr 30, 2008 9:45 pm
Posts: 693
Location: Saudi Arabia
My walkthrough (just the two interesting bits):

R1: hidden pair at c56 r1c56 = {12}
C1: hidden pair at r45 r45c1 = {12}
r1c1=3 row and column
Overlaps on r1, r9, c1, c9 total = 3+3+38+39+39+36 = 158
180-158 = 22
r1c1+r1c9+r9c1+r9c9+r5c9+r6c9+r9c4+r9c5 = 22
r1c9+r9c1+r9c9+r5c9+r6c9+r9c4+r9c5= 19
the first 4 must be different as on diagonals

min r1c9,r9c1 = 45
max r9c9+r5c9+r6c9+r9c4+r9c5= 10
r9c9 = 1/2
if 1 min r56c9 and r9c45 are both 23 total 11
so r9c9 =2 and r56c9 = {13} and r9c45 = {13}

Lots of bits ending with:
4 locked in cage at r23c6
r4c679 = 22
R3c7 = 4c9
r3c7r4c67 = 22
r4c6 =6789
r4c7=569
r4c9=r3c7=689
if r3c7 =8 -> r4c7=5 r4c6=9 but r4c7=5-> r23c6={49} fail
r3c7=r4c9 = 69

Straightforward from here


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