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HATMAN
 Post subject: Human Solvable 8 Boxes
PostPosted: Sat Oct 02, 2010 7:39 pm 
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HS 8 Boxes

I've had this human solvable for a while - however I have versions with different levels of difficulty and could not decide which to publish. This is the second hardest.

JSudoku solves it with trouble and fishes.

Interesting box interactions - as they say enjoy.

Image
Original link failed; image link replaced by moderator.


JS killer code:
3x3::k:13:14:3589:3589:6406:6406:3586:3586:15:16:17:3589:3589:6406:6406:3586:3586:18:6660:6660:4620:4620:19:20:5385:5385:21:6660:6660:4620:4620:22:23:5385:5385:24:3847:3847:5640:5640:6657:6657:25:26:27:3847:3847:5640:5640:6657:6657:28:29:30:6659:6659:31:32:4107:4107:4618:4618:33:6659:6659:34:35:4107:4107:4618:4618:36:37:38:39:40:41:42:43:44:45:

JS twin killer code:
3x3::k:10:11:12:13:14:15:16:17:18:19:20:21:22:23:24:25:6657:6657:26:27:28:29:30:31:32:6657:6657:33:34:35:6406:6406:36:37:4359:4359:38:39:40:6406:6406:41:42:4359:4359:43:6404:6404:44:45:6658:6658:3592:3592:46:6404:6404:47:48:6658:6658:3592:3592:49:3593:3593:5893:5893:6659:6659:50:51:52:3593:3593:5893:5893:6659:6659:53:54:
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Ed
PostPosted: Thu Oct 07, 2010 6:56 am 
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HATMAN wrote:
Interesting box interactions
This is still a HUS puzzle for me. Here are examples of the interesting interactions I found of which there are other eliminations but nothing to crack the puzzle. Hope someone else can see something. If we can't crack it, perhaps HATMAN can give us a hint after the weekend.

HS8 Boxes
prelims
i. 14(4) at r1c3, r1c6, r6c8 & r8c2: no 9
ii. 26(4) at r2c8, r3c1, r5c5, r6c6, r7c1 & r8c6: no 1

1. "45" on n5: 25(4)+26(4): 1 overlap cell r5c5 - 6 = 2 innies r4c6 + r6c4
1a. -> r5c5 = 9
1b. r4c6+r6c4 = {12}: both locked for n5
1c. 22(4)r5c3 must have 1/2 for r6c4 -> no 1,2 elsewhere in 22(4)
1d. deleted wrong step: thanks Andrew

2. "45" on n78: 26(4)+16(4) -> 10 innies = 48
2a. "45" on c34: 14(4)+18(4)+22(4) -> 6 innies = 36
2b. step 2 - step 2a -> 4 outies r9c1256 = 12 = {1236/1245}
2c. 1 and 2 both locked for r9

3. "45" on r56: 6 innies r56c789 = 27
3a. -> "45" on n6: 3 remaining innies r4c789 = 18 (no eliminations yet)
3b. "45" on n3: 5 innies = 31
3c. adding steps 3a + 3b = 49
3d. then subtracting the 21(4)r3c7 -> 4 remaining outies r1234c9 = 28
3e. h28(4) = {4789/5689}(no 1,2,3): 8 and 9 locked for c9

4. From step 3e, 5 remaining innies c9 r56789c9 = 17
4a. "45" on n9: 5 innies = 27
4b. -> from step 4 and 4a, r9c78 - 10 = r56c9
4c. -> min. r9c78 = 13 (no 3)
4d. and max. r56c9 = 7 (no 7)

5. deleted: thanks to Andrew for picking up a flaw and showing how to rework the next couple.

6. 9 in c6 is in r123c6 or 16(4)n8
6a. however, can't be in 16(4): like this.
6b. 26(4) at r5c5 must have 9 = 9{368/458/467} = at least one of 4/6 in c56
6c. the only combo in 25(4)n2 without 9 is {4678} = both 4&6
6d. when 16(4)n8 has 9 it must be {1249} = one of 4/6
6e. ie, all 4 & 6 taken for c56
6f. but this forces 16(4)n8 to clash with h12(4)r9 since r9c56 can only be {35}
6g. -> no 9 in r78c6

7. 9 in c6 only in r123c6: locked for n2

8. "45" on n7: 5 innies = 19
8a. 3 of those cells r8c3 + r9c23 overlap the 14(4)n7 -> r7c3 + r9c1 - 5 = r8c2
8b. -> no 5 in r7c3 nor r9c1 (IOU)
[Andrew pointed out a much simpler way to do the previous step;
“45” on n7 26(4)+14(4) 1 overlap cell r8c2 + 5 = 2 innies r7c3 + r9c1 -> no 5 in ... (IOU).
Nice!]

Perhaps the same sort of IOU step is available when 3 cages overlap. Just can't see it!

Cheers
Ed
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HATMAN
PostPosted: Thu Oct 07, 2010 1:30 pm 
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I'll try and think of a cryptic clue description over the weekend and post on Monday/Tuesday.

I'm happy that the HS part was not seen immediately.
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simon_blow_snow
PostPosted: Thu Oct 07, 2010 3:31 pm 
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Thanks for the nice puzzle. Not too easy, not too hard! ;)

12 steps walkthrough:
1:
Overlap 25(4)+26(4) N5: R5C5-D/46=6
--> R5C5=9, D/46={12} [N5]

2:
Overlap innies R12+C12: R12C9-R9C12=14
--> R12C9=17={89} [C9,N3], R9C12=3={12} [R9,N7]

3:
Innies N7: R789C3=16
Innies C34: R789C4=20 <>{12}
Innies N8: R9C56=9={36/45}

4:
26(4) R2C8 <>{123} (or max sum=3+6+7+9=25<26)
Also must include {67} (or max sum=4+5+7+9=25<26)
--> 26(4) R2C8={49/58}{67} [{67} N3]

5:
14(4) R8C2 <>{789} (or min sum=1+3+4+7=15>14)
Also must include {34} (or min sum=1+3+5+6=15>14)
--> 14(4) R8C2={16/25}{34} [{34} N7]

6:
Overlap innies R89+C89: R1C89-R89C1=6
Max R1C89=13 (cannot be 5+9=14 for 26(4) R2C8)
Min R89C1=7 (cannot be 1+5=6 for 14(4) R8C2)
--> R1C89=13=[49/58], R89C1=7=[52/61]
Hidden triple N3: R123C7={123} [C7]
Hidden triple N7: R8C123={789} [R7]

7:
26(4) R6C6: min R6C67+R7C6=26-6=20 <>{12}
26(4) R8C6: min R8C67+R9C7=26-6=20 <>{12}
Hidden pair N8: R78C5={12} [C5]
Hidden triple N8: R8C46+R9C4={789}
Innies N8: R7C46=9={36/45}
16(4) R7C5: R78C6=16-1-2=13=[49/58/67] --> R7C4<>6, R7C6<>3
23(4) R8C4: min R9C5=23-8-9-2=4 --> R9C5<>3, R9C6<>6

*8:
26(4) R6C6: R6C67<>{3456} (or max sum=4+5+6+9=24<26)
--> 26(4) R6C6={4958/4967/5867}
But R67C6<>13 (because R78C6=13)
--> R67C7<>13 cannot be [94] --> R7C7<>4

9:
Innies R7: R7C5789=12
--> R7C89<>{56} (or min sum=1+2+5+6=14>12)
Innies R89: R89C89=20
Innies N9: R7C789+R89C7=25
--> R89C7<>{4} (or max sum=2+3+4+6+9=24<25)
18(4) R7C7: R8C8<>{56789} (or min sum=1+5+6+7=19>18)

*10:
26(4) R8C6: R89C7 must include 5 or 6
(or min sum=3+7+8+9=27>26)
--> R789C7 must include {56} [C7,N9]
Hidden quad N9: R789C7+R9C8={5689}

11:
Intersection C7: R456C7 must include {47} [N6]
Innies R56: R56C789=27
Innies N36: R34C9=11={56} [C9]
Intersection C9: R789C9 must include {4} [N9]

*12:
18(4) R7C7: R8C7<>{56} (or max sum=2+3+5+6=16<18)
But R9C67<>9 (because R9C56=9)
26(4) R8C6: R8C67<>17 cannot be {89} --> R78C6=13=[67]

Easy to follow


Hopefully the "hardest version" is also human easily solvable like this one. 8-)
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HATMAN
PostPosted: Fri Oct 08, 2010 2:38 pm 
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When I re-solved it I forgot the overlap in N5 - which made it a bit harder.

For the harder version leave out the twin cage 26(4) at r6c6.

Overlap innies seems a resonable description of the technique.

Maurice


Last edited by HATMAN on Sat Oct 09, 2010 2:22 pm, edited 1 time in total.

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simon_blow_snow
PostPosted: Sat Oct 09, 2010 8:23 am 
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Turns out the harder version is just a little bit harder:

(Edited thx to Andrew) 14 steps walkthrough:
Steps 1-6 same as last time

7:
16(4) R7C5 must include {12} of N8, ={12}{49/58/67} (no 3)
Overlap 16(4)+23(4) N8: R7C4+R9C6-R8C5=6
--> R7C4+R9C6 <>{6} --> R7C4+R9C6 <10 --> R8C5<>{45678}
--> R9C5<>3 (R9C56=9)
Hidden triple N8: R8C46+R9C4={789}

*8:
R789C4=20={389/479/578} must include one of {37}
--> R89C6 must include one of {37}
--> 26(4) R8C6={3689/4967/5867} must include {6}
Intersection: R89C7 must include {6} [C7,N9]
Intersection R7: R7C56 must include 6 [N8]
R9C56=9={45} [R9,N8] --> 16(4) R7C5={1267} --> R8C6=7
R7C4=3, R89C4={89} [C4]
Intersection R7: R7C789 must include {45} [N9]

9:
18(4) R7C7: R8C8<>{89} (or min sum=1+4+6+8=19>18)
Hidden pair R8: R8C47={89}
Hidden single N9: R9C7=6 --> R9C39=[37]
Innies R56: R56C789=27
Innies N36: R34C9=11={56} [C9]
Intersection N3: R123C8 must include {47} [C8]

*10:
R7C89 must include one of {45} of N9
14(4) R6C8: R6C8<>{89} (or min sum=1+2+4+8=15>14)
Also R123C8 must include one of {56} of N3
--> R67C8 cannot include both {56}
--> 14(4) R6C8={16/25}{34} must include {34}
--> R6C89 must include {3} [R6,N6]
--> R67C9 must include {4} [C9]

11:
Innies N6: R4C789=18
--> R4C8<>{12} (or max sum=2+6+9=17<18)
17(4) R4C8: R5C8<>{568} (or min sum=1+5+6+8=20>18)
--> R5C89=3={12} [R5,N6], R4C89=17-3=14=[86/95]
--> R4C7=18-14=4 --> R7C7=5, 14(4) R6C8: R67C9=[34]
Hidden single N9: R8C8=3

*12:
R5C34 cannot include both {78} of R5 for R5C7
22(4) R5C3: R6C3<>{12456} (or max sum=2+5+6+8=21<22)
--> 25(4) R6C2={1789} --> R6C2=1 --> R6C4=R9C2=2
--> R4C6=1, R89C1=7=[61]

13:
15(4) R5C1 <>{89} (or min sum=1+3+4+8=16>15)
Also must include {3} [R5,N4] (or min sum=1+4+5+6=16>15)
Hidden single N5: R4C5=3
25(4) R4C4: R45C4=25-3-9=13={67} [C4,N5]

14:
Hidden single R6: R6C8=6
--> 14(4) R6C8=[6314] --> 17(4) R4C8=[9521]
Hidden single R6: R6C3=9
--> R789C3=16=[853] --> 25(4) R6C2=[1978]
--> 22(4) R5C3=[4792] --> 18(4) R3C3=[1476]
--> 26(4) R3C1=[7928], 21(4) R3C7=[3549]

Finished


Last edited by simon_blow_snow on Sat Nov 20, 2010 8:13 am, edited 2 times in total.

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Børge
PostPosted: Sat Oct 09, 2010 10:29 am 
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HATMAN wrote:
For the harder version leave out the twin cage 24(4) at r6c6.
I presume you mean the 26(4) twin cage at r6c6?
I have added a v2 image in the Images with "udosuk Style Killer Cages" thread.
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Ed
PostPosted: Mon Oct 11, 2010 9:50 am 
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HATMAN wrote:
I'm happy that the HS part was not seen immediately.
You're welcome! You'll be even happier that I gave up! I would have given up way earlier if that easy placement wasn't available.

Welcome simon_blow_snow! I didn't really understand how you did step 2 but think it must work like this. It's only just clicked now doing this post!!

Hidden Text:
...the three boxes in r1 sum to 53. In c1 sum to 67...so the difference is 53-67 = -14. They share the four cells at r12c12 which must have the same total so the difference of 14 has to be reflected in the remaining blank cells to get the 2 rows or 2 columns up to 90 each. The only way to get a difference of 14 is with r12c9 = 17 and r9c12 = 3. Very cool indeed!!
Thanks very much!
Ed
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HATMAN
PostPosted: Mon Oct 11, 2010 1:43 pm 
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Glad you enjoyed it Ed

Note Simon has been doing my puzzles for a while on DJApe's (defunct) site.
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simon_blow_snow
PostPosted: Wed Oct 13, 2010 10:04 am 
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Ed, HATMAN


I don't know about you guys, but I am very disappointed about how djape totally abandoned his forum. A lot of useful information and great puzzles just vanished for nothing. He did not even make an offer for someone else to try taking over the forum, at least preserving the 4 years worth of valuable data. I guess selling books is all that matters now. :-(


Disgruntled Simon
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