SudokuSolver Forum

A forum for Sudoku enthusiasts to share puzzles, techniques and software
It is currently Tue Mar 19, 2024 5:26 am

All times are UTC




Post new topic Reply to topic  [ 4 posts ] 
Author Message
 Post subject: Assassin 345
PostPosted: Sat Dec 30, 2017 8:44 pm 
Offline
Grand Master
Grand Master

Joined: Wed Apr 16, 2008 1:16 am
Posts: 1039
Location: Sydney, Australia
Attachment:
a345.JPG
a345.JPG [ 67.43 KiB | Viewed 10037 times ]
Assassin 345
Hatman in A344 intro wrote:
I pulled up JSudoku to do a final check on an assassin
Please post that Assassin Hatman! In the mean time, here's A345. My first since May 2015. It's based on the lovely cage structure for A337. I used one complicated step - nothing elegant unfortunately. Took me a long time to find it so a serious challenge for me. SudokuSolver gives it 1.30 and JSudoku has to work hard with some fishes.

I've put the image inline rather than hosting to an image hosting place. One day that will charge me serious coin so it might go black. Inline might be safer in the long run.

Note: it has a disjoint 13(4) cage at r1c34 + r67c9. One of these in a killer is quite interesting. More than one wrecks too many "45"s in my experience.

code: paste into solver:
3x3::k:3072:3072:3329:3329:4866:4866:5891:5891:5891:3072:6404:6404:2821:4866:4866:3846:3846:5891:5895:6404:3336:2821:7177:7177:7177:3846:5891:5895:6404:3336:6410:6410:6410:7177:4107:4107:5895:2060:3336:2829:1550:6410:7177:4107:4107:2060:2060:2829:1550:3087:6410:3856:3856:3329:4881:7954:0000:3087:4115:4115:4115:5652:3329:4881:7954:7954:7954:7954:5652:5652:5652:6165:4881:4881:4881:7954:2582:2582:2582:6165:6165:
solution:
Code:
+-------+-------+-------+
| 1 4 3 | 7 8 6 | 2 5 9 |
| 7 9 5 | 2 1 4 | 6 8 3 |
| 6 8 2 | 9 3 5 | 7 1 4 |
+-------+-------+-------+
| 8 3 4 | 6 7 1 | 9 2 5 |
| 9 1 7 | 5 2 8 | 4 3 6 |
| 2 5 6 | 4 9 3 | 8 7 1 |
+-------+-------+-------+
| 5 7 8 | 3 6 9 | 1 4 2 |
| 3 2 9 | 1 4 7 | 5 6 8 |
| 4 6 1 | 8 5 2 | 3 9 7 |
+-------+-------+-------+
Happy New Year!
Ed


Top
 Profile  
Reply with quote  
 Post subject: Re: Assassin 345
PostPosted: Fri Jan 05, 2018 12:03 am 
Offline
Addict
Addict

Joined: Mon Apr 28, 2008 10:58 pm
Posts: 47
Location: Victoria, B.C., Canada
Wow! A new assassin! Happy new year Ed and many thanx for this.

Here is how I did it:
Innies of n3 -> r3c7 = 7.
Innies of the entire assassin -> r7c3 = 405 - 397 = 8.
Outies of n236 -> r1c3 + r7c9 = 5, r1c4 + r6c9 = 8.
Innies of n5 -> r5c4 + r6c5 = 14.
Outies of n5 -> r6c3 + r7c4 = 9.
Innies of n14 -> r1c3 + r6c3 = 9.
Innies of n789 -> r7c4 + r7c9 = 5.

So either
r16c3 = [36], r7c49 = [32],
or
r16c3 = [45], r7c49 = [41].

The latter fairly easily leads to a clash. The former cracks the puzzle :).

Cheers - Frank


Top
 Profile  
Reply with quote  
 Post subject: Re: Assassin 345
PostPosted: Mon Jan 08, 2018 8:59 pm 
Offline
Grand Master
Grand Master

Joined: Wed Apr 16, 2008 1:16 am
Posts: 1039
Location: Sydney, Australia
Here's my start. Glad you enjoyed it Frank!

A345
start: 14 steps:
Preliminaries courtesy of SudokuSolver
Cage 6(2) n5 - cells only uses 1245
Cage 15(2) n6 - cells only uses 6789
Cage 12(2) n58 - cells do not use 126
Cage 11(2) n2 - cells do not use 1
Cage 11(2) n45 - cells do not use 1
Cage 23(3) n14 - cells ={689}
Cage 24(3) n9 - cells ={789}
Cage 8(3) n4 - cells do not use 6789
Cage 10(3) n89 - cells do not use 89
Cage 13(4) n1269 - cells do not use 89

1. "45" on n3: 1 innie r3c7 = 7

2. "45" on whole grid: 1 innie r7c3 = 8

3. "45" on r789: 2 innies r7c49 = 5 = [41/32]
3a. r6c5 = (89)

4. "45" on n236: 2 outies r1c3 + r7c9 = 5 = [41/32]

5. "45" on n14: 2 innies r16c3 = 9 = [36/45]
5a. r5c4 = (56)

6. 8(3)n4 = {125/134}: 1 locked for n4
6a. 23(3)r3c1 = {689}: all locked for c1

The key. Complicated.
7. "45" on n1: 1 innie r1c3 + 28 = 5 outies r4c123+r5c13
7a. 3 of the 5 outies must have {789} = 24 for n4
7b. -> r1c3 + 4 = remaining two outies
7c. but [4](r1c3) + {35}(outies) blocked by [45] in h9(2)r16c3 (step 5)
(edit: Andrew felt, "Your step 7c would be simpler as [4] + {35} blocked by 8(3) n4". I saw that way as well but didn't feel simpler to me. Take your pick.)
7d.= [3]+{25};[3]+[34]};[4]+{26}
7e. ie. no 3 r45c3, no 4 r4c2

8. 13(3)r3c3: {139} blocked by 1,3 only in r3c3;{256} blocked by r6c3; {346} blocked by r1c3
8a. = {157/247}(no 3,6,9)
8b. must have 7: locked for c3 and n4
8c. 1 in {157} must be in r3c1 -> no 5 in r3c1

Nice recursive cascade now. Also a condensed way to do this (see following)
9. "45" on n1: 3 innies r1c3 + r3c13 = 1 outie r4c2 + 8
9a. max. 3 innies = 16 -> max. r4c2 = 8

10. 9 in n4 only in c1 -> locked for c1

11. "45" on n1: 3 innies r1c3 + r3c13 = 1 outie r4c2 + 8
11a. max. 3 innies = 15 -> max. r4c2 = 7

12. 8 in n4 only in c1: locked for c1
12a. r3c1 = 6

(alternative step 9-12
9. r4c2 sees all 6,8,9 in n1 apart from r3c1 -> must repeat there -> from IOD = +8, the remaining two innies must sum to 8 but this is not possible for r13c3 -> no 6,8,9 in r4c2 -> 8,9 both only in c1 in n4 -> r3c1 = 6)

13. "45" on n1: 1 outie r4c2 + 2 = 2 remaining innies r13c3
13a. max. 2 innies = 7 -> max r4c2 = 5

14. hidden single 6 in n4 -> r6c3 = 6

Much easier now.
Cheers
Ed


Top
 Profile  
Reply with quote  
 Post subject: Re: Assassin 345
PostPosted: Thu Jan 18, 2018 4:43 am 
Offline
Grand Master
Grand Master

Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
Location: Lethbridge, Alberta, Canada
Great to see that you've become active again Ed, first by solving several fairly recent puzzles and now for posting another Assassin! :D

It looks like we've now got a good chance of reaching another milestone with A350 fairly soon. Whether we'll ever reach a really major milestone of A500 I'm less sure about, at least while I'm still an active solver.

A good point about the cost of posting images. One hosting site wants me to upgrade to its premium level which costs several hundred dollars per year; as a retired person and someone whose image postings so far have mainly been for archives in the Other Variants forum, I'm not prepared to spend that amount of money.

Returning to the main topic, I found this a hard puzzle mainly because I was looking in the wrong place for a long time.
I knew:
that the disjoint 13(4) cage was an important feature of this puzzle but I wasn't looking at it the right way.

Here is my walkthrough for Assassin 345:
Prelims

a) R23C4 = {29/38/47/56}, no 1
b) 11(2) cage at R5C4 = {29/38/47/56}, no 1
c) 6(2) cage at R5C5 = {15/24}
d) 12(2) cage at R6C5 = {39/48/57}, no 1,2,6
e) R6C78 = {69/78}
f) 23(3) cage at R3C1 = {689}
g) 8(3) cage at R5C2 = {125/134}
h) 24(3) cage at R8C9 = {789}
i) 10(3) cage at R9C5 = {127/136/145/235}, no 8,9
j) 13(4) disjoint cage at R1C3 = {1237/1246/1345}, no 8,9
k) 19(5) cage at R7C1 = {12349/12358/12367/12457/13456}

Steps resulting from Prelims
1a. Naked triple {689} in 23(3) cage at R3C1, locked for C1
1b. 8(3) cage at R5C2 = {125/134}, 1 locked for N4
1c. Naked triple {789} in 24(3) cage at R8C9, locked for N9
1d. 13(4) disjoint cage at R1C3 = {1237/1246/1345}, CPE no 1 in R1C9
1e. 19(5) cage at R7C1 = {12349/12358/12367/12457/13456}, 1 locked for N7
1f. 7 in N4 only in R4C2 + R456C3, CPE no 7 in R2C3

2. 45 rule on N3 1 innie R3C7 = 7, clean-up: no 4 in R2C4, no 8 in R6C8

3. 45 rule on full grid 1 innie R7C3 = 8, clean-up: no 3 in R5C4, no 4 in R6C5
3a. 45 rule on N7 3 remaining innies R7C2 + R8C23 = 18 = {279/369/459/567}
3b. 45 rule on N7 3 outies R8C45 + R9C4 = 13 = {139/148/157/238/247/346} (cannot be {256} which clashes with R7C2 + R8C23)

4. 45 rule on N236 2 outies R1C3 + R7C9 = 5 = {14/23}
4a. R1C3 + R7C9 = 5 -> R1C4 + R6C9 = 8 = {17/26/35}, no 4

5. 45 rule on N5 2 innies R5C4 + R6C5 = 14 = [59/68/95] -> R6C3 = {256}, R7C4 = {347}

6. 45 rule on N14 2 innies R16C3 = 9 = [36/45], clean-up: no 9 in R5C4, no 3,4 in R7C9 (step 4)
6a. Naked pair {56} in 11(2) cage at R5C4, CPE no 5,6 in R5C123 + R6C456, clean-up: no 1 in R5C5, no 7 in R7C4
6b. Killer pair 8,9 in R6C5 and R6C78, locked for R6
6c. R23C4 = {29/38}/[74] (cannot be {56} which clashes with R5C4), no 5,6

7. 22(4) cage at R7C8 = {2569/3469/3568/4567} (other combinations contain more than one of 7,8,9), no 1, R8C6 = {789}, 6 in R7C8 + R8C78, locked for N9

8. 45 rule on R89 3 outies R7C128 = 16 = {169/259/367/457} (cannot be {349} which clashes with R7C4)
8a. 9 of {259} must be in R7C2 -> no 2 in R7C2

9. 13(3) cage at R3C3 = {139/157/247} (cannot be {256} which clashes with R6C3, cannot be {346} which clashes with R1C3), no 6
9a. 1 of {139/157} must be in R3C3 -> no 3,5,9 in R3C3

10. R8C45 + R9C4 (step 3b) = {139/148/157/238/247} (cannot be {346} which clashes with R7C4), no 6
10a. 16(3) cage at R7C5 = {169/259/367/457} (cannot be {349} which clashes with R7C4)
10b. Killer triple 7,8,9 in 16(3) cage, R8C45 + R9C4 and R8C6, locked for N8

11. 45 rule on N1 3 innies R1C3 + R3C13 = 1 outie R4C2 + 8
11a. Max R1C3 + R3C13 = 16 -> max R4C2 = 8

12. 45 rule on C12 3 outies R289C3 = 15 = {159/249/267} (cannot be {456} which clashes with R6C3, cannot be {357} which clashes with R16C3), no 3

[Just spotted.]
13. R1C4 + R6C9 (step 4a) = {17/26}/[53] (cannot be [35] which clashes with R16C3)
13a. 5 in R6 only in R6C123, locked for N4
13b. 13(3) cage at R3C3 (step 9) = {139/247}
13c. Killer pair 3,4 in R1C3 and 13(3) cage, locked for C3

[Then, while looking for a way to also remove [53], I found …]
14. R16C3 (step 6) = [36/45], 13(3) cage at R3C3 (step 13b) = {139/247}
14a. Consider combinations for 8(3) cage at R5C2 = {125/134}
8(3) cage = {125} => R16C3 = [36] => 13(3) cage = {247}
or 8(3) cage = {134} => 13(3) cage = 4{27}
-> 13(3) cage = {247}, locked for C3, 7 also locked for N4 -> R16C3 = [36], R5C4 = 5, R7C9 = 2 (step 4), R6C5 = 9 (step 5) -> R7C4 = 3, R6C78 = [87]
[An alternative way to look at step 15a is
Combined cages R16C3 + 8(3) cage at R5C2 = [36]{125/134}/[45]{134}
13(3) cage at R3C3 = {247} (cannot be {139} = 1{39} which clashes with R16C3 + 8(3) cage), locked for C3, 7 also locked for N4 …]
14b. R1C4 + R6C9 (step 4a) = 8 = [71]
14c. 5 in R6 only in 8(3) cage at R5C2 -> R5C2 = 1, R6C12 = {25}, locked for R6 and N4 -> 6(2) cage at R5C5 = [24], R6C6 = 3
14d. R3C13 = [62] (hidden singles in C1 and C3), R23C4 = [29] (only remaining permutation)
14e. R4C2 = 3 (hidden single in N4) -> R2C23 + R3C2 = 22 = {589}, locked for N1 -> R1C12 = [14], R2C1 = 7
14f. R9C3 = 1 (hidden single in C3) -> 10(3) cage at R9C5 = [523] (only remaining permutation)
14g. R7C7 = 1 (hidden single in N7) -> R7C56 = 15 = [69]

15. 45 rule on N6 2 remaining innies R45C7 = 13 = {49}, locked for C7, N6 and 28(5) cage at R3C5
15a. R345C7 = 20 -> R3C56 = 8 = [35]

and the rest is naked singles.

Rating Comment:
I'll rate my WT for A345 at Easy 1.5 because I used a short forcing chain; the alternative way that I then found for that breakthrough step involved a complex combination of cages, so probably also gets the same rating.


Top
 Profile  
Reply with quote  
Display posts from previous:  Sort by  
Post new topic Reply to topic  [ 4 posts ] 

All times are UTC


Who is online

Users browsing this forum: No registered users and 14 guests


You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot post attachments in this forum

Search for:
Jump to:  
cron
Powered by phpBB® Forum Software © phpBB Group