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 Post subject: Assassin 300
PostPosted: Thu Sep 04, 2014 10:36 pm 
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Grand Master
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Joined: Mon Apr 21, 2008 9:44 am
Posts: 310
Location: MV, Germany
We've reached another landmark with this Assassin - Number 300! :applause:

I don't want to embarass myself with my painting skills, so I hope that even without colours you can see why I've chosen this pattern. In comparison to the last Assassins this is a bit easier, so I also made a V2 Killer. But first things first, post your walkthroughs for V1!
Assassin 300

Attachment:
a300.png
a300.png [ 10.62 KiB | Viewed 16634 times ]

Note the remote 9(2) cage in R6C58.
SumoCue PS-Code:
3x3::k:2056:2056:1287:1287:3334:3334:5381:5381:3097:3337:3082:3082:3348:2581:5381:5381:3097:3097:3337:10497:10497:3348:2581:1559:1559:2842:2331:3083:3083:10497:2326:2326:4376:4376:2842:2331:3083:10497:10497:9730:9730:9730:10243:10243:10243:4108:4108:10497:9730:2308:9730:10243:2308:10243:4108:10497:10497:9730:9730:9730:10243:10243:10243:3597:3597:5902:5902:1296:1296:3346:3346:2579:3597:5902:5902:3087:3087:3345:3345:3345:2579:
Estimated rating:
1.0 - Hard 1.0
Solution:
6 2 1 4 8 5 7 9 3
4 5 7 6 9 3 2 8 1
9 8 3 7 1 2 4 6 5
1 3 6 2 7 8 9 5 4
8 7 9 3 5 4 1 2 6
5 4 2 1 6 9 8 3 7
7 1 5 8 2 6 3 4 9
3 9 8 5 4 1 6 7 2
2 6 4 9 3 7 5 1 8


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 Post subject: Re: Assassin 300
PostPosted: Sun Sep 14, 2014 7:27 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 Afmob! I've been travelling as well so only just got to this. I didn't find a really nice way to do it so am looking forward to alternatives. It seems to be a puzzle with various possible lines of attack. Here was mine...

Hidden Text:
1. 17(2)@r4 = {89}
-> 9(2)@r4 not {18}
Whichever of (89) is in r4c6 must go in 38(8) in r7c45
Whichever of (89) is in r4c7 must go in 40(8) in r7c89
-> (89) in r7 locked in r7c4589

2. 38(7) does not contain a 7
-> Either 9(2)@r4 = {27} or 9(2)@r6c58 = [72]
40(8) does not contain a 5
-> Either 5 in n6 in r4c89 or 9(2)@r6c58 = [45]
-> Either 9(2)@r4 = {27} or 5 in r4c89 or both.
-> 9(2)@r4 not {45}
-> 9(2)@r4 either {27} or {36}
-> At least one of (23) in the 38(8) in r7c456
-> 5(2)@n8 = {14}

3. Outies c123 -> r18c4 = +9
Since r1c4 from (1234) -> r8c4 from (5678)
-> 3 of the numbers (5678) in r8 are in r8c478

If whatever was in r8c4 was in r9c1 in n7 this puts r8c39 = +13 which is impossible given that 4 and three of (5678) are already elsewhere in r8.
-> whatever is in r8c4 must go in r7c123 in n7.
Since (89) already elsewhere in r7 -> r8c4 is not 8.

4. Following on from that
-> r1c4 is not 1.
-> r1c3 is not 4.
-> Since 41/8 does not contain a 4 -> 12(2) and 13(2) in n1 must between them contain a 4.
-> 12(2) and 13(2) in n1 are either {48}&{67} or {57}&{49}.
-> 8(2)@n1 cannot be {17}
-> HS 1 in r1 -> r1c3 = 1 (Since innies r1 -> r1c789 = +19)
-> r1c4 = 4
-> r8c4 = 5
-> 12(2)@n8 = {39}
-> HS 7 in n8 -> r9c6 = 7
Also 13(2)@n9 = {67}
Also 5 in n9 only in r9c78
-> r9c78 = {15}
-> 10(2)@n9 = {28}
-> r7c789 = {349}
Also r7c456 = {268}
-> r7c123 = {157}
Also 13(2)@r2c4 = {67}
-> 13(2)@r1 = {58}
-> 10(2)@n2 = {19}
Also r23c6 = [32]
Also 8(2)@n1 = {26}

etc.


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 Post subject: Assassin 300 V2
PostPosted: Fri Sep 26, 2014 3:17 pm 
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Joined: Mon Apr 21, 2008 9:44 am
Posts: 310
Location: MV, Germany
If you thought V1 was too easy, here is V2. I've merged some cages which eliminates the "easy" solving path of first version. Though it can be cracked quite early it proved to be quite stubborn until the end.
Assassin 300 V2

Attachment:
a300v2.png
a300v2.png [ 10.56 KiB | Viewed 16586 times ]

Note the remote 9(2) cage in R6C58.
SumoCue PS-Code:
3x3::k:5136:5136:4625:4625:8714:8714:8714:8714:3081:3343:5136:5136:4625:4118:8714:8714:3081:3081:3343:10497:10497:4625:4118:4118:4118:2823:2312:7179:7179:10497:2310:2310:4357:4357:2823:2312:7179:10497:10497:9730:9730:9730:10243:10243:10243:7179:7179:10497:9730:2308:9730:10243:2308:10243:7179:10497:10497:9730:9730:9730:10243:10243:10243:3602:3602:5907:5907:1292:1292:3349:3349:2580:3602:5907:5907:3085:3085:3342:3342:3342:2580:

Estimated rating:
1.5 - Hard 1.5

Same solution as V1.


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 Post subject: Re: Assassin 300
PostPosted: Sun Sep 28, 2014 11:19 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 Afmob for posting another Assassin after a long break, although you have been posting walkthroughs in recent months.

I usually use a maximum of 4 colours (except when there are diagonally connected or disjoint cages) but I've made an exception this time to highlight the theme of this landmark puzzle.

I hope this site will continue until at least Assassin 500! We've got 4 excellent creators of Assassins and other puzzles, Afmob, Ed, HATMAN and wellbeback; maybe others will be inspired to create puzzles. I apologise that I've never found myself interested in creating puzzles; I'm happy to post my walkthroughs for Assassins and variants and the TJKs plus continuing the archives which Ed started.

Image

I've been on holiday for two and a half weeks, with no access to the internet and not doing any sudokus; however I don't think that made any difference to how I solved this puzzle. I found some interesting steps before I found the way in to get any placements.

Here is my walkthrough for Assassin 300:
Prelims

a) R1C12 = {17/26/35}, no 4,8,9
b) R1C34 = {14/23}
c) R1C56 = {49/58/67}, no 1,2,3
d) R23C1 = {49/58/67}, no 1,2,3
e) R2C23 = {39/48/57}, no 1,2,3
f) R23C4 = {49/58/67}, no 1,2,3
g) R23C5 = {19/28/37/46}, no 5
h) R3C67 = {15/24}
i) R34C8 = {29/38/47/56}, no 1
j) R34C9 = {18/27/46/45}, no 9
k) R4C45 = {18/27/46/45}, no 9
l) R4C67 = {89}
m) Disjoint cage R6C58 = {18/27/46/45}, no 9
n) R8C56 = {14/23}
o) R8C78 = {49/58/67}, no 1,2,3
p) R89C9 = {19/28/37/46}, no 5
q) R9C45 = {39/48/57}, no 1,2,6
r) 41(8) cage at R3C2 = {12356789}, no 4
s) 38(8) cage at R5C4 = {12345689}, no 7
t) 40(8) cage at R5C7 = {12346789}, no 5

1. Naked pair {89} in R4C67, locked for R4, clean-up: no 2,3 in R3C8, no 1 in R3C9, no 1 in R4C45

2. 45 rule on R1 3 innies R1C789 = 19 = {289/379/469/568} (cannot be {478} which clashes with R1C56), no 1

3. 45 rule on R12 3 innies R2C145 = 19 = {289/379/469/478/568}, no 1, clean-up: no 9 in R3C5

4. 1 in R2 only in R2C6789
4a. 45 rule on R1 4 outies R2C6789 = 14 = {1238/1247/1256/1346}, no 9

5. 45 rule on C123 2 outies R18C4 = 9, R1C4 = {1234} -> R8C4 = {5678}

6. 45 rule on N2 3 innies R1C4 + R23C6 = 9 = {126/135/234}, no 7,8

[Maybe these are over-the-top steps to use at this stage ?! … but they fit the theme of the cage pattern.]

7. 7 in N5 only in R4C45 = {27} or R6C58 = [72] (locking cages) -> 2 in R4C45 or R6C8, CPE no 2 in R4C89, clean-up: no 9 in R3C8, no 7 in R3C9
7a. 9 in N3 only in R1C789, locked for R1, clean-up: no 4 in R1C56
7b. R1C789 (step 2) contains 9 = {289/379/469}, no 5

8. 5 in N6 only in R4C89 or in R6C58 = [45] (locking cages) -> R4C45 = {27/36} (cannot be {45} because 5 in R4C89 or 4 in R6C5), no 4,5 in R4C45

9. 9 in N5 only in R4C6 + R5C456 + R6C46, CPE no 9 in R7C6
9a. 1,4,5 in N5 only in R56C456, CPE no 1,4,5 in R7C5

10. 9 in N6 only in R4C7 + R5C789 + R6C79, CPE no 9 in R7C7

11. 45 rule on N1 3 innies R1C3 + R3C23 = 12
11a. 45 rule on R1234 3 innies R3C23 + R4C3 = 1 outie R5C1 + 9
11b. R4C3 and R5C1 cannot be equal because both are in N4 -> R3C23 cannot total 9 -> no 3 in R1C3, clean-up: no 2 in R1C4, no 7 in R8C4 (step 5)
[I think step 11b is IOE; Ed will correct me if I’m wrong about this. Is there a more direct way to do this step?]
11c. 7 in N8 only in R9C789, locked for R9, clean-up: no 3 in R8C9

12. 9 in N2 only in R23C4 = {49} or R23C5 = [91] -> R23C5 = {28/37}/[91] (cannot be {46}, locking-out cages), no 4,6 in R23C5

13. R1C3 + R3C23 (step 11) = 12 = {129/138/147/237/246/345} (cannot be {156} which clashes with R1C12)
13a. 4 in R1C3 + R3C23 only in R1C34 = [41] when only remaining 1 in N1 must be in R3C23 -> R1C3 + R3C23 = {129/138/147/237} (cannot be {246/345}, blocking cages), no 5,6 in R3C23
13b. Killer triple 7,8,9 in R1C3 + R3C23, R23C1 and R2C23, locked for N1, clean-up: no 1 in R1C12
13c. 1 in N1 only in R1C3 + R3C23 = {129/138/147}
13d. Killer pair 5,6 in R1C12 and R1C56, locked for R1

14. R1C34 = {14} (cannot be [23] which clashes with R1C12), locked for R1, clean-up: no 6 in R8C4 (step 5)
14a. R8C78 = {49/67} (cannot be {58} which clashes with R8C4), no 5,8
14b. R89C9 = {19/28}/[73] (cannot be {46} which clashes with R8C78), no 4,6
14c. 6 in N8 only in R7C456 + R9C6, CPE no 6 in R45C6
14d. 5 in R8 only in R8C1234, CPE no 5 in R9C23

15. R1C4 + R23C6 (step 6) = {126/135/234}
15a. R1C4 = {14} -> no 1,4 in R23C6, clean-up: no 2,5 in R3C7
15b. 3,6 only in R2C6 -> R2C6 = {36}
15c. 4 in N2 only in R123C4, locked for C4, clean-up: no 8 in R9C5

[It took me a long time to find this step, which is effective because R1C4 only contains 1,4.]
16. 45 rule on N23 3 innies R1C4 + R3C89 = 15 = {168/348/456} (cannot be {258/267/357} because R1C4 only contains 1,4), no 7 in R3C8, clean-up: no 4 in R4C8
16a. R1C4 = {14} -> no 4 in R3C89, clean-up: no 7 in R4C8, no 5 in R4C9
16b. 5 in C9 only in R23C9, locked for N3, clean-up: no 6 in R4C8
16c. 5 in N6 only in R46C8, locked for C8

17. R9C7 = 5 (hidden single in N9), clean-up: no 7 in R9C45
17a. R9C6 = 7 (hidden single in R9) -> R9C8 = 1 (cage sum), clean-up: no 6 in R1C5, no 8 in R6C5, no 9 in R89C9
17b. Killer pair 3,4 in R8C56 and R9C45, locked for N7
17c. R34C9 = {36}/[54/81] (cannot be [27] which clashes with R89C9), no 2,7

18. 38(8) cage at R5C4 = {12345689}, 3,4 locked for N5, clean-up: no 6 in R4C45, no 5,6 in R6C8
18a. Naked pair {27} in R4C45, locked for R4 and N5, clean-up: no 2,7 in R6C8
18b. 38(8) cage at R5C4 = {12345689}, 2 locked for R7 and N8, clean-up: no 3 in R8C56
18c. Naked pair {14} in R8C56, locked for R8 and N8, clean-up: no 9 in R8C78, no 8 in R9C4
18d. Naked pair {67} in R8C78, locked for R8 and N9, clean-up: no 3 in R9C9
18e. Naked pair {39} in R9C45, locked for R9 and N8
18f. Naked pair {28} in R89C9, locked for C9 and N9, clean-up: no 1 in R4C9
18g. Naked triple {349} in R7C789, locked for R7 and 40(8) cage at R5C7, no 3,4,9 in R5C79 + R6C789
18h. 6 in N8 only in R7C456, locked for R7 and 38(8) cage at R5C4, no 6 in R5C45 + R6C4

19. R6C5 = 6 (hidden single in N5) -> R6C8 = 3, R4C8 = 5 -> R3C8 = 6, R8C78 = [67], clean-up: no 7 in R2C1, no 7 in R2C4

20. R3C8 = 6 -> R1C4 + R3C89 (step 16) = {456} (only remaining combination) -> R1C4 = 4, R3C9 = 5 -> R1C3 = 1, R4C9 = 4, R3C6 = 2 -> R3C7 = 4 -> R7C789 = [349], clean-up: no 8,9 in R2C1, no 8,9 in R2C4, no 9 in R3C4
20a. Naked quad {5678} in R1C56 + R23C4, locked for N2 -> R2C6 = 3, R23C5 = [91], R8C56 = [41], R9C45 = [93]

21. R4C7 = 9 (hidden single in N6) -> R4C6 = 8, R5C5 = 5, R7C6 = 6, R1C6 = 5 -> R1C5 = 8, R7C45 = [82], R8C4 = 5, R23C4 = [67], R4C45 = [27]
21a. Naked triple {389} in R3C123, locked for N1, clean-up: no 4 in R2C23
21b. Naked pair {57} in R2C23, locked for R2
21c. R2C1 = 4 -> R3C1 = 9

22. R2C9 = 1, R1C9 = 3 (hidden single in C9) -> R2C8 = 8 (cage sum)

23. Naked pair {38} in R3C23, locked for 41(8) cage at R3C2 -> R4C3 = 6

24. 41(8) cage at R3C2 = {12356789}, 1 only in R57C2, locked for C2 -> R4C12 = [13], R5C1 = 8 (cage sum)

25. Naked pair {57} in R27C3, locked for C3, R5C8 = 2

26. 14(3) cage at R8C1 = {239} (only remaining combination) -> R9C1 = 2

and the rest is naked singles.

Rating Comment:
I'll rate my walkthrough for A300 at 1.5. I used some locking cages and locking-out cages, which would be Easy 1.5, but step 11b is harder so I've gone for the full 1.5


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 Post subject:
PostPosted: Wed Oct 01, 2014 7:26 am 
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Grand Master
Grand Master

Joined: Mon Apr 21, 2008 9:44 am
Posts: 310
Location: MV, Germany
Good thing I checked my walkthrough again this morning since I discovered a logical mistake. I was able to fix it without changing the rating, so be surprised how you can crack this Killer with a couple of "simple" moves.
A300 Walkthrough:
1. R123 !
a) Outies R12 = 17(3): R3C5 <> 8 since {45}8 blocked by Killer pair (45) of 6(2)
b) 10(2): R2C5 <> 2
c) ! Innies+Outies R1: 2 = R2C67 - R1C9 -> R2C6 <> 2 (IOU @ N3)
d) 2 locked in R2C789 @ R2 for N3
e) Innies R1 = 19(3) <> 1
f) ! Hidden Killer pair (12) in 8(2) for R1 since 5(2) can only have one of them -> 8(2) = {17/26}
g) 13(2) @ R1 <> {67} since it's a Killer pair of 8(2)
h) Innies N2 = 9(3) <> 7,8,9

2. N58
a) Hidden Single: R9C6 = 7 @ C6
b) 13(3) = 7{15/24}
c) 12(2) <> 5
d) Killer pair (34) locked in 5(2) + 12(2) for N8
e) 3,4 locked in 38(8) for N5
f) 17(2) = {89} locked for R4
g) 9(2) @ R4 = {27} locked for R4+N5
h) 2 locked in 38(8) for R7+N8;
i) 5(2) = {14} locked for R8+N8
j) 12(2) = {39} locked for R9+N8

3. R789
a) 1,9 locked in 38(8) for N5
b) R4C6 = 8, R4C7 = 9
c) 8 locked in 38(8) for R7+N8
d) 13(2) <> 9
e) Killer pair (56) locked in R8C4 + 13(2) for R8
f) 9 locked in 40(8) for R7+N9
g) 10(2) = {28} locked for C9+N9
h) 13(2) = {67} locked for R8+N9
i) 13(3) = {157} -> 1,5 locked for R9+N9
j) R8C4 = 5
k) 23(4) = {4568} since R9C23 <> 3,9 -> R8C3 = 8; 4,6 locked for N7

4. R123+N5
a) Outie C123 = R1C4 = 4
b) Cage sum: R1C3 = 1
c) 13(2) @ R1 = [85] -> R1C6 = 5, R1C5 = 8
d) 13(2) @ C4 = {67} locked for C4+N2
e) 10(2) = {19} locked for C5+N2
f) R3C6 = 2 -> R3C7 = 4, R9C1 = 2
g) 8(2) = [62] -> R1C1 = 6, R1C2 = 2
h) R2C6 = 3, R1C7 = 7, R1C8 = 9 -> R2C7 = 2, R1C9 = 3, R9C9 = 8
i) 9(2) @ N3 = [54] -> R3C9 = 5, R4C9 = 4
j) 12(3) = {138} -> R2C9 = 1, R2C8 = 8
k) 12(2) = {57} locked for R2+N1
l) R2C5 = 9, R2C1 = 4 -> R3C1 = 9
m) 9(2) @ R6 = [63] -> R6C5 = 6, R6C8 = 3

5. Rest is singles.

Rating:
(Hard?) 1.0. I used IOU and a Hidden Killer pair.


Last edited by Afmob on Sat Oct 04, 2014 7:20 am, edited 1 time in total.

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 Post subject: Re: Assassin 300
PostPosted: Thu Oct 02, 2014 7:58 pm 
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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
Here's my WT for V2. I started off similarly to V1, but did it using a couple of short contradiction chains shortcuts. Thanks Afmob!

Hidden Text:
1. 17(2)r4c6 = {89}
40/8 does not contain a 5.
-> 5 in n6 either in r4c89 or in r6c8 which puts 9(2)r6c58 = [45]
Either way 9(2)r4c4 cannot be {45}
-> 9(2)r4c4 from {27} or {36}
-> At least one of (23) in 38(8) in r7c456
-> 5(2)r8c5 = {14}
-> (14) in 38(8) in n5
-> 9(2)r6c58 not [45]
-> 5 in n6 in r4c89

2. Outies n23 = r1c3+r4c89 = +10
-> (Since 5 in r4c89) -> r1c3 from (1234)
-> (Innies - Outies c123) -> r8c4 from (5678)

3. Whichever of (89) is in r4c6 must go in 38(8) in r7c45
Similarly Whichever of (89) is in r4c7 must go in 40(8) in r7c89
-> (89) locked in r7 in r7c4589

4. 13(2)n9 from {58} or {67}
Whatever is in r8c4 (from 5678) must go in n7 in r7c123 or r9c1
But if the latter this puts r8c124 = +14 and -> r8c39 = +13
But this is impossible since 4 and three of (5678) are already placed elsewhere in r8.
-> Whatever is in r8c4 (from 5678) must go in n7 in r7c123
-> r8c4 cannot be an 8. -> r8c4 from (567)

5. Trying 9(2)r4c4 = {36}
Puts (36} in 38(8) in r7c456
Puts 12(2)n8 = {57}
Which contradicts r8c4 from (567)
-> 9(2)r4c4 = {27}
-> 2 in 38(8) in r7c456
Also 9(2)r6c58 cannot be {27}
-> (27) in 40(8) in n6
-> (27) in n9 in r89c789

6. 5 in n9 only in r89c78
Trying 8 in r9c6 puts 8 in r7 in r7c89 leaves no place for 5 in n9.
-> 8 in n8 in r7c45
-> 17(2)r4 = [89]
-> 9 in n9 in r7c89

7. Trying 9 in r9c6 puts 2 in n9 in 10(2) = {28} which leaves no place for 5 in n9.
-> 9 in n8 in 12(2) = {39}

8. Since 9(2)r4c4 = {27} -> outies n23 cannot be [3{25}]
-> r1c3 cannot be 3
-> r8c4 cannot be 7.
-> HS 7 in n8 -> r9c6 = 7
-> 7 in n9 only in r8c78 -> 13(2)n9 = {67}
-> r9c78 = {15}
-> 10(2)n9 = {28}
-> r7c789 = {349}
Also r8c4 = 5
-> r7c456 = {268}
-> r7c123 = {157}
Also r1c3 = 1
-> r4c89 = {45}
-> r4c123 = {136}
Also HS 6 in n5 -> r6c5 = 6
-> r6c8 = 3

9. In n7 (39) are in r8 and (46) are in r9. The other two in r89 in n7 are (28).
Only possible combination for 14(3) and H18(3) are 14(3) = [{39}2] and H18(3)=[8{46}]
-> 10(2)n9 = [28]

10. Given r1c3 = 1 -> 1 in r4 in r4c12
-> HS 1 in r7 -> r7c2 = 1
-> r4c1 = 1

11. 41(8) does not contain a 4 -> 4 in 28(6) in n4
r7c13 = {57}. Whichever is in r7c3 must go in 28(6) in n4
-> 28(6)n4 contains (1574) + ((29) or (38)) -> 6 not in 28(6)!
-> r4c3 = 6
-> r4c2 = 3
-> 28(6) = [13{4578}] with r7c1 from (57)
-> r8c12 = [39] and r9c23 = [64]
Also r3c23 = [83]!

12. r4c89 = {45} -> r3c89 = [74] or [65]
-> 9 in n3 in r12c89
Since r1c3 = 1 and r8c9 = 2 trying 9 in 12(3)n3 means it must be [921]
-> 9 in n3 in r1c89 - locked for r1
Since 9 in n4 locked in r45c3 -> 9 in n1 in r23c1 -> 13(2)n1 = {49}
-> 20(4) n1 = [6{257}]

13. Trying r1c9 = 9 puts 12(3)n3 = [921] means (12) not in 34(6) means 8 not in 34(6) leaves no place for 8 in n3.
-> r1c8 = 9
-> r7c9 = 9
-> r7c78 = [34]
-> r4c89 = [54]
-> r3c89 = [65]
-> 13(2)n9 = [67] and r9c78 = [51]
Also 3 in n3 in r12c9 and 4 in n3 in r123c7

13. 1 in r3 in r3c567 - i.e., in the 16(4)n2
and given (356) already in r3 and r6c5 = 6 there is no possible combination for 16(4)n2 which includes a 7.
-> HS 7 in r3 -> r3c4 = 7
-> Innies - outies r12 -> r2c5 = r3c1 (from 49)

Also -> r12c4 = [46] or {28}
-> NT (139) in c4 -> r569c4 = {139}
-> 4 in c4 only in r12c4
-> r12c4 = [46]
-> r2c5 = 9
-> 13(2)n1 = [49]

Also r3c567 = [{12}4]
-> HS 8 in n2 -> r1c5 = 8
-> r12c6 = {35}
-> (34(6)) -> r12c7 = {27}
-> 12(2)n3 = [381]

Also 9(2)r4c4 = [27]
Also r7c456 = [826]
-> r3c56 = [12]
-> 5(2)n8 = [41]
Also 12(2)n8 = [93]
-> r56c4 = [31]
Also r5c5 = 5
-> r56c6 = {49}

etc.


Last edited by wellbeback on Sun Nov 16, 2014 12:10 am, edited 1 time in total.

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 Post subject: Re: Assassin 300
PostPosted: Fri Oct 03, 2014 10:40 pm 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
Location: Lethbridge, Alberta, Canada
Thanks Afmob for showing us the way you solved your Assassin using simple steps.

wellbeback and I were both attracted by technically slightly harder steps which seemed to fit the theme of your '300' cage pattern.

I hope solvers liked my coloured diagram. :)


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 Post subject: Re: Assassin 300
PostPosted: Sat Oct 04, 2014 2:21 am 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
Location: Lethbridge, Alberta, Canada
Combining 4 pairs of cages took away most of the 45s which I'd used for the basic A300. Even so I didn't find this variant much harder; that's because I'd used some interesting 'over the top' steps for the original version. I've learned from wellbeback's solving path for the V1 and used a couple of his steps.

Afmob pointed out that I hadn't spotted that the 10(2) cage at R2C5 and the 6(2) cage at R3C6 were also combined. It's easy to accidentally miss that a pair of cages have been combined when one is modifying an existing diagram, rather than starting with an empty grid and setting up a new diagram.

As a result my original walkthrough was only valid as far as step 19d. I've now re-worked my solving path; that extra 16(4) combined cage made things a lot harder for me than any of the other combined cages.

Combining 5 pairs of cages must be a record for converting a V1 into a V2!


SudokuSolver found this puzzle difficult, see Rating Comment.

Here is my re-worked walkthrough for Assassin 300 V2:
I originally solved this puzzle with 4 pairs of cages from A300 combined. I hadn’t spotted that the 10(2) cage R23C5 and the 6(2) cage R3C67 had also been combined. Thanks Afmob for pointing that out and telling me that my steps were valid as far as step 19d! I’ve re-worked from there; a few earlier steps, or parts of steps have been deleted and others renumbered.

Prelims

a) R23C1 = {49/58/67}, no 1,2,3
b) R34C8 = {29/38/47/56}, no 1
c) R34C9 = {18/27/46/45}, no 9
d) R4C45 = {18/27/46/45}, no 9
e) R4C67 = {89}
f) Disjoint cage R6C58 = {18/27/46/45}, no 9
g) R8C56 = {14/23}
h) R8C78 = {49/58/67}, no 1,2,3
i) R89C9 = {19/28/37/46}, no 5
j) R9C45 = {39/48/57}, no 1,2,6
k) 41(8) cage at R3C2 = {12356789}, no 4
l) 38(8) cage at R5C4 = {12345689}, no 7
m) 40(8) cage at R5C7 = {12346789}, no 5

1. Naked pair {89} in R4C67, locked for R4, clean-up: no 2,3 in R3C8, no 1 in R3C9, no 1 in R4C45

2. 45 rule on C123 1 outies R8C4 = 1 innie R1C3 + 4, R1C3 = {12345}, R8C4 = {56789}

[Now I’ll use some of the early steps from wellbeback’s walkthrough for A300. I hope that, if necessary, I’d have found this step if I hadn’t seen his walkthrough.]
3. R4C67 = {89}, 38(8) cage at R5C4 and 40(8) cage at R5C7 each contain both of 8,9 -> R7C4589 must contain both of 8,9 (cannot be in R7C67 because of CPEs in N5 and N6), locked for R7
3a. R7C4589 contains both of 8,9 -> the remaining parts of the 38(8) and 40(8) cages must each contain one of 8,9
3b. Killer pair 8,9 in R4C6 and 38(8) cage, locked for N5
3c. Killer pair 8,9 in R4C7 and 40(8) cage, locked for N6
3d. Clean-up: no 1 in R6C58
[Note that R7C45 and R7C89 must each contain one of 8,9.]

4. 7 in N5 only in R4C45 = {27} or R6C58 = [72] (locking cages) -> 2 in R4C45 or R6C8, CPE no 2 in R4C89, clean-up: no 9 in R3C8, no 7 in R4C9

5. 5 in N6 only in R4C89 or in R6C58 = [45] (locking cages) -> R4C45 = {27/36} (cannot be {45} because 5 in R4C89 or 4 in R6C5), no 4,5 in R4C45

[And another of the early steps from wellbeback’s walkthrough for A300.]
6. 38(8) cage at R5C4 = {12345689}
6a. R7C456 must contain at least one of 2,3 (they cannot both be in R5C456 + R6C46, which would clash with R4C45)
6b. R8C56 = {14} (cannot be {23} which clashes with R7C456, which contains at least one of 2,3), locked for R8 and N8, clean-up: no 9 in R8C78, no 8 in R9C45, no 6,9 in R9C9
6c. 38(8) cage at R5C4 = {12345689}, 4 locked for N5, clean-up: no 5 in R6C8
6d. 5 in N6 only in R4C89, locked for R4

[Extending step 4, which I missed when solving A300.]
7. 7 in N5 only in R4C45 = {27} or R6C58 = [72] (locking cages) -> 2 in R4C45 or R6C8, CPE no 2 in R6C46
7a. 7 in N5 only in R4C45 = {27} or R6C58 = [72] -> no 2 in R6C5 (locking-out cages), clean-up: no 7 in R6C8

8. 4 in N4 only in R4C12 + R5C1 + R6C12, locked for 28(6) cage at R4C1, no 4 in R7C1
8a. 4 in R7 only in R7C789, locked for N9 and 40(8) cage at R5C7, no 4 in R5C789 + R6C79, clean-up: no 6 in R8C9

9. 45 rule on N23 2 innies R3C89 = 1 outie R1C3 + 10
9a. Min R3C89 = 11, max R3C8 = 8 -> min R3C9 = 3, clean-up: no 7 in R4C9
9b. 7 in N6 only in R4C8 + R5C789 + R6C79, CPE no 7 in R7C8

10. 7 in N8 only in R8C4 + R9C456, CPE no 7 in R9C23

11. 13(3) cage at R9C6 = {139/157/238/256}
11a. Killer pair 3,5 in R9C45 and 13(3) cage, locked for R9, clean-up: no 7 in R8C9

12. R7C89 contains one of 8,9 (from step 3)
12a. Killer triple 7,8,9 in R7C89, R8C78 and R89C9, locked for N9
12b. Hidden killer triple 7,8,9 in R7C89, R8C78 and R8C89 for N9, R8C78 contains one of 7,8, R89C9 contains one of 7,8,9, R7C89 contains one of 8,9 -> no 7 in R7C9
[Steps 12a and 12b slightly simplified; thanks Afmob.]
12c. 7 in R7 only in R7C123, locked for N7
12d. 40(8) cage at R5C7 = {12346789}, 7 locked for N6, clean-up: no 4 in R3C8
12e. 9 in N9 only in R7C89 +R8C9, CPE no 9 in R56C9

13. 14(3) cage at R8C1 = {158/239/248/356} (cannot be {149} because 1,4 only in R9C1)
13a. 1,4 of {158/248} must be in R9C1 -> no 8 in R9C1
13b. 6 of {356} must be in R9C1 -> no 6 in R8C12

14. 13(3) cage at R9C6 (step 11) = {139/157/238/256}
14a. 8,9 in {139/238} must be in R9C6 -> no 3 in R9C6
14b. 8 of {238} must be in R9C6, 2 of {256} must be in R9C78 (R9C78 cannot be {56} which clashes with R8C78) -> no 2 in R9C6
14c. 2 in N8 only in R7C456, locked for R7 and 38(8) cage at R5C4, no 2 in R5C456

15. 2 in N5 only in R4C45 = {27}, locked for R4 and N5, clean-up: no 2 in R6C8

16. 45 rule on N23 3(1+2) outies R1C3 + R4C89 = 10
16a. 5 in R4 only in R4C89 -> the remaining two cells must total 5 -> no 5 in R1C3, no 6 in R4C89, clean-up: no 5 in R3C8, no 3 in R3C9, no 9 in R8C4 (step 2)
16b. R4C89 cannot total 7 (because it contains 5) -> no 3 in R1C3, clean-up: no 7 in R8C4 (step 2)
16c. 6 in R4 only in R4C123, locked for N4

17. 7 in N8 only in R9C456, locked for R9, clean-up: no 3 in R8C9
17a. 7 in N9 only in R8C78 = {67}, locked for R8 and N9, clean-up: no 2 in R1C3 (step 2)
17b. 13(3) cage at R9C6 (step 11) = {139/157/238/256}
17c. Killer pair 1,2 in R89C9 and 13(3) cage, locked for N9
17d. 1 in N9 only in R9C789, locked for R9

18. 40(8) cage at R5C7 = {12346789}, 1 locked for N6, clean-up: no 8 in R3C9
18a. 1 in R4 only on R4C123, locked for N4

19. R1C3 + R4C89 = 10 (step 16)
19a. 5 in R4 only in R4C89 -> R1C3 = 1, R4C89 = {45}, locked for R4 and N6, R8C4 = 5 (step 2), clean-up: no 8 in R3C8, no 6 in R3C9, no 5 in R6C5, no 7 in R9C45
19b. Naked pair {67} in R38C8, locked for C8 -> R6C8 = 3, R6C5 = 6
19c. Naked pair {45} in R34C9, locked for C9
19d. Naked pair {39} in R9C45, locked for R9 and N8

20. 41(8) cage at R3C2 = {12356789} -> R7C2 = 1
20a. Naked pair {36} in R4C23, locked for N4 -> R4C1 = 1
20b. Naked pair {36} in R4C23, CPE no 3,6 in R3C2

21. Naked triple {268} in R7C456, locked for R7, N8 and 38(8) cage at R5C4, no 8 in R5C456 + R6C46 -> R9C6 = 7
21a. Naked triple {349} in R7C789, locked for R7, N9 and 40(8) cage at R5C7, no 9 in R5C78 + R6C7, clean-up: no 1 in R9C9
21b. Naked pair {28} in R89C9, locked for C9 and N9
21c. 1 in C4 only in R56C4, locked for N5

22. R4C67 = [89] (hidden singles in N5 and N6)

23. R8C4 = 5 -> 23(4) cage at R8C3 = {4568} (only remaining combination, cannot be {3569} because 3,9 only in R8C3) -> R8C3 = 8, R9C23 = {46}, locked for N7-> R9C1 = 2
23a. 41(8) cage at R3C2 = {12356789}, 8 locked for C2, 3 locked for C3

24. 45 rule on N1 2 remaining innies R3C23 = 11 = [29/56/83/92], no 7 in R3C2, no 5,7 in R3C3

[There’s a larger amount of re-work from here.]

25. 41(8) cage at R3C2 = {12356789} -> R34C3 = {36}, locked for C3
25a. R9C23 = [64] -> R4C2 = 3, R34C3 = [36], R3C2 = 8 (step 24), R8C12 = [39], clean-up: no 5 in R23C1

26. 41(8) cage at R3C2 = {12356789}, 2 locked for N4, 9 locked for C3 and N4

27. 18(4) cage at R1C3 contains 1 = {1269/1368/1467} (cannot be {1278} which clashes with R4C3), 6 locked for C3 and N2
27a. R7C6 = 6 (hidden single in R7)

28. 12(3) cage at R1C9 = {129/138/237} (cannot be {147} which clashes with R6C9, cannot be {156} which clashes with R56C9, ALS block, cannot be {246/345} because 2,4,5 only in R2C8), no 4,5,6
28a. 9 of {129} must be in R1C9 -> no 9 in R2C89
28b. 2, 8 of {129/138} must be in R2C8 -> no 1 in R2C8
28c. R5C9 = 6 (hidden single in C9)

29. 2 in N1 only in 20(4) cage at R1C1 = {2459/2567}
29a. 6,9 only in R1C1 -> R1C1 = {69}
29b. 5 in C1 only in R567C1, locked for 28(6) cage at R4C1, no 5 in R6C2

30. R3C89 = R1C3 + 10 (step 9), R1C3 = 1 -> R3C89 = 11 = [65/74]
30a. 1 in R3 only in R3C567, locked for 16(4) cage at R2C5, no 1 in R2C5
30b. 16(4) cage = {1249/1258/1267/1456} (cannot be {1348} because 3,8 only in R2C5, cannot be {1357} which clashes with R3C89), no 3
30c. 7 of {1267} must be in R2C5 (R3C567 cannot contain both of 6,7 which would clash with R3C8) -> no 7 in R3C57
30d. 8 of {1258} must be in R2C5, 4 of {1456} must be in R2C5 (R3C567 cannot contain both of 4,6 which would clash with R3C89), no 5 in R2C5

31. 9 in N3 only in R1C89, locked for R1 -> R1C1 = 6, clean-up: no 7 in R23C1
31a. Naked pair {49} in R23C1, locked for C1 and N1
31b. R6C2 = 4 (hidden single in N4)

32. 45 rule on R12 2 innies R2C15 = 1 outie R3C4 + 6
32a. R2C15 cannot total 8,10,15 -> no 2,4,9 in R3C4
32b. R3C4 = {67} -> R2C15 = 12,13 = [48]/{49}, no 2,7 in R2C5, 4 locked for R2
32c. Naked pair {67} in R3C48, locked for R3

33. 2 in C6 only in R123C6, locked for N2
33a. 16(4) cage at R2C5 (step 30b) = {1249/1258}
33b. 9 of {1249} must be in R23C5 (R23C5 cannot be [41] which clashes with R8C5) -> no 9 in R3C6
[Thanks Afmob for pointing out detail errors here and in step 38a; step 33 moved forward to make step 33b work.]

34. 18(4) cage at R1C3 (step 27) = {1368/1467}, no 9
34a. 4 of {1467} must be in R1C4 -> no 7 in R1C4

35. 45 rule on N2 4 remaining outies R1C78 + R23C7 = 22 = {1489/2479/2569/3478/3568} (cannot be {1678/4567} which clash with R3C8, cannot be {2389} which clashes with R2C8, cannot be {2578/3459} which clash with R3C89, cannot be {1579} which clashes with 12(3) cage at R1C9)
35a. Hidden killer pair 3,4 in R123C7 and R7C7 for C7, R7C7 = {34} -> R123C7 must contain one of 3,4 -> R1C78 + R23C7 = {1489/2479/3478/3568} (cannot be {2569} which doesn’t contain 3 or 4)
35b. R1C78 + R23C7 = {1489/2479/3568} (cannot be {3478} because 3,7,8 in C7 only in R12C7 and both of 3,4 in R123C7 would clash with R7C7)
35c. 9 of {1489/2479} must be in R1C8 -> no 2,4 in R1C8

36. 18(4) cage at R1C3 (step 34) = {1368/1467}, R3C89 (step 30) = [65/74]
36a. Consider combinations for 16(4) cage at R2C5 (step 33a) = {1249/1258}
16(4) cage = {1249}, caged X-Wing for 4,9 in R23, no other 4,9 in R23 => R3C89 = [65] => R3C4 = 7 => 18(4) cage = {1467}
or 16(4) cage = {1258} => R2C5 = 8 => 18(4) cage = {1467}
-> 18(4) cage = {1467} -> R1C4 = 4, R23C4 = {67}, locked for C4 and N2 -> R4C45 = [27], R7C45 = [82]

37. 3 in N2 only in R1C56 + R2C6, locked for 34(6) cage at R1C5, no 3 in R12C7
37a. 3 in N3 only in R12C9, locked for C9 -> R7C789 = [349], R4C8 = 5 -> R3C8 = 6

38. R1C8 = 9 (hidden single in C8)
38a. 9 in N2 only in R23C5, locked for C5 -> R9C45 = [93], R56C4 = [31], R6C9 = 7, R12C9 = [31], R2C8 = 8 (cage sum)

and the rest is naked singles.

Rating Comment:
I'll also rate my walkthrough for A300 V2 at Hard 1.5. I used a forcing chain, as well as some of the locking steps which I'd used for the V1.

SudokuSolver gave this puzzle a high rating of 1.95 and said that it used Extensive T&E. Maybe it couldn't do steps 3 and 6, which require putting steps together, something which humans do more easily.

I had a look at what the SS score would be if the 10(2) cage at R2C5 and the 6(2) cage at R3C6 had been retained, the puzzle I first solved. Surprisingly SS gives that an even higher score of 2.35!


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