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PostPosted: Fri Mar 30, 2012 2:06 pm 
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Torygraph 18th March - uncaged killer

This is one from the Sunday Telegraph 18th March 2012.

It took me a long time with iteration between cages and numbers - very enjoyable.

Perhaps I missed something or took things in the wrong order.

Anyway as I had loaded it in Excel I though I'd put it up for those that do not use that paper.

I've put the initial lines in to help and to make it clear to those who have not done the puzzle type before.

The cages cover all 81 cells with no overlaps.

Singleton cages are not allowed.

As before the cell sum is in the top lefthand corner of the cage - i.e. the leftmost cell of the top row. Note as Simon says this one might be doable without this constraint; but I have not tried it.

Thinking on writing sum in the previous sentance an algebraic killer should be doable this way.

I've attached my Exel file - note it contains the solution as well - corrected thanks Andrew.


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Last edited by HATMAN on Sun Apr 01, 2012 11:16 am, edited 3 times in total.
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PostPosted: Fri Mar 30, 2012 3:02 pm 
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Do we still assume the convention that each cell with a sum must be the leftmost cell of the topmost row of its cage?

If that's the case then the lines are obsolete (but very helpful), and this has the same rule as the previous uncaged killers posted here.

If not then this is a new variant for me.

E.g. R3C3 might be part of the 8(3) cage from R4C2, so the puzzle is harder with more possibilities on the cage structure.


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PostPosted: Sat Mar 31, 2012 3:00 pm 
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Just solved it using the assumption that a numbered cell is the leftmost topmost cell of its cage. Like HATMAN said very enjoyable, but perhaps not extremely hard as HATMAN found (perhaps BRAIN/IQ in the djape ratings system). It is probably paper solvable (justifying its appearance in a newspaper).

(However, if we don't assume the leftmost topmost property then it can probably get insanely difficult, that is if the solution is still unique. Perhaps Richard can try to incorporate this variant in SudokuSolver to help check uniqueness using brute force? ;-))

The connectivity/area theme reminds me a little bit of the Island genre which I chanced upon in Para's blog, which I fell madly in love lately. (I even opened a thread in the "Recreational Area" section in this forum which has a copy in the "Other logic puzzles" section in the enjoysudoku forum.)


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PostPosted: Sun Apr 01, 2012 4:21 am 
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Thanks HATMAN for posting this fun puzzle from the Sunday Telegraph. Living in Canada we just get the Weekly Telegraph (an overseas version of the Telegraph) which only has a couple of regular Sudokus, no killers.

My walkthrough has been withdrawn because of Simon's total lack of respect. See his posts further down this thread.

Whenever I send feedback with corrections to any posted walkthroughs, I will follow normal forum etiquette and send them by PM.


Last edited by Andrew on Tue Apr 10, 2012 6:58 pm, edited 4 times in total.

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PostPosted: Sun Apr 01, 2012 1:21 pm 
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Andrew wrote:
4. N7 cannot have any innies because R7C2 is required for 17(3 or 4) cage at R7C1
4a. Two or more outies from N7 total 13 must include R7C4 as part of 11(?) cage at R7C3 and R9C4 as part of 21(3 or 4) cage at R8C3
4b. R9C1 must be part of 7(2 or 3) cage at R8C1, no 7,8,9 in R89C1
4c. R9C2 must be part of 7(3) cage at R8C1 or 19(4) cage at R7C1

...

6. One outie for N14 = 3 (cannot be {12} because of 3(2) cage at R4C4) -> either R3C4 = 3 or R6C4 = 3, no other 3 in C4, clean-up: no 9 in R1C5


Your solution is correct but the logic is faulty here:

Step 4c: R9C2 at that stage could also be part of 21(4) cage at R8C3.

Step 6: You have overlooked the possibility that R3C3 at that stage could be part of 12(3) cage at R2C4, so instead of outie for N14 it could be innie-outies for N14: R6C4-R3C3=3. You need more in-depth analysis to eliminate that scenario.

Also R3C4 must be part of 12(?) cage at R2C4 so cannot be outie for N14, so the only possible outie for N14 is R6C4.


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PostPosted: Sun Apr 08, 2012 6:29 am 
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Here is my walkthrough to this puzzle. I managed to follow this walkthrough without using pencilmarks at all.

Walkthrough:
R2C9 belongs to R1C9 8(?), R3C8 belongs to R3C7 7(?)
--> R3C9 must belong to either of these two cages
(Note here we assume R3C9 cannot belong to R4C7 30(?),
otherwise the whole logic falls apart and the puzzle becomes insanely difficult!)
--> R12C9+R3C789 = 15(5) = {12345}
--> R4C89 must belong to R4C7 30(?)

R1C4 12(?) = 12(2) in R1C45
R3C4 belongs to R2C4 12(?) (which potentially extends to R3C3)
--> R2C5 21(?) has at least 3 cells, including R3C56
--> R2C6 13(?) has at least 2 cells, including R2C7
--> R1C6 25(?) has at least 4 cells, = 25(4) in R1C678+R2C8
--> R2C6 13(?) = 13(2) in R2C67

R7C1 19(?) has at least 3 cells, including R78C2
R7C3 11(?) has at least 2 cells, including R7C4
--> R6C23 must belong to R6C1 19(?) (which potentially extends to R6C4)
R5C4 belongs to R4C4 3(?) = 3(2) = {12}
--> R5C1 20(?) has at least 3 cells, = 20(3) in R5C123
--> R4C2 8(?) = 8(2) in R4C23
N4: sum of the 3 cages = 8+20+19 = 47 > 45
--> R6C1 19(?) has at least 4 cells, = 19(4) in R6C1234

R1C1 10(?) = 10(2) in R12C1
R1C2 8(?) has 2 or 3 cells, including R1C3 and possibly R2C3
R2C2 22(?) has 3 or 4 cells, including two or all three of R2C3+R3C23
--> R2C3 must belong to either R1C2 8(?) or R2C2 22(?)
Recall that R12C9+R3C789 = 15(5) = {12345}
--> R3C7 7(?) = {25/34/124} must include [2/3]
--> Innies N1: R3C1/R3C12/R3C13 = 5 <> {23}, = [5]/{14}
--> R3C1 6(?) cannot be 6(2) in R3C12 or 6(3) in R3C12+R4C1
--> R3C1 6(?) = 6(2) in R34C1 <> [24], = [51/15/42]

Outies R123: R4C16 = 9 (R3C1 6(2): cannot be R4C1 alone)
--> R4C6 must belong to R2C5 21(?), = 21(4) in R23C5+R34C6
--> R4C5 10(?) = 10(2) in R45C5, = {37/46} has [4/7] (R4C4 3(2) = {12})
--> Outies N1234: R4C6+R6C4 = 11 <> {47}, = {29/38/56}
--> R4C16 = 9 = [18]
--> R3C1 6(2) = [51], R4C4 3(2) = [21], R4C6+R6C4 = 11 = [83]

R4C2 8(2) = {35}
R5C1 20(3) cannot have [2]
--> R6C1 19(4): R6C123 = 16 must include [2] of N4, = {268}
--> R5C1 20(3) = {479}
--> R4C5 10(2) = [46]
--> R5C6 9(?) = 9(3) in R5C67+R6C7 = [531]

[5] of N3 must be in R12C9 of R1C9 8(?) = {35/125}
R4C789 = {679} belongs to R4C7 30(?)
--> the remaining 8 in R4C7 30(?) can only be formed by R5C9 = [8]
--> R4C7 30(?) = 30(4) in R4C789+R5C9 = {679}+[8]
--> R5C8 11(?) = 11(3) in R5C8+R6C89 = [254]

R9C6 belongs to R9C5 10(?) (which potentially extends to R9C7 and beyond)
N9: sum of the 3 cages = 17+14+16 = 47 > 45
--> R8C6 must belong to R7C7 17(?) (which potentially extends to R9C7 etc)
N7: sum of the 3 cages = 19+7+21 = 47 > 45
--> R9C4 must belong to R8C3 21(?) (which potentially extends to R9C2)
--> R8C4 17(?) = 17(2) in R8C45 = {89}

Outies N3: R12C6 = 8 <> [4]
[4] of N3 must be in R3C78 of R3C7 7(?) = {34/124}
--> [4] of N2 must be in R12C4
--> R79C4 <> [4], = {56/57/67}
--> Innies N23: R23C4 = 12 <> [57]
--> R2C4 12(?) = 12(2) in R23C4 = [48]
--> R8C4 17(2) = [98]
--> R1C4 12(2) <> {39/48}, = {57}
--> R12C6 = 8 <> {17/35}, = {26}
--> R2C6 13(2) = [67], R1C6 = [2]

R1C23 <> [2/5/7] cannot sum to 8
--> R1C2 8(?) = 8(3) in R1C23+R2C3 = {134}
--> R1C1 10(2) <> {19/37/46}, = [82]
--> R1C6 25(4) = [2]+{69}+[8], R2C8 = [8]
--> R2C2 22(?) = 22(3) in R2C2+R3C23 = [9]+{67}, R2C2 = [9]
--> R6C1 19(4) = [6]+{28}+[3], R6C1 = [6]
Recall that R78C2 belongs to R7C1 19(?)
--> R8C1 7(?) = 7(2) in R89C1 = {34} (minimum sum for R89C1)

Recall that R7C4 belongs to R7C3 11(?)
R6C56 = {79} belongs to R6C5 19(?)
--> R6C5 19(?) has at least 3 cells, including R7C56 = [21] or R7C6 = [3]
--> R7C6 must belong to R6C5 19(?), <> [4]
--> [4] of N8 must be in R89C6
--> Innies N578: R8C6+R9C56 = 12 must include [4] and have no [6]
--> R9C5 <> [2]
Hidden single N8: R7C5 = [2]

R7C3 11(?) cannot include R7C6 which belongs to R6C5 19(?)
Now R7C34 <> [2/3/4] and R7C4 <> [1/8]
--> R7C34 cannot sum to 9
--> R7C3 11(?) cannot include R7C5 = [2]
--> R7C3 11(?) = 11(2) in R7C34, = {56}
--> R6C5 19(?) = 19(4) in R67C56, = {79}+[21], R7C6 = [1]

Innies N578: R8C6+R9C56 = 12 = {345}
--> R9C5 = [5], R89C6 = {34}
--> R9C5 10(?) = 10(3) in R9C567 = [532]
--> R1C4 12(2) = [57], R7C3 11(2) = [56], R8C1 7(2) = [34], R8C6 = [4]
--> R4C2 8(2) = [53], R6C5 19(4) = [9721], R9C4 = [7]
--> R1C2 8(3) = [341]
--> R2C5 21(4) = [3198]
--> R1C9 8(?) = 8(3) in R123C9 = [152]
--> R3C7 7(?) = 7(2) in R3C78 = [43]

Hidden singles N9: R78C7 = [85], R7C89 = [43]
--> R7C7 17(?) = 17(3) in R7C7+R8C67 = [845]
--> R7C1 19(?) = 19(4) in R7C12+R8C23 = [9721]
--> R8C3 21(?) = 21(3) in R8C3+R9C34 = [687]
--> R2C2 22(3) = [967], R5C1 20(3) = [749], R6C1 19(4) = [6823]

Hidden single N9: R8C8 = [1]
--> R7C8 14(?) = 14(4) in R7C89+R89C8 = [4316]
--> R8C9 16(?) = 16(2) in R89C9 = [79]
--> R1C6 25(4) = [2698], R4C7 30(?) = [9768]


PS code:
3x3::k:2563:2066:2066:3076:3076:6412:6412:6412:2070:2563:5652:2066:3081:5386:3339:3339:6412:2070:1537:5652:5652:3081:5386:5386:1813:1813:2070:1537:2053:2053:770:2568:5386:7696:7696:7696:5126:5126:5126:770:2568:2318:2318:2831:7696:4871:4871:4871:4871:4888:4888:2318:2831:2831:4890:4890:2835:2835:4888:4888:4377:3612:3612:1815:4890:5403:4365:4365:4377:4377:3612:4125:1815:4890:5403:5403:2577:2577:2577:3612:4125:


Solution:
834572691
291436785
567819432
153248976
749165328
682397154
975621843
326984517
418753269


If anyone spot any typo or logical error in my walkthrough, please feel free to post in public about them. Once I read the replies and if I find they are legitimate corrections I will edit the text in my walkthrough accordingly.

Here I have to thank Richard for providing such a nice public forum for Sudoku players to share and communicate freely. It is nice to see that the freedom of speech is generally respected here, and everyone is allowed to express his/her own views in the format he/she thinks as appropriate, and members generally treat others and their own personal choices of what and how they like to express with due respect.

In the Universal Declaration of Human Rights (1948), they wrote:

Everyone has the right to freedom of opinion and expression; this right includes freedom to hold opinions without interference and to seek, receive and impart information and ideas through any media and regardless of frontiers.


:sun: :cheers: :ok:


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PostPosted: Tue Apr 10, 2012 6:14 pm 
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One thing I hate is logically wrong material.

In one of the walkthroughs posted above, there is at least a very obvious logical error, which affects the logical soundness of all steps afterwards.

To prevent readers (if any) from being confused and wasting time pondering about this error, I have to list it below in plain text. It is by no mean personal towards anyone, just my opinion is that a correction of a mistake is also a precious piece of information to the public audience. It serves as a good example for future solvers on how to avoid similar mistakes, on this or other similar puzzles in the future.

So without further ado, here goes:

In one of the walkthroughs above, someone wrote:
10. 45 rule on N4 1 outie R6C4 = 1 innie R4C1 + 2
10a. R4C1 is part of a 6(2 or 3) cage at R3C1
10b. If 6(2) then R34C1 = {15/24}
or if 6(3) then R3C3 must be part of 22(3) cage at R2C2 => R6C4 = 3 (step 6, only outie for N14) => R4C1 = 1
-> R4C1 = {1245}, R6C4 = {3467}

...

20. 45 rule on N1234 2 outies R4C6 + R6C4 = 11 = [56/83] (cannot be [47/74] which clash with 10(2) cage at R4C5), R4C6 = {58}, R6C4 = {36}, R4C1 = {15} (step 10)
20a. Killer pair 3,6 in 10(2) cage at R4C5 and R6C4, locked for N5

...

22. R4C1 is part of a 6(2 or 3) cage at R3C1 (step 10a) -> 6(2) cage = {15} or 6(3) cage = {23}1 -> R3C1 = {1235}, 1 locked for C1, clean-up: no 9 in R12C1


The glaring mistake here is:

From step 10 it is established that R6C4 = R4C1 + 2. Then in step 20 it is established that R6C4 = [3] or [6]. But somehow R4C1 is wrongfully deduced to be [1] or [5].

This incorrect claim subsequently leads to a faulty conclusion in step 22 about the 6(?) cage at R3C1 and thus all steps after these ones (from 23 all the way to 56) are built on false logical ground.

The lesson to learn here is when using innie-outies it is very important to subtract accurately, otherwise potentially a lot of time and effort will be wasted. I hope this serves as a good piece of learning material for all solvers of these types of puzzles.


For a walkthrough with completely sound and solid logical steps, just refer to the one in my previous post. There could be minor typo errors (and I will be grateful to anyone who replies and posts about them), but at least I am 100% certain all the logic is 100% correct before I post it out.


Also here I have to state that in the future my PM box will be reserved for only genuinely private matters (for example sharing private materials with my friends). So anything considered as public (such as what I posted in the forum to the public audience), I will not reply. Such messages will be considered spam and will be promptly ignored. I think every member has a right to decide how he/she would like to use the PM box, not to mention how he/she would like to express his/her opinions in public.


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PostPosted: Wed Apr 11, 2012 4:12 am 
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Logical errors and typos happen quite often since one is usually blind to one's own mistakes. That's why I'm always grateful if my walkthroughs are checked by members of this forum, especially Andrew pretty much goes through every of my walkthroughs and I did the same with his walkthroughs.

Even though we mainly use PMs to suggest corrections, I'm fine with someone posting my typos in public so that I can correct them. But your posts, espcially your last one, go beyond that and lack any respect which you presumedly like about this forum. They are just plain rude. If you really respect the rules and members of this forum, you apologize to Andrew in public.


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