The New Sudoku Players' Forum

[Para]

Okay, here is another simple variation on a variant. Again haven't seen this like this before. I have seen Frame sudokus where you get the sums of the first 3 digits from the outside in each row and column. Also have seen puzzles where they varied the amount of digits.

for this one I have changed it a little though. In this puzzle I won't tell you how many digits will add to the sum. That is for you to figure out. You'll just know that if you add the digits from that side, some amount will add to the given sum.

Because Frame Sudoku always mark till what point the digits add to the sum, I've called this type Frameless Sudoku as there is no frame marking the length of the sums.

Frameless Sudoku

Place the digits 1-9 once in every row, column and marked 3x3 area. The clues on the outside are the sums of the first digits you see from that side. The amount of digits in the sum can vary from 1 digit to 9 digits. They can differ from sum to sum.



Hope the rules are clear to everyone. any questions about it, just let me know.

Anyways, enjoy this puzzle. I'm happy with how it turned out.

Bram

[HATMAN]

Bram

Lovely idea - "frameless" is a good name.

Maurice

[rechuk]

Frame Sudoku (also known as "Outside Sum Sudoku" and "Sum Frame Sudoku") consists of an empty grid. Numbers in the outside frame equal the sum of the first three numbers in the corresponding row or column in the given direction.

Sudoku may contain additional four regions 3 x 3. This format was first introduced in the Dutch newspaper "NRC Handelsblad" by Peter Ritmeester. It is called Windoku ("Four-Box Sudoku", "Hyper Sudoku").

In some sudokus main diagonals also contain the digits 1 through 9. This variant is called "Diagonal Sudoku" or Sudoku X. Use the check box Diagonals for such sudokus.

Other http://frame-sudoku.blogspot.com/

[Smythe Dakota]

Frameless Sudoku (and, for that matter, Frame Sudoku) is an interesting way of combining Sudoku with Kakuro. As such, it has huge possibilities, especially for those of us who like both.

I note that, in your puzzle, all the clue pairs (top vs bottom, and left vs right) add up to 45 (so there is no middle section) or less than 45 (so that the middle section adds to 45-X-Y). There is no particular reason why the clue pairs couldn't add up to more than 45. This would simply mean that, instead of a gap between the two end sections, there would be an overlap. The middle section (overlap) would then add to X+Y-45.

Has anybody solved your puzzle yet? My impression is that it is wa-a-a-a-ay too hard. Then again, I consider myself a mere recreational solver -- I don't like it when solving a puzzle is too much like work. In this case, after an hour or two all I could solve was r1c3, r1c9, a naked pair in r1c7 and r2c7, and a few other minor observations.

Bill Smythe

[simon_blow_snow]

Has anybody solved your puzzle yet? My impression is that it is wa-a-a-a-ay too hard. Then again, I consider myself a mere recreational solver -- I don't like it when solving a puzzle is too much like work. In this case, after an hour or two all I could solve was r1c3, r1c9, a naked pair in r1c7 and r2c7, and a few other minor observations.

Bill, thanks for digging this thread up from 1 year ago. If you are a beginner in Killer Sudoku solving than this must be an impossible puzzle to tackle. However for us more advanced killer solvers this is about just right in terms of difficulty, which makes it a very enjoyable challenge. A few years ago in the (now vanquished) djape forum there were once this bunch of very talented solvers (e.g. Frank, HATMAN, H3lix, JC, Udosuk, Wellbeback as well as Para - the author of this puzzle) who would thrive on these types of challenges, exchanging the most creative walkthroughs for the delights of the small but exciting community there. Too bad the golden age is all but gone now.

Here I will list the solution of this puzzle as well as the first few steps of my walkthrough, to give a feel on what kind of techniques and logical analyses are required to solve puzzles of this level of difficulty. If necessary I will complete the walkthrough when I have more free time later:

(Added later: I've finished the complete walkthrough, so just replaced it below.)

Solution and complete walkthrough: Show

Code: Select all

   14 29  4   21 17 17    4 13 30
28 .. .. .. | .. .. .. | .. .. ..  6
34 .. .. .. | .. .. .. | .. .. .. 11
19 .. .. .. | .. .. .. | .. .. .. 26
   ---------+----------+---------
28 .. .. .. | .. .. .. | .. .. .. 17
19 .. .. .. | .. .. .. | .. .. .. 26
23 .. .. .. | .. .. .. | .. .. .. 22
   ---------+----------+---------
25 .. .. .. | .. .. .. | .. .. .. 10
12 .. .. .. | .. .. .. | .. .. .. 17
24 .. .. .. | .. .. .. | .. .. .. 21
   15 16 38   24  5 13   33 16  8

594821376
863475129
172639845
749352681
318764952
625198437
251983764
486217593
937546218


Step 1:
C3 is (4)+(3)+(38), the (4) part must be [4] instead of {13} (otherwise no way to form the (3) part) --> R1C3=[4] --> R2C3=[3] or R23C3={12} --> R2C3=[1/2/3]

Step 2:
C7 is (4)+(8)+(33), the (4) part must now be {13} --> R12C7={13} (C7,N3) --> R3C7=[8] or R34C7={26}

Step 3:
R1 is (28)+(11)+(6), the (6) part must now be [6] --> R1C9=[6] --> R1C8<>[9] --> (step 2) R3C7=[8] or R34C7=[26]

Step 4:
R2 is (34)+(11), the (11) part can't be {1235} (R2C3) or {128} (R3C7), must be of length 2 --> R2C89={29/47}

Step 5:
C8 is (13)+(16)+(16), the (13) part can't be {49/58/1345}, must be of length 3 --> R123C8={247} (C8,N3) --> (step 4) R2C89=[29] --> R13C8=[74] --> (step 2) R3C79=[85] --> (step 3) R1C67={13} (R1) --> R2C37={13} (R2) --> now the 2 (16) parts must be {169/358}, both of length 3

Step 6:
C9 is (30)+(7)+(8), the (8) part can't be {134} (R789C8), must be [8] or {17} --> (step 5) R789C9 must include exactly one of [1/3] and [1/8] --> the (7) part must be [7] or {34} --> R456789C9 must be {124}+[378] or {127}+{34}+[8] or {28}+[34]+{17} --> R7C9=[3/4]

Step 7:
R7 is (25)+(10)+(10), the rightmost (10) part can't be {1234} (otherwise no way to form the other (10) part) --> (step 5) R7C789 can't be [253/613] --> R7C9=[4] --> R7C8=[6] or R7C78=[51] --> (step 5) R789C8={169} (N9) with R7C8<>[9], R456C8={358} (N6) --> (step 6) R89C9=[38], R456C9={127} (N6)

Step 8:
R9 is (24)+(21) --> the (21) part: R9C8<>[9] --> R8C8=[9]

Step 9:
R8 is (12)+(16)+(17), the (17) part must be [593] --> R8C7=[5] --> (step 7) R79C8=[61] --> the (12) part must be {48/147/246} --> [4] is locked in R8C12 (R8,N7)

Step 10:
C5 is (17)+(23)+(5) --> (step 8) R9C5<>[5] --> R89C5=[14/23] --> (step 8) R9C67=[57] or R9C567=[462] --> R9C6=[5/6] --> (step 7) the middle (10) part of R7 can't be {12}+[7] or {35}+[2] --> R7C67=[37] or R7C567={17}+[2] --> R7C6=[1/3/7]

Step 11a:
C6 is (17)+(15)+(13), (step 10) the (13) part can't be [4315] or {17}+[5] --> R89C6=[85/76] --> collectively the middle (10) part of R7, R89C5, R89C6 must be [37]+[14]+[76] or {17}+[2]+[23]+[85] --> R789C6=[376/185/785] must have [1/7],[3/5],[3/8],[7/8]

Step 11b:
Now the (17) part can't be [179/359/386], the (15) part can't be [87], so they must have length 4 and 3 respectively --> R567C6=15 can't be {59/68}+[1] or {57}+[3] or {35}+[7], must be {48}+[3] or {26}+[7] --> R7C6=[3/7], R56C6={48/26} --> R1234C6=17={1925/1934} --> R2C6=[4/5], R134C6={129/139}

Step 11c:
Now R789C6+R89C5+R7C5+R789C4=[376]+[14]+{2589} or [785]+[23]+[1]+[964] --> [1] is locked in R78C5 (C5,N8) --> (step 11b) R2789C6+R89C5=[4785]+[23] or [5376]+[14] --> R123C5 can't have [5] and any of [2/3/4] together --> (step 10) the (17) part of C5 can't have [5] in R12C5 --> R12C5<>[5]

Step 12a:
C4 is (21)+(24) --> the (24) part: R6789C4<>{2589} (R1C4) --> R789C4<>{589} --> (step 11bc) R56C6+R789C4={26}+[964] or {48}+{258/259/289} --> the (24) part can't be {23}+[964] or {14}+{289} --> R6789C4=[5964] or R789C4={258/259} --> [5] is locked in R679C4 (C4) --> hidden single N2: R2C6=[5]

Step 12b:
R789C6=[376], R89C5=[14], R79C7=[72], R1C67=[13], R2C37=[31], R56C6={48} (N5), hidden single N2: R2C4=4, hidden triple N2: R2C5+R3C45={367} --> (step 10) R23C5<>{67} --> R3C5=[3]

Step 13:
R3 is (19)+(26) --> the (26) part: R34C6=[92] --> R18C4={28} (C4) --> R79C4={59} (C4,N8) --> (step 12a) R8C4=2, R56C4={17} (C4,N5) --> R134C4=[863] --> (step 10) R1247C5=[2758]

Step 14:
R4 is (28)+(17) --> the (17) part: R4C6789=[2681] --> R4C123={479} (N4)

Step 15:
R6 is (23)+(22) --> R6: 4 is locked in R6C67 --> the (22) part: R6C6789=[8437] --> R5C456789=[764952], R6C45=[19]

Step 16:
C1 is (14)+(16)+(15) --> the (14) part: R123C1=[581] --> (step 9) R8C123=[486] --> the (16) part: R456C1=[736] --> R9C1234=[9375] --> R8C14=[29] --> R34567C3=[29851] --> R1234567C2=[9674125]

Comments are welcomed!

[Smythe Dakota]

Interesting. My reasoning was exactly the same as yours, through step 3 and about half of step 4, before I gave up. Does this mean that, not only is there only one solution, but also only one way of getting there?! This would be highly unusual, I would think.

Bill Smythe

[simon_blow_snow]

Interesting. My reasoning was exactly the same as yours, through step 3 and about half of step 4, before I gave up. Does this mean that, not only is there only one solution, but also only one way of getting there?! This would be highly unusual, I would think.

Bill, according to Para (author of this puzzle), this puzzle was "progressively hand designed", and the solution path is "tight and designed", meaning we are "forced to go his way" to solve the puzzle. So your assessment of "only one way of getting there" could be Para's original intention.

That said, and even though the first 10 steps or so of my walkthrough seem in line of the "intended path", I guess my step 11c and 12a (where I narrow down the possibility of [5] in nonet 2) could be somewhat different to what is intended, because they are marginally "bifurication" or "forcing chain" to me, in constrast to the "elegancy level" of previous steps. Therefore I strongly speculate that there is a better way to get around those steps via other regions of the grid, which has unfortunately escaped from my vision. I would love for others to have better luck in finding that "intended way", and pointing it out to me.

There is another thread in another forum talking about this puzzle, where a poster named "goooders" has also solved that puzzle (but he never bothered to post his walkthrough). The original author Para also gave some more of his thoughts about this puzzle in there. Perhaps you would like to take a look. Here is the link:

http://www.rcbroughton.co.uk/sudoku/forum/viewtopic.php?f=13&t=795

[Para]

I agree it is pretty hard, probably a bit too hard. I noticed this when I tried to solve it again a little while ago and had trouble getting through it myself. I knew it was on the hard side when I made it, didn't mean it to come out this hard. Just sometimes when you're working on a puzzle, you'll know what you intend to do with it and then can overlook how hard a certain step is, if you don't instantly see it. I was trying to test how flexible this idea was when making it. I wasn't at all sure if this idea would work as I thought a lot of clues couldn't be used. I then found out I was wrong and there's a lot more flexibility in this idea than I would have thought.
It might have also contributed that I used to do a lot of really hard Killer Sudokus, like Simon mentioned, so I've got this inbuilt way of thinking where certain steps, that are considered very hard Killer Sudoku steps by most, are just part of my basic solving strategy. Nowadays that has subsided a bit though. If I look through some of the walkthroughs I have written years ago now, I now wonder how I ever came up with complex steps like that.
The solving path is very constrained because the further along you get in designing the puzzle, the less specific your clues can get.The more digits you place, the more the sudoku element also comes into play. That's why for example the clues in the middle three rows all are split into 2 sections with highly variable sums. Such sections really leave no room for an opening and will always be forced to the end of the puzzle. I think there's one nice path through it. Depending on how much trial and errory steps you want to use, the easier it is to get off the path.

There's some more of these puzzles that appeared in the UK Sudoku Championship and the Indian Puzzle championship made by Deb Mohanty. You'll see that those puzzles do have the feature of sums in the same row adding up to more than 45. They take a bitbof a different approach then I do with mine.

http://logicmastersindia.com/lmitests/?test=ipc2011 (Round 3)
http://www.ukpuzzles.org/contests.php?contestid=6

I also made these 2. The smaller one is not too hard. The bigger one has a tricky opening (which you really have to see to get started) but after that it's an easier solve than the above one I made.

puzzlepicnic.com/community/posts/list/848.page#4276

[simon_blow_snow]

I also made these 2. The smaller one is not too hard. The bigger one has a tricky opening (which you really have to see to get started) but after that it's an easier solve than the above one I made.

puzzlepicnic.com/community/posts/list/848.page#4276

Thanks so much Para. I solved these 2, and the bigger one (9x9 framelesssudoku #2) is especially fabulous. You are right that the original #1 is perhaps more challenging but this #2 gives the solver a very enjoyable solving experience. Each step is not so evil but altogether they compose an excellent adventure/joyride. I guess this is the charm of hand made puzzles that computer generated puzzles simply can't compete with.

Anyway, here is my complete walkthrough. I think all steps are "elegant logic" (i.e. no bifurication/forcing chains) but would love to see if others disagree or not:

17-step walkthrough for 9x9 framelesssudoku #2: Show

Step 1:
The left (10) parts of R123: max R1C12,R2C12,R3C12=10 --> min R123C3=45-10-10-10=15

Step 2:
C3 is (12)+(33) --> the (12) part is R12C3={39/48/57}
(Step 1) max R12C123=10+10+12=32 --> min R3C123=45-32=13

Step 3:
R3 is (10)+(5)+(30) --> the (10) part is R3C12 (<>[5]) --> the (5) part: max R3C3=5 (<>{6789})
(Step 1) max R13C123,R23C123=10+9+10+5=34 --> min R1C123,R2C123=45-34=11
(Step 1) the left (10) parts of R12 are R1C12,R2C12 (<>[5]) --> R3C3=45-10-10-10-12=[3] --> the (5) part is R3C34=[32]
(Step 2) [5] of N1 is locked in R12C3={57} (C3,N1)

Step 4:
R1 is (10)+(10)+(25) --> the middle (10) part: max R1C34=10 --> max R1C4=4 (<>{56789})

Step 5:
C4 is (8)+(15)+(22) --> the (8) part: R1C4<>[4], R2C4<>{134689} --> R2C34={57} (R2) --> R1C3=R2C4 --> R1C34=R12C4
(Step 4) R1C34<>8 --> R12C4<>8 -> the (8) part is R123C4=[152]
(Step 4) the middle (10) part of R1 is R1C345=[514], R2C3=[7]
(Step 4) the left (10) part of R1 is R1C12={28} (R1,N1)

Step 6:
C6 is (13)+(32) --> the (13) part is R12C6=[76] --> R3C56={89} (R3,N2) --> R2C5=3
(Step 3) R2C12={19}, R3C12={46} (R23)

Step 7:
C5 is (20)+(25) --> the (20) part is R1234C5=[4385] --> R3C6=[9]

Step 8:
R2 is (10)+(25)+(10) --> the right (10) part is R2C89={28} --> R2C7=[4]

Step 9:
C7 is (12)+(21)+(12) --> the top (12) part is R123C7=[345] --> the bottom (12) part is R789C7={129} (C7,N9) --> R456C7={678} (N6)

Step 10:
C9 is (35)+(10) --> the (10) part is R89C9={37/46} (<>{58})

Step 11:
C8 is (34)+(11) --> the (11) part is R89C8={38/56} (step 10: <>{47}) --> R7C89=45-12-11-10=12={48/57} (<>{36})

Step 12:
R7 is (14)+(17)+(14) --> the right (14) part: R7C7=[2]

Step 13:
R8 is (34)+(11) --> the (11) part must have length 2/3/4
(Step 11) R89C8=11 --> R8C89<>11 --> not length 2
(Step 10) R89C9=10 --> R8C789<>1+10 --> not length 3
--> the (11) part must have length 4, is R8C6789=[2153]
(Step 11) R7C89={48} (R7,N9) --> R9C789=[967] --> R1C89=[96], R3C89=[71], R9C5=[1]
(Step 12) the (17) part of R7: [3] is locked in R7C46 (R7,N8)

Step 14:
R9 is (15)+(30) --> the (30) part is R9C6789=[8967] --> R9C34=[24] --> (hidden singles N8) R7C46=[35]
(Step 12) the left (14) part of R7 is R7C123={167} (N7) --> R7C5=[9]

Step 15:
C2 is (15)+(13)+(17) --> the (17) part: R78C2<>[69/79]
R2C2: R78C2<>[19] --> R8C2<>[9]
the (15) part: max R123C2=15

Step 16:
C1 is (20)+(25) --> the (20) part: max R123C1=20 --> min R123C2=45-20-15=10
(Step 15) R123C2=11/13/15 (must be odd with 2 even and 1 odd values)
R13C2<>10 --> not 11
R8C2: R123C2<>[814] --> not 13
--> R123C2=15=[294/816] --> R123C1=45-15-15=15=[294/816]
the (20) part: R45C1=20-15=5={23} (C1,N4) (R123C1: <>{14})
--> R123C1=[816], R123C2=[294], R9C12=[53] --> R7C1=[7], R8C2=8

Step 17:
R6 is (13)+(32) --> the (13) part is R6C123=[418]
(Step 5) the (22) part of C4 is R6789C4=[9364]

Naked singles to finish.

Code: Select all

   20 15 12    8 20 13   12 34 35
10 .. .. .. | .. .. .. | .. .. .. 25
10 .. .. .. | .. .. .. | .. .. .. 10
10 .. .. .. | .. .. .. | .. .. .. 30
   ---------+----------+---------
26 .. .. .. | .. .. .. | .. .. .. 19
27 .. .. .. | .. .. .. | .. .. .. 18
13 .. .. .. | .. .. .. | .. .. .. 32
   ---------+----------+---------
14 .. .. .. | .. .. .. | .. .. .. 14
34 .. .. .. | .. .. .. | .. .. .. 11
15 .. .. .. | .. .. .. | .. .. .. 30
   25 17 33   22 25 32   12 11 10

825147396
197536482
643289571
279851634
356724819
418963725
761395248
984672153
532418967

[Para]

Thanks.

That's pretty much the path I designed. If I had to explain it, I wouldn't have put everything like that. It always looks a bit weird when I see these logic paths, you have designed in your head, written out. When you only see in your head, they always look a bit more static when written down in a way that people are able to follow. It's usually easier to just show by hand, sitting next to them. Makes the logic seem more elegant than written down.

My favourite partsI built in were the opening of course (step 1-3), i thought it was elegant and unexpected to most solvers, and step 13, which I thought was just a funny force which probably wasn't what you'd think to see
happen there either at first.

[simon_blow_snow]

My favourite parts I built in were the opening of course (step 1-3), i thought it was elegant and unexpected to most solvers, and step 13, which I thought was just a funny force which probably wasn't what you'd think to see happen there either at first.

For me the highlight is step 16 where a range "squeeze" and then an odd/even analysis are used to narrow down the possibilities of 3 cells.

This prompts me for another attempt to puzzle #1. Hopefully I can find an elegant alternative path to avoid the ugly step 11c in my walkthrough above, using the new found insights from this delightful #2.