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 Post subject: Frameless Sudoku
PostPosted: Sun Jun 27, 2010 7:44 pm 
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I haven't been here in ages. Haven't done much with killer sudokus much lately. Handmade a killer a while back which shockingly got a rating of 1.83 in my version of SS. Didn't think I could pull that off without some computer help.

But I made a puzzle which is pretty much a killer sudoku variant. So I figured I'd post it here too.

Here is a 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.

Image

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


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 Post subject: Re: Frameless Sudoku
PostPosted: Fri Jul 02, 2010 1:36 pm 
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Para! Long time no see!

I gave this one a shot, but I had written two of the sums wrong so I'll have to start over, but I like the mechanics of this variant.

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 Post subject: Re: Frameless Sudoku
PostPosted: Fri Jul 02, 2010 3:24 pm 
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Brilliant logic.But it did take ages!There again I couldn't do a74v3.Congratulations on a very clever idea.


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 Post subject: Re: Frameless Sudoku
PostPosted: Mon Jul 05, 2010 9:46 am 
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Great to hear from you again Para! It does feel very killer like.

This is the decent chunks I could find but every avenue from here goes dead. Perhaps I'm missing some nonet wide overlap...anyway, really enjoyed trying. Perhaps goooders (good to hear from you again too!) can help me out.

1. "45" on c3: innie/s = 3(1/2) = [3]/{12} = [1/3..]
1a. -> r2c3 from (123)
1b. and no 3 in r3c3

2. 4(1/2) at r1c3 can't be 4(2) = {13} (step 1)
2a. -> r1c3 = 4

3. 4(1/2) at r1c7 can't be 4(1) -> must be 4(2) = {13}: both locked for c7 & n3

4. 6(1/2/3) at r1c9 can't be 6(3) = {123} since only 1 of 1,3 is available
4a. can't be 6(2) = {15/24} since no 1 or 4 in r1c89
4b. -> r1c9 = 6

5. "45" on r1: 2/3/4 innies = 11 -> no 9 in r1c8

6. 11(2/3/4) at r2c9: can't be 11(4) = {1235}, blocked by r2c3
6a. = 11(2) = {29/47}
6b. or 11(3) which must have 1 or 3 for r2c7 = [1]{28} only
6c. = [2/4,4->7,7->4..]
6d. no 5 in r2c89

7. 13(2/3/4) at r1c8
7a. can't be 13(4) since min. r123c8 = {247} = 13 (can't have {245}[2])
7b. if 13(2), = [58]
7c. if 13(3), [2]{47} & [742] clash with step 6c; = [724] only works
7d. ie = [58/724] -> r1c8 from (57), r2c8 from (28)
7e. -> 11(2/3) at r2c9 (from step 6a,b)= [29]/[182]
7f. 2 locked for r2 and n3
7g. r2c9 = (29)

8. naked pair {13} at r2c37: both locked for r2

9. "45" on c7: 1,2,3 innies = 8:
9a. can't be 8(3) = {1..} since no 1
9b. can't be 8(2) since no 1,2,3,6 in r3c7 and no 1,3 in r4c7
9c. -> r3c7 = 8

10. r2c89 = [29]
10a. r13c8 = [74] step 7d
10c. r3c9 = 5

11. "45" on r1: 2,3 innies = 11 (can't be 4 innies since 7 is already in r1c8)
11a. can't be 11(2) because no 4 in r1c7
11b. = 11(3) = {13}[7]
11c. 1&3 locked in r1c67 for r1

Code:
.-------------------------------.-------------------------------.-------------------------------.
| 2589      2589      4         | 2589      2589      13        | 13        7         6         |
| 5678      5678      13        | 45678     45678     45678     | 13        2         9         |
| 123679    123679    12679     | 123679    123679    123679    | 8         4         5         |
:-------------------------------+-------------------------------+-------------------------------:
| 123456789 123456789 12356789  | 123456789 123456789 123456789 | 245679    135689    123478    |
| 123456789 123456789 12356789  | 123456789 123456789 123456789 | 245679    135689    123478    |
| 123456789 123456789 12356789  | 123456789 123456789 123456789 | 245679    135689    123478    |
:-------------------------------+-------------------------------+-------------------------------:
| 123456789 123456789 12356789  | 123456789 123456789 123456789 | 245679    135689    123478    |
| 123456789 123456789 12356789  | 123456789 123456789 123456789 | 245679    135689    123478    |
| 123456789 123456789 12356789  | 123456789 123456789 123456789 | 245679    135689    123478    |
'-------------------------------.-------------------------------.-------------------------------'
Cheers
Ed


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 Post subject: Re: Frameless Sudoku
PostPosted: Mon Jul 05, 2010 6:27 pm 
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Basically there is an incredibly tight path. Go to column 8 and realise that 45 can only be split 13, 16 and 16.That means 7,9(ie 2 digits) is impossible(because of the 7 placement in row 1).That means row 8 can only be split 358 and 169.You will soon see that nonet 9 has to be 169 and then rows 8 and 9 of column 9 can only be 3 and 8.Then go along nonets 7,8 and 9.


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 Post subject: Re: Frameless Sudoku
PostPosted: Tue Jul 06, 2010 9:34 am 
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goooders wrote:
You will soon see that nonet 9 has to be 169
Thanks for your outline. This is the part I'm stuck on. I still have {358} in r789c8 with r789c9 = [4]{17} (step 16c below). The 10(2/3/4) at r7c9 could then be 10(4) = [1234]. How did you eliminate this scenario in n9?

goooders outline
12. 16(2/3/4) at r9c8 & "45" on c8 -> innies c8 = 16
12a. neither of the 16s can be 16(2) because no 7 available
12b. -> both must be 16(3) = {169/358} = [1/3,1/8..]

13. 8(1/2/3) at r9c9
13a. can't be 8{3} since {134} clashes with 16(3) at r7c8 (step 12b)
13b. either 8(2) = {17}
13c. or 8(1)
13d. r89c9 = [1/8..]
13e. -> r9c9 from (178)

14. Killer pair 1,8 in r89c7 (step 13d) and 16(3) at r7c8 -> no 1,8 in r7c9, no 8 in r8c9

15. Outies of 30r1c9 must be 10(2/3/4)
15a. if 10(4) = {1234} must be {124}[3] so it will not clash with 16(3) at r4c8.

16. "45" on c9: 1,2,3 innies = 7
16a. if 7(3) at r678c9 (can't be at r567c9 since need at least two cells to make up outies of 30 at r1c9) = [1]{24} -> r789c9 = {24}[8]
16b. either 7(2) at r78c9 = {34} -> r789c9 = {34}[8]
16c. or 7(2) at r67c9 (with 8(2) at r89c9 = {17}) = [34] ([43] blocked by 16(3) at r7c8 = [1/3..] step 12b) -> r789c9 = [4]{17}
16d. 7(1) at r7c9 is blocked by 8(2) at r89c9 = {17} (step 13b)
16e. 7(1) at r8c9 -> 8(1) at r9c9 -> innies c9 = 10(4) = {124}[3](step 15a)-> r789c9 = [378]
16f. in summary r789c9 = {24/34}[8]/[4]{17}/[378]
16g. ->no 7 in r7c9!

Code:
.-------------------------------.-------------------------------.-------------------------------.
| 2589      2589      4         | 2589      2589      13        | 13        7         6         |
| 5678      5678      13        | 45678     45678     45678     | 13        2         9         |
| 123679    123679    12679     | 123679    123679    123679    | 8         4         5         |
:-------------------------------+-------------------------------+-------------------------------:
| 123456789 123456789 12356789  | 123456789 123456789 123456789 | 245679    135689    123478    |
| 123456789 123456789 12356789  | 123456789 123456789 123456789 | 245679    135689    123478    |
| 123456789 123456789 12356789  | 123456789 123456789 123456789 | 245679    135689    123478    |
:-------------------------------+-------------------------------+-------------------------------:
| 123456789 123456789 12356789  | 123456789 123456789 123456789 | 245679    135689    234       |
| 123456789 123456789 12356789  | 123456789 123456789 123456789 | 245679    135689    12347     |
| 123456789 123456789 12356789  | 123456789 123456789 123456789 | 245679    135689    178       |
'-------------------------------.-------------------------------.-------------------------------'


Cheers
Ed


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 Post subject: Re: Frameless Sudoku
PostPosted: Tue Jul 06, 2010 1:39 pm 
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Row 7 has to be split 25,10 and 10.If you want to put in 1234 on the right hand side 10 it is not possible to find any numbers for the middle 10.As a basic premise much of the analysis depends upon thinking about the middle number if there is one.


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 Post subject: Re: Frameless Sudoku
PostPosted: Tue Jul 06, 2010 6:21 pm 
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Tight path, designed path. About the same. The further you get with your design, the less specific your sums have to be. which really leaves little room for openings in those parts. so you're kinda forced to go my way :twisted:

Ed, you're following the path so far. If you try your step 9 before step 6, you get a simpler 6 and 7. That was at least why I put that bit in there.
But yeah, row 7 is you next starting point. I realise you're not too use with overlaping cages/sums from mostly doing killers, but there's a lot of logic gained from them that you wouldn't normally see in killers. Especially overlapping cages which stay within the same nonet.

Not too shabby for a progressively hand designed puzzle, right?


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 Post subject: Re: Frameless Sudoku
PostPosted: Mon Jul 04, 2011 9:26 am 
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Posts: 170
Dear guys, found this puzzle from another forum (forum.enjoysudoku.com), and after a brief struggle managed to solve it completely. Some of the steps are a little bit boarish (to me), namely steps 11c and 12a, but would love to know how you great minds think. Since this puzzle is more than 1 year old, I would just list my walkthrough and solution out without hidden tags:

This is the complete walkthrough I wrote:
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]

Puzzle and solution in text format:
Code:
   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


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