(Non) consecutive X

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(Non) consecutive X

Postby Jean-Christophe » Thu Sep 21, 2006 10:05 pm

Here is a consecutive / non consecutive diagonal sudoku
Numbers in adjacent cells separated by a thick line must be consecutive.
Numbers in adjacent cells without a thick line must be non consecutive.

Image

Have fun.
Jean-Christophe
 
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Joined: 22 January 2006

Postby Jean-Christophe » Sat Sep 23, 2006 12:01 pm

Hint : for me the key to these cons/non cons is spotting "chains" of adjacents consecutive cells.

Maybe you used some other consecutive specific techniques. Feel free to tell them.

Here is my walkthrough.


Step 1
Chain of consecutives form R7C2 = 4 -> R8C2 = {35}, R9C2 = {26}, R9C3 = {17}
R9C234 <> [212] -> R8C2 = 5, R9C234 = [678], R7C3 = 3

Step 2
R8C456 = 3 consecutive numbers without {58} -> {123|234} -> R8C5 = {23}, R8C46 = {1234}
{23} locked in R8C456, not elsewhere in N8, R8
R7C56 = 2 consecutive numbers without {2348} -> {56|67}
6 locked in R7C56, not elsewhere in N8, R7

Step 3
Since R79C3 = [37] -> R8C3 <> {28} = {19}
{28} of N7 locked in R789C1 -> not elsewhere in C1
R12C1 & R345C forms a sequence of consecutive numbers -> R12345 <> {19}

Step 4
Since R8C1 <> 2 -> R9C1 <> 1, R6C1 <> 2 -> R7C1 <> 1
-> 9 of C1 locked in R69C1, 1 of C1 locked in R68C1
1 on N7 locked in R8C13 -> R8C456 = {234}, R8C5 = 3, R8C46 = {24}

Step 5
R45C3, R56C2, R6C3 forms a chain of consecutive numbers
Since R789C2 = [456] -> R45C3, R56C2, R6C3 = [1..5|5..9]
-> R46C3 = {159}
Since R6C2 <> {46} -> R6C3 <> 5 = {19}, naked pair with R8C3
-> R4C3 = 5, R5C3 = {46}, R5C2 = {37}, R6C2 = {28}
R46C2 = {28} (hidden pair in N4) -> naked pair in C2

Step 6
R345C1 are consecutive = {345|567} = {5...}
-> R3C1 = 5, R4C1 = {46}, R5C1 = {37}
R5C12 = naked pair on {37}, R6C13 = naked pair on {19}
-> R5C4 = 5, R6C4 = {46}

Step 7
Since R7C56 = {56|67} -> R6C56 = {67|78} = {7...}
R4C4 = 3 (hidden single in N5)
4 of D/ locked in N5
-> R45C5 = [21]
R46C2 = [82], R5C23 = [34], R6C13 = [91], R45C1 = [67]
R789C1 = [812], R8C3 = 9
R12C1 = [43], R1C2 = 1, R2C2 = {79} -> R2C3 = {68}, R3C3 = 2

Step 8
R9C5 = 5, R9C9 = 9, R9C678 = [143]
R8C46 = [24], R7C4 = 9
R6C4 = 4, R4C6 = 9, R567C6 = [687], R67C5 = [76]
...

Solution:
418795326
376142985
592683741
685329174
734516298
921478653
843967512
159234867
267851439
Jean-Christophe
 
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Postby Smythe Dakota » Sat Sep 23, 2006 6:37 pm

Jean-Christophe wrote:.... for me the key to these cons/non cons is spotting "chains" of adjacents consecutive cells. ....

Yes. I had actually solved your puzzle a few days ago, before you posted your hint. If I could solve it without a hint, I'm surprised nobody else stepped forward.

Nice puzzle, in any case. Just the right degree of difficulty. Some authors, when they post variants, insist on making them killers at the same time. This ought to be a no-no -- the interesting properties (and new solving techniques) of the variant should be the main feature, as in this example.

The 4 given in the lower left box immediately generates five results in its vicinity. Then, the chain in the box above it generates just two possibilities, both of which have a 5 in r4c3. Each of these possibilities generates a l-o-n-g sequence of consequences, one of which eventually leads to a contradiction. After that it's reasonably simple -- but the rich variety of types of reasoning available makes things interesting.

(By the way, I didn't look at your fine print to see if that's the same thing you said.)

Bill Smythe
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Postby udosuk » Sun Sep 24, 2006 7:04 am

Bill, once again this puzzle was posted in other forums and people (for example, I) have replied and commented elsewhere. It seems the majority of habitants here are not particularly interested in variants, so I wouldn't be too surprised to see nobody "step forward"...:)

This puzzle isn't too difficult you're right. If you like tough challenges, JC has sent me a "ruudiculous" version. PM me if you want a try...:D
udosuk
 
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Postby Smythe Dakota » Sun Sep 24, 2006 7:49 am

udosuk wrote:.... If you like tough challenges, JC has sent me a "ruudiculous" version. PM me if you want a try...:D

No, thanks. As I said above, I'm not into killers, and variants aren't interesting to me if they're also killers. Variants should stick to just being variants.

Bill Smythe
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Postby Jean-Christophe » Sun Sep 24, 2006 8:43 am

Bill, the "Ruudiculous" one isn't a killer, just a plain cons/non cons.
It was a test I made to see how hard these can be. Since they are so much constrained with so few givens, they can be so "Ruudiculously" hard to solve, no "normal" human beeing can solve them. I found it hard to desing such highly constrained puzzle with just the "right" level of difficulty.

Here is another cons/non cons, this time with Argyle diagonals
Numbers cannot be repeated in the diagonals.

Image

Have fun:)
Jean-Christophe
 
Posts: 149
Joined: 22 January 2006

Postby udosuk » Sun Sep 24, 2006 8:48 am

No, it's not a killer (as in having cages and sums), it's the same variant as the one on this thread, but with higher difficulty... By "ruudiculous" I mean the level of difficulty that most players would need to use T&E to solve it... But it seems you're not allergic to T&E, so it shouldn't be an issue for you...

PM me if you're still interested...:)

Edit: Just beaten by JC of a few minutes for the post... He must have submitted the post just as I click the "reply" button!:)
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Postby Pyrrhon » Sun Sep 24, 2006 7:43 pm

Here my walkthrough, not the shortest, but a possible one:

Consecutive Neighbors of R5C5 => R4C5, R5C4={12}R4C5, R5C4 forms a naked Pair on {24} within N5 -> not elsewhere in D/ R1C8 .. R8C1, N5, R3C4, R4C3
Naked Pair on {24} within D/ R1C8 .. R8C1 -> D/ R1C8 .. R8C1 = {24...}
R45C4 are consecutive => R4C4={15}
analog R5C56 and R5C34 => R4C6, R5C3={15}
R4C46 forms a naked Pair on {15} within R4 -> not elsewhere in N5, R4
conseuctive chain R3C1234 => R3C23<>1,9
consecutive chain R6C345 => R5C4 <> 9R6C4 = {678}
consecutive pair R6C34 =>R6C3<>1,3
R56C6, R6C45 forms a naked Quad on {6789} => R6C7={12345}
Consecutive Chain R9C5-R8C567 => R8C56<>19
Consecutive Chain R8C123=>R8C2<>19
Consecutive Pair R67C7=> R7C7 <7
Consecutive Pair R67C4 => R7C4={56789}
Consecutive Pair R67C5 => R7C5={56789}
9 of N5 locked in R56C6, R6C5 -> R7C6 <> 9
Consecutive Chain R54C9-R45C8-R5C7 => R4C9, R5C8<> 1,9, R4C8<>1289
difference of R5C89 = 3 =>
R5C9<>6
Consecutive Chain R1C45-R2C5 =>
R1C5<>19
Consecutive Chain R23C7 - R3C6 (R2C7 and R3C6 are weak linked) => R3C7<>19
Consecutive Chain R8C123=> difference R8C13=2 => R8C3<> 2
Consecutive Chain R7C5-R6C543 => difference R7C5-R6C3=3 =>
R7C5,R6C3<>7
Overlapping Consecutive Chains=>
R7C5-R6C54-R6C3/R7C4 => R6C3=R7C4 => R7C4<>7
Possible Combinations in Consecutive Chain R1C45-R2C5
R1C45 => R1C4<> 12, R1C5<>2
Conseutive Chain R8C567 => difference R8C57 = 2 => R8C7<>1
Overlapping Consecutive Chains R8C56-R8C7/R9C6 => R8C7=R9C6 => R9C6 <> 1
Possible Combnations Consecutive Chain R9C6-R8CC65-R9C5
=> R9C5 <> 2, R8C6<> 2
Consecutive Pair R8C12 =>
R8C2<> 3
Overlapping Consecutive Chain R3C32-R4C2/R3C1 (R3C3-R4C2 are weak linked) => R3C1=R4C2 => R3C1 <>15
Possible Combinations of Consecutive Chain R5C78R4C89-R5C9 => R5C7<>2, 7,8, R5C8<>7,R5C9<>7
Two consecutive pairs of naked quad in N5 => R6C45 = 67,76 or 89
with R6C3 follows R6C3={58}, R6C4={67},
R6C5={67}
Naked Pair 67 in N5
Naked Pair 89 in C6
In both cases of R6C45 (in one case forbidden consecutive in the other R6C3=8) is R6C6<>8
R5C6=8, R7C4=5,R4C4=6,R4C5=7
consecutive cells: R4C3=5
7 of R5 locked in N4
1 of N8 locked in R7C6, R9C5 => R7C7, R9C8 <> 1
Non-Consecutive R4C45=>R4C5=4
R5C4 = 2
consecutive R67C5 =>R7C5<>6
R8C7=R9C6=> R9C6 <> 27, R8C7 <> 589
Possible Combinations
Consecutive Chain R8C765-R9C4
=>R8C7={46},
R8C6={37}, R8C5={28}, R9C5={19}
R8C7=R9C6 => R9C6 <> 3
Nonconsecutive neighbors of 46 in R8C7 => R8C8, R9C7 <> 5
consecutive cells R67C7 => R7C7 <> 6R7C7 = {234}
noncons of R6C7={123} => R6C8<> 2
possible combinations in consecutive chain
R5C78-R4C89-C5C9 =>
R5C9=9, R4C9=8,R4C8=7,R5C8=6, R5C7=5
noncons of R4C7 => E3C7<>23
noncons of naked singles => R3C5 <> 5, R3C6 <> 6, R6C2<>4
Consecutive Chains => R9C5=1,R8C5=2, R8C5 = 2,R8C6=3,R8C7=4,R9C6=4 ...

+-------+-------+-------+
| 9 2 8 | 4 5 1 | 7 3 6 |
| 1 3 7 | 8 6 2 | 9 5 4 |
| 6 5 4 | 3 9 7 | 8 2 1 |
+-------+-------+-------+
| 2 6 9 | 1 4 5 | 3 7 8 |
| 7 4 1 | 2 3 8 | 5 6 9 |
| 3 8 5 | 6 7 9 | 1 4 2 |
+-------+-------+-------+
| 4 1 3 | 5 8 6 | 2 9 7 |
| 8 7 6 | 9 2 3 | 4 1 5 |
| 5 9 2 | 7 1 4 | 6 8 3 |
+-------+-------+-------+
Pyrrhon
 
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Postby Jean-Christophe » Mon Sep 25, 2006 10:19 am

You went the hard way, Pyrrhon ...
Here is my walkthrough for the 2nd one


Step 1
R4C5 & R5C4 = {24}, naked pair, also in D/
R4C46 = {15}, naked pair
R5C3 = {15}

Step 2
R56C6 & R6C45 = two pairs of consecutives -> {67} & {89}, not {78}
R6C345 = chain of 3 consecutives -> R6C4 <> 9
R67C45 = chain of 4 consecutives
R7C45 are non consecutive -> R6C5 <> 9 because R6C45 = [89] & R7C45 = [78|98] can't work
R6C45 <> {89} = {67}, R6C3 = {58}, R56C6 = {89}
R6C45 = naked pair on {67}. Diagonal connection -> R7C4 <> {67} = {58}
R7C45 <> [56|87] = {58}

Step 3
R89C56 = chain of 4 consecutives without {58} = {1234}
R8C56 = [23], R9C56 = [14], R8C7 = 4
R5C4 = 2, R4C456 = [145], R5C3 = 1

Step 4
R4C89&R5C789 = chain of 5 consecutives = {56789}
R5C6 = {89} -> R5C7 <> 9, R5C78 = [56], R4C89 = [78], R5C9 = 9
R56C6 = [89], R7C45 = [58], R6C345 = [567], R7C67 = [62], R4C7 = 3, R6C789 = [142]
R7C89 = [97], R8C89 = [15], R9C789 = [683]
...
Jean-Christophe
 
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Postby Pyrrhon » Mon Sep 25, 2006 10:56 am

The key steps are the same, but there was many tribbling in my walk.

Pyrrhon
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Postby Smythe Dakota » Tue Sep 26, 2006 2:33 am

Well, I solved it, but the path was strange and twisted.

First I stared it for an hour without placing even one digit. Then I tried out a few possibilities for the chains in the center and right-center boxes. Eventually I eliminated all but one. Thereafter, I used new and strange logic of all sorts, almost forgetting that I could also use standard Sudoku logic. (What?! There's a naked pair here? Why did I bother to notice?) I found that, after about half of the puzzle was filled in, there were a gazillion ways to skin each remaining cat. I was constantly in fear that I had made a mistake somewhere, and would soon stumble against a contradiction. But it didn't happen (as far as I've noticed, although I'll admit I didn't check absolutely every Argyle diagonal) and after a slow start, the last half came quickly.

Nice puzzle!

Bill Smythe
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Postby emm » Tue Sep 26, 2006 11:48 pm

I like this variant too. I like the way the puzzle doesn’t unravel after one or two speed bumps, you have to keep thinking all over the grid.

I used a bit of T&E without the hints - it didn’t seem so much like cheating when the ordering of consecutives needs T&E - for me anyway. But I don’t like those l-o-n-g ones and the walk throughs are great for getting the idea of what to look for … for the next one.

I hope you did check your Argyles, Bill - there was a bug in there:D
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Postby Smythe Dakota » Wed Sep 27, 2006 1:55 am

emm wrote:.... I hope you did check your Argyles, Bill - there was a bug in there:D

Is that some kind of inside joke? In any case, I did go back (just now) and check the diagonals, and found no bug.

Bill Smythe
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Postby emm » Thu Sep 28, 2006 12:10 am

Good on you for checking. I’ve just checked mine again too. It wasn’t a joke, but did I mean BUG?

Didn’t you end up with 79 pairs in r89c24 that needed the diagonal constraint?
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Postby udosuk » Thu Sep 28, 2006 1:31 am

emm wrote:Didn’t you end up with 79 pairs in r89c24 that needed the diagonal constraint?

That rectangle of {79} in r89c24 didn't need the diagonal constraint. Since r8c23 are consecutive and r8c3=6, r8c2 must be 7 and those 4 cells are fixed...
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