one trick challenge for SE 4.5

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one trick challenge for SE 4.5

Postby yzfwsf » Mon Feb 22, 2021 9:59 pm

Code: Select all
.1......582.....3...9.4.......6.8..4.3.12.9.66......2.....8..1...85.....7....2..3
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Re: one trick challenge for SE 4.5

Postby AnotherLife » Tue Feb 23, 2021 10:50 am

After the basics the sudoku state becomes as follows:
Code: Select all
.--------------.----------------.--------------.
| 34  1    46  | 8   3679  3679 | 2   79   5   |
| 8   2    56  | 79  156   156  | 4   3    79  |
| 35  7    9   | 2   4     35   | 1   6    8   |
:--------------+----------------+--------------:
| 1   9    2   | 6   57    8    | 3   57   4   |
| 45  3    457 | 1   2     457  | 9   8    6   |
| 6   8    457 | 3   579   4579 | 57  2    1   |
:--------------+----------------+--------------:
| 29  456  3   | 47  8     67   | 56  1    29  |
| 29  46   8   | 5   13    13   | 67  479  279 |
| 7   456  1   | 49  69    2    | 8   45   3   |
'--------------'----------------'--------------'

It is possible to prove that r2c6=6 and get a one-step solution, but we need to construct a forcing net, that is, a series of chains. Can we consider this as a one-step solution? I am not sure. I also do not understand why we have to search for a complicated one-step solution when a simple human two-step solution is available.
I found this one manually:
1. Uniqueness Test 1: 2/9 in r7c19,r8c19 => r8c9<>2, r8c9<>9.
2. XY-Wing: 3/5/1 in r2c5,r38c6 => r2c6,r8c5<>1 (after a series of singles).
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Re: one trick challenge for SE 4.5

Postby Hajime » Tue Feb 23, 2021 12:04 pm

After Singles Only you can come to:
Code: Select all
   +-----'----'-----+----'-----'-----+----'----'---+ 
   |  34    1   3467|  8   379   379 |  2  4679  5 | 
   |  8     2   457 | 79   1579   6  | 147   3  179| 
   |  35   567   9  |  2    4    1357| 167  67   8 | 
   +-----'----'-----+----'-----'-----+----'----'---+ 
   | 1259  579  1257|  6   3579   8  |1357  57   4 | 
   |  45    3   457 |  1    2    457 |  9    8   6 | 
   |  6     8   1457|3479  3579 34579|1357   2   17| 
   +-----'----'-----+----'-----'-----+----'----'---+ 
   |23459 4569 23456|3479   8    3479|4567   1  279| 
   |12349  469   8  |  5  13679 13479| 467 4679 279| 
   |  7   4569  1456| 49   169    2  |  8  4569  3 | 
   +-----'----'-----'----'-----'-----'----'----'---+ 
                                                     
#1//B4
.1.8..2.582...6.3...924...8...6.8..4.3.12.98668.....2.....8..1...85.....7....28.3

From here the puzzle is solvable with Basics: Naked/Hidden Pair & Triplets and 3x Pointing/Claiming

With Singles Only and Forcing Net, I need 2x Forcing Nets, not once ....
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Re: one trick challenge for SE 4.5

Postby Cenoman » Tue Feb 23, 2021 1:07 pm

Don't know if the following net is considered as one-step (I guess, in TDP it would be...)
To me, it's a disguised two-stepper :(
Code: Select all
 +-------------------+---------------------+-------------------+
 |  34   1    c6-4   |  8    3679 AB3679   |  2    79    5     |
 |  8    2    b56    |  79   156   a156    |  4    3     79    |
 |  35   7     9     |  2    4      35     |  1    6     8     |
 +-------------------+---------------------+-------------------+
 |  1    9     2     |  6    57     8      |  3    57    4     |
 |  45   3     457   |  1    2      457    |  9    8     6     |
 |  6    8   zC457   |  3   y579  zC4579   |  57   2     1     |
 +-------------------+---------------------+-------------------+
 |  29   56-4  3     |b'47   8   vb'67     |  56   1     29    |
 |  29 z'46    8     |  5    13     13     |  67   479   279   |
 |  7  y'456   1     |  49 xw69     2      |  8    45    3     |
 +-------------------+---------------------+-------------------+
Kraken row (6)r127c6

(6)r1c6 - (9)r1c6 = (94)r6c36 *
 ||   \
 ||   (6=74)r7c46 **
 ||   /
(6)r2c6 - r2c3 = (6)r1c3 *
 ||                       
 ||                 (9)r9c5 - r6c5 = (94)r6c36 *
 ||               //
(6)r7c6 - (6)r9c5
                  \\
                    (6)r9c2 - (6=4)r8c2 **
------------------------
=> -4 r1c3*, r7c2**; ste


Attempt to write it in lines:
Kraken row (6)r127c6
(6)r1c6 - (9r1c6 | 6r7c6) = (94r6c36 & 74r4c46)
(6)r2c6 - (6r2c3 | 6r7c6) = (6r1c3 & 74r4c46)
(6)r7c6 - r9c5 = (9r9c5 & 6r9c2) - (9r6c5 | 6r8c2) = (94r6c36 & 4r8c2)
------------------------
=> -4 r1c3*, r7c2**; ste
Last edited by Cenoman on Tue Feb 23, 2021 2:32 pm, edited 1 time in total.
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Re: one trick challenge for SE 4.5

Postby AnotherLife » Tue Feb 23, 2021 1:18 pm

Hajime wrote:After Singles Only you can come to:
Code: Select all
   +-----'----'-----+----'-----'-----+----'----'---+ 
   |  34    1   3467|  8   379   379 |  2  4679  5 | 
   |  8     2   457 | 79   1579   6  | 147   3  179| 
   |  35   567   9  |  2    4    1357| 167  67   8 | 
   +-----'----'-----+----'-----'-----+----'----'---+ 
   | 1259  579  1257|  6   3579   8  |1357  57   4 | 
   |  45    3   457 |  1    2    457 |  9    8   6 | 
   |  6     8   1457|3479  3579 34579|1357   2   17| 
   +-----'----'-----+----'-----'-----+----'----'---+ 
   |23459 4569 23456|3479   8    3479|4567   1  279| 
   |12349  469   8  |  5  13679 13479| 467 4679 279| 
   |  7   4569  1456| 49   169    2  |  8  4569  3 | 
   +-----'----'-----'----'-----'-----'----'----'---+ 
                                                     
#1//B4
.1.8..2.582...6.3...924...8...6.8..4.3.12.98668.....2.....8..1...85.....7....28.3

From here the puzzle is solvable with Basics: Naked/Hidden Pair & Triplets and 3x Pointing/Claiming

With Singles Only and Forcing Net, I need 2x Forcing Nets, not once ....

As far as I see, the basic steps such as naked/hidden pairs and triples, locked candidates are always ignored on the forum when the task is to find a one-step solution. If the task were to find a one-step solution after Singles Only then one-step solutions might not exist even in very simple cases. As to this example, after Singles Only we can apply a hidden triple (1,2,9 in r478c1 => r47c1<>5, r78c1<>3, r78c1<>4) then, after a sequence of singles, we have a forcing net, and finally the last sequence of singles. I do not see a reason to apply two forcing nets.

By the way, this looks ridiculous when we apply two forcing nets to a simple puzzle.
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Re: one trick challenge for SE 4.5

Postby Hajime » Tue Feb 23, 2021 1:34 pm

AnotherLife wrote:As to this example, after Singles Only we can apply a hidden triple (1,2,9 in r478c1 => r47c1<>5, r78c1<>3, r78c1<>4) then, after a sequence of singles....
you allow methods like naked/hidden subsets (and pointing/claiming needed) and then you can finish the puzzle: btte
No Forcing Net needed
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Re: one trick challenge for SE 4.5

Postby AnotherLife » Tue Feb 23, 2021 4:44 pm

Hajime wrote: you allow methods like naked/hidden subsets (and pointing/claiming needed) and then you can finish the puzzle: btte
No Forcing Net needed

I do not understand what you mean. Will you write the whole one-step solution from the start?
Code: Select all
.1......582.....3...9.4.......6.8..4.3.12.9.66......2.....8..1...85.....7....2..3
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Re: one trick challenge for SE 4.5

Postby Hajime » Tue Feb 23, 2021 7:19 pm

AnotherLife wrote:I do not understand what you mean. Will you write the whole one-step solution from the start?

I see your point. And now I cannot repeat the solving process. I am not able to solve 6r2c6 anymore with basic methods. So all my above statements are void.
I will look again into it tomorrow.
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Re: one trick challenge for SE 4.5

Postby Cenoman » Tue Feb 23, 2021 8:43 pm

Hi AnotherLife,

Welcome to this forum !

You wonder how it works. Some months ago, daily simple puzzles were proposed. By simple, I mean solvable in one step after basics, (with "singles to the end" finish as an extra challenge). Basics are the techniques you have in mind: naked and hidden singles, naked and hidden subsets, claiming and pointing, that I call these "lcls" (Locked Candidares, Locked Sets), because the term "basics" is sometimes misused as encompassing other techniques (fishes, wings,...). Recently, other types of puzzles have been proposed, sometimes with an extra challenge (one-step, ste finish, many fish-steps, finding symmetries or exotic patterns), from easy to very hard.

Another predominant practise in the forum is to use Eureka conventions to write chains.
Showing pencimarks is also a common practise.

As regards the puzzle of this thread,
AnotherLife wrote:It is possible to prove that r2c6=6 and get a one-step solution, but we need to construct a forcing net, that is, a series of chains. Can we consider this as a one-step solution? I am not sure. I also do not understand why we have to search for a complicated one-step solution when a simple human two-step solution is available.
I found this one manually:
1. Uniqueness Test 1: 2/9 in r7c19,r8c19 => r8c9<>2, r8c9<>9.
2. XY-Wing: 3/5/1 in r2c5,r38c6 => r2c6,r8c5<>1 (after a series of singles).

Your two-step solution is nice!
Proving +6r2c6 is not easy, I'm curious to see the forcing net.
AnotherLife wrote:As far as I see, the basic steps such as naked/hidden pairs and triples, locked candidates are always ignored on the forum when the task is to find a one-step solution. If the task were to find a one-step solution after Singles Only then one-step solutions might not exist even in very simple cases. As to this example, after Singles Only we can apply a hidden triple (1,2,9 in r478c1 => r47c1<>5, r78c1<>3, r78c1<>4) then, after a sequence of singles, we have a forcing net, and finally the last sequence of singles. I do not see a reason to apply two forcing nets.

By the way, this looks ridiculous when we apply two forcing nets to a simple puzzle.

Basic steps are omitted by a large part of players, at the start and between significant steps. Readers are supposed to be able to process these themselves, or at least they are supposed to trust the author.

The reason to apply two forcing nets to a simple puzzle here was in yzfwsf's expectation ("One trick pony"). Of course the puzzle is simple and forcing nets are "a sledgehammer to crack a nut". Just a bit more fun !
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Re: one trick challenge for SE 4.5

Postby AnotherLife » Tue Feb 23, 2021 10:53 pm

Cenoman wrote:Proving +6r2c6 is not easy, I'm curious to see the forcing net.

Thanks for your reply, sir.
I will try to show that all three possibilities in row 2 r2c3=6, r2c5=6 and r2c6=6 lead to r2c6=6. See the picture: https://disk.yandex.ru/i/OyvUOCzpX6k-jg
1. The red chains in the picture. r2c3=6 => r1c3<>6 => r1c3=4 => r6c3<>4 => r6c6=4 => r6c6<>9 => r1c6=9 =>
1.1. r1c6<>6
1.2. r2c4<>9 => r9c4=9 => r9c5<>9 => r9c5=6 => r7c6<>6 => (because of 1.1) r2c6=6.
2. The black chains in the picture. r2c5=6 =>
2.1. r1c6<>6
2.2. r2c6<>6 => (because of 2.1) r7c6=6 => r7c7=5 => r6c7<>5
2.3. r2c3<>6 => r2c3=5 =>
2.3.1. r6c3<>5
2.3.2. r3c1<>5 => r3c6=5 => r6c6<>5 => (because of 2.2, 2.3.1) r6c5=5 => r4c5<>5 => r4c8=5 => r9c8<>5 => r9c2=5 =>
r9c2<>6 => r9c5=6 => r2c5<>6 => (because of 2.1) r2c6=6.
3. r2c6=6 => r2c6=6.

I doubt that this forcing net can be taken as a one-step solution. Frankly speaking, this is not my solution, and I just tried to explain the computer inferences in more intelligible terms.

EDIT
I think it is easier to show that the premise r2c6<>6 leads to a contradiction (my above picture remains valid). If r2c6<>6 then either r2c3=6 or r2c5=6.
1. r2c3=6 => (see above) r2c6=6 => r2c3<>6.
2. r2c5=6 => (see above) r2c5<>6.
Last edited by AnotherLife on Wed Feb 24, 2021 3:44 pm, edited 1 time in total.
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Re: one trick challenge for SE 4.5

Postby jco » Wed Feb 24, 2021 1:23 pm

Hello,

I only found solution with two moves (first one using colouring).

After basics

Code: Select all
  1   2    3     4   5     6      7   8    9
.--------------+----------------+--------------.
| 34  1    46  | 8   3679  3679 | 2   79   5   | 1
| 8   2    56  | 79  156   156  | 4   3    79  | 2
| 35  7    9   | 2   4     35   | 1   6    8   | 3
|--------------+----------------+--------------|
| 1   9    2   | 6   57    8    | 3   57   4   | 4
| 45  3    457 | 1   2     457  | 9   8    6   | 5
| 6   8    457 | 3   579   4579 | 57  2    1   | 6
|--------------+----------------+--------------|
| 29 c456  3   |b47  8    a7-6  | 56  1    29  | 7
| 29  46   8   | 5   13    13   | 67  479  279 | 8
| 7  d456  1   | 49 e69    2    | 8   45   3   | 9
'--------------+----------------+--------------'

1. (7)r7c6=(7-4)r7c4=(4-5)r7c2=(5-6)r9c2=(6)r9c5 => -6 r7c6 (+ 22 placements)


Code: Select all
  1   2  3    4   5     6      7  8  9
.-----------+----------------+---------.
| 34  1  46 | 8  b39    369  | 2  7  5 | 1
| 8   2  56 | 7   1-5   156  | 4  3  9 | 2
| 35  7  9  | 2   4    c35   | 1  6  8 | 3
|-----------+----------------+---------|
| 1   9  2  | 6   7     8    | 3  5  4 | 4
| 45  3  7  | 1   2     4-5  | 9  8  6 | 5
| 6   8  45 | 3  a59    49-5 | 7  2  1 | 6
|-----------+----------------+---------|
| 9   6  3  | 4   8     7    | 5  1  2 | 7
| 2   4  8  | 5   13    13   | 6  9  7 | 8
| 7   5  1  | 9   6     2    | 8  4  3 | 9
'-----------+----------------+---------'

2. XY-Wing (5=9)r6c5-(9=3)r1c5-(3=5)r3c5 => -5 r2c5, -5 r56c6; ste

Also from colouring: the following two moves do not need placements in between.
If I could join them somehow ...

Code: Select all
       1  2    3     4    5     6      7   8    9
    .--------------+-----------------+--------------.
    |e34  1    46  | 8   d3679  3679 | 2   79   5   | 1
    | 8   2   g56  | 79 hb156  a156  | 4   3    79  | 2
    |f35  7    9   | 2    4    a35   | 1   6    8   | 3
    |--------------+-----------------+--------------|
    | 1   9    2   | 6   i57    8    | 3   57   4   | 4
    | 45  3    457 | 1    2     47-5 | 9   8    6   | 5
    | 6   8    457 | 3   i579   4579 | 57  2    1   | 6
    |--------------+-----------------+--------------|
    | 29  456  3   | 47   8     67   | 56  1    29  | 7
    | 29  46   8   | 5   c13    13   | 67  479  279 | 8
    | 7   456  1   | 49   69    2    | 8   45   3   | 9
    '--------------+-----------------+--------------'


1. (5)r13c6=(5-1)r2c5=(1-3)r8c5=(3)r1c5-(3)r1c1=(3-5)r3c1=(5)r2c3-(5)r2c5=(5)r46c5 => -5 r5c6
2. (7)r7c6=(7-4)r7c4=(4-5)r7c2=(5-6)r9c2=(6)r9c5 => -6 r7c6; ste

Regards,
jco
EDIT: added second solution
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