An alternative way to solve the hard 17 clue

Advanced methods and approaches for solving Sudoku puzzles

An alternative way to solve the hard 17 clue

Postby eleven » Mon May 24, 2010 11:25 pm

When i tried to reconstruct Steve K's ingenious solution to this puzzle, i saw that it can be solved quite moderate the following way.

This is it after simple moves:
Code: Select all
 *-----------------------*
 | 1 . 3 | . 2 . | . 4 5 |
 | . 5 . | 4 . 1 | . 2 3 |
 | 2 . 4 | 5 6 3 | 1 . . |
 |-------+-------+-------|
 | . . 2 | . 5 4 | 3 . 1 |
 | 5 3 7 | 1 8 6 | 4 9 2 |
 | 4 1 . | . . 2 | 5 . . |
 |-------+-------+-------|
 | . 4 . | 6 1 . | 2 3 . |
 | . 2 . | . . . | . 1 4 |
 | 3 . 1 | 2 4 . | . 5 6 |
 *-----------------------*

It's a very special puzzle, because its that hard though 3 numbers are solved and the 3 and 5 only remain in empty x-wings. Note that 789 only are given in one box of the center band each. Now i remembered that Steve treated them as a group. So i call them x,y and z now. Starting with box 9 we easily get with basic moves for x
Code: Select all
 *-----------------------*
 | 1 . 3 | . 2 . | . 4 5 |
 | . 5 . | 4 . 1 | . 2 3 |
 | 2 x 4 | 5 6 3 | 1 . . |
 |-------+-------+-------|
 | . . 2 | . 5 4 | 3 . 1 |
 | 5 3 7 | 1 8 6 | 4 9 2 |
 | 4 1 . | . . 2 | 5 . . |
 |-------+-------+-------|
 | . 4 . | 6 1 . | 2 3 x |
 | . 2 . | . . . | y 1 4 |
 | 3 . 1 | 2 4 x | z 5 6 |
 *-----------------------*

Now in r9c2 there must be y in row 9, this gives directly r7c6=y, r7c3=5, r8c6=5, r7c1=z
Code: Select all
 *-----------------------*
 | 1 . 3 | . 2 . | . 4 5 |
 | . 5 . | 4 . 1 | . 2 3 |
 | 2 x 4 | 5 6 3 | 1 . . |
 |-------+-------+-------|
 | . . 2 | . 5 4 | 3 . 1 |
 | 5 3 7 | 1 8 6 | 4 9 2 |
 | 4 1 . | . . 2 | 5 . . |
 |-------+-------+-------|
 | z 4 5 | 6 1 y | 2 3 x |
 | . 2 . | . . 5 | y 1 4 |
 | 3 y 1 | 2 4 x | z 5 6 |
 *-----------------------*

Further with simple moves r1c4=y, r2c5=x, r1c6=z, r2c3=z, r2c1=y, r1c2=6, r2c7=6, r1c7=x
Code: Select all
 *-----------------------*
 | 1 6 3 | y 2 z | x 4 5 |
 | y 5 z | 4 x 1 | 6 2 3 |
 | 2 x 4 | 5 6 3 | 1 . . |
 |-------+-------+-------|
 | . . 2 | . 5 4 | 3 . 1 |
 | 5 3 7 | 1 8 6 | 4 9 2 |
 | 4 1 . | . . 2 | 5 . . |
 |-------+-------+-------|
 | z 4 5 | 6 1 y | 2 3 x |
 | . 2 . | . . 5 | y 1 4 |
 | 3 y 1 | 2 4 x | z 5 6 |
 *-----------------------*
Thus z cannot be 7 (sees it in c3), x cannot be 8 (c5).
Now y in box 4 is either 7 or in r6c3, in box 5 it is either 8 or in r6c5, so it cant be 9 and it must be in r4c8, then in r3c9 and z in r3c8.
Code: Select all
 *-----------------------*
 | 1 6 3 | y 2 z | x 4 5 |
 | y 5 z | 4 x 1 | 6 2 3 |
 | 2 x 4 | 5 6 3 | 1 z y |
 |-------+-------+-------|
 | . . 2 | . 5 4 | 3 y 1 |
 | 5 3 7 | 1 8 6 | 4 9 2 |
 | 4 1 . | . . 2 | 5 . . |
 |-------+-------+-------|
 | z 4 5 | 6 1 y | 2 3 x |
 | . 2 . | . . 5 | y 1 4 |
 | 3 y 1 | 2 4 x | z 5 6 |
 *-----------------------*

So z also sees 9 and must be 8, then y is 7 and x is 9.

This solves the puzzle without having any hard move.
eleven
 
Posts: 1583
Joined: 10 February 2008

Re: An alternative way to solve the hard 17 clue

Postby aran » Wed May 26, 2010 5:28 pm

Eleven, I don't know the original puzzle, nor SK's move but your solution is very elegant.
btw you could even finish earlier : in the third of your grids (ie excluding the starting grid) r6c5=y, so y like x cannot be 8, hence z=8
aran
 
Posts: 334
Joined: 02 March 2007

Re: An alternative way to solve the hard 17 clue

Postby eleven » Wed May 26, 2010 10:38 pm

aran wrote:Eleven, I don't know the original puzzle, nor SK's move but your solution is very elegant.

Hi Aran,

this is the 17 clue
Code: Select all
 *-----------*
 |...|...|...|
 |...|..1|.23|
 |..4|56.|...|
 |---+---+---|
 |...|...|...|
 |..7|.8.|4..|
 |.1.|..2|5..|
 |---+---+---|
 |...|6..|...|
 |.2.|...|.1.|
 |3..|.4.|...|
 *-----------*

So it needs just a skyscraper (3) to get to my starting grid.

I did not try longer to reconstruct Steve's solution (which i had called the "solution of the year", but it was lost with the forum crash). It did not need normal pencilmarks either, but was a guardian thingy for 789 simultaneously, and combined i think with the DP in r68c45.
Maybe someone can ask Steve on the Eureka forum or elsewhere, if he still has a copy.

btw you could even finish earlier : in the third of your grids (ie excluding the starting grid) r6c5=y, so y like x cannot be 8, hence z=8

No, i dont think so, you have to be careful with the boxes containing 789. At that point i dont know, if y isnt 8 in r5c5.
eleven
 
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Joined: 10 February 2008

Re: An alternative way to solve the hard 17 clue

Postby aran » Thu May 27, 2010 12:44 pm

Eleven,
I think this
in the third of your grids (ie excluding the starting grid)
was not clear : it was intended to mean the third grid of your workings...just writing "fourth" grid would have avoided the confusion.
Anyway from this point :
Code: Select all
Further with simple moves r1c4=y, r2c5=x, r1c6=z, r2c3=z, r2c1=y, r1c2=6, r2c7=6, r1c7=x
     *-----------------------*
     | 1 6 3 | y 2 z | x 4 5 |
     | y 5 z | 4 x 1 | 6 2 3 |
     | 2 x 4 | 5 6 3 | 1 . . |
     |-------+-------+-------|
     | . . 2 | . 5 4 | 3 . 1 |
     | 5 3 7 | 1 8 6 | 4 9 2 |
     | 4 1 . | . . 2 | 5 . . |
     |-------+-------+-------|
     | z 4 5 | 6 1 y | 2 3 x |
     | . 2 . | . . 5 | y 1 4 |
     | 3 y 1 | 2 4 x | z 5 6 |
     *-----------------------*

the shortened version is : r1c4+r7c6=y =>r6c5=y => c5 contains x, y, and 8 => xy <8> =>z=8.
aran
 
Posts: 334
Joined: 02 March 2007

Re: An alternative way to solve the hard 17 clue

Postby eleven » Thu May 27, 2010 5:20 pm

Ok, what i meant is:
There is no direct contradiction, that y could not be 8 in this grid, like
Code: Select all
         *-----------------------*
         | 1 6 3 | y 2 z | x 4 5 |
         | y 5 z | 4 x 1 | 6 2 3 |
         | 2 x 4 | 5 6 3 | 1 . y |
         |-------+-------+-------|
         | . . 2 | . 5 4 | 3 y 1 |
         | 5 3 7 | 1 8 6 | 4 9 2 |
         | 4 1 y | . . 2 | 5 . . |
         |-------+-------+-------|
         | z 4 5 | 6 1 y | 2 3 x |
         | . 2 . | . . 5 | y 1 4 |
         | 3 y 1 | 2 4 x | z 5 6 |
         *-----------------------*

So i could not say directly, that it must be in r6c5.
eleven
 
Posts: 1583
Joined: 10 February 2008

Re: An alternative way to solve the hard 17 clue

Postby aran » Thu May 27, 2010 7:21 pm

Eleven, you're quite right.
Here is this time a solid (!) alternative : from that same position : hidden single r4c8=y =>r3c9=y =>r3c8=z. So z <79> from c38 =>z=8.
Anyway this is only a distraction from your very elegant solution, the idea behind which :
- unit with a naked triple C1 C2 C3 (xyz)
- one of whose member cells is fully "congruent" to another cell C4 (xyz) in another unit
- simple consequences leading to simple solution of difficult problem
...surely deserves a name
aran
 
Posts: 334
Joined: 02 March 2007

Re: An alternative way to solve the hard 17 clue

Postby eleven » Tue Jun 01, 2010 3:11 pm

Now i could find (this part of) Steve K's solution:
Steve K wrote:
Code: Select all
 *-----------*
 |1G3|.2.|A45|
 |.5.|4.1|A23|
 |2A4|563|1AA|
 |---+---+---|
 |.G2|.54|3.1|
 |537|186|492|
 |41.|..2|5..|
 |---+---+---|
 |.4.|61.|23.|
 |.2.|...|G14|
 |3A1|24G|A56|
 *-----------*
=> sis [r1c2, r4c2, r8c7, r9c6]
 
 *-----------*
 |1.3|.2.|.45|
 |.5.|4.1|.23|
 |2.4|563|1..|
 |---+---+---|
 |..2|U54|3.1|
 |537|186|492|
 |41.|XX2|5..|
 |---+---+---|
 |.4.|61U|23.|
 |.2.|XXU|.14|
 |3.1|24U|.56|
 *-----------*

=> sis [r4c4, r789c6] for 79


The G marked cells are guardians for 79 (and 8): for each number at least one G must be true, otherwise there is an impossible loop marked by the A's.
To avoid the deadly pattern X for 37 and 39 resp. both 7 and 9 must be either in r4c4 or in r789c6.

So for x in {7,9} there are 4 cases:
r1c2=x -> r1c6<>x -> x in r789c6
r4c2=x -> r4c4<>x -> (UR) x in r789c6
r8c7=x
r9c6=x

In all cases x cannot be in r8c45, i.e. r8c45<>79

The rest ist not hard (finned x-wing 8, remote pairs 79).
eleven
 
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Joined: 10 February 2008

Re: An alternative way to solve the hard 17 clue

Postby ttt » Tue Jun 01, 2010 4:03 pm

eleven wrote:I did not try longer to reconstruct Steve's solution (which i had called the "solution of the year", but it was lost with the forum crash)

Hm..., I don’t care that – the “solution of the year”: 2009, I like this – the “solution of the century”... :D

ttt
ttt
 
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Re: An alternative way to solve the hard 17 clue

Postby aran » Tue Jun 01, 2010 4:25 pm

Following on from ttt, I always thought that Allan Barker's presentation of the SK loop was one of the finest illustrations of the power, simplicity, and elegance of base sets/covers sets.
http://sudokuone.com/sweb/extra/steaster/steaster.htm
Just admire the overall symmetry of the 16 base sets
1267r28
1267c28
and the 16 cover sets
r2c56
c2r56
r8c45
c8r45
27b19
16b37
aran
 
Posts: 334
Joined: 02 March 2007

Re: An alternative way to solve the hard 17 clue

Postby eleven » Tue Jun 01, 2010 7:36 pm

No question, that i like the SK loop, but did it solve any puzzle?
I only know examples, where it lowered the difficulty of a puzzle from the hardest family to one, which is still too hard for a manual solver Also if Steve solved it completely manually, i have no interest to read the solution.

Btw the eliminations in the sample can be made from the 16 cells in the 4 edge boxes alone. I dont know, why all the illustrations use the middle boxes too (i did not read the explanations, too complicated for my simple mind).
eleven
 
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Joined: 10 February 2008

Re: An alternative way to solve the hard 17 clue

Postby aran » Tue Jun 01, 2010 10:14 pm

In fact Allan Barker does show two other base/cover solutions for the loop, including the one which I think you have in mind (probably the most pleasing to the eye).http://sudokuone.com/sweb/old/a_em.htm
No question, that i like the SK loop, but did it solve any puzzle?

From my perspective, generally moves are what are memorable, rather than solutions except that is for solutions consisting only of memorable moves, or solutions based on a memorable idea (such btw as yours in the initial post to this thread)
aran
 
Posts: 334
Joined: 02 March 2007

Re: An alternative way to solve the hard 17 clue

Postby eleven » Wed Jun 02, 2010 8:37 am

Allans graphics look very nice, but for non experts in base/cover they hide the loop more than show it. So personally i prefer something like this:
Code: Select all
+-------------------------+-------------------------+-------------------------+
| 1       48+7    34578   | 3567    3689    5678    | 3489    6+39    2       |
| 2+38    9       7+38    | 4       12368   12678   | 38+1    5       38+6    |
| 23458   48+2    6       | 1235    12389   1258    | 7       1+39    3489    |
+-------------------------+-------------------------+-------------------------+
| 2468    5       1478    | 9       1246    3       | 128     1267    678     |
| 234689  12468   13489   | 126     7       1246    | 123589  12369   35689   |
| 2369    1267    1379    | 8       5       126     | 1239    4       3679    |
+-------------------------+-------------------------+-------------------------+
| 7       1+48    14589   | 1235    12348   12458   | 6       39+2    3459    |
| 45+6    3       45+1    | 12567   1246    9       | 2+45    8       7+45    |
| 45689   6+48    2       | 3567    3468    45678   | 3459    39+7    1       |
+-------------------------+-------------------------+-------------------------+

So its easy to see the SK loop itself (counter clockwise, starting in box 1)
48-(16)45-(27)39-(16)/38-(27)/48
And you can eliminate in the boxes the numbers common to the minirow and -column (box 1 27),
and in the (middle minirow/column of the) middle boxes the numbers common to the outer minicolums/rows (column 2 48).
eleven
 
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Joined: 10 February 2008

Re: An alternative way to solve the hard 17 clue

Postby aran » Wed Jun 02, 2010 9:57 am

In that view, how do you readily deal with the case where there is no starting pair ?
aran
 
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Re: An alternative way to solve the hard 17 clue

Postby eleven » Wed Jun 02, 2010 11:05 am

I dont know more than this basic SK loop. Can you give me a link please ?
eleven
 
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Re: An alternative way to solve the hard 17 clue

Postby aran » Wed Jun 02, 2010 12:05 pm

I mean that the loop is fine for the case PAIR-....-PAIR ie 48r13c2=>48r13c2 or 27r13c2=>27r13c2.
That is if r13c2=48 or if r13c2=27 then we know where we stand and all those eliminations can be made.
But r13c2 may well be (at this stage of the proceedings) {48}{27} ie one from each.
So we have to deal with this - the NOT PAIR case - before any conclusions can be drawn.
That was the whole difficulty for the chain approach, or so it seemed to me.
Whereas base/cover deals with the whole thing at once, which is why I particularly admire it.
aran
 
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