## An alternative way to solve the hard 17 clue

Advanced methods and approaches for solving Sudoku puzzles

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

aran wrote:So we have to deal with this - the NOT PAIR case - before any conclusions can be drawn.

Ah, i see what you mean. I know, that it has been proved, that it is sufficient to find a loop like i described it to make the eliminations.

The proof can be made by case distinction.
Either you have the loop described or the "opposite" loop as you said or none of the both. In the 3rd case it can be shown, that (for each pair like 48 and 27) always one of the pair numbers must be in the minirow and the other in the minicolumn (or opposite minirow/column resp.), leading to the same eliminations.

So i may use the loop like above (in the same way i use a swordfish).
eleven

Posts: 1907
Joined: 10 February 2008

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

Yes that is true.
For example not pair r13c2=> one (and only one) of 2/7in r13c2 which working round the clock shows that one (and hence the other) of that two-some must be in r2c13.
So either there's a pair 27r13c2 or 2 and 7 lie on the cross r13c2+r2c13. Hence <27> outside the cross.
But I wouldn't want to take that for granted in a loop looking like it might be SK. So I'd have to do the work.
The nice thing about base/cover, as mentioned, is that it does all that at once.
Of course those bases and covers have to be located, it's not as if they just fall from the sky, so there is work.
But once found, it looks (to me) more satisfying than having to consider pair loops, then not-pair loops.
And again the symmetry within the 16 base sets of the first Allan Barker presentation (1267r2/r8/c2/c8) remains spectacular (ps crucial to there being precisely 16 base sets is the fact that there is no overlap, which is of course what drives the loop, so yes these things do all tie up)
Last edited by aran on Wed Jun 02, 2010 2:53 pm, edited 1 time in total.
aran

Posts: 334
Joined: 02 March 2007

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

aran wrote:But I wouldn't want to take that for granted in a loop looking like it might be SK. So I'd have to do the work.

Of course the proof is (can be made) general for any pair loop over these cells (just replace the numbers by variables).
The nice thing about base/cover, as mentioned, is that it does all that at once.
Of course those bases and covers have to be located, it's not as if they just fall from the sky, so there is work.

base/cover is a more general approach, as i have seen and i dont doubt its benefits. I just dont know much about it and have no practice at all, how to find the sets. For the special case of SK loops, they are easy to spot with my method.
And again the symmetry within the base sets of the first Allan Barker presentation (1267r2/r8/c2/c8) remains spectacular.

Yes.
eleven

Posts: 1907
Joined: 10 February 2008

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

eleven wrote : Of course the proof is (can be made) general for any pair loop over these cells (just replace the numbers by variables).
Agreed (despite which I'd probably check the non-pair loop should I ever come across one of those animals)
base/cover is a more general approach, as i have seen and i dont doubt its benefits. I just dont know much about it and have no practice at all, how to find the sets. For the special case of SK loops, they are easy to spot with my method.

Part of the trick is locating non-overlapping sets (btw I was editing my previous post to put in this very point when your last post arrived).
For loops non-overlapping sets are crucial, just as overlapping sets are hopeless for loops (the overlap means that the number of truths is not fixed at the outset, since it all depends on what happens in the overlaps, equivalent to "almost" logic).
In this grid we have the rectangle of cells r2c28 / r8c28 assigned with givens 9538. This is a very strong hint that the-therefore-non-overlapping r2r8c2c8 rows and columns are of great interest with respect to the remaining candidates (ie excluding those givens) particularly so (in fact this may be a necessary condition) as those givens include no repeats.
=> consider 12467 in r28c28
but 4 already appears in r2 and c8, so reducing its potential.
=>consider 1267 in r28c28.
aran

Posts: 334
Joined: 02 March 2007

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

I believe that I did something like what follows. I probably used some sort of AIC argument. Perhaps it is more clear just upon inspection?

Code: Select all
`  *-----------* |1.3|.2.|.45| |.5.|4.1|.23| |2.4|563|1..| |---+---+---| |..2|.54|3.1| |537|186|492| |41.|..2|5..| |---+---+---| |.4.|61.|23.| |.2.|...|.14| |3.1|24.|.56| *-----------*   `
Let v be the vertices of an oddagon. We are using grouped vertices in box 3. g marks the guardian cells for this oddagon. Note that three candidates, (789) share this guardian restriction.

Code: Select all
` *-----------* |1g3|.2.|v45| |.5.|4.1|v23| |2v4|563|1vv| |---+---+---| |.g2|.54|3.1| |537|186|492| |41.|..2|5..| |---+---+---| |.4.|61.|23.| |.2.|...|g14| |3v1|24g|v56| *-----------*`

Each of (789) must fit into g. (6) is already locked into (g). We know then that exactly one of (789) exists at g. Consider that r1c6 is already limited to one of (789). We can label it Z. r1c6 => r8c45, and together they will see all the g's but one, so, Z's will exist as follows. Note we have a group node for Z in box 8.

Code: Select all
`*-----------* |1g3|.2Z|v45| |.5.|4.1|v23| |2v4|563|1vv| |---+---+---| |.Z2|.54|3.1| |537|186|492| |41.|..2|5..| |---+---+---| |.4.|61.|23.| |.2.|ZZ.|g14| |3v1|24g|v56| *-----------*`

Since (3)XW exists at r68c45, If Z is either one of (79), we have UR (3Z) at r68c45. Therefor, Z is (8). ste.
SteveK
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