The New Sudoku Players' Forum

[ravel]

For a longer time i wondered, if it is possible, that a puzzle becomes harder, when another clue is added. It was clear for me that it might be possible, when the additional clue would destroy a uniqueness pattern. Now i think, i found such an example, a puzzle posted by tso in January:

Code: Select all

8 . . | 2 . 5 | . . 1
. . . | 1 . 3 | . . .
. . 3 | . 7 . | 8 . .
------+-------+------
6 3 . | . . . | . 7 5
. . 8 | . . . | 2 . .
9 1 . | . . . | . 4 8
------+-------+------
. . 5 | . 9 . | 1 . .
. . . | 7 . 6 | . . .
3 . . | 5 . 2 | . . 9


A UR (type 3) in r35c46 allows to eliminate 4 from r35c4 and the puzzle is easily solved. But if you add a 9 in r3c6, you cannot use this UR and the puzzle is very hard.
Is that correct?

[Ruud]

I tested your theory with SudoCue.

The version with 28 clues you posted scores 1409 in difficulty rating.
The version with 9 in r2c6 (29 clue) scores 3719 in difficulty rating.

Ruud.

[vidarino]

Huh. It seems you nailed one, ravel.

Ruud: I'm not sure of the scale of your difficulty measurements, but the original puzzle seems awfully over-scored to me. Nothing but singles, a box/line elimination and the mentioned Unique Rectangle are needed. (There are a couple of finned fishes to be found just before the UR, but they're not needed.)

The one with the extra clue is indeed much harder.

Hmm, it makes sense, though, since the Uniqueness techniques relies on an ambiguous pattern, and adding that clue removed that ambiguousity. Puzzling.

ravel: By the way, you're describing a type 4 UR. It is a type 3, too, however; the extra candidates in R35C4 (6 and 36) form a pair with the 36 in R6C4, eliminating 3 from R7C4. The effect is the same, though; the rest of the puzzle falls to singles.

[ravel]

Ruud and Vidar, thanx for the quick confirmations.

By the way, you're describing a type 4 UR.

Yes, many type 3 UR's are also type 4 (and normally i spot them as type 4 easier)

[Havard]

Hmm, it makes sense, though, since the Uniqueness techniques relies on an ambiguous pattern, and adding that clue removed that ambiguousity. Puzzling.

Makes you think about what the Uniqueness really is! Is it not so that we are trying to avoid the "deadly pattern" by all means? Could we not then say that:

Code: Select all

8      4679   679    | 2      46     5      | 3467   369    1
2457   245679 679    | 1      468    3      | 4567   2569   2467
1245   2456   3      | 46     7      9      | 8      256    246
---------------------+----------------------+---------------------
6      3      4      | 8      2      1      | 9      7      5
57     57     8      | 369    356    4      | 2      136    36
9      1      2      | 36     356    7      | 36     4      8
---------------------+----------------------+---------------------
247    2467   5      | 34     9      8      | 1      236    23467
124    2489   19     | 7      134    6      | 345    2358   234
3      4678   167    | 5      14     2      | 467    68     9


In r3c46 and r5c46 we have a Possible Deadly Pattern (PDP) of:

Code: Select all

4---9
|   |
9---4


In other words, if you took out the 6 from r3c4 and the 36 from r5c4 you would get this pattern...? And would we not be able to perform that same type 3 elimination with those two as "quantum" cells combined with r6c4 and make the elimination in r7c4?

Is not the whole uniqeness-theory about avoiding these patterns? You can do the elimination, because you might end up with one of these?

I am really confused now, so can someone please tell me wheter this is just bullshit or not?

Havard

[RW]

In r3c46 and r5c46 we have a Possible Deadly Pattern (PDP) of:

Code: Select all

4---9
|   |
9---4


In other words, if you took out the 6 from r3c4 and the 36 from r5c4 you would get this pattern...? And would we not be able to perform that same type 3 elimination with those two as "quantum" cells combined with r6c4 and make the elimination in r7c4?

Not if number 9 in r3c6 would be a given clue. That's why we cannot make the reduction and the puzzle is harder with one more clue. Very nice example!

RW

[keith]

Havard,

The uniqueness logic does not apply if you have not reduced the possibilities to a minimum set. (From experience.)

In this example, R5C4 = <9>, it is pinned to be so. Then, R3C4 = <6> to avoid the deadly pattern is the correct observation, I believe.

Keith

Edit: But I agree, if one of the solved squares is given in the initial pattern, UR logic should not be used*.

It would be interesting if someone modified their solver to look for the deadly pattern when the rectangle has solved squares (not given in the initial pattern).

* Or, you can cheat: Knowing that the puzzle is made by adding an initial value to a valid puzzle, the UR elimination will give you the correct answer, even though the logic to apply it is unsound.

[ravel]

Havard,

a unique rectangle is a deadly pattern, when a solution with 4-9-4-9 also always allows a solution with 9-4-9-4, what means, that there are at least 2 solutions for the puzzle.
This is (as RW said), only the case, when none of the numbers in the 4 cells are given, and when the rectangle lies in 2 boxes only (otherwise another 4 or 9 in one of the 4 boxes could fix the pattern to a single possibility).

[Animator]

For those of you who wonder where it gets harder:

Code: Select all

8      47     69     | 2      46     5      | 367    39     1
57     2457   69     | 1      8      3      | 67     259    246
1      245    3      | 46     7      9      | 8      25     246
---------------------+----------------------+---------------------
6      3      4      | 8      2      1      | 9      7      5
57     57     8      | 9      36     4      | 2      1      36
9      1      2      | 36     5      7      | 36     4      8
---------------------+----------------------+---------------------
24     6      5      | 34     9      8      | 1      23     7
24     9      1      | 7      34     6      | 5      8      23
3      8      7      | 5      1      2      | 4      6      9

(A possible move: color the number 6)

[Smythe Dakota]

.... it makes sense, though, since the Uniqueness techniques relies on an ambiguous pattern, and adding that clue removed that ambiguousity. Puzzling. ....

A math purist would probably say that uniqueness techniques should not be considered valid, and that part of the task in solving any puzzle is to establish uniqueness.

Bill Smythe

[Ruud]

A math purist would probably say that uniqueness techniques should not be considered valid, and that part of the task in solving any puzzle is to establish uniqueness.

Many websites and newspapers that publish Sudokus explicitly state that "every puzzle has a unique solution that can be found by logic".

A math purist would IMO stick to the problem definition and not try to prove something given as an axiom.

Ruud.

[tarek]

Against a set of techniques that you are using now, a puzzle may get harder as you add a clue......

But consider 2 things:

1. New techniques are discovered with time, & that particular puzzle that gets more difficult with a set of known techniques would then become simpler....

2. Remeber that by adding a clue, you are 1 step closer towards solving the puzzle.

tarek

[vidarino]

I'd like to add that you don't get any less information by adding that clue, though, despite that being what it looks like.

Without the extra clue, you have the information that the rectangle in question does *not* contain 4-9-9-4 or 9-4-4-9, and therefore you use that information to make a deduction.

With the added clue, however, you lose this information, replacing it with the knowledge about a single cell. Therefore you can not make the above deduction any more, and have to find alternative ways around the obstacle. The alternative way being much harder is coincidental.

[RW]

This seems to be a quite common phenomenom. This SOLO-unreasonable can be solved with one UR and singles:

Code: Select all

 *-----------*
 |2..|9..|...|
 |.3.|.51|82.|
 |.5.|...|..7|
 |---+---+---|
 |.49|...|.6.|
 |.6.|8.2|.4.|
 |.7.|...|58.|
 |---+---+---|
 |6..|...|.5.|
 |.93|28.|.7.|
 |...|..4|..9|
 *-----------*


You would have to add r5c9=3, r6c1=3 or r6c9=1 to the givens, to satisfy the title "unreasonable".

And a similar BUG-lite example I found in another unreasonable (this might be something for the benchmarklist, Ruud):

Code: Select all

 *-----------*
 |..8|...|..6|
 |93.|...|.2.|
 |.62|.1.|9..|
 |---+---+---|
 |.8.|9.2|7..|
 |.7.|...|.9.|
 |..9|7.6|.1.|
 |---+---+---|
 |..7|.5.|46.|
 |.9.|...|.35|
 |1..|...|8..|
 *-----------*


Starting grid with candidates:

Code: Select all

   
 *-----------------------------------------------------------------------------*
 | 457     145     8       | 2345    23479   34579   | 135     457     6       |
 | 9       3       145     | 4568    4678    4578    | 15      2       1478    |
 | 457     6       2       | 3458    1       34578   | 9       4578    3478    |
 |-------------------------+-------------------------+-------------------------|
 | 3456    8       13456   | 9      *34      2       | 7       45     *34      |
 | 23456   7       13456   | 13458  *348     13458   |-2356    9      *348+2   |
 | 2345    245     9       | 7      *348     6       |-235     1      *348+2   |
 |-------------------------+-------------------------+-------------------------|
 | 238     2       7       | 1238    5       1389    | 4       6      -129     |
 | 2468    9       46      | 12468   24678   1478    | 12      3       5       |
 | 1       245     3456    | 2346    234679  3479    | 8       7      -279     |
 *-----------------------------------------------------------------------------*


Make the very obvious BUG-lite reductions and the whole puzzle can be solved with nothing but singles in no time. If we added r4c5=3 or r4c9=4 to the given clues, the puzzle would be a lot harder. If we wanted, we could even preserve symmetry by adding r5c5=4 and the puzzle would still be in a totally different league than the initial puzzle.

RW

[vidarino]

Hmm, another side-effect of this phenomenon is that you could theorhetically find a hard puzzle, remove a clue, and make it easier.

However, you'd have to somehow verify that it still has a unique solution after the removal. Otherwise the use of uniqueness techniques would be invalid.