The New Sudoku Players' Forum

[urhegyi]

2020-07-28 (2).png (133.49 KiB) Viewed 985 times

The left picture is the same as the right one with 1 more extra clue , the 7 at R6C4. I would think giving one clue extra makes the sudoku easier to solve but in this case hodoku rates the left one Hard(1016) and the right one Hard(990), as you can see on the picture. Any ideas how this is possible?

[Ajò Dimonios]

Hi urhegyi.


After basics the two schemes (without r6c4 = 7 and with r6c4 = 7) lead to the following situation:

Code: Select all

+---------------+-------------+----------+
| 479 6    247  | 1   279 8   | 3  27 5  |
| 5   239  237  | 379 4   237 | 8  6  1  |
| 8   1    237  | 5   237 6   | 4  27 9  |
+---------------+-------------+----------+
| 2   7    9    | 6   1   34  | 5  8  34 |
| 6   34   8    | 2   5   9   | 7  1  34 |
| 34  5    1    | 37  8   347 | 6  9  2  |
+---------------+-------------+----------+
| 79  8    5    | 4   279 1   | 29 3  6  |
| 1   2349 2347 | 379 6   237 | 29 5  8  |
| 39  239  6    | 8   239 5   | 1  4  7  |
+---------------+-------------+----------+


Code: Select all

+---------------+-----------+----------+
| 479 6    247  | 1  279 8  | 3  27 5  |
| 5   239  237  | 39 4   27 | 8  6  1  |
| 8   1    237  | 5  237 6  | 4  27 9  |
+---------------+-----------+----------+
| 2   7    9    | 6  1   34 | 5  8  34 |
| 6   34   8    | 2  5   9  | 7  1  34 |
| 34  5    1    | 7  8   34 | 6  9  2  |
+---------------+-----------+----------+
| 79  8    5    | 4  279 1  | 29 3  6  |
| 1   2349 2347 | 39 6   27 | 29 5  8  |
| 39  239  6    | 8  239 5  | 1  4  7  |
+---------------+-----------+----------+


The 7 in r6c4 does not make a great contribution, the difficulty is about the same. Both schemes can be solved easily with two finned fishes.
Finned x-wing, candidate 3; r3; r9; finned r9c12 = 3 => - 3r8c3
and
Finned x-wing, candidate 9; r1; r9; finned r9c1 = 9 => - 9r7c1 => stte

[tarek]

A sudoku solver will work according to how it was programmed and there could be instances like your 2nd example where different techniques are used that give the impression that a puzzle is more difficult when it shouldn't.

Paolo has nicely demonstrated that the 2nd example isn't more difficult. Other solvers may show that it had become easier depending on how they were programmed.

Logically, there is more information in the 2nd example and therefore should be simpler!

Tarek

[eleven]

Long time ago such a puzzle was found, see here.

[Leren]

Try these two puzzles in Hodoku:

.9...2...6..4....3..5..628....8....4.1...........27........341...65.43.2..9.1..76
.9...2...6..4....3..5..628....8....4.1...........27........341...65.43.24.9.1..76

The second puzzle is the same as the first, but has an extra clue 4 in r9c1. The Hodoku score I got for the first puzzle is 390 and for the second 2816.

The reason is, of course, that the first puzzle uses a UR and in the second puzzle the UR is blocked by adding the extra clue. (I see that eleven has found an old similar example.)

In the first puzzle you "know" that both r3c1 = 3, r3c2 = 4, r9c1 = 4 , r9c1 = 3 and r3c1 = 4, r3c2 = 3, r9c1 = 3 , r9c1 = 4 can't lead to a valid solution, so you can use a UR for an "easy" solution.

When you put 4 in r9c1 you "lose" this information and can't use the "easy" solution.

This just illustrates the fact that the "UR information" you use in the first puzzle is, strictly speaking, not logically deducible from the clues. It relies on the "convention" that a puzzle has only one solution.

Leren

[tarek]

Long time ago such a puzzle was found, see here.

These examples always expose the weaknesses in the automated resolution rules. The missing digits in uniqueness patterns was a prime example. Because we are trained in using the resolution rules given to us, any deviation from them can make the solution path appear less obvious or more difficult. The fact is: (uniqueness or not) by adding a clue you would have more information than before & you may choose to revert to the previous sate if you need to, so logically, the puzzle can't be more difficult

[yzfwsf]

For the computer solver, it is easy to implement the UR rule, which is to detect whether there are redundant clues when initializing the puzzle.The problem is that the human solver is not so easy to find out the redundant clues, so the question comes, is it meaningful to implement such a function in the program?

[Ajò Dimonios]

leren wrote:
Try these two puzzles in Hodoku:

.9...2...6..4....3..5..628....8....4.1...........27........341...65.43.2..9.1..76
.9...2...6..4....3..5..628....8....4.1...........27........341...65.43.24.9.1..76

The second puzzle is the same as the first, but has an extra clue 4 in r9c1. The Hodoku score I got for the first puzzle is 390 and for the second 2816.

The reason is, of course, that the first puzzle uses a UR and in the second puzzle the UR is blocked by adding the extra clue. (I see that eleven has found an old similar example.)

In the first puzzle you "know" that both r3c1 = 3, r3c2 = 4, r9c1 = 4 , r9c1 = 3 and r3c1 = 4, r3c2 = 3, r9c1 = 3 , r9c1 = 4 can't lead to a valid solution, so you can use a UR for an "easy" solution.

When you put 4 in r9c1 you "lose" this information and can't use the "easy" solution.

This just illustrates the fact that the "UR information" you use in the first puzzle is, strictly speaking, not logically deducible from the clues. It relies on the "convention" that a puzzle has only one solution.

You can easily refute what Hodoku reports . Having an extra clue cannot make the scheme more difficult because it is in contradiction with the fact that it is always possible to adopt a method of solving the difficulty identical to that of the scheme with a clue in less (always with single solution) simply using the elimination of the candidate as the first resolution move.

Paolo

[Leren]

Well I did find (using my own solver) an easy solution to my second puzzle using the ALS-XZ rule eliminating a different candidate. Hodoku didn't find this easy solution, so maybe I need to re-tune it someway.

So I used eleven's example instead:

8..2.5..1...1.3.....3.7.8..63.....75..8...2..91.....48..5.9.1.....7.6...3..5.2..9
8..2.5..1...1.3.....3.798..63.....75..8...2..91.....48..5.9.1.....7.6...3..5.2..9

The puzzles are identical except that the second puzzle has an extra clue 9 in r3c6.

The UR based OTP solution for the first puzzle is :

Code: Select all

*---------------------------------------------------*
| 8    4679   679 | 2     46   5  | 3467 369  1     |
| 2457 245679 679 | 1     8    3  | 4567 2569 2467  |
| 1    2456   3   |*49+6  7   *49 | 8    256  246   |
|-----------------+---------------+-----------------|
| 6    3      4   | 8     2    1  | 9    7    5     |
| 57   57     8   |*49+36 346 *49 | 2    1    36    |
| 9    1      2   |#36    5    7  | 36   4    8     |
|-----------------+---------------+-----------------|
| 247  2467   5   | 4-3   9    8  | 1    236  23467 |
| 24   2489   19  | 7     134  6  | 345  2358 234   |
| 3    4678   167 | 5     14   2  | 467  68   9     |
*---------------------------------------------------*

UR Type 3 (43 +36) r35c46 + r6c4 => - 3 r7c4; stte

The post basics position for the second puzzle is:

Code: Select all

*----------------------------------------------*
| 8    4679   679 | 2  46   5 | 3467 369  1    |
| 2457 245679 679 | 1  8    3 | 4567 2569 2467 |
| 1    2456   3   | 46 7   *9 | 8    256  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    |
|-----------------+-----------+----------------|
| 247  2467   5   | 34 9    8 | 1    236  2367 |
| 24   2489   19  | 7  134  6 | 345  2358 234  |
| 3    4678   167 | 5  14   2 | 467  68   9    |
*----------------------------------------------*

Can anyone see an easy solution from there ?

Leren

[champagne]

The post basics position for the second puzzle is:

Code: Select all

*----------------------------------------------*
| 8    4679   679 | 2  46   5 | 3467 369  1    |
| 2457 245679 679 | 1  8    3 | 4567 2569 2467 |
| 1    2456   3   | 46 7   *9 | 8    256  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    |
|-----------------+-----------+----------------|
| 247  2467   5   | 34 9    8 | 1    236  2367 |
| 24   2489   19  | 7  134  6 | 345  2358 234  |
| 3    4678   167 | 5  14   2 | 467  68   9    |
*----------------------------------------------*

Can anyone see an easy solution from there ?

Leren

Hi Leren,
No doubt that the first puzzle with the uniqueness property prevent to put r1c4=4. This not any more true here, and you have to produce a tougher logical rule.
This is not an answer to your question, but shows again that the uniqueness property is source of much shorter paths.
Using it or not to solve a puzzle is a matter of taste. this has been long discussed.

[eleven]

The fact is: (uniqueness or not) by adding a clue you would have more information than before & you may choose to revert to the previous sate if you need to, so logically, the puzzle can't be more difficult

Having an extra clue cannot make the scheme more difficult because it is in contradiction with the fact that it is always possible to adopt a method of solving the difficulty identical to that of the scheme with a clue in less (always with single solution) simply using the elimination of the candidate as the first resolution move.

No.
If you remove a clue from a unique puzzle, you cannot know, if it still has a single solution. You would have to prove it, before you use the UR. But to prove it, you must solve it without uniqueness techniques.
So the puzzle with the extra clue is definitely harder.

[Leren]

Just to explain why adding a clue can sometimes produce the (seemingly crazy) result that the puzzle becomes harder.

Using eleven's puzzle, when you add the extra clue you 1. Add some information ie that r3c6 = 9. and 2. You also remove some "information" ie that r3c4 = 4, r3c6 = 9, r5c4 = 9, r5c6 = 4 can't be part of a valid solution.

So how could you know that 2. was True in the original puzzle ? Well, strictly speaking, you can't know this a priori, but you trust that the puzzle maker has a puzzle with just 1 solution and that can't happen if 2. was False.

Notice that I put the removed "information" in inverted commas. This is because it's not strictly information that you can deduce. Assuming it's validity requires what I call a "Leap of Faith" - ie that the puzzle maker has made a puzzle with a unique solution. Of course it's not such a big leap of faith. That's because unique solution puzzles are overwhelmingly popular. In fact on this site there is a thread called Sudokus of Shame here, which is a list of sites that have been known to produce multiple solution puzzles, so they are definitely not popular on this forum. Nevertheless, it puts uniqueness moves in a different class of proof than other moves.

Leren

[Leren]

To answer the original question, the quirky scores for the two puzzles just means that they are in the noise area of the difficulty scores for the moves used.

The main quirk I can see is that the fewer clue puzzle uses a Swordfish - worth 150 points and the more clues puzzle uses an X Wing - worth 140 points.

Well, you could argue all day about the relative difficulty score of different moves, but a Swordfish has 3 Strong links whereas an X Wing has only 2.

The stingy 10 point difference between the two moves seems a bit rich, or is that poor, to me.

Also the fewer clue puzzle uses an Empty Rectangle whereas the more clues puzzle uses a Skyscraper. Both moves use 2 Strong links but the Skyscraper is worth 10 more points. Shock horror !

Leren

[champagne]

To tell it in a different way, we are here in the example of "leren" in a very classical link between an unavoidable set of size 4 and a UR.

Assuming that the solution would have the UR forbidden pattern eg:4949, one of the four must be a given. This is the unavoidable set.
If the puzzle has a unique solution with none of the 2 possible values of the unavoidable set, then we have a UR.

With no given in the UR area, both possibilities are forbidden;
With one given in the UR area, one of the 2 options remains possible.

As a consequence, adding a clue, we loose part of the information available in the original puzzle.
That's it.

[tarek]

Here is another link to add to the discussion from Andrew Stuart about Avoidable Rectangles:
https://www.sudokuwiki.org/Avoidable_Rectangles
Here you need to know that the puzzle has a single solution & that the information you are taking comes from a solved cell and not from a given clue