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Neuer (?) Fund über Sudoku-Geometrie
Ich habe es gestern auch gelöst und gebe Ulrich recht, es ist auch ohne "das Argument" zu lösen. Man trägt nur längere Zeit keine Zahlen ein. Bin aber eher bei 4 Sternen, also schwer. Wundert mich das etwa?.... Nicht wirklich. ;-)
(13.04.2020, 13:48)uvo schrieb: Es geht auch ohne jegliches T&E, und hat meines Erachtens auch einen sehr schönen Lösungsweg - stell es ruhig ins Portal, dann skizziere ich meinen Lösungsweg als versteckten Kommentar.

Gemessen an den sonstigen Portalrätseln finde ich es auch nicht übermäßig schwer, sicher nicht mehr als drei Sterne.
Das klingt großartig. Ich wäre ziemlich an einer Lösung ohne den Trick interessiert. Nachdem ich prinzipiell keine Rätsel ins Portal stelle, von denen ich die Lösung nicht kenne, hab ich nur eine kleine Bitte an dich: Könntest du mir bitte eine Lösungsskizze per PN schicken?
Klar, gerne.
Wow, a new sudoku technique ?
That's pretty impressive, given that I don't think lot of progress has been done concerning sudoku solving technique since 2006.
I'm not sure that it's really new, meaning that this technique would help you to solve very hard classic. My intuition seems to tell me that as the argumentation is simple, it could be transpose in simple argument for single cells, too, i.e. easy technique.
But I'm not sure and if someone achieve to solve for example a weekly unsolvable: using this technique along with basic technique, I would be very impressed.
I would like to generalize this technique for all latin squares (regions doesn't play a role if you formulate a bit differently) and choice of x rows and y columns.
The technique can be seen as dissections of the grid in 4 sets of cells, and it is interesting to see how the latin square rules act.

Make a choice of x rows and y columns. Let's take your example: rows 1289 and columns 1289. Let's say the Red (R ) set of cells for intersecting cells of these rows and columns.
Then you can define the Green (G) set as intersecting cells of complementary rows and complementary columns (rows and colmuns 34567 in our example).
Then the set R and G are the same up to a number of whole set of digits 1-9 (Let's call this set S).
But we can also define the Yellow (Y) set: intersecting cells of chosen rows and complementary columns (in our example rows 1289 and columns 34567),
And Blue (B) set as intersecting cells of complementary rows with chosen columns (rows 34567 columns 1289).
Then we have the same conclusion: Y=B mod(S) Yellow set of cells contain same digits as Blue set, modulo whole set of digits 1-9.

It is very easy to understand: R+Y give complete rows R+Y=nS and G+Y give complete columns G+Y=mS, then R=G+(n-m)S or R=G modulo S. Same for Y and B: Y=B modulo S.
The number of combination of different dissections in the grid is impressively high (15'876 combinations if you chose 4 rows and 4 columns, but you can chose any number of rows and any number of columns). This is why probably this technique is not practicable for human solvers (You can't check such big number of dissections).
You can make fun dissections of the grid:
Like first example: Red set of digits is the same as Green set, and Yellow set contains Blue set + complete set of digits 1-9.

Another funny observation is that you can build X-wing and fish techniques (swordfish, jellyfish) with this technique:
If a digit is not in the red set, it is not in the green set (X-wing), in other words, it appears twice in yellow set.

Same for swordfish here.

That is a really cool generalization and an elegant proof! It is a neat observation that one can show all of the fish techniques with this argument.
Thank you, Fred! :-)
I agree that the technique is not necessarily helpful in a classic sudoku. I tried to apply it to some of the Weekly Unsolvables, but it never seemed like I actually gained any useful information from it. I think it is rather a different perspective on a classic sudoku than an advanced technique. Still, it might be useful in a killer sudoku.
(12.05.2020, 16:43)cdwg2000 schrieb: Maybe this killer can apply this.

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Thanks for sharing the puzzle! The trick can be applied in this puzzle to conclude that the two 15-cages have to be 456-triples, as the cells in the four corner squares cannot contain any digit higher than 6. This is nice, thank you! Smile
(11.04.2020, 17:00)Phistomefel schrieb: 大家好,


[Bild: bild.php?data=be82f74e-6191-3030303343332d31]
8 * 45-ABCD。
8 * 45蓝色区域。
因此我们得到:蓝色区域中的数字总和= A + B + C +D。

[Bild: bild.php?data=b2a6bb64-6192-3030303343332d32]
为了显示索赔,请考虑第3行和第7行,第3行和第7列以及四个角的3x3块。为了便于描述事实,将列和行简称为线。现在,让“ n”为1到9之间的一个数字。然后,“ n”必须在4行和4个块中的每一个中恰好出现一次。因此,如果“ n”出现在

因此,只有绿色单元格才能覆盖它们覆盖的行数和框数。因此,每当“ n”出现在红色单元格中时,“ n”也必须出现在蓝色或橙色单元格中,反之亦然。由于这适用于1到9之间的每个数字,因此红色区域必须包含与蓝色和橙色区域相加的数字相同的数字。


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(02.08.2020, 02:28)光碟2000 schrieb:
(11.04.2020, 17:00)hist schrieb: 大家好,


[Bild: bild.php?data=be82f74e-6191-3030303343332d31]
1.将第1、2、8和9行以及第1、2、8和9列相加,并从中到深色红色的笼A, B,C和D,即
8 * 45-
ABCD。2 . 将所有3x3块加到中间的一个,然后再连接中间的蓝色区域,即
8 * 45蓝色区域。
因此我们得到了:蓝色区域中的,数字的总和= A + B + C +D。

[Bild: bild.php?data=b2a6bb64-6192-3030303343332d32]
为了显示,请考虑第3行和第7行,第3行和第7列以及四个角的现在,让“ n”为1到9之间的一个数字。然后,“ n”必须在4行和4个块中的每个一个中。恰好出现一次。因此,如果“ n”出现在



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