The New Sudoku Players' Forum

[udosuk]

This puzzle is adopted from one of Pyrrhon's failed attempts...



Could be of interest to anyone who likes Hanji/Chinese characters...

PS: It could assist a little if you know the meaning of the characters...

Here is the list of the characters in Chinese font (if your computer is compatible with the language):

前列在陣者鬥兵皆臨 (not in order)...

[nj3h]

Wouldn't the puzzle with the pencil marks entered be really festive?

[tso]

Translation of single characters out of context can be iffy.

Chinese to English translation from Babelfish:

前 -- Front
列 -- Row
在 -- In
陣 -- ??
者 -- ??
鬥 -- Fights
兵 -- Soldier
皆 -- All
臨 -- Near

Japanese to English translation from Babelfish:

前 -- Before
列 -- Line
在 -- Resident in
陣 -- Position
者 -- Person
鬥 -- ??
兵 -- Soldier
皆 -- Everyone
臨 -- Overlooking

[udosuk]

前 -- Front
列 -- Arrangement / List / Line / Row
在 -- In
陣 -- Array
者 -- Person
鬥 -- Fighting
兵 -- Soldiers
皆 -- All / Everyone
臨 -- Approaching / Arriving / Near

If you understand Chinese/Japanese, you should notice the middle column (c5) contains a famous sentence reading from top to bottom... Knowing this sentence, the puzzle can be solved with singles only, without the need of candidate marking.

If you don't understand the language, you'd have to replace the symbols by digits or positioned dots and the puzzle would be a bit more difficult...

[udosuk]

I should have said "Kanji" (the Japanese term for "Chinese characters") instead of "Hanji" (the Korean term)... In Chinese it's spelt "Hanzi" (漢字)...

It's all very confusing... (Confuciusing?)

[Pyrrhon]

I guess I haven't solved the translation act. How works the shorter way?

Pyrrhon

[udosuk]

Okay, the 9 characters, when arranged in a particular order, will form a famous Buddhist/Zen/Ninja incantation phrase... Putting this phrase on the middle column (c5) from top to bottom will reduce the puzzle to singles only...

There is info in "another place" about where to find this phrase...

[udosuk]

This is the puzzle in numerical format:

Code: Select all

.........
1.47.83.5
.2..3..9.
7.83.95.6
.........
24.86.97.
.62.73.49
..3..4..2
.........

After basic moves (singles/pairs/locked candidates) you need a complex turbot fish/multiple colouring, a naked quad/hidden pair, an x-wing, and finally an xy-wing to solve it... But of course all these are not necessary if you know the middle column from the characters...

[Pyrrhon]

You can solve it with naked and hidden singles, one box-line interaction, two naked paira and one xyz-wing (the almost naked pair is in B8 (R9C5, R7C4) and the helping cell is R1C5).

All the other techniques you used are not necessary.

Pyrrhon

[udosuk]

You can solve it with naked and hidden singles, one box-line interaction, two naked paira and one xyz-wing (the almost naked pair is in B8 (R9C5, R7C4) and the helping cell is R1C5).

All the other techniques you used are not necessary.

Pyrrhon

I couldn't see the xyz-wing...

Code: Select all

 *-----------------------------------------------------------*
 | 3     58    67    | 149  *19    56    | 2     18    1478  |
 | 1     9     4     | 7     2     8     | 3     6     5     |
 | 58    2     67    | 14    3     56    | 478   9     1478  |
 |-------------------+-------------------+-------------------|
 | 7     1     8     | 3     4     9     | 5     2     6     |
 | 6     3     9     | 2     5     7     | 48    18    148   |
 | 2     4     5     | 8     6     1     | 9     7     3     |
 |-------------------+-------------------+-------------------|
 | 58    6     2     |*15    7     3     | 18    4     9     |
 | 9     78    3     | 16    18    4     | 1678  5     2     |
 | 4     578   1     | 569  *89    2     | 678   3     78    |
 *-----------------------------------------------------------*

Could you explain a little please? Thanks!

[Pyrrhon]

You are right and we can take the turbot chain with strong links R7C1-R7C7 in R7, R3C1-R1C2 in B1, and R1C8-R5C8 in C8 0> R5C7 <> 8 and a x-wing of 8 with strong links in R3 R7 and weak links in C1 C7. Another way is the simple forcing loop

R1C5 -> R1C8 -> R1C2 -> R9C2 -> R9C5 -> R1C5

It follows that:

R1C45 <> 1 because of R1C58
R1C9 <> 8 because of R1C82
R9C79 <> 8 because of R9C25

[udosuk]

Awww... My head is blowing up... I can't follow a bit of that "simple forcing loop"... But one thing is certain, in the solution r1c5=1 so your assertion that "r1c45<>1" is clearly false...

Are you deliberately writing false things to mess with my brain?

[Pyrrhon]

I need better glasses (its to hot in Germany): From the forcing loop follows that R1C5=1 or R1C8 = 1 and so R1C4 <> 1 and R1C9 <> 1.

[udosuk]

Let's recap... This is the state you can achieve with basic techniques:

Code: Select all

 3     58    67    | 149   19    56    | 2     18    1478
 1     9     4     | 7     2     8     | 3     6     5   
 58    2     67    | 14    3     56    | 478   9     1478
-------------------+-------------------+------------------
 7     1     8     | 3     4     9     | 5     2     6   
 6     3     9     | 2     5     7     | 48    18    148 
 2     4     5     | 8     6     1     | 9     7     3   
-------------------+-------------------+------------------
*58    6     2     |*15    7     3     | 18    4     9   
 9    -78    3     | 16   *18    4     | 1678  5     2   
 4     578   1     | 569   89    2     | 678   3     78   

There is a simple xy-wing in r7c14 and r8c25 => r8c2<>8, therefore r8c2=7...

Pyrrhon's "simple forcing loop" works like this:

Code: Select all

 3    *58    67    | 149  *19    56    | 2    *18    1478
 1     9     4     | 7     2     8     | 3     6     5   
 58    2     67    | 14    3     56    | 478   9     1478
-------------------+-------------------+------------------
 7     1     8     | 3     4     9     | 5     2     6   
 6     3     9     | 2     5     7     | 48    18    148 
 2     4     5     | 8     6     1     | 9     7     3   
-------------------+-------------------+------------------
 58    6     2     | 15    7     3     | 18    4     9   
 9     7     3     | 16    18    4     | 168   5     2   
 4    *58    1     | 569  *89    2     | 678   3     78   

[r1c5]-1-[r1c8]-8-[r1c2]=58=[r9c2]-8-[r9c5]=9=[r1c5]

It follows that:
r1c49=1|8 (together with r1c8) will force r1c5=9 & r1c2=5 => r9c2=8 => r9c5=9, so r1c4<>1, r1c9<>1|8
r9c79=8 will force r9c5=9 & r9c2=5 => r1c2=8 => r1c8=1 => r1c5=9, so r9c79<>8

Alternatively, you can work on the following turbot chain:

Code: Select all

 3    *58    67    | 149   19    56    | 2    *18    1478
 1     9     4     | 7     2     8     | 3     6     5   
*58    2     67    | 14    3     56    | 478   9     1478
-------------------+-------------------+------------------
 7     1     8     | 3     4     9     | 5     2     6   
 6     3     9     | 2     5     7     |-48   *18    148 
 2     4     5     | 8     6     1     | 9     7     3   
-------------------+-------------------+------------------
*58    6     2     | 15    7     3     |*18    4     9   
 9     7     3     | 16    18    4     | 168   5     2   
 4     58    1     | 569   89    2     | 678   3     78   

[r5c7]-8-[r7c7]=8=[r7c1]=8=[r3c1]=8=[r1c2]-8-[r1c8]=8=[r5c8]-8-[r5c7]

It follows directly that r5c7<>8 and thus r5c7=4... And then a naked quad on r3 and an x-wing on 8 solves it...