The New Sudoku Players' Forum

[JeJ]

Today I was playing a "hard" game generated by Hodoku which I was able to take to:

Code: Select all

.--------------.-------------.-------------.
| 9    8   16  | 3   26   7  | 4    15  25 |
| 3    2   67  | 15  46   45 | 167  8   9  |
| 5    4   167 | 9   268  18 | 67   3   27 |
:--------------+-------------+-------------:
| 1    57  8   | 4   3    6  | 9    2   57 |
| 2    57  4   | 8   1    9  | 3    57  6  |
| 6    9   3   | 7   5    2  | 8    4   1  |
:--------------+-------------+-------------:
| 47   1   9   | 2   47   3  | 5    6   8  |
| 478  6   2   | 15  478  45 | 17   9   3  |
| 78   3   5   | 6   9    18 | 2    17  4  |
'--------------'-------------'-------------'


Until I got stuck. I asked for a clue and the program sugested a hidden rectangle that I wasn't able to find, so when I asked to show the step it said: "Hidden Rectangle: 6/7 in r2c37,r3c37 => r2c3<>6" and shows in green (valid) both values on r3c7 as well as the 6 and 7 on r3c3 and r2c7 as well as the 7 in r2c3, producing a red (elimination) 6 on r2c3.

I tested both values of r3c7 and if that cell is 7, then r2c7 would be 6 and therefore r2c3 could not be 6. If r3c7 were a 7, then r3c3 could not be 7 so therefore r2c3 would have to be 7. Even though both values lead to the elimination of 6 on r2c3, I don't see the hidden rectangle, it seems to me more like a "forcing chains" method. Am I right or is hodoku telling the truth?

[David P Bird]

Examining the 1s in r2 & r3, they must either occupy r1c4 and r3c3 or r2c7 and r3c6 so there can't be a (67)UR:r23c37. Every inference chain based on the UR that I can find is merely an alternative way of proving that either (1)r3c3 or (1)r2c7 must be true.

For me the (6)r2c3 exclusion follows from this chain which proves that if (7)r2c3 is false (6)r1c3 must be true and vice versa.

(7)r2c3 = (7-1)r2c7 = (1)r1c8 - (1=6)r1c3 => r2c3 <> 6

[daj95376]

From my notes on UR patterns: x, y, and z represent possible extra candidates

Code: Select all

===== ===== ===== ===== Hidden Unique Rectangle
                        (*) bivalue w/ 2x strong links in an "L" pattern

+--------------+
|  .   .   .   |
| 12x  .  1y-2 |<  strong link on <1>
|  .   .   .   |
+--------------+
|  .   .   .   |
|*12   .  12z  |
|  .   .   .   |
+--------------+
           ^       strong link on <1>


Code: Select all

 +-----------------------------------------------------+
 |  9    8    16   |  3    26   7    |  4    15   25   |
 |  3    2   *67   |  15   46   45   |  67+1 8    9    |
 |  5    4    67+1 |  9    268  18   | *67   3    27   |
 |-----------------+-----------------+-----------------|
 |  1    57   8    |  4    3    6    |  9    2    57   |
 |  2    57   4    |  8    1    9    |  3    57   6    |
 |  6    9    3    |  7    5    2    |  8    4    1    |
 |-----------------+-----------------+-----------------|
 |  47   1    9    |  2    47   3    |  5    6    8    |
 |  478  6    2    |  15   478  45   |  17   9    3    |
 |  78   3    5    |  6    9    18   |  2    17   4    |
 +-----------------------------------------------------+
 # 33 eliminations remain


There is a strong link in [r2] and [c3] on <7> with a bivalue <67> in r3c7. This would lead to r2c3<>6 from a Hidden Unique Rectangle. My solver returned the following UR eliminations as present:

Code: Select all

 r23c37  <67> UR via s-link              <> 6    r2c3,r3c3
 r23c37  <67> UR via s-link              <> 7    r2c7,r3c3

 r78c15  <47> UR via s-link              <> 4    r7c5,r8c1
 r78c15  <47> UR via s-link              <> 7    r7c1,r8c5


[David P Bird]

My previous post was made just before I went to bed and reading it again I see I failed to make my concluding point.

It seems that Hoduku is using UR logic to prove the inferences between the 1s in the top tier of boxes, but those inferences are direcly available without needing to invoke the UR pattern.

[ronk]

Today I was playing a "hard" game generated by Hodoku which I was able to take to:

Code: Select all

.--------------.-------------.-------------.
| 9    8   16  | 3   26   7  | 4    15  25 |
| 3    2   67  | 15  46   45 | 167  8   9  |
| 5    4   167 | 9   268  18 | 67   3   27 |
:--------------+-------------+-------------:
| 1    57  8   | 4   3    6  | 9    2   57 |
| 2    57  4   | 8   1    9  | 3    57  6  |
| 6    9   3   | 7   5    2  | 8    4   1  |
:--------------+-------------+-------------:
| 47   1   9   | 2   47   3  | 5    6   8  |
| 478  6   2   | 15  478  45 | 17   9   3  |
| 78   3   5   | 6   9    18 | 2    17  4  |
'--------------'-------------'-------------'


Until I got stuck. I asked for a clue and the program sugested a hidden rectangle that I wasn't able to find, so when I asked to show the step it said: "Hidden Rectangle: 6/7 in r2c37,r3c37 => r2c3<>6" and shows in green (valid) both values on r3c7 as well as the 6 and 7 on r3c3 and r2c7 as well as the 7 in r2c3, producing a red (elimination) 6 on r2c3.

Well, it does match hobiwan's "example on the right" here, but IMO it's better suited to solving without pencilmarks. With two UR cells without extra candidates and two bilocal links (7r2 and 7c3), there are enough strong links to make an inclusion (placement), rather than an exclusion (elimination).

In NL notation: r2c3 =7= AUR(67)r23c37:[r2c7 =1= r3c3] =7= r2c3 ==> r2c3=7
In AIC notation: (7)r2c3 = (7-1)r2c7 =[(67)AUR:r23c37]= (1-7)r3c3 = (7)r2c3 ==> r2c3=7

Conjecture: For an almost-uniqueness-rectangle with N cells with extra candidates, only N-1 bilocal links ... properly located between AUR cells and involving only the UR digits ... are required to make a valid exclusion.

Using only one of the available bilocal links above at a time, for AUR(67)r23c37 ...

with 7r2 => r3c3<> 7
with 7c3 => r2c7<>7
with 6c7 => r3c3<>6

[JeJ]

Well thanks for all the answers, this UR thing really involves more that what is seen at first.