The New Sudoku Players' Forum

[rliu]

Here is the o/s grid after applying all my previous solving techniques:

Code: Select all

7     5     1     | 8     4     9     | 2     6     3
9     6     8     | 12    12    3     | 7     4     5
3     4     2     | 6     5     7     | 18    18    9
------------------+-------------------+-----------------
1     27    357   | 25    78    4     | 38    9     6
256   9     3457  | 125   78    126   | 348   1258  12
256   8     45    | 3     9     126   | 14    125   7
------------------+-------------------+-----------------
25    127   57    | 9     3     8     | 6     12    4
4     12    6     | 7     12    5     | 9     3     8
8     3     9     | 4     6     12    | 5     7     12


I solved it by using the following new (at least to me) technique:
In the rectangle r5c6, r5c9, r9c6 and r9c9, if r9c9 is 2, both r5c9 and
r9c6 are 1. Then, in Box 5 (middle box), there are no empty cells with
candidate 1 left. Hence, r9c9 cannot be 2.

However, this technique seems to be a bit of trial-and-error. Is this
puzzle can be solved by other techniques?

[MCC]

I wouldn't call this T&E but a splendid piece of logic.

But you could use an Unique Rectangle (UR) and simple colouring to arrive at the same answer.

Code: Select all

7     5     1     | 8     4     9     | 2     6     3
9     6     8     | 12    12    3     | 7     4     5
3     4     2     | 6     5     7     | 18    18    9
------------------+-------------------+-----------------
1     27    357   | 25    78    4     | 38    9     6
256   9     3457  | 125   78    126*  | 348   1258  12*
256   8     45    | 3     9     126   | 14    125   7
------------------+-------------------+-----------------
25    127   57    | 9     3     8     | 6     12    4
4     12    6     | 7     12    5     | 9     3     8
8     3     9     | 4     6     12*   | 5     7     12*

The *'s point out the UR, if r5c6 is not a 6 then each of r5c6, r5c9, r9c6 and r9c9 has the candidates 1,2.

This will give two solutions:
r5c6=1, r5c9=2, r9c6=2 and r9c9=1 or,
r5c6=2, r5c9=1, r9c6=1 and r9c9=2.
and mean the grid will not have an unique solution.
Therefore r5c6 must be a 6.

Simple colouring on the 1's:

Code: Select all

7     5     1     | 8     4     9     | 2     6     3
9     6     8     | 12    12    3     | 7     4     5
3     4     2     | 6     5     7     | 18    18    9
------------------+-------------------+-----------------
1     27    357   | 25    78    4     | 38    9     6
256   9     3457  | 1+25  78    6     | 348   1258  1-/+2
256   8     45    | 3     9     1-26  | 14    125   7
------------------+-------------------+-----------------
25    127   57    | 9     3     8     | 6     12    4
4     12    6     | 7     12    5     | 9     3     8
8     3     9     | 4     6     1+2   | 5     7     1-2

If r5c4 is a 1 then the 1 in r5c9 sees both a (+) an a (-), contradiction, therefore r5c9 cannot be a 1.


MCC

[rliu]

The *'s point out the UR, if r5c6 is not a 6 then each of r5c6, r5c9, r9c6 and r9c9 has the candidates 1,2.

This will give two solutions:
r5c6=1, r5c9=2, r9c6=2 and r9c9=1 or,
r5c6=2, r5c9=1, r9c6=1 and r9c9=2.
and mean the grid will not have an unique solution.
Therefore r5c6 must be a 6.

According to the solution from pappocom, r5c6 is 1. In fact, I used the
similar reasoning to solve other puzzles. However, when I filled in 6 in
r5c6 to solve this one, I found I was wrong!

[Hud]

I think the UR is invalid since the candidates reside in 4 different boxes.

There's a naked quad in row 5 (1256) which can eliminate those in the rest of row 5.

[rliu]

There's a naked quad in row 5 (1256) which can eliminate those in the rest of row 5.

Thanks. I missed that one - finding quad is not good for me.

[Pat]

at 51:

Code: Select all

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

r5 duo 3,4
gives r5 7

[rliu]

r5 duo 3,4
gives r5 7

Never think of it. I learnt a new way on solving. Thank you very much.

[re'born]

There is a standard trick for this type of grid called remote naked pairs (a type of coloring but terribly easy to spot)

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7     5     1     | 8     4     9     | 2     6     3
9     6     8     | 12A   12a   3     | 7     4     5
3     4     2     | 6     5     7     | 18    18    9
------------------+-------------------+-----------------
1     27    357   | 25    78    4     | 38    9     6
256   9     3457  | 125-  78    126   | 348   1258  12a
256   8     45    | 3     9     126   | 14    125   7
------------------+-------------------+-----------------
25    127   57    | 9     3     8     | 6     12    4
4     12    6     | 7     12A   5     | 9     3     8
8     3     9     | 4     6     12a   | 5     7     12A


An easy argument shows that r2c4 and r5c9 do not have the same value. Hence, r5c4 <> [12], i.e., r5c4 = 5. The rest is easy from here.

[rliu]

There is a standard trick for this type of grid called remote naked pairs (a type of coloring but terribly easy to spot)

I heard of coloring but I haven't paid any attention on it. It seems the use of it here makes this problem easy to solve. I will put this technique in my tool box. Many thanks.

[re'born]

Incidentally, your solving technique is an example of an alternating inference chain. In the words of the creator, Myth Jellies, "AIC's are all guaranteed to be pattern-based, theoretical, and not brute force." So you needn't worry about your method being T & E. You should read the definitions at the beginning of that thread before reading my explanation.

Code: Select all

7     5     1     | 8     4     9     | 2     6     3
9     6     8     | 12    12    3     | 7     4     5
3     4     2     | 6     5     7     | 18    18    9
------------------+-------------------+-----------------
1     27    357   | 25    78    4     | 38    9     6
256   9     3457  | 125C  78    126C  | 348   1258  12B
256   8     45    | 3     9     126D  | 14    125   7
------------------+-------------------+-----------------
25    127   57    | 9     3     8     | 6     12    4
4     12    6     | 7     12    5     | 9     3     8
8     3     9     | 4     6     12E   | 5     7     12A


A1=B1-C1=D1-E1=A1

At least one of the two endpoints of an AIC is 'true'. In your case, the two endpoints are the same point, hence A=1. A shorter way to get an equivalent result is

Code: Select all

7     5     1     | 8     4     9     | 2     6     3
9     6     8     | 12    12    3     | 7     4     5
3     4     2     | 6     5     7     | 18    18    9
------------------+-------------------+-----------------
1     27    357   | 25    78    4     | 38    9     6
256   9     3457  | 125A  78    126A  | 348   1258  12-
256   8     45    | 3     9     126B  | 14    125   7
------------------+-------------------+-----------------
25    127   57    | 9     3     8     | 6     12    4
4     12    6     | 7     12    5     | 9     3     8
8     3     9     | 4     6     12C   | 5     7     12D


A1=B1-C1=D1

As either A=1 or D=1, we have r5c9 <> 1.

Historically, I think this would be an example of 'grouped coloring' (which is why the cells r5c46 are grouped in all of the above AIC's.

[rliu]

Thank you for pointing me to the post. I tried to use AIC to solve a puzzle. It makes me a little bit dizzy (switching true, false, true, ...). Though I cannot completely solve it (it has more than 1 solutions), I can further reduce it. Without AIC, I think I cannot go further. I have posted that puzzle for help at http://forum.enjoysudoku.com/viewtopic.php?t=4994