## New technique?

Advanced methods and approaches for solving Sudoku puzzles

### New technique?

First off, I should say that I came across this in a non-"standard" puzzle, but I think it's worth bringing up here as it may be a new technique.

Consider the following partial grid:

Code: Select all
`17* 935 *46**6 47* 3*1*34 16* 57*6** 324 *174*3 716 **5217 859 463769 243 158*** 681 7****1 597 6**`

Looking at row one, we have a 2/8 pair at C3R1 and C7R1
Looking at the centre block, we have a 2/8 pair at C6R2 and C6R3.
Looking at row three, we have 8/9 at C1R3 the 2/8 pair at C6R3, and 2/9 at C9R3.

Ignore any other possible candidate reductions from the rest of the grid, and just look at the distribution of 2,8 & 9 in rows two and three.

We have a "right handed triangle" of 2's, and a "left handed triangle" of 8's, with the "common axis" of the triangle being a 2/8 pair, and the two other points of the triangle having a common third candidate (the 9). The pair on the common axis is the same pair as that currently unresolved in the remaining row.

Whichever way you try out the 2/8 pair in column 6, the top row is forced to the same solution.

Two questions I suppose.

1. Am I just describing an existing technique in a different way (remember I'm ignoring any other potential candidate reduction from the rest of the grid here)?
2. Would you consider this to be a "technique", or do you consider this as "guesswork"?
whohe

Posts: 32
Joined: 28 May 2005

I did a quick test with another puzzle and followed the logical conclusions from two assumptions (two 29 pairs).

Following one of the assumptions means I run into trouble after a while, but when I compare the results with the finished solution I noticed that some of the numbers were the same. So, to paraphrase your original post :

"Whichever way you try out the [a/b] pair in column [Xi], [cell XpYq] is forced to the same solution."

While this may be so, I would be hard (if at all possible) to derive a generic logical rule / algorithm from such an occurrence.

Until such a rule is formulated I would consider this guesswork.

The question is : are the "triangles" you describe a cause, an effect or a rationalisation after guesswork?
Hedgegarlic

Posts: 1
Joined: 22 June 2005

whohe wrote:Am I just describing an existing technique in a different way

It sort of resembles nishio in the sense that the correct arrangement is uncovered quickly and with little effort. But instead of searching for a contradiction -- as nishio does -- your technique seeks to reveal that there is only one arrangement possible. I had a delightful ah-ha moment when I followed your logic and understood the 2 could only be at r1c3 and the 8 could only be at r1c7.

Here is another way of looking at this. Consider only the cells r1c3, r1c7, r3c1, and r3c9. Place an 8 in r1c3. That makes r3c1 a 9, which in turn makes r3c9 a 2, which in turn makes r1c7 an 8. Because of the contradiction of two 8's in row 1, r1c3 must be a 2.

Where did you find this puzzle?
scrose

Posts: 322
Joined: 31 May 2005

### Re: New technique?

whohe wrote:
Code: Select all
`17* 935 *46**6 47* 3*1*34 16* 57*6** 324 *174*3 716 **5217 859 463769 243 158*** 681 7****1 597 6**`

This is solved by a technique I call "extended coloring", which is an extension of the coloring algorithm.

In the standard coloring algorithm, you pick a value and then look on the grid for all possible rows, column, boxes where the value can be placed in only two cells, and color them accordingly.

In the extended coloring algorithm, you don't limit to a single value. Instead, you look for every case where two possibilities are excluding each other; and you don't color the cells, but the pairs cell/value.

So in the example problem one would do:

cell (1,3) can only contain 2 and 8, therefore:
- (1,3)=2 A
- (1,3)=8 B
row 1 can contain 2 in only two places, therefore
- (1,7)=2 B (forced)
row 1 can contain 8 in only two places, therefore
- (1,7)=8 A (forced)

cell (2,6) can contain only 2 and 8, therefore:
- (2,6)=2 C
- (2,6)=8 D
box 2 can contain 2 in only two places, therefore:
- (3,6)=2 D (forced)
box 2 can contain 8 in only two places, therefore:
- (3,6)=8 C (forced)

row 3 can contain 8 in only two places, therefore:
- (3,1)=8 D (forced)
row 3 can contain 2 in only two places, therefore:
- (3,9)=2 C (forced)

now look at box 1. Here you have
(1,3)=8 B
(3,1)=8 D
these two possibilities exclude each other. Therefore B and D cannot be true at the same time.

now look at box 3. Here you have
(1,7)=2 B
(3,9)=2 C
these two possibilities exclude each other. Therefore B and C cannot be true at the same time.

Therefore both C and D exclude B. But one of C and D must be true; therefore B must be false, and A true, which means we can assume true all possibilities to which we assigned color A:
(1,3)=2 A
(1,7)=8 A
Nick70

Posts: 156
Joined: 16 June 2005

scrose wrote:
whohe wrote:Am I just describing an existing technique in a different way

It sort of resembles nishio in the sense that the correct arrangement is uncovered quickly and with little effort. But instead of searching for a contradiction -- as nishio does -- your technique seeks to reveal that there is only one arrangement possible. I had a delightful ah-ha moment when I followed your logic and understood the 2 could only be at r1c3 and the 8 could only be at r1c7.

Here is another way of looking at this. Consider only the cells r1c3, r1c7, r3c1, and r3c9. Place an 8 in r1c3. That makes r3c1 a 9, which in turn makes r3c9 a 2, which in turn makes r1c7 an 8. Because of the contradiction of two 8's in row 1, r1c3 must be a 2.

Where did you find this puzzle?

I realised about the approach by just considering the 4 cells you mentioned, but it was the "pattern" from the triangles that got me thinking about whether or not this would qualify as guesswork or a technique.

The puzzle was one from the Bunnydoku program for Pocket PC's.
whohe

Posts: 32
Joined: 28 May 2005

(r1c3 and r1c7 have candidates 28.
r3c1 and r3c9 have candidates 89 and 29, respectively,
and r3c5 has candidates 28.)

I look at such situations as follows:
in row 3, there are just two possibilities for r3c1-r3c5-r3c9,
8-2-9, and 9-8-2. In row 1, there obviously
are only 2-8 and 8-2 for r1c3-r1c7.

These candidates have to be combined such that box 1 contains
only one 8, and box 3 contains only one 2. That leaves only
two possible combinations:
2-8 for r1c3-r1c7, 8-2-9 for r3c1-r3c5-r3c9
and
2-8 for r1c3-r1c7, 9-8-2 for r3c1-r3c5-r3c9
Thus, there is a 2 in r1c3, and an 8 in r1c7.

Thomas
rrabbit

Posts: 9
Joined: 22 June 2005