The New Sudoku Players' Forum

by **Condor** » Mon Jul 04, 2005 4:09 am

Recently while trying to solve one problem I came up with something that I realised could help as a general solving technique for sudokus.

It relys on a characteristic of all rows of boxes. (and columns) To show this technique, I will first show the characteristic, then an example.

I will represent the first row as: XXX YYY ZZZ, where the order of the digits in each triplet is ignored. With this in mind, as soon as a number, represented by one of those letters, is placed in 2 of the boxes, the box and row for the third digit is immediately decided.

The second row depends on how many digits are common between the first triplet and YYY. There are four possibilities.

First, no digits in common. That means the first triplet is ZZZ. Filling in the rest of rows 2 and 3 gives:

XXX YYY ZZZ
ZZZ XXX YYY
YYY ZZZ XXX

This shows that each triplet reading from left to right, moves 1 box right - 1 row down. If it moves below the third row, it raps around to the first row.

Next, all three digits in common. Filling in the same way gives:

XXX YYY ZZZ
YYY ZZZ XXX
ZZZ XXX YYY

This time the triplets move 1 box right - 1 row up.

1 or 2 digits in common is slightly harder. 2 of the digits move together as a pair while the other 1 moves in the opposite direction. I will represent the single digit with the letters Q, R and S. Note: To make it a bit easier to read the next grid I have written those letters last. Q, R and S could just as easily represent the first or second digits in the triplet.

First 1 digit in common.

XXQ YYR ZZS
ZZR XXS YYQ
YYS ZZQ XXR

Remember, the order in each triplet is ignored. XXQ could just as easily be written as XQX, or QXX. Likewise all the other triplets.

This time 2 of the numbers moves 1 box right - 1 row down, while the other number do the opposite, 1 box right - 1 row up.

Finally 2 digits in common.

XXQ YYR ZZS
YYS ZZQ XXR
ZZR XXS YYQ

Now for an example. Consider:

374 965 812
569 821 347
281 374 569

648 593 271
157 682 493
923 417 685

732 148 956
416 259 738
895 736 124

Take the second row of boxes:

648 593 271
157 682 493
923 417 685

Showing just the pairs:

6.8 .93 .71
1.7 68. .93
9.3 .17 68.

And now just the singles:

.4. 5.. 2..
.5. ..2 4..
.2. 4.. ..5

We see each pair (68, 17, 39) moving right 1 box - down 1 row and the singles doing the opposite moving right 1 box - up 1 row.

In the example above, the first row of boxes contain triplets, (347,569,128) while the third row contains pairs, (37,14, and 59) and singles (2,6,and 8).

Likewise, the columns. The first column of boxes contain pairs (25, 78, and 19) and singles (3, 6, and 4). The second column contain pairs (89, 27, and 45) and singles (3, 6, and 1). And the third column contain pairs (35, 46, and 79) and singles (8, 1, and 2).

Thus by following the numbers, we can see if a number is a single or pair and know which row it will be on. This gets easier as more digits are found.

by **Condor** » Sun Jul 17, 2005 9:44 pm

I should have given an example to show how this method works, so here are 2.

I will start with an easy one and show how following the numbers can help solve it.

Example 1

Code: Select all: 1 8 7 6 4 2 9 5 3 2 9 4 7 3 5 8 6 1 6 3 5 8 9 1 2 4 7 . 4 . 2 8 6 . 1 . . . . 1 7 4 . . 2 . 1 . 9 5 3 . 8 4 . . 3 4 6 9 1 2 8 4 6 1 3 2 8 . . . 9 2 8 5 1 7 4 3 6

A 7 can only exist in box 6 in column 7, so (8,8)=7.

Now lets look at the right stack with the 7 added, and the pattern it matches.

Code: Select all: 9 5 3 X Y Z 8 6 1 X Y Z 2 4 7 Q R S . 1 . Z X Y . . 2 Z X Y . 8 4 R S Q 1 2 8 Y Z X . 7 . Y Z X 4 3 6 S Q R

Column 8 box 9 has the digits 237 and column 9 box 3 has the digits 137. 3 and 7 are in both boxes so are represented by ZZ and Q represents 2. In column 7 box 3 are the digits 289. 2 is a single so 8 and 9 are a pair represented by XX.

There is an XX in column 8 box 6 and column 9 box 9, so 8 and 9 are in those 2 places. Inserting gives:

Code: Select all: 9 5 3 X Y Z 8 6 1 X Y Z 2 4 7 Q R S . 1 . Z X Y . 9 2 Z X Y . 8 4 R S Q 1 2 8 Y Z X . 7 9 Y Z X 4 3 6 S Q R

We can add in (8,7)=5 and (4,9)=5.

Thus:
XX represents 8 and 9
YY represents 4 and 5
ZZ represents 3 and 7

Q represents 2
R represents 6
S represents 1

Example 2

Code: Select all: . 8 4 . . . . 2 . . . 9 . . . 7 . . 7 . . . 2 . . . 6 1 . . 7 5 3 8 6 4 . . . 8 1 2 . . . 8 3 . 9 4 6 . . 7 5 . . . 9 . . . 8 . . 7 . . . 3 . . . 1 . . . . 9 7 .

The puzzle has just been started with (4,7)=8, (4,6)=3, (6,4)=9, and (5,5)=1 added.

Now lets look at the middle band and the pattern it matches.

Code: Select all: 1 . . 7 5 3 8 6 4 . . . 8 1 2 . . . 8 3 . 9 4 6 . . 7 X X Q Y Y R Z Z S Z Z R X X S Y Y Q Y Y S Z Z Q X X R

Row 4 box 6 has the digits 468 and row 6 box 5 has the digits 469, so 4 and 6 are represented by ZZ and S represents 8 and Q represents 9.

Also the digit 8 goes right one box up one row as you go from left to right. 7 does the same so must be a single. In row 4 box 5 we have the digits 357, 7 is a single so 3 and 5 are a pair. In row 6 box 4 is a 3 so we can now enter 5 - Thus (6,3)=5.

Thus:
XX represents . and .
YY represents 3 and 5
ZZ represents 4 and 6

Q represents 9
R represents 7
S represents 8

Only two digits remain so XX represents 1 and 2.

We can now say the following:
Row 4 box 4 contains the digits 129 in some order.
Row 5 box 4 contains the digits 467 in some order.

Row 5 box 6 contains the digits 359 in some order.
Row 6 box 6 contains the digits 127 in some order.

by **Mike Barker** » Thu Mar 29, 2007 10:33 pm

It looks like Mr. Hamilton's travelling pairs/triples and David P Bird's Braid Analysis have been around a lot longer that I thought. Based on these posts I'd say credit should go to Condor as the discoverer (at least until someone comes up with an earlier post). I wish I had the time to go back and mine all the gold I'm sure exists within this forum. I know I enjoy the old posts that Pat uncovers on fish.

by **r.e.s.** » Fri Mar 30, 2007 3:09 am

Interesting find.

For reference, I've posted a link to the present thread in the following one on "Braid Analysis" (on a different forum, having discussion by both T. G. Hamilton and D. P. Bird):
http://www.sudoku.org.uk/cgi-bin/discus/show.cgi?tpc=29&post=4735#POST4735

MrHamilton's thread here on this forum is at
http://forum.enjoysudoku.com/viewtopic.php?t=3522

by **Condor** » Sat Apr 21, 2007 12:07 am

I've been looking at the threads on travelling pairs and braid analysis, catching up on what other people have developed. While it is interesting, the way it is applied seems to me to be a little complicated, so I thought I would show how I use this technique to solve a puzzle and hopefully it would make it easier for other people.

I will use the following puzzle to illustrate this technique. (Published New Zealand Herald 3 April 2007. Rated easy) There is a couple of methods I use.

Code: Select all: *-----------------------* | . . . | . . . | . 8 . | | 2 4 . | 7 . . | 9 . 1 | | 6 . 1 | 5 . . | . . 7 | |-------+-------+-------| | . . 8 | 1 . . | . 4 . | | . . 3 | 2 . 7 | 6 . . | | . 1 . | . . 3 | 8 . . | |-------+-------+-------| | 5 . . | . . 1 | 2 . 8 | | 8 . 2 | . . 6 | . 5 4 | | . 3 . | . . . | . . . | *-----------------------*

First, I write an '/' and '\' to the right of each band (just above the rows) with a line between them. I also do likewise under each stack. It then looks like this.

Code: Select all: *-----------------------* / | \ | . . . | . . . | . 8 . | | | 2 4 . | 7 . . | 9 . 1 | | | 6 . 1 | 5 . . | . . 7 | | |-------+-------+-------| / | \ | . . 8 | 1 . . | . 4 . | | | . . 3 | 2 . 7 | 6 . . | | | . 1 . | . . 3 | 8 . . | | |-------+-------+-------| / | \ | 5 . . | . . 1 | 2 . 8 | | | 8 . 2 | . . 6 | . 5 4 | | | . 3 . | . . . | . . . | | *-----------------------* / / / ------------------------ \ \ \

It helps to have some blank space on the page for this purpose. I use a cardboard overlay with a rectangle cutout to photocopy the puzzles in the paper.

The '/' to the right of the bands is for the numbers that travel diagonally up as they move from boxrow to boxrow from left to right. The '\' is for those that move diagonally down.

Next I write the numbers for which I can identify as how they are travelling, in the same row as the box they are next to. With the columns, I write the numbers in the same column they would be in the bottom box. Thus:

Code: Select all: *-----------------------* / | \ | . . . | . . . | . 8 . | | | 2 4 . | 7 . . | 9 . 1 | | 1 | 6 . 1 | 5 . . | . . 7 | | 7 |-------+-------+-------| / | \ | . . 8 | 1 . . | . 4 . | | | . . 3 | 2 . 7 | 6 . . | | 1 | . 1 . | . . 3 | 8 . . | | 8 |-------+-------+-------| / | \ | 5 . . | . . 1 | 2 . 8 | | 2 8 | 8 . 2 | . . 6 | . 5 4 | 5 | | . 3 . | . . . | . . . | | *-----------------------* / / / 1 3 7 1 8 ------------------------ 2 4 \ \ \

From this we can see that the left stack has pairs and singles, but can not tell which are which yet. Also in the bottom band we can see that 5 is a single while 2 and 8 are a pair.

With some solving, I arrived at the following position.

Code: Select all: *-----------------------* / | \ | . . . | . 1 . | . 8 . | | | 2 4 . | 7 . . | 9 . 1 | | 1 | 6 . 1 | 5 . . | . . 7 | | 7 |-------+-------+-------| / | \ | . 2 8 | 1 . . | . 4 . | | 4 3 | 4 5 3 | 2 8 7 | 6 1 9 | | 1 6 | . 1 6 | . . 3 | 8 . . | | 8 2 |-------+-------+-------| / | \ | 5 6 . | . . 1 | 2 3 8 | | 2 8 | 8 . 2 | . . 6 | 1 5 4 | 5 | 1 | 1 3 . | . . . | 7 9 6 | | 6 *-----------------------* / 5 6 / / 7 6 1 3 7 1 1 9 8 ------------------------ 8 2 4 \ \ \

I now noticed that r6c8=7 (and therefore r6c9=2). As boxrow 66 contains the same numbers as boxrow 55, (box number first) we now know that the middle band contains triples. Because row 5 is filled in, we can see that the triples are 453, 287, and 619. from this we can immediately fill in most of the rest of the band to arrive at the following:

Code: Select all: *-----------------------* / | \ | . . . | . 1 . | . 8 . | | | 2 4 . | 7 . . | 9 . 1 | | 1 | 6 . 1 | 5 . . | . . 7 | | 7 |-------+-------+-------| / | \ | 7 2 8 | 1 6 9 | . 4 . | | 4 3 5 | 4 5 3 | 2 8 7 | 6 1 9 | = | 1 6 9 | 9 1 6 | 4 5 3 | 8 7 2 | | 8 2 7 |-------+-------+-------| / | \ | 5 6 . | . . 1 | 2 3 8 | 3 | 2 8 | 8 . 2 | . . 6 | 1 5 4 | 5 | 1 | 1 3 . | . . . | 7 9 6 | | 6 *-----------------------* / 5 6 / / 7 6 1 3 4 7 1 1 9 8 ------------------------ 8 2 5 4 \ \ 6 \

I have put an '=' in the left side to indicate the band contains triples. From noticing 1 number I was able to place 8 numbers straight into the grid.

From the information in the middle stack we can see 1 and 7 are singles while 5 and 6 are a pair. Looking at the middle box we can immmediately deduce that the other two pairs are 2&4, and 9&3, and the remaining single is 8. Placing these numbers gives:

Code: Select all: *-----------------------* / | \ | . . . | 6 1 . | . 8 . | | | 2 4 . | 7 3 8 | 9 . 1 | | 1 | 6 . 1 | 5 9 . | . . 7 | | 7 |-------+-------+-------| / | \ | 7 2 8 | 1 6 9 | . 4 . | | 4 3 5 | 4 5 3 | 2 8 7 | 6 1 9 | = | 1 6 9 | 9 1 6 | 4 5 3 | 8 7 2 | | 8 2 7 |-------+-------+-------| / | \ | 5 6 . | 9 4 1 | 2 3 8 | 3 | 2 8 | 8 . 2 | 3 7 6 | 1 5 4 | 5 | 1 | 1 3 . | 8 2 5 | 7 9 6 | | 6 *-----------------------* / 5 6 / / 7 6 1 3 4 8 7 1 1 9 8 ------------------------ 8 2 9 2 5 4 \ \ 3 4 6 \

Completing the puzzle gives:

Code: Select all: *-----------------------* / | \ | 3 7 9 | 6 1 2 | 4 8 5 | 8 | 4 5 | 2 4 5 | 7 3 8 | 9 6 1 | 9 | 1 6 | 6 8 1 | 5 9 4 | 3 2 7 | 2 | 7 3 |-------+-------+-------| / | \ | 7 2 8 | 1 6 9 | 5 4 3 | | 4 3 5 | 4 5 3 | 2 8 7 | 6 1 9 | = | 1 6 9 | 9 1 6 | 4 5 3 | 8 7 2 | | 8 2 7 |-------+-------+-------| / | \ | 5 6 7 | 9 4 1 | 2 3 8 | 3 | 2 8 | 8 9 2 | 3 7 6 | 1 5 4 | 5 | 1 4 | 1 3 4 | 8 2 5 | 7 9 6 | 9 | 6 7 *-----------------------* / 5 6 7 / / 7 3 6 1 3 4 8 7 1 1 9 8 ------------------------ 8 9 2 9 2 5 2 5 4 \ \ 3 4 6 \

Being an easy puzzle, this technique looked a bit more effective than in harder puzzles, but is an example of what is possible.

The second method I use is just an abridged version of the above. I just write the singles to the left of the rows and above the columns. The same puzzle would look like this when finished:

Code: Select all: 2 8 9 7 1 8 4 2 5 *-----------------------* 9 | 3 7 9 | 6 1 2 | 4 8 5 | 2 | 2 4 5 | 7 3 8 | 9 6 1 | 8 | 6 8 1 | 5 9 4 | 3 2 7 | |-------+-------+-------| | 7 2 8 | 1 6 9 | 5 4 3 | = | 4 5 3 | 2 8 7 | 6 1 9 | | 9 1 6 | 4 5 3 | 8 7 2 | |-------+-------+-------| 5 | 5 6 7 | 9 4 1 | 2 3 8 | 9 | 8 9 2 | 3 7 6 | 1 5 4 | 3 | 1 3 4 | 8 2 5 | 7 9 6 | *-----------------------*

I would recomend getting familiar with the first method before moving on to the second.

by **ronk** » Sat Apr 21, 2007 12:58 am

Nice explanation. I too use traveling pairs/triples and find it helpful even if I've just overlooked a hidden single somewhere.

A couple of comments:

Condor wrote:
Code: Select all
*-----------------------* / | \ | . . . | . . . | . 8 . | | | 2 4 . | 7 . . | 9 . 1 | | 1 | 6 . 1 | 5 . . | . . 7 | | 7 |-------+-------+-------| / | \ | . . 8 | 1 . . | . 4 . | | | . . 3 | 2 . 7 | 6 . . | | 1 | . 1 . | . . 3 | 8 . . | | 8 |-------+-------+-------| / | \ | 5 . . | . . 1 | 2 . 8 | | 2 8 | 8 . 2 | . . 6 | . 5 4 | 5 | | . 3 . | . . . | . . . | | *-----------------------* / / / 1 3 7 1 8 ------------------------ 2 4 \ \ \

From this we can see that the left stack has pairs and singles, but can not tell which are which yet. Also in the bottom band we can see that 5 is a single while 2 and 8 are a pair.

Based on just the bottom band, why can't the 5 be paired with 4, for example?

Code: Select all
*-----------------------* / | \ | . . . | . 1 . | . 8 . | | | 2 4 . | 7 . . | 9 . 1 | | 1 | 6 . 1 | 5 . . | . . 7 | | 7 |-------+-------+-------| / | \ | 7 2 8 | 1 6 9 | . 4 . | | 4 3 5 | 4 5 3 | 2 8 7 | 6 1 9 | = | 1 6 9 | 9 1 6 | 4 5 3 | 8 7 2 | | 8 2 7 |-------+-------+-------| / | \ | 5 6 . | . . 1 | 2 3 8 | 3 | 2 8 | 8 . 2 | . . 6 | 1 5 4 | 5 | 1 | 1 3 . | . . . | 7 9 6 | | 6 *-----------------------* / 5 6 / / 7 6 1 3 4 7 1 1 9 8 ------------------------ 8 2 5 4 \ \ 6 \

Looking at the middle box we can immmediately deduce that the other two pairs are 2&4, and 9&3, and the remaining single is 8.

Seems to me it takes a bit more than the middle box to deduce the pairing. By looking at the middle stack, don't you have to first recognize that digit 1 and 7 are traveling singles?

by **Condor** » Mon Apr 23, 2007 4:47 am

Hi ronk. Thanks for the reply. It really helps when people post a reply about something they don't understand. Maybe this topic would have been known in the sudoku community a lot sooner if someone had replied to my original postings.

Something I take for granted or think is obvious can be missed in my explanations. Anyway to answer your questions.

In general, we only need 1 number on each side to know we have singles and pairs. We need to know at least one pair to know which side is singles and which is pairs.

Once we know which are singles, any other number in the same boxrow (or boxcol for stacks) must be one of a pair, even if we don't know its partner.

Pairs do not travel with singles. Instead each pair will share a boxrow with each of the singles.

With that in mind, lets look at your questions.

ronk wrote:
Condor wrote:
Code: Select all
*-----------------------* / | \ | . . . | . . . | . 8 . | | | 2 4 . | 7 . . | 9 . 1 | | 1 | 6 . 1 | 5 . . | . . 7 | | 7 |-------+-------+-------| / | \ | . . 8 | 1 . . | . 4 . | | | . . 3 | 2 . 7 | 6 . . | | 1 | . 1 . | . . 3 | 8 . . | | 8 |-------+-------+-------| / | \ | 5 . . | . . 1 | 2 . 8 | | 2 8 | 8 . 2 | . . 6 | . 5 4 | 5 | | . 3 . | . . . | . . . | | *-----------------------* / / / 1 3 7 1 8 ------------------------ 2 4 \ \ \

From this we can see that the left stack has pairs and singles, but can not tell which are which yet. Also in the bottom band we can see that 5 is a single while 2 and 8 are a pair.

Based on just the bottom band, why can't the 5 be paired with 4, for example?

I write the numbers as I identify how they are travelling in the same row as the box they are next to. In this case box 9.

On the right side we have a pair, 2&8, and on the left side the number 5. 5 therefore must be a single and any number that occurs in the same boxrow as 5 occurs in must therefore be one of a pair.

So 4 is one of a pair. We are now able to eliminate 4 as a candidate from b8r9 (thats box 8 row 9) and b7r7. We can also write it on the right side. Thus:

Code: Select all: *-----------------------* / | \ | . . . | . . . | . 8 . | | | 2 4 . | 7 . . | 9 . 1 | | 1 | 6 . 1 | 5 . . | . . 7 | | 7 |-------+-------+-------| / | \ | . . 8 | 1 . . | . 4 . | | | . . 3 | 2 . 7 | 6 . . | | 1 | . 1 . | . . 3 | 8 . . | | 8 |-------+-------+-------| / | \ | 5 . . | . . 1 | 2 . 8 | | 2 8 | 8 . 2 | . . 6 | . 5 4 | 5 | 4 | . 3 . | . . . | . . . | | *-----------------------* / / / 1 3 7 1 8 ------------------------ 2 4 \ \ \

It is being able to infer the behaviour of 1 number (i.e. how its travelling) from the behaviour of other numbers that is key to understanding how this method works.

ronk wrote:
Code: Select all
*-----------------------* / | \ | . . . | . 1 . | . 8 . | | | 2 4 . | 7 . . | 9 . 1 | | 1 | 6 . 1 | 5 . . | . . 7 | | 7 |-------+-------+-------| / | \ | 7 2 8 | 1 6 9 | . 4 . | | 4 3 5 | 4 5 3 | 2 8 7 | 6 1 9 | = | 1 6 9 | 9 1 6 | 4 5 3 | 8 7 2 | | 8 2 7 |-------+-------+-------| / | \ | 5 6 . | . . 1 | 2 3 8 | 3 | 2 8 | 8 . 2 | . . 6 | 1 5 4 | 5 | 1 | 1 3 . | . . . | 7 9 6 | | 6 *-----------------------* / 5 6 / / 7 6 1 3 4 7 1 1 9 8 ------------------------ 8 2 5 4 \ \ 6 \

Looking at the middle box we can immmediately deduce that the other two pairs are 2&4, and 9&3, and the remaining single is 8.

Seems to me it takes a bit more than the middle box to deduce the pairing. By looking at the middle stack, don't you have to first recognize that digit 1 and 7 are traveling singles?

When we look at the stacks, everything is in columns including my notes. 5 and 6 are in the same column because they are a pair. 1 and 7 are on the other side, so we know they are singles. This is how I knew that they were traveling singles. I then simply applied that information to box 5.

Ps If you want to know what I was working on when I first discovered what is now known as Braid analysis and Traveling Pairs, have a look at the second half of this posting.

by **ravel** » Mon Apr 23, 2007 9:45 am

Condor wrote:It really helps when people post a reply about something they don't understand.

To be honest, i never was interested in your method. You show some samples, that can be solved with singles only and make complicated notations and conclusions. When i remember right, Mr.Hamiltons traveling pairs only found eliminations, that are catched by singles and locked candidates also. So what is the purpose of your analysis ?

by **ronk** » Mon Apr 23, 2007 10:52 am

Ah so, I wasn't taking full advantage of the fact that traveling pairs and traveling singles traverse chutes in opposite directions. Thanks.

BTW I don't use -- and have no plans to use -- any side notation for traveling pairs/triples.

by **MrHamilton** » Wed Apr 25, 2007 10:47 am

ravel wrote:
Condor wrote:It really helps when people post a reply about something they don't understand.
To be honest, i never was interested in your method. You show some samples, that can be solved with singles only and make complicated notations and conclusions. When i remember right, Mr.Hamiltons traveling pairs only found eliminations, that are catched by singles and locked candidates also. So what is the purpose of your analysis ?

Doesn't Sudopedia provide a worked out example, which Simple Sudoku cannot solve using basic methods such as locked candidates and pairs by the way, under this link:

http://www.sudopedia.org/wiki/Braid_Analysis

I didn't post the example, and i might have done it differently. But Ravel's objection is not quite correct.

by **ravel** » Wed Apr 25, 2007 11:28 am

Ok, lets take the sample grid from this site:

Code: Select all: 005000000004570000690800000001000000470100020000060570300004058000030200109020300 '---------------------'---------------------'---------------------' | 278 1238 5 | 2369 149 12369 | 146789 134689 234679| | 28 1238 4 | 5 7 12369 | 1689 13689 2369 | | 6 9 237 | 8 14 123 | 147 134 5 | :---------------------+---------------------+---------------------: | 2589 23568 1 | 2379 589 235789| 4689 34689 3469 | | 4 7 368 | 1 589 3589 | 689 2 369 | | 289 238 238 | 4 6 2389 | 5 7 1 | :---------------------+---------------------+---------------------: | 3 26 267 | 679 19 4 | 1679 5 8 | | 578 4568 678 | 679 3 156789| 2 1469 4679 | | 1 4568 9 | 67 2 5678 | 3 46 467 | '---------------------'---------------------'---------------------'

With basic methods (ignoring an xy-wing) i get here:

Code: Select all: *---------------------------------------------------* | 278 13 5 | 23 9 1236 | 1467 468 2467 | | 28 13 4 | 5 7 1236 | 169 689 269 | | 6 9 27 | 8 4 12 | 17 3 5 | |----------------+---------------+------------------| | 259 256 1 | 23 8 7 | 469 469 3469 | | 4 7 36 | 1 5 39 | 8 2 369 | | 29 28 238 | 4 6 239 | 5 7 1 | |----------------+---------------+------------------| | 3 26 267 | 679 1 4 | 679 5 8 | | 57 4 678 | 679 3 58 | 2 1 679 | | 1 58 9 | 67 2 58 | 3 46 467 | *---------------------------------------------------*

Can someone demonstrate, how to proceed with Braiding now ?

by **re'born** » Wed Apr 25, 2007 12:24 pm

ravel wrote:
With basic methods (ignoring an xy-wing) i get here:
Code: Select all
*---------------------------------------------------* | 278 13 5 | 23 9 1236 | 1467 468 2467 | | 28 13 4 | 5 7 1236 | 169 689 269 | | 6 9 27* | 8 4 12 | 17 3 5 | |----------------+---------------+------------------| | 259 256 1 | 23 8 7 | 469 469 3469 | | 4 7 36 | 1 5 39 | 8 2 369 | | 29 28* 238- | 4 6 239 | 5 7 1 | |----------------+---------------+------------------| | 3 26* 267* | 679 1 4 | 679 5 8 | | 57 4 678* | 679 3 58 | 2 1 679 | | 1 58- 9 | 67 2 58 | 3 46 467 | *---------------------------------------------------*

Can someone demonstrate, how to proceed with Braiding now ?

I can't see it and it almost seems to be more trouble than its worth to look. But I can see a nice subset counting deduction (or ALS xz-rule, if you prefer). The 5 cells r67c2, r378c3 contain the candidates (2,6,7,8) with max multiplicities (2,1,1,2). So we may eliminate 8 from r6c3 and r9c2, solving the puzzle.

by **ronk** » Wed Apr 25, 2007 1:45 pm

ravel wrote:Can someone demonstrate, how to proceed with Braiding now ?

Shouldn't we be looking at it without pencilmarks?

Code: Select all: . . 5 | . 9 . | . . . . . 4 | 5 7 . | . . . 6 9 . | 8 4 . | . 3 5 -------+-------+------- . . 1 | . 8 7 | . . . 4 7 . | 1 5 . | 8 2 . . . . | 4 6 . | 5 7 1 -------+-------+------- 3 . . | . 1 4 | . 5 8 . 4 . | . 3 . | 2 1 . 1 . 9 | . 2 . | 3 . .

[edit: However, I doubt there is a next placement using braids.]

by **MrHamilton** » Wed Apr 25, 2007 10:53 pm

ronk wrote:
ravel wrote:Can someone demonstrate, how to proceed with Braiding now ?

Shouldn't we be looking at it without pencilmarks?
Code: Select all
. . 5 | . 9 . | . . . . . 4 | 5 7 . | . . . 6 9 . | 8 4 . | . 3 5 -------+-------+------- . . 1 | . 8 7 | . . . 4 7 . | 1 5 . | 8 2 . . . . | 4 6 . | 5 7 1 -------+-------+------- 3 . . | . 1 4 | . 5 8 . 4 . | . 3 . | 2 1 . 1 . 9 | . 2 . | 3 . .

[edit: However, I doubt there is a next placement using braids.]

In an actual solving situation, I would have no reason to ignore the X-Y wing, a technique that even Simple Sudoku is programmed to find. And I suppose we agree that Braids are useful both before and after the diagram position of the puzzle, if X-Y wing is permitted.
However, suppose I missed or ignored this X-Y wing for any reason, I think I would still find the answer fairly easily since the next thing I would try to do is rule out triples (which I can do easily in the 3 stacks and two lowest bands). If I then determined what triples are possible in the top band, and in this case there is only one possible distribution of them (357/148/269 on the N strands), and then I considered this as my hypothesis, and was able to totally rule it out in any segment of the third row, for any reason, I would solve the third row, and shortly the puzzle.
Making a 692 in r3c1-r3c2-r3c3 creates a quick contradiction in the left stack, and is therefore not possible.
r3c3=2>
r2c1=8>
r1c1=7>
r8c1=5>
r9c2=8, but then r6c2 and r7c2 are each supposed to be 2 and can't be.

by **ronk** » Thu Apr 26, 2007 12:01 am

So r3c3<>2 by contradiction? I thought braid analysis was 100% direct deduction, without an elimination-by-contradiction component.

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