The New Sudoku Players' Forum

[RyanStorm]

I know that in some situations that where if in 1 column (up to down), there is a box that can only be 4 or 7, and if you go up to a box in same column and has 3, 4 or 7, that if another box in same row (left to right) to the right also has a setup of only 4 or 7, that I can erase the 4 and 7 from the box that could only be 2, 4 or 7. Because it is in same path as two boxes that can only have 4 or 7.

I know this technique only erases like numbers in some situations, in this case only the 4 and 7. But, I was wondering if the exact same technique works, but the 4 or 7, is in same row, but the other in same 9x9 box. See picture:

I want to know if I can erase the 7 from the 3 or 7, because a 4 or 7 is in same row (to the right), and a 4 or 7 is in same 9x9 box? In this particual case, there happens to be a 4 or 7 in the bottom middle 9x9 box, and 2nd one in top left 9x9, I only added those to see if they can possibly be used somehow?

I am 90% sure I can erase the 3 from the top middle box that says 3 or 7, because there are two boxes with 4 or 7 in its path? Am I right? If need to see rest of boxes I posted the second photo below with more info. Sorry it is a little cut off, the top right is 4 or 7, below it is 8, and below it is 9.

[JasonLion]

If you have a pair (two cells in the same house both of which can only contain the same two digits) you can't have those two digits anywhere else in that same house. A house is a row, column, or box.

In you first example, the pair on 47 eliminates 4 & 7 from all of the other cells in that column, but doesn't tell you anything about the 37, which is not in the same column.

[eleven]

RyanStorm,

note that both the 47 box in the 1st and the left 47 box in the 3rd row could be 4 in your first picture. So the 7 in the 37 box is still possible.

But your picture shows a "remote pair" pattern.
If the box with 4 or 7 in the first row is 4, then the one in row 7 must be 7, and vice versa, if the one is 7, the other is 4.
That means, that (taking the row/column numbering of the second picture) the box in row 7 and column 9 cannot be 4 or 7, because 4 and 7 are in these 2 boxes, in the same column or row resp.

However we cannot see, if this would help you in this puzzle.

[RyanStorm]

[RyanStorm]

That is my board as it stands, I just didn't know the technique used to solve from where I am. I read of remote pairs, but I just didn't grasp it the first time I read it.

I thought I had some kind of triplet, or some thing going on with the 6 or 9's.

[daj95376]

_

My solver treats a Remote Pair as two XY-Chains using the same cells. (see below)

Code: Select all

 +-----------------------------------------------------------------------+
 |  234    24     5      |  9      37     8      |  1      6     a47     |
 |  34     7      9      |  34     6      1      |  5      2      8      |
 |  1      8      6      |  2      5     c47     |  3     b47     9      |
 |-----------------------+-----------------------+-----------------------|
 |  9      126    7      |  5      24     3      |  8      14     246    |
 |  268    1256   3      |  78     2478   69     |  2679   14579  24567  |
 |  268    256    4      |  1      278    69     |  2679   579    3      |
 |-----------------------+-----------------------+-----------------------|
 |  7      46     2      |  34     39     5      |  69     8      1      |
 |  45     9      8      |  6      1     d47     |  27     3      25-7   |
 |  56     3      1      |  78     789    2      |  4      579    567    |
 +-----------------------------------------------------------------------+
 # 65 eliminations remain

 RP (7=4)r1c9 - (4=7)r3c8 - (7=4)r3c6 - (4=7)r8c6  =>  r8c9<>7
 RP (4=7)r1c9 - (7=4)r3c8 - (4=7)r3c6 - (7=4)r8c6  =>  nada


However, you are now stuck on where to go after the Remote Pair. There are two choices that I would recommend.

(1) Look for other XY-Chains present. Especially those that contain a <4> in the starting cell.

(2) Learn about "Fish" patterns. There are several for <4> and <6> that crack your puzzle.

Spoiler: Show

Code: Select all

 XY-Chain: (4=3)r2c1 - (3=4)r2c4 - (4=7)r3c6 - (7=4)r8c6  =>  r8c1<>4

 Fish:     (4)r2c4 = r3c6 - r8c6 = (4)r8c1  =>  r2c1<>4


_

[RyanStorm]

I have never read anything like that before. I faintly understand what your saying and I faintly understand any of it.

Just to make sure I am 100% clear, "row" aka "r" means left to right aka horizontal, and "column" aka "c" means up to down aka vertical?

I need to go do some research, because I know which boxes your XY Chains are talking about but I don't get how it knows to solve it.


https://www.brainbashers.com/sudokuremotepairs.asp
^On the very bottom sudoku, that is what I was asking. In the bottom right "house", you got 2, 6, or 7, with a 2 or 6 in the same house but that one is not in same row or column. But there is another set of 2 or 6,that is in same column as the 2, 6 or 7. That alone isn't enough, but the fact that the two 2 or 6's (one in same house, and one in same column) can be linked through other sets of 2 or 6, allows it to be possible.

Just like on mine, the 3 or 7, has a 4 or 7 in same house (top middle house), and there is a 4 or 7 in same row as the 3 or 7, and both those 4 or 7's are not directly connected, but they are connected through the other 4 or 7's, like the second one in the top left house. Does make sense? The only difference between the two is, in the 2 or 6 example, by going through the connected, when you get to end you end up with opposite number, mine, the locked pairs I am talking about, end with same number.

[RyanStorm]

Wait I think I get it.

Based on "deduction" or the concepts of XY chains , the top right box of 4 or 7, no matter if it is 4 or 7, it will take out the 3 in the very top left box of 2, 3, or 4. (which is tells me r2c1, is the only 3 left in c1, and that 3 is only one in row 1.)

If 7, then 3 or 7 in r1c5, is 3, which makes r1c1 not possible of a 3.

If 4, and you go through the 4 or 7's, you end up at r8c7, the 2 or 7, then it is 2, meaning r8c9 is 5, which makes r1c8 a 4, then of course makes r2c1 a 3! That 3 also says r1c1 can't be 3!

[RyanStorm]

I DID IT!

By proving that 3 can't be in that 1 square r1c1, it allowed me to solve the 3 in row 1 and the 3 column 1, only because they were the 3's were the only ones left in their column or row. Which cracked the entire puzzle. That technique was awesome. I will look into what you posted, but I learned the XY method either way, which is actually something I really liked doing.