barking up the wrong tree

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barking up the wrong tree

Postby petertohen » Wed Dec 13, 2017 9:56 am

Number I have read somewhere about having three strong links that can be used to solve an X-Wing. With that, I have saved various puzzles where I have come across X-Wings that fit this description. I cannot find anything written about this but I think I have seen a posting on this forum relating to using three strong as a method of solving for a candidate. I may be all wrong, but please look at each of the below and let me know if I “barking up the wrong tree”.


The puzzles have the rectangle’s candidates are
labeled as (a)b (b),(c) and (d).

Number 1: On (4’s)
(4) Locked on bd
(4) Locked on cd
(7) Locked on cd
(d) can be made a (4)
(a) can be made a (4)
Code: Select all
*--------------------------------------------------*
 | 6    249  7    | 1    8    5    | 24a   3    49b  |
 | 3    459  459  | 69   2    7    | 46   8    1    |
 | 29   1    8    | 69   4    3    | 256  7    59   |
 |----------------+----------------+----------------|
 | 489  789  49   | 47   5    6    | 3    1    2    |
 | 24   27   1    | 47   3    9    | 8    5    6    |
 | 5    3    6    | 2    1    8    | 47c   9    47d  |
 |----------------+----------------+----------------|
 | 7    489  349  | 5    69   2    | 1    46   38   |
 | 1    6    359  | 8    79   4    | 57   2    357  |
 | 48   458  2    | 3    67   1    | 9    46   578  |
 *--------------------------------------------------*

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Re: barking up the wrong tree

Postby StrmCkr » Thu Dec 14, 2017 2:37 am

Welcome to the forum

the only way i see and know of 3 strong links on 1 candidate performs an elimination is when 2 sets of corners share linked bivalves aka xy- wing (uses the cells relation ship instead of the strong weak links)
{edit: either this or you are thinking of an Unique Rectangle type 1 .}

Code: Select all
ab       bc    {strong link on b towards bd corner}
           |
           |
da----- bd   {strong link on d towards da cell}


when AB is solved as A or B the BD cell solves as B
when bd is solved as d then ab = blank. { invalid state}
when da is solved as A then bd = B &D {contradiction}
when bC is solved as C then bd = b and da = d and Ab = AB { undetermined}
when bc is solved as B then bd = b and da = a and AB = blank. {invalid state}

thus:
bc = C
BD = b
Da = D
ab = Ab

clear as mud?
Last edited by StrmCkr on Fri Dec 15, 2017 3:20 am, edited 1 time in total.
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Re: barking up the wrong tree

Postby Leren » Thu Dec 14, 2017 10:40 am

Code: Select all
*----------------------------------------*
| 6    249   7    | 1  8  5 | 24  3  49  |
| 3    459   45-9 | 69 2  7 | 46  8  1   |
|a29   1     8    | 69 4  3 | 256 7  59  |
|-----------------+---------+------------|
| 48-9 4789 c49   | 47 5  6 | 3   1  2   |
|b24   247   1    | 47 3  9 | 8   5  6   |
| 5    3     6    | 2  1  8 | 47  9  47  |
|-----------------+---------+------------|
| 7    489   349  | 5  69 2 | 1   46 38  |
| 1    6     359  | 8  79 4 | 57  2  357 |
| 48   458   2    | 3  67 1 | 9   46 578 |
*----------------------------------------*

Perhaps you are talking about an XY Wing : (9=2) r3c2 - (2=4) r5c1 - (4=9) r4c4 => - 9 r2c3, r4c1 This move has 3 strong links.

After that you are reduced to BUG+1 in r7c3 and the puzzle is solved

Code: Select all
*--------------------------------*
| 6  2   7    | 1 8  5 | 4 3  9  |
| 3  4   5    | 9 2  7 | 6 8  1  |
| 9  1   8    | 6 4  3 | 2 7  5  |
|-------------+--------+---------|
| 48 89  49   | 7 5  6 | 3 1  2  |
| 2  7   1    | 4 3  9 | 8 5  6  |
| 5  3   6    | 2 1  8 | 7 9  4  |
|-------------+--------+---------|
| 7  89 *9-43 | 5 69 2 | 1 46 38 |
| 1  6   39   | 8 79 4 | 5 2  37 |
| 48 5   2    | 3 67 1 | 9 46 78 |
*--------------------------------*

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