Post the puzzle or solving technique that's causing you trouble and someone will help

In the position below, if I use the y-wing (in yellow) to solve the orange square, this gives the orange box a solution of 7, and this unravels the puzzle and leads to a final solution.

HOWEVER: If you first use the same y-wing to point to the RED square, this technique should eliminate the possibility of 5 from the red square. However, 5 is actually the correct solution to the red square!

What am I missing here? Or am I somehow using y-wing incorrectly? Or is this some paradox? HELP!

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xscientist

Posts: 3
Joined: 28 January 2009

xscientist wrote:In the position below, if I use the y-wing (in yellow) to solve the orange square, this gives the orange box a solution of 7, and this unravels the puzzle and leads to a final solution.

HOWEVER: If you first use the same y-wing to point to the RED square, this technique should eliminate the possibility of 5 from the red square. However, 5 is actually the correct solution to the red square!

What am I missing here? Or am I somehow using y-wing incorrectly? Or is this some paradox? HELP!

1) Y-Wing is passe ... XY-Wing is contemporary.

Code: Select all
` +-----------------------------------------------------+ | a59   7    359  |  6    8    1    | b25   23   4    | |  2    6    35   |  4    7    9    |  1    8    35   | |  1    4    8    |  5    2    3    |  6    9    7    | |-----------------+-----------------+-----------------| |  6    8    7    |  9    5    4    |  3    1    2    | |  3    9    1    |  7    6    2    |  4    5    8    | |  4    5    2    |  1    3    8    |  7    6    9    | |-----------------+-----------------+-----------------| |  57   1    4    |  2    9    6    |  8    37   35   | |  7-9  3    6    |  8    4    5    | c29   27   1    | |  8    2    59   |  3    1    7    |  59   4    6    | +-----------------------------------------------------+ # 15 eliminations remain`

2) Your XY-Wing a-b-c has vertex/pivot cell "b" and eliminates <9> in [r8c2] because cells "a" and "c" both see it.

3) However, cell "c" doesn't see cell [r1c3], so the XY-Wing can't perform an elimination in that cell.

4) Finally, you might want to read up on BUG+1 because it will resolve cell [r1c3] directly.
daj95376
2014 Supporter

Posts: 2624
Joined: 15 May 2006

1) I will read up on XY-wing and BUG for future reference. however:

2) This does not eliminate the paradox in my original question. Why does my y-wing produce a false elimination?
xscientist

Posts: 3
Joined: 28 January 2009

Firstly, your usage of the term "y-wing" is not acceptable to most players here, because it has been widely used to describe other techniques before. In the future please call that technique "xy-wing" (which is universally accepted everywhere without any confusion).

Secondly, I don't think you have fully understood the mechanism of applying this technique. Your first application works like this:

r1c7 must be 2 or 5.
=> Either r1c1=9, or r8c7=9, or both.
=> r8c1, seeing both r1c1 and r8c7, can't be 9, must be 7.

But I can't see anything similar allows you to eliminate 5 from the red cell, r1c3. To do so, you must find 2 cells commonly "seen" by r1c3, one or both of which must be forced to be 5. But I don't see how you can find such 2 cells.

On the other hand, there are other moves which allow you to eliminate 3 and 9 from r1c3, although they're not "xy-wings", but instead are the more complicated "xy-chains":

r9c7 must be 5 or 9.
=> Either r1c7=2, or r9c3=5, or both.
=> Either r1c8=3, or r2c3=3, or both.
=> r1c3, seeing r1c8+r2c3, can't be 3.

r8c7 must be 2 or 9.
=> Either r1c7=5, or r9c7=5, or both.
=> Either r1c1=9, or r9c3=9, or both.
=> r1c3, seeing r1c1+r9c3, can't be 9.

http://www.sudopedia.org/wiki/XY-Wing
http://www.sudopedia.org/wiki/XY-Chain

udosuk

Posts: 2698
Joined: 17 July 2005

intresting grid:

as daj put it the
xy-wing
you are trying to build doesnt remove any candidate from the marked cell.
because it is not where the xy wing eliminates the candiate:

as the red cell doesnt directly see the same dgit elimination from all views of the marked cells.

think of it this way
i find its the easiest:(when looking for them as well)

if you have a pattern of 3 sperate cells the middle connection
when active as
a or b,
the cells it see = the same digit expressed
the elimination is in a cell that see both the wings and is the expressed digit.

example

Code: Select all
`B(12)   ->   X|            ||            vA(23) -  (C)31`

when
A=2 => B=1
A=3 => C=1
thus X <>1

the same thing applies to both C & D
in all cases x <>1

am example of this in work is shown above by daj as well.

there is however a complex move:

als-xy
that does accomplish what you are asking though.
{there is many varations that could be used here is an example}

A=r1c17 - {259}, B=r7c1,r9c3 - {579}, C=r8c17 - {279}, X,Y=2,7, Z=9 => r1c3<>9

another way to do it is an

xy-chain:
(as noted by daj) here is the easiest one.
(both operate off the same cells)

9 r9c3 -5- r9c7 -9- r8c7 -2- r1c7 -5- r1c1 => r1c3,r8c1<>9
5 r9c7 -9- r8c7 -2- r1c7 -5- r1c1 -9- r8c1 -7- r7c1 => r7c9,r9c3<>5

or: easier approaches..

any one of the following solves the puzzle as well.

Skyscrapers:
5 in r1c1,r2c9 (connected by r7c19) => r1c7,r2c3<>5
5 in r2c9,r9c7 (connected by r29c3) => r1c7,r7c9<>5
5 in r2c3,r7c1 (connected by r27c9) => r1c1,r9c3<>5
5 in r7c1,r9c7 (connected by r1c17) => r7c9,r9c3<>5

or
Empty rectangles:
5 in b1 (r9c37) => r1c7<>5
5 in b1 (r19c7) => r9c3<>5

or:
2-string kytes
5 in r1c7,r7c1 (connected by r7c9,r9c7) => r1c1<>5
5 in r1c1,r9c7 (connected by r7c1,r9c3) => r1c7<>5
5 in r2c9,r9c3 (connected by r7c9,r9c7) => r2c3<>5
5 in r2c9,r7c1 (connected by r1c1,r2c3) => r7c9<>5
5 in r2c3,r9c7 (connected by r1c7,r2c9) => r9c3<>5
Some do, some teach, the rest look it up.

StrmCkr

Posts: 889
Joined: 05 September 2006