The New Sudoku Players' Forum

by **Kent** » Sat Mar 11, 2006 6:38 am

Need help.

1)I tried forcing chain.Can i deduce like this : r1c4=3->r2c4=5-> r3c6=2->r3c7=5->r2c9=3->r1c9=2 and therefore r1c4=3.So r1c4=3.Can i deduce like this??

2) Can I form a finned X-wing using candidate 4 in cells r4r6,r4r7,r6c6,c6c7??

549l.87l61.
276l.91l84.
183l46.l.97
.61l.3.l.79
.35l..9l1..
792l.1.l...
327l156l984
914l.2.l.56
658l9..l..1

by **MCC** » Sat Mar 11, 2006 2:56 pm

1) No. What happens if you make r1c4=2?

2) No. The 4's in cells r4r6,r4r7,r6c6,c6c7 make up an x-wing, with this you can make an elimination in r4c1.

MCC

by **Kent** » Sat Mar 11, 2006 4:06 pm

MCC

The 4's in cells r4r6,r4r7,r6c6,c6c7 make up an x-wing, with this you can make an elimination in r4c1.

1)How can those cells make an X-wing??Because there's 3 candidate 4 in column 6 and 3 candidate at row4.

X-wing must be

..*........*......

..*........*.....

2) how do u continue by r1c4=2?? Can u please demonstrate?? I didnt know where i was heading towerds.
Thanks

by **MCC** » Sat Mar 11, 2006 5:43 pm

Your right Kent, what I took to be a 8 turns out to be a 4. I tend to do these puzzles on P&P and my scribbles are sometimes indescribable.

So there is no x-wing or finned x-wing in 4's.

Concerning point 1)
If r1c4=3 then r1c9=2
If r1c4=2 then r1c9=3

Nothing can be deduced from this.

You could form an x-wing in 3's in cells:
(r12c4)(r12c9)
This will eliminate the 3 in r6c9

Or that the 3's in box 3 are locked in column 9.
This will also eliminate the 3 in r6c9.

The 3's in box 2 are also locked in column 4
eliminating the 3 in r8c4

MCC

by **Kent** » Sat Mar 11, 2006 7:01 pm

I already eliminated those 3 and i'm stuck now.Need to know how to procede

by **tso** » Sat Mar 11, 2006 7:38 pm

Kent wrote:Need help.

1)I tried forcing chain.Can i deduce like this : r1c4=3->r2c4=5-> r3c6=2->r3c7=5->r2c9=3->r1c9=2 and therefore r1c4=3.So r1c4=3.Can i deduce like this??

You cannot discuss forcing chains and x-wings without first posting a candidate grid. There's no way of knowing that we're all looking at the same position.

You've posted 50 times already in less than two weeks. It's time you learned how to consistantly post legible candidate grids. PLEASE read the instructions here. PLEASE click PREVIEW before SUBMIT to make sure your diagrams are legible. Make sure DISABLE BBCODE IN THIS POST is NOT checked.

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

Code: Select all: *-----------------------------------------------------------* | 5 4 9 | 23 8 7 | 6 1 23 | | 2 7 6 | 35 9 1 | 8 4 35 | | 1 8 3 | 4 6 25 | 25 9 7 | |-------------------+-------------------+-------------------| | 48 6 1 | 258 3 25 | 245 7 9 | | 48 3 5 | 2678 47 9 | 1 26 28 | | 7 9 2 | 568 1 48 | 345 36 58 | |-------------------+-------------------+-------------------| | 3 2 7 | 1 5 6 | 9 8 4 | | 9 1 4 | 78 2 38 | 37 5 6 | | 6 5 8 | 9 47 34 | 237 23 1 | *-----------------------------------------------------------*

You ask so many questions at once and do not seem to be reading the answers.

To use a forcing chain, ALL possibilites -- not ONE, not SOME -- must lead the the SAME conclusion. You specifically asked this same question before and were told "no".

Consider this:

[12][23]
[14][34]

r1c1=1 -> r2c1=4 -> r2c2=3 -> r1c2=2 -> r1c1=1

From this, you want to conclude that r1c1=1. You cannot. You have followed only ONE of two paths. You have found no contradiction. If we allow this sort of leap, then what do we make of this:

r1c1=2 -> r1c2=3 -> r2c2=4 -> r2c1=1 -> r1c1=2

IF an assumption leads to a contradiction, that assumption can be excluded. If there is no contradiction, there is no deduction. That's basic proof by contradiction.

Here is one possible forcing chain from the given position:

Code: Select all: *-----------------------------------------------------------* | 5 4 9 | 23 8 7 | 6 1 [23] | | 2 7 6 | 35 9 1 | 8 4 [35] | | 1 8 3 | 4 6 25 | 25 9 7 | |-------------------+-------------------+-------------------| | 48 6 1 | 258 3 25 | 245 7 9 | |[48] 3 5 | 2678{x47} 9 | 1 26 [28] | | 7 9 2 | 568 1 [48] | 345 36 [58] | |-------------------+-------------------+-------------------| | 3 2 7 | 1 5 6 | 9 8 4 | | 9 1 4 | 78 2 38 | 37 5 6 | | 6 5 8 | 9 47 34 | 237 23 1 | *-----------------------------------------------------------*

ANY value in ANY of the bracketed cells will lead to an excusion of the candidate 4 from r5c5. For example:

Row 5 column 1 can have only two values -- 4 or 8.

If r5c1=4, then r5c5=7
But if r5c1=8 -- the only other possibility -- then r5c9=2, r1c9=3, r2c9=5, r6c9=8, r6c6=4 and finally, r5c5=7.

Since ALL POSSIBLE VALUES in r5c1 lead to r5c5=7, then r5c7 must be 7. This does NOT directly tell you what the value of r5c1 (or any other cell) is.

If all roads lead to Rome, all you know is that you'll end up in Rome, NOT what road you're on. If SOME roads lead to Rome, and you don't know what road you're on -- you don't know anything at all.

2) Can I form a finned X-wing using candidate 4 in cells r4r6,r4r7,r6c6,c6c7??

No one candidate appears in all four cells -- at least not in my candidate grid. Therefore, no x-wing, finned or otherwise can exist. However, maybe your candidate grid looks like this:

Code: Select all: *-----------------------------------------------------------* | 5 4 9 | 23 8 7 | 6 1 23 | | 2 7 6 | 35 9 1 | 8 4 35 | | 1 8 3 | 4 6 25 | 25 9 7 | |-------------------+-------------------+-------------------| |x48 6 1 | 258 3 *2458 |*245 7 9 | | 48 3 5 | 2678 47 9 | 1 26 28 | | 7 9 2 | 568 1 *458 |*345 36 58 | |-------------------+-------------------+-------------------| | 3 2 7 | 1 5 6 | 9 8 4 | | 9 1 4 | 78 2 38 | 37 5 6 | | 6 5 8 | 9 47 #34 | 237 23 1 | *-----------------------------------------------------------*

If so, I assume that you're asking if there is a finned x-wing in 4s. No. Here's why:

If r9c6<>4, then r46c67 form an x-wing in columns, then r4c1<>4
Unfortunately, r9c6=4 does NOT lead to r4c1<>4
Only one path leads to Rome -- you can make no conclusion.

OR

If r4c1<>4, then r46c67 form an x-wing in rows, then r9c6<>4
Unfortunately, r4c1=4 does NOT lead to r9c6<>4
Again, since the two choices diverge, there is no deduction.

Another answer would be "Yes, there is a finned x-wing in 4s -- unfortunately, it doesn't allow any exclusions."

by **Kent** » Sat Mar 11, 2006 8:04 pm

Thanks a lot tso!! I finally understand everything.You are saying that it is a finned x-wing but it doesn't help because the fin doesn't MEET the cell that u were gonna exclude.Don't worry i'm gonna refer and study carefully before i post anything.

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