The New Sudoku Players' Forum

by **ravel** » Tue Apr 10, 2007 2:02 pm

Nice solution again, rep'nA.

Code: Select all: .---------------.---------------.---------------. | 129 129 6 | 5 34 34 | 7 12 8 | | 8 5 7 | 16 9 2 |B16 3 4 | | 12 4 3 | 7 16 8 | 9 5 126 | :---------------+---------------+---------------: | 3 8 12 | 4 167 5 |B126 9 1267| | 1679 179 5 | 8 2 16 | 3 4 167 | | 1267 127 4 | 3 167 9 | 8 1267 5 | :---------------+---------------+---------------: | 5 A237 8 | 16 B34 16 |B24 27 9 | |A127 6 A12 | 9 5 34 |-124 8 1237| | 4 13 9 | 2 8 7 | 5 16 136 | '---------------'---------------'---------------'

You can also make the elimination with an (xy?) ALS:
The 1 in r8c7 locks the 3 in A to r7c2, in B to r7c5.

by **udosuk** » Tue Apr 10, 2007 3:46 pm

ravel wrote:You can also make the elimination with an (xy?) ALS:
The 1 in r8c7 locks the 3 in A to r7c2, in B to r7c5.

I don't like ALSs which don't reside in a single house (like your set B here)...
It's valid logically but I feel it's less elegant than the "pure" ALS-xz...

Here is such a "pure" ALS-xz, for that elimination:

Code: Select all: *-----------------------------------------------------------* | 129 129 6 | 5 34 34 | 7 12 8 | | 8 5 7 | 16 9 2 | 16 3 4 | | 12 4 3 | 7 16 8 | 9 5 126 | |-------------------+-------------------+-------------------| | 3 8 12 | 4 167 5 | 126 9 1267 | | 1679 179 5 | 8 2 16 | 3 4 167 | | 1267 127 4 | 3 167 9 | 8 1267 5 | |-------------------+-------------------+-------------------| | 5 1237 8 | 16 34 1346 | 124 127 9 | |A127 6 A12 | 9 5 134 |-124 8 A1237 | | 4 13 9 | 2 8 7 | 5 B16 B136 | *-----------------------------------------------------------*

A: r8c139={1237}
B: r9c89={136}
x=3
z=1

Therefore r8c7<>1. :idea:

by **wapati** » Tue Apr 10, 2007 8:12 pm

Perhaps, too easy. Too many unique patterns. Ignore them and suffer!

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

by **wapati** » Tue Apr 10, 2007 8:17 pm

This too. Unique patterns are short cuts. (This too may be too easy.)

Well, they may be the only pattern way. You decide?

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

by **wapati** » Tue Apr 10, 2007 9:32 pm

This is harder.

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

by **re'born** » Tue Apr 10, 2007 11:10 pm

wapati wrote:This too. Unique patterns are short cuts. (This too may be too easy.)

Well, they may be the only pattern way. You decide?

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

With 3D-multicoloring from:

Code: Select all: .------------.------------.------------. | 8 9 12 | 124 3 5 | 24 6 7 | | 56 56 127| 124 9 147| 8 3 14 | | 37 237 4 | 8 6 17 | 5 129 19 | :------------+------------+------------: | 4 678 789| 5 1 3 | 69 78 2 | | 1 258 289| 7 28 6 | 49 45 3 | | 56 278 3 | 49 28 49 | 1 78 56 | :------------+------------+------------: | 9 14 6 | 3 5 2 | 7 14 8 | | 2 14 5 | 6 7 8 | 3 149 149| | 37 378 78 | 19 4 19 | 26 25 56 | '------------'------------'------------'

we get the chain:

[r5c2]=5=[r5c8]-5-[r9c8]-2-[r3c8]=2=[r3c2]-2-[r5c2]

implying r5c2<>2. Continuing with this same line of coloring, one then obtains:

[r5c3]-9-[r5c7]-4-[r1c7]-2-[r3c8]=2=[r3c2]-2-[r6c2]=2=[r5c3]

and so r5c3<>9, which solves the puzzle.

Alternatively, to avoid a deadly pattern in r39c12[37], one obtains r39c2<>7. Locked candidates then implies r4c3<>7. The potential deadly pattern in r46c28[78] then implies that r46c2<>8. To avoid yet another deadly pattern, this time in r56c25[28], we cannot have r5c2=8 and r6c2=2, so we get:

[r5c2]-8|2-[r6c2]-7-[r6c8]-8-[r6c5]-2-[r5c5]-8-[r5c2]

giving us r5c2<>8, which solves the puzzle.

by **Carcul** » Wed Apr 11, 2007 12:35 pm

Wapati wrote:This is harder.

Code: Select all: *------------------------------------------------------------------* | 139 1239 6 | 5 139 7 | 4 8 23 | | 45 123459 13 | 123468 13489 123469 | 16 7 23 | | 7 1234 8 | 12346 134 12346 | 9 16 5 | |--------------------+------------------------+--------------------| | 56 13 2 | 9 13 8 | 7 56 4 | | 456 45 13 | 1237 1357 1235 | 256 9 8 | | 8 7 9 | 24 6 245 | 25 3 1 | |--------------------+------------------------+--------------------| | 2 139 4 | 137 13579 1359 | 8 15 6 | | 13 6 7 | 1348 13458 1345 | 15 2 9 | | 19 8 5 | 16 2 169 | 3 4 7 | *------------------------------------------------------------------*

[r2c7]-1-[r3c8]-6-[r4c8](-5-[r7c8]-1-[r7c2]=1=[r89c1]-1-[r1c1])=6=
=[r4c1]-6-[r5c1]=6|4=[r3c2]-4-[r3c5]=4|9=[r1c5]-9-[(r1c1)]-3-[r2c3]-1-
-[r2c7], => r2c7<>1.

Then: [r4c2]-3-[r3c2]=3=[r3c46]-3-[r1c5]=3=[r4c5]-3-[r4c2], => r4c2<>3.

Carcul

by **Carcul** » Wed Apr 11, 2007 12:46 pm

Wapati wrote:This too. Unique patterns are short cuts. (This too may be too easy.) Well, they may be the only pattern way. You decide?

Code: Select all: *--------------------------------------------------* | 8 9 12 | 124 3 5 | 24 6 7 | | 56 56 127 | 124 9 147 | 8 3 14 | | 37 237 4 | 8 6 17 | 5 129 19 | |----------------+----------------+----------------| | 4 678 789 | 5 1 3 | 69 78 2 | | 1 258 289 | 7 28 6 | 49 45 3 | | 56 278 3 | 49 28 49 | 1 78 56 | |----------------+----------------+----------------| | 9 14 6 | 3 5 2 | 7 14 8 | | 2 14 5 | 6 7 8 | 3 149 149 | | 37 378 78 | 19 4 19 | 26 25 56 | *--------------------------------------------------*

[r3c8]=2|4=[r8c9]-4-[r78c8]-1,9-[r3c8], => r3c8=2 solving the puzzle.

Carcul

by **Mike Barker** » Wed Apr 11, 2007 1:40 pm

Carcul, I am always impressed at how briefly you solve these problems, but I have to admit when I try to go through the logic I often get lost. In this last example:

Code: Select all: [r3c8]=2|4=[r8c9]-4-[r78c8]-1,9-[r3c8], => r3c8=2

does [r3c8]=2|4=[r8c9] refer to the UR r38c89? Is there an easy way to tell when you are refering to an ALS vs a UR? After that I get really confused because if it is the UR then there are overlapping cells with r78c8 which I don't understand. I do understand if r78c8=4 then r8c8 doesn't equal 1 or 9. This isn't the first time I've had problems understanding you notation and would appreciate some help because I see your advanced nice loops as a way for me to solve even tougher problems.
TIA
Mike

by **tarek** » Wed Apr 11, 2007 1:49 pm

ravel wrote:Nice solution again, rep'nA.
Code: Select all
.---------------.---------------.---------------. | 129 129 6 | 5 34 34 | 7 12 8 | | 8 5 7 | 16 9 2 |B16 3 4 | | 12 4 3 | 7 16 8 | 9 5 126 | :---------------+---------------+---------------: | 3 8 12 | 4 167 5 |B126 9 1267| | 1679 179 5 | 8 2 16 | 3 4 167 | | 1267 127 4 | 3 167 9 | 8 1267 5 | :---------------+---------------+---------------: | 5 A237 8 | 16 B34 16 |B24 27 9 | |A127 6 A12 | 9 5 34 |-124 8 1237| | 4 13 9 | 2 8 7 | 5 16 136 | '---------------'---------------'---------------'
You can also make the elimination with an (xy?) ALS:
The 1 in r8c7 locks the 3 in A to r7c2, in B to r7c5.

I tried to tackle this hurdle with finned fish & UR but still that "1! in r8c7 still needed an ALS-xz (the one udosuk describes) to remove..... from there there was another finned x-wing (or the skyscraper) to seal the deal.....

tarek

by **re'born** » Wed Apr 11, 2007 2:22 pm

Carcul wrote:
Wapati wrote:This too. Unique patterns are short cuts. (This too may be too easy.) Well, they may be the only pattern way. You decide?

Code: Select all
*--------------------------------------------------* | 8 9 12 | 124 3 5 | 24 6 7 | | 56 56 127 | 124 9 147 | 8 3 14 | | 37 237 4 | 8 6 17 | 5 129 19 | |----------------+----------------+----------------| | 4 678 789 | 5 1 3 | 69 78 2 | | 1 258 289 | 7 28 6 | 49 45 3 | | 56 278 3 | 49 28 49 | 1 78 56 | |----------------+----------------+----------------| | 9 14 6 | 3 5 2 | 7 14 8 | | 2 14 5 | 6 7 8 | 3 149 149 | | 37 378 78 | 19 4 19 | 26 25 56 | *--------------------------------------------------*

[r3c8]=2|4=[r8c9]-4-[r78c8]-1,9-[r3c8], => r3c8=2 solving the puzzle.

Carcul

Or, alternatively, UR type 1 in r78c28[14] implies r8c8=9 and then r3c9=9. Then the potential deadly pattern in r38c89[19], gives

[r3c8]=2|4=[r8c9]-4-[r2c9]-1-[r3c8]

implying r3c8=2, solving the puzzle.

by **Carcul** » Wed Apr 11, 2007 4:08 pm

Mike Barker wrote:does [r3c8]=2|4=[r8c9] refer to the UR r38c89?

That link refers to an almost TIUR (two incompatible unique rectangles) situation. Let's suppose that r3c8 is not "2" and r8c9 is not "4": then we would have the following grid,

Code: Select all: *--------------------------------------------------* | 8 9 12 | 124 3 5 | 24 6 7 | | 56 56 127 | 124 9 147 | 8 3 14 | | 37 237 4 | 8 6 17 | 5 19 19 | |----------------+----------------+----------------| | 4 678 789 | 5 1 3 | 69 78 2 | | 1 258 289 | 7 28 6 | 49 45 3 | | 56 278 3 | 49 28 49 | 1 78 56 | |----------------+----------------+----------------| | 9 14 6 | 3 5 2 | 7 14 8 | | 2 14 5 | 6 7 8 | 3 149 19 | | 37 378 78 | 19 4 19 | 26 25 56 | *--------------------------------------------------*

Do you see now what those two URs in r78c28 and r38c89 imply? So we must have r3c8=2 or r8c9=4.

Carcul

by **Carcul** » Wed Apr 11, 2007 5:24 pm

Wapati wrote:I didn't see any shortcuts. Lots of patterns, swordfish or finned swordfish depending on your solving order. Some unique fun.

Code: Select all: *------------------------------------------------------* | 8 35 6 | 4 2 35 | 9 1 7 | | 9 4 35 | 1 7 8 | 2 35 6 | | 12 7 12 | 35 9 6 | 8 4 35 | |------------------+------------------+----------------| | 137 9 158 | 3578 6 1357 | 15 2 4 | | 13 6 4 | 2 135 9 | 7 8 35 | | 1237 158 1258 | 3578 4 1357 | 156 356 9 | |------------------+------------------+----------------| | 6 138 1389 | 357 135 1357 | 4 59 2 | | 5 2 79 | 6 8 4 | 3 79 1 | | 4 13 137 | 9 135 2 | 56 567 8 | *------------------------------------------------------*

We have r6c7=5 or r7c5=5 or r9c8=5, and r6c2=1 or r6c6=1 or r6c7=1. But:

[r4c7]=5=[r5c9]-5-[r3c9]=5=[r2c8]-5-[r79c8]=5=[r9c7]-5-[r4c7], => r6c7<>5.

[r7c5]-5-[r59c5]([r6c2]-1-[r5c1]=1=[r5c5]-1-[r9c5]=1=[r9c2]-1-[r6c2])
=5=[r5c9](-5-[r4c7]-1-[r4c3|r6c7])-5-[r3c9](=5=[r3c4]-5-[r46c4])=5=
=[r2c8]-5-[r2c3]=5=[r1c2]-5-[(r6c2)]-8-[(r4c3)]-5-[r4c6]=5=[r6c6], => r7c5<>5.

The final conclusion is that r9c8=5 and the puzzle is solved.

Carcul

by **Carcul** » Wed Apr 11, 2007 5:25 pm

Wapati wrote:There is at least one finned fish here, and some uniqueness!

Code: Select all: *------------------------------------------------------------* | 8 17 9 | 567 256 27 | 4 15 3 | | 25 25 3 | 4 1 9 | 7 6 8 | | 4 6 17 | 578 3 78 | 15 9 2 | |---------------------+-------------------+------------------| | 12356 134 8 | 567 256 247 | 9 125 15 | | 9 25 256 | 3 256 1 | 8 7 4 | | 7 14 1245 | 58 9 248 | 3 125 6 | |---------------------+-------------------+------------------| | 15 8 47 | 2 47 6 | 15 3 9 | | 256 9 2567 | 1 8 3 | 26 4 57 | | 1236 1347 12467 | 9 47 5 | 26 8 17 | *------------------------------------------------------------*

[r8c9]-7-[r9c9](-1-[r4c9]-5-[r4c15])-1-[r7c7](-5-[r7c1]-1-[r4c1|r9c13])-5-
-[r3c7]-1-[r3c3]-7-[r789c3]=5=[r8c3](-5-[r8c1|r5c3])=3=[r9c1]-3-[r4c1]
=3|5=[r5c5]-5-[r5c2], => r8c9<>7.

Carcul

by **wapati** » Wed Apr 11, 2007 8:28 pm

The way I do this one is using big finned fish.

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

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