Can this be solved without using long chains?

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Can this be solved without using long chains?

Postby 999_Springs » Thu Oct 30, 2008 8:26 am

Code: Select all
Original puzzle:
...|...|...
...|841|...
.64|...|53.
---+---+---
..2|7.3|6..
7..|...|..5
.9.|.2.|.7.
---+---+---
..1|...|3..
.4.|.7.|.6.
9..|.8.|..7

My current state:
128  28  79 |56  3  56  |789 1489 1489
35   35  79 |8   4  1   |279 29   6
18   6   4  |2   9  7   |5   3    18
------------+-----------+-------------
458  158 2  |7   15 3   |6   489  489
7    138 38 |469 16 469 |28  248  5
456  9   56 |45  2  8   |1   7    3
------------+-----------+-------------
2568 7   1  |459 56 249 |3   589  289
2358 4   358|13  7  259 |89  6    129
9    235 36 |13  8  256 |4   15   7

So far I have used a few rather long chains to find some numbers and some shorter ones to find a few completely useless eliminations.
Now I am stuck.

(I would also prefer not to use uniqueness.)
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Postby Draco » Thu Oct 30, 2008 8:56 am

From your PM's, one, not-too-long contradiction net cracks the puzzle to SSTS:

r5c5=6 r5c2=1 r5c3=3 + r5c5=6 r9c3=3 ==> r5c5 <> 6

I suck at turning these into nice loops and/or chains, but I have no doubt someone else in the forum can easily explain how to do that. I think it might be represented as:

r5c5-6-r9c3-3-r5c3=8 ... which leaves r5c2 as either 3 or 1. Since r5c2 is the only square in r5 that has a 3 or 1 you have your contradiction, hence r5c5<>6.

A couple of XY wings and a nice swordfish are the hardest steps before you are left with nothing but singles.

Cheers...

- drac
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Postby Glyn » Thu Oct 30, 2008 9:15 am

Draco an AIC for that is
(1)r5c5=(1-3)r5c2=(3)r5c3-(3=6)r9c3-(6)r9c6=(6)r7c5 => r5c5<>6
The Nice Loop I'd better leave to someone else.
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Postby hobiwan » Thu Oct 30, 2008 9:34 am

Draco wrote:From your PM's, one, not-too-long contradiction net cracks the puzzle to SSTS:

r5c5=6 r5c2=1 r5c3=3 + r5c5=6 r9c3=3 ==> r5c5 <> 6


Not your loop, but a possible loop for your elimination:

r5c5 -6- r7c5 =6= r7c1 -6- r9c3 -3- r5c3 =3= r5c2 =1= r5c5 => r5c5<>6
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Postby hobiwan » Thu Oct 30, 2008 10:00 am

999_springs,

there are a number of moves that have the same effects as draco's chain, but if you want a fast solution you will need uniqueness:

Code: Select all
Finned Franken Swordfish: 8 c37b1 r15b7 fr3c1 fr8c7 => r8c1<>8
Uniqueness Test 3: 7/9 in r1c37,r2c37 => r8c7<>8
Singles

The finned franken swordfish can be replaced by a Forcing Chain:
Code: Select all
Forcing Chain Contradiction in c7 => r8c1<>8
  r8c1=8 r3c9=8 r1c7<>8
  r8c1=8 r5c3=8 r5c7<>8
  r8c1=8 r8c7<>8
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Re: Can this be solved without using long chains?

Postby Luke » Thu Oct 30, 2008 10:52 am

999_Springs wrote:
Code: Select all

My current state:
128  28  79 |56  3  56  |789 1489 1489
35   35  79 |8   4  1   |279 29   6
18   6   4  |2   9  7   |5   3    18
------------+-----------+-------------
458  158 2  |7   15 3   |6   489  489
7    138 38 |469 16 469 |28  248  5
456  9   56 |45  2  8   |1   7    3
------------+-----------+-------------
2568 7   1  |459 56 249 |3   589  289
2358 4   358|13  7  259 |89  6    129
9    235 36 |13  8  256 |4   15   7

So far I have used a few rather long chains to find some numbers and some shorter ones to find a few completely useless eliminations.
Now I am stuck.

(I would also prefer not to use uniqueness.)

So you have no interest the type 4 UR on <79> that solves the puzzle?
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Postby DonM » Thu Oct 30, 2008 11:44 am

Nice catch Luke. It's like a naked pair just sitting there in a little camouflage.:)
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Postby Glyn » Thu Oct 30, 2008 12:09 pm

Luke451 Doesn't just avoiding the (79)URr12c37 => r8c7=9 still leave some work to be done.
Did you follow it with avoid (49)URr57c46 => r5c8=4 to leave singles.
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Postby Draco » Thu Oct 30, 2008 12:28 pm

hobiwan wrote:Not your loop, but a possible loop for your elimination:

r5c5 -6- r7c5 =6= r7c1 -6- r9c3 -3- r5c3 =3= r5c2 =1= r5c5 => r5c5<>6

I don't quite follow why r5c2=3; it oculd just as easily be r5c2=1 since both a 3 and 1 are required for the row, which is what leads to the contradiction. Are you relying on r5c2=3 as the only 3 for b4 for your chain?

Asking because I am never quite sure how to express these contradiction chains with ambiguous forces. My supposition after looking at your chain is one is free to pick whichever "legal" value most clearly illustrates the contradiction, but want to validate that against your thinking.

Cheers...

- drac

PS: 999_Springs - thanks for a great puzzle!
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Postby susume » Thu Oct 30, 2008 1:22 pm

In nice loop notation, "=" doesn't mean "equals;" rather it is the symbol used for a strong inference. The snippet r5c3 =3= r5c2 does not mean r5c2 equals 3 or is assumed to equal 3; it means 3 must be in either r5c3 OR r5c2. The overlapping snippet r5c2 =1= r5c5 means 1 must be in either r5c2 OR r5c5. The two inferences connecting to r5c5 in this loop are a strong one (on 1) and a weak one (on 6); by nice loop rules the digit with the weak inference (6) can be eliminated.
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Postby Luke » Thu Oct 30, 2008 1:45 pm

Glyn wrote:Luke451 Doesn't just avoiding the (79)URr12c37 => r8c7=9 still leave some work to be done.
Did you follow it with avoid (49)URr57c46 => r5c8=4 to leave singles.
Exactly, but with a little triple in between. I figured mentioning a second UR would be pushing it, given Springs's stated preference.
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Postby hobiwan » Fri Oct 31, 2008 12:54 am

Draco,

susume's explanation is correct. In Nice Loop Notation only links between cells are written, so you have to add the links within the cells yourself.

The original loop:
r5c5 -6- r7c5 =6= r7c1 -6- r9c3 -3- r5c3 =3= r5c2 =1= r5c5 => r5c5<>6

The loop with all inferences written out (the parts in read happen inside the respective cells and should ommitted, the notation therefore is incorrect):
r5c5 -6- r7c5 =6= r7c1 -6- r9c3 =3= r9c3 -3- r5c3 =3= r5c2 -1- r5c2 =1= r5c5 -6- r5c5 => r5c5<>6

read from left to right that means:
r5c5=6 => r7c5<>6 => r7c1=6 => r9c3<>6 => r9c3=3 => r5c3<>3 => r5c2=3 => r5c2<>1 => r5c5=1 => r5c5<>6

So r5c5=6 => r5c5<>6 (and of course r5c5<>6 => r5c5<>6) therefore r5c5<>6.

[edited to make things clearer]
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Postby Draco » Fri Oct 31, 2008 7:47 pm

Thanks Susume & Hobiwan... I knew there was a good reason for me to resist NL notation:) . Seriously though, I do appreciate your taking the time to run through the details.

Cheers...

- drac
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Postby 999_Springs » Sat Nov 01, 2008 10:24 am

Thank you everyone for your help - but from the original puzzle I have had to use some rather long chains to get to my current state (in particular, to solve r6c7 and then r6c6). Is there a solution using shorter chains?
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Postby Glyn » Sat Nov 01, 2008 11:18 am

999_Springs With the benefit of hindsight
If you did allow Uniqueness then this would work
After the initial singles we have
Code: Select all
.---------------------.---------------------.---------------------.
| 1258   1258   5789  | 56     3      56    | 124789 12489  12489 |
| 235    235    3579  | 8      4      1     | 279    29     6     |
| 18     6      4     | 2      9      7     | 5      3      18    |
:---------------------+---------------------+---------------------:
| 1458   158    2     | 7      15     3     | 6      1489   1489  |
| 7      138    368   | 1469   16     4689  | 12489  12489  5     |
| 14568  9      568   | 1456   2      4568  | 148    7      3     |
:---------------------+---------------------+---------------------:
| 2568   7      1     | 4569   56     24569 | 3      24589  2489  |
| 2358   4      358   | 1359   7      259   | 1289   6      1289  |
| 9      235    356   | 13456  8      2456  | 124    1245   7     |
'---------------------'---------------------'---------------------'
9's of Box 5 locked in row 5.
(79)URr12c37 =>r8c7=9
1's of Box 8 locked in column 4.
Hidden pair {13}r89c4
Naked pair {56}r1c46
Naked triple {128}r1c12,r3c1
2's of Box 1 locked in Row 1
2's of Box 9 locked in Column 9
The AIC (4=1)r9c7-(1)r6c7=(1-6)r6c1=(6)r7c1-(6)r9c3=(6)r9c6 => r9c6<>4
4's of Box 8 locked in Row 7
(49)URr57c46 => 4's in r5c78
Hidden singles r1c9=4, r4c9=9.
XY-wing on r4c8 => r7c5<>5
Then Singles
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