## bored one night and...

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

### bored one night and...

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`. . .|. 2 3|. . 7. 9 .|. . .|. 6 .. . 8|9 . .|. . .-----+-----+-----. . 4|. 8 .|. . 53 . .|2 . 4|. . 61 . .|. 9 .|8 . .-----+-----+-----. . .|. . 2|4 . .. 6 .|. . .|. 1 .5 . .|4 7 .|. . .`

sometimes I will give the random generator on Sudocue a spin and this gem was displayed.
I liked Donm's thread so much that I thought it needed a "part 2."
sadly, anything I threw at this one hasn't crack it yet. its on my weekend to do list
any good moves are welcome.
norm
storm_norm

Posts: 85
Joined: 27 February 2008

Maybe a good first move would be applying the SSTS:
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`  *--------------------------------------------------------------------* | 6      145    15     | 158    2      3      | 59     4589   7      | | 27     9      135    | 1578   145    1578   | 235    6      12348  | | 27     1345   8      | 9      1456   1567   | 235    2345   1234   | |----------------------+----------------------+----------------------| | 9      27     4      | 1367   8      167    | 1237   237    5      | | 3      8      57     | 2      15     4      | 179    79     6      | | 1      257    6      | 357    9      57     | 8      2347   234    | |----------------------+----------------------+----------------------| | 8      137    1379   | 156    1356   2      | 4      57     39     | | 4      6      2379   | 58     35     589    | 57     1      2389   | | 5      13     1239   | 4      7      189    | 6      238    2389   | *--------------------------------------------------------------------*`
This is no trivial puzzle at SE 8.4. It's like a "Weekly Extreme."

Luke
2015 Supporter

Posts: 435
Joined: 06 August 2006
Location: Southern Northern California

Luke451 wrote:Maybe a good first move would be applying the SSTS:
Code: Select all
`  *--------------------------------------------------------------------* | 6      145    15     | 158    2      3      | 59     4589   7      | | 27     9      135    | 1578   145    1578   | 235    6      12348  | | 27     1345   8      | 9      1456   1567   | 235    2345   1234   | |----------------------+----------------------+----------------------| | 9      27     4      | 1367   8      167    | 1237   237    5      | | 3      8      57     | 2      15     4      | 179    79     6      | | 1      257    6      | 357    9      57     | 8      2347   234    | |----------------------+----------------------+----------------------| | 8      137    1379   | 156    1356   2      | 4      57     39     | | 4      6      2379   | 58     35     589    | 57     1      2389   | | 5      13     1239   | 4      7      189    | 6      238    2389   | *--------------------------------------------------------------------*`
This is no trivial puzzle at SE 8.4. It's like a "Weekly Extreme."

A solution as I happened upon it, so there could be redundancy in there somewhere...
1. 79r5c89=1r5c7-(1=5)r5c5-(5=7)r5c3-(7=2)r4c2-(27=5)r6c2-(5=7)r6c6 : <7>r6c8
2. 135r125c3=7r5c3-(7=9)r5c8-(79=1)r5c7-(1=5)r5c5-(5=3)r8c5 : <3>r8c3
3. 24r6c89=3r6c89-(3=57)r6c46-(57=2)r6c2-(2=7)r4c2-(7=5)r5c3-(5=1)r5c5-(17=6)r4c6-(671=3)r4c4-(37=2)r4c8 : <2>r4c7
4. 15r1c34=8r1c4-(8=35)r8c45-5r8c7=5r7c8 : <5>r1c8
5a. 58r18c4=1r1c4-(1=5)r1c3-(5=7)r5c3-(7=2)r4c2-(27=5)r6c2 : <5>r6c4
5b. chain continues : -(5=7)r6c6-(7=3)r6c4-(137=6)r4c4-(16=5)r7c4 : <5>r2c4
6. 589(r8c46+r9c6)=
i). (1-9)r9c6=(9-8)r8c6=8r2c6-8r2c9=8r1c8
ii). 1r9c6-(1=3)r9c2
=>2r9c8
=>8r9c9
=>3r8c9
=>5r8c5
: <5>r7c45 : placements 5r7c8 7r8c7 + resulting eliminations
7. 239r789c9=8r89c9-8r9c8=234r369-(23=7)r4c8-(7=9)r5c8-(9=1)r5c7-(1=3)r4c7-(3=25)r23c7 : <3>r6c9 <2>r23c9
8. 16r7c45=3r7c5-(3=5)r8c5-(5=1)r5c5-(1=9)r5c7-(9=5)r1c7-(5=1)r1c3 : <1>r7c3
9. 39r7c39=7r7c3-(7=5)r5c3-(5=1)r5c5-(1=9)r5c7-(9=5)r1c7-(5=1)r1c3-(15=8)r1c4-(8=5)r8c4-(5=3)r8c5 : <3>r7c5 => placement 3r8c5, eliminations resulting (including from pair 16r7c45).
10. 157(r5c5+r6c6+r4c6)=6r4c6-6r4c4=6r8c4-(6=1)r7c5-(1=5)r5c5-(5=7)r6c6 : <7>r6c4 : placement 3r7c4, resulting eliminations+pair24r6c89=>placement 2r4c2
11. 6r3c6=6r3c5-(6=1)r7c5-(1=5)r5c5-(5=7)r6c6 : <7>r3c6 =>placement 7r3c1 and 2r2c1, eliminations resulting
12. 5r6c6=(5-1)r5c5=1r5c7-(1=3)r4c7-(3=5)r2c7 : <5>r2c6
13. 5r2c7=3r2c7-(3=1)r4c7-(1=9)r5c7-(9=7)r5c8-(7=5)r5c3 : <5>r2c3
14. disc loop : 5r1c7-(5=3)r2c7-(3=1)r2c3-(1=5)r1c3-5r1c7 : <5>r1c7 : placement 9r1c7
singles to finish
Last edited by aran on Sun Mar 01, 2009 1:41 am, edited 3 times in total.
aran

Posts: 334
Joined: 02 March 2007

Quick work, aran! I guess what's extreme to one might be trivial to another. If it's not too tedious for you, I have several questions, and I'm only half way through your path!
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` *--------------------------------------------------------------------* | 6      145    15     | 158    2      3      | 59     4589   7      | | 27     9      135    | 1578   145    1578   | 235    6      12348  | | 27     1345   8      | 9      1456   1567   | 235    2345   1234   | |----------------------+----------------------+----------------------| | 9      27     4      | 1367   8      167    | 1237   237    5      | | 3      8      57     | 2      15     4      | 179    79     6      | | 1      257    6      | 357    9      57     | 8      2347   234    | |----------------------+----------------------+----------------------| | 8      137    1379   | 156    1356   2      | 4      57     39     | | 4      6      2379   | 58     35     589    | 57     1      2389   | | 5      13     1239   | 4      7      189    | 6      238    2389   | *--------------------------------------------------------------------* `

1. 79r5c89=1r5c7-(1=5)r5c5-(5=7)r5c3-(7=2)r4c2-(27=5)r6c2-(5=7)r6c6 : <7>r6c8
2. 135r125c3=7r5c3-(7=9)r5c8-(79=1)r5c7-(1=5)r5c5-(5=3)r8c5 : <3>r8c3
Steps 1 and 2 make sense to me, but only if I use "chain memory" (all highlights mine.)

3. 24r6c89=3r6c89-(3=57)r6c46-(57=2)r6c2-(2=7)r4c2-(7=5)r5c3- (5=1)r5c5-(67=1)r4c6-(671=3)r4c4-(37=2)r4c8 : <2>r4c7
This sequence lost me. Would you have any problem looking at it like this? (5=1)r5c5-(16)r4c46=(3)r4c4 (mindful of the memory on 7.)

6. 589(r8c46+r9c6)=
i). (1-9)r9c6=(9-8)r8c6=8r2c6-8r2c9=8r1c8
ii). 1r9c6-(1=3)r9c2
=>2r9c8
=>8r9c9
=>3r8c9
=>5r8c5
: <5>r7c45 : placements 5r7c5 7r8c7 + resulting eliminations.
I follow i) and ii) but the vertical notation after that is new to me. I gather that either the (589) group is true or the (3) and (8) are, but can't take that onward to the eliminations.

Hopefully you won't have to spend more time explaining mysteries to me than you did solving the puzzle...

Luke
2015 Supporter

Posts: 435
Joined: 06 August 2006
Location: Southern Northern California

Luke451 wrote:Quick work, aran! I guess what's extreme to one might be trivial to another. If it's not too tedious for you, I have several questions, and I'm only half way through your path!
Code: Select all
` *--------------------------------------------------------------------* | 6      145    15     | 158    2      3      | 59     4589   7      | | 27     9      135    | 1578   145    1578   | 235    6      12348  | | 27     1345   8      | 9      1456   1567   | 235    2345   1234   | |----------------------+----------------------+----------------------| | 9      27     4      | 1367   8      167    | 1237   237    5      | | 3      8      57     | 2      15     4      | 179    79     6      | | 1      257    6      | 357    9      57     | 8      2347   234    | |----------------------+----------------------+----------------------| | 8      137    1379   | 156    1356   2      | 4      57     39     | | 4      6      2379   | 58     35     589    | 57     1      2389   | | 5      13     1239   | 4      7      189    | 6      238    2389   | *--------------------------------------------------------------------* `

1. 79r5c89=1r5c7-(1=5)r5c5-(5=7)r5c3-(7=2)r4c2-(27=5)r6c2-(5=7)r6c6 : <7>r6c8
2. 135r125c3=7r5c3-(7=9)r5c8-(79=1)r5c7-(1=5)r5c5-(5=3)r8c5 : <3>r8c3
Steps 1 and 2 make sense to me, but only if I use "chain memory" (all highlights mine.)

3. 24r6c89=3r6c89-(3=57)r6c46-(57=2)r6c2-(2=7)r4c2-(7=5)r5c3- (5=1)r5c5-(67=1)r4c6-(671=3)r4c4-(37=2)r4c8 : <2>r4c7
This sequence lost me. Would you have any problem looking at it like this? (5=1)r5c5-(16)r4c46=(3)r4c4 (mindful of the memory on 7.)

6. 589(r8c46+r9c6)=
i). (1-9)r9c6=(9-8)r8c6=8r2c6-8r2c9=8r1c8
ii). 1r9c6-(1=3)r9c2
=>2r9c8
=>8r9c9
=>3r8c9
=>5r8c5
: <5>r7c45 : placements 5r7c5 7r8c7 + resulting eliminations.
I follow i) and ii) but the vertical notation after that is new to me. I gather that either the (589) group is true or the (3) and (8) are, but can't take that onward to the eliminations.

Hopefully you won't have to spend more time explaining mysteries to me than you did solving the puzzle...

Luke
Delighted to respond to your questions.

Step 1 : Yes that was chain memory of 7r5c3 (5r6c2 placed in the chain results from the immediate previous step 2r4c2 and the one preceding that 7r5c3, which is therefore the "remembered" element : little effort of memory required !).
Step 2 : Yes indeed 7r5c3 is remembered in reaching 1r5c7 in the chain.
It occurs to me here that it might be a good idea to do this in notation involving memory : place first "expected" candidate ie the immediate previous placement, then next the nearest remembered element, and so on. An example will make that clear :
in step 2 : this (7=9)r5c8-(79=1)r5c7 as presented, would become (7=9)r5c8-(97=1)r5c7.
Step 3 : two points here
i : there is a little typo in my chain...this : (5=1)r5c5-(67=1)r4c6 should read (5=1)r5c5-(17=6)r4c6. Mea culpa (will edit after this post). ie 1r5c5 and the remembered 7 from 57r6c46 produce (-17=6)r4c6. This 6r4c6 with remembered 1r5c5 and remembered 7 from 57r4c6 produce 3r4c4
From the foregoing I would now write that as (-617=3) to respect the order of memory recall.
ii : my objection to what you propose is that the memory doesn't appear in the transcription. So far as I know I don't use memory without referring to it in the chain : ie if I am remembering b and c having justed placed a, so as to produce d : then I will always write (-abc=d)
Step 6 :
As you have seen (i) and (ii) are separate strands emanating from 1r9c6. (i) serves first to assist in the placement of 2r9c8 as you have also followed, but also later with 9r8c6.
First of all the logic, then comment on the notation.
Logic : with 1r9c6 and 2r9c8 placed in the chain, 8r9c9 now follows being the only remaining possibilty for 8 in row 9.
We now use 2r8c9+8r9c9+9r8c6 (from the first strand) to establish 3r8c9 which therefore establishes 5r8c5 and we are home ie <5>r7c45 resulting from the strong link now proven 589(r8c46+r9c6)=5r8c5.
As to the notation well I just I thought it was simpler than using chain notation : because
- I would have had to explain that r9c9 was the only place left for 8, which I could have done for example by (-132=9)r9c2-(329=8)r9c9
- then I would have had to remember 2r9c8 from the "current" strand and 9r8c6 from the first strand which I could have done as (-829=3)r8c9 without any reference to the first strand or maybe with an asterisk to remind (-829*=3)r8c9.

The foregoing is not I hope too indigestable.
I consider in any case that step 6 is the best move in the solution even if it does involve two strands (in other words a net, but far from a tortuous one, with a clear objective).
aran

Posts: 334
Joined: 02 March 2007

Code: Select all
`After SSTS *--------------------------------------------------------------------* | 6      145    15     | 158    2      3      | 59     4589   7      | | 27     9      135    | 1578   145    1578   | 235    6      12348  | | 27     1345   8      | 9      1456   1567   | 235    2345   1234   | |----------------------+----------------------+----------------------| | 9      27     4      | 1367   8      167    | 1237   237    5      | | 3      8      57     | 2      15     4      | 179    79     6      | | 1      257    6      | 357    9      57     | 8      2347   234    | |----------------------+----------------------+----------------------| | 8      137    1379   | 156    1356   2      | 4      57     39     | | 4      6      2379   | 58     35     589    | 57     1      2389   | | 5      13     1239   | 4      7      189    | 6      238    2389   | *--------------------------------------------------------------------*`

Don’t know, why I’m here…
My path for this one, on using Eureka/AIC notation.

01: (1)r4c7=(1)r5c7-(1=5)r5c5-(5=7)r5c3-(7=2)r4c2 => r4c7<>2
02: (1)r4c7=(1)r5c7-(1=5)r5c5-(5=7)r5c3-(7)r8c3=(7)r8c7 => r4c7<>7
03: (3)r2c3=(3-hp45)r13c2=(5)r6c2-(5)r5c3=(5-1)r5c5=(1)r5c7-(1=3)r4c7 => r2c7<>3
04: (5=2)r2c7-(2)r2c1=(2-7)r3c1=(7)r3c6-(7=5)r6c6-(5)r5c5=(5)r5c3 => r2c36<>5
05: (7)r3c6=(7-2)r3c1=(2)r2c1-(2)r2c7=(2-3)r3c7=(3-1)r4c7=(1)r5c7-(1=5)r5c5-(5=7)r6c6 => r2c6<>7
06: (4)r2c5=(4-6)r3c5=(6-7)r3c6=(7-2)r3c1=(2)r2c1-(2=5)r2c7 => r2c5<>5
07: (hp58=1)r18c4-(1=5)r1c3-(5)r5c3=(5)r5c5-(5=7)r6c6-(7)r3c6=(7)r2c4 => r2c4<>5, singles to the end

Edit: I studied more and see that step 5 is not necessary, in that case step 7 can write as:
(hp58=1)r18c4-(1=5)r1c3-(5)r5c3=(5)r5c5-(5=7)r6c6-(7)r23c6=(7)r2c4 => r2c4<>5, singles to the end

ttt
ttt

Posts: 185
Joined: 20 October 2006
Location: vietnam

ttt wrote:
Code: Select all
`After SSTS *--------------------------------------------------------------------* | 6      145    15     | 158    2      3      | 59     4589   7      | | 27     9      135    | 1578   145    1578   | 235    6      12348  | | 27     1345   8      | 9      1456   1567   | 235    2345   1234   | |----------------------+----------------------+----------------------| | 9      27     4      | 1367   8      167    | 1237   237    5      | | 3      8      57     | 2      15     4      | 179    79     6      | | 1      257    6      | 357    9      57     | 8      2347   234    | |----------------------+----------------------+----------------------| | 8      137    1379   | 156    1356   2      | 4      57     39     | | 4      6      2379   | 58     35     589    | 57     1      2389   | | 5      13     1239   | 4      7      189    | 6      238    2389   | *--------------------------------------------------------------------*`

Don’t know, why I’m here…
My path for this one, on using Eureka/AIC notation.

01: (1)r4c7=(1)r5c7-(1=5)r5c5-(5=7)r5c3-(7=2)r4c2 => r4c7<>2
02: (1)r4c7=(1)r5c7-(1=5)r5c5-(5=7)r5c3-(7)r8c3=(7)r8c7 => r4c7<>7
03: (3)r2c3=(3-hp45)r13c2=(5)r6c2-(5)r5c3=(5-1)r5c5=(1)r5c7-(1=3)r4c7 => r2c7<>3
04: (5=2)r2c7-(2)r2c1=(2-7)r3c1=(7)r3c6-(7=5)r6c6-(5)r5c5=(5)r5c3 => r2c36<>5
05: (7)r3c6=(7-2)r3c1=(2)r2c1-(2)r2c7=(2-3)r3c7=(3-1)r4c7=(1)r5c7-(1=5)r5c5-(5=7)r6c6 => r2c6<>7
06: (4)r2c5=(4-6)r3c5=(6-7)r3c6=(7-2)r3c1=(2)r2c1-(2=5)r2c7 => r2c5<>5
07: (hp58=1)r18c4-(1=5)r1c3-(5)r5c3=(5)r5c5-(5=7)r6c6-(7)r3c6=(7)r2c4 => r2c4<>5, singles to the end

Edit: I studied more and see that step 5 is not necessary, in that case step 7 can write as:
(hp58=1)r18c4-(1=5)r1c3-(5)r5c3=(5)r5c5-(5=7)r6c6-(7)r23c6=(7)r2c4 => r2c4<>5, singles to the end

ttt

ttt
You made short work of that
You can even reduce your solution to 4 moves with a final discontinuous loop :
4. (5=2)r2c7-2r2c1=(2-7)r3c1=
(i) (7-6)r3c6=(6-4)r3c5=(4-5)r2c5
(ii)7r3c6-(7=5)r6c6-5r5c5=5r5c3-(5=1)r1c3-(1=58r18c4)-5r2c4
=5r2c7
: => placement 5r2c7
(underlined items=removal/removal mechanism of 5s in row 2)
aran

Posts: 334
Joined: 02 March 2007

aran wrote:ttt
You made short work of that
You can even reduce your solution to 4 moves with a final discontinuous loop :
4. (5=2)r2c7-2r2c1=(2-7)r3c1=
(i) (7-6)r3c6=(6-4)r3c5=(4-5)r2c5
(ii)7r3c6-(7=5)r6c6-5r5c5=5r5c3-(5=1)r1c3-(1=58r18c4)-5r2c4
=5r2c7
: => placement 5r2c7
(underlined items=removal/removal mechanism of 5s in row 2)

I don’t understand what you meant…
I only tried to reduce steps to minimum without AAICs or nets. On my experiences, most (or all?) of puzzles that has rank up to ER8.4 can solve without AAICs or nets.

ttt
ttt

Posts: 185
Joined: 20 October 2006
Location: vietnam

ttt wrote:
aran wrote:ttt
You made short work of that
You can even reduce your solution to 4 moves with a final discontinuous loop :
4. (5=2)r2c7-2r2c1=(2-7)r3c1=
(i) (7-6)r3c6=(6-4)r3c5=(4-5)r2c5
(ii)7r3c6-(7=5)r6c6-5r5c5=5r5c3-(5=1)r1c3-(1=58r18c4)-5r2c4
=5r2c7
: => placement 5r2c7
(underlined items=removal/removal mechanism of 5s in row 2)

I don’t understand what you meant…
I only tried to reduce steps to minimum without AAICs or nets. On my experiences, most (or all?) of puzzles that has rank up to ER8.4 can solve without AAICs or nets.

ttt

Move 4 as I gave it is neither a net nor an AAIC...
it is a two strand chain, so you can regard that as 2 moves if you prefer
4(i) which establishes <5>r2c5
4(ii) which establishes <5>r2c346
aran

Posts: 334
Joined: 02 March 2007

ttt wrote:03: (3)r2c3=(3-hp45)r13c2=(5)r6c2-(5)r5c3=(5-1)r5c5=(1)r5c7-(1=3)r4c7 => r2c7<>3
I had trouble figuring out the start of this move. Would you consider it the same as this?
Code: Select all
`(3)r2c3=(15)r12c3-(5)r13c2=(5-1)etc.`
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`*--------------------------------------------------------------------* | 6      145    15     | 158    2      3      | 59     4589   7      | | 27     9      135    | 1578   145    1578   | 25     6      12348  | | 27     1345   8      | 9      1456   1567   | 235    2345   1234   | |----------------------+----------------------+----------------------| | 9      27     4      | 1367   8      167    | 13     237    5      | | 3      8      57     | 2      15     4      | 179    79     6      | | 1      257    6      | 357    9      57     | 8      2347   234    | |----------------------+----------------------+----------------------| | 8      137    1379   | 156    1356   2      | 4      57     39     | | 4      6      2379   | 58     35     589    | 57     1      2389   | | 5      13     1239   | 4      7      189    | 6      238    2389   | *--------------------------------------------------------------------*`

ttt also wrote:04: (5=2)r2c7-(2)r2c1=(2-7)r3c1=(7)r3c6-(7=5)r6c6-(5)r5c5=(5)r5c3 => r2c36<>5

Could you have continued? I recently learned about a criss cross group link (as applied by aran in b3 here) that seems to work:
Code: Select all
`=(5-7)r5c3=(7)r5c78-(7)r46c8=(7)r7c8-(7=5)r8c7-(5)r2c7=loop`

This softens things up considerably.

Luke
2015 Supporter

Posts: 435
Joined: 06 August 2006
Location: Southern Northern California

Luke451 wrote:
ttt also wrote:04: (5=2)r2c7-(2)r2c1=(2-7)r3c1=(7)r3c6-(7=5)r6c6-(5)r5c5=(5)r5c3 => r2c36<>5

Could you have continued? I recently learned about a criss cross group link (as applied by aran in b3 here) that seems to work:
Code: Select all
`=(5-7)r5c3=(7)r5c78-(7)r46c8=(7)r7c8-(7=5)r8c7-(5)r2c7=loop`

Strong inference (7)r46c8=(7)r7c8 should be (7)r456c8=(7)r7c8. Then (7)r5c8 being part of two strong inferences stops this from working.
ronk
2012 Supporter

Posts: 4764
Joined: 02 November 2005
Location: Southeastern USA

ronk wrote:
Luke451 wrote:
ttt also wrote:04: (5=2)r2c7-(2)r2c1=(2-7)r3c1=(7)r3c6-(7=5)r6c6-(5)r5c5=(5)r5c3 => r2c36<>5

Could you have continued? I recently learned about a criss cross group link (as applied by aran in b3 here) that seems to work:
Code: Select all
`=(5-7)r5c3=(7)r5c78-(7)r46c8=(7)r7c8-(7=5)r8c7-(5)r2c7=loop`

Strong inference (7)r46c8=(7)r7c8 should be (7)r456c8=(7)r7c8. Then (7)r5c8 being part of two strong inferences stops this from working.

Looking back at the cited example, I notice that aran's "criss-cross" has one group in a row, the other in a column with no interaction. Mine has col/row interaction. Mine is actually a "T," not a cross. Ah, well, another day, another lesson. Thanks for the clarification.

Luke
2015 Supporter

Posts: 435
Joined: 06 August 2006
Location: Southern Northern California

Busy week- finally got a little time to work on this. Thanks storm_norm- this type of puzzle (circa ER=8.4) is one of my favorite types of puzzles. It gives you a nice challenge to use everything at your disposal short of a net. I wish I had more time to optimize my solution because there's a fair amount of redundancy (ie. re-use of the same patterns) and I think I'd be able to chop off 2 or 3 chains. Nothing too fancy- lots of groups... hopefully no typos. Btw: tf=turbofish ie. 2-string kite and the like, linebox=locked candidates:

1. grp(7)r5c78=r5c3-(7=2)r4c2-r6c2=hp(24)r6c89 => r6c8<>7
-> als xz-rule/2-set chain: (237)r4c28->rc=3<=(234)r6c89 => r4c7<>2 -> linebox(2)r23c7: r3c8<>2, r23c9<>2
2. (7)r8c7=r7c8-r7c2=grp(7)r46c2-r5c3=grp(7)r5c78 => r4c7<>7
3. (3=1)r4c7-grp(1)r4c46=(1-5)r5c5=(5-7)r5c3=grp(7)r46c2-als(7=13)r79c2-(3)r3c2=grp(3)r3c789 => r2c7<>3
4. (5=2)r2c7-(2=7)r2c1-r2c4=grp(7)r46c4-(7=5)r6c6-(5=1)r5c5-r5c7=(1-3)r4c7=(3)r3c7 => r2c6<>5, r3c7<>5
5. (2)r3c7=r2c7-(2=7)r2c1-r2c4=grp(7)r46c4-(7=5)r6c6-(5=1)r5c5-r5c7=(1-3)r4c7=(3)r3c7-loop => r24c6<>7
6. (5=2)r2c7-(2=3)r3c7-(3=1)r4c7-grp(1)r4c46=(1-5)r5c5=(5)r5c3 => r2c3<>5
7. grp(5)r2c45=(5-2)r2c7=(2-3)r3c7=(3-1)r4c7=grp(1)r4c46-(1=5)r5c5-r5c3=(5)r1c3 => r1c4<>5
-> np(18)r1c4/r2c6:r2c45<>1, r3c56<>1, r2c4<>8 -> nq(2457)r2c1457: r2c9<>4 -> r2c5=4
8. (1=5)r5c5-r5c3=r6c2-(5)r1c2=tf(5)r36c2/r5c35-(5=6)r3c5-r3c6=(6)r4c6 => r4c6<>1=6 -> r3c5=6, r7c4=6
9. xy-chain: (2)r2c7->r2c1->r2c4->r8c4->r1c4->(1->5)r1c3 => r1c7<>5=9 -> r5c8=9 -> xy-wing(157)r5c7/r5c5/r8c7: r8c5<>5=3
-> xy-chain: (7)r3c6->r2c4->r2c7->r3c7->r4c7->r5c7->r5c3->r1c3->r1c4->r8c4->r7c5->(1->5)r5c5 => r6c6<>5=7

STTE
DonM
2013 Supporter

Posts: 475
Joined: 13 January 2008

Luke451 wrote:
ttt wrote:03: (3)r2c3=(3-hp45)r13c2=(5)r6c2-(5)r5c3=(5-1)r5c5=(1)r5c7-(1=3)r4c7 => r2c7<>3
I had trouble figuring out the start of this move. Would you consider it the same as this?
Code: Select all
`(3)r2c3=(15)r12c3-(5)r13c2=(5-1)etc.`

Yes, nice find... . The reason..., I found the net to eliminate r2c3=5 first that require r2c7<>3 based on hp(45)r13c2 and after that downgrade this puzzle.

Couldnot extend that step... Yes, that's T (not cross...)

ttt
ttt

Posts: 185
Joined: 20 October 2006
Location: vietnam

DonM wrote:1. grp(7)r5c78=r5c3-(7=2)r4c2-r6c2=hp(24)r6c89 => r6c8<>7
-> als xz-rule/2-set chain: (237)r4c28->rc=3<=(234)r6c89 => r4c7<>2 -> linebox(2)r23c7: r3c8<>2, r23c9<>2
2. (7)r8c7=r7c8-r7c2=grp(7)r46c2-r5c3=grp(7)r5c78 => r4c7<>7
3. (3=1)r4c7-grp(1)r4c46=(1-5)r5c5=(5-7)r5c3=grp(7)r46c2-als(7=13)r79c2-(3)r3c2=grp(3)r3c789 => r2c7<>3
4. (5=2)r2c7-(2=7)r2c1-r2c4=grp(7)r46c4-(7=5)r6c6-(5=1)r5c5-r5c7=(1-3)r4c7=(3)r3c7 => r2c6<>5, r3c7<>5
5. (2)r3c7=r2c7-(2=7)r2c1-r2c4=grp(7)r46c4-(7=5)r6c6-(5=1)r5c5-r5c7=(1-3)r4c7=(3)r3c7-loop => r24c6<>7
6. (5=2)r2c7-(2=3)r3c7-(3=1)r4c7-grp(1)r4c46=(1-5)r5c5=(5)r5c3 => r2c3<>5
7. grp(5)r2c45=(5-2)r2c7=(2-3)r3c7=(3-1)r4c7=grp(1)r4c46-(1=5)r5c5-r5c3=(5)r1c3 => r1c4<>5
-> np(18)r1c4/r2c6:r2c45<>1, r3c56<>1, r2c4<>8 -> nq(2457)r2c1457: r2c9<>4 -> r2c5=4
8. (1=5)r5c5-r5c3=r6c2-(5)r1c2=tf(5)r36c2/r5c35-(5=6)r3c5-r3c6=(6)r4c6 => r4c6<>1=6 -> r3c5=6, r7c4=6
9. xy-chain: (2)r2c7->r2c1->r2c4->r8c4->r1c4->(1->5)r1c3 => r1c7<>5=9 -> r5c8=9 -> xy-wing(157)r5c7/r5c5/r8c7: r8c5<>5=3
-> xy-chain: (7)r3c6->r2c4->r2c7->r3c7->r4c7->r5c7->r5c3->r1c3->r1c4->r8c4->r7c5->(1->5)r5c5 => r6c6<>5=7
STTE

Hi, Don. You'd think I'd spend my time coming up with my own solution path rather than scoping out everyone else's, but NOooo... at this point in my career I find the latter much more educational.

Code: Select all
`1. grp(7)r5c78=r5c3-(7=2)r4c2-r6c2=hp(24)r6c89 => r6c8<>7 *--------------------------------------------------------------------* | 6      145    15     | 158    2      3      | 59     4589   7      | | 27     9      135    | 1578   145    1578   | 235    6      12348  | | 27     1345   8      | 9      1456   1567   | 235    2345   1234   | |----------------------+----------------------+----------------------| | 9      27     4      | 1367   8      167    | 1237   237    5      | | 3      8      57     | 2      15     4      | 179    79     6      | | 1      257    6      | 357    9      57     | 8      2347   234    | |----------------------+----------------------+----------------------| | 8      137    1379   | 156    1356   2      | 4      57     39     | | 4      6      2379   | 58     35     589    | 57     1      2389   | | 5      13     1239   | 4      7      189    | 6      238    2389   | *--------------------------------------------------------------------*`
I'm not surprised you found this one to start your solution. If it seems familiar, it is: it's the same first move that you used in ASI #2c Extreme #118. Is it my imagination, or did you "target" this elimination after noting the conjugate 4's in b6 (hidden pair potential)?

Later you were here: 8.(1=5)r5c5-r5c3=r6c2-(5)r1c2=tf(5)r36c2/r5c35-(5=6)r3c5-r3c6=(6)r4c6 => r4c6<>1=6 -> r3c5=6, r7c4=6
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` *-----------------------------------------------------------* | 6     145   15    | 18    2     3     | 59    4589  7     | | 27    9     13    | 57    4     18    | 25    6     138   | | 27    1345  8     | 9     56    567   | 23    2345  1234  | |-------------------+-------------------+-------------------| | 9     27    4     | 1367  8     16    | 1237  237   5     | | 3     8     57    | 2     15    4     | 179   79    6     | | 1     257   6     | 357   9     57    | 8     2347  234   | |-------------------+-------------------+-------------------| | 8     137   1379  | 156   1356  2     | 4     57    39    | | 4     6     2379  | 58    35    589   | 57    1     2389  | | 5     13    1239  | 4     7     189   | 6     238   2389  | *-----------------------------------------------------------*`

If the highlighted section is gone would you have the same result?

Luke
2015 Supporter

Posts: 435
Joined: 06 August 2006
Location: Southern Northern California

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