The New Sudoku Players' Forum

Website · by **denis_berthier** » Wed Sep 07, 2022 10:31 am

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Code: Select all: +-------+-------+-------+ ! . . . ! . . 6 ! . . . ! ! . . . ! 1 8 . ! . 3 6 ! ! . . 6 ! . 2 3 ! 1 . 4 ! +-------+-------+-------+ ! . 4 5 ! . . . ! . . . ! ! . . . ! . . . ! . 4 1 ! ! 7 1 9 ! . . . ! . . . ! +-------+-------+-------+ ! . 3 . ! 6 . . ! 4 . 8 ! ! . . . ! 8 3 . ! . 1 2 ! ! . 8 . ! . 4 . ! 3 6 . ! +-------+-------+-------+ .....6......18..36..6.231.4.45.............41719.......3.6..4.8...83..12.8..4.36.;100;28435 SER = 11.6

Code: Select all: Resolution state after Singles and whips[1]: +-------------------------+-------------------------+-------------------------+ ! 1234589 2579 123478 ! 4579 579 6 ! 25789 25789 579 ! ! 2459 2579 247 ! 1 8 4579 ! 2579 3 6 ! ! 589 579 6 ! 579 2 3 ! 1 5789 4 ! +-------------------------+-------------------------+-------------------------+ ! 2368 4 5 ! 2379 1679 12789 ! 26789 2789 379 ! ! 2368 26 238 ! 23579 5679 25789 ! 256789 4 1 ! ! 7 1 9 ! 2345 56 2458 ! 2568 258 35 ! +-------------------------+-------------------------+-------------------------+ ! 1259 3 127 ! 6 1579 12579 ! 4 579 8 ! ! 4569 5679 47 ! 8 3 579 ! 579 1 2 ! ! 1259 8 127 ! 2579 4 12579 ! 3 6 579 ! +-------------------------+-------------------------+-------------------------+ 200 candidates

by **shye** » Wed Sep 07, 2022 6:25 pm

this was fun!
after hidden pair at the start:

Code: Select all: .-----------------.--------------------.-------------------. | 1 *2579 3 | 4 #579 6 | 25789 25789 #579 | |*2459 *2579 *247 | 1 8 #579 |#2579 3 6 | | 8 579 6 |#579 2 3 | 1 #579 4 | :-----------------+--------------------+-------------------: | 236 4 5 |*2379 1 8 | 2679 279 379 | | 236 6-2 8 |*23579 5679 *2579 | 25679 4 1 | | 7 1 9 |*235 56 4 | 2568 258 35 | :-----------------+--------------------+-------------------: | 259 3 127 | 6 #579 12579 | 4 #579 8 | | 4569 5679 47 | 8 3 #579 |#579 1 2 | | 259 8 127 |#2579 4 12579 | 3 6 #579 | '-----------------'--------------------'-------------------'

trivalue oddagon with dual ER
# marked cells cannot all be 579
2b1p2 = 2b1p456 - 2r2c7 = 2r9c4 - 2b5p147 = 2b5p6
=> -2r5c2

after singles and some lc's:

Code: Select all: .-------------.-------------------.-------------------. | 1 259 3 | 4 579 6 | 25789 25789 579 | | 4 259 7 | 1 8 59 | 259 3 6 | | 8 59 6 | 579 2 3 | 1 579 4 | :-------------+-------------------+-------------------: | 23 4 5 | 2379 1 8 | 6 279 379 | | 23 6 8 | 23579 579 2579 | 2579 4 1 | | 7 1 9 | 235 6 4 | 258 258 35 | :-------------+-------------------+-------------------: | 59 3 #12 | 6 579 #1579-2| 4 579 8 | | 6 7 4 | 8 3 59 | 59 1 2 | | 59 8 #12 | 2579 4 #1579-2| 3 6 579 | '-------------'-------------------'-------------------'

12UR in r79c36
=> -2r79c6
stte

by **SerTerence** » Thu Sep 08, 2022 9:33 am

+-------------------------+-------------------------+-------------------------+
! 1 2579 3 ! 4 579 6 ! 25789 25789 579 !
! 2459 2579 247 ! 1 8 579 ! 2579 3 6 !
! 8 579 6 ! 579 2 3 ! 1 5789 4 !
+-------------------------+-------------------------+-------------------------+
! 236 4 5 ! 2379 1 8 ! 2679 279 379 !
! 236 26 238 ! 23579 5679 2579 ! 25679 4 1 !
! 7 1 9 ! 235 56 4 ! 2568 258 35 !
+-------------------------+-------------------------+-------------------------+
! 259 3 127 ! 6 579 12579 ! 4 579 8 !
! 4569 5679 47 ! 8 3 579 ! 579 1 2 !
! 259 8 127 ! 2579 4 12579 ! 3 6 579 !
+-------------------------+-------------------------+-------------------------+
(c1r8) & (c2r8) are the only two places a 6 can go.
If (c2r8) = 5,7 or 9. (c1r8) = 6.
If (c2r8) = 6 then (c3r8) = 4.
Therefore the 4 can be removed from (c1r8).
(c3r8) = 4. so far

by **totuan** » Thu Sep 08, 2022 2:33 pm

SerTerence wrote: (c1r8) & (c2r8) are the only two places a 6 can go.
If (c2r8) = 5,7 or 9. (c1r8) = 6.
If (c2r8) = 6 then (c3r8) = 4.
Therefore the 4 can be removed from (c1r8).
(c3r8) = 4. so far

Code: Select all: *--------------------------------------------------------------------* | 1 2579 3 | 4 579 6 | 25789 25789 579 | | 2459 2579 *247 | 1 8 579 | 2579 3 6 | | 8 579 6 | 579 2 3 | 1 579 4 | |----------------------+----------------------+----------------------| | 236 4 5 | 2379 1 8 | 2679 279 379 | | 236 c26 8 | 23579 5679 *2579 | 25679 4 1 | | 7 1 9 | 235 56 4 | 2568 258 35 | |----------------------+----------------------+----------------------| | 259 3 #127 | 6 579 #12579 | 4 579 8 | |a569-4 b5679 @47 | 8 3 579 | 579 1 2 | | 259 8 #127 | 2579 4 #12579 | 3 6 579 | *--------------------------------------------------------------------*

Hi SerTerence, I can explain your move as below:
Look at UR(12)r79c36 => (2)r5c6=(2)r2c3
(6)r8c1=r8c2-(6=2)r5c2-(2)r5c6==r2c3-(2=147)r789c3 => r8c1<>4, r8c3=4
From here, without tridragon the puzzle is still very very hard

Thanks for the puzzle!
totuan

by **eleven** » Thu Sep 08, 2022 3:27 pm

Welcome SerTerence,
to format the grid, mark it and click the "Code" button.

SerTerence wrote:If (c2r8) = 6 then (c3r8) = 4.

Why ? Because of the 12 UR ? You should explain that.

----
Using the TH (Thor's hammer or tridagon) plus the UR, you get a one-stepper:

Code: Select all: +----------------------+-----------------------+-----------------------+ | 1 2579 3 | 4 #579 6 | 25789 25789 #579 | | 2459 2579 c247 | 1 8 #579 |b#579+2 3 6 | | 8 579 6 | #579 2 3 | 1 #579 4 | +----------------------+-----------------------+-----------------------+ | 236 4 5 | 2379 1 8 | 2679 279 379 | | 236 26 28 | 23579 5679 e*2579 | 25679 4 1 | | 7 1 9 | 235 56 4 | 2568 258 35 | +----------------------+-----------------------+-----------------------+ | 259 3 d@127* | 6 #579 @12579 | 4 #579 8 | | 4569 5679 c47 | 8 3 #579 | #579 1 2 | | 259 8 d@127* |a#579+2 4 @12579 | 3 6 #579 | +----------------------+-----------------------+-----------------------+

TH 579 in b2389: 2r2c7=2r9c4
UR 12r79c36: 2r5c6 = 7r79c3
2r9c4 == 2r2c7 - (2=47)r28c3 - r79c3 == 2r5c6 - r79c6 = 2r9c4 => 2r9c4, stte

by **SerTerence** » Thu Sep 08, 2022 3:40 pm

Hello eleven

As a newbie I would appreciate some help in formatting the grid and gaining familiarity with some of the notation used.

What is 12UR?

Unique rectangle on 1 & 2 ?

Many thanks.

by **totuan** » Thu Sep 08, 2022 3:44 pm

Code: Select all: *--------------------------------------------------------------------* | 1 @2579 3 | 4 #579 6 |*5789-2*5789-2#579 | | 2459 2579 %247 | 1 8 #579 |#2579 3 6 | | 8 579 6 |#579 2 3 | 1 #579 4 | |----------------------+----------------------+----------------------| | 236 4 5 | 2379 1 8 | 2679 279 379 | | 236 26 8 | 23579 5679 %2579 | 25679 4 1 | | 7 1 9 |@235 56 4 |*2568 *258 35 | |----------------------+----------------------+----------------------| | 259 3 &127 | 6 #579 &1579-2 | 4 #579 8 | | 4569 5679 47 | 8 3 #579 |#579 1 2 | | 259 8 &127 |#2579 4 &1579-2 | 3 6 #579 | *--------------------------------------------------------------------*

Anothers view to solve this one

Look at tridagon (579)B2389 => (2)r2c7=(2)r9c4 and UR(28)r16c78 => (2)r6c4=(2)r1c2

01: (2)r2c7==r9c4-r6c4==r1c2 => r1c78<>2, many singles
And the puzzles is solved with UR(12)r79c36 (as by shye)

Or like eleven’s - with a bit change

Look at tridagon (579)B2389 => (2)r9c4=(2)r2c7 and UR(12)r79c36 => (2)r2c3=(2)r5c6

01: (2)r9c4==r2c7-r2c3==r5c6 => r79c6<>2, stte

Again, thanks for the puzzle!
totuan

by **eleven** » Thu Sep 08, 2022 6:44 pm

SerTerence wrote:As a newbie I would appreciate some help in formatting the grid and gaining familiarity with some of the notation used.

What is 12UR?

Unique rectangle on 1 & 2 ?

ok, inside the code tags your grid would look like this:

Code: Select all: +-------------------------+-------------------------+-------------------------+ ! 1 2579 3 ! 4 579 6 ! 25789 25789 579 ! ! 2459 2579 247 ! 1 8 579 ! 2579 3 6 ! ! 8 579 6 ! 579 2 3 ! 1 5789 4 ! +-------------------------+-------------------------+-------------------------+ ! 236 4 5 ! 2379 1 8 ! 2679 279 379 ! ! 236 26 238 ! 23579 5679 2579 ! 25679 4 1 ! ! 7 1 9 ! 235 56 4 ! 2568 258 35 ! +-------------------------+-------------------------+-------------------------+ ! 259 3 127 ! 6 579 12579 ! 4 579 8 ! ! 4569 5679 47 ! 8 3 579 ! 579 1 2 ! ! 259 8 127 ! 2579 4 12579 ! 3 6 579 ! +-------------------------+-------------------------+-------------------------+

So you would have to add/remove some spaces. Inside the tags they are formatted like in editors like notepad (or the courier new font - same space for each character). Many solvers can copy the candidates in a similar format.
Clicking the Code button inserts the code tags

Code: Select all: [code]..[/code]

before and after the marked (highlighted) text.
You can see the raw text of other posts, if you click on the response button.

Yes, with 12 UR i meant a unique rectangle for 1 & 2. Note, that you can use both the "internals", i.e. the extra candidates in the UR pattern (here 7r79c9 and 579 in r79c6) and the "externals", i.e. all UR candidates in the same unit as UR cells (if they were missing you would get a hidden pair there), e.g. in the rows you would only have 2r7c1 and 2r9c14, so one of them must be true. As well one of the internals 7r79c3 and the external 2r5c6 must be true to avoid the UR.

Concerning the AIC notation there should be many links explaining it.
Here's just, how you could read totuan's for your elimination:
(6)r8c1=r8c2-(6=2)r5c2-(2)r5c6==r2c3-(2=147)r789c3 => r8c1<>4, r8c3=4
Either 6r8c1 OR 6r8c2 => not 6r5c2 => 2r5c2 => not 2r5c6 => 2r2c3 => not 2r789c3 => 147r789c3
So you have 6r8c1 OR the 3 digits 147 in r789c3 (implying 4r8c3). In both cases r8c1 cannot be 4.

-
@totuan: for my chain i had a grid with another 2 in r5c3

by **SerTerence** » Thu Sep 08, 2022 10:08 pm

Code: Select all: +-------------------------+-------------------------+-------------------------+ ! 123456789 123456789 123456789!123456789 123456789 123456789! 123456789 123456789 123456789! ! 123456789 123456789 123456789 ! 123456789 123456789 123456789 ! 123456789 123456789 123456789! ! 123456789 123456789 123456789 ! 123456789 123456789 123456789 ! 123456789 123456789 123456789! +-------------------------+-------------------------+-------------------------+ ! 123456789 4 5 ! 2379 1 8 ! 2679 279 379 ! ! 123456789 26 238 ! 23579 5679 2579 ! 25679 4 1 ! ! 123456789 1 9 ! 235 56 4 ! 2568 258 35 ! +-------------------------+-------------------------+-------------------------+ ! 123456789 3 127 ! 6 579 12579 ! 4 579 8 ! ! 123456789 5679 47 ! 8 3 579 ! 579 1 2 ! ! 123456789 8 127 ! 2579 4 12579 ! 3 6 579 ! +-------------------------+-------------------------+-------------------------+

Test

by **eleven** » Thu Sep 08, 2022 11:00 pm

Note the Preview button.

by **SerTerence** » Thu Sep 08, 2022 11:05 pm

Yeah that was a bit scruffy but I'm confident I can tidy that up for next time.

It's late in London now so tomorrow, tackle the notation!

Thanks Eleven

Website · by **denis_berthier** » Fri Sep 09, 2022 5:45 am

.
Thanks for your solutions.
Here is mine, based on ORk-Forcing-Whips (http://forum.enjoysudoku.com/ork-forcing-whips-and-ork-contrad-whips-t40189.html)

First, some easy cleaning, allowing to reduce the number of guardians and using only short reversible chains:

Code: Select all: hidden-pairs-in-a-column: c3{n3 n8}{r1 r5} ==> r5c3≠2, r1c3≠7, r1c3≠4, r1c3≠2, r1c3≠1 singles ==> r1c1=1, r1c3=3, r5c3=8, r3c1=8, r1c4=4, r6c6=4, r4c6=8, r4c5=1 z-chain[3]: c1n4{r2 r8} - r8n6{c1 c2} - c2n5{r8 .} ==> r2c1≠5 whip[1]: c1n5{r9 .} ==> r8c2≠5 z-chain[3]: c1n4{r2 r8} - r8n6{c1 c2} - c2n9{r8 .} ==> r2c1≠9 whip[1]: c1n9{r9 .} ==> r8c2≠9 +-------------------+-------------------+-------------------+ ! 1 2579 3 ! 4 579 6 ! 25789 25789 579 ! ! 24 2579 247 ! 1 8 579 ! 2579 3 6 ! ! 8 579 6 ! 579 2 3 ! 1 579 4 ! +-------------------+-------------------+-------------------+ ! 236 4 5 ! 2379 1 8 ! 2679 279 379 ! ! 236 26 8 ! 23579 5679 2579 ! 25679 4 1 ! ! 7 1 9 ! 235 56 4 ! 2568 258 35 ! +-------------------+-------------------+-------------------+ ! 259 3 127 ! 6 579 12579 ! 4 579 8 ! ! 4569 67 47 ! 8 3 579 ! 579 1 2 ! ! 259 8 127 ! 2579 4 12579 ! 3 6 579 ! +-------------------+-------------------+-------------------+

The Tridagon part, using only short OR2-Forcing-Whips:

Code: Select all: OR2-anti-tridagon[12] (type antidiag) for digits 5, 7 and 9 in blocks: b2, with cells: r1c5, r2c6, r3c4 b3, with cells: r1c9, r2c7, r3c8 b8, with cells: r7c5, r8c6, r9c4 b9, with cells: r7c8, r8c7, r9c9 with 2 guardians: n2r2c7 n2r9c4 OR2-forcing-whip-elim[2] based on OR2-anti-tridagon[12] for n2r2c7 and n2r9c4: || n2r2c7 - || n2r9c4 - partial-whip[1]: c6n2{r9 r5} - ==> r5c7≠2 OR2-forcing-whip-elim[3] based on OR2-anti-tridagon[12] for n2r2c7 and n2r9c4: || n2r2c7 - || n2r9c4 - partial-whip[2]: c6n2{r9 r5} - b4n2{r5c2 r4c1} - ==> r4c7≠2 OR2-forcing-whip-elim[3] based on OR2-anti-tridagon[12] for n2r2c7 and n2r9c4: || n2r2c7 - || n2r9c4 - partial-whip[2]: c6n2{r9 r5} - b4n2{r5c2 r4c1} - ==> r2c1≠2 singles ==> r2c1=4, r8c3=4 OR2-forcing-whip-elim[3] based on OR2-anti-tridagon[12] for n2r2c7 and n2r9c4: || n2r2c7 - partial-whip[1]: r1n2{c8 c2} - || n2r9c4 - partial-whip[1]: c6n2{r9 r5} - ==> r5c2≠2

The end is easy:

Code: Select all: singles ==> r5c2=6, r8c2=7, r2c3=7, r8c1=6, r4c7=6, r6c5=6 whip[1]: c3n2{r9 .} ==> r7c1≠2, r9c1≠2 naked-pairs-in-a-column: c6{r2 r8}{n5 n9} ==> r9c6≠9, r9c6≠5, r7c6≠9, r7c6≠5, r5c6≠9, r5c6≠5 hidden-pairs-in-a-row: r7{n1 n2}{c3 c6} ==> r7c6≠7 finned-x-wing-in-rows: n7{r7 r3}{c8 c5} ==> r1c5≠7 hidden-single-in-a-block ==> r3c4=7 whip[1]: r4n7{c9 .} ==> r5c7≠7 singles ==> r1c7=7, r1c8=8, r2c7=2, r1c2=2, r6c7=8 finned-swordfish-in-rows: n5{r2 r3 r8}{c6 c2 c8} ==> r7c8≠5 finned-swordfish-in-rows: n9{r2 r3 r8}{c6 c2 c8} ==> r7c8≠9 stte

by **yzfwsf** » Fri Sep 09, 2022 6:53 am

Code: Select all: Hidden Pair: 13 in r1c1,r1c3 => r1c1<>24589,r1c3<>2478 Hidden Single: 4 in r1 => r1c4=4 Hidden Single: 4 in r6 => r6c6=4 Hidden Single: 8 in c3 => r5c3=8 Hidden Single: 8 in c1 => r3c1=8 Hidden Single: 3 in c3 => r1c3=3 Hidden Single: 1 in r1 => r1c1=1 Hidden Single: 8 in c6 => r4c6=8 Hidden Single: 1 in r4 => r4c5=1 UR Forcing Chain: Each true guardian of UR 12{r79c36} will all lead to: r456c4,r9c136,r7c6<>2 7r7c3 - (7=42)r28c3 - 2r2c7 = 2r9c4 5r7c6 - (5=792)b8p267 7r7c6 - (7=592)b8p267 9r7c6 - (9=572)b8p267 7r9c3 - (7=42)r28c3 - 2r2c7 = 2r9c4 5r9c6 - (5=792)b8p267 7r9c6 - (7=592)b8p267 9r9c6 - (9=572)b8p267 stte

by **Cenoman** » Fri Sep 09, 2022 9:53 am

After so many solutions using UR (12)r79c36, what could I post else, but a poor solution w/o uniqueness ?

Code: Select all: +----------------------+-------------------------+------------------------+ | 1 2579 3 | 4 579* 6 | 25789 25789 579* | | 2459 2579 247 | 1 8 579* | 2579* 3 6 | | 8 579 6 | 579* 2 3 | 1 579* 4 | +----------------------+-------------------------+------------------------+ | 236 4 5 | 2379 1 8 | 2679 279 379 | | 236 26 8 | 23579 5679 2579 | 25679 4 1 | | 7 1 9 | 235 56 4 | 2568 258 35 | +----------------------+-------------------------+------------------------+ | 259 3 127 | 6 579* 12579 | 4 579* 8 | | 4569 5679 47 | 8 3 579* | 579* 1 2 | | 259 8 127 | 2579* 4 12579 | 3 6 579* | +----------------------+-------------------------+------------------------+

1. TH(579)b2389 (*)having two guardians: 2r2c7, 2r9c4
(2)r1c2 = r1c78 - r2c7 == r9c4 - r79c6 = (2)r5c6 => -2 r5c2; lcls, 8 placements

Code: Select all: +------------------+----------------------+------------------------+ | 1 259 3 | 4 59-7 6 | 25789 25789 579 | | 4 259 7 | 1 8 59 | 259 3 6 | | 8 59 6 | 579* 2 3 | 1 579* 4 | +------------------+----------------------+------------------------+ | 23 4 5 | 2379 1 8 | 6 279 379 | | 23 6 8 | 23579 579 27 | 2579 4 1 | | 7 1 9 | 235 6 4 | 258 258 35 | +------------------+----------------------+------------------------+ | 59 3 12 | 6 579* 12 | 4 579* 8 | | 6 7 4 | 8 3 59 | 59 1 2 | | 59 8 12 | 259-7 4 127 | 3 6 579 | +------------------+----------------------+------------------------+

2. Skyscraper: (7)r7c5 = r7c8 - r3c8 = r3c4 => -7 r1c5, r9c4; lcls, 6 placements

Code: Select all: +-----------------+---------------------+-------------------+ | 1 2 3 | 4 59 6 | 7 8 59 | | 4 59* 7 | 1 8 59* | 2 3 6 | | 8 59* 6 | 7 2 3 | 1 59* 4 | +-----------------+---------------------+-------------------+ | 23 4 5 | 239 1 8 | 6 279 379 | | 23 6 8 | 2359 579 27 | 59 4 1 | | 7 1 9 | 235 6 4 | 8 25 35 | +-----------------+---------------------+-------------------+ | 59 3 12 | 6 579 12 | 4 7-59 8 | | 6 7 4 | 8 3 59* | 59* 1 2 | | 59 8 12 | 259 4 127 | 3 6 579 | +-----------------+---------------------+-------------------+

3. Remote pair (59) r3c8, r8c7 (r3c2, r2c2, r2c6, r8c6) => -59 r7c8; ste

by **eleven** » Fri Sep 09, 2022 10:20 am

yzfwsf, this is the long version of totuan's
(2)r9c4==r2c7-r2c3==r5c6 => r79c6<>2
which uses the externals 2r2c3,r5c6.

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