The New Sudoku Players' Forum

[P.O.]

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for this one my solution is 2 forcing chains.

9 . .   3 . .   . 5 .
. 1 .   . . 2   9 . .
. . 7   . 1 .   . . 8
. 2 .   . . 8   6 . .
. . 3   1 . .   . . 5
5 . .   . 2 .   . 4 .
. 3 .   . . 6   5 . .
4 . .   . . .   . . 3
. . 9   . . .   . 7 .

9..3...5..1...29....7.1...8.2...86....31....55...2..4..3...65..4.......3..9....7.

9      468    2468   3      4678   47     1247   5      12467           
368    1      4568   45678  45678  2      9      36     467             
236    456    7      4569   1      459    234    236    8               
17     2      14     4579   34579  8      6      139    179             
678    46789  3      1      4679   479    278    289    5               
5      6789   168    679    2      379    1378   4      179             
1278   3      128    24789  4789   6      5      1289   1249           
4      5678   12568  25789  5789   1579   128    12689  3               
1268   568    9      2458   3458   1345   1248   7      1246       


[jco]

Code: Select all

.-----------------------------------------------------------------------.
| 9      68-4   268-4  |  3       678-4 Gw47     | 127-4  5    Bz12(4)67|
| 368    1      4568   |Fv45678 Fv45678   2      | 9     C36  EBu(4)67  |
| 236    456    7      |  4569    1       459    | 234   C236    8      |
|----------------------+-------------------------+----------------------|
| 17     2      14     |  4579    34579   8      | 6     D139   D179    |
| 678    46789  3      |  1       4679    479    | 278    289    5      |
| 5      6789   168    |  679     2       379    | 1378   4     D179    |
|----------------------+-------------------------+----------------------|
| 1278   3      128    |  24789   4789    6      | 5      1289  A12(4)9 |
| 4      5678   12568  |  25789   5789    1579   | 128    12689  3      |
| 1268   568    9      |  2458    3458    1345   | 1248   7     A12(4)6 |
'-----------------------------------------------------------------------'

1. Kraken Column (4)r1279c9
(4)r1c9
||
(4-7)r2c9 = (7)r2c45 - (7=4)r1c6
||
(4-26)r79c9 = (2|6)r12c9 - (26=3)r23c8 - (3=917)b6p239 - (7)r2c9 = (7)r2c45 - (7=4)r1c6

=> -4 r1c2357 [& HP(45)b1p68]
----

Code: Select all

.--------------------------------------------------------------------.
| 9      68     268    | 3      678  hb47     | 127    5     i12467  |
| 368    1      45     |g45678 g45678  2      | 9     e36   fA67(4)  |
| 236   a45     7      | 4569   1     b459    | 23-4   236    8      |
|----------------------+----------------------+----------------------|
| 17     2      14     | 4579   34579  8      | 6     d139    179    |
| 678    46789  3      | 1      4679  b479    | 278    289    5      |
| 5      6789   168    | 679    2     b379    |c1378   4      179    |
|----------------------+----------------------+----------------------|
| 1278   3      128    | 24789  4789   6      | 5      1289   1249   |
| 4      5678   12568  | 25789  5789   1579   | 128    12689  3      |
| 1268   568    9      | 2458   3458   1345   | 1248   7      1246   |
'--------------------------------------------------------------------'

2. (4)r2c9 = [ (4=5)r3c2 - (5=4793)r1356c6 - r6c6 = r3c8 - (3=6)r2c8 - (6*=*7)r2c9 - (7)r2c45 = (7-4)r1c6 = (4)r1c9 ]

=> -4 r3c7; lclste

[DEFISE]

whip[5]: r5n2{c7 c8}- b6n8{r5c8 r6c7}- c7n3{r6 r3}- r2c8{n3 n6}- r3c8{n6 .} => -7r5c7
whip[7]: c7n3{r3 r6}- r4n3{c8 c5}- r4n5{c5 c4}- r4n4{c4 c3}- c2n4{r5 r1}- r1c6{n4 n7}- c7n7{r1 .} => -4r3c7
Naked triplets: 236b3p578 => -2r1c7 -2r1c9 -6r1c9 -6r2c9
STTE

N.B: I deleted an unnecessary "Naked Triplets".

[eleven]

P.O., just a note:
you seem to generate puzzles with as much remaining candidates as possible. This is a good method to keep manual solvers away from solving them, because normally it is a boring long way (though the puzzles are not really hard). Watching at program generated "shortened" solutions (especially those using memory chains) is about as interesting as watching chess games between programs.

[P.O.]

hi Eleven, i won't disagree with you that they must be boring for manual solvers; my idea is that since the resolution of a puzzle starts after the basics are out of the way why not offer puzzles already rid of the basics at the start: no singles no intersections no subsets no fish; so what's left?
on the other hand when i started looking for puzzles of that sort there were still many contributors of puzzles, including you; in such a context those that i propose would only add to the diversity of the offer, not too difficult but with a little challenge all the same, and more oriented towards people using algorithms;
concerning their difficulty i evaluate it by means of my algorithm: 1 to 3,4 steps, with chains neither too long nor too complicated.
however, i disagree with you regarding chess engine tournaments, for people involved in writing chess algorithms they are exciting and a very good motivation to improve your own algorithm, i watched a lot of it, maybe i keep this mindset.

[eleven]

Thanks for your answer, Fair enough. We just have different goals.

[P.O.]

Code: Select all

thank you for your answers, so here is my solution with 2 forcing chains as i'm a bit biased towards them at the moment:

9      468    2468   3      4678   47     1247   5      12467           
368    1      4568   45678  45678  2      9      36     467             
236    456    7      4569   1      459   d×2+34  236    8               
17     2      14     4579   34579  8      6      139    179             
678    46789  3      1      4679   479   r+278   q2+89  5               
5      6789   168    679    2     c+379  d1-378  4      179             
1278   3      128    24789  4789   6      5     q12-89  1249           
4      5678   12568  25789  5789  b-1579 a1A2p8 q126-89 3               
1268   568    9      2458   3458  b+1345  1248   7      1246         

r8c7{n1n2n8} => r3c7 <> 2
 r8c7{n2n8 n1} - c6n1{r8 r9} - c6n3{r9 r6} - c7n3{r6 r3}
 r8c7{n1n8 n2}
 r8c7{n1n2 n8} - c8n8{r7r8 r5} - r5n2{c8 c7}

 9     a+468   2468   3       4678   b4+7    1247   5         12467           
 368    1      4568  c456-78 c456-78  2      9      36       c46+7             
 236   A+456   7      4569    1       459   B+34    236       8               
 17     2      14     4579    34579   8      6     C×(19)+3  d(19)7   
 678   p+46789 3      1       4679    479    278    289       5               
 5      6789   168    679     2       379   C1-378  4        d(19)7
 1278   3      128    24789   4789    6      5      1289      1249           
 4      5678   12568  25789   5789    1579   128    12689     3               
 1268   568    9      2458    3458    1345   1248   7         1246   

c2n4{r1r3r5} => r4c8 <> 1,9
 r1c2{n6n8 n4} - r1c6{n4 n7} - r2n7{c4c5 c9} - c9{r4r6}{n1n9}
 r3c2{n5n6 n4} - r3c7{n4 n3} - b6n3{r6c7 r4c8}
 r5c2{n6n7n8n9 n4} - r4{c1c3c8c9}{n1n3n7n9}
ste.

well, annotating the pm has its limits, here are duplicates of row 4 and 5:
q1+7    2       +14   4579    34579   8      6     q+3×(19)  q17+9
 678   p+46789  3     1       4679    479    278    289      5       


[DEFISE]

r1c2{n6n8 n4} - r1c6{n4 n7} - r2n7{c4c5 c9} - c9{r4r6}{n1n9}
r3c2{n5n6 n4} - r3c7{n4 n3} - b6n3{r6c7 r4c8}
r5c2{n6n7n8n9 n4} - r4{c1c3c8c9}{n1n3n7n9}
ste.

Hi P.O.
I agree with the first 2 lines but I believe that the 3rd does not make it possible to conclude that r4c8 <> 1,9
I would prefer this:
r5c2{n6n7n8n9 n4}- r4c3(n4 n1) - r4c1(n1 n7) - r4c9(n1n7 n9)

That said, I think I've seen this kind of shortcut in AICs before.

[P.O.]

hi DEFISSE, i don't disagree it can be written as 3 links as you did, i just wrote it as a quad: if r5c2=4 then r4c3=1 r4c1=7 r4c9=9 r4c8=3

[denis_berthier]

Code: Select all

r8c7{n1n2n8} => r3c7 <> 2
 r8c7{n2n8 n1} - c6n1{r8 r9} - c6n3{r9 r6} - c7n3{r6 r3}
 r8c7{n1n8 n2}
 r8c7{n1n2 n8} - c8n8{r7r8 r5} - r5n2{c8 c7}

Instead of this forcing-chain of length 9, this elimination can be done by a whip[6]:
whip[6]: c7n3{r3 r6} - c6n3{r6 r9} - c6n1{r9 r8} - r8c7{n1 n8} - c8n8{r8 r5} - r5n2{c8 .} ==> r3c7≠2

Code: Select all

c2n4{r1r3r5} => r4c8 <> 1,9
 r1c2{n6n8 n4} - r1c6{n4 n7} - r2n7{c4c5 c9} - c9{r4r6}{n1n9}
 r3c2{n5n6 n4} - r3c7{n4 n3} - b6n3{r6c7 r4c8}
 r5c2{n6n7n8n9 n4} - r4{c1c3c8c9}{n1n3n7n9}
ste.

Instead of this forcing-chain of length 14, these eliminations can be done by a whip[8] and a whip[10]
whip[8]: b6n3{r4c8 r6c7} - c6n3{r6 r9} - c6n1{r9 r8} - c8n1{r8 r7} - c7n1{r8 r1} - c7n7{r1 r5} - r6c9{n7 n1} - r4c9{n1 .} ==> r4c8≠9
whip[10]: b6n3{r4c8 r6c7} - b5n3{r6c6 r4c5} - r4n5{c5 c4} - r4n9{c4 c9} - r6c9{n9 n7} - b3n7{r1c9 r1c7} - r1c6{n7 n4} - b5n4{r5c6 r5c5} - c2n4{r5 r3} - r3c7{n4 .} ==> r4c8≠1
stte


Considering the puzzle is solvable in W4, I wouldn't look for a solution involving anything more complicated that chains of length 6.

[P.O.]

so in summary the grid has too many candidates, i am not writing the chains correctly, this is not the right way to solve the grid.
oh but it's an april fool, you guys almost got me.

hi Denis,

that said, here's what comes to mind:
as can be seen whips work a lot by contradiction, build on a target, between z and t candidates they empty a cell in some 2d space and conclude that the target is invalid; it works and you made the theory of it.
when an ordinary chain encounters an OR situation and cannot continue a whip solves the problem with its z candidate: this preventer from building the chain is linked to the target and can be ignored: it works as soon after a contradiction is reached.
but there is something that is still bothering me with whips and i can't put my finger on it yet;

you measure the complexity of a chain by means of its length and you have elaborate a rating system: the W rating; it is fine by me, i find it consistent and reliable: seeing the W rating of a puzzle i know the difficulty i will have to solve it, not so consistently with the SER rating, and you have integrated forcing chains in it: http://forum.enjoysudoku.com/the-tridagon-rule-t39859-30.html#p319410
regarding this puzzle, based on their length as you measure it and as you said in the quoted post you have no choice, you find the whips less complex than the forcing chains: for the first one it can be argued, but seriously for the second one from anybody else but you i would take that has a joke:

that
whip[8]: b6n3{r4c8 r6c7} - c6n3{r6 r9} - c6n1{r9 r8} - c8n1{r8 r7} - c7n1{r8 r1} - c7n7{r1 r5} - r6c9{n7 n1} - r4c9{n1 .} ==> r4c8≠9
whip[10]: b6n3{r4c8 r6c7} - b5n3{r6c6 r4c5} - r4n5{c5 c4} - r4n9{c4 c9} - r6c9{n9 n7} - b3n7{r1c9 r1c7} - r1c6{n7 n4} - b5n4{r5c6 r5c5} - c2n4{r5 r3} - r3c7{n4 .} ==> r4c8≠1
is simpler than that
c2n4{r1r3r5} => r4c8 <> 1,9
r1c2{n6n8 n4} - r1c6{n4 n7} - r2n7{c4c5 c9} - c9{r4r6}{n1n9}
r3c2{n5n6 n4} - r3c7{n4 n3} - b6n3{r6c7 r4c8}
r5c2{n6n7n8n9 n4} - r4{c1c3c8c9}{n1n3n7n9}
it is a joke.

right now i'm coding forcing chains and i sort of like them and looking for them i find them everywhere, i think i'm going to use them a lot, and i'm not going to stick to the trivalues.
a puzzle in W4 may need lot of eliminations to be solved, if with one or two W8 chains only two or three eliminations are sufficient i prefer this second solution.
in (another) summary: length is not all there is to it.

[denis_berthier]

i would take that has a joke:
that
whip[8]: b6n3{r4c8 r6c7} - c6n3{r6 r9} - c6n1{r9 r8} - c8n1{r8 r7} - c7n1{r8 r1} - c7n7{r1 r5} - r6c9{n7 n1} - r4c9{n1 .} ==> r4c8≠9
whip[10]: b6n3{r4c8 r6c7} - b5n3{r6c6 r4c5} - r4n5{c5 c4} - r4n9{c4 c9} - r6c9{n9 n7} - b3n7{r1c9 r1c7} - r1c6{n7 n4} - b5n4{r5c6 r5c5} - c2n4{r5 r3} - r3c7{n4 .} ==> r4c8≠1
is simpler than that
c2n4{r1r3r5} => r4c8 <> 1,9
r1c2{n6n8 n4} - r1c6{n4 n7} - r2n7{c4c5 c9} - c9{r4r6}{n1n9}
r3c2{n5n6 n4} - r3c7{n4 n3} - b6n3{r6c7 r4c8}
r5c2{n6n7n8n9 n4} - r4{c1c3c8c9}{n1n3n7n9}
it is a joke.

What "that" means is unclear. n4 should be the left candidate of the first 3 cells. The meaning of several n's in the 3 starting cells is undefined.

Anyway, you have 3 streams of reasoning instead of 1 and a larger total length. So yes, obviously, the whips are simpler.
I've defined forcing whips long ago and shown that they don't bring anything useful wrt whips. The case of Tridagons is unrelated to this discussion.

in (another) summary: length is not all there is to it.

If you don't care about length, Forcing-T&E will solve it in 1 step.
Having a small number of steps is irrelevant if this implies using absurdly long chains (be they forcing or not).
.

[P.O.]

the first that means what you say is a joke
the second and third designate what follow
r1c2{n6n8 n4} r3c2{n5n6 n4} r5c2{n6n7n8n9 n4} means respectively when r1c2=4 r3c2=4 r5c2=4
three short reasoning are simpler than a long one
the quoted post begin by: OR-forcing-chains in general
length is not all there is to it means length is not a good measure of complexity
your last judgment is purely subjective
you have a serious comprehension problem

[denis_berthier]

three short reasoning are simpler than a long one

Not when they must be considered simultaneously! In order to be able to make any elimination, you have to make them converge towards the same candidate. So, obviously, when it comes to ascribing some complexity to your forcing network, you have to take the sums of their complexities (+1 for the trivalue cell at the start).

If you think that the 3 whips I gave was my solution of the problem, you've totally misunderstood me. I was just stating that each of your eliminations could be done by a shorter whip.

Starting form the resolution state after Singles and whips[1]:

Code: Select all

   +-------------------+-------------------+-------------------+
   ! 9     468   2468  ! 3     4678  47    ! 1247  5     12467 !
   ! 368   1     4568  ! 45678 45678 2     ! 9     36    467   !
   ! 236   456   7     ! 4569  1     459   ! 234   236   8     !
   +-------------------+-------------------+-------------------+
   ! 17    2     14    ! 4579  34579 8     ! 6     139   179   !
   ! 678   46789 3     ! 1     4679  479   ! 278   289   5     !
   ! 5     6789  168   ! 679   2     379   ! 1378  4     179   !
   +-------------------+-------------------+-------------------+
   ! 1278  3     128   ! 24789 4789  6     ! 5     1289  1249  !
   ! 4     5678  12568 ! 25789 5789  1579  ! 128   12689 3     !
   ! 1268  568   9     ! 2458  3458  1345  ! 1248  7     1246  !
   +-------------------+-------------------+-------------------+
204 candidates.

My various solutions are as follows.

1) the simplest-first one in W4:

Code: Select all

z-chain[4]: c6n1{r8 r9} - c6n3{r9 r6} - r4n3{c5 c8} - c8n1{r4 .} ==> r8c7≠1
z-chain[3]: r8c7{n8 n2} - r5n2{c7 c8} - c8n8{r5 .} ==> r9c7≠8
t-whip[3]: r8c7{n2 n8} - c8n8{r8 r5} - r5n2{c8 .} ==> r9c7≠2, r3c7≠2, r1c7≠2
biv-chain[3]: r3n2{c8 c1} - c1n3{r3 r2} - r2c8{n3 n6} ==> r3c8≠6
t-whip[3]: r5n2{c8 c7} - r8c7{n2 n8} - b6n8{r5c7 .} ==> r5c8≠9
biv-chain[3]: r5c8{n8 n2} - r3c8{n2 n3} - b6n3{r4c8 r6c7} ==> r6c7≠8
whip[1]: r6n8{c3 .} ==> r5c1≠8, r5c2≠8
hidden-pairs-in-a-row: r5{n2 n8}{c7 c8} ==> r5c7≠7
biv-chain[4]: b3n2{r3c8 r1c9} - r1n1{c9 c7} - c7n7{r1 r6} - b6n3{r6c7 r4c8} ==> r3c8≠3
singles ==> r3c8=2, r5c8=8, r5c7=2, r8c7=8, r1c3=2, r8c4=2
finned-swordfish-in-columns: n6{c3 c8 c4}{r6 r8 r2} ==> r2c5≠6
biv-chain[4]: c2n9{r5 r6} - b4n8{r6c2 r6c3} - r7c3{n8 n1} - r4c3{n1 n4} ==> r5c2≠4
hidden-single-in-a-block ==> r4c3=4
biv-chain[3]: b1n4{r1c2 r3c2} - r3c7{n4 n3} - r3c1{n3 n6} ==> r1c2≠6
biv-chain[3]: c5n6{r1 r5} - b5n4{r5c5 r5c6} - r1c6{n4 n7} ==> r1c5≠7
whip[3]: r1c6{n4 n7} - r2n7{c5 c9} - r2n4{c9 .} ==> r3c6≠4
whip[3]: r1c6{n4 n7} - r2n7{c5 c9} - r2n4{c9 .} ==> r3c4≠4
whip[3]: r1c6{n4 n7} - r2n7{c5 c9} - r2n4{c9 .} ==> r1c5≠4
biv-chain[4]: r3n4{c2 c7} - c7n3{r3 r6} - c7n7{r6 r1} - r1c6{n7 n4} ==> r1c2≠4
stte

Many steps (the number of steps can probably be reduced in W4), but each step is elementary.


2) A 1-step solution based on Forcing-T&E:

Code: Select all

FORCING[3]-T&E(SFin+TridFW) applied to trivalue candidates n4r2c9, n6r2c9 and n7r2c9 :
===> 15 values decided in the three cases: n2r1c3 n2r7c9 n2r8c4 n2r9c1 n3r9c5 n3r6c6 n2r3c8 n8r9c4 n7r8c2 n9r4c9 n8r5c8 n7r5c1 n2r5c7 n8r8c7 n9r7c8
===> 106 candidates eliminated in the three cases: n4r1c2 n6r1c2 n4r1c3 n6r1c3 n8r1c3 n4r1c5 n7r1c5 n2r1c7 n4r1c7 n2r1c9 n6r1c9 n6r2c1 n6r2c3 n8r2c3 n5r2c4 n6r2c4 n8r2c4 n4r2c5 n6r2c5 n7r2c5 n2r3c1 n6r3c2 n4r3c4 n4r3c6 n2r3c7 n3r3c8 n6r3c8 n7r4c1 n4r4c4 n7r4c4 n9r4c4 n3r4c5 n4r4c5 n9r4c5 n1r4c8 n9r4c8 n1r4c9 n7r4c9 n6r5c1 n8r5c1 n4r5c2 n7r5c2 n8r5c2 n7r5c5 n7r5c6 n7r5c7 n8r5c7 n2r5c8 n9r5c8 n6r6c2 n7r6c2 n1r6c3 n7r6c4 n7r6c6 n9r6c6 n3r6c7 n8r6c7 n9r6c9 n2r7c1 n7r7c1 n2r7c3 n2r7c4 n8r7c4 n9r7c4 n8r7c5 n9r7c5 n1r7c8 n2r7c8 n8r7c8 n1r7c9 n4r7c9 n9r7c9 n5r8c2 n6r8c2 n8r8c2 n1r8c3 n2r8c3 n8r8c3 n5r8c4 n7r8c4 n8r8c4 n9r8c4 n7r8c5 n8r8c5 n7r8c6 n1r8c7 n2r8c7 n2r8c8 n8r8c8 n9r8c8 n1r9c1 n6r9c1 n8r9c1 n8r9c2 n2r9c4 n4r9c4 n5r9c4 n4r9c5 n5r9c5 n8r9c5 n3r9c6 n4r9c6 n2r9c7 n8r9c7 n1r9c9 n2r9c9
stte

Those were the two extremes: minimum lengths of the chains vs minimum number of steps.
Next come intermediate cases:


3) A 5-step solution in W6:

Code: Select all

whip[6]: b6n3{r6c7 r4c8} - r2c8{n3 n6} - r3c8{n6 n2} - r5n2{c8 c7} - r8c7{n2 n1} - c8n1{r7 .} ==> r6c7≠8
whip[1]: r6n8{c3 .} ==> r5c1≠8, r5c2≠8
hidden-pairs-in-a-row: r5{n2 n8}{c7 c8} ==> r5c7≠7, r5c8≠9
whip[5]: r1n1{c9 c7} - c7n7{r1 r6} - c9n7{r6 r2} - b3n4{r2c9 r3c7} - c7n3{r3 .} ==> r1c9≠2
whip[1]: c9n2{r9 .} ==> r7c8≠2, r8c7≠2, r8c8≠2, r9c7≠2
z-chain[4]: c6n1{r8 r9} - c6n3{r9 r6} - r4n3{c5 c8} - c8n1{r4 .} ==> r8c7≠1
singles ==> r8c7=8, r5c7=2, r5c8=8, r3c8=2, r1c3=2, r8c4=2
z-chain[6]: r3c1{n6 n3} - r2n3{c1 c8} - c8n6{r2 r8} - c3n6{r8 r2} - r3n6{c2 c4} - r6n6{c4 .} ==> r5c1≠6
w1-tte

4) A 4-step solution in W7:
see François's solution


5) A 2-step solutions in W9:

Code: Select all

whip[6]: c7n3{r3 r6} - c6n3{r6 r9} - c6n1{r9 r8} - r8c7{n1 n8} - c8n8{r8 r5} - r5n2{c8 .} ==> r3c7≠2
whip[9]: c7n3{r3 r6} - r4n3{c8 c5} - r4n5{c5 c4} - r4n4{c4 c3} - c2n4{r5 r1} - r1c6{n4 n7} - c7n7{r1 r5} - b6n8{r5c7 r5c8} - r5n2{c8 .} ==> r3c7≠4
stte

Which solution is the best? There's no answer to this question. It depends on personal preferences. But I would say that in any case, all these solutions are simpler than yours, whether you consider the number of steps or the total length of the hardest step.

[P.O.]

hi Denis,
i know that the three whips you gave are not your solution, you have repeated enough that your are not for 'absurdly long chains' to solve a puzzle solvable in many short steps, but you imply they are simpler than the two forcing chains i gave but they are not;
let’s consider the second which is composed of three short chains;
the reasoning of the three components are not taken simultaneously but successively, one at a time you check they do the eliminations stated and at the end of the verification you conclude at the correctness of the statement made, a human brain can do that without much effort, but to follow the path of the whip[8] and whip[10] need a greater effort due to their length and complexity;
the forcing chain is divide and conquer, it breaks down the elimination of 1,9 from r4c8 into three short pieces of reasoning, their length does not add up they simplify a much more complex proof like the one given by the two whips.