## Almost ER?

Advanced methods and approaches for solving Sudoku puzzles

### Almost ER?

If you look at this example (#224 from the 1465 list) you won't get much farther with basic techniques (at least not me):

Code: Select all
` *-----------* |...|.7.|94.| |.7.|.9.|..5| |3..|..5|.7.| |---+---+---| |.87|4..|1..| |463|.8.|...| |...|..7|.8.| |---+---+---| |8..|7..|...| |7..|...|.28| |.5.|268|...| *-----------*  *--------------------------------------------------------------------* | 1256   12     1258   | 1368   7      1236   | 9      4      1236   | | 126    7      1248   | 1368   9      12346  | 2368   136    5      | | 3      1249   12489  | 168    124    5      | 268    7      126    | |----------------------+----------------------+----------------------| | 259    8      7      | 4      235    2369   | 1      3569   2369   | | 4      6      3      | 159    8      129    | 257    59     279    | | 1259   129    1259   | 3569   235    7      | 23456  8      23469  | |----------------------+----------------------+----------------------| | 8      12349  12469  | 7      1345   1349   | 456    1569   1469   | | 7      1349   1469   | 1359   1345   1349   | 456    2      8      | | 19     5      149    | 2      6      8      | 347    139    13479  | *--------------------------------------------------------------------*`

But whatever you choose as a 2 in r5 you will eliminate the candidate 2 in r3c9. I don't know if this observation belongs to a certain technique or if it is just a random constellation without a specific name.
Neunmalneun

Posts: 52
Joined: 22 December 2005

Hi Neunmalneun,
This is Tough puzzle and I don’t know how can eliminate r3c9=2 if r5c7=2, not easy…
ttt

Posts: 185
Joined: 20 October 2006
Location: vietnam

### Re: Almost ER?

Neunmalneun wrote:If you look at this example (#224 from the 1465 list) you won't get much farther with basic techniques (at least not me)

This is indeed a hard puzzle (SER=9.3)
With nrczt-chains, we get the following resolution path, which shows that there are very few easy eliminations. An nrczt13 chain can hardly be considered as accessible to a human solver.

Code: Select all
`column c1 interaction-with-block b1 ==> r3c3 <> 6, r2c3 <> 6, r1c3 <> 6row r9 interaction-with-block b9 ==> r8c7 <> 3, r7c9 <> 3, r7c8 <> 3, r7c7 <> 3, r6c5 <> 1, r6c4 <> 1nrct4-chain n2{r7c3 r7c2} - {n2 n1}r1c2 - {n1 n9}r6c2 - n9{r3c2 r3c3} ==> r7c3 <> 9, r3c3 <> 2nrczt4-chain {n9 n1}r9c1 - {n1 n4}r9c3 - n4{r2c3 r3c2} - n9{r3c2 r3c3} ==> r8c3 <> 9nrczt5-chain n6{r3c9 r3c4} - n6{r6c4 r6c9} - n4{r6c9 r6c7} - {n4 n5}r7c7 - {n5 n6}r8c7 ==> r2c7 <> 6nrczt8-chain n2{r6c1 r4c1} - n2{r4c5 r3c5} - n4{r3c5 r2c6} - n2{r2c6 r5c6} - n1{r5c6 r5c4} - n1{r3c4 r1c6} - {n1 n2}r1c2 - n2{r2c3 r2c7} ==> r6c7 <> 2nrczt9-chain {n1 n2}r1c2 - {n2 n9}r6c2 - {n9 n4}r3c2 - n4{r2c3 r2c6} - n2{r2c6 r3c5} - n1{r3c5 r7c5} - {n1 n3}r7c2 - {n3 n9}r7c6 - n9{r8c6 r8c2} ==> r8c2 <> 1nrczt9-chain n1{r7c5 r3c5} - n4{r3c5 r2c6} - n2{r2c6 r1c6} - {n2 n9}r5c6 - {n9 n3}r8c6 - n1{r8c6 r8c3} - n6{r8c3 r7c3} - n2{r7c3 r7c2} - n3{r7c2 r7c6} ==> r7c6 <> 1nrczt9-chain n2{r6c3 r4c1} - n2{r4c5 r3c5} - n4{r3c5 r2c6} - n2{r2c6 r5c6} - n1{r5c6 r5c4} - n1{r1c4 r1c6} - n6{r1c6 r4c6} - n6{r4c8 r6c7} - n4{r6c7 r6c9} ==> r6c9 <> 2nrczt9-chain n2{r7c3 r7c2} - {n2 n1}r1c2 - {n1 n9}r6c2 - n9{r4c1 r9c1} - n1{r9c1 r6c1} - n2{r6c1 r6c5} - n2{r5c6 r2c6} - n4{r2c6 r2c3} - {n4 n2}r3c2 ==> r1c3 <> 2nrczt11-chain n7{r5c7 r9c7} - n7{r9c9 r5c9} - n2{r5c9 r5c6} - n2{r6c5 r3c5} - n2{r3c7 r2c7} - n3{r2c7 r6c7} - n4{r6c7 r6c9} - n6{r6c9 r6c4} - n5{r6c4 r8c4} - n9{r8c4 r5c4} - {n9 n5}r5c8 ==> r5c7 <> 5nrczt13-chain n2{r7c3 r7c2} - {n2 n1}r1c2 - {n1 n6}r2c1 - {n6 n5}r1c1 - n5{r6c1 r6c3} - n1{r6c3 r6c1} - {n1 n9}r6c2 - n2{r6c2 r6c5} - n2{r5c6 r1c6} - n6{r1c6 r4c6} - {n6 n3}r6c4 - n3{r1c4 r2c6} - n4{r2c6 r2c3} ==> r2c3 <> 2nrczt9-chain n4{r3c5 r2c6} - n2{r2c6 r1c6} - {n2 n1}r1c2 - {n1 n8}r2c3 - n8{r2c7 r3c7} - {n8 n6}r3c4 - {n6 n2}r3c9 - {n2 n3}r2c7 - {n3 n1}r2c4 ==> r3c5 <> 1column c5 interaction-with-block b8 ==> r8c6 <> 1, r8c4 <> 1nrc4-chain n1{r8c5 r8c3} - n6{r8c3 r7c3} - n2{r7c3 r7c2} - n3{r7c2 r8c2} ==> r8c5 <> 3nrczt4-chain n4{r8c2 r3c2} - n4{r3c5 r8c5} - n1{r8c5 r8c3} - n6{r8c3 r7c3} ==> r7c3 <> 4nrczt6-chain n1{r8c5 r8c3} - {n1 n9}r9c1 - {n9 n4}r9c3 - n4{r8c2 r3c2} - n4{r3c5 r7c5} - n1{r7c5 r8c5} ==> r8c5 <> 5nrc2-chain n5{r5c8 r5c4} - n5{r8c4 r7c5} ==> r7c8 <> 5column c8 interaction-with-block b6 ==> r6c7 <> 5nrczt5-chain n5{r7c7 r7c5} - n1{r7c5 r8c5} - n4{r8c5 r3c5} - n4{r2c6 r8c6} - n4{r8c2 r7c2} ==> r7c7 <> 4nrczt6-chain n5{r7c5 r8c4} - n3{r8c4 r8c2} - n9{r8c2 r8c6} - {n9 n4}r7c6 - n4{r7c2 r3c2} - n4{r3c5 r7c5} ==> r7c5 <> 3column c5 interaction-with-block b5 ==> r6c4 <> 3, r4c6 <> 3nrczt8-chain n6{r1c6 r4c6} - n6{r4c8 r7c8} - {n6 n5}r7c7 - n5{r7c5 r8c4} - n3{r8c4 r1c4} - n8{r1c4 r1c3} - n5{r1c3 r1c1} - n6{r1c1 r2c1} ==> r2c4 <> 6nrczt7-chain {n6 n5}r7c7 - {n5 n4}r8c7 - n4{r9c7 r9c3} - n4{r2c3 r2c6} - n6{r2c6 r2c1} - n2{r2c1 r2c7} - n8{r2c7 r3c7} ==> r3c7 <> 6nrc2-chain n6{r3c9 r3c4} - n6{r6c4 r4c6} ==> r4c9 <> 6nrct5-chain n6{r4c6 r6c4} - n6{r6c9 r4c8} - n5{r4c8 r5c8} - n5{r5c4 r8c4} - n9{r8c4 r5c4} ==> r4c6 <> 9nrct3-chain {n6 n2}r4c6 - n2{r1c6 r3c5} - n4{r3c5 r2c6} ==> r2c6 <> 6nrczt5-chain n2{r6c3 r6c5} - n3{r6c5 r4c5} - n5{r4c5 r4c8} - {n5 n9}r5c8 - n9{r4c8 r4c1} ==> r4c1 <> 2block b4 interaction-with-row r6 ==> r6c5 <> 2nrczt5-chain {n1 n2}r1c2 - {n2 n6}r2c1 - {n6 n5}r1c1 - {n5 n9}r4c1 - {n9 n1}r6c2 ==> r3c2 <> 1nrczt5-chain {n2 n1}r1c2 - {n1 n6}r2c1 - {n6 n5}r1c1 - {n5 n9}r4c1 - {n9 n2}r6c2 ==> r3c2 <> 2nrczt5-chain n2{r1c2 r2c1} - n6{r2c1 r1c1} - n6{r1c6 r4c6} - n2{r4c6 r5c6} - n2{r5c9 r4c9} ==> r1c9 <> 2nrct6-chain n7{r9c9 r9c7} - {n7 n2}r5c7 - n2{r5c9 r3c9} - {n2 n4}r3c5 - n4{r2c6 r2c3} - n4{r9c3 r9c9} ==> r9c9 <> 9, r9c9 <> 3, r9c9 <> 1nrczt6-lr-lasso {n3 n5}r6c5 - n5{r6c3 r1c3} - n8{r1c3 r1c4} - n3{r1c4 r1c6} - n3{r8c6 r8c4} - n5{r8c4 r5c4} ==> r6c9 <> 3nrczt6-lr-lasso n1{r7c9 r9c8} - n3{r9c8 r9c7} - n3{r6c7 r6c5} - n5{r6c5 r4c5} - {n5 n9}r4c1 - {n9 n1}r9c1 ==> r7c5 <> 1hidden-single-in-a-block ==> r8c5 = 1nrczt3-chain n4{r7c5 r8c6} - n4{r2c6 r2c3} - n4{r9c3 r9c9} ==> r7c9 <> 4nrczt5-lr-lasso n4{r3c5 r7c5} - n4{r7c2 r8c2} - n3{r8c2 r7c2} - {n3 n9}r7c6 - n9{r8c4 r8c2} ==> r3c3 <> 4nrct7-chain n4{r6c9 r6c7} - n4{r9c7 r9c9} - n7{r9c9 r9c7} - {n7 n2}r5c7 - n2{r5c9 r3c9} - n6{r3c9 r3c4} - n6{r6c4 r6c9} ==> r6c9 <> 9nrczt7-chain n6{r1c6 r4c6} - n6{r6c4 r6c7} - {n6 n5}r7c7 - {n5 n4}r7c5 - {n4 n2}r3c5 - n2{r4c5 r4c9} - n3{r4c9 r1c9} ==> r1c9 <> 6nrczt5-chain n6{r3c9 r3c4} - n1{r3c4 r3c3} - {n1 n2}r1c2 - {n2 n6}r2c1 - n6{r2c8 r3c9} ==> r3c9 <> 2column c9 interaction-with-block b6 ==> r5c7 <> 2naked and hidden singles ==> r5c7 = 7, r9c9 = 7, r6c9 = 4hidden-pairs-in-a-column {n2 n8}{r2 r3}c7 ==> r2c7 <> 3xy3-chain {n5 n6}r7c7 - {n6 n3}r6c7 - {n3 n5}r6c5 ==> r7c5 <> 5naked and hidden singles ==> r7c5 = 4, r3c5 = 2, r3c7 = 8, r2c7 = 2, r2c6 = 4, r3c2 = 4, r3c3 = 9, r8c4 = 5, r5c8 = 5, r7c7 = 5column c4 interaction-with-block b2 ==> r1c6 <> 3, r5c6 <> 9naked-pairs-in-a-block {n1 n6}{r1c6 r3c4} ==> r2c4 <> 1, r1c4 <> 6, r1c4 <> 1swordfish-in-rows n6{r1 r2 r4}{c6 c1 c8} ==> r7c8 <> 6nrc3-chain n3{r1c9 r2c8} - n6{r2c8 r4c8} - {n6 n3}r6c7 ==> r4c9 <> 3hidden-single-in-a-column ==> r1c9 = 3naked and hidden singles ==> r1c4 = 8, r2c4 = 3, r2c3 = 8naked-pairs-in-a-block {n2 n9}{r4c9 r5c9} ==> r4c8 <> 9column c8 interaction-with-block b9 ==> r7c9 <> 9nrc3-chain n9{r7c8 r9c8} - {n9 n1}r9c1 - n1{r2c1 r2c8} ==> r7c8 <> 1...(naked and hidden singles)...GRID top1465#224 SOLVED AT LEVEL = L13, MOST COMPLEX RULE = NRCZT13215876943678394215349125876587432169463981752192657384826743591734519628951268437`
denis_berthier
2010 Supporter

Posts: 1261
Joined: 19 June 2007
Location: Paris

This is the Ancient Thoughest Known Puzzle.
Carcul

Posts: 724
Joined: 04 November 2005

Code: Select all
`. . .|. 7 .|9 4 .. 7 .|. 9 .|. . 53 . .|. . 5|. 7 .-----+-----+-----. 8 7|4 . .|1 . .4 6 3|. 8 .|. . .. . .|. . 7|. 8 .-----+-----+-----8 . .|7 . .|. . .7 . .|. . .|. 2 8. 5 .|2 6 8|. . .After SSTS.------------------.------------------.------------------.| 1256  12    1258 | 1368  7     1236 | 9     4     1236 || 126   7     1248 | 1368  9     12346| 2368  136   5    || 3     1249  12489| 168   124   5    | 268   7     126  |:------------------+------------------+------------------:| 259   8     7    | 4     235   2369 | 1     3569  2369 || 4     6     3    | 159   8     129  | 257   59    279  || 1259  129   1259 | 3569  235   7    | 23456 8     23469|:------------------+------------------+------------------:| 8     12349 12469| 7     1345  1349 | 456   1569  1469 || 7     1349  1469 | 1359  1345  1349 | 456   2     8    || 19    5     149  | 2     6     8    | 347   139   13479|'------------------'------------------'------------------'`

Hi All,
This puzzle is very difficult for me, I can’t understand Denis’s solution by his notations so I have to solve by myself and I don’t know this puzzle have other solutions or not.
My apologize, I’m not good for using NL notations (like my English - most popular here ), therefore I present my solution by Eureka notations.

1) (9)r3c3=(9-4)r3c2=(hp34-2)r78c2=(2-6)r7c3=(6)r8c3 => r78c3<>9
2) (hp27)r5c79=(2-1)r5c6=(1-59)r5c4=(hp59-6)r67c4=(hp46)r6c79–(35)r6c7=(ht456)r678c7
=> r5c7<>5, r6c79<>2
3) (ht456)r678c7=(3)r6c7–(hp46)r6c79=(6)r6c4–(6)r3c4=(6)r3c79 => r2c7<>6
4) A complex step :
(9)r3c3=(9-4)r3c2=(hp34)r78c2-(4=hp19)r9c13-(19)r9c89=(hp19)r7c89-(19=hp34)r7c26-(134=5)r7c5-[(5)r4c5 & (6)r7c8]={(6)r1c1=(6)r2c1-(6)r2c8=(6-5)r4c8=(5)r4c1}-(5)r1c1=(ht126)r12c1/r1c2
=> r3c23<>12
Above step meant :
If r3c3<>9 => r4c5<>5 & r7c8<>6 => r1c1<>5 => (ht126)r12c1/r1c2
The same if start from r3c2<>4. I don’t know how to present it as Diagram, too complex...

5) (4)r2c6=(4)r2c3-(4)r3c2=(hp34)r78c2-(4=hp19)r9c13-(19=3)r9c8-(3)r9c7={(3)r2c7=(3)r6c7-(hp46)r6c79=(6)r6c4-(6=ht138)r123c4}
=> r2c6<>3
Present as Diagram :
Code: Select all
`(4)r2c6=(4)r2c3-(4)r3c2=(hp34)r78c2-(4=hp19)r9c13-(19=3)r9c8-(3)r9c7                                                             ||                                                             (3)r2c7                                                             ||                                                             (3)r6c7-(hp46)r6c79=(6)r6c4-(6=ht138)r123c4`

6) Present as Kraken Column
Code: Select all
`(2)r1c6||(2-6)r4c6=(6)r6c4-(6=ht138)r123c4||(2-1)r5c6=(1)r5c4-(1=ht368)r123c4||(2-4)r2c6=(4)r2c3 - (hp49=8)r3c23-(8)r1c3=(8)r1c8-(8)r2c34=(8)r2c7-(3)r2c7                   \                                               ||                     (4)r3c2=(hp34)r78c2-(4=hp19)r9c13-(19=3)r9c8– (3)r2c8                                                                   ||                                                                  (3)r1c9`
=> r1c6<>3
7) (3)r2c4=(3-8)r1c4=(8-5)r1c3=(5-6)r1c1=(6)r2c1 => r2c4<>6
8) (8)r1c4=(8-5)r1c3=(5-6)r1c1=(6)r2c1-(6)r2c8={(3)r1c4=(3)r2c4-(3=1)r2c8-(1)r2c13=(1)r1c123}
=> r1c4<>1
Present as Kraken Cell :
Code: Select all
`(1)r2c8-(1)r2c13=(1)r1c123||(3)r2c8-(3)r2c4=(3)r1c4||(6)r2c8-(6)r2c1=(6-5)r1c1=(5-8)r1c3=(8)r1c4`

9) A complex step, present it as Diagram :
Let A : (4)r3c2=(hp34)r78c2-(4=hp19)r9c13-(19)r9c89=(hp19)r7c89-(19=hp34)r7c26
Code: Select all
`  A – (34) AUR r78c26       ||      (1)r8c6–(1)r78c5=(1-4)r3c5=(4)r2c6      ||      (9)r8c6-(9)r8c4                ||              (1)r8c4-(1)r78c5=(1-4)r3c5=(4)r2c6               ||              (5)r8c4-(5)r8c7                             ||                      (4)r8c7-(4)r9c79=(4)r9c3                       ||                      (6)r8c7-(6)r8c3=(6-2)r7c3=(2-hp34)r78c2=(4)r3c2`

=> r2c3<>4 => single r2c6=4
10) (hp14)r78c5=(1-2)r3c5=(2-6)r1c6=(6-3)r4c6=(3)r78c6 => r78c5<>3
11) (hp26)r14c6=(2-1)r5c6=(1-59)r5c4=(hp59-6)r68c4=(6)r4c6 => r4c6<>9
12) (2=6)r4c6-(6)r6c4=(hp46-3)r6c79=(3)r6c5 => r6c5<>2
13) (5=9)r5c8-(9)r4c89=(9-5)r4c1=(5)r6c13 => r6c7<>5
14) (2)r7c3=(2)r7c2-(hp12=9)r16c2-(9=5)r4c1-(5)r6c3=(5-8)r1c3=(8)r2c3 => r12c3<>2
15) (2)r2c7=(2)r1c1-(2=1)r1c2-(1=8)r2c3-(8)r1c3=(8-3)r1c4=(3)r2c4
=> r2c7<>38 => some singles
16) (2)r7c2=(2-6)r7c3=(6-1)r8c3=(1-4)r8c5=(4)r7c5 => r7c2<>4 => some singles
17) (39) AUR r78c26 => r7c2<>39, SINGLES TO THE END
Please correct me if something wrong or typo.
My energy down to empty after finish this one together EURO2008, need times for recovering…
Waiting : Germany vs Turkey tonight…
Thanks to All
ttt
ttt

Posts: 185
Joined: 20 October 2006
Location: vietnam

Here is my stab at this monster. Not as short as ttt's - so next on my todo list is to study his:
start, ss

1) TM counting method (TM's are generalized predecessor to *****tchains, matrix counting is an alg to find them (and all matrices.))
aur sis using naked sets: [(1)r78c5, (3)r78c5, (6)r78c7]
sis used and matrix counts:
(4)Box2 r3c5=r2c6 +2 (-1) / (2)Box 2 r3c5,r2c6, r1c6 +0/(12)r1c2 +0/(6)c6 r124 +0/(3)c6 r124[78] +0 assuming group/aur sis: (1)r78c5, (3)r78c5,(6)r78c7: +1, total +3(-1) / (6)c8r247 +0 / (1)Box3 r1c9,r2c8,r3c9 +0 (-1)
Find target (1)r3c5 see counting adjustments in bold above. total 0. done. => r3c5<>1 key to this step really is almost hidden pair 24 Box 2, then linking that to a few sis.
2) (1)r78c5 => r8c4,r78c6<>1
3) using aur 34 r78c26: (9=np34)r78c6-aur(hp34)r782=(4)r3c2-(4)r2c3=(4)r2c6-(4=np39)r78c6 => r8c4,r56c6<>9
4) (4)r3c2=(4)r78c2-[(4)r9c3=(4)r9c79-(4=*6)r8c7-(6)r8c3=(nt149)r9c13.r8c3]=*(5)r8c7-(5)r7c78=(5)r7c5-(5=np23)r46c5-(2=4)r3c5 partial loop =>
r7c3<>4, r8c5<>5, r78c5<>3, r3c3<>4, r4c6,r6c4<>3
5) typ als: (2=6)r4c6-(6)r4c4=(nq1259)r6c1234 => r4c1,r6c5<>2
6) (2)r6c123 => r6c79<>2
7) using AUR 39 r89c26 => sis: [(4)r89c6,(4)r89c2, (12)r89c2] =>
(nt126)r12c1.r1c2=(5)r1c1-(5=9)r4c1-(9)r6c2=(np12)r16c2pause-(12*)r78c2=[(4)r2c3=(4)r2c6-(4)r89c6=*(4)r89c2-(4)r3c2=(4)r2c3] => r2c3<>12, r3c2<>12
8) (3)r8c4=(hp39-hp34)r78c6=(4)r2c6-(4=8)r2c3-(8)r1c3=(8)r1c4 => r1c4<>3
9) (3)r1c9=(3)r1c6-(3=np49)r78c6-(4)r2c6=(4)r2c3-(4)r9c3=(hp47-hp73)r9c79=(3)r9c8=>r2c8,r9c9<>3
10) (4)r2c6=(4-2)r3c5=(2-3)r4c5=(3)r6c5-(3*)r6c7=[(3)r2c7=*(hp37-hp74)r9c79=(4)r9c3-(4)r2c3=(4)r2c6] => r2c6<>3
11) (8=4)r2c3-(4)r6c2=(hp49-hp93)r78c6=(3)r1c6-(3)r2c4=(3)r2c7 => r2c7<>8 =>(8)r3c7
12) (6)r3c9=(6)r3c4-(6)r6c4=[(hp64)r6c79&nt(456)r678c7] => r2c7<>6
13) (2=3)r2c7-(3)r1c9=(3)r1c6-(3=np49)r78c6-(4)r2c6=(4)r2c3-(4=9)r3c2-(9)r3c3=(np12)r1c2.r3c3 => r2c1<>2 Also shorter (less sis) if we used ahp 24Box 2, but less well-liked as it happens to be krakenese. i.e.:
(2*)r2c6-[(2)r1c6=*(2-4)r3c5=(4)r2c6-(4=np39)r78c6]-(3)r1c6=(3)r1c9-(3=2)r2c7
compare sis, 6 using kraken 2 box 2, versus 8 sis using typical AIC.
14) (2)r7c3=(hp23-hp34)r78c2=(4-9)r3c2=(9)r3c3 => r3c3<>2
15) (2)r1c123 => r1c69<>2
16) (np16)r2c18 => r2c46<>16
17) (1)r1c46=(1-6)r3c4=(6)r1c46-(6)r1c1=(6)r2c1-(6=1)r2c8 => r9c9<>1
18)(np36)r1c69=(1)r1c6-(1)r5c6=(1-9)r5c4=(9)r6c4-(9)r6c123=(9-5)r4c1=(hp25)r16c1 => r1c1<>6 solve 2 cells
19) (6)r13c9 => r467c9<>6
20) (3)r6c5=(3-2)r4c5=(2)r3c5-(2=6)r3c9-(6=3)r1c9 => r6c9<>3
21) (2)r4c5=(2)r3c5-(2=4)r2c6-(4=8)r2c3-(8)r1c3=(nt125)r1c123-mug(nt125)r6c123=(9)r6c123-(9=5)r4c1=> r4c5<>5
22) (2=3)r4c5-(3)r6c5=(3-6)r6c7=(6)r4c8-(6=2)r4c6 loop => r6c7<>45, r5c6<>2,r4c9<>2
23) (2)r2c7=(2-7)r5c7=(7-4)r9c7=(4)r9c3-(4)r2c3=(4)r2c6 => r2c6<>2 singles to end

Please forbear the likely typos.... hopefully not too many. Max sis in any one step, assuming aur/mug sis is native: 10.

Edited in the promised bold markers.
Last edited by Steve K on Thu Jul 03, 2008 6:21 pm, edited 2 times in total.
Steve K

Posts: 98
Joined: 18 January 2007

Hi Steve K,
Your first step is very difficulty for me to understand so I try to “translate” it to Kraken with 6’s on column 6 together AUR 45 at r78c57 then still too complex to present it even by diagram
Can you explain more detail?

Thanks again,
ttt
ttt

Posts: 185
Joined: 20 October 2006
Location: vietnam

ttt: using aur 45 r89c57 => sis [(1)r89c5, (3)r89c5, (6)r89c7], one can build a TM:
Code: Select all
` *--------------------------------------------------------------------* | 1256   12     1258   | 1368   7      1236   | 9      4      1236   | | 126    7      1248   | 1368   9      12346  | 2368   136    5      | | 3      1249   12489  | 168    124    5      | 268    7      126    | |----------------------+----------------------+----------------------| | 259    8      7      | 4      235    2369   | 1      3569   2369   | | 4      6      3      | 159    8      129    | 257    59     279    | | 1259   129    1259   | 3569   235    7      | 23456  8      23469  | |----------------------+----------------------+----------------------| | 8      12349  12469  | 7      1345   1349   | 456    1569   1469   | | 7      1349   1469   | 1359   1345   1349   | 456    2      8      | | 19     5      149    | 2      6      8      | 347    139    13479  | *--------------------------------------------------------------------*`

Code: Select all
`4B2  r3c5  r2c6                  2B2  r3c5  r2c6  r1c6               6c6         r2    r1    r4            r1c2               2           1         1B3  r3c9                     r1c9  r2c8      6c8                     r4           r2   r7   3c6               r1    r4                    r89aur 1c5B8                               6c7B9 3c5B8`

=> sis[(24)r3c5,(1)c5Box8,(1)r3c9] => r3c5<>1. I may have changed the order of the sis from the original post in the TM. The order can be somewhat arbitrary, albeit often not completely with a TM. The order of sis consideration with a MBM, however, is arbitrary. This is also true with the alg, MBM counting.
Steve K

Posts: 98
Joined: 18 January 2007

Hi Steve K,
I present your step 1 by diagram:
Code: Select all
`(45)AUR r78c57   => r3c5<>1 ||(1)r78c5 ||(3)r78c3-(3)r78c6 ||       || ||      (3-2)r1c6=(hp24)r2c6/r3c5  ||       || ||      (3-4)r2c6=(4)r3c5  ||       || ||      (3-6)r4c6-  ||               | (6)r78c7-(6)r7c8  |          ||      |         (6)r4c8-(6)r4c6          ||      ||            ||     (6-2)r1c6=(hp24)r2c6/r3c5          ||      ||          ||     (6-4)r2c6=(4)r3c5          ||         (6)r2c8-(1)r2c8                  ||                 (1)r3c9                  ||                 (1)r1c9-(1=2)r1c2-(2)r1c6=(hp24)r2c6/r3c5`
This is correct what you meant?
Thanks
ttt
ttt

Posts: 185
Joined: 20 October 2006
Location: vietnam

Hi to all.

I thought it might be interesting to add a set logic solution to the others for comparision. I have posted brief set logic descriptions along with references to diagrams and images with more detail. Eliminations are of the form:

Rank N(n sets+ l linksets), [ cover sets ]=[ linksets ] reason for elimination => result

Sets: 9r3 is the set of 9s in row 3, and 4n5 is the set of candidates in cell r4c5. A finned X-wing in the rows would be: rank 1 (2,3) [5r25] = [5c25 5b1] 5c2*5b1 => r1c2<>5, where the * indicates overlap sets. For more on sets and notation, see this short reference or a full description at sudokuone.com. I apologize for the troublesome notation, which is partly due to differences between 1st and 2nd order logic. The referenced diagrams should help.

ttt wrote:
Step 4) A complex step : (9)r3c3=(9-4)r3c2=(hp34)r78c2-(4=hp19)r9c13-(19)r9c89=(hp19)r7c89-(19=hp34)r7c26-(134=5)r7c5-[(5)r4c5 & )r7c8] = {(6)r1c1=(6)r2c1-(6)r2c8=(6-5)r4c8=(5)r4c1}-(5)r1c1=(ht126)r12c1/r1c2
=> r3c23<>12 .......how to present it as Diagram, too complex...

Ttt, below is a set diagram (click to enlarge) for elimination 4 where the solid bars are strong sets. Your solution is very elagant and this eliminationmakes a good comparison with set logic. It uses 33 sets, 14 strong sets, and has a rank of 5 because it needs 5 extra linksets to cover the sets. A finned fish has a rank of 1. Several triplets make up for the extra linksets (red arrows).

3D view

I also tried drawing Steve's move 1 (another nice move) here, as drawn by ttt.

This solution has two complex steps (7, 11) each with 34 sets.

(1) Rank 1(5,6), [9r3 2r7 6c3 16n2] = [129c2 78n3 9c3] 9c2*(78n3) => r78c3<>9 Short ALS chain
(2) Rank 1(4,5), [9r3 2r7 16n2] = [129c2 2c3 3n3] 2c2*3n3 => r3c3<>2 Short ALS chain
(3) Rank 1(6,7), [12r5 46r6 59c4] = [5n46 6n479 8n4 2b6] 2b6*5n46 => r6c7<>2, r6c9<>2 3D kraken with ALS chain
(4) Rank 1(5,6), [6r36 78n7 4b6] = [6c4 456c7 6n9 6b3] 6c7*6b3 => r2c7<>6 Finned X-wing + ALS chain
(5) Rank 1(8,7), [46r6 59c4 78n7 5n8] = [59r5 456c7 6n49 8n4] 5c7*5r5 => r5c7<>5 Rank 1 multiloops
(6) Rank 2(12,14), [8r1 46r6 6c1 5c3 3c7 356c8 23n4 6b2] = [6r1 36r2 5r6 3r9 168c4 1n3 4n8 6n79 7n8 5b6] 5r6*6n7*5b6 => r6c7<>5, (T)5c8*7c8 => r7c8<>5 Rank 1 multiloops
(7) Rank 2(16,18), [13568r1 6r2 1r6 5c3 9n1 1236n4 1b3 9b4 6b5] = [1r3 59r6 169c1 1c2 368c4 6c6 1n139 2n8 6n3 13b2], (T)1r3*1b2 => r3c5<>1 Complex structure
(8) Rank 0(1,1), [1c5] = [1b8] 1b8 => r7c6<>1, r8c4<>1, r8c6<>1 Locked candidates
(9) Rank 2(8,10), [5r7 4c256 7n89 8n7 1b8] = [4r3 149r7 4r8 78n5 4b2 56b9] (TT)4r7 => r7c3<>4, r7c7<>4, (TT)8n5 => r8c5<>3, r8c5<>5 Intertwined ALSs and chain make rank 0 sets
(10) Rank 2(8,10), [2r6 19c1 136n2 3n5 2b7], [4r3 129c2 2c3 2c5 69n1 1b1 9b4] (TT)2c3 => r1c3<>2, r2c3<>2 Intertwined ALSs make rank 0 sets
(11) Rank 4(15,19) [4r269 6r36 5c48 7c9 3n5 5n7 7n7 8n357 6b6] = [2r3 57r5 156r8 4c3 6c4 4c5 2456c7 6c9 4n8 6n49 9n9 4b2] (TTT)2r3*2c7 => r3c7<>2 Complex structure

(12) Rank 4(10,14) [5r147 6c1 8c3 14c5 6c8 3n7 4b1] = [6r2 48r3 6r7 5c15 6c7 1n13 2n3 4n8 7n57 8n5] (TT)6r7*6c7*7n7 => r7c7<>6 => 5r7c7 => r7c57<>5 => 5r8c5 => r56c4<>5 r8c4<>39 => 5r5c8 => r4c8<>5 r5c8<>9 Multiloop with 3-way chain overlap,
(13) Rank 0(2,2), [39r8] = [8n26] 8n26 => r8c2<>1, r8c2<>4 8c6<>4 Hidden pair,
(14) Rank 1(2,3), [4c2 4b8] = [4r37 4c5] 4r3*4c5 => r3c5<>4 => 4r2c6 => r2c3<>4 r2c6<>1236 r7c6<>4 => 2r3c5 => r3c29<>2 r46c5<>2 r1c6<>2 Box ended chain,
(15) Rank 1(4,5), [2r7 6c3 4c5 8n7], [4r7 46r8 7n23] 4r7*7n2 => r7c2<>4 => 4r3c2 => r3c2<>19, r3c3<>4 => 9r3c3 => r3c3<>18 r79c3<>9, r5c7<>2 => 8r3c7 => r3c7<>6, r3c4<>8=> 7r5c7 => r5c9<>7, r9c7<>7 => 7r9c9 => r9c9<>1349 Simple chain,
(16) Rank 2(8,10), [1r5 2c167 5c1 6c6 9c4 9b4] = [2r25 6r1 9r6 146n1 5n4 45n6] (T)6r1*1n1 => r1c1<>6 => 6r2c1 => r2c1<>12, r2c48<>6 => 2r2c7, r2c7<>38, r1c9<>2 AALS like multiloop + converging chains
(17) Rank 1(8,9), [6r3 19r5 26c6 4n8 36b3] = [6r4 3c8 1n9 3n9 4n6 5n46 6b2 9b6] 6r3*6b3 => 6r3c9, r1c9<>6, r3c4<>6 Discontinuous Nice Multi-loop with ALS

(singles to the end)
Allan Barker

Posts: 266
Joined: 20 February 2008

As I've been away for some time, here's a late answer.

ttt wrote:I can’t understand Denis’s solution by his notations

I don' think that's a matter of notations but of knowing the basic concepts of the chains used in my solution.
The most general chains I use are nrczt-chains, which are a straightforward, but very powerful, generalisation of xy-chains.
You can find all the definitions (and notations) in my book or in the first posts of following threads:
- for the general concept of a resolution rule: http://forum.enjoysudoku.com/viewtopic.php?t=5600
- for the 3D chains: http://forum.enjoysudoku.com/viewtopic.php?t=5591
and for the specific 2D cases:
- for the 2D chains: http://forum.enjoysudoku.com/viewtopic.php?t=5562
- for the hidden chains: http://forum.enjoysudoku.com/viewtopic.php?t=5555
nrczt-chains (under the imprecise name of t-chains) are used everyday in the French forum: http://www.sudoku-factory.com/forumsudoku/viewforum.php?f=3

Anyway, you're right to look for your own solution with your preferred methods.

Steve K wrote:TM's are generalized predecessor to *****tchains, matrix counting is an alg to find them (and all matrices.))

As TM's are the mere translitteration of the basic four Sudoku axioms in matrix notation, it shouldn't be a surprise that any resolution rule (which must be provable from the Sudoku axioms) can be reformulated as TM's. The real problem is finding interesting special cases of TM's.
Stating that TM's are generalisations of nrczt-chains is thus vacuous.
Stating that TM's are predecessors of these chains is thus either vacuous if understood in the sense of "they happen to be abstract generalisations of" or mere goofing if understood as "I knew nrczt-chains even before they were formulated".
TM's include very complex nets (in particular nrczt-nets). Moreover, the algorithm implemented in Steve's solver, "matrix counting" as he calls it, has nothing to do with the linear structure of nrczt-chains and the way they can be found.
Unfortunately, the extent of TM's remains rather fuzzy; e.g. I've never seen any precise final definition of them.
As this discussion already took place in the un-moderated Eureka forum, one year ago, I won't insist.

But the interesting point in this discussion is what we are looking for:
- are we looking for the most general pattern, the RuleOfEverything, a pattern so complex that nobody can find its occurrences without a computer?
- or are we looking for a hierarchy of patterns of increasing complexities?
Said otherwise: should we use a bazooka to solve a puzzle that can be solved with a small hammer?
As for me, I wouldn't look for nets when I can get a solution with chains.

As each of us has his own, often unstated, goals, I've lost any hope that any agreement can be reached on this. Instead, what I've tried to do is launch a classification program according to the mean complexity of each proposed rule (here: http://forum.enjoysudoku.com/viewtopic.php?t=5995). As of now, only nrczt-chains and their specialisations have been studied in a systematic way from this POV.
A typical case of a total lack of information is about AIC's (or NL's, as you prefer) with ALS's: how does their mean complexity vary wrt to the a priori main two parameters: length of the chain vs maximun size of the ALS's. Is there any meaningful means of aggregating these two parameters into a single one (e.g. total number of candidates)?
Said otherwise: what is the main parameter of ALS's complexity?
Second question: how can we classify AIC's with ALS's and nrczt-chains wrt each other?

Allan Barker wrote:I thought it might be interesting to add a set logic solution to the others for comparision.

Comparison seems difficult as the resolution paths are different from the start.
This brings me back to my previous point: as comparisons on specific puzzles prove to be impossible most of the time, because there are many possible resolution paths, except on elementary ones, shouldn't we try to evaluate the mean complexity of each pattern (which supposes we define a classification of the associated rules)?
denis_berthier
2010 Supporter

Posts: 1261
Joined: 19 June 2007
Location: Paris

Steve K wrote:
TM's are generalized predecessor to *****tchains, matrix counting is an alg to find them (and all matrices.))

Context here is important. After this statement, I present a chain, which is a *****t chain that happens to use Uniqueness to create "cells". I also present an alg that finds this chain.

Denis B wrote:
As TM's are the mere translitteration of the basic four Sudoku axioms in matrix notation, it shouldn't be a surprise that any resolution rule (which must be provable from the Sudoku axioms) can be reformulated as TM's. The real problem is finding interesting special cases of TM's.

TM's cannot reformulate ANY resolution rule. If, in fact, one knows otherwise, please supply a proof of same. It would be quite exciting! Sadly, I fear the proof will be illusive. For example, TM's cannot derive a hidden pair loop. One can formulate a resolution rule for that pattern.
Denis B wrote:
Stating that TM's are generalisations of nrczt-chains is thus vacuous.
Stating that TM's are predecessors of these chains is thus either vacuous if understood in the sense of "they happen to be abstract generalisations of" or mere goofing if understood as "I knew nrczt-chains even before they were formulated".

An interesting twist, but it assumes facts not in evidence. The claim of "Vacuous" is deliciously vacuous . My intent was only to help readers understand an alternate way to view the elimination. Historically, I was introduced to TM's before *****t chains were published. "predecessor" only indicates that one happened before the other. This is a fact, and it is in evidence at the sudoku.com.au forum.
Denis B wrote:
TM's include very complex nets (in particular nrczt-nets). Moreover, the algorithm implemented in Steve's solver, "matrix counting" as he calls it, has nothing to do with the linear structure of nrczt-chains and the way they can be found.

One might think that alternate ways to find a pattern would be a good thing. I see nothing in the resolution rules defining *****t chains that limits how one may choose to search for them.
Importantly, the fact that a particular technique can be viewed as part of a greater (meaning larger) whole is not meant as a criticism of that technique. If, in fact, an alg can "catch" more than one technique, that proves exactly that the alg must have "something" to do with each of those techniques. The claim, therefor, of no relationship between the two has no basis.
TM's, in general, are no more "nets" than any ****t chain.

Denis B wrote:
Unfortunately, the extent of TM's remains rather fuzzy; e.g. I've never seen any precise final definition of them.

The precise final defintion has been in existence for quite some time. Here is a link: http://www.sudoku.org.uk/SudokuThread.asp?fid=4&sid=9059&p1=3&p2=11
One can also read the forums at sudoku.com.au. Triangular matrices were defined there in late Winter of 2006.
The definition has never been in flux. How the "extent is fuzzy" is unclear (another very beautiful irony!). I suppose, since few of us are divine, the extent of most things is fuzzy.

I am a bit confused, however. If the definition of TM's is unclear, how could one know that TM's are a mere translitteration of the sudoku axioms? In fact, the definition of TM has nothing directly to do with sudoku, thus the claim Denis B. makes is, in the most generous possible interpretation, misleading. Suppose it is true that TM's are a mere translitteration of the sudoku axioms. In mathematics, adding zero and multiplying by one are popular tools.

Denis B wrote
But the interesting point in this discussion is what we are looking for:
- are we looking for the most general pattern, the RuleOfEverything, a pattern so complex that nobody can find its occurrences without a computer?
- or are we looking for a hierarchy of patterns of increasing complexities?
Said otherwise: should we use a bazooka to solve a puzzle that can be solved with a small hammer?
As for me, I wouldn't look for nets when I can get a solution with chains.

I can only speak for one human. I am looking for patterns that humans can find. There is, however, no demonstration that I am aware of which connects the "complexity of pattern" with "human findability". It seems unlikely that one can achieve such a goal. For example, one human may find a hidden single easier to locate than a naked single. Another may find the opposite. No mathematical analysis of complexity deals well with this inconsistency in human judgement. If failure of mathematical analysis of "pattern location difficulty" exists for even the simplest of possible patterns, it seems unlikely that mathematical analysis of complexity can successfully predict "findability" for more complex patterns.

Although Denis presents us with exactly two possibilities, some human solvers probably seek neither of those two possibilities.

Often, one is looking for tricks to place in the tool bag. One is free then to accept or reject those tricks as they are found useful. Associating one group of tricks with another is nothing more than exactly what it seems to be: noting the association. Use of any trick can be determined by the individual solver, according to the needs implied by the particular problem at hand.

****t chains are a very powerful, and very general tool. That they overlap with other techniques in one direction has been noted by Denis numerous times. The fact that they overlap with another tool in another direction is equal in relevance. Understanding the "markers" for both the more specific tools and the more general tools could be helpful.

Denis B wrote:
Stating that TM's are generalisations of nrczt-chains is thus vacuous
The reason I made this post.

Finally, no offense was meant by my very short sentence to which Denis has made a rather long response. Nothing in my original short sentence is false.
Steve K

Posts: 98
Joined: 18 January 2007

Steve K wrote:
denis_berthier wrote:As TM's are the mere translitteration of the basic four Sudoku axioms in matrix notation, it shouldn't be a surprise that any resolution rule (which must be provable from the Sudoku axioms) can be reformulated as TM's. The real problem is finding interesting special cases of TM's.

TM's cannot reformulate ANY resolution rule.

True; my formulation was sloppy. They are a PARTIAL reformulation. As a result, my considerations about the generalisation being vacuous were not logically justified.

Steve K wrote:Historically, I was introduced to TM's before *****t chains were published. "predecessor" only indicates that one happened before the other.

Fine. I wasn't debating this point, which everyone can check.

Steve K wrote:"predecessor" only indicates that one happened before the other.

Then if by "predecessor" you mean nothing more, no problem for me.
What should be clear, because it is important if one wants to understand them, is that there is no logical anteriority of TM's: the logic of xyt, nrczt and other chains was inspired by that of xy-chains - not by any matrix idea. (The fact is, I didn't know TM's when I introduced these chains. My book has been published in May 2007, which entails that the xyt, xyz and xyzt chains - making 100 pages of the book - must have been in my papers and in SudoRules much before this)

Steve K wrote:
denis_berthier wrote:TM's include very complex nets (in particular nrczt-nets). Moreover, the algorithm implemented in Steve's solver, "matrix counting" as he calls it, has nothing to do with the linear structure of nrczt-chains and the way they can be found.

One might think that alternate ways to find a pattern would be a good thing. I see nothing in the resolution rules defining *****t chains that limits how one may choose to search for them.

The purpose of a resolution rule is not to be a search algorithm. The same is true for matrices: in and by themselves, they don't define any algorithm.

Steve K wrote:Importantly, the fact that a particular technique can be viewed as part of a greater (meaning larger) whole is not meant as a criticism of that technique. If, in fact, an alg can "catch" more than one technique, that proves exactly that the alg must have "something" to do with each of those techniques. The claim, therefor, of no relationship between the two has no basis.

Still a problem of vocabulary:
- as neither nrczt-chains nor matrices are techniques or algorithms, one being more general than the other can't be stated in terms of algorithms but of logic;
- as for the techniques possibly associated to each of them, your counting algorithm can hardly be considered as any natural generalisation of the techniques for nrczt-chains, which are based on their LINEAR structure and need no counting.
But the real question is: are you suggesting that your counting algorithm is simpler (for a human, not for your computer) than the algorihms I've described for finding nrczt-chains? That'd be the only interesting meaning of "alternate way to find" them.
Now, if I take your statement litterally, recursive Trial and Error (rT&E) is an algorithm that can "catch" more than any other technique (as it can solve any puzzle). Following your line of reasoning, rT&E should therefore have "something" to do with any other technique. Which is a vacuous statement, unless "something" is specified.

Steve K wrote:TM's, in general, are no more "nets" than any ****t chain.

Absurd. Your counting algorithm is the proof you're using branching (several right-linking candidates), which never happens in ****t-chains. I wonder why you need so much to maintain such confusion.

Steve K wrote:
denis_berthier wrote:But the interesting point in this discussion is what we are looking for:
- are we looking for the most general pattern, the RuleOfEverything, a pattern so complex that nobody can find its occurrences without a computer?
- or are we looking for a hierarchy of patterns of increasing complexities?
Said otherwise: should we use a bazooka to solve a puzzle that can be solved with a small hammer?
As for me, I wouldn't look for nets when I can get a solution with chains.

I can only speak for one human. I am looking for patterns that humans can find. There is, however, no demonstration that I am aware of which connects the "complexity of pattern" with "human findability". It seems unlikely that one can achieve such a goal. For example, one human may find a hidden single easier to locate than a naked single. Another may find the opposite. No mathematical analysis of complexity deals well with this inconsistency in human judgement. If failure of mathematical analysis of "pattern location difficulty" exists for even the simplest of possible patterns, it seems unlikely that mathematical analysis of complexity can successfully predict "findability" for more complex patterns.

"one human may find a hidden single easier to locate than a naked single": is this your best example?
It is a fact that there exists several ratings for puzzles and that these ratings, based on different approaches, are reasonably well correlated (see this thread: http://forum.enjoysudoku.com/viewtopic.php?t=5995). Negating any possibility of mathematical analysis is thus provably absurd.
From a more intuitive POV, it seems obvious that, if there are more USELESS partial chains of some kind, it will be more difficult to find the useful ones.

Steve K wrote:****t chains are a very powerful, and very general tool. That they overlap with other techniques in one direction has been noted by Denis numerous times. The fact that they overlap with another tool in another direction is equal in relevance. Understanding the "markers" for both the more specific tools and the more general tools could be helpful.

If (and only if) things are stated clearly, I agree that comparisons can be useful. But comparing nrczt-chains with a countig algorithm and trying at the same time to deny their fundamentally linear structure can only be misleading.
denis_berthier
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Steve K wrote:
denis_berthier wrote:Unfortunately, the extent of TM's remains rather fuzzy; e.g. I've never seen any precise final definition of them.

The precise final defintion has been in existence for quite some time. Here is a link: http://www.sudoku.org.uk/SudokuThread.asp?fid=4&sid=9059&p1=3&p2=11
The definition has never been in flux. How the "extent is fuzzy" is unclear (another very beautiful irony!). I suppose, since few of us are divine, the extent of most things is fuzzy.

A special post for this.
Steve K wrote:Pigeonhole matrix definition:
Always nxn. Thus number of columns = number of rows.
First column is the result column.
Each Row is a strong inference set. Thus in each row there must exist at least one truth.
Each column, except the first column is a weak inference set. Thus in each column except column 1, there can be no more than one truth.

All is said: the definition is not given in factual terms, but in terms of weak and strong inference sets. The way these inference sets are related to facts on the grid is undefined. E.g. do these strong inference sets allow ALSs?

Now where are TM's defined?
denis_berthier
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Denis, further down the page is the TM definition. It is unclear what you think is undefined. Everything that needs to be defined is defined.

Denis, let me preface this post with the following: It is your habit to use words such as "absurd", "vacuous", etc. This perhaps is your style, but it is not becoming. Another habit is to presume meaning into words that does not exist. This is, of course, counterproductive.

Never have I suggested, in any post in any forum that you did not arrive at your techniques independently.

Never have I submitted, any solution that was computer generated. Thus, my experience with algs is from a human perspective. Admittedly, I am only one human.

The lack of care in responding to me is extraordinary. Denis B wrote
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Steve K wrote:
Denis, further down the page is the TM definition. It is unclear what you think is undefined. Everything that needs to be defined is defined.

Steve K wrote:
Tri-angular matrix definition:
nxn
Each row contains at least one truth
The top entry of each column is in conflict with each item below it.
For row i, items i+2 and greater are empty. This can be translated, in Booleans, as False.

I guess you've just forgotten to give the condition on rows ("Each Row is a strong inference set. Thus in each row there must exist at least one truth."). Or am I wrong to suppose this?

It is extraordinarily tiresome to read a post that mutilates what someone has written with falsehoods, wild extrapolations, and innuendo. I could respond to your posts, but you have proven before, and here again, that you have no interest in an actual discussion.

In my original post, I was careful to not write nrczt-chains specifically to avoid this type of interchange. Instead I wrote *****t chains. The way that I choose to use that particular deductive path and your preferred use of nrczt chains will forever be different. I respect your reasons for wishing to do so, whatever they may be.

In my next post, I forgot to maintain the distinction between ***** t chains from nrczt-chains. I apologize for allowing your wildly inaccurate statements to allow me to lose my discipline. Rather than pollute this thread with more garbage, I shall discontinue. You are free to do as you please, including declaring victory.

Edited in a cross post, then edited again to reflect on the next post.
Last edited by Steve K on Fri Jul 25, 2008 2:46 am, edited 2 times in total.
Steve K

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