The New Sudoku Players' Forum

[Max Beran]

I don't know if I'm having my knuckles rapped by QBasicMac or whether he's just trying to restart the old debate on what is and what isn't T&E. For myself, I take the hedonic (if such a word exists) approach - if it's fun, it's okay. I'm totally with tso and em in decrying Guest's foolish posts but presumably you, QBasicMac, don't have much problem with them (now who's being provocative). And finding a forcing chain or a non-repetitive loop is fun to the point of orgasmic! Doubly so if I can get the right answer on the board before anyone else does.

Having said that, I do notice a resistance to the pathway approach and to that end I will add a post to Jeff's Bilocation bivalue topic at http://forum.enjoysudoku.com/viewtopic.php?t=2143&highlight=
to show how the task of finding these pathways can be simplified – basically by converting them from algebra into geometry.

Whenever I see the term “T&E”, I automatically translate it in my head as “S&M” – am I alone?

[QBasicMac]

knuckles rapped

Knuckles rapped? I wondered how you might imagine that until I noted that my post directly follows yours. Pure coincidence, Max. I was not responding to anyone. I just took the original puzzle and tried to see how far I could get before bogging down. The last move I could find was an X-Wing which didn't help make any more moves.

So I am asking if there are any more along the hidden-quad line.

You seem to equate T&E with sadism and masochism. I find T&E quite a lot of fun. Instead of only one puzzle to solve, you get many. In this case, I tried pencilmark 1 in cell r9c9 and after a lot of singles, proved that puzzle was invalid. So next comes pencilmark 2: After singles, got a solution. Finally comes pencilmark 7: That puzzle was also not valid.

So I got to solve three SuDokus instead of one! And they were fast and easy.

As only one of the trials led to a solution and all other trials led to invalid puzzles, the puzzle is legal and the solution has been found.

If I enjoy that so much, why not only solve singles and when stuck, go immediately into T&E? Why fiddle with locked-candidates, etc.? Why ever make any pencilmarks at all?

I've done that. But singles-only puzzles, while being a lot of fun to solve, can become boring after a while. So I like to use pencilmark techniques, too, at times. But just as some people like apple pie, whereas others insist cherry pie is better, so I like T&E and feel it is easier and more fun than complicated and, to me, boring and time-consuming coloring and chaining stuff. A matter of taste.

It was not my point to argue about T&E. I just mentioned it in an attempt to get an answer to my question that would be useful.

Q. Is there a non-chaining-coloring continuation (even if for only one step) after my current position? If "no", then let me know that. If "yes" then point out the Swordfish, Triple or Quad, etc. that I missed.

Mac

P.S. I am also in the who-wants-to-see-the-solution camp. I even wonder why solutions are published in newspapers and books at all. The solution is boring and not instructive (unlike word puzzles, where the solution can be used to learn something).

[emm]

Max B - I’m struggling to see the connection between T&E and S&M. I may not be alone. It seems about as far-fetched as an orgasmic forcing chain - but maybe you have a bigger picture!

I have to say that Mac’s solving method has less appeal to me than a tofu noodle salad, but I have learnt to appreciate his forthrightness and determination to stick to his guns in this forum despite criticism and total lack of agreement from anyone else. One thing I get from him - and why I still need to learn this after all the lessons I’ve had, is beyond me - is that other people don’t see things my way. Crazy I know, but true!

Mac - I think it’s hilarious that you see your method as value for money – 3 puzzles in one! Hey you might need to find only one thousand solution puzzle and it’d do you for life!

Seriously though - there are 2 simple moves to get beyond your posted candidates. A naked triple in row three and an Xwing in r69c15. After that it’s colouring or chains I’m afraid.

[QBasicMac]

There are 2 simple moves to get beyond your posted candidates. A naked triple in row three and an Xwing in r69c15. After that it’s colouring or chains I’m afraid.

THANKS!!!!

Mac

[rubylips]

please tell me in great detail where it is.

I hope this is sufficient. The chain introduced in Move 5 probably won't interest you.

Code: Select all

1. The values 7, 8 and 9 occupy the cells r3c2, r3c3 and r3c5 in some order.
- The moves r3c2:=2, r3c3:=2, r3c5:=1, r3c5:=2 and r3c5:=3 have been eliminated.
Consider the chain r8c5~1~r1c5-1-r1c9-1-r3c8-1-r8c8.
When the cell r8c5 contains the value 1, so does the cell r8c8 - a contradiction.
Therefore, the cell r8c5 cannot contain the value 1.
- The move r8c5:=1 has been eliminated.
Consider the chain r1c5-1-r1c9-1-r3c8-1-r8c8-1-r8c6-1-r9c5.
The cell r9c5 must contain the value 1 if the cell r1c5 doesn't.
Therefore, these two cells are the only candidates for the value 1 in Column 5.
- The move r6c5:=1 has been eliminated.
Consider the chain r6c5-5-r6c1-5-r9c1-5-r9c5.
The cell r9c5 must contain the value 5 if the cell r6c5 doesn't.
Therefore, these two cells are the only candidates for the value 5 in Column 5.
- The moves r4c5:=5 and r8c5:=5 have been eliminated.
The values 1 and 5 occupy the cells r8c6 and r9c5 in some order.
- The moves r8c6:=2, r8c6:=6 and r9c5:=2 have been eliminated.
The cell r8c7 is the only candidate for the value 6 in Row 8.
2. The value 4 is the only candidate for the cell r7c7.
3. The cell r7c6 is the only candidate for the value 6 in Row 7.
4. Consider the chain r5c1~2~r9c1-2-r9c9-2-r8c8-2-r3c8~2~r3c6-2-r5c6.
When the cell r5c1 contains the value 2, so does the cell r5c6 - a contradiction.
Therefore, the cell r5c1 cannot contain the value 2.
- The move r5c1:=2 has been eliminated.
The value 7 is the only candidate for the cell r5c1.
5. Consider the chain r8c5+<7|5>+r9c5-1-r8c6+<5|2>+r8c8.
When the cell r8c5 contains the value 2, so does the cell r8c8 - a contradiction.
Therefore, the cell r8c5 cannot contain the value 2.
- The move r8c5:=2 has been eliminated.
The value 7 is the only candidate for the cell r8c5.


[Guest]

Man, there's always something; sorry.

Could someone explain this bit to me...

[rubylips]

tso, you're right and the crow isn't as tasty as the turkey.

Could you explain that bit while you're at it?

[QBasicMac]

I'll try to help tso out here.

> tso, you're right and the crow isn't as tasty as the turkey.

Part A. To admit someone is right is often called "eating crow".
Part B. It was Thanksgiving in America and traditionally people enjoy eating turkey on that day.
Combining these parts, we get a joke. Get it?

> Meaning I copied the grid to a paper having another one on it and dubbed the wrong one in. Man, there's always something

"Man, " - Just a general expression denoting frustration.
"there's always something" - meaning that there is always a reason why things go wrong.

Welcome to the world of American English, guys.

"guys" - people, whether male or female.

Mac

[rubylips]

Thanks Mac. Unfortunately, Google doesn't offer UK English-to-US English translation.

[rmadhusudanan]

Using notation: Top row is R1 and Bottom Row is R9. Left Column is C1 and Right most Column is C9.

The possibility of number 1 in R8C5 is false because 1. If R8C5 is true R8C8 is false 2. R3C8 is true. 3.R1C9 is false 4. R1C5 is true. This cannot be as the Column 5 can contain 1 in only one cell. Therefore R8C5 DOESNOT CONTAIN 1.

When you eliminate this candidate you will easily see that R7C6 CAN BE ONLY 6.

Best of luck to finish the puzzle.

R Madhusudanan
India

[QBasicMac]

When you eliminate this candidate you will easily see that R7C6 CAN BE ONLY 6.

Sorry, I cannot "easily see" that.

Below is my position after making your R8C5 DOESNOT CONTAIN 1 change.

Mac


Solution So Far

Code: Select all

63- 5-4 87-
1-- 6-7 349
4-- --- 5-6
91- 8-- 26-
--- -6- 95-
-63 --9 -8-
8-1 --- -95
3-- 9-- --8
-96 4-8 -3-


Pencilmarks So Far

Code: Select all

-         -         29        -         129       -         -         -         12       
-         258       258       -         28        -         -         -         -       
-         278       2789      123       12389     123       -         12        -       
-         -         45        -         3457      35        -         -         347     
27        248       248       1237      -         123       -         -         1347     
257       -         -         127       12457     -         17        -         147     
-         247       -         237       237       236       46        -         -       
-         2457      2457      -         1257      1256      46        12        -       
25        -         -         -         125       -         17        -         127     


[Jeff]

Code: Select all

-         -         29        -         129       -         -         -         12       
-         258       258       -         28        -         -         -         -       
-         278       2789      123       12389     123       -         12        -       
-         -         45        -         3457      35        -         -         347     
27        248       248       1237      -         123       -         -         1347     
257       -         -         127       12457     -         17        -         147     
-         247       -         237       237       236       46        -         -       
-         2457      2457      -         1257      1256      46        12        -       
25        -         -         -         125       -         17        -         127

Mac, may I ask how did you create the above grid. Was it an output from your own solver or Did you type out each number? It must have been painstaking. Reason I ask is that no solvers known to me would output a format like that.

[QBasicMac]

How did you create the above grid?
Was it an output from your own solver?

Heh - well, not a "solver". I wrote a scratchpad program to help me avoid paperwork. I have to solve the puzzle myself.

(Well, it does provide hints if I stall too long before making any progress, and it does naked singles automatically, if I ask it to.)

I think the total number of people in the Universe who use this program is two.

Anyway, it saves current progress in that format, which I merely copy to here. If you think there is a superior format, let me know. I can either change my scratchpad or, at a minimum, write a quickie program to convert my format to one you prefer to look at.

Mac

[Jeff]

If you think there is a superior format, let me know. I can either change my scratchpad or, at a minimum, write a quickie program to convert my format to one you prefer to look at.

Thanks Mac. Your current format is sufficient. I don't want to impose extra work for you, but the format below would be a little bit better. Feel free to ignore.

Code: Select all

+------------------+------------------+------------------+
| 6     3     29   | 5     129   4    | 8     7     12   |
| 1     258   258  | 6     28    7    | 3     4     9    |
| 4     278   2789 | 123   12389 123  | 5     12    6    |
+------------------+------------------+------------------+
| 9     1     45   | 8     3457  35   | 2     6     347  |
| 27    248   248  | 1237  6     123  | 9     5     1347 |
| 257   6     3    | 127   12457 9    | 17    8     147  |
+------------------+------------------+------------------+
| 8     247   1    | 237   237   236  | 46    9     5    |
| 3     2457  2457 | 9     1257  1256 | 46    12    8    |
| 25    9     6    | 4     125   8    | 17    3     127  |
+------------------+------------------+------------------+

[rmadhusudanan]

My pencil marks are slightly different in the following manner
1. Row 3 has 3 cells having 123,123 & 12 and hence the number 1,2&3 cannot figure in any other cell in that row. Consequently the pencil marks now read as 78,789,123,89,123,12 in those cells in R3.
2. I do not have 5 in my pencil marking in R8C5. Logic being- If 5 is true in R8C5 then it is false in R6C5-true in R6C1 and false in R9C1 which means there is a 5 in either R8C2 OR R8C3. This cannot be true and hence there is no 5 in R8C5.
3. Now if you eliminate 1 in R8C5 the result is 27 which makes a trio of 237,237&27 in that Block.Therefore 2&3 cannot be in R7C6.
Hope I have helped