Solving without pencilmarks
by Finlip » Fri Aug 11, 2006 12:47 pm
I like solving without pencilmarks. I have been doing it regularly nowadays. I feel that all the solving with pencilmarks has made it difficult for me to spot things now. I was able to solve one simple sudoku extreme under ten minutes today but I generally feel happy when I do one in half an hour.
- Finlip
- Posts: 49
- Joined: 15 July 2005
Re: Solving without pencilmarks
by drsuds » Mon Sep 04, 2006 3:02 pm
Hello RW,
I'm a newbie and my interests are in solving without pencilmarks so I am very grateful for your post.
In trying to understand, your techniques I think I've found a couple typos which I present below and then I ask for some clarifications.
First what I believe to be the first typo.
In the last line above, the following makes more sense to me:
if r1c1=9 => r3c2=3 (B) => r9c2=4 (Co)
Basically, it seems row and column are swapped.
The next typo I believe is in the following.
It seems that the 3rd find just above should be r5c4=2 (B) instead of r5c1=2 (B).
Please forgive me (newbie) if these were pointed out before. I did not see corrections / notifications in the thread.
And now the request for clarification:
In Puzzle 2, following your trail, I get the following:
You said this results in a double solution.
Somehow I can't see it. Please clarify this point. Where is the double solution in the encoding above.
Much Thanks,
drsuds
I'm a newbie and my interests are in solving without pencilmarks so I am very grateful for your post.
In trying to understand, your techniques I think I've found a couple typos which I present below and then I ask for some clarifications.
First what I believe to be the first typo.
RW wrote:Puzzle 1
- Code: Select all
*-----------*
|.2.|1.8|..5|
|1..|.76|.3.|
|7..|4..|...|
|---+---+---|
|..7|...|2..|
|48.|...|.59|
|..2|...|6..|
|---+---+---|
|...|..9|..1|
|.5.|74.|..6|
|6..|8.1|.7.|
*-----------*
First the obvious ones:
r7c2=7 (B)
r1c7=7 (B)
r6c9=7 (B)
r5c6=7 (B)
r8c3=1 (B)
I can’t see anymore obvious cells by drawing lines, but I can see two pairs: r78c1={2,8} (R) and r23c3={5,8} (B). Next I start checking rows, starting with row 1. Two pairs there also: c1&5={3,9} and c3&8={4,6}. This is enough to solve the next number:
[edit: typo there, the first pair should read r78c1={2,8} (Co) = the two numbers cannot go anywhere else in the column.]
r9c3=9 (Co)
This gives me another pair: r7c3 and r9c2={3,4} – so far 5 pairs to remember. Now I see a very obvious trail:
if r1c1=9 => r3c2=3 (B) => r2c9=4 (Co) => r1c3=4 (B) => nowhere to place 6 in box 1. This gives me:
In the last line above, the following makes more sense to me:
if r1c1=9 => r3c2=3 (B) => r9c2=4 (Co)
Basically, it seems row and column are swapped.
The next typo I believe is in the following.
RW wrote:
Now I have:
- Code: Select all
*-----------*
|324|198|765|
|19.|.76|.3.|
|76.|4..|...|
|---+---+---|
|..7|...|2..|
|486|..7|.59|
|..2|...|6.7|
|---+---+---|
|.73|..9|..1|
|.51|74.|..6|
|649|8.1|.7.|
*-----------*
Still can’t see any obvious so I have a look at column 9, missing numbers 2,3,4 and 8. This gives me an idea that I have to check out and this is what I see:
if r3c5 or r3c6=2 =>r2c9=2 (B) and
if r2c4=2 => r2c3=5 (R) => r3c3=8 => r3c9=2 (Ce)
Anyway, r9c9 cannot be 2. This gives me a 2 in either r7c8 or r8c8 that not only forms an x-wing with c1 but also lets me exclude 8 from both squares (uniqueness). Thanks to this I can go on:
r8c6=3 (Ce)
r9c5=2 (B)
r5c1=2 (B)
r3c6=2 (B)
r2c9=2 (B)
It seems that the 3rd find just above should be r5c4=2 (B) instead of r5c1=2 (B).
Please forgive me (newbie) if these were pointed out before. I did not see corrections / notifications in the thread.
And now the request for clarification:
In Puzzle 2, following your trail, I get the following:
- Code: Select all
*-----------*
|1..|.32|578|
|58.|174|.92|
|7.2|589|461|
|---+---+---|
|8..|.2.|71.|
|67.|418|9..|
|921|75.|68.|
|---+---+---|
|..8|365|1.7|
|31.|297|8.6|
|..7|841|..9|
*-----------*
You said this results in a double solution.
RW wrote:if r5c7=9 => r5c1=6 (Ce) => r6c7=6 (B) => r3c8=6 (B) => r3c6=9 (Ce) => r6c1=9 (R) => Double solution. I can remove that option and go on:
Somehow I can't see it. Please clarify this point. Where is the double solution in the encoding above.
Much Thanks,
drsuds
- drsuds
- Posts: 2
- Joined: 04 September 2006
Is this one hard with no marks?
by fermat » Wed Sep 06, 2006 8:18 pm
This puzzle is pretty medium in technique but icky in fact.
Too many pencil-marks!
RW, is this pretty easy without? (It is a bit tedious with. <sigh> )
Too many pencil-marks!
RW, is this pretty easy without? (It is a bit tedious with. <sigh> )
- Code: Select all
. 7 . | 6 . . | 4 . .
8 . . | . 1 . | . 9 .
. . . | . . . | 1 . 7
---------------------
9 . . | . 2 . | . . .
. 5 . | 7 . 4 | . 2 .
. . . | . 3 . | . . 5
---------------------
3 . 7 | . . . | . . .
. 9 . | . 5 . | . . 3
. . 5 | . . 1 | . 4 .
- fermat
- Posts: 105
- Joined: 29 March 2006
by RW » Wed Sep 06, 2006 9:34 pm
Thank you drsuds for finding the typos, typos corrected.
If you look at r56c17 you can see this configuration:
This is called an unavoidable set. An unavoidable set is a set of clues that if removed gives multiple solutions. If you remove all of these digits you'll notice that there is one alternative way to solve it:
Now look back at the original puzzle, none of these cells had a clue in it from the beginning, therefore these four cells cannot contain an unavoidable set if the puzzle has an unique solution. A good introduction to uniqueness technique can be foundhere, then there is off course all the links to different uniqueness related techniques in Mike Barker's list.
Funny, apparently I got past the "hard" part almost immediately, but then spent almost 20 minutes finding the remaining singles... I haven't been solving very much in the last months and apparently I'm losing my singles-eye. Wasn't hard to get here:
I actually first saw:
If r9c7=9 => r5c9=9 => r7c9=1 => both row 5 and column 2 only have possible cells for digit 1 in box 4 and r5c2 is already filled => r9c6<>9
=> r9c9=9
Then I realized that I could do the straight elimination
If r7c9=1 => both row 5 and column 2 only have possible cells for digit 1 in box 4 and r5c2 is already filled => r7c9<>1 => r5c9=1
This move was so simple that there ought to be some kind of formal description of it already, but I'm too tired right now to figure it out.
[Edit: I think this particular pattern would be called a grouped Turbot fish.]
RW
drsuds wrote:
- Code: Select all
*-----------*
|1..|.32|578|
|58.|174|.92|
|7.2|589|461|
|---+---+---|
|8..|.2.|71.|
|67.|418|9..|
|921|75.|68.|
|---+---+---|
|..8|365|1.7|
|31.|297|8.6|
|..7|841|..9|
*-----------*
You said this results in a double solution.
Somehow I can't see it. Please clarify this point. Where is the double solution in the encoding above.
If you look at r56c17 you can see this configuration:
- Code: Select all
6..|...|9..
9..|...|6..
This is called an unavoidable set. An unavoidable set is a set of clues that if removed gives multiple solutions. If you remove all of these digits you'll notice that there is one alternative way to solve it:
- Code: Select all
9..|...|6..
6..|...|9..
Now look back at the original puzzle, none of these cells had a clue in it from the beginning, therefore these four cells cannot contain an unavoidable set if the puzzle has an unique solution. A good introduction to uniqueness technique can be foundhere, then there is off course all the links to different uniqueness related techniques in Mike Barker's list.
fermat wrote:RW, is this pretty easy without? (It is a bit tedious with. <sigh> )
- Code: Select all
. 7 . | 6 . . | 4 . .
8 . . | . 1 . | . 9 .
. . . | . . . | 1 . 7
---------------------
9 . . | . 2 . | . . .
. 5 . | 7 . 4 | . 2 .
. . . | . 3 . | . . 5
---------------------
3 . 7 | . . . | . . .
. 9 . | . 5 . | . . 3
. . 5 | . . 1 | . 4 .
Funny, apparently I got past the "hard" part almost immediately, but then spent almost 20 minutes finding the remaining singles... I haven't been solving very much in the last months and apparently I'm losing my singles-eye. Wasn't hard to get here:
- Code: Select all
*-----------*
|.7.|6..|4..|
|8..|.17|.9.|
|...|...|1.7|
|---+---+---|
|9..|.2.|..4|
|.5.|7.4|.2.|
|7..|.3.|..5|
|---+---+---|
|3.7|...|...|
|.9.|.5.|..3|
|..5|371|.4.|
*-----------*
I actually first saw:
If r9c7=9 => r5c9=9 => r7c9=1 => both row 5 and column 2 only have possible cells for digit 1 in box 4 and r5c2 is already filled => r9c6<>9
=> r9c9=9
Then I realized that I could do the straight elimination
If r7c9=1 => both row 5 and column 2 only have possible cells for digit 1 in box 4 and r5c2 is already filled => r7c9<>1 => r5c9=1
This move was so simple that there ought to be some kind of formal description of it already, but I'm too tired right now to figure it out.
[Edit: I think this particular pattern would be called a grouped Turbot fish.]
RW
- RW
- 2010 Supporter
- Posts: 1010
- Joined: 16 March 2006
by drsuds » Sat Sep 16, 2006 3:03 am
RW wrote:Now look back at the original puzzle, none of these cells had a clue in it from the beginning, therefore these four cells cannot contain an unavoidable set if the puzzle has an unique solution. A good introduction to uniqueness technique can be foundhere, then there is off course all the links to different uniqueness related techniques in Mike Barker's list.
RW,
Much thanks for the clarification.
Kind Regards,
drsuds
- drsuds
- Posts: 2
- Joined: 04 September 2006
by Finlip » Sun Oct 15, 2006 12:54 pm
Friday the thirteenth.
I hate solving sudokus by looking at the pencilmarks. But without pencilmarks, it may sometimes take way too long. I am now going to try one puzzle everyday. I am going to solve it without pencilmarks but I will keep a log of all the puzzles as I solve it.
For a start, I will take an extreme from Simple Sudoku.
This is the puzzle that came. When I paste it, the pencilmarks may be pasted, but I have the pencilmarks turned off in my program. So, I'll paste at the end of my solution. I will only type the log first. When I'm thinking about the puzzle, I will have the sudoku window activated. I think it will give me an accurate idea of how long I take to solve the puzzle.
1. r1c3=3 The time bar shows 00:11 and I spotted this quite easily. 3 is the only number that goes to block 1 and row 1 though there are alternatives in the column.
2. r6c2=3 I naturally looked for the next 3 and found it. Time: 00:22
I have also seen that the other two 3s are in two columns, and two rows of the remaining two boxes.
3. r1c8=4 This is the only place in the row and the block where 4 goes.
4. r6c9=4 It was quite difficult to see this. The time is: 10:14. After some looking arround, I saw that putting a 4 at r9c2 would force 4s at r8c5, r7c7, r6c9, and r5c3. A 4 in r78c3 would force a 4 at r5c2. So, I could see that the 4 in block 4 has to be in the fifth row. This proves that the last option to put a 4 in row 6 is r6c9.
5. r6c5=8 This was much faster. Time: 10:20. Looking at the sixth row, this was the only place on the row that 8 could go to.
6. r6c1=7
7. r6c3=9 These two were easy to see since there is already a 9 in column 1.
8. r3c3=7 It took me some time to see this easy thing. The time is 11:59. I drew the lines through the 7s in blocks 3, 4, and 7 to locate the 7 in block 1.
9. r7c8=1 Its either 1 or 9. If it is 9, there will be an error in trying to fill c9. 1 will go to r7c7, r5c8, and r3c9. The last two squares on the column can only contain 7 now as 8 and 9 are both already in the block. Time: 25:00
10. r9c9=1 A glance at column 9 gave me this one. 1, 5, 8, and 9 are missing from c9. 5,8, and 9 are already in r9. Time: 27:49
11. r5c8=1 Searching for the next 1, this was easy to see. I drew a line with the last 1. It showed me that block 3 has a 1 in column 7. Block 6 now had to have a 1 in column 8 which gave me the result.
12. r5c2=5 Seaching for 5s, I found this cell which is the only in the row that 5 can go to. I first looked at the 5 in column 7 which showed that block 6 would have a 5 in row 4. Since there is a five in row 6 as well, the 5 of block 4 had to be in row 5. The 5 in block 7 confirmed the cell as well.
13. r2c1=5
14. r3c4=5 (28:45) These two fives are pretty easy to see by drawing lines through the last few 5s.
15. r4c1=2
16. r5c3=4 (28:55) This was as a result of trying to fill the last two cells of block 4.
17. r9c2=4 (29:21) Looking for 4 in column 2
18. r7c1=1
19. r2c3=1 (29:48) Looking at block 7: 1, 2, and 8 remained. 2 and 8 are already on column 1. So, 1 had to be at the two places mentioned.
20. r1c1=6 (29:54) The last cell in column 1.
21. r1c4=7 (30:15) I looked at row 1. 1, 7, and 9 remained but 1 and 9 were already in column 1.
22. r3c7=1 (30:50) The lines drawn from the 1s in blocks 1 and 9 intersected leaving only this square.
23. r9c8=5 (31:15) I looked at column 9 and counted: 5, 8, and 9. I looked at block 9 and knew where to put 5.
24. r4c8=5 (31:24) This didn't take too long after drawing lines from 5 in blocks 3 and 6.
25. r8c8=7 (31:32) This is the last cell in the column.
26. r7c7=4. (32:22) I looked at row 7 and saw that there was nowhere else that 4 could go to.
27. r8c5=4 (32:28) This last 4 could easily be seen by drawing lines.
29. r7c8=2
30. r7c9=2 (32:58) Attempt to fill block 9 was successful, thanks to the 3 at r8c1.
31. r7c3=2
32. r8c3=8 (33:05) This was very easy. I tried to fill the last two cells in block 7.
33. r8c6=1 (33:11) The last cell in the row.
34. r1c5=1 (33:17) Drawing two lines from blocks 1 and 8 would have been enough to see the only position for 1 in the block.
35. r7c4=3
36. r7c5=8 (33:31) It was easy to fill row 7.
37. r2c4=8
38. r3c9=8
39. r2c7=9 (33:51) These were easy to see. I drew lines for 8 from blocks 5 and 8 to locate the first 8 which easily helped complete block 3.
40. r3c9=9 (34:00) This is the last cell in the column.
41. r4c7=8
42. r5c7=7 (34:09) It wasn't that hard to fill block 6.
43. r4c5=7 (34:12) This is the last digit in the row, but I saw it by drawing lines from other 7s.
44. r5c5=9 (34:21) The only possibility in column However, I saw it in this way. The nine in block 3 forced 9s in block 2 to be in column 6 which forced the 9 in block 5 to be in the middle with the help of the 9 in block 8.
45. r9c6=7 (34:40) Drawing lines frem columns 4 and 5.
46. r2c2=2
47. r3c2=9 (35:01) The last two in block 1 could be filled easily as a result of r2c7.
48. r3c6=2
49. r5c4=2
50. r9c5=2(35:09) The three 2s confused me
51. r9c4=6 (35:17) The last cell in the block.
52. r1c6=9
53: r2c5=6
54: r5c6=6 (35:41) This finished the puzzle in thirtyfive minutes though I'm quite slow.
Now let me try the same puzzle with pencilmarks on and see how long it takes. I am really feeling sleepy. I will post more logs and make my logs better in te past.
I hate solving sudokus by looking at the pencilmarks. But without pencilmarks, it may sometimes take way too long. I am now going to try one puzzle everyday. I am going to solve it without pencilmarks but I will keep a log of all the puzzles as I solve it.
For a start, I will take an extreme from Simple Sudoku.
This is the puzzle that came. When I paste it, the pencilmarks may be pasted, but I have the pencilmarks turned off in my program. So, I'll paste at the end of my solution. I will only type the log first. When I'm thinking about the puzzle, I will have the sudoku window activated. I think it will give me an accurate idea of how long I take to solve the puzzle.
1. r1c3=3 The time bar shows 00:11 and I spotted this quite easily. 3 is the only number that goes to block 1 and row 1 though there are alternatives in the column.
2. r6c2=3 I naturally looked for the next 3 and found it. Time: 00:22
I have also seen that the other two 3s are in two columns, and two rows of the remaining two boxes.
3. r1c8=4 This is the only place in the row and the block where 4 goes.
4. r6c9=4 It was quite difficult to see this. The time is: 10:14. After some looking arround, I saw that putting a 4 at r9c2 would force 4s at r8c5, r7c7, r6c9, and r5c3. A 4 in r78c3 would force a 4 at r5c2. So, I could see that the 4 in block 4 has to be in the fifth row. This proves that the last option to put a 4 in row 6 is r6c9.
5. r6c5=8 This was much faster. Time: 10:20. Looking at the sixth row, this was the only place on the row that 8 could go to.
6. r6c1=7
7. r6c3=9 These two were easy to see since there is already a 9 in column 1.
8. r3c3=7 It took me some time to see this easy thing. The time is 11:59. I drew the lines through the 7s in blocks 3, 4, and 7 to locate the 7 in block 1.
9. r7c8=1 Its either 1 or 9. If it is 9, there will be an error in trying to fill c9. 1 will go to r7c7, r5c8, and r3c9. The last two squares on the column can only contain 7 now as 8 and 9 are both already in the block. Time: 25:00
10. r9c9=1 A glance at column 9 gave me this one. 1, 5, 8, and 9 are missing from c9. 5,8, and 9 are already in r9. Time: 27:49
11. r5c8=1 Searching for the next 1, this was easy to see. I drew a line with the last 1. It showed me that block 3 has a 1 in column 7. Block 6 now had to have a 1 in column 8 which gave me the result.
12. r5c2=5 Seaching for 5s, I found this cell which is the only in the row that 5 can go to. I first looked at the 5 in column 7 which showed that block 6 would have a 5 in row 4. Since there is a five in row 6 as well, the 5 of block 4 had to be in row 5. The 5 in block 7 confirmed the cell as well.
13. r2c1=5
14. r3c4=5 (28:45) These two fives are pretty easy to see by drawing lines through the last few 5s.
15. r4c1=2
16. r5c3=4 (28:55) This was as a result of trying to fill the last two cells of block 4.
17. r9c2=4 (29:21) Looking for 4 in column 2
18. r7c1=1
19. r2c3=1 (29:48) Looking at block 7: 1, 2, and 8 remained. 2 and 8 are already on column 1. So, 1 had to be at the two places mentioned.
20. r1c1=6 (29:54) The last cell in column 1.
21. r1c4=7 (30:15) I looked at row 1. 1, 7, and 9 remained but 1 and 9 were already in column 1.
22. r3c7=1 (30:50) The lines drawn from the 1s in blocks 1 and 9 intersected leaving only this square.
23. r9c8=5 (31:15) I looked at column 9 and counted: 5, 8, and 9. I looked at block 9 and knew where to put 5.
24. r4c8=5 (31:24) This didn't take too long after drawing lines from 5 in blocks 3 and 6.
25. r8c8=7 (31:32) This is the last cell in the column.
26. r7c7=4. (32:22) I looked at row 7 and saw that there was nowhere else that 4 could go to.
27. r8c5=4 (32:28) This last 4 could easily be seen by drawing lines.
29. r7c8=2
30. r7c9=2 (32:58) Attempt to fill block 9 was successful, thanks to the 3 at r8c1.
31. r7c3=2
32. r8c3=8 (33:05) This was very easy. I tried to fill the last two cells in block 7.
33. r8c6=1 (33:11) The last cell in the row.
34. r1c5=1 (33:17) Drawing two lines from blocks 1 and 8 would have been enough to see the only position for 1 in the block.
35. r7c4=3
36. r7c5=8 (33:31) It was easy to fill row 7.
37. r2c4=8
38. r3c9=8
39. r2c7=9 (33:51) These were easy to see. I drew lines for 8 from blocks 5 and 8 to locate the first 8 which easily helped complete block 3.
40. r3c9=9 (34:00) This is the last cell in the column.
41. r4c7=8
42. r5c7=7 (34:09) It wasn't that hard to fill block 6.
43. r4c5=7 (34:12) This is the last digit in the row, but I saw it by drawing lines from other 7s.
44. r5c5=9 (34:21) The only possibility in column However, I saw it in this way. The nine in block 3 forced 9s in block 2 to be in column 6 which forced the 9 in block 5 to be in the middle with the help of the 9 in block 8.
45. r9c6=7 (34:40) Drawing lines frem columns 4 and 5.
46. r2c2=2
47. r3c2=9 (35:01) The last two in block 1 could be filled easily as a result of r2c7.
48. r3c6=2
49. r5c4=2
50. r9c5=2(35:09) The three 2s confused me
51. r9c4=6 (35:17) The last cell in the block.
52. r1c6=9
53: r2c5=6
54: r5c6=6 (35:41) This finished the puzzle in thirtyfive minutes though I'm quite slow.
- Code: Select all
*-----------*
|.8.|...|5.2|
|...|..4|.37|
|4..|.3.|.6.|
|---+---+---|
|.16|4.3|...|
|8..|...|..3|
|...|1.5|62.|
|---+---+---|
|.7.|.5.|..6|
|36.|9..|...|
|9.5|...|.8.|
*-----------*
Now let me try the same puzzle with pencilmarks on and see how long it takes. I am really feeling sleepy. I will post more logs and make my logs better in te past.
- Finlip
- Posts: 49
- Joined: 15 July 2005
by Finlip » Sun Oct 15, 2006 12:59 pm
Continuing with the rule I started yesterday.
Even though it took me quite some time, I was able to complete yesterday's puzzle. I am afraid today's may be more difficult but let me see.
1. r7c7=6 That is easy to see by drawing lines.
(00:14)
2. r1c6=6 Looking for more 6s, I found this.
(00:23)
3. r3c3=5 Couldn't find the other 6s, but then saw this 5 by drawing lines again.
(00:32)
4. r6c4=5
5. r7c6=5
6. r5c8=5 I spotted these three together. One 5 followed another and I finished all 5s.
(00:49)
7. r8c5=9 It wasn't easy to spot this though there may be an easier way. I'll explain it. Look at column 4. Can you see 3 and 7. This means that 3 and 7 go to row 2 in block 2. Of the remaining numbers in column 4, we know which go to which block. Now, since the 9 of column 4 goes to block 2, r8c5 is the position of 9 in block 8.
(02:51)
8. r2c5=7
9. r2c6=3
I knew that 3 and 7 were the remaining digits in these cells. When I checked out row 5, I could see that r2c5 was the only spot where 7 could go to. Thus, this was filled.
(03:08)
10. r4c6=2
Before I figured out that 2 went to this cell, I knew that I could easily find what goes to this cell. Of the remaining cells of block 5, the other two were the last cells in the almost filled column 5. I just had to look for a while before I saw that 2 was missing in block 5 an couldn't be in column 5 anymore.
(03:28)
11. r4c5=1
12. r6c5=3
I looked at other almost filled blocks before looking at block 5. Here, it was easy to notice that 1 couldn't go to row 6 and so, I filled my first block.
(??:??) Forgot to note the time!
13. r4c2=4
14. r4c3=8
I looked at the fourth row and started counting. 3 was absent and I saw that it had to be in block 6. 4 was also absent but could go to any of three possible cells in the row. I counted 8 but didn't see that it could only go to block 4. Then, when I saw that 9 could also only go to block 6, I checked block 6 to see if I could locate the 3 and 9 that had to go there. When I was not successful at that, I came back to block 4 and found the 4 and 8 which had to be there.
(04:30)
15. r6c3=6
I was looking at column 3 and finally noticed that there was only one cell in the column where I could place a 6.
(05:22)
16. r32c2=6
The last 6 easily lead to this.
(05:39)
17. r3c4=9
18. r3c1=8
I looked at row 3 and saw that 3,7,8, and 9 were missing. 3, and 7 couldn't go to the first two empty cells. So, I had to fill these cells with 8, and 9. That's what I've done. I couldn't have done this the other way round because of the 9 already present in column 1.
(06:01)
19. r9c2=8
The last 8 lead to the discovery of this one.
(06:09)
20. r1c8=8
Though this is the only cell in the block where this number can go, I had to check the remaining dgits in block 3 to find this 8.
(06:31)
21. r8c9=8
22. r2c4=8
23. r1c4=1
I went looking for the last 8s. Then, I saw that block 2 was almost full and filled it with 1.
(06:46)
24. r4c8=9
This again took some effort to see. There were 9s in rows 4 and 6 already. So, I saw that the 9 of block 9 would go to column 9. After I noticed that the 9 of block would have to go to column 7, I could easily fill 9 in column 8 of row 4 where I remembered that 9 and 3 were the remaining digits.
(09:29)
25. r4c7=3
Oh, I did this in 2 seconds.
(09:31)
26. r2c1=4
This is easy to prove though I couldn't see it so easily.
(11:39)
27. r1c1=2
28. r2c7=2
I looked at the remaining digits in column 1. 1 and 7 were already in row 1. So, r1c1 had to be 2. It took me almost no time to see the next two in row 2.
(12:01)
29. r1c2=3
30. r1c7=9
When I saw that there were only two unfilled cells in row 2 and they were both in block 1, I knew I could fill row 1 by interaction of row 1 and block 1. That's exactly what I did here.
(12:31)
Actually, I've now filled up the rest of the puzzle, I just need to fill the boxes. Even though, I was first able to fill all the remaining 24 cells starting with r2c2=1, I still decided to double check it. Putting a 9 there, would make me put 1 at r2c3 and then, I would have to put 2 and 9 respectively at r89c3. That would force, r89c4 to be 4 and 2 respectively. Looking for places to put 2 and 4 in block 9, I could see that 4 would have to go to r9c8 and 2 would have to go to r7c8 since 9 would go to r7c9. Looking at the remaining cells in block 9, I could count 1, 2, and 7. Since 7 was the only of these that could go to r8c7, I could fill column 6 in block 8. That would first force 1 in the eigth row of that column. But sadly, that would leave no space for 1 in block 9. Thus I can prove that r2c2 is actually 1.
(20:13)
I was indeed able to solve the rest of the puzzle easily after this. The time taken was 22:12.
Here is a list of the remaining solutions.
r2c2=1
r2c3=9
r3c8=7
r3c9=3
r5c1=1
r5c2=2
r5c9=7
r6c2=7
r6c7=4
r6c8=2
r7c1=7
r7c2=9
r7c8=1
r7c9=2
r8c3=2
r8c4=4
r8c6=1
r8c7=7
r8c8=3
r9c3=1
r9c4=2
r9c6=7
r9c8=4
r9c9=9
Even though it took me quite some time, I was able to complete yesterday's puzzle. I am afraid today's may be more difficult but let me see.
- Code: Select all
*-----------*
|..7|.5.|..4|
|...|...|.65|
|...|.24|1..|
|---+---+---|
|5..|7..|..6|
|..3|649|8..|
|9..|..8|..1|
|---+---+---|
|..4|38.|...|
|65.|...|...|
|3..|.6.|5..|
*-----------*
1. r7c7=6 That is easy to see by drawing lines.
(00:14)
2. r1c6=6 Looking for more 6s, I found this.
(00:23)
3. r3c3=5 Couldn't find the other 6s, but then saw this 5 by drawing lines again.
(00:32)
4. r6c4=5
5. r7c6=5
6. r5c8=5 I spotted these three together. One 5 followed another and I finished all 5s.
(00:49)
7. r8c5=9 It wasn't easy to spot this though there may be an easier way. I'll explain it. Look at column 4. Can you see 3 and 7. This means that 3 and 7 go to row 2 in block 2. Of the remaining numbers in column 4, we know which go to which block. Now, since the 9 of column 4 goes to block 2, r8c5 is the position of 9 in block 8.
(02:51)
8. r2c5=7
9. r2c6=3
I knew that 3 and 7 were the remaining digits in these cells. When I checked out row 5, I could see that r2c5 was the only spot where 7 could go to. Thus, this was filled.
(03:08)
10. r4c6=2
Before I figured out that 2 went to this cell, I knew that I could easily find what goes to this cell. Of the remaining cells of block 5, the other two were the last cells in the almost filled column 5. I just had to look for a while before I saw that 2 was missing in block 5 an couldn't be in column 5 anymore.
(03:28)
11. r4c5=1
12. r6c5=3
I looked at other almost filled blocks before looking at block 5. Here, it was easy to notice that 1 couldn't go to row 6 and so, I filled my first block.
(??:??) Forgot to note the time!
13. r4c2=4
14. r4c3=8
I looked at the fourth row and started counting. 3 was absent and I saw that it had to be in block 6. 4 was also absent but could go to any of three possible cells in the row. I counted 8 but didn't see that it could only go to block 4. Then, when I saw that 9 could also only go to block 6, I checked block 6 to see if I could locate the 3 and 9 that had to go there. When I was not successful at that, I came back to block 4 and found the 4 and 8 which had to be there.
(04:30)
15. r6c3=6
I was looking at column 3 and finally noticed that there was only one cell in the column where I could place a 6.
(05:22)
16. r32c2=6
The last 6 easily lead to this.
(05:39)
17. r3c4=9
18. r3c1=8
I looked at row 3 and saw that 3,7,8, and 9 were missing. 3, and 7 couldn't go to the first two empty cells. So, I had to fill these cells with 8, and 9. That's what I've done. I couldn't have done this the other way round because of the 9 already present in column 1.
(06:01)
19. r9c2=8
The last 8 lead to the discovery of this one.
(06:09)
20. r1c8=8
Though this is the only cell in the block where this number can go, I had to check the remaining dgits in block 3 to find this 8.
(06:31)
21. r8c9=8
22. r2c4=8
23. r1c4=1
I went looking for the last 8s. Then, I saw that block 2 was almost full and filled it with 1.
(06:46)
24. r4c8=9
This again took some effort to see. There were 9s in rows 4 and 6 already. So, I saw that the 9 of block 9 would go to column 9. After I noticed that the 9 of block would have to go to column 7, I could easily fill 9 in column 8 of row 4 where I remembered that 9 and 3 were the remaining digits.
(09:29)
25. r4c7=3
Oh, I did this in 2 seconds.
(09:31)
26. r2c1=4
This is easy to prove though I couldn't see it so easily.
(11:39)
27. r1c1=2
28. r2c7=2
I looked at the remaining digits in column 1. 1 and 7 were already in row 1. So, r1c1 had to be 2. It took me almost no time to see the next two in row 2.
(12:01)
29. r1c2=3
30. r1c7=9
When I saw that there were only two unfilled cells in row 2 and they were both in block 1, I knew I could fill row 1 by interaction of row 1 and block 1. That's exactly what I did here.
(12:31)
Actually, I've now filled up the rest of the puzzle, I just need to fill the boxes. Even though, I was first able to fill all the remaining 24 cells starting with r2c2=1, I still decided to double check it. Putting a 9 there, would make me put 1 at r2c3 and then, I would have to put 2 and 9 respectively at r89c3. That would force, r89c4 to be 4 and 2 respectively. Looking for places to put 2 and 4 in block 9, I could see that 4 would have to go to r9c8 and 2 would have to go to r7c8 since 9 would go to r7c9. Looking at the remaining cells in block 9, I could count 1, 2, and 7. Since 7 was the only of these that could go to r8c7, I could fill column 6 in block 8. That would first force 1 in the eigth row of that column. But sadly, that would leave no space for 1 in block 9. Thus I can prove that r2c2 is actually 1.
(20:13)
I was indeed able to solve the rest of the puzzle easily after this. The time taken was 22:12.
Here is a list of the remaining solutions.
r2c2=1
r2c3=9
r3c8=7
r3c9=3
r5c1=1
r5c2=2
r5c9=7
r6c2=7
r6c7=4
r6c8=2
r7c1=7
r7c2=9
r7c8=1
r7c9=2
r8c3=2
r8c4=4
r8c6=1
r8c7=7
r8c8=3
r9c3=1
r9c4=2
r9c6=7
r9c8=4
r9c9=9
- Finlip
- Posts: 49
- Joined: 15 July 2005
Save the Blankness
by Jumper » Sun Oct 15, 2006 5:54 pm
The longer I work noteless or WPM (without pencil marks) the more valuable I find it to not even write in the quickly-solved numerals. Often enough the hardest part of the puzzle is there in front of me before I write in anything at all. Study the puzzle for a few minutes before filling in. I find this helps. After all, the puzzle will only be "clean" once: at the beginning.
- Jumper
- Posts: 8
- Joined: 23 September 2005
by RW » Sun Oct 15, 2006 9:22 pm
Interesting to see your solution Finlip. I had a shot on the first puzzle myself in Simple Sudoku without pencilmarks to compare my solution to yours. The first thing I notice in your solution is that you don't seem to be looking for naked singles very efficiently. You spent 10 minutes looking for a forcing chain when there was a naked single in the most obvious place, a row with 5 cells already filled:
When I got to this stage and couldn't spot any more hidden singles I immediately started checking the rows and columns as I explained in the original post in this thread. Starting from a row with many filled cells, I could see that row 6 missed only digits 4789, r6c1 can see 489 => r6c1=7. Looking for naked singles and subsets like this, one row/column at a time, is very efficient and I recommend you to try it. Good puzzles to practise on can be found among the high-steppers in the inferior thread if you restrict yourself to only using singles. Of course you can practise on any other puzzle as well, there's naked singles everywhere.
After the naked single and a hidden single it revealed, I found an interesting simple forcing chain:
I had noticed that in box 9 digit '9' has to go in row 7 and in box 6 digit '1' has to go in row 5. Also in box 3, at least one of the two digits have to go in column 7. This is so far quite useless, but when I did my column check on column 8 I saw that there's only two possible values for r7c8, '1' or '9'. This is when I found this interesting move:
if r5c7=1 => r23c7=9 => r7c8=9
if r5c8=1 => r7c8=9
After this I saw some interesting links on digits '4'. In column 2 it can only go in row 5 or 9, in box 6 only in r5c7 or r6c9. After a little thinking I could see the implication r9c9<>4 (if r9c2<>4 => r5c2=4 => r6c9=4). With this new knowledge fresh in mind I could easily see the naked single, r9c9=1. After this it was quite straight forward. After solving the whole puzzle (finished in 12:29) I looked at SS's solution path, and it turned out that I had used exactly the same move as it used to crack the puzzle, my elimination using digit '4' can also be described as multiple colors. I mentioned in my original post of the thread that multiple colors can be seen as a forcing chain also, you had used that forcing chain as step 4 of your solution.
Anyway, this was definitely not a very easy puzzle to solve without pms, and I think your solution using the forcing chain was very nice. Looking at your log it only confirms what I've noticed in my solving also: the hardest thing is not to find 'extreme eliminations'. The solution actually only required the elimination you did in step 4 plus singles, so out of the 35 minutes you used to solve the puzzle, 25 minutes was required to find the singles. I often notice the same myself, basic moves are not always the fastest to find. So for anybody who wishes to improve their solving, I suggest that you mostly solve puzzles that you know can be solved with basic techniques (singles&subsets) and really make an effort to find these simple solutions to all puzzles, don't start looking for forcing chains immediately when you come to a halt. You can never get too good at finding these most basic patterns.
I might have a look at the other puzzle tomorrow, need some sleep now.
RW
- Code: Select all
*-----------*
|.83|...|542|
|...|..4|.37|
|4..|.3.|.6.|
|---+---+---|
|.16|4.3|...|
|8..|...|..3|
|.3.|1.5|62.|
|---+---+---|
|.7.|.5.|..6|
|36.|9..|...|
|9.5|...|.8.|
*-----------*
When I got to this stage and couldn't spot any more hidden singles I immediately started checking the rows and columns as I explained in the original post in this thread. Starting from a row with many filled cells, I could see that row 6 missed only digits 4789, r6c1 can see 489 => r6c1=7. Looking for naked singles and subsets like this, one row/column at a time, is very efficient and I recommend you to try it. Good puzzles to practise on can be found among the high-steppers in the inferior thread if you restrict yourself to only using singles. Of course you can practise on any other puzzle as well, there's naked singles everywhere.
After the naked single and a hidden single it revealed, I found an interesting simple forcing chain:
- Code: Select all
*-----------*
|.83|...|542|
|...|..4|.37|
|4.7|.3.|.6.|
|---+---+---|
|.16|4.3|...|
|8..|...|..3|
|73.|1.5|62.|
|---+---+---|
|.7.|.5.|..6|
|36.|9..|...|
|9.5|...|.8.|
*-----------*
I had noticed that in box 9 digit '9' has to go in row 7 and in box 6 digit '1' has to go in row 5. Also in box 3, at least one of the two digits have to go in column 7. This is so far quite useless, but when I did my column check on column 8 I saw that there's only two possible values for r7c8, '1' or '9'. This is when I found this interesting move:
if r5c7=1 => r23c7=9 => r7c8=9
if r5c8=1 => r7c8=9
After this I saw some interesting links on digits '4'. In column 2 it can only go in row 5 or 9, in box 6 only in r5c7 or r6c9. After a little thinking I could see the implication r9c9<>4 (if r9c2<>4 => r5c2=4 => r6c9=4). With this new knowledge fresh in mind I could easily see the naked single, r9c9=1. After this it was quite straight forward. After solving the whole puzzle (finished in 12:29) I looked at SS's solution path, and it turned out that I had used exactly the same move as it used to crack the puzzle, my elimination using digit '4' can also be described as multiple colors. I mentioned in my original post of the thread that multiple colors can be seen as a forcing chain also, you had used that forcing chain as step 4 of your solution.
Anyway, this was definitely not a very easy puzzle to solve without pms, and I think your solution using the forcing chain was very nice. Looking at your log it only confirms what I've noticed in my solving also: the hardest thing is not to find 'extreme eliminations'. The solution actually only required the elimination you did in step 4 plus singles, so out of the 35 minutes you used to solve the puzzle, 25 minutes was required to find the singles. I often notice the same myself, basic moves are not always the fastest to find. So for anybody who wishes to improve their solving, I suggest that you mostly solve puzzles that you know can be solved with basic techniques (singles&subsets) and really make an effort to find these simple solutions to all puzzles, don't start looking for forcing chains immediately when you come to a halt. You can never get too good at finding these most basic patterns.
I might have a look at the other puzzle tomorrow, need some sleep now.
RW
- RW
- 2010 Supporter
- Posts: 1010
- Joined: 16 March 2006
by Finlip » Mon Oct 16, 2006 12:22 pm
Many thanks for you suggestions. RW. I'll have a deatailed study when I reach home. I've already copied this thread into my pendrive as I do my internet surfing at cyber cafes.
I was unsuccessful at the solution yesterday. I will now copy and paste from the text document I prepared.
Without Pencilmarks 3
In this series I try to solve one sudoku puzzle a day without using any pencilmarks. I do not necessarily succeed in solving. Here are the conventions that I will use from today.
R: Only possible cell in the row for the candidate.
C: Only possible cell in the column for the candidate.
B: Only possible cell in the block for the candidate.
N: Naked Single.
T: Other Techniques (An explanation will also be included in this case)
The time after each move shows the time elapsed since start of puzzle.
Let me start.
r5c2=4 R,B 00:12
r9c1=4 R,B 00:12
r7c1=3 R,C,B 00:42
r2c1=1 R,C 01:02
r3c9=1 R,C,B 01:05
r4c5=1 R,B 05:46
r8c5=4 N 08:01
r2c8=4 R 08:19
r2c5=2 R 08:35
r5c6=2 T 18:18
In order to put the digit 2 in column 9, we have to put it in block 6. So, in order to put the digit 2 in row 5, we have to put it in this cell.
I gave up today after time crossed 44 minutes.
I failed but my failure is now even useless since I forgot to copy the puzzle itself.
So instead, I opened a Simple Sudoku Easy and finished it in 07:55
Here is the puzzle:
I will now undo this puzzle and show you the steps which lead to the solution.
r2c3
r8c9
r7c2
r2c1
r5c9
r6c9
r3c7
r3c8
r8c7
r3c5
r3c6
r3c2
r2c8
r1c7
r4c5
r4c1
r4c4
r4c8
r5c8
r5c7
r9c7
r5c4
r6c5
r6c6
r6c2
r9c3
r9c4
r1c3
r8c1
r8c2
r5c2
r5c3
r5c1
r7c3
r9c1
r7c4
r8c4
r2c4
r7c5
r8c5
r8c6
r9c5
r5c6
r5c5
r1c6
r2c6
r1c9
r2c9
r1c2
r2c2
r1c5
r2c5
r7c8
r8c8
r9c8
I was unsuccessful at the solution yesterday. I will now copy and paste from the text document I prepared.
Without Pencilmarks 3
In this series I try to solve one sudoku puzzle a day without using any pencilmarks. I do not necessarily succeed in solving. Here are the conventions that I will use from today.
R: Only possible cell in the row for the candidate.
C: Only possible cell in the column for the candidate.
B: Only possible cell in the block for the candidate.
N: Naked Single.
T: Other Techniques (An explanation will also be included in this case)
The time after each move shows the time elapsed since start of puzzle.
Let me start.
r5c2=4 R,B 00:12
r9c1=4 R,B 00:12
r7c1=3 R,C,B 00:42
r2c1=1 R,C 01:02
r3c9=1 R,C,B 01:05
r4c5=1 R,B 05:46
r8c5=4 N 08:01
r2c8=4 R 08:19
r2c5=2 R 08:35
r5c6=2 T 18:18
In order to put the digit 2 in column 9, we have to put it in block 6. So, in order to put the digit 2 in row 5, we have to put it in this cell.
I gave up today after time crossed 44 minutes.
I failed but my failure is now even useless since I forgot to copy the puzzle itself.
So instead, I opened a Simple Sudoku Easy and finished it in 07:55
Here is the puzzle:
- Code: Select all
*-----------*
|5..|1..|.7.|
|...|...|8..|
|9.8|4..|..2|
|---+---+---|
|.37|..2|6.5|
|...|...|...|
|1.2|6..|73.|
|---+---+---|
|4..|..1|5.9|
|..5|...|...|
|.1.|..7|..3|
*-----------*
*-----------*
|543|128|976|
|721|396|854|
|968|475|312|
|---+---+---|
|837|912|645|
|654|783|291|
|192|654|738|
|---+---+---|
|476|831|529|
|385|249|167|
|219|567|483|
*-----------*
I will now undo this puzzle and show you the steps which lead to the solution.
r2c3
r8c9
r7c2
r2c1
r5c9
r6c9
r3c7
r3c8
r8c7
r3c5
r3c6
r3c2
r2c8
r1c7
r4c5
r4c1
r4c4
r4c8
r5c8
r5c7
r9c7
r5c4
r6c5
r6c6
r6c2
r9c3
r9c4
r1c3
r8c1
r8c2
r5c2
r5c3
r5c1
r7c3
r9c1
r7c4
r8c4
r2c4
r7c5
r8c5
r8c6
r9c5
r5c6
r5c5
r1c6
r2c6
r1c9
r2c9
r1c2
r2c2
r1c5
r2c5
r7c8
r8c8
r9c8
- Finlip
- Posts: 49
- Joined: 15 July 2005
by Finlip » Tue Oct 17, 2006 12:46 pm
After yesterday's failure, I decided to go at this differently today. I have already finished solving todays puzzle, another simple sudoku extreme, in record time. I don't think I've ever solved a simple sudoku extreme in less time when solving without pencilmarks. This is the time in which I would be happy to finish even easy puzzles. It took me 06:34 to finish this one.
Now, I am going to undo and then explain my steps.
Oh my goodness me. I should have pressed Ctrl + Z but instead pressed Ctrn + N and did it twice. The first time it opened a new puzzle straight away. The second time, it asked me if I really wanted to start a new one. But I just couldn't paste the puzzle that I was able to finish in record time. I had used quite a few interesting moves there.
I'll now solve a new one from start. I am sure it will take longer.
I think, I should log and solve simultaneously so that I can see which steps took how long.
Here is the random simple sudoku extreme.
Now, let me start solving.
*5 of C1 and R7 both have to be in B7
The time is now 33:55 and I feel really sleepy.
Now, I am going to undo and then explain my steps.
Oh my goodness me. I should have pressed Ctrl + Z but instead pressed Ctrn + N and did it twice. The first time it opened a new puzzle straight away. The second time, it asked me if I really wanted to start a new one. But I just couldn't paste the puzzle that I was able to finish in record time. I had used quite a few interesting moves there.
I'll now solve a new one from start. I am sure it will take longer.
I think, I should log and solve simultaneously so that I can see which steps took how long.
- Code: Select all
*-----------*
|...|.9.|.5.|
|6..|4.3|.8.|
|1.8|..5|...|
|---+---+---|
|..4|8..|.15|
|..1|...|7..|
|76.|..4|9..|
|---+---+---|
|...|9..|4.6|
|.8.|6.7|..1|
|.7.|.4.|...|
*-----------*
Here is the random simple sudoku extreme.
Now, let me start solving.
- Code: Select all
1. r7c2=1 C,B 00:11
2. r7c8=7 R,B 00:21
3. r1c6=8 R,B 00:55
4. r7c5=8 R,C 01:00
5. r7c6=2 N 01:18
6. r9c3=6 R,C,B 01:33
7. r8c1=4 R,B 01:47
8. r9c6=1 C 02:46
9. r9c7=8 C 03:28
10.r5c1=8 C,B 03:53
11.r6c9=8 R,C,B 03:58
12.r4c5=7 R,B 04:52
13.r7c1=5 T1 07:03
14.r7c3=3 R 07:05
15.r8c7=5 C,B 07:25
16.r8c5=3 N 07:33
17.r9c4=5 R,B 07:39
18.r3c5=6 B 12:57
19.r1c7=6 R,B 13:06
20.r5c8=6 C,B 13:14
21.r4c6=6 R,C,B 13:19
22.r5c6=9 C,B 13:25
23.r2c7=1 C,B 13:59
24.r1c4=1 R,B 14:21
25.r3c4=7 C,B 14:30
26.r2c5=2 B 14:32
27.r6c5=1 R,C,B 14:37
28.R5C5=5 C,B 14:42
29.r6c3=5 R,B 14:48
30.r5c9=4 R,B 15:26
31.r1c2=4 R 18:22
32.r3c8=4 R,C,B 18:25
33.r2c2=5 R,C,B 20:10
*5 of C1 and R7 both have to be in B7
The time is now 33:55 and I feel really sleepy.
- Finlip
- Posts: 49
- Joined: 15 July 2005
by RW » Tue Oct 17, 2006 2:38 pm
Finlip wrote:
- Code: Select all
*-----------*
|...|.9.|.5.|
|6..|4.3|.8.|
|1.8|..5|...|
|---+---+---|
|..4|8..|.15|
|..1|...|7..|
|76.|..4|9..|
|---+---+---|
|...|9..|4.6|
|.8.|6.7|..1|
|.7.|.4.|...|
*-----------*
Here is the random simple sudoku extreme.
Well,well, just as we talked about how rare the reverse BUG is in the reverse BUG-lite thread, you post this nice example! After solving the singles:
- Code: Select all
*-----------------------------*
| . 4 . | 1 9 8 | 6 5 . |
| 6 5 . | 4 *2 *3 | 1 8 . |
| 1 . 8 | 7 6 5 | . 4 . |
|---------+---------+---------|
| . . 4 | 8 7 6 | . 1 5 |
| 8 . 1 | . 5 9 | 7 6 4 |
| 7 6 5 | . 1 4 | 9 . 8 |
|---------+---------+---------|
| 5 1 *3 | 9 8 *2 | 4 7 6 |
| 4 8 . | 6 *3 7 | 5 . 1 |
| . 7 6 | 5 4 1 | 8 . . |
*-----------------------------*
Digit 2 in r8c3 would complete an unavoidable set that includes all solved instances of digits 2 and 3 => r8c3<>2. That's the second randomly generated Reverse BUG I've managed to catch!
RW
- RW
- 2010 Supporter
- Posts: 1010
- Joined: 16 March 2006
by Finlip » Thu Oct 19, 2006 6:37 am
- Code: Select all
1. r2c3=1 C,B 00:06
2. r3c6=1 R,B 00:09
3. r7c4=1 R,C,B 00:14
4. r8c7=1 R,C,B 00:17
5. r4c5=4 C,B 00:36
6. r7c5=8 R,B 01:00
7. r7c9=2 N 01:45
8. r2c5=6 R,B 05:58
9. r8c9=5 C,B 12:12
10.r6c9=9 N 12:16
11.r2c9=4 C 12:18
12.r3c8=9 C,B 12:22
13.r2c1=9 C 12:32
14.r8c1=4 C 12:38
15.r1c2=4 R,C,B 15:22
16.r4c8=8 T1 22:56
17.r6c6=8 R,B 23:06
18.r6c7=2 N 23:28
19.r1c8=2 C,B 23:35
20.r4c3=6 R 24:29
21.r5c7=6 T2 25:34
22.r5c8=4 R,B 25:41
23.r9c7=4 R,C,B 25:44
24.r4c1=2 T3 28:25
25.r3c1=3 C 28:28
26.r1c8=3 R,C,B 28:36
27.r2c8=8 C,B 28:41
28.r3c8=7 C,B 28:44
29.r1c4=8 R,C,B 29:03
30.r4c4=3 R 29:26
31.r6c2=3 R,B 29:29
32.r6c5=5 R 29:39
33.r3c5=2 C 29:58
34.r8c5=3 C,B 30:01
35.r1c6=5 B 30:20
36.r2c4=7 B 30:24
37.r5c6=7 R,C,B 30:27
38.r5c4=2 R,B 30:30
39.r9c4=5 R,C,B 30:39
40.r1c3=7 R,B 30:55
41.r2c2=2 R,B 30:58
42.r2c2=5 R,B 31:00
43.r5c3=5 R,C,B 31:04
44.r5c2=9 R,B 31:07
45.r9c3=9 R,C,B 31:10
46.r8c3=2 C,B 31:15
47.r9c6=2 R,C,B 31:18
48.r9c8=3 R 31:23
49.r9c2=6 R 31:26
50.r7c3=3 R,C,B 31:30
51.r7c2=7 C,B 31:32
52.r8c8=7 R,C,B 31:36
53.r7c8=6 R,C,B 31:39
54.r8c6=6 R,C,B 31:41
T1: Whether I put 8 in c4 or c7 of r2, I get this 8 (by working through b5 in the first case)
T2: 4 can only go to r59c78. 6 can go to r59c78 plus r78c8. To avoid the deadly pattern, 6 has to go to r78c8. This leads to the next 6.
T3: If 2 goes to any other cell in block 4 (r5c2 or r5c3), it leads to r4c4, r3c5, and r1c2 being 2. In either case, column 1 doesn't have a 2 which proves that 2 can't go to these other cells.
After yesterday's mishap, I have remembered to copy today's puzzle before closing it. It wasn't that I forgot to copy yesterday's puzzle, I only typed Ctrl+N instead of Ctrl+Z. That was a result of my usual practice of solving many puzzles.
Today's puzzle took five times more time than did yesterday's puzzle. But my times are not continuous when I keep a log simultaneously. However, it helps me see which move took relatively longer to find. However, when I shift to typing in the middle of solving it may shift my concentration on the puzzle.
I am glad to be succesfully able to post a puzzle after a two day gap.
- Finlip
- Posts: 49
- Joined: 15 July 2005
How do you solve this one without PMs?
by DreamCH » Sun Oct 22, 2006 10:55 am
*-------------------*
|1 . . |. 6 . |. . 5|
|. 8 . |9 . 3 |. .1 .|
|. . 7 | . . .|8 . .|
|------|------|-----|
|. 7 . |. 4 . |.9. .|
|5 . . |7 . 1 |. . 6|
|. 6 . |. 3 . |. 7 .|
|------|------|-----|
|. . 3 |. . . |9 . .|
|. 1 . |3 . 9 |. 4 .|
|9 . . |. 2 . |. . 1|
*-------------------*
I wonder how long you need to go beyond the first 11 (or 12) numbers...this one is beyond my skills for now
. Any hidden things you see?
|1 . . |. 6 . |. . 5|
|. 8 . |9 . 3 |. .1 .|
|. . 7 | . . .|8 . .|
|------|------|-----|
|. 7 . |. 4 . |.9. .|
|5 . . |7 . 1 |. . 6|
|. 6 . |. 3 . |. 7 .|
|------|------|-----|
|. . 3 |. . . |9 . .|
|. 1 . |3 . 9 |. 4 .|
|9 . . |. 2 . |. . 1|
*-------------------*
I wonder how long you need to go beyond the first 11 (or 12) numbers...this one is beyond my skills for now
. Any hidden things you see?- DreamCH
- Posts: 10
- Joined: 10 June 2006
