The New Sudoku Players' Forum

[Carcul]

Now is my time to offer some interesting puzzles (for the manual solvers):

Puzzle #1

Code: Select all

 . 2 . | . . . | 6 . .
 . . . | 1 . . | . . .
 7 . . | . 3 . | . 5 .   
-------+-------+-------
 . . 8 | . 4 . | 9 . 3
 3 . . | . . . | . . 8
 . . . | . 2 . | 4 . 6
-------+-------+-------
 6 3 . | 5 . . | . . .
 . . 5 | . . 9 | . . 2
 . 1 . | . 8 4 | . . .


Puzzle #2

Code: Select all

 3 1 . | 6 . . | . . .
 . . 2 | . . . | . . .
 . 5 . | . 9 . | 8 6 .
-------+-------+-------
 . . . | . . 5 | . . .
 . 6 . | . 1 . | . 9 .
 . . . | 4 . . | . . .
-------+-------+-------
 . 9 4 | . 6 . | . 7 .
 . . . | . . . | 3 . .
 . . . | . . 7 | . 1 2


Puzzle #3

Code: Select all

 . . . | . . . | 5 8 .
 . 3 7 | . 8 5 | . . .
 . . . | 3 . . | 9 . .   
-------+-------+-------
 . . . | . . . | . 7 4
 2 . 6 | 9 . . | . . .
 3 . . | . 6 . | . . .
-------+-------+-------
 . 9 . | . . . | 3 . 1
 5 . 4 | 2 . 7 | . . .
 . . . | . . . | . . .


This last puzzle is the easiest of the three, but the challenge that I propose here is to solve this grid with only one step after the basic steps.
Btw, this last grid is from the collection of Henk here, where some very good puzzles for practise advanced techniques can be found.

Have fun.

Carcul

[gsf]

Now is my time to offer some interesting puzzles (for the manual solvers):

hey carcul, attribution for these puzzles?

[Wolfgang]

Hi Carcul,

tried the first one. Forgive me the notation, but i hope, its correct:
A.r9c1=9:
If r9c1=2
1. r4c1=1,r4c8=2
2. r6c3=1,r6c7=7,r5c8=2
[edit: Thanks to Carcul, this is obviously not correct delete the rest]

[Carcul]

Hi Gsf.

What do you mean by "attribution"?

Hi Wolfgang.

tried the first one. Forgive me the notation, but i hope, its correct:
A.r9c1=9:
If r9c1=2
1. r4c1=1,r4c8=2
2. r6c3=1,r6c7=7,r5c8=2

I don't understand this: if r9c1=2, how can r6c3=1 if r4c1=1?

Regards, Carcul

[Wolfgang]

I don't understand this: if r9c1=2, how can r6c3=1 if r4c1=1?

Thanks for the hint that r6c3 and r4c1 are in the same box
Try it this way now:
A r8c8=4:
A1. r8c8<>3:
If r8c8=3,
1.r8c4<>3,r9c4=3,r9c4<>2
2.r9c7=7,r8c7=1,r7c7=8,r3c7=2,r3c4<>2
no 2 in col 4 => r8c8<>3

A2. r8c8<>6:
r8c6=6:
1.r8c45<>6,r9c4=6,r4c7=7,r8c4=3,r8c7<>3,r9c7=3,(r9c8<>6)r9c8=9,r9c1=2,r4c1=1
2.(r9c1=2)r79c3<>2,(r789c89<>4)r79c3<>4,r79c3=79,r6c1=1

B r2c9=7:
[edited (had another mistake and typos here):]
r1c9=7,r2c9=9,r2c5=5,r23c2=89,r3c3=6,r3c6=8,r1c6=7

[gsf]

Hi Gsf.
What do you mean by "attribution"?

where did you get the puzzles

[tarek]

Nice puzzles Carcul, especially puzzle #2

what is everybody's next step at this position in puzzle #2 ?

Code: Select all

*-----------------------------------------------------------------*
| 3      1      9     | 6      578    48    | 247    25     457   |
| 6      8      2     | 157    357    134   | 147    35     9     |
| 4      5      7     | 123    9      123   | 8      6      13    |
|---------------------+---------------------+---------------------|
| 1278   4      138   | 9      2378   5     | 6      238    137   |
| 2578   6      358   | 2378   1      238   | 247    9      3457  |
| 9      27     1358  | 4      2378   6     | 127    2358   1357  |
|---------------------+---------------------+---------------------|
| 12     9      4     | 123    6      123   | 5      7      8     |
| 12578  27     158   | 1258   258    9     | 3      4      6     |
| 58     3      6     | 58     4      7     | 9      1      2     |
*-----------------------------------------------------------------*

Tarek

[gsf]

also you changed the subject from "2 puzzles" to "3 puzzles" underfoot
and I thought I was losing my mind

[Carcul]

Hi Wolfgang.

Try it this way now:...

Very good: a solution in only two steps. Mine has a lot more (simple) steps: personally, I only try more complex deductions like those two when I cannot spot simpler ones. Just a question: how did you spot these deductions without some sort of T&E?
Btw, if you can make these kind of deductions, then you should be able to solve your "old puzzle".

Regards, Carcul

[Carcul]

Hi Gsf.

where did you get the puzzles

Puzzle #1 I get it from a post in this forum (I don't remember which). Puzzle #2 belongs (or used to belong) to the three hardest puzzles generated by Nick70.

Regards, Carcul

[Wolfgang]

Just a question: how did you spot these deductions without some sort of T&E?

I cannot spot them fully in advance. What i was doing is to play around with the puzzle in an editor (more than an hour for this one), until i think, i have found a cell where i have good chances that
(i) i can show that only one candidate remains
(ii) it helps me further as much as possible
So you can call this also a sort of T&E.

[gsf]

Puzzle #1 I get it from a post in this forum (I don't remember which). Puzzle #2 belongs (or used to belong) to the three hardest puzzles generated by Nick70.

puzzle #1 is posted here http://www.research.att.com/~gsf/sudoku/
it was part of a ~225M randomly generated puzzle collection
and is currently one of only two known 2-constrained up to simple coloring sudokus
2-constrained means the backdoors / magic cells come in pairs
i.e., it will take 2 lucky guesses to crack the puzzle using techniques up to simple coloring
the other 2-constrained puzzle is

Code: Select all

7 . . | . . . | 4 . .
. 2 . | . 7 . | . 8 .
. . 3 | . . 8 | . . 9
------+-------+------
. . . | 5 . . | 3 . .
. 6 . | . 2 . | . 9 .
. . 1 | . . 7 | . . 6
------+-------+------
. . . | 3 . . | 9 . .
. 3 . | . 4 . | . 6 .
. . 9 | . . 1 | . . 5


I lost the attribution for this one -- its not mine though
(edit) its #77 from http://magictour.free.fr/top1465

[TKiel]

Tarek,

In response to your question about the next step for puzzle #2, it seems somebody (read: not me) aught to be able to use the fact that both r3c9 and r7c1 can't be 1 as that would make a rectangle of death or rectangle of non-uniqueness or whatever it's called in r3c4,6 and r7c4,6 (they would all have 2,3 as their only values).

Tracy

[Carcul]

In response to your question about the next step for puzzle #2, it seems somebody (read: not me) aught to be able to use the fact that both r3c9 and r7c1 can't be 1 as that would make a rectangle of death or rectangle of non-uniqueness or whatever it's called in r3c4,6 and r7c4,6 (they would all have 2,3 as their only values).

Tracy, those deductions are wrong. Perhaps what you are trying to say is that r3c9 and r7c1 cannot be "1" at the same time, but each one by their own will not create a poli-solution situation. In fact, in the solution of the puzzle r3c9 turns out to be "1".

Regards, Carcul

[tarek]

Try it this way now:
A r8c8=4:
A1. r8c8<>3:
If r8c8=3,
1.r8c4<>3,r9c4=3,r9c4<>2
2.r9c7=7,r8c7=1,r7c7=8,r3c7=2,r3c4<>2
no 2 in col 4 => r8c8<>3

hi Wolfgang,

What is this technique called, I know that the base is a bifurcation that leads to contradiction elimination, but is there a specific name for it.....

Tarek