A deduction from uniqueness

Advanced methods and approaches for solving Sudoku puzzles

A deduction from uniqueness

Postby Scott H » Fri Jul 29, 2005 4:54 am

This puzzle:

Code: Select all
...|..7|51.
..2|...|..9
3.9|..2|...
-----------
.7.|.6.|..8
6..|9.5|..7
2..|.1.|.5.
-----------
...|1..|3.5
1..|...|7..
.67|4..|...


has an interesting combination. After some normal deductions the candidate sets are:

Code: Select all
{8}   {4}   {6}   {3}   {9}   {7}   {5}   {1}   {2}   
{7}   {5}   {2}   {6}   {48}  {1}   {48}  {3}   {9}   
{3}   {1}   {9}   {58}  {458} {2}   {48}  {7}   {6}   
{45}  {7}   {135} {2}   {6}   {34}  {19}  {49}  {8}   
{6}   {38}  {13}  {9}   {38}  {5}   {12}  {24}  {7}   
{2}   {9}   {48}  {7}   {1}   {48}  {6}   {5}   {3}   
{49}  {2}   {48}  {1}   {7}   {689} {3}   {689} {5}   
{1}   {38}  {35}  {58}  {258} {69}  {7}   {69}  {4}   
{59}  {6}   {7}   {4}   {235} {389} {29}  {289} {1}   


The four cells in rows 2/3 and columns 5/7 contain just 48, except r3c5 has candidates 458.

If we trust that the puzzle has a unique solution (we can, this is from Angus's Simple Sudoku), we can deduce r3c5 = 5 (since if not, the 48's in an X-wing cannot possibly have a unique solution).

This deduction is not needed (simpler techniques suffice). But I saw this idea mentioned in another thread, and was interested to see it actually occur.
Scott H
 
Posts: 73
Joined: 28 July 2005

Postby Doyle » Fri Jul 29, 2005 4:45 pm

In my view, it is not one of the givens that the puzzle has a unique solution, so the puzzle should be soved without recourse to that. The uniqueness is just a reassurance, an aside, from the puzzle maker, and should not be used as a factor in any legitimate logical process. The only givens are that there is only one of each value in each row, column and box. As you said, it is not a necessary element of the solution of this particular puzzle, anyway.
Doyle
 
Posts: 61
Joined: 11 July 2005

Postby PaulIQ164 » Fri Jul 29, 2005 5:07 pm

I've used this technique myself (doubtless it wasn't strictly necessray though) in some Times puzzles (notably one of the 'super-fiendshes'). I disagree, it's a valid technique in my opinion. Making use of the fact that the solution is unique is perfectly valid as far as I can see, as it is one of the prerequisites of a proper sudoku puzzle.
PaulIQ164
 
Posts: 533
Joined: 16 July 2005

Postby PaulIQ164 » Sat Jul 30, 2005 8:51 pm

I've been thinking some more about this tactic, and I wondered: what if there were a sudoku puzzle for which you had to use this technique to find the unique solution? Then finding the unique solution would be reliant on being told there was one. If someone gave you the puzzle and told you there was a unique solution, you'd be able to use this tactic and find it. Whereas, if someone gave you the puzzle and told you there were multiple solutions, you wouldn't be able to use this tactic and so, lo and behold, you would indeed find multiple solutions. It would be some kind of bizarre self-fulfilling sudoku prophecy.

Does this make any sense?
PaulIQ164
 
Posts: 533
Joined: 16 July 2005

Postby tso » Sat Jul 30, 2005 9:48 pm

I don't see any basis for ignoring information when solving a puzzle. The puzzle is a combination of the rules and the specific clues given. The rules include that there is a unique solution. In his example, as soon as Scott had this realization, he had *already* proved that r3c5 = 5.

A Sudoku with two solutions is no more or less invalid as one with no solution, so his deduction is no more or less valid than assuming that he can't place two 5's in one row.

Imagine you found an old puzzle that was not refered to as a Sudoko, in fact, predated the coining of the word. This puzzle had a 9x9 grid. The rules are: "Place the digits 1-9 in this grid so that there is no duplication in any row, column or box. What numbers go in the four corners?" In this case you couldn't use the uniqueness tactic -- but NOT because it is unfair, but because you couldn't be sure it would work. (I've seen this type of puzzle many times over the years. I think the point is to trip up the solver who is intent on solving the puzzle -- not realizing that there isn't enough information to solve it, only enough to answer the question.)
tso
 
Posts: 798
Joined: 22 June 2005

Postby PaulIQ164 » Sat Jul 30, 2005 9:53 pm

And at any rate, it's not like anyone would seriously notice this, but then think "aah, that's not really a valid tactic - I'd better not write in that 5", is it? (Or is it?)
PaulIQ164
 
Posts: 533
Joined: 16 July 2005

Postby stuartn » Sat Jul 30, 2005 10:33 pm

It's up to you Paul - you play the game - you decide what rules you want to stick to. However, you can only compare performance realistically if all players stick to the same rules.

By the way - are there really 163 other PaulIQ's out there?:D

www.brightonandhove.org
stuartn
 
Posts: 211
Joined: 18 June 2005

Postby Karyobin » Sat Jul 30, 2005 10:54 pm

Like the 163 bit Stuart. The thing is, I don't see how the initial data should make a difference. Consider (sorry to sound pompous)...

1. You were given this starting point and told there was only one solution, therefore - you would find it using this method.

2. You were given this starting point and told there was more than one solution - paradox! This combination of candidates is dependent upon there being one solution: ergo: (sorry - height of pomposity) this situation could necessarily not arise.

Therefore again, should you stumble upon this combination, you have confirmed that your current puzzle has a unique solution.

All Praise the Stella goblin.
Karyobin
 
Posts: 396
Joined: 18 June 2005

Postby PaulIQ164 » Sat Jul 30, 2005 11:13 pm

So, are you saying that in a solution with multiple solutions, the kind of situation where you have a {24}, {24}, {24}, {245} or whatever, cannot arise? Why not?
PaulIQ164
 
Posts: 533
Joined: 16 July 2005

Formal rules based on uniqueness as a given

Postby laogui32 » Tue Sep 06, 2005 4:42 am

I was thinking along these lines too, working backwards from a completed puzzle, removing cell values, the simplest pattern which gives a non-unique solution removes a and b numbers from 4 four cells in a square pattern - ie remove 2 a's, 2 b's restricted to two rows, two columns and two blocks
Then we would have four cells with candidates {a,b} where a and b are interchangeable.
Then we can use this pattern to generate some elimination rules such as:
If we have this pattern plus a third candidate such as:
row x columns 123: {a,b},{abc},{bc}
row y columns 123: {a,b},{a,b},d
We know by uniqueness that we cannot have
row x columns 12 containing both a and b
therefore c must be in column 1 or 2
therefore row x col 3 is b
Ie. we can eliminate c from row x col 3 obtaining:
row x: a,c,b
row y: b,a,x

I believe there is a substantial set of such patterns we can use to generate deductive rules of this kind
laogui32
 
Posts: 3
Joined: 02 September 2005

Postby stuartn » Tue Sep 06, 2005 1:41 pm

This was interesting.... ONLY by knowing that the puzzle was unique could I solve it. There are 4 eliminations needed on the basis of 'if I leave this candidate it won't be a unique grid.

Code: Select all
127658493
 [36]  [63] 4729581
598314726
 [37]  [48] 1 [98] 6 [35] 2 [573]  [489]
 [2367]  [638] 5 [289] 4 [23] 1 [73]  [89]
 [23]  [438] 9 [1258] 7 [1235] 6 [53]  [48]
8 [15] 6 [15] 97342
972436815
4 [15] 3 [125] 8 [125] 967


Which seems to point at the uniqueness strategy being a valid one......

or have I had too many at lunch?

This was the start grid.....



Code: Select all
12 .  . 5 .  .  . 3
 .  .  . 7 . 958 .
 . 98 .  .  .  .  .  .
 .  . 1 .  .  . 2 .  .
 .  . 5 . 4 . 1 .  .
 .  . 9 .  .  . 6 .  .
 .  .  .  .  .  . 34 .
 . 724 . 6 .  .  .
4 .  .  . 8 .  . 67



stuartn
stuartn
 
Posts: 211
Joined: 18 June 2005

Postby PaulIQ164 » Tue Sep 06, 2005 1:59 pm

I only got this far before the Uniqueness Tactic couldn't take me any further (that I could see):

Code: Select all
+-------+-------+-------+
| 1 2 7 | 6 5 8 | 4 9 3 |
| 6 3 4 | 7 2 9 | 5 8 1 |
| 5 9 8 | 3 1 4 | 7 2 6 |
+-------+-------+-------+
| . 8 1 | 9 6 . | 2 . 4 |
| . 6 5 | 8 4 . | 1 . 9 |
| . 4 9 | . 7 . | 6 . 8 |
+-------+-------+-------+
| 8 . 6 | . 9 7 | 3 4 2 |
| 9 7 2 | 4 3 6 | 8 1 5 |
| 4 . 3 | 2 8 . | 9 6 7 |
+-------+-------+-------+
PaulIQ164
 
Posts: 533
Joined: 16 July 2005

Postby stuartn » Tue Sep 06, 2005 2:09 pm

Paul - look at R4 C6+8, and R6, C6 +8... if they were all 35 then it wouldn't be unique -

stuartn
stuartn
 
Posts: 211
Joined: 18 June 2005

Postby PaulIQ164 » Tue Sep 06, 2005 2:30 pm

I see that, but I have the candidates:

[35] [357]
[1235] [35]

so how do you know whether it's r4c8 or r6c6 (or both) that isn't 3 or 5?
PaulIQ164
 
Posts: 533
Joined: 16 July 2005

Postby tso » Tue Sep 06, 2005 3:34 pm

Unless you're proposing a variant rule set, this does not work.

It's valid to say "If this cell is x, the puzzle will be ambiguous, therefore it isn't x." if and only if the puzzle has one and only one solution. It doesn't PROVE that the puzzle has a unique solution, it REQUIRES that it has one.

As given, this puzzle has multiple solutions.

Code: Select all
 1 2 . | . 5 . | . . 3
 . . . | 7 . 9 | 5 8 .
 . 9 8 | . . . | . . .
-------+-------+------
 . . 1 | . . . | 2 . .
 . . 5 | . 4 . | 1 . .
 . . 9 | . . . | 6 . .
-------+-------+------
 . . . | . . . | 3 4 .
 . 7 2 | 4 . 6 | . . .
 4 . . | . 8 . | . 6 7


However, if what you are proposing is an additional rule -- I don't know how it would work. One could construct a grid in which the solver could step-by-step think "if I leave this candidate, the puzzle will have multiple solutions, therefore I'll elliminate it" -- that could lead to any number of solutions -- each of which would be perceived to be unique upon arrival.

There's a difference between:

(A) "This puzzle has a unique solution."

and

(B) "This puzzle has a unique solution if and only if the correct number is placed in r5c5."


Puzzles like (B) are perfectly valid, but cannot be solved without giving the additional stipulation.
tso
 
Posts: 798
Joined: 22 June 2005

Next

Return to Advanced solving techniques