The New Sudoku Players' Forum

[maria45]

Hello tso,

I just solved your example by forcing chains.

the point where it gets interesting:

8 4679 679 2 46 5 3467 369 1
2457 245679 679 1 8 3 4567 2569 2467
1 2456 3 469 7 49 8 256 246
6 3 4 8 2 1 9 7 5
57 57 8 3469 346 49 2 1 36
9 1 2 36 5 7 36 4 8
247 2467 5 34 9 8 1 236 23467
24 2489 19 7 134 6 345 2358 234
3 4678 167 5 14 2 467 68 9

let a-k without j denote rows, 1-9 columns:

k8=6 or k8=8, h2=8, h3=9, k3=1 > k3 != 6 > abc2 != 6
k8=6 or k8=8, h2=8, h3=9, k3=1, k5=4, a5=6, b3=6, e5=3, e9=6, c8=6 > ab8!=6, g89!=6 > g2=6
f4=6 or f4=3, e9=3, g4=4, f7=6, k8=6, h8=8, h5=3, g8=3, h19=24, h2=9, a8=9, b3=9, a3=6, c4=6 > e4!=6
c4=6 or a5=6, b3=6, f4=6, e9=6 > c9!=6
b9=6 or e9=6, f4=6, a5=6, b3=6, c8=6 > ab7!=6
c9=2 or c9=4, a5=4, e5=6, e9=3, h9=2 > bg9!=2
f4=3 or f4=6, f7=3, a8=3, a5=6, c9=2, c8=6, g8=2, h8=5, h7=4, h9=3, g4=3, e5=3 > e4!=3
e46=49 > e5!=4
g8=2 or g8=3, g4=4, g9=7, g1=2, h1=4, k7=4, e9=3, b9=6, c9=4, h9=2 > h8!=2
k7=6 or k8=6, b9=6, g9=7 > k7!=7 > g9=7
gh1=24 > b1=57, hk2!=24
g8=3 or g8=2, g1=4, g4=3, e5=3, f7=3, a8=3, h9=3 > h78!=3
naked quad hk78=4568 > h9!=4 > ab7!=4
b17=57
naked triple abg8=239 > c8!=2
c9=4 or c9=2, h9=3, g4=3, f4=6, c46=49 > c2!=4
a7=7 or b7=7, c8=5, b9=6, a3=6 > a3!=7 > k3=7 and the remaining numbers fall into place.

Sure, I grant that the unique rectangle was tempting in plain view and much shorter of course, and normally I would not refrain from taking advantage of it, I only wanted to give you a proof of concept...

Kind regards, Maria

[tso]

Maria,

I guess your labeling the 9 rows using the first 11 letters minus 'i' and 'j'.
Not completely random, but ...

Parsing your first statement...

k8=6 or k8=8, h2=8, h3=9, k3=1 > k3 != 6 > abc2 != 6

...I believe you are saying:

k8=6 -> k23!=6 -> g2=6 -> abc2!6
k8=8 -> k2!=8 -> h2=8 -> h3=9 -> k3=1 -> k3!=6 -> ab3=6 -> abc2!=6
Therefore, abc2!=6

While this is certainly a legitimate logical deduction, it is, as I said, beyond what most people consider forcing chains, or at least, beyond simple (exclusivlely bivalue or bilocation) or comprehensive (bivalue AND bilocation) forcing chains, as it requires multiple cell links (k23 and ab3). I should have said "beyond simple or typical" forcing chains -- or maybe "beyond forcing chains that *I* can find! If we drop all restrictions, *all* deductions can be written as forcing chains.

[Wolfgang]

I just solved your example by forcing chains.

wow, you can solve really tough puzzles. Hope you dont do it every day, its a bit eery for me (like what Carcul is doing)
The term i use for deductions like yours is forcing net.

[Jeff]

I just solved your example by forcing chains.

Hi maria, You are indeed in the premier league for forcing chain solving. You are invited to share you technique for identification of forcing chains here.

[aeb]

I don't think there's any question that a valid sudoku has to have a unique solution. The question is, what if a sudoku has multiple solutions, but if you assume it has a unique solution and using techniques based on that assumption you can arrive at a unique answer?

One can formulate the uniqueness argument not assuming that the solution is unique. Then the conclusion becomes that one loses an even number of solutions by the argument. It follows that if the total number of solutions was even, the uniqueness argument will not give a unique solution. On the other hand, it is not difficult to see that if the total number of solutions is odd, one can pick any solution as the desired outcome, and eliminate the others using uniqueness arguments.

[maria45]

Hello tso,

I guess your labeling the 9 rows using the first 11 letters minus 'i' and 'j'.

my, you got me there...

Parsing your first statement...

k8=6 or k8=8, h2=8, h3=9, k3=1 > k3 != 6 > abc2 != 6

...I believe you are saying:

k8=6 -> k23!=6 -> g2=6 -> abc2!6
k8=8 -> k2!=8 -> h2=8 -> h3=9 -> k3=1 -> k3!=6 -> ab3=6 -> abc2!=6
Therefore, abc2!=6

well, I thought you would be able to decipher my shortened code, and you proved me nearly right...

the > sign denotes a conclusion.
so the first line of my solution contains 2 conclusions. I only added the second conclusion as it is the simple single-column technique. Simple conclusions are just added inline.

but I only meant to say: k8=6 > k3 != 6. That is the first half of the chain and obviously doesnt imply multiple cells, there is only one bivalue "starting cell", k8, and one "goal cell", k3. The first conclusion does only come to the result that k3 cannot contain a 6.

The second half of the chain you got completely right:
k8=8 then h2=8 then h3=9 then k3=1.

This I believe to be a simple forcing chain. Summarizable like this:
If k8=6, k3 cannot be 6.
AND If k8=8, k3 can also not be 6.

So I exclude 6 as possibility for k3.

More complex forcing chains run multiple steps on both halves of the chain, but also reach a single "goal cell". The one thing I didnt try until now is multivalue starting cells, as I could solve all puzzle with bivalue starting cells. Well, bivalue is not quite correct. Better: a binary chain:

Either k8=6 OR k8=8

Further down in my solution is another example:

Either b9=6 OR e9=6

In the first example, the binary logic involves one cell with two mutually exclusive values, in the second it involves two cells with a value which is mutually exclusive.

After elimination of 6 in k3, then the second, easy step follows with the single column in ab3=6. But that has nothing to do with forcing chains.

[Moschopulus]

But I'm afraid I don't see how that answers my question. I realise you can't make a multiple-solution puzzle magically have a single answer, but all I'd like to know is, can you make a multiple-solution puzzle that arrives at a single solution puzzle if you treat it as a single-solution one ans use techniques dependent on that erroneous assumption. I know it would be the wrong thing to do, and I know it wouldn't stop the original puzzle being multiple-solutioned, but I just think it'd be an interesting puzzle to have.

Here is an example. Consider this pseudoku puzzle

Code: Select all

876 453 219
394 182 675
251 .6. 438

647 2.8 .53
512 .36 .84
938 5.4 .26

763 841 592
42. 3.. 861
18. 62. 347


This (pseudo) puzzle has 3 solutions.

Now look at r3c4. This is either 7 or 9. If you put a 9 here, the resulting puzzle has 2 solutions. Therefore, if you assume there is a unique solution, you will put a 7 here, and then the puzzle does have a unique solution.

Unfortunately, someone else might not look at r3c4 first, they might look at r8c3 first. This is either 5 or 9. If they put a 5 here, the resulting puzzle has 2 solutions. Therefore, if this person assumes that there is a unique solution, they will put a 9 here, and then the puzzle does have a unique solution. It is a different solution to the one you found.