The New Sudoku Players' Forum

[Myth Jellies]

Code: Select all

 *-----------------------------------------------------------*
 | 689   3689  389   | 146   1346   7    | 2     5    14689  |
 |N679   1    A29    | 5     8     G24   |M4679 MF46   3     |
 | 5    Q23678 4     | 126   1236   9    |L67   H168  H168   |
 |-------------------+-------------------+-------------------|
 |C468  C468   7     |K1248' K1'24  3    | 5     9    D26    |
 | 3     4689 B89    |J248   5     J248  | 1    E26    7     |
 | 2     5     1     | 9     7      6    | 34    348   48    |
 |-------------------+-------------------+-------------------|
 |O479  P2479  6     | 3     249    124  | 8     124   5     |
 | 489   23489 23589 | 2468  2469  12458 | 3469  7     12469 |
 | 1     23489 23589 | 7     2469   2458 | 3469  2346  2469  |
 *-----------------------------------------------------------*


For the 2nd, r5c2<>4 is also a valid exclusion.

A2 = A9 - B9 = B8 - C8 = C(4&6) - D6 = E6 - F6 = F4 - G4 = G2 (- A2); closed loop => r189c3 <> 9, r5c2 <> 8, r39c8 <> 6, and r2c7 <> 4
Can you explain, because for some reason I don't see this one.

For the 3rd, defining set J = {r4c5,r5c46} would simplify the expression a little.

Yeah, you caught me going a bit nuts there. Sometimes you just don't see the easiest path. Could have used strong links C8 = K8, and then between 1's in box 5 as well.

In the 4th, I would've missed the r2c1<>9 exclusion. How do you think to even look for that?

L7 = L6 - M6 = N6 - N7 = O7 - P7 = Q7 (- L7); closed loop => r1c9 & r3c89 <> 6, and r2c1 <> 9
With a closed AIC loop, either one side or the other of every weak inference must be true; so you just check them all. It's pretty obvious from the bolded part of the chain that N just equals 67

[r.e.s.]

IsThisLineWrappedInInternetExplorer ...?

No, but then it doesn't contain the hyphens common in Carcul's chains (which I think is what makes the difference -- IE wraps on a hyphen, Firefox doesn't).

[ravel]

Fine, so i will keep it there for a day to give Carcul a chance to see, how his chains look for Firefox users

[Carcul]

I have tried to fix the problem in the notation. How does it look now? (However, the grid between the "code" still looks very long: I have never seen this)

Carcul

[ravel]

Pretty fine now, many thanks.

[daj95376]

I have tried to fix the problem in the notation. How does it look now? (However, the grid between the "code" still looks very long: I have never seen this)

Carcul

Thanks from us Netscape users as well !!! The code field is fine for me.

[ronk]

For the 2nd, r5c2<>4 is also a valid exclusion.

A2 = A9 - B9 = B8 - C8 = C(4&6) - D6 = E6 - F6 = F4 - G4 = G2 (- A2); closed loop => r189c3 <> 9, r5c2 <> 8, r39c8 <> 6, and r2c7 <> 4
Can you explain, because for some reason I don't see this one.

Reading your chain left-to-right, if A=9 then C=4&6 ... right-to-left, if A=2 then C=4&8. Either way, C contains a 4. For a closed loop, therefore, 4s may be excluded from any cells that see all the 4s in C, meaning r5c2<>4 and r4c45<>4. Your notation obscures the case where C=4&8, so I'm not surprised you missed it.

I've used something like ... -a-C(a|bc|d)-d- ... in a handful of places. From the left, the weak link "peels away" the 'a' and you're left with "bcd". From the right, the other weak link peels away the 'd' and you're left with "abc". When the loop is continuous (closed), the digits between the vertical lines ('|'), "bc" in this example, are locked in the set.

[Myth Jellies]

That's cool.

So, if I understand you correctly, in my notation for a closed loop any unused "and-ed" digits must also be true?!

That will be useful. Gonna keep my eye out for that.

[ronk]

So, if I understand you correctly, in my notation for a closed loop any unused "and-ed" digits must also be true?!

For "unused and-ed digits" meaning those digits not used in weak links directly connected to the ALS containing the "unused and-ed digits"... "yes"!