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https://imgur.com/a/YvELsKX in this puzzle, I've stuck here for a while and then I saw 4s can help me. In box 1, r2c2 and r3c1 could be 4 so first, I assumed r2c2 is 4. In that case, r3c9 would be 4 aswell in box 3. Then I assumed r3c1 is 4. In that case, r1c9 would be 4. So, there was no way that r1c8 could be 4 in box 3. I've tried that and it was right, I can definitely eliminate 4 from r1c8 and probably solve this but obviously trial and error is not my favorite method. I want to forget this information and solve this puzzle in other way. Is there anything that can help me and I'm missing?

- 813554
**Posts:**16**Joined:**05 September 2018

- Code: Select all
`*---------------------------------------------*`

| 368 5 36 | 14 7 9 | 1238 3-4 12348 |

| 38 c14 7 | 14 6 2 | 5 39 389 |

|b14 9 2 | 5 3 8 | 7 6 a14 |

|---------------+---------+-------------------|

| 56 8 56 | 9 1 3 | 4 2 7 |

| 349 2 349 | 6 8 7 | 39 1 5 |

| 139 7 139 | 2 5 4 | 369 8 369 |

|---------------+---------+-------------------|

| 2 16 159 | 37 4 56 | 1389 3579 1389 |

| 1459 d146 8 | 37 2 56 | 139 *34579 139-4 |

| 7 3 45 | 8 9 1 | 26 *45 26 |

*---------------------------------------------*

A 2 Stringed Kite solves this puzzle. In the diagram one of cells a or d must be 4, so r8c9 <> 4. A pointing pair of 4's in r89c8 solves r1c8 and you are home.

You can read about Kites here.

Leren

- Leren
**Posts:**3319**Joined:**03 June 2012

Leren already gave you what I would have recommended too. If you really want to take out the (4)r1c8 directly, you can do that by adding two nodes to the Kite chain (among other routing possibilities):

(4)r3c9 = r3c1 - r2c2 = r8c2 - r9c3 = (4)r9c8 => -4 r1c8, r8c9

- Code: Select all
`.-----------------.-----------.--------------------.`

| 368 5 36 | 14 7 9 | 1238 3-4 12348 |

| 38 c14 7 | 14 6 2 | 5 39 389 |

|b14 9 2 | 5 3 8 | 7 6 a1(4) |

:-----------------+-----------+--------------------:

| 56 8 56 | 9 1 3 | 4 2 7 |

| 349 2 349 | 6 8 7 | 39 1 5 |

| 139 7 139 | 2 5 4 | 369 8 369 |

:-----------------+-----------+--------------------:

| 2 16 159 | 37 4 56 | 1389 3579 1389 |

| 1459 d146 8 | 37 2 56 | 139 34579 139-4 |

| 7 3 e45 | 8 9 1 | 26 f(4)5 26 |

'-----------------'-----------'--------------------'

(4)r3c9 = r3c1 - r2c2 = r8c2 - r9c3 = (4)r9c8 => -4 r1c8, r8c9

- Code: Select all
`* | | | | *`

* |=()=| / _ \ |=()=| *

* | | |-=( )=-| | | *

* \ ¯ / *

- SpAce
**Posts:**681**Joined:**22 May 2017

Wow it's a great solution. I guess I could never see this but now, I know what to look for. Thank you so much.

- 813554
**Posts:**16**Joined:**05 September 2018

813554 wrote:I've stuck here for a while and then I saw 4s can help me. In box 1, r2c2 and r3c1 could be 4 so first, I assumed r2c2 is 4. In that case, r3c9 would be 4 aswell in box 3. Then I assumed r3c1 is 4. In that case, r1c9 would be 4. So, there was no way that r1c8 could be 4 in box 3. I've tried that and it was right, I can definitely eliminate 4 from r1c8 and probably solve this but obviously trial and error is not my favorite method.

I wouldn't call your original method trial and error. Trial and error would be if you tried to place 4 in r8c1 and then found a contradiction (i.e. "error") which would tell you that r8c1 can't be 4. Your method works like our chains in that it proves r8c1<>4 without ever assuming that value in that cell. It's basically a verity type forcing chain (or actually a forcing net, because it seems to use branching) which does exactly what an AIC does. Leren's 2-String Kite is an AIC, just like my longer chain. A non-branching forcing chain can be easily converted into an AIC as well, so they're basically equivalent techniques (just different points of view).

Your logic could be expressed as a forcing net like this:

(4)r2c2 - r3c1 = (4)r3c9

||

(4*)r3c1 - r2c2 = r8c2 - r9c3 = r9c8 - (r1c8|*r3c9) = (4)r1c9

=> -4 r1c8, r8c9

It proves that no matter which 4 in box 1 is true (and one must be), r1c8 (or r8c9) can't be 4 because the chains force either r3c9 or r1c9 to be 4 (and they both see the victims). It's perfectly valid logic, and it's not t&e or "assumptive" any more than an AIC is (even though some might say otherwise). Its only problem is that it's a bit too complicated for this situation, especially because the second chain uses branching. There's no reason to prove (4)r3c1 -> (4)r1c9 because (4)r3c1 -> (4)r9c8 does the trick just as well more simply:

(4)r2c2 - r3c1 = (4)r3c9

||

(4)r3c1 - r2c2 = r8c2 - r9c3 = (4)r9c8

=> -4 r1c8, r8c9

Again, both end points see the victims. That forcing chain is equivalent to the X-Chain (a single-digit AIC) I presented in the previous post. To turn it into an AIC you can just "pull" one of the end points to straighten it into a single chain:

(4)r3c9 = r3c1 - (4)r2c2 = (4)r3c1 - r2c2 = r8c2 - r9c3 = (4)r9c8

Once the duplicate nodes are removed we have my X-Chain:

(4)r3c9 = r3c1 - r2c2 = r8c2 - r9c3 = (4)r9c8 => -4 r1c8, r8c9

Similarly Leren's 2-String Kite could be expressed as a forcing chain:

(4)r2c2 - r3c1 = (4)r3c9

||

(4)r3c1 - r2c2 = (4)r8c2

=> -4 r8c9

But the AIC format is preferable:

(4)r3c9 = r3c1 - r2c2 = (4)r8c2 => -4 r8c9

I just wanted to say that you were on the right track, so don't beat yourself too hard!

- Code: Select all
`* | | | | *`

* |=()=| / _ \ |=()=| *

* | | |-=( )=-| | | *

* \ ¯ / *

- SpAce
**Posts:**681**Joined:**22 May 2017

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