Another technique (help)

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Another technique (help)

Postby 813554 » Fri Sep 21, 2018 8:12 am

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
 
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Joined: 05 September 2018

Re: Another technique (help)

Postby Leren » Fri Sep 21, 2018 9:08 am

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
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Re: Another technique (help)

Postby SpAce » Fri Sep 21, 2018 9:23 am

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):

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
 
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Re: Another technique (help)

Postby 813554 » Fri Sep 21, 2018 9:29 am

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

Re: Another technique (help)

Postby SpAce » Fri Sep 21, 2018 10:46 am

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
 
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