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storm_norm

Joined: 18 Oct 2007
Posts: 1741

Posted: Mon Dec 31, 2007 12:25 am Post subject: danger...enter at your own risk...danger...warning

Code:

. . . | 1 . . | 2 . .
. 8 . | . . . | . 1 5
. . 3 | . 8 . | . 4 7
------+-------+------
. 6 . | . . 9 | . . .
7 . . | . 5 . | . . 3
. . . | 8 . . | . 6 .
------+-------+------
6 1 . | . 9 . | 5 . .
5 2 . | . . . | . 7 .
. . 4 | . . 1 | . . .

krazy dad, book 100 puzzle 2 for those who want to spend a couple days over new years looking at a challenge.
after basics I was stuck.

Asellus

Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA

Posted: Wed Jan 02, 2008 11:52 pm Post subject:

I have been out of commission for a while, so decided to tackle this puzzle no one else has taken on as a way of getting back into things. I believe this is the most difficult puzzle I have ever solved. It did indeed keep me busy for a couple of days. My solution path is very long and not all that elegant. While it is mostly a "slog," there are some interesting points along the way.

The grid after basics:

Code:

+-------------------+----------------------+---------------------+
| 49 4579 5679 | 1 3467 34567 | 2 389 689 |
| 249 8 2679 | 34679 3467 3467 | 369 1 5 |
| 1 59 3 | 269 8 256 | 69 4 7 |
+-------------------+----------------------+---------------------+
| 2348 6 128 | 2347 12347 9 | 1478 5 1248 |
| 7 49 1289 | 246 5 246 | 1489 289 3 |
| 2349 3459 1259 | 8 12347 2347 | 1479 6 1249 |
+-------------------+----------------------+---------------------+
| 6 1 78 | 2347 9 23478 | 5 238 248 |
| 5 2 89 | 346 346 3468 | 134689 7 14689 |
| 389 379 4 | 5 267 1 | 689 289 2689 |
+-------------------+----------------------+---------------------+

While some complex chaining could accomplish the same thing, I resorted to a rather obvious "forcing" to crack into this by noting the great number of inter-related Almost Locked Sets.

In particular, note {3468} in r8 and {3467} in c5. r8c3 plays a pivotal (no pun intended) role:
IF r8c3=8 then r8c456={346}, r7c3=7, r7c4=2, r7c6=8, r7c9=4, r4c1=8, and r4c3=r4c9={12}.
But also (continuing from r8c3=8), r9c5=7, r128c5={346}, and r4c5={12}.

But, there can't be three cells in r4 that are {12}. So, r8c3<>8.

That doesn't get us very far, but allows some Medusa coloring starting in box 7:

Code:

+--------------------+---------------------+-------------------+
| 49 457G9 567 | 1 {346}7 34567 | 2 389 689 |
| 249 8 267 | 34679 {346}7 3467 | 369 1 5 |
| 1 59 3 | 269 8 256 | 69 4 7 |
+--------------------+---------------------+-------------------+
| 2348R 6 128 | 2347 {12}347 9 | 1478 5 1248 |
| 7 49g 128 | 2{46} 5 2{46} | 1489 289 3 |
| 2349 @3R45-9 125 | 8 {12}347 2347 | 1479 6 1249 |
+--------------------+---------------------+-------------------+
| 6 1 7G8R | 2347 9 23478 | 5 238 248 |
| 5 2 9 | 346 {346} 3468 | 13468 7 1468 |
| 3R8G 3G7R 4 | 5 267G 1 | 689 289 2689 |
+--------------------+---------------------+-------------------+

I use RG (upper case) for the conventional Medusa coloring. Then, I exploit the <7> in r9c5 once again by noting that IF it is true (i.e., G is true) then the Locked Sets indicated by braces in c5 and b5 result, leaving <9> in r5c2. So, this <9> is colored "g", the lower case indicating that it is dependent upon the assumption that G is true. It can be used for eliminations against R, and thus eliminates <9> from r6c2.

Continuing the "if green is true" approach, it is possible to eliminate <5> from r6c2 as well:

Code:

+--------------------+---------------------+-------------------+
| 49 457G9 567 | 1 {346}7 34567 | 2 389 689 |
| 249 8 267 | 34679 {346}7 3467 | 369 1 5 |
| 1 5g9 3 | 269 8 256 | 69 4 7 |
+--------------------+---------------------+-------------------+
| 2348R 6 128 | 2347 {12}347 9 | 1478 5 1248 |
| 7 49g 128 | 2{46} 5 2{46} | 1489 289 3 |
| 2349 @3R4-5 125g | 8 {12}347 2347 | 1479 6 1249 |
+--------------------+---------------------+-------------------+
| 6 1 7G8R | 2347 9 23478 | 5 238 248 |
| 5 2 9 | 346 {346} 3468 | 13468 7 1468 |
| 3R8G 3G7R 4 | 5 267G 1 | 689 289 2689 |
+--------------------+---------------------+-------------------+

Next, the "8G" at r9c1 produces a {69} Locked Pair in c7. In turn, this means that the <3>s at r2c7 and r7c8 are "g". This eliminates <8> from r7c8:

Code:

+--------------------+---------------------+--------------------+
| 49 457G9 67 | 1 {346}7 34567 | 2 389 689 |
| 249 8 267 | 34679 {346}7 3467 | 3g69 1 5 |
| 1 5g9 3 | 269 8 256 |{69} 4 7 |
+--------------------+---------------------+--------------------+
| 2348R 6 128 | 2347 {12}347 9 | 1478 5 1248 |
| 7 49g 128 | 2{46} 5 2{46} | 1489 289 3 |
| 2349 3R4G 5 | 8 {12}347 2347 | 1479 6 1249 |
+--------------------+---------------------+--------------------+
| 6 1 7G8R | 2347 9 23478 | 5 @23g-8 248 |
| 5 2 9 | 346 {346} 3468 | 13468 7 1468 |
| 3R8G 3G7R 4 | 5 267G 1 |{6}8{9} 289 2689 |
+--------------------+---------------------+--------------------+

Next, the "7G" in r1c2 makes "5g" in r1c6 and "6g" in r1c3. This in turn makes an {89} Locked Pair in r1c89 and "4g" in r1c1, eliminating <4> in r4c1, marked @. At the same time, we are left with "3g" in r1c5, eliminating <3> in r6c5, marked #:

Code:

+--------------------+----------------------+--------------------+
| 4g9 457G9 6g7 | 1 3g467 345g67 | 2 3{89} 6{89} |
| 249 8 267 | 34679 3467 3467 | 3g69 1 5 |
| 1 5g9 3 | 269 8 256 |{69} 4 7 |
+--------------------+----------------------+--------------------+
|@23-48R 6 128 | 2347 12347 9 | 1478 5 1248 |
| 7 49g 128 | 246 5 246 | 1489 289 3 |
| 2349 3R4G 5 | 8 #12-347 2347 | 1479 6 1249 |
+--------------------+----------------------+--------------------+
| 6 1 7G8R | 2347 9 23478 | 5 23g 248 |
| 5 2 9 | 346 346 3468 | 13468 7 1468 |
| 3R8G 3G7R 4 | 5 267G 1 |{6}8{9} 289 2689 |
+--------------------+----------------------+--------------------+

Next, I shifted to the cell r5c2, coloring two chains of implications (r and g) from this (R and G) bivalue. Note that this is not Medusa coloring and so does not exploit alternates such as bivalues along the two chains.

A key point to see is the exploitation of the Type 4 {26} UR in R35c46 along the "g" chain. If "4G" is true in r5c2, then there is a {26} pair in r5c46. To avoid the Deadly Pattern, r4c7 must be <6>, hence it is colored "g".

The {26} pair in b5 also means that the only <2> remaining in c5 is r9c5, which is thus "2g". Together with the b5 pair, this creates an {89} pair in c8 along the "g" chain, resulting in "3g" in r1c8 and eliminating <3> from r1c6, marked @ below. A "9g" in r2c7 removes <9> from r5c7, marked #, since it also "sees" the "9R" in r5c2.

Code:

+--------------------+-----------------------+--------------------+
| 49 4579 67 | 1 3467 @-345r67 | 2 3g89 689 |
| 249 8 267 | 34679 3467 3467 | 369g 1 5 |
| 1 5r9 3 | 269 8 256 | 6g9 4 7 |
+--------------------+-----------------------+--------------------+
| 238 6 128 | 2347 12347 9 | 1478 5 1248 |
| 7 4G9R 128 |{2}4{6} 5 {2}4{6} |#148-9 2{89} 3 |
| 2349 3g4 5 | 8 1247 2347 | 1479 6 1249 |
+--------------------+-----------------------+--------------------+
| 6 1 78g | 2347 9 23478 | 5 23 248 |
| 5 2 9 | 346 346 3468 | 13468 7 1468 |
| 3g8 37g 4 | 5 2g67 1 | 689 2{89} 2689 |
+--------------------+-----------------------+--------------------+

Continuing, the "4G" and "7g" in c2 create a {59} pair in b1, making "4g" in r1c1 and eliminating <4> from r1c6, marked @:

Code:

+----------------------+-----------------------+--------------------+
| 4g9 4{5}7{9} 67 | 1 3467 @-45r67 | 2 3g89 68g9 |
| 2g49 8 267 | 34679 3467 3467 | 369g 1 5 |
| 1 {5r9} 3 | 269 8 256 | 6g9 4 7 |
+----------------------+-----------------------+--------------------+
| 238g 6 128 | 2347 12347 9 | 1478 5 1248 |
| 7 4G9R 128 |{2}4{6} 5 {2}4{6} | 148 2{89} 3 |
| 2349g 3g4 5 | 8 1247 2347 | 1479 6 1249 |
+----------------------+-----------------------+--------------------+
| 6 1 78g | 2347 9 23478 | 5 2g3 24g8 |
| 5 2 9 | 346 346 3468g | 13468 7 1468 |
| 3g8 37g 4 | 5 2g67 1 | 689 2{89} 2689 |
+----------------------+-----------------------+--------------------+

The "g" <2>, <4> and <8> in r7 create a "g" {37} pair in r7c46 and, in turn, a {46} pair in r8c45. This leads to more "g" coloring in boxes 6 and 9. Now, there are two "g" <9>s in r6, marked #. So, the "G" <4> in r5c2, marked @, must be false:

Code:

+-------------------+---------------------------+--------------------+
| 4g9 4579 67 | 1 3467 5r67 | 2 3g89 68g9 |
| 2g49 8 267 | 34679 3467 3467 | 369g 1 5 |
| 1 5r9 3 | 269 8 256 | 6g9 4 7 |
+-------------------+---------------------------+--------------------+
| 238g 6 128 | 2347 12347 9 | 1478 5 12g48 |
| 7 @-4G9R 128 | 246 5 246 | 148 28g9 3 |
|#2349g 3g4 5 | 8 1247 2347 | 1479 6 #1249g |
+-------------------+---------------------------+--------------------+
| 6 1 78g | 2{3}4{7} 9 2{3}4{7}8 | 5 2g3 24g8 |
| 5 2 9 | 3{46} 3{46} 3468g | 13g468 7 1g468 |
| 3g8 37g 4 | 5 2g67 1 | 68g9 289g 26g89 |
+-------------------+---------------------------+--------------------+

The grid now:

Code:

+---------------+-------------------+------------------+
| 49 47 67 | 1 3467 5 | 2 389 689 |
| 249 8 267 | 34679 3467 3467 | 369 1 5 |
| 1 5 3 | 269 8 26 | 69 4 7 |
+---------------+-------------------+------------------+
| 238 6 128 | 2347 12347 9 | 1478 5 1248 |
| 7 9 128 | 246 5 246 | 148 28 3 |
| 234 34 5 | 8 127 237 | 179 6 129 |
+---------------+-------------------+------------------+
| 6 1 78 | 2347 9 23478 | 5 23 248 |
| 5 2 9 | 346 -346 3468 | 13468 7 1468 |
| 38 37 4 | 5 267 1 | 689 289 2689 |
+---------------+-------------------+------------------+

An ER in b9 and strong link in r1 eliminates <3> from r8c5.

Back to coloring:

Code:

+---------------------+---------------------+---------------------+
| 4r9 4G7R 6r7 | 1 @3-467 5 | 2 389 689 |
|@2-49 8 267 | 34679 3467 3467 | 369 1 5 |
| 1 5 3 | 269 8 26 | 69 4 7 |
+---------------------+---------------------+---------------------+
| 238G 6 128 | 2347 12347 9 | 1478 5 1248 |
| 7 9 128 | 246 5 246 | 148 28 3 |
| 234G 3G4R 5 | 8 127 237 | 179 6 129 |
+---------------------+---------------------+---------------------+
| 6 1 7R8G | 2347 9 23478 | 5 23 248 |
| 5 2 9 | 346 46 3468 | 13468 7 1468 |
| 3G8R 3R7G 4 | 5 267 1 | 689r 289 2689 |
+---------------------+---------------------+---------------------+

As before, "RG" is standard Medusa and "rg" are implications from R-true or G-true, respectively. Eliminations are possible between R-G, r-G and R-g, but not between r-g. (Apologies for reversing the colors from the previous grids; it was an oversight when writing this up that I didn't notice until much later.)

If R is true, then r1c3 is 6 and thus (via b3) r9c7 is 9, both now marked "r". This in turn (again via b3) means r1c1 must be "4r". Now, <4> is eliminated from r2c1 and r1c5, marked @, due to "4G" in r6c1 and r1c2. This creates a strong link on <4> in c1 (or r1) and some coloring can be promoted to upper case and extended.

Code:

+---------------------+---------------------+---------------------+
| 4R9G 4G7R 6r7 | 1 367 5 | 2 389r @68-9 |
| 2G9R 8 2R67 | 34679 3467 3467 | 369 1 5 |
| 1 5 3 | 269 8 26 | 69 4 7 |
+---------------------+---------------------+---------------------+
| 238G 6 128 | 2347 12347 9 | 1478 5 1248 |
| 7 9 128 | 246 5 246 | 148 28 3 |
| 234G 3G4R 5 | 8 127 237 | 179 6 129r |
+---------------------+---------------------+---------------------+
| 6 1 7R8G | 2347 9 23478 | 5 23 248 |
| 5 2 9 | 346 46 3468 | 13468 7 1468 |
| 3G8R 3R7G 4 | 5 267 1 | 689r 289 2689 |
+---------------------+---------------------+---------------------+

Next, "9r" in r9c7 results in "9r" in r6c9 and r1c8. This, in turn, eliminates <9> from r1c9. More coloring promotion results. Specifically, we get "9R" in r1c8 and "9G" in r9c8, eliminating <9> from r9c9 and determining r6c9 as <9> (the only <9> remaining in c9).

Code:

+---------------------+---------------------+---------------------+
| 4R9G 4G7R 6r7 | 1 367 5 | 2 389R 68 |
| 2G9R 8 2R67 | 34679 3467 3467 | 369 1 5 |
| 1 5 3 | 269 8 26 | 69 4 7 |
+---------------------+---------------------+---------------------+
| 238G 6 128 | 2347 12347 9 |@147-8 5 @124-8 |
| 7 9 @12-8 | 246 5 246 | 148 28r 3 |
| 234G 3G4R 5 | 8 127 237 | 17 6 9 |
+---------------------+---------------------+---------------------+
| 6 1 7R8G | 2347 9 23478 | 5 23 248 |
| 5 2 9 | 346 46 3468 | 13468 7 1468 |
| 3G8R 3R7G 4 | 5 267r 1 | 689R @2r-89G 26r8 |
+---------------------+---------------------+---------------------+

Because we have 8R and 9R in r9, we must have 2r in r9c8 and 6r in r9c9. <8> is eliminated from r9c8. The 2r in r9c8 produces 8r in r5c8, eliminating <8> from r4c79 and, as a consequence, from r5c3. There is also an XY Wing (believe it or not) with pivot at r1c3 that removes <8> from r7c9.

We are now here:

Code:

+---------------------+---------------------+---------------------+
| 4R9G 4G7R 6r7 | 1 3r67 5 | 2 389R 68r |
| 2G9R 8 2R67 | 34679 3467 3467 | 369 1 5 |
| 1 5 3 | 269 8 26 | 69 4 7 |
+---------------------+---------------------+---------------------+
| 238G 6 128R | 2347 12347 9 | 147 5 124 |
| 7 9 12 | 246 5 246 | 148 28r 3 |
| 234G 3G4R 5 | 8 127 237 | 17 6 9 |
+---------------------+---------------------+---------------------+
| 6 1 7R8G | 2347 9 23478R| 5 23 24 |
| 5 2 9 | 346 46 3468G | 13468 7 @146-8 |
| 3G8R 3R7G 4 | 5 267r 1 | 689R 2R9G 26r8 |
+---------------------+---------------------+---------------------+

The 6r at r9c9 produces 8r at r1c9. And, the <8>s in b8 can now be colored. This eliminates <8> from r8c9.

Coloring further:

Code:

+---------------------+---------------------+---------------------+
| 4R9G 4G7R 6r7 | 1 3r67 5 | 2 389R 68r |
| 2G9R 8 2R67 | 34679 3467 3467 | 369 1 5 |
| 1 5 3 | 269 8 26 | 69 4 7 |
+---------------------+---------------------+---------------------+
|@-238G 6 128R | 2347 12347 9 | 147 5 12r4 |
| 7 9 1r2 | 246 5 246 | 14r8 28r 3 |
| 234G 3G4R 5 | 8 127 237 | 17 6 9 |
+---------------------+---------------------+---------------------+
| 6 1 7R8G | 2347 9 23478R| 5 23 24r |
| 5 2 9 | 346 46 3468G | 13468R 7 1r46 |
| 3G8R 3R7G 4 | 5 267r 1 | 689R 2R9G 26r8 |
+---------------------+---------------------+---------------------+

We must have 1r in r5c3 and 4r in r5c7. The resulting "r" {17} pair in r46c7 results in 2r in r4c9, then 4r in r7c9 and 1r in r8c9. <2> is eliminated from r4c1, resulting in 2R in r6c1, eliminating <3> from r6c1.

Next, there is more <3> coloring and elimination, plus more "r" coloring in b9 and beyond:

Code:

+---------------------+----------------------+---------------------+
| 4R9G 4G7R 6r7 | 1 3r67 5 | 2 389R 68r |
| 2G9R 8 2R67 | 34679 3467 3467 | 3r69 1 5 |
| 1 5 3 | 269r 8 2r6 | 6r9 4 7 |
+---------------------+----------------------+---------------------+
| 3R8G 6 128R | 2347 12347 9 | 147 5 12r4 |
| 7 9 1r2 | 246r 5 2r46 | 14r8 28r 3 |
| 2R4G 3G4R 5 | 8 127 23R7 | 17 6 9 |
+---------------------+----------------------+---------------------+
| 6 1 7R8G | 2r347 9 23478R | 5 23r 24r |
| 5 2 9 | 3r46 46 @-3468G | 13468R 7 1r46 |
| 3G8R 3R7G 4 | 5 267r 1 | 689R 2R9G 26r8 |
+---------------------+----------------------+---------------------+

Note particularly that r7c4 must have 2r, resulting in 6r in r5c4 and 2r in r5c6. But, this is a contradiction: there are no red <2>s possible in c5. So, all the "R" values are false and all the "G" values true. (Note that the "r" values are not necessarily false so cannot automatically be eliminated.) Amazingly, this does not solve the puzzle. The grid is now:

Code:

+--------+------------------+-------------+
| 9 4 67 | 1 @-367 5 | 2 38 68 |
| 2 8 67 | 34679 3467 3467 |@-369 1 5 |
| 1 5 3 | 269 8 26 | 69 4 7 |
+--------+------------------+-------------+
| 8 6 12 | 2347 12347 9 | 147 5 124 |
| 7 9 12 | 246 5 246 | 148 28 3 |
| 4 3 5 | 8 127 27 | 17 6 9 |
+--------+------------------+-------------+
| 6 1 8 | 2347 9 2347 | 5 23 24 |
| 5 2 9 | 346 46 8 | 1346 7 146 |
| 3 7 4 | 5 26 1 | 68 9 268 |
+--------+------------------+-------------+

There is a Skyscraper on <3> in c68 that makes the eliminations shown and FINALLY(!) solves the puzzle.

storm_norm

Joined: 18 Oct 2007
Posts: 1741

Posted: Thu Jan 03, 2008 12:36 am Post subject:

whew, I think I got tired just reading that

ravel

Joined: 21 Apr 2006
Posts: 536

Posted: Sat Jan 05, 2008 2:26 pm Post subject:

Asellus, congratulations on solving this monster. The first step is great and easy to follow (though hard to find). You also showed the coloring steps in a way i could follow, but it is not easy without graphics (or at least colors). The disadvantage of this powerful method is, that therefore mistakes can be made easily, if you dont be very careful. But i only found one typo: To avoid the Deadly Pattern, r4c7 must be <6>, hence it is colored "g", should say r3c7.

ravel

Joined: 21 Apr 2006
Posts: 536

Posted: Sun Jan 06, 2008 8:37 pm Post subject:

Asellus wrote:

As before, "RG" is standard Medusa and "rg" are implications from R-true or G-true, respectively. Eliminations are possible between R-G, r-G and R-g, but not between r-g.

Hm, i cannot see, why not. Either all r must be true or all g. So i should be able to eliminate each number safely, that both sees the same number colored r and colored g (in another cell). Or a number that has another one colored r in the same cell and sees the same number colored g and so on.

I only understand that if R is true, then not all g must be false (but all G) and vice versa.
Do you have an example ?

Asellus

Joined: 05 Jun 2007
Posts: 865
Location: Sonoma County, CA, USA

Posted: Sun Jan 06, 2008 10:14 pm Post subject:

ravel wrote:

So i should be able to eliminate each number safely, that both sees the same number colored r and colored g (in another cell). Or a number that has another one colored r in the same cell and sees the same number colored g and so on.

ravel,

After pondering this, I believe you are correct. For some reason, I believed that I had gotten myself into trouble once or twice with "r-g" trapping. But, thinking about it carefully, I can't see any reason that it would not be valid. I must have been confusing that detail with something else.

So, this "extending" technique is even easier to use than I already thought it was!

Thanks for reading the post so carefully! I'm amazed that there was only one typo.

In re-reading the solution above, I've discovered that some of the eliminations can be explained more easily. For instance, toward the end of the solution, I refer to a "r" {17} pair in r46c7 resulting in 2r in r4c9, etc., for a <2> elimination in r4c1. But, it's not necessary to invoke the {17} pair: there are only 2 <2>s in b6 and one of them shares a cell with 8r; therefore, the other one must be 2r. (And, I found a couple other such instances. Such strong link situations allow the coloring to be extended quite easily.)

Thanks for the helpful responses!

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