dailysudoku.com

[keith]

This one is worthy of attention, if you like them a little harder than "Very Hard".

[Asellus]

I used the {45} UR to get rid of the <5>s in R5C46, followed by a Skyscraper on <9> in C57.

[nataraj]

Without the UR, this is as far as I got:

[Asellus]

nataraj,

One way to crack that nut is with an XY Chain with a twist. Start at R9C4 and note that if it is <7> then there is a {45} pair in C4 making R7C4 a {69} bivalue. We will exploit that in the chain:

(9=7)R9C4-(7=9)R9C9-(9=1)R2C9-(1=6)R2C1-(6=1)R3C3-(1=6)R7C3-(6=9)R7C4

<9> is eliminated from R7C5 and from R2C4.

After that, some more XY Chains solve the puzzle.

This example illustrates something I often find useful on really tough puzzles: exploiting an ALS that can provide a locked set in chaining.

[Marty R.]

Keith, is "Freep" a Keith creation?

I used a Type 2 rectangle and three Finned X-Wings.

[nataraj]

Thanks, Asellus, for the "nutcracker". I couldn't log onto this site earlier so I tried some more coloring (starting from those "1"s in r2 and c9 and came up with a contradiction (color wrap) that made r2c9=1. The contradiction comes from assuming r2c9=9

a1) r2c9=9 => r2c5=2 => r7c5 not 2.
a2) r2c9=9 => r9c9=7 => r9c4=9 => r7c5 not 9

=> a) r2c9=9 => r7c5 = 5

but ...

b) r2c9=9 => r9c9=7 => r5c9=5 => r5c46 not 5 => r4c5=5

it is not possible that both r7c5 and r4c5 are 5, therefore r2c9=1

After that I did not encounter any more chains ... the puzzle solved easily
I will have a longer look into your technique of using an ALS in a chain because I feel this method could be useful quite often

Thx

edit 1403 GMT: "color wrap" is definitely wrong. Since I used different numbers in the chain it should probably be called Medusa wrap, plus I'm not even sure if it is really a wrap or rather a trap. Someday there will be a standard textbook on all this terminology ...

edit 1949 GMT:
come to think of it, probably even Medusa is incorrect, since the elimination is not based on strong links alone, but uses weak links or even branched AICs.
In this case it is definitely easier to stick with UR.

BTW, Asellus, your pointing out my reasoning was ER got me interested in ERs and I read up on the topic. Neat.

[Asellus]

nataraj,

I agree with your suspicions: it is not coloring, Medusa or otherwise, in any conventional sense. I'd say it is implication chaining that leads to a contradiction. Some would probably call it "forcing." Note that in your third chain, you made use of the {457} ALS in R5C46. (It becomes a {47} locked pair if R5C9 is <5>.)

Usually, we use an AIC to create pincers at the ends that effect eliminations. Sometimes, in very difficult cases, the AIC will have a branch or two. Sometimes, we discover a contradiction instead of pincers. I can't see any reason to ignore the contradiction just because it isn't pincers. For me, it's still logic, not forcing.

It is certainly possible to use colors to mark an implication chain. However, down this path it is only a short tumble to what even I would consider to be forcing.

By the way... your reference elsewhere about GEM sent me to look it up. Turns out I've been doing it without knowing it. I use dots (rather than digits) in a 9-grid pattern to pencil mark. To color, I use the "^" character, oriented up and down, in place of the dot. If I need a second color cluster, I use the same shape oriented left and right. Admittedly, this approach isn't for everyone. (It's sort of like reading Braille!) But, it has worked for me.

I'm going to try exploring this puzzle further using Medusa.

[Asellus]

Okay! Medusa does it. It's a nice example of the power of Medusa and only requires a single cluster (no Medusa multi-coloring... so only EM needed, not GEM, for P&P types).

Start coloring in R2C9 and quickly get here:

[keith]

[nataraj]

I really appreciate your taking the time, Asellus, to point out the (for me) previously unthought-of possibilities in finding clever and solid (as opposed to doubtful forcing) inferences.

To really understand what is going on I have to ask a few more questions, please bear with me:

1) I always thought coloring (and also Medusa) uses only strong links, so that ALL candidates of the same parity are either true or false and there is always a strong link between any two opposite parities (like in your grid, if r2c9<>9 then any of the greens must be true). All equivalence marks work in both directions. Pls confirm or correct this interpretation.

2) The reasoning in box 9, on the other hand, looks one-directional to me: if red then 5 must be true. (as it turns out, r is false and any implications based on r2c9=9 must be discarded - ex falso quodlibet). In my GEM coloring the 5 in r8c8 gets a "super" mark meaning (if red then r8c8=5) but not a parity mark. I must have missed a crucial step here.

And - as I said last night - I am not opposed to using the uniqueness argument, I just tried to avoid it if possible. Of course, it is SOOO much faster. Also, I don't mind the extra mile if the scenery is worth it, and I must say this time it was worth a hundred miles (learnt so much about solving techniques I wouldn't have looked into otherwise...)

[Asellus]

nataraj,

I'm glad you share my appreciation of the scenery!

To answer your questions...

Going back and pondering some more, I'd say that you've got me!

Yes, in basic Coloring, whether single digit or Medusa (multi-digit), the Color Cluster is limited to strong ("conjugate") links. However, it is sometimes possible to assign polarities due to implied conjugate relationships, such as can occur in a Box-Line interaction (a form of "group coloring") or when a bivalue pair can each see an instance of opposite polarities.

I thought something like that was happening here in Box 9 ... something akin to "survivor promotion" from super to par parity in GEM. But, I went and scrutinized those GEM rules and I believe you are correct: it isn't the case here.

However, I have convinced myself that the reasoning is otherwise sound, it just isn't Medusa Coloring. That alternate implication has sneaked back in the door! If red is true, there is definitely a contradiction; so red can't be true.

So, while it isn't strictly Medusa, it's still seems to be a useful way to proceed when other approaches are running dry.

[Earl]

Asellus,

There must be something in this discussion that I am missing.
Starting with your grid, the UR eliminates the 7 in R5C9, making it a 5 which eliminates other 5's and leaves a pair (29) in C5, eliminating the 9 in R3C5 which makes R3 in Box 2 a triple and the rest is singles.

Did I miss something? Are you trying to solve it without using the UR?

Earl

[nataraj]

yep.

Stictly speaking, Asellus didn't but I did (try to solve w/o UR)
Asellus is being helpful as always but just the same we are (ahem: I am* ) reaching the conclusion that it is pretty tedious to do it without UR.

As much as I enjoy the scientific debate about Medusa and other colorings - it is clear that the puzzle is already solved.

(still trying to digest the Asellus postings)

nataraj
_________

*: cf. "we are Bates" - Top Hat, 1935

[keith]

[Marty R.]