Easter Monster Round 5


The following is the fifth page of an illustrated solution of the Easter Monster.

If this is your first visit to this blog, WELCOME!!

Previous blog pages may be helpful. Links to these pages are found to the right, under Previous Entries. Specifically, it may be helpful to have visited the following pages:

The illustrations of steps shown in this proof will share the new key style:

  • black line = strong inference performed upon a set (strong link)
  • red line = weak inference performed upon a set (weak link)
  • black containers define a partioning of a strong set(s)
  • candidates crossed out in red = candidates proven false
  • Orange labels mark derived inferences
  • Blue circles indicate proven strong inference set result
  • Green circles indicate intermediate strong inference points
  • Brown numbers indicate approximate chain order
  • Other marks provided prn
Please be aware that, for me, strong and weak need not be mutually exclusive properties.

Previous information

Previously, many derived inference groups were proven. This page will use the following sis groups:

  • Each set in Group Z below contains exactly one truth:
    • (27)r2c13,(16)r2c56,(16)r2c79
    • (16)c8r13,(16)c8r45,(27)c8r45,(27)c8r79
    • (27)r8c79,(27)r8c45,(16)r8c45,(16)r8c13
  • The result column sis from step 2g
    • (2):[r4c5,r6c7,r4c1]

Step 2j prelude using Kraken Cell Transport.

Since this puzzle resists most normal attempts to crack it, it seemed likely that some research might be required into relationships to locate good attacks. On the previous page, the guardian sis was used to locate a likely chain. Here, the relative symmetry of candidates 345 in relation to cells r7c9 and r9c7 caused me to create a transport list out of each cell.

Basically, the goal is take the sis inherent in a cell and find some equivalent sis at least partially out of the cell. Although I created many such lists, some of them seemed more likely to be fruitful than others. In this case, the transport list for r7c9 seemed more likely to be fruitful than the one for r9c7, primarily because candidate 67 interactions uncovered in the guardian step.

Kraken Cell partial transport

Above, I am illustrating three short transportation chains.

  1. (349=5)r7c9-(5)r7c46=(5)r9c46
  2. (459=3)r7c9-[(3):r79c8=r3c8-r3c4=r7c7]=sis(3):[r1c48,r9c4]
    • Note that there are many such chains possible, but I wanted one that interacted strongly with canidates 67.
  3. (359=4)r7c9-[(4):[Xwing r79c25]=r1c2-r1c7=r3c9]=(4)r3c2
    • This one is nice, as it requires only a single value transported out
PUtting all three of these together we get:
  • Old sis (3459)r7c9 =>
  • sis[(35)r9c4,(5)r9c6,(3)r1c4,(3)r1c8,(4)r3c2,(9)r7c9]
This new sis seems a bit unwieldy, but a short look at some of the previous chains considered makes it very promising. The only item that cannot be resolved easily is that darn (9)r7c9. However, one may see a potential logjam primarily because of sis[9 column 3].

Step 2j (2)r2c1=(6)r6c1

Below, the transported portions of cell r7c9 are labelled with a green k. The rough order of chained sis are marked with brown numbers. Only strong inference sets are marked. The Z argument group is used twice, and labelled accordingly.

Kraken Cell used in a chain/net

The easiest way to describe such a long chain is probably a Triangular Matrix

2B1 r2c1 r3c2
Z27 r65c27r6c2
7r1 c2c4
7c6 r2 r9
6r6 c1c2 c6
1r6 c2 c6c3
1r8 c3c45
6B8 r9c6Zc8 r9c4
6r1 c4 c6 c8
2jK ** * **9r7c9
r6c9 7 93
r2c9 63 8
r2c1 2 83
3B4 r6 c1r5c3
9c3 r6 r7 r5

It is a bit unusual for a Triangular matrix of this length to have only two items listed in the first column. The result column reads:

  • (2)r2c1=(6)r6c1
  • =>r6c1≠2
This leads to a few relatively easy eliminations, although all of them use proven sis.

Step 2k as an immediate result of <2j> and <2g>

Below, recall step 2j. Proven in that step was sis(2)[r6c1,r7c6,r4c5]. However, we have just shown that (2)r6c1 is false, therefor we can update the strong inference set to:

  • 2g 2:[r7c6,r4c5]

Trivial step

The logic for this step is fairly simple

  • (2)r4c6=(2)r6c7 => r4c7≠2
Remembering what has been already proven certainly makes that one a breeze.

Step 2L as a Y Wing Style using 2j

Below, recall 2j sis [(2)r2c1=(6)r6c1]. Add a pair of native strong inferences.

Y Wing Style

The AIC for this one could be written:

  • (1=6)r6c6-(6)r6c1=2j(2)r2c1-(2)r4c1=(2)r4c5 => r4c5≠1

Step 2m as a depth 4 chain using 2j and Z

Below, recall 2j sis [(2)r2c1=(6)r6c1]. Add one native strong inference and use strong inferences from the Z group twice.

Chain of length four using primarily derived inferences

The AIC for this one could be written:

  • (1)r8c3=Z(6)r8c1-(6)r6c1=2j(2)r2c1-(2)r4c1=(2)r4c5-(2)r8c5=Z(27)r8c4
  • => (1)r8c3=(2=7)r8c4 => r8c4≠1
After just a little practice, using derived strong inference sets seems just as natural as using native strong inference sets. Although the structure of a derived strong inference set may be a tad unconventional, the logic is precisely the same.

Digression

Some large and unconventional weaponry seems warranted with this particular puzzle. The practical use of most of these ideas will be very small. However, mini-cases of the same logic type are possible. In all but a very few puzzles, the depth and complexity found here will not be encountered.

I have a few more unconventional attacks to attempt. On the next page, I probably will employ an sis derived from an Almost Unique Rectangle - more aptly named, I think, a Forbidden Rectangle.

Thanks for your patience!




4 Comments
Indicate which comments you would like to be able to see

Hi STEVE,

Nobody work at others Site when you start...EM!
For me, I have one question... (forgive me, please) : How many rounds for this...? X'mas come in...
05/Dec/07 3:47 AM
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Steve,

I don't recall seeing the terms transport list or transportation chain in any of your other blogs. Can you please tell me where can I read about these?
05/Dec/07 2:22 PM
 |  |
Hi ttt!

We are getting close. Next round will solve some cells. Then the puzzle starts to collapse, especially recognizing some of the absurd symmetry. Hopefully, it will finish long before Christmas!
05/Dec/07 6:01 PM
 |  |
Hi Jeff!

I am probably not the best person to ask. I generally read the Eureka forum and a little bit in the Players Forum. I sort of take what is written and translate into my own way of thinking.

To my way of thinking, every native strong set has an effect on the puzzle. The idea More...
05/Dec/07 6:07 PM
 |  |
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