I like the new square sizes for Fractional. Mad Max and Full scare me...

* Digitail Tapa: "There cannot be a zero in a multi-number clue cell." not true as you seen in the example. This sentence should be " There may be one zero or more zeros in a multi-number clue cell.

In the power of tapa, the 2 1 2 clue doesn't seem to be correct.

2^1^2 clue may be 2 or 212. I think there is not a problem in the example.

When I first replied to this problem, I couldn't remember why I made the puzzle that way, so I said there was a problem. But when I attempted for a fix I remembered that: I made the puzzle that way because large cells counting as one cell could be confusing for neighbouring boxes' types. So I counted them as multiple cells. But we accidentally used the actual puzzle's instruction. So this puzzle type is not Irregular, maybe it can be Ir-irregular, thus making it a regular Tapa with different cells.

Meiosis: There is a possibility of single digit(even) clues breaking down to a single digit again(even or odd). Is that allowed? This will let two-digit clues breaking down to 2-digit clues again.the example does not show such behaviour.

As I understand correctly, you are asking this: If there is "2" in a cell, can it be only "1" after division, instead of "1-1"? The answer is no. Division here is not mathematical division, it is biological cell division, so something splits into two and both elements should be there.

Now I'm well and truly confused .
If a "2" splits into "1-1", shouldn't a "1" split into "0.5-0.5" i.e. "0-5-0-5"? According to the way "1" is divided only into "0-5", "2" should be divided only into "1" I suppose.

alright, thats clear now.However,the practice puzzle i made follows my earlier understanding of splitting an even number into a single half.An even digit splitting into 2 parts might feel something like math tapa with the + operator.
looks like Tapa is slowly transforming into applied tapa.Reproduction,cell division. Physics is next i think(i have one variation on that theme).

Edited by anurag 2012-03-14 11:14 PM

I'm sorry, I'm still not clear on Meiosis Tapa. As I understand things now,
-- a 1 can either be used as a 1 or substituted for 0 5 (aka just 5)
-- a 2 can either be used as a 2 or substituted for 1 1
-- a 3 can either be used as a 3 or substituted for 1 5
-- a 4 can either be used as a 4 or substituted for 2 2
etc.
Is this correct? Also, can a clue divide twice? (4 becoming 2 1 1 for example)

Also, for Balance Tapa, for a puzzle of the size in the IB, is it only saying that the total number of black squares in columns 1,2,3 is the same as the total number of black squares in columns 5,6,7?

Edited by MellowMelon 2012-03-15 12:09 AM

But I just finished the Combined Tapa that way........

I figured it would just work the same for any edge clue. Although I didn't have any clues bordering the irregular tapa part.

I'll wait with the meiosis when it's completely confirmed.

Edited by Para 2012-03-15 12:28 AM

Dear puzzlers, Gulce speaking. Serkan is so busy nowadays and I was the one to write the comment of Serkan above, didn't notice I was logged in with his account :) There is indeed a confusion with the rules, we have to set a constraint. Current example allows 2 becoming 1. Now I talked to him, he'll fix the rules when he gets back home. Sorry for causing trouble

I've posted the first 3 practise puzzles. Tomorrow the next 3 or 4 will be posted. The last few will come out on friday.

http://puzzleparasite.blogspot.com/2012/03/tvc-xi-practise-part-1.h...

No worry. No real hurry for it. As long as it gets resolved.

Edited by Para 2012-03-15 2:25 AM

Is relative order preserved in the Visionary Tapa clues? e.g. Is a 1234 different from a 1324?

Order of the digits is not important, 1234 and 1324 are the same.

Corrections on the instructions are stated below. Meiosis Tapa will have a new example puzzle, edited version of the IB will be posted soon.

1. Combined Tapa:
Digital Tapa: Any digit has the possibility of being zero.
Ir-irregular Tapa: Each irregular shape counts as many cells as the unit squares it contains.

10. Meiosis Tapa:
Some of the given clue digits may be divided in half. This may happen in two different ways: Digit splits into two and creates two identical digits; or digit is only divided by two and results in a single number. Multi-digit clue cells may have divided and undivided digits together. If a digit divides in half and results in a decimal, each digit in the result counts as a new Tapa clue (e.g. if the original clue is 3, it becomes 1-5 after the division). Resulting digit of a division cannot be divided again.

So if I get it right a 22 clue can be a either of the following:
22 (no division)
21 (one divided by two)
211 (one divided in two)
11 (two divided by two)
111 (one divided by two, one divided in two)
1111 (two divided in two)

Edited version of the IB is uploaded.

Yes, all cases are possible.

Sorry,this is a bit ridiculous,in my opinion.Having so many possibilities for a clue may only increase confusion.Its hard to figure out possibilities from the large mappings during the test.I believe it should be as simple as possible.

Maybe there should be a small clarification to the ir-iregular Tapa rules. Currently it reads that each shape counts as as many cells it contains. But that's not exactly correct as it could be interpretted that a 3 cell area which touch the Tapa clue with 1 cell is counted as 3 cells, even though it's only meant to be counted as 1 cell.
It's supposed to mean each cell in a shape counts towards a tapa clue if it's touching the clue.


On a side note. Can anyone explain how to approach the Full Tapa example? I don't see a way to get started.

Second set of practise puzzles have now been posted. Hope they help.

http://puzzleparasite.blogspot.com/2012/03/tvc-xi-practise-part-2.h...

How exactly does the answer key work for Fractional cells? If you look at row 4 or row 7 for example,it can get a bit confusing.


For row 7 is it
1,2,4 (count all squares that are touching as 1)
1,2,3 (the bottom of the fractional cell only)
1,2,1,1 (the top of the fractional cells only)
1,2,2.75 (all cells touching, but fractional cells counted as fractions like in the clues)

For row 4 it's even so that the fractioned cell has two seperate pathways running off the same cell.

Deb edited to add the solution image.

Edited by Para 2012-03-16 12:13 PM

If R6C6 is used by a vertical word,it can be One ,zero,two,six or four.One,two,six fail as it leaves a white space of 2 cells in the top-right corner.Four or zero leave R6C3-4 white,so they fail too.that leaves r6C6 with a horizontal word,which is atleast 4 long to satisfy the black block in the last row.A 4-letter word would leave a V-shaped 5 which leaves a white space in R5 that cant be filled.So only three can use the 6th row.This completes 6 and the entire right half.R1C6 must be black.Done!

Edited by anurag 2012-03-16 5:52 PM