Friday, 16 March 2012

TVC XI Practise Part 3

Here are the last few examples for TVC XI. They are a bit on the late side, but I had only 3 days. I did what I could. These genres were the trickiest to construct, so that's why they come last. I don't know how representative of the test they will be as they can be a lot harder than what I made now. I just needed puzzles that I could logically construct. I am also learning the logic involved in these genres.

First up is the Full Tapa. I had a lot of trouble to come up with an idea for this puzzle. When I settled on the 3 and 4 word lengths it slowly got started. I like how it worked out. Most of the puzzle runs on the word lengths, but you need to place the words to get to a unique solution. I assume it's easier than the actual test puzzle, considering the amount of points it's worth.
Secondly is the Power of Tapa puzzle. I don't think it's especially hard. Tried to get a mix of power and normal clues in, but didn't succeed as well as I had hoped. It's still a nice puzzle. The puzzle in the test is worth a the most points of all puzzles, so will most likely be harder.
Lastly the Meiosis Tapa. It's really tricky to construct because there are a lot of options for multiclue cells. Which is also why I avoided multiclue cells for the most part of the puzzle. I at least succeeded in implementing both types of division in the puzzle. It's not too hard, but as the points table goes, it could be of equal difficulty of the test puzzle.

Puzzles can be found below.

Rules for Tapa

6. Full Tapa

Use standard Tapa rules. Additionally all empty cells will be filled with the given words. All words have to be written across or down. Each word has to be used. All words formed by consecutive letters has to be in the list. No other words can be formed.

Click to enlarge

8. Power of Tapa

Use standard Tapa rules. Additionally each clue either represents a normal Tapa clue or an exponential operation, where the result of this operation will be used as a Tapa clue.
Note: 0^0 is undefined and won’t be used. Otherwise, a^0=1; 1^b=1; 0^c=0; d^1=d; e^f^g =e^(f^g).

10. Meiosis Tapa

Use regular Tapa rules. Additionally each clue in a cell can undergo meiosis. This means that the clue can either be divided by two or split into two equal halves. These results are then used as Tapa clues. In case of a decimal result of division, each digit in the result functions as a Tapa clue. Multiclue cells may have divided and undivided digits together. Digits as a result of meisosis can't undergo meiosis again.


  1. Thanks for all the practice puzzles. You put this batch just as I was about to take the test, and then I did this before taking. They all helped quite a lot, even though I've messed up yet another test overall. At least I wasn't completely clueless :P

  2. (I can't tell if my other comment got through, actually, so I'll partly repeat myself.)

    For Meiosis Tapa, the instructions are a bit unclear on what "two equal halves" will look like. For your puzzle, I figured out that this means a clue X might be replaced by two X/2's, but I think the intent could be that X can be replaced by two X's. The example on the test doesn't even use the rule in question, and neither variation creates an extra solution.

    1. Sorry. I guess when I say split into half, it means the result is twice half of something. But for the test it was meant to be x/2+x/2 as well. I actually completely verified how it should work before making this puzzle, which is why it's one of the last.
      And yes, your other reply somehow was conceived as spam by blogspot. I'll just leave it there as you already repeated yourself in this one.