This week's Teaser is something a little different, more along the lines of an "exercise" than a full puzzle. Actually two, as I seem to have gotten in the habit of providing multiple puzzles out of guilt.

This is a polyking: Polykings are groups of squares joined at the edges or corners and are also called polyplets. They are less-discussed than polyominoes, so I'm not solid on the terminology, but I believe this one can be called a J tri-king.

In general, polykings are allowed to cross at places where corners meet.

1. Tile a 5x6 grid with ten one-sided J tri-kings. "One-sided" means you are allowed to rotate the tri-kings, but not reflect them. For example, you can tile a 2x3 rectangle, but only if you flip one of the J's over first; that's what I'm disallowing here.


2. An "irrep-tile" is a shape that can be fully tiled with similar copies of itself, usually of different size. (This is different from a "rep-tile," wherein all copies are of one size.) The picture below shows that the J tri-king is an irrep-tile with a tiling that includes 12 copies. (Note that one-sidedness is no longer in effect.) The exercise is to find a similar tiling that uses only 10 copies, or perhaps fewer.

I figured out at least part of my worries: size. I want puzzles I put here to be large enough to be interesting, but I'm lazy enough that I worry about whether I can muster the effort. Then again, with school having started yesterday, this isn't the time to lack energy. I don't know, it's always something, unless it's nothing.

The first Tuesday Teasers had 107 spaces and 162 spaces, respectively, and this one has 108 spaces, about 63 spaces per puzzle. But before I started TT, my puzzles averaged closer to 40 when they even had spaces. I think I should do a 6x6 something next week to balance things out.

Anyway, these Heyawakes were tricky for me to make. I used to wonder why there were no L-shaped rooms in normal Heyawake, but after making the second triangular one I now think rectangular rooms are hard enough. So I ended up making the simpler one you see on top.

I'm nervous. Did I set the bar too high last week? What are the high expectations for, specifically?

Okay, I shouldn't worry that much. My viewers (readers? visitors?) want me to do a good job, but they won't call me out on doing just an okay job. I have plenty of material, and I'm sure you'd be just as happy to see it if it weren't on a schedule.

So without further ado-

These are two Fillomino puzzles. They have the same clues, and there's a reason for that. The left one is normal, while the one on the right, helpfully labeled "Toric," is one of grandpascorpion's variants, Wraparound Fillomino. (I was also a little worried about having broken the symmetry, but none of those in his entry have perfect symmetry either.)

One reason people start blogs is because they're bored. I wouldn't say that applied to me when I started, but I do run out of things to do, at least voluntarily.

So in an attempt to start posting more regularly, I'm putting up a new feature, a counterpart to motris's Friday Puzzle: the "Tuesday Teaser." (I know, it's late.) I'll try posting a puzzle here every week for a while, see if I can handle it.

Up first are two of a Masyu variant I submitted to nickbaxter for his consideration. Of course, he made no promises about where/when/whether the idea would appear, so I finally decided to make it appear myself.

A loop drawn on a hexagonal grid by connecting adjacent hexagons will have a combination of straight segments, soft 120-degree turns, and hard 60-degree turns, so it would make sense to have three kinds of pearl in the grid. The loop makes a hard turn when going through a black pearl, but must not turn in the neighboring cells (those immediately before and after on the loop, as distinct from the up to six adjacent cells). Also, the loop goes straight through white pearls, but makes the same size turn in both neighbors (the directions don't need to match). Finally, the loop makes a soft turn through gray pearls, makes any turn through one neighbor, and goes straight through the other. Here are the puzzles.

So my mom is on FaceBook and saying I should join. I'm wondering how many sites is appropriate to join, and how many is too many. I like furry art, so I browse FurAffinity; should I join that site? Either way is free, but one allows access to poarm, and that is not an advantage to me.

But I'm wandering. The question is, when does joining a site make sense, and when is it not worth the effort?

That second part is a little tricky to codify, in part because the amount of effort tends to be quite small. I think I'm a member of eBay, but haven't used the account because I don't have a credit card. Should I spend more effort and get one, or did I already overinvest?

Right, so I don't get penalized for not using a site, but if I'm a non-using member, I'm a slight drain on bandwidth. Is that a good enough reason not to be a member, or is there a better reason? Or should I join anyway?

I'm trading puzzles with grandpascorpion. I chose to do his Slitherlink variant, OctaLink, and as you may guess, I themed it on the octagonal stop sign.



In normal Slitherlink, some squares have numbers ranging from 0 to 3 that tell how many of their edges are part of a loop. OctaLink adds octagons with numbers from 0 to 7 with the same meaning. And to be sure, shapes along the grid edge always have the same number of edges as the internal shapes; the cell with the green 6, for example, has five internal edges and three (not 1) external edges.

Happy 4th of July!

This week, I played matchmaker between Erich Friedman and Alexandre Muñiz. (These links contain some necessary background on weightominoes and sumominoes, respectively.) As sort of a thank-you, Alexandre sent me a problem reminiscent of Erich's extension tilings (problem 5) last month.

Is there a n-omino for which all the weightominoes with weight n+1 formed from it can tile a rectangle with equal weight in each square?

To be clear, a basic n-omino has weight 1 in each square cell. Let's define a "weight-extension" as a polyomino where 1 of its cells weighs 2 and the rest of its cells weigh 1. Then there may be up to n weight-extensions of a given n-omino (less if the original is symmetrical), and the problem is to cover a rectangle with these weight-extensions so that every square weighs the same.

Of course, I don't know whether there's an answer for any polyomino (except for the trivial 1x1x2 for the monomino). But my intuition suggests that an asymmetrical octomino with extensions of weight 9 would be a good entry point.

I may not be old enough to remember the good old days, and there probably weren't that many, but I get nostalgic for when the stock market meant something to people. I mean, money makes the world go 'round, don't get me wrong, but recently money's made the world go pear-shaped.

See, markets are places to buy and sell goods, and the stock market used to be similar, except it funded the production of those goods. More recently, it became entirely about making money and thus the collapse. You can't just hedge your funds and siphon out cash indefinitely.

The problem is based in human nature, though it manifests in money's fungibility. What that means is that people get greedy for money precisely because it lacks inherent meaning. So what's the answer?

An incomplete answer is that folks should invest only in things they have interest in. This is one of the assumptions Adam Smith made in conceptualizing the Invisible Hand of the Marketplace. The theory is that popular businesses will automatically survive; this theory doesn't take into account the possibility of impossible or unsustainable models, like making money from nothing in a zero-sum game. But please keep in mind that inasmuch as we can determine what's "right," it won't always match what's popular.

EDIT, Disclaimer: I have not read Smith, and I really should have.

Introducing Charles Babbage and his Amazing Calculating Machines!



This is a cryptic KenKen. Each digit is replaced uniformly by a different letter from A-K (skipping I). If you choose to print out the puzzle to solve it, you can write the decrypted clue numbers to the right of the operations.

Edit: To be sure, the spectrum (the numbers in each row/column of the Latin square) is indeed 1-7.

This one gave me a bit of trouble in the making. I finally got the right dividend in the shower.

    _    PA
WASH)PLASMA
     RWYWR
      ARSAA
      LAWUR
       PLLS