Had a puzzle idea for yesterday. Didn't get it drawn and up because... I couldn't think of a name? Maybe it could go into the Hunt? I don't know, I just have excuses. Here's a different puzzle.

Backgammon has five dice: two for each player (though it can be played with a single pair of dice) and a "doubler die." The doubler die is used for scoring over multiple games, but what you need to know here is that the numbers on its faces are 2, 4, 8, 16, 32, and 64. So 2^1 through 2^6.

1. Take a large doubler die with its numbers written small enough for this task to make sense. Write some more digits so the faces display, in some order, the positive integers x, 2x, 3x, 4x, 5x, and 6x. Minimize x.
2. Take a large standard die with small writing instead of pips. So 1-6 are written on the faces. Add digits so the faces display the positive integers y, y^2, y^3, y^4, y^5, and y^6. Minimize y.

(Tricks optional.)

Take a six-letter word, convert it to Braille, and write it in two rows of three letters. You'll notice this forms a 6x6 array of dots. Is there a starting word such that turning this array 90 degrees results in another word?

Braille and other codes can be found here (enlarged Braille in lower right). A 180-degree rotation doesn't work, since only T and X match themselves and only R and W match each other.

(Program solutions welcome.)

One item on my current roster of excuses for not posting here more often is my new presence on Facebook. A more relevant item is the 2012 MIT Mystery Hunt, which happened over MLK Jr. Day weekend. It's more relevant because I'm on the Manic Sages, this year's winning team. And so, tradition calls for us to write next year's Hunt, which would in theory reduce my availability to post puzzles.

Yeah, not so much. I have ideas I can't share, sure, but they tend to be grand, impractical things. This isn't one of those.

In each heavily-outlined region, shade four squares to make a tetromino such that all shaded squares form a single polyomino. Lines of four consecutive squares in one row or column are not allowed (though 2x2 squares are).



This is a LITS variant by way of the Tapa Variations Contest. For Tapa Line, lines of four are not allowed, but neither are 2x2 squares. I figured, why not use just the new restriction?

(By the way, the other three tetrominoes don't work. When there's more than one shape, forbidding O4 is equivalent to forbidding P5, which would be automatic if L4, S4, or T4 were forbidden.)

Cross-posted from FurAffinity, for those of you who fancy yourselves detectives.


I was walking with the triplets to the Dollar Theater for a screening of "A Night at the Roxbury," when Russell asked, "What's wrong?"

I guess I must've been looking at them funny, but I dunno. "I just can't get over--" I sputtered. "I mean, you have three heads on one body. I keep expecting you to break into a true-false puzzle, where I have to figure out who's lying."

They chuckled. "What," Beryl said, you mean like..."

Cyril: "At least one of us is telling the truth."
Beryl: "At least two of us are telling the truth."
Russell: "All three of us are telling the truth."

"That?" finished Cyril.

"Well, yeah, but that's pretty simple," I complained. "Four possible answers, depending on how many of you are actually telling the truth."

It was then they decided to indulge me.

Head A: "Well, at least one of us is a liar."
Head B: "At least two of us are liars."
Head C: "All three of us are liars."

I was gazing ahead at that time, and they all sound the same, but I turned to smile at them. "Okay, that's getting interesting. In these puzzles, Liars always lie, and Knights always tell the truth, but there's an odd number of you, so that can't be the whole picture." I scratched under my horn. "But sometimes there are Knaves in these puzzles, and they strictly alternate between truths and lies. How many of you are Knaves?"

Head I: "Exactly one."
Head II: "Exactly two."
Head III: "Exactly three."

I mulled this over. Finally, I stepped in front of them. "Okay, I've figured out how many of you are which kind, but I don't remember exactly who made which statement," I confessed. "So now I ask each of you: among the three statememts you just made, did you repeat a number?"

Cyril: "...Nooo..."
Beryl: "...Yeeesss..."
Russell: "...Nooo..."

"All right, end of puzzle!" I pulled out my notepad and scribbled excitedly.

Just as we got to the theater, I told the three dogs which statements I thought each made. "Is that correct?"

"...Yeeesss..." they said in unison.

"Really?" I perked up my ears.

"...No."

"Aw." I put my pad away. Of *course* they hadn't decided what to be ahead of time, and they couldn't all have been Knaves the whole time. "Well, it makes a good puzzle, at least."

Since it's likely that people missed this, I'm repeating myself and changing the date.

On Sunday, January 22, the price of a commissioned ambigram will go up from $10 to $15. If you purchase one on or before Saturday, January 21, you can pay the current price of $10. Other details are consistent with what's here.

Some of you may be wondering whether I have any New Year's resolutions, like posting Tuesday Teasers on a tighter schedule. Well, I'm sorry to say, not really.

At least, nothing I wasn't already thinking about pretty continuously. I have trouble actually resolving to do things, and it doesn't carry special weight around the 31st and 1st. (The 3/1st?) Which isn't to say there's nothing special I'd like to resolve to do, just that I know I'd only have fits and starts of interest.

Eh, more after the puzzle. And sorry, this one isn't really punchy, so it's probably not that good.


Somewhere around here, I have a die with eight sides (a d8, an octahedron), and those sides are naturally labeled 1-8. As with the die from my first puzzle, this die is loaded so that the numbers from 1 to 8 are not all equally likely to come up.

Question 1. The die has four powers of two on it-- W, X, Y, and Z, in some order. The probability that rolling the die results in one of these numbers is WX/YZ. If that fraction is already in lowest terms, what is it?

Now, to determine eight values, you need eight equations. Question 1 gives the first, and of course, P(1, 2, 3, 4, 5, 6, 7, or 8) = 1 is the second. The other six give the expected value (EV) of a subset, so I'll have to explain what I mean by that.

Say the probabilities of 1, 2, 4, and 8 coming up are 3/20, 1/10, 1/4, and 1/20. If we know the roll is a power of two, we can just use their relative probabilities, which have ratios of 3:2:5:1. Then the EV of a power of two is (1*3+2*2+4*5+8*1)/(3+2+5+1) = 35/11, or 3 2/11.

As it happens, I chose the weights so that the EVs of certain subsets would be integers.

The EV of a prime is a.
The EV of a composite is b.
The EV of a Fibonacci number is c.
The EV of a triangle is d.
The EV of a square is e.
The EV of a pentagonal number is f.

...Yeah, I couldn't decide right away. But I came up with a-f independent of W-Z, and I can tell you some other things.

Question 2. Each of a, b, ... f is a distinct integer, and each one can be said to "represent" a different one of the six types in the list above. That is, each has at least one of the six properties, and representation is assigned so that each type is represented by a different number with its property. For example, if 8 represents the Fibonacci numbers, then none of 1, 2, 3, or 5 represents them, and 4 or 6 must represent the composites. Given all that, what are the six representatives? (You don't have the information for letter assignment yet.)

Here we go.

Question 3. Now, a is not prime and doesn't represent the primes. It represents another type, and I've picked a-f so that there is one EV-representation loop connecting all six types. If none of the subsets has an EV of the same type, much less its representative, that means three possible loops. What are they?

Question 4. Well, only one of the loops actually worked. So which was it, and what are the probabilities of each number 1*8?


...All right, let me get back to my non-resolutions. My hopes for the new year fall into two broad categories: be more productive, and get paid more for it. The first includes ambigrams and puzzles, but also posts here and elsewhere, other writing, chores and other work, and perhaps some art. So... not that focused. That automatically makes resolving to do it difficult, or at least squishy.

But the second one is looking better. I actually got commissioned for an ambigram at a New Year's Eve/Day party and finished it on the 3rd. Still, Mom suggested the commissioner might easily have paid $25, so I'm thinking of raising the price.

If you commission me on or before the 16th, you can pay the current price of $10. After that, the price will be $15 (not $25). Comment or note me for coordination.

This puzzle is dedicated to terrific coincidences, rather than to any one person's body of work.



I wish I could have done more with this variant, but it's a good proof-of-concept. The top grid is two Spiral Galaxies puzzles overlaid, with no two clues coinciding. To solve this puzzle, move each galaxy symbol to the same relative position in one (and only one) of the two grids below, so you can solve each of them as a normal Spiral Galaxies.

Right. Some of you might need directions for solving a normal Spiral Galaxies. Well, here's how it goes: Each gray swirl is the center of a "spiral galaxy," a polyomino with rotational symmetry. On a square grid, this might be 90-degree rotation or 180-degree rotation. So, given that the galaxies take up the full area of the grid and have the given centers of rotation, divide the grid and identify the galaxies.

I just sent a message to OpenID asking them to fix my inability to comment with my LJ profile. In the meantime, I have comment on another of Anurag's puzzles.

Minesweeper Tapa
It's a pretty interesting idea. Unfortunately, there's ambiguity about which of R7C6 and R7C9 is a minesweeper clue, and which is a tapa. One choice gives one answer, and the other gives 18.

I'm getting another idea, though. I should take a whack at a Tapa Repossession (actually, I think I made that name up myself) where each tapa clue has an extra number, a Minesweeper clue. When's your birthday, Anurag?

Ayup, that's a hiatus alright. So here's a puzzle!

     _     YO
FUNNY)WHATNOT
      WCONUN
       HNWCNT
       HUHWOA
        HHFYH

Also, I've been unable to comment on Anurag's puzzles because OpenID is throwing a hissy fit. There aren't many workarounds for that, but I remember that he visits my blog too. So it might work to comment on those puzzles here.

http://anuragthefirst.blogspot.com/2011/12/fillominogarbage-collector-medium.html
That went well. I did get a little stuck halfway in, but I broke through with the 4 in the lower left.

http://anuragthefirst.blogspot.com/2011/12/lilits-hard_03.html
This puzzle currently has five solutions. If the polyomino in C10 occupies rows 5-7, there are two solutions; if it's rows 6-9, there are two more; and for rows 8-10, one more.

http://anuragthefirst.blogspot.com/2011/12/fillomino-connected-medium.html
Sorry, but this one's a mess. For starters, the rules would confuse me if I weren't pretty sure I'd formulated the rules. I mean, "1s may or may not be connected"? Fillomino connectivity is defined by edge-adjacency, so 1's aren't connected to each other. Perhaps rewrite it as "1s, being neither prime nor composite, contribute to neither region."
Even without that confusion, the 2 and 3 in R4-6, C9-10 can perform a small flip between R5C9 and R6C10. There are multiple solutions! If I ignore that region and then assume R7C4=2, there's still about a dozen solutions.