# Puzzle 93: IVAN

Scrabble Fill a letter in some squares such that they form a Scrabble: all cells with letters are connected, every word on the right appears in the grid as a contiguous sequence of letters (not broken with other letters or empty cells) reading right (in the same row) or down (in the same column), and every such contiguous sequence of two or more letters form a word.

Expected difficulty MediumAnswerComment/E-mail if you want a solution to be published

Puzzle 93: IVAN
Scrabble

Surprise, a 5×4 puzzle having a medium difficulty. Actually I’m not sure you can solve this without brute force, but given the very small grid I think it should be easy enough to carefully enumerate all possibilities.

# Puzzle 92: Word Puzzle?!

Bonza Word Puzzle Arrange the pieces given into a crossword pattern, like in the game (or puzzle genre) Scrabble, such that every contiguous sequence of two or more letters read left-to-right or top-to-bottom spells out a word, which are thematically linked. Also see Grant’s take on this.

Expected difficulty EasyAnswerComment/E-mail if you want a solution to be published

Puzzle 92: Word Puzzle?!
Bonza Word Puzzle

Let’s say I’m not too inspired.

On the other hand, I actually have the game. Of course, I’m always biased, preferring free games way more than games that include necessary in-app purchases (this includes Bonza for locking some of its puzzles, even if there are packs available with coins), but I suppose I should stop here before trashing more on the business model which I myself can’t understand why I loathe so much. The idea itself about “jigsaw crossword” is amazing. (By the way, I should have put the genre name as “Jigsaw Crossword” if I want to be neutral, but eh.) I might tinker with the idea again some time in the future.

# A Good Snowman Is Especially Hard To Build

Yesterday I bought A Good Snowman Is Hard To Build, which is probably one of the rare occasions I actually bought a game. When I’m going this far to write a blog post about it, it’s either very good or very bad; A Good Snowman is undoubtedly the former.

On the other hand, because I cannot write a review (and not that this post is supposed to be a review anyway), I’ll just recommend everyone reading this blog for its puzzle contents to go buy it. Or not right away; it is sold according to the temperature for the first two weeks, and I got it when it was $7 yesterday. (Now it’s$10, so hope for London’s temperature, where the price is taken, goes down soon. After two weeks, it seems like the price becomes a flat \$12.) Stock market kind-of thingy yay. If you’ve bought it and don’t think a 1.5-hour game (I completed the first thirty puzzles in 1.5 hours) doesn’t worth the price, then you haven’t found the second half of the game…

The rules of the game are unfortunately not explained explicitly (but you can actually figure them out pretty easily). But because this blog is mostly for deductive puzzle enthusiasts, which loathe MIT Mystery Hunt-style puzzles with absolutely no instructions, the rules of the first part follows. The rules for the second half of the game won’t be written here, because they are just so amazing.

You’re in a room with various walls, Sokoban-style, as some sort of featureless monster thingy. You can move in four cardinal directions. There are several snowballs on the grid, as well as some snow on the ground. You can push snowballs around on the ground, but cannot pull them, just like in Sokoban. You also cannot push snowballs to the wall.

Snowballs come in three distinct sizes: small, medium, and large. Rolling a snowball over a snow on the ground increases its size and removes the snow from the ground; a large snowball simply “absorbs” the snow, remaining a large snowball. A smaller snowball can be rolled on top of a larger snowball, Tower of Hanoi-style, and can be also pushed down. You cannot roll a larger snowball to a smaller (or equal-sized) snowball.

To win, use all snowballs to form snowman(s): a stack of three snowballs (which must necessarily be large, medium, and small from the bottom to the top). In case there are multiple snowmen to be built, once you form a snowman it cannot be disassembled any more. Example follows.

Sample puzzle follows. Solution is right beneath, so be careful.

A Good Snowman sample puzzle

In the above, the black circle is you. The green squares are normal ground, without snow, while the white squares have snow. The gray squares are walls. The part to the right is a reference for snowball sizes, and what you’re aiming for. This puzzle is created by me, so you won’t be spoiled with any of the puzzles in the game, although you might be spoiled on some of the tricks.

# Kalah Always Ends

Kalah is a game in the mancala family, whose rules can be seen in the given Wikipedia link.

Theorem A game of Kalah always ends, no matter how the players play. In fact, this can be generalized heavily: no matter what the initial distribution of stones is, no matter how many houses are in each row, no matter how the players play, a game of such generalized Kalah always ends (although probably will take a long time).

The proof is actually pretty simple, and is inspired from Goodstein’s theorem.

First, observe that a stone that falls to a house will never move back out. Thus we have a monovariant: the total number of stones in the houses will never increase. For simplicity, let the number of stones in the houses (not in the store) be called the free count (for count of free stones); the monovariant says that the free count will never increase.

Also, one seemingly moot but important point is that the free count is a natural number (a non-negative integer). Why does that matter? Because armed with that, we can invoke a powerful theorem, the well-ordering theorem: any set of natural numbers has a smallest element. Or in other words, any decreasing sequence of natural numbers will eventually reach zero.

Why does that matter? Well, let’s suppose we can show that the free count cannot last the same forever. That is, at some point it must change. It doesn’t matter how long it is until the change, and how big the change, as long as it eventually changes. Due to the monovariant above, it cannot become larger, and thus it must become smaller. Now we can use the same argument again to show that the free count will become smaller again, and then even smaller, and so on. This naturally is a sequence of free counts, which means a sequence of natural numbers, and a decreasing sequence at that! Which by well-ordering theorem above means it will eventually reach zero. The free count will eventually reach zero. This means there are no stones in play, so the game ends. (The game could have ended earlier, when one player has more stones than the opponent or when one player has no stones in houses, but that doesn’t matter, since it still means the game ends.)

Good, now we get almost everything! Only that we’re missing the key argument: how do we prove that the free count cannot last the same forever? Well, now here’s where my inspiration from Goodstein’s theorem comes in.

Take one player. Label their houses as $0, 1, 2, \ldots, k$ from the closest to their store, and the stones in it $a_0, a_1, a_2, \ldots, a_k$ respectively. Now suppose $\omega$ is some sufficiently large number (in fact, two suffices, but I originally thought about the smallest infinite ordinal (the next number that comes after all natural numbers; you can go scour ordinal number about it)). Consider the number $A = a_0 \omega^0 + a_1 \omega^1 + a_2 \omega^2 + \ldots + a_k \omega^k$. What happens after a move by this player?

There are two cases. The first case, the move passes the store or makes a capture, so the free count decreases and we’re done (remember that we now want to prove that the free count must decrease at some point). The second case, the move doesn’t pass the store. This means $A$ decreases by $a_n \omega^n$, where $n$ is the hole where the move is taken, and increases by $\omega^{n-1} + \omega^{n-2} + \omega^{n-3} + \ldots + \omega^{n-a_n}$ (because the move doesn’t reach the store). The key is, the former is greater than the latter! You can show this easily:
$\omega^{n-1} + \omega^{n-2} + \omega^{n-3} + \ldots + \omega^{n-a_n}$
$\le \omega^{n-1} + \omega^{n-1} + \omega^{n-1} + \ldots + \omega^{n-1}$
$= a_n \omega^{n-1}$
$< a_n \omega^n$

Thus $A$ decreases.

Remember about well-ordering theorem above? True: after each of this player’s move, $A$ decreases, and it is a natural number! Thus at some point is must be zero, which means all stones are removed. But this means some move has sent some stone to the store, decreasing the free count.

Of course, both players play alternately when no stone comes to the store. But just do the similar thing for the opponent; at some point, one of the two $A$‘s must decrease low enough for some stone to be forced to the store, decreasing the free count.

Thus we’re done!

Clearly, the inspiration from Goodstein’s theorem is the above, with the $\omega$‘s appearing. Had I noticed that the simple number two works when I first thought about it, I wouldn’t have used $\omega$‘s. But eh, why not.

And now I need to sleep…

# Same Number Hunt

Used as Final Match of Season 3, Game 2.

Rules

Score points by correctly forming mathematical expressions resulting on the given number. Win by scoring 10 points first.

There is a grid of 16 letters. Behind each letter, there are the twelve numbers 1-12 and the four arithmetic operations +, -, ×, ÷. At the beginning, the back sides are shown for 5 seconds to the players, after which they are hidden again, only showing the letters. Then the game begins, playing round after round, until someone scores 10 points to be the winner.

In each round, a number is revealed; this number is a natural number that can be formed using the given numbers and operations. (For example, 23 can appear as 11+12 can be formed from the grid, but 29 will not appear.) A player that thinks they can form an equation should press the buzzer and recite three letters, corresponding to the numbers and operations. For example, if 11 is behind the letter C, + is behind the letter P, and 12 is behind the letter K, on a target number of 23 a player can recite CPK (or KPC) to score.

The expression must be in the form “number, operator, number”, and may not repeat any letter (thus 7+7 is not permitted). Upon pressing the buzzer, the player has 5 seconds to recite the expression, otherwise it’s considered as an incorrect answer. After reciting the letters, they are revealed; if the answer is incorrect, the turn is passed to the opposing player, who gets no time limit to form the equation. The turns are passed back and forth until a correct answer is made, which scores 1 point and begins the next round. The first player to score 10 points win the game.

As this is a Final Game, there are three items available (in addition to Item Copy and Item Disabler):

• Priority: Allows the user to answer the next round first without pressing the buzzer first. (This can only be activated at the start of a round, before the target number is revealed.)
• Double Chance: Allows the user to construct a 5-letter expression (number, operation, number, operation, number), which if correct, gives 2 points.
• Peek: Allows the user to check the back of one letter before answering.