# Puzzle 95: Logicsmith v2.0

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

Puzzle 95: Logicsmith v2.0
Fillomino

Four years ago (okay, 45 months ago), there’s Logicsmith Exhibition, back when Grant was still active on his blog (before he migrated to Grandmaster Puzzles). Back then, I was still starting at constructing logic puzzles; you can see my submission there. Here is a “revamped” version: exactly the same layout as before, and each of 1-9 appears exactly four times as in the competition.

…okay, I just got this theme idea and toyed with it, to liven up my blog again.

# Six-Player Card Games

Recently, I’ve been playing cards with my siblings and cousins. The total number of players is six. There’s not a lot of games that can be played with six people but only uses one standard deck of cards (maybe with paper and pencil, but preferably not), so I’ve been inventing random stuff. Here’s a (probably incomplete, if I missed stuff) list of those random inventions. Note that these are certainly prototypes, although some are of better shape (read: playable) than others. I might add a few more ideas here later.

## Perudo

(In fact, works pretty well with 2-7 players. With more players, reduce each player’s starting life count; keep the total less than 40.)

Initially, each player has five “lives”. Usually people will just be honest with their life count, but if necessary, keep track with paper and pencil. Your objective is to be the last person with lives.

In each round, deal each player as many cards as they have lives. They can see the cards, but shouldn’t show them to anyone else.

Each player, starting from some agreed person (usually the one after the last person to make a bid in the previous round, or just choose somebody for the first round), declares a bid containing a number and a suit; this is a bid that “among all cards dealt, there are at least these many cards of this suit”. Bids may not go smaller. A bid with higher number beats a bid with lower number; for bids of equal number, a bid of “stronger” suit beats a bid of “weaker” suit. (Suits are ranked in some predetermined fashion; I use Big Two’s diamond < club < heart < spade, but there’s actually no difference.)

At any time, a player may challenge a bid. In this case, everyone opens their cards. If the bid is met (there’s enough cards of the given suit), the challenger loses a life; otherwise, the bidder loses a life. Only the latest bid can be challenged.

Optional: To each bid, you may also declare “exact”. If this is challenged and there are exactly as many cards of the suit that you stated, you also gain one life. If there are more cards, you lose one life. A normal bid and an exact bid are of equal strength; you cannot reply “eight hearts” with “exactly eight hearts”.

## Partners

(Better for 4.)

Just a trick-taking game. The objective is to meet a bid contract.

Discard one rank (for a 48-card deck) and agree on some scoring system. Examples:

• Aces worth 4, Kings worth 3, Queens worth 2, Jacks worth 1. Discard Twos.
• Aces worth 1, Twos worth 2, Threes worth 3, Fours worth 4. Discard Tens.
• Each trick is worth 1 point (read: each card is worth 1/6 points). Discard Twos.

Deal 8 cards for each player. (For 4 playing, don’t discard a rank and deal 13 to each. It’s possible to have 5 playing, although I’ve yet to figure out a good way for the cards; maybe remove two Twos and deal 10 each? Add 3 jokers and deal 11 each?)

The first phase is the auction. Each player may make a bid that beats the previous bid, or passes; if a player passes, they cannot bid again. A bid is an amount of points, together with a declaration of a trump suit (or that the deal is played at no trumps). Like above, larger number beats smaller number; for equal numbers, no trump beats trump, and suits are ranked in some fashion (again, I use Big Two). The bid is a declaration of getting at least that many points, with the given trump.

After everyone but one player passes, the last bidder is the declarer. Their target is to meet the given contract. They have partners, however; they declare two cards that they don’t hold themselves. (If 4 playing, declare 1 card.) The holders of these cards will be the declarer’s teammates (although the cards are not revealed); at the end, points counted by them will be counted for the declarer. Each player plays a card just like in trick-based games (follow suit if possible, but anything if not; highest trump wins, otherwise highest in the suit led wins).

After all tricks are played, count the points in the cards won, and add up the points in the partnership. If the contract is met, then the team is successful, otherwise the opposing team is successful.

If both called cards are on the same person, they score double (they only need half of the points to reach the contract). (Optional: bad luck; they need to work around having a two-man team for a contract designed for three.)

# Puzzle 94: Writer Block

Cross the Streams Shade some of the cells black so that all black cells are connected and no 2×2 square is entirely shaded black. The clues outside the grid gives the contents of the corresponding row/column, reading from left to right and from top to bottom. A number means a group of consecutive black cells; two different groups in the same row/column must be separated by at least one white cell. A question mark indicates a single group of unknown size; an asterisk indicates an unknown number of groups (which may differ in size, and there might be no group at all).

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

Puzzle 94: Writer Block
Cross the Streams

Yay, I’m back. Actually, this might be temporary as well; I’m not sure why I don’t feel the same interest on making puzzles as I had a few years ago, but let’s hope I can still trickle out puzzles occasionally. (The title is a reference on that. Just back from writer block, if this is considered writing. Better than a title like “Logical”, at least.)

Meanwhile, I’ve been playing several “programming” games. I’ve completed SpaceChem (finished all obligatory puzzles for the story) a few months back; I got my hands on TIS-100 which I recently completed (but with the upcoming bonus campaign I’ll have several more puzzles to do); I’m redoing Manufactoria after I realized I haven’t completed it. Those might be not exactly the kind of puzzles that you (as in people that enjoy pencil puzzles like this) like, but who knows.

# 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…