Beguiling Games III: Splotch!

In this installment of Beguiling Games, we’ll learn how to play Splotch!

Day119Splotch1

But first, I’ll start off by giving you the solution to the last puzzle I presented in the last installment.  The statement of the puzzle is too long to repeat, so you can refresh your memory at Beguiling Games II.

First, it is important to note that Lucas’s statement actually provides no information!   Suppose Ophelia’s card was Truthteller.  Then she would have told the truth in Round 1, passed her card to Lucas, and he would truthfully have stated that she told the truth in Round 1.

But what if Ophelia’s card was Liar?  Then her statement “I am a Truthteller” in Round 1 would in fact have been a lie.  Now she passes her card to Lucas.  He lies and says she told the truth in Round 1!  What this means is Lucas could have made his statement in Round 2 regardless of what card Ophelia passed him.

Note that the same logic applies to Mordecai’s statement.  He could have said “Lucas also told the truth in the first round” regardless of whether Lucas passed him a Truthteller or a Liar card.

Now let’s examine Nancy’s statement in some detail.  She said that there is at least one liar at the table.  Could she have lied?

Well, if the fact that the there is at least one liar at the table is false, that means everyone is a Truthteller.  But there is no way Nancy could have known this, since the only two cards she saw were hers and the card passed to her by Mordecai.

That means Nancy must have told the truth in Round 2, and Mordecai must have passed her a Truthteller card.  But in order for her to have sufficient information to say there is at least one Liar at the table, she must have been holding a Liar card in Round 1.

Now this information could be deduced by anyone at the table.  In other words, anyone would know that Nancy held a Liar card in Round 1, and Mordecai held a Truthteller card.  That leaves Lucas’s and Ophelia’s cards in Round 1.

The only person who could know both these cards would be Lucas — he knew his own card in Round 1, and he knew Ophelia’s card because she passed it to him in Round 2.  So he had enough information to declare after Nancy’s statement.

It is important to point out that there is no way to know what Lucas’s and Ophelia’s cards actually were.  All we need to know is that Lucas knew what both of them were.

Did you figure it out?  It takes a little bit of reasoning, but all the facts were there.

Now as I mentioned in the last installment, this time I’d give you a geometrical two-player game to work out.  I call it Splotch!

In the game of Splotch!, players alternate coloring in squares on a 4 x 4 grid.  The goal is to create a target shape, called a splotch.  A player wins when he or she colors in a square which completes a splotch, and then announces the win.

So if a player completes a splotch but doesn’t notice it and doesn’t announce the win, then play continues until someone completes another splotch.  You can only call Splotch! on your turn, so you cannot win by calling it if your opponent fails to.  It is also important to remember that the square you color in on your turn must be part of a splotch you announce.

In the version of Splotch! I’m sharing with you this week, the target splotch is:

Day119Splotch1

Just so you know how it works, let’s look at a sample game, shown below.

Day119Splotch2

There are two players, A and B.  A goes first, and plays in the lower left corner (labelled with A1).  B colors in a square in the top row (B1).  Then A’s second move is in the second row.  B wins on the next play by completing — and announcing! — a splotch.  The splotch may be rotated (as in this example), reflected, or even both!  So you’ve got to watch carefully.

Now admittedly, A was not a very clever player in this round of Splotch!  In fact, B could also have won by playing as follows:

Day119Splotch3

But here is the question:  In this version of Splotch!, which player can force a win every time?  Remember:  you must be able to provide a response to every move by your opponent!  I’ll reveal which player can force a win in my next installment of Beguiling Games, and I’ll share my particular strategy for winning.

One final remark — a bit of a tangent, but related to puzzles and games.  If you’ve been following my blog for a long while, you might remember my (fiendishly diabolical) number searches.  (These were posted well over a year ago.)

Day031Fig1

Here, the numbers in the right are in base 10, but you have to convert them to another base to find them in the grid!  You can read more about these puzzles here and here.

I submitted these puzzles to MAA Focus, the newsmagazine of the Mathematical Association of America.  I am happy to report that they will be featured on the Puzzle Page in the December 2017/January 2018 issue!

Now I will admit that this doesn’t exactly make me famous, given the number of people who subscribe to MAA Focus.  But I composed these puzzles specifically for my blog — so if I hadn’t decided to write a blog, these puzzles might never have been created.  Perhaps another reason to write a blog!

Stay tuned for the next round of Beguiling Games, where you’ll learn (if you didn’t already figure it out for yourself) who has a winning strategy in the game of Splotch!

Beguiling Games II: Nuh-Uh!

First, we’ll look at the game I left you with in the last installment of Beguiling Games.  These were the rules of yet another variation of Tic-Tac-Toe:  if during the game either play gets three-in-a-row, then X wins.  If at the end, no one has three in a row, then O wins.  Does X have a winning strategy?  Does O?  Why?

We’ll show that X has a winning strategy.  As with usual Tic-Tac-Toe, X starts at the center.  If O goes in any square which is not a corner, then X will win as a result of the winning strategy in Tic-Tac-Toe.  So all that’s left to do is look at the case when O goes in a corner.  Let’s suppose it’s the upper right corner.

Then X goes in the lower right corner.  O must block, or else X will get three-in-a-row and win (since X wins if either player gets three-in-a-row).  So now, the board looks like this:

Day114a

Take a moment to see if you can figure out X’s winning strategy.  Do you see it?  All X needs to do is place in the center of the bottom row.

Day114b

Look at the top row.  Someone eventually has to place an X or O in this location.  No matter which one, a three-in-a-row will be created, and X will win!  So in this version of Tic-Tac-Toe, X has a winning strategy.

Today’s game will involve a different dynamic — cards and logic.  But first, let me give a little bit of context.  I have often used the book Problem Solving in Recreational Mathematics by Averbach and Chien when I’ve taught an introduction to proofs course.  There are many interesting problems in this book, and the chapter on logic has several problems involving Truthtellers and Liars.

So you have to solve problems by figuring out who are the Truthtellers and who are the Liars.  I like to use these problems to get students writing arguments in complete sentences since not only are they fun to work out, but they don’t involve notation.  Using notation correctly is another issue entirely, and I prefer to deal with that later on in the course.

As before, there’s a story to go along with the game….

Lucas, Mordecai, Nancy, and Ophelia decided that enough was enough, and so ended their Saturday afternoon’s cramming for the National Spelling Bee.  Wondering what they might do to pass the time, Lucas suggested, “Let’s play Nuh-Uh!”  Everyone enthusiastically agreed.

And so a game of Nuh-Uh! — whose creators, incidentally claim that it is the only game on the market which “Tortures your Mind, Warps your Character, and Impoverishes your Soul — all at the Same Time!” — commenced.

So Lucas shuffled the deck of eight cards; four of the cards had the word Truthteller on them, and four had the word Liar.  Dealing from the left, Lucas first dealt a card to Mordecai, then one to Nancy, one to Ophelia, and then finally one to himself.  Cards are dealt face down so a player can only see his or her card.

Per the rules of Nuh-Uh!, Lucas, then Mordecai, Nancy, and finally Ophelia made the statement “I am a Truthteller.”  Of course, such a statement was consistent with each player’s card, regardless of what was written on it.  Thus ended the first round.

In subsequent rounds, each player passes his or her card to the player on the left, and makes a statement consistent with the new card.  Thus, if Lucas passed Mordecai a card with Liar written on it, Mordecai would have to make a statement which is false.  The statement a player makes is based on the knowledge of the cards he or she has seen, and any other information which may be deduced from the statements of the previous players.

The same player begins each round of making statements.  So Lucas began each round in this game.  Once the game is over, the deal passes to the left, so Mordecai would deal and begin each round with a statement.

After a player makes a statement, players may, beginning with the player on the left of the one who just spoke, “Declare” — that is, say what card each of the players held during the first round.  The declaring player then looks at the cards held by the players to decide if he or she has won.  If an error is made in declaring, the player drops out and play continues; otherwise the cards are turned over and the deal passes to the next player on the left.

So the players passed their cards to the left, and thus the second round of making statements began.  Lucas started off with “Ophelia told the truth in the first round.”  After a polite few seconds of pause to give someone a chance to declare, Mordecai said, “Lucas also told the truth in the first round.”

Another brief pause ensued before Nancy stated, “There is at least one liar at the table.”  Note:  there are four Truthteller and four Liar cards, so it is possible that all players were dealt a Truthteller card.

After Nancy made her statement, one of the players had enough information to declare and win the game.  Which player declared?

Yes, there are enough clues to solve this!  But a word of caution — the solution to the puzzle only involves answering the question, “Which player declared?”  There may not be enough information to give a more complete answer….

Good luck!  In the next installment of Beguiling Games, I’ll give the solution to this logic puzzle, and give a geometrical two-player game to analyze as well.  Happy solving!

Beguiling Games I: Nic-Nac-No

It has been some time since I’ve posted any puzzles or games.  In going through some boxes of folders in my office, I came across some fun puzzles I created for a class whose focus was proofs and written solutions to problems.  I’d like to share one this week.

For the assignments, I sometimes wrote stories around the puzzles.  So here is one such story.  The date on the assignment, if you’re interested in such things, is January 16, 2003.  (I assume that you are familiar with the game Tic-Tac-Toe, as well as the fact that if both players play intelligently, the game ends in a draw.)  I called the game “Nic-Nac-No.”

Betty and Clyde, after their favorite breakfast of blueberry pancakes one sunny Saturday morning, began a Tic-Tac-Toe tournament.  They were reasonably bright children — taking turns going first, the initial 73 games ended in a draw.

“Just once, Clyde, couldn’t you try putting your O first on a side instead of in a corner?” prodded Betty.  “That way, it wouldn’t be the same boring game every time.”

“Well, it’s my turn to go first this time,” said Clyde, putting an X in the center.  “OK, now you show me how you want me to play so I can do it that way next time.”

“Oh, shut up, Clyde,” sighed Betty, putting her O in the upper left corner.  And so game #74 ended in a draw.

“Hey, I’ve got an idea!” exclaimed Clyde.  “Let’s make up different rules.  How about this:  the first one who gets three-in-a-row loses.  Whaddya think, Betty?”

“That’s so random, Clyde,” said Betty, secretly excited by the suggestion.

“No, it’s not.  And besides,” reasoned Clyde, “it’s got to be better than playing another game of Tic-Tac-Toe since you won’t ever try anything different.

“OK, potato brain.  Let’s try.  You go first.”

“Great!” exclaimed Clyde, until he realized Betty was trying to outmaneuver him.  He just realized that it you’re trying to avoid three-in-a-row, the fewer squares you own, the better.

Assuming Betty and Clyde play optimally, will the game be a win for Betty, a win for Clyde, or a draw?

I should remark that the idea of a misere game — where you turn the winning condition into a losing condition — is not original with me.  But most students have not considered this type of game, so misere versions of games often make for engaging problems.

Before I discuss the solution, you might want to try it out for yourself!  There are likely many strategies possible to produce the desired result; I’ll just show you the ones I thought were the most straightforward.

In my solution, Betty and Clyde use different strategies, but the end result is the same:  the game must end in a draw.  Let’s look at what strategies they might use.

It turns out that if Clyde starts in the center, he can use a strategy where he does not lose.  It’s fairly simple:  always play opposite Betty.  Thus, when Betty plays a corner/side, Clyde takes the opposite corner/side.

Why can’t Clyde lose?  First, it should be clear that Clyde can never make a three-in-a-row that passes through the center.  Since he always plays opposite Betty, any line of three passing through the center must contain two X’s and one O (recall Clyde started with an X in the center), and so is not a three-in-a-row.

What about a three-in-a-row along a side?  Since Clyde plays opposite Betty, if he ever placed an X to make three-in-a-row along a side, that would mean Betty already had three O’s in a row on the opposite side, and would have already lost!  So it’s impossible for Clyde to lose this way.

Since any three-in-a-row must pass through the center or be along a side, this means that Clyde — if he plays intelligently — can never lose Nic-Nac-No.

Now let’s look a non-losing strategy for Betty.  There is no guarantee she will be able to take the center square on her first move, so we’ve got to consider something different.  And we can’t just rely on playing opposite Clyde, since there is no opposite move if the takes the center first.  Moreover, it may be the case that Clyde uses some other strategy than the one I mentioned, so we can’t even assume that he does take the center on his first move!

To see a strategy for Betty, consider the following diagram:

Day109NicNanNo

Betty’s strategy is simple:  place an O in one of the squares marked A, one marked B, one marked C, and one marked D.

Note that this is always possible.  Even if Clyde does not play in the center on his first move, he can only occupy one square labelled A, B, C, or D.  Then Betty places her O on the other square with the same letter.  If Clyde does begin in the center, then Betty has her choice of first move.

Since it is always possible — and since Betty only has four moves — these comprise all of Betty’s moves.  But note that since Betty never has an O on two of the same letter, she can never get three-in-a-row on a side.  Further, since Betty’s strategy never involves a move in the center, she can never get three-in-a-row in a line going through the center square.  This means that Betty can never lose!

So if the two players play their best games, then Nic-Nac-No ends up in a draw.  And while these strategies do indeed work, I would welcome someone to find simpler strategies.

I’ll leave you with another version of Tic-Tac-Toe to think about.  Here are the rules:  if during the game either play gets three-in-a-row, then X wins.  If at the end, no one has three in a row, then O wins.  Does X have a winning strategy?  Does O?  Note that in this game, there cannot be a draw!  I’ll give you the answer in my next installment of Beguiling Games….