mathematics - 20 coins on the table

Two boys , Brian and Gabe, decide to play a game.

They lay 20 coins on the table. Each player took turns taking 1, 2, or 3 coins. Whoever took the last coin won.

Brian thought if he went first, he'd win for sure since he had the advantage of making the first move. So Brian did go first. But Gabe won every time!

Obviously Gabe had a winning strategy, but what was it? Could Brian have outsmarted it?

Answer

Always leave a multiple of 4 coins. So if Brian takes 1, Gabe takes 3. 2-2, 3-1.

Starting with a multiple of 4, player 2 can always win with perfect play. Starting with any other number, player 1 can win with perfect play.

logical deduction - Heyacrazy: Crosses

This is a Heyacrazy puzzle.

Heyacrazy is an original genre, inspired by the "two border" rule of Heyawake and Heyawacky.

Rules of Heyacrazy:

• 
  

  Shade some cells of the grid.

  
  


• 
  

  Shaded cells cannot be orthogonally adjacent; unshaded cells must be orthogonally connected.

  
  
  


• 
  

  When the puzzle is solved, you may not be able to draw a line segment that passes through two borders, and does not pass through any shaded cells or grid vertices.

  
  

Example:

Here, the top solution is valid. The first three wrong solutions all break the third rule with the given red segments, and the other two break the two parts of the second rule in the indicated areas.

The puzzle:

Answer

Solution:

Reasoning:

There are a couple of key observations that help us solve this:

1. All $+$'s and $\times$'s must be shaded diagonally in one of the two possible ways ($▚$ or $▞$), otherwise they would internally violate rule #3
2. Once a $+$ has been shaded, its internal borders are irrelevant since no line can pass through them any more.

3. This means that we only have to be careful about rule #3 for the $\times$'s and central borders

With those observations out of the way, we can start solving.
The top-right cell must be shaded (otherwise it would be an enclosed unshaded cell). Being careful to obey rule #2, we can deduce half the board from just that one cell:

The lower-right quarter must have its right side connected to its left side (otherwise the right side is an unconnected group of unshaded cells, since it's already closed off in the upper-right quarter). If we shade the $+$ in the $▚$ direction, a connection is impossible - therefore we must shade the $+$ in the $▞$ direction.

From here it's pretty straight forward to solve the lower-left quarter, and get our final solution.

Old answer (before puzzle was updated)

There are multiple solutions.

Reasoning:

The top-right cell must be shaded (otherwise it would be an enclosed unshaded cell). This gives us most of the top-right square.

All of the +'s can be shaded diagonally in one of two ways. This includes the central +. If we shade the central + in the \ direction, we end up with a situation like this:

However if we shade the + marked "1" in the \ direction, we close off the unshaded cells in each half, which violates rule #2. If we instead mark it in the / direction, then we always end up being able to connect the central borders which violates rule #3.
For this reason we know that the central + must be shaded in the / direction. Knowing that, we can make it this far:

From here it's guesswork. We can shade the + in the lower right square in either of the two directions. I am not sure that the solutions I've given in this answer are the only ones possible.

no computers - Don't bully your little sister at chess

This puzzle was inspired by @RewanDemontay's puzzle De-zephyr The Solution!. Rewan kindly granted me permission for making this puzzle and citing his as my source of inspiration. I thought of naming it "Zephyr that!" but titles can't be that short on SE.

---

You're a little girl playing chess with your brother. He knows that you have just learned the rules and he's been toying with you for the whole game, setting up half of his pawns for promotion and getting his pieces into a nice square shape while your king was constantly under checkmate threat. He was hardly paying attention to your moves and didn't see that you were cooking up a little treat of your own.

Please wipe out the entire set of black pieces and deliver checkmate in as few moves as possible.

Your brother is playing optimally and making every move he can to make the game last longer if he cannot force a win or a draw.

Bonus: at some point, your brother grins in delight, thinking you've lost control of the situation. You just captured a rook and four black pieces are left on the board: a pawn, a knight, a queen and the king. You did not choose the shortest way to a complete cleaning of the board, but you are still pretty much in control of the situation. Prove it to him and deliver checkmate to a lone king with your 49th move.

FEN: 1q1N3k/2prN1R1/Kn3Qn1/1p4p1/1r4b1/1p4b1/2pppp2/8 w - - 0 1

cipher - Real life puzzle: Unknown alphabet or shorthand

User JellicleCat posted this on linguistics stackexhange a few years ago. None of us were able to come up with a satisfying answer but most seemed to agree that it was an idiosyncratic cipher or shorthand. It means that puzzlers may be better equipped to solve it than linguists.

Anyway, the original question was:

I've wondered about this script since I saw it years ago. I imagine it's an English cipher. Can anyone tell me?

He later added in the comments section:

an intern left it on his notepad on his last day in the office

Answer

This appears to be Elian Script. I'm not sure I can read the writer's handwriting entirely (and they seem to have added some nonstandard things like a zigzag for T), but the first few lines read:

PAR?NRE RENR

?HE C PROGRAMMING LANGUAGE!

I REALLY WANT MY PEN!

Edit by OP: I spend some time to fully decipher it but I think Deusovi deserves the real credit for an answer so I present it as an edit:

park nre renr
the c programming langua#ge!
i really want my pen!
create a c[=b?]etter interface for reports
hey mr flava! flava
i reall#y should practice my elian s#cript more
the d language is fun to use indeed!
i really should practice my elian script
mayby # i should standerdiz ze

{ changed pens }

testing this pen

it writes # mu#a [=much?] faster

geometry - Dissect the pixel-heads

Here are two figures, each composed of 54 pixels. Cut each figure into six nonominoes having the same shape and size.

• Both figures use the same nonomino



• The nonominoes could be rotated, reflected, or both


• The four colors (blue, green, pink, gray) are for artistic effect only. These colors have no relationship to the solution.

Answer

Found it!

Very nice! I wasted 30min until I've realized I mis-drew my black & white copy with only 53 squares! :c)

---

Edit: As requested, and for the purpose of this site (teaching how to build/solve puzzles) I'm adding how I found the solution to this puzzle.
General considerations:

• Mostly by trial & error. I'm using a graphic program with layers/objects (i.e. Powerpoint) such that I can easily modify & clone tiles while trying fitting together the puzzle.
  


• The tile-size of the nonominoes is known if all are of same size, the total field size is known, and the number of nomoninoes is known.
  


• Then I start with one (in this case close to a hole) and copy/tilt it to see if they fit. Rule-out conditions are:
  
  
  
  • two noominoes overlap
  
  
  
  • isolated 'holes' are created
  
  
  
  


• From there, it's a process of reduction, guided by intuition. (i.e. usually good puzzles have rather 'complicated' nonomino-shapes. So I try those first.)


• Sometimes it helps to do the process with a partial nonomino (i.e. of smaller size) first. This rules out whole sets of shapes. (If you have a problem with a 5-sized shape, no bigger shape will work out.)

As for this particular puzzle:

- The tile-size of the nonominoes is known to be 9, because we have to fit 6 tiles into 54 fields, hence: 6 x 9 = 54.

In the particular puzzle, it was soon clear that nonominoes had to partly wrap around a one-square hole. Few tries showed that it had to be a "5-tiles around hole" solution.
- So I made this 5-sized nonomino and copied it 6 times, arranging it around the head-holes.
- Then I successively added 1 square (i.e. made all 6-sized, 7-sized...). One could immediately see, what fits, and what wouldn't. As always: You must not create overlaps and you must not create isolated holes. - I actually only solved the left head. It was then immediately visible, that the right one has just a mirror in it's left half.

riddle - I'm but a speck of fairy dust

A simple riddle of mine, focused towards meter and rhyme.

I'm but a speck of fairy dust
A glitt'ring jew'l, but only just
So small I scarcely can be seen
A trinity in me convenes

And yet without me you can't see
Fine art and words are made of me
I'm resolute; through me you can

View realms beyond the mortal ken

I shrink along with passing years
My number grows, I gain more peers
But treat me well 'tis very true
I sleep and live and die like you

Please include explanations for as many lines as possible in your answer.

Answer

You are:

A pixel

--

I'm but a speck of fairy dust / A glitt'ring jew'l, but only just

Pixels are small and bright

So small I scarcely can be seen / A trinity in me convenes

RGB

And yet without me you can't see

You can't see content without them. Assuming specific to the context of its realm.

Fine art and words are made of me

Again, assuming the context of its realm.

I'm resolute; through me you can / View realms beyond the mortal ken

Pretty obvious, via the internet, games, etc.

I shrink along with passing years / My number grows, I gain more peers

DPI increases as computers get more advanced.

But treat me well 'tis very true / I sleep and live and die like you

"Sleep" possibly referring to monitor state, and pixels can die.

lateral thinking - To answer this riddle, what will you open first?

It's 7:00 AM. You are asleep and there is a sudden knock on the door. Behind the door are your parents who came to have breakfast. In your fridge are bread, milk (pasteurized), juice, and a jar of jam.

What will you open first?

logical deduction - Four cups on a table

I didn't invent this myself, I heard it a long time ago and it stuck with me.

Four glasses are placed on the corners of a square rotating table. Some of the glasses are facing upwards and some upside-down. Your goal is to arrange the glasses so that they are all facing up or all facing down. Here are the rules:

1. You must keep your eyes closed at all times. (No tricks or lateral thinking, this is a pure logic puzzle)


2. In a single turn, any two glasses may be inspected. After feeling their orientation, you may reverse the orientation of either, neither, or both glasses.


3. After each turn, the table is rotated through a random angle.



4. At any point, if all four glasses are of the same orientation a bell will ring.

Find a solution to ensure that all glasses have the same orientation (either up or down) in a finite number of turns. The algorithm must not depend on luck.

Edit: The bell will not ring mid-turn, it will only ring if the glasses are all of the same orientation at the end of your turn.

Answer

I have solution that can guarantee the end goal being reached in a maximum of four five steps. (Many thanks for @dmg and @Taemyr for their comments to fix this solution.) The trick is to:

Reduce the cups into the following configuration:
$\begin{bmatrix} \circ & \bullet \\ \bullet & \circ \end{bmatrix} $
where turning any two cups along the diagonal will satisfy the end goal

For a constructive proof, we first consider the following two three CASES:

CASE 1) the orientation of a single cup is different from the others, e.g. $\begin{bmatrix} \circ & \bullet \\ \circ & \circ \end{bmatrix} $ or $\begin{bmatrix} \bullet & \circ \\ \bullet & \bullet \end{bmatrix} $
CASE 2a) two cups are in each orientation along the diagonal, e.g. $\begin{bmatrix} \circ & \bullet \\ \bullet & \circ \end{bmatrix} $
CASE 2b) two cups are in each orientation along two sides, e.g. $\begin{bmatrix} \circ & \bullet \\ \circ & \bullet \end{bmatrix} $

Step 1

Take two cups along the diagonal.
- If they are different, flip $\circ$ to $\bullet$. If this doesn't sound the bell, we either have $\begin{bmatrix} \circ & \bullet \\ \bullet & \circ \end{bmatrix}$ or $\begin{bmatrix} \circ & \bullet \\ \circ & \circ \end{bmatrix}$, with the majority orientation being $\circ$. In either case, proceed to step 2.
- If they are the same, flip both. If this doesn't sound the bell, we either have $\begin{bmatrix} \bullet & \circ \\ \bullet & \bullet \end{bmatrix}$ or $\begin{bmatrix} \circ & \bullet \\ \circ & \circ \end{bmatrix}$. Either way, the majority orientation is known, and we can proceed to step 3.

Step 2

Take two cups along the diagonal.
- If they are different, we now know we have $\begin{bmatrix} \circ & \bullet \\ \circ & \circ \end{bmatrix}$. Flip $\bullet$ to $\circ$ to win.
- If they are the same, flip both. If this doesn't sound the bell, we now know we have $\begin{bmatrix} \circ & \bullet \\ \circ & \circ \end{bmatrix}$. Proceed to Step 3.

Step 3

At this point, we have already figured out what the majority orientation is, so we assume we have $\begin{bmatrix} \circ & \bullet \\ \circ & \circ \end{bmatrix} $ WLOG. Then:

Take two cups along the diagonal.
- If the cups are different, flip $\bullet$ to $\circ$ to win.
- If the cups are both $\circ$, flip one of them. We now know we have $\begin{bmatrix} \circ & \bullet \\ \circ & \bullet \end{bmatrix}$.

Step 4

Take two adjacent cups along an edge and flip both.
- If the cups chosen are the same, we win.
- If they are different, we now know we have $\begin{bmatrix} \circ & \bullet \\ \bullet & \circ \end{bmatrix} $

Step 5

Take two cups along a diagonal and flip both. We win!

cipher - Rendez-vous with your eccentric uncle

Your eccentric uncle is a world-renowned cryptologist and he sees in you a high potential to follow in his footsteps. You are of course very flattered and can't wait to get going with the special training program he has set up for you. Your training will commence tomorrow at a rendez-vous point that your uncle has communicated to you by means of an encrypted message. To prove that you are worthy, decrypt the message and find out where the rendez-vous will take place!

The message reads:

Note: The puzzle contains multiple layers and is best solved one step at a time. Only the message itself is needed to solve this puzzle, all the rest is just flavour text.

Since it seems to be unclear how to deal with the different layers, I will add a little hint in the form of an example here:

Say the message were: "hI, this Is a HiNT MeSsAge", then this can be seen as a combination of two layers. In one layer some characters have been capitalized: "hI, this Is a HiNT MeSsAge" and in another layer some letters where made bold and italicized (both were done at the same time, just to make the letters jump out a little more): "hi, this is a hint message".
When working on solving one layer, you can (and should) ignore all other layers. The solution will then provide a clue for the next layer. The layers must be solved in a specific order, though.

Hint for the first layer:

For the first layer the exact characters are not important, just the pattern of highlighted vs. non-highlighted characters. The solution is 43 characters long.

Hint 2:

For the first layer should be read as a binary string.

Answer

Final Answer

Rendez-vous will take place at -

CENTRAL PARK

Taking the clue from the Hint "For the first layer should be read as a binary string."

If the first layer mean - uppercase treated as 1's and lower case as 0's we get - 215 characters turned into binary as - 101001001001001100000000001000110010010101110000010000 00010100001101000000001011010011011101001000000010100001001110110000 10100000110010000110111011100000001001100100111101110010010001100000 1001110100110010110000101

Another hint says the first layer gives 43 characters hence-

We need to split the 215 binary into group of 5 to get 43 characters.

So..

Grouping them into 5-bits and checking the corresponding letter we get
10100 = 20 = T

10010 = 18 = R
01001 = 9 = I
....
We get "TRIP HYENA EAT KIWI EAGLE YAWN IRONIC STYLE" (Treating 00000 as space")

So...

Taking first letters we get - > THE KEY IS and Thanks to Ankoganit for figuring out the key. which are last letters PATIENCE

Knowing this(Thanks to @Sconibulus, @Gareth) -

PATIENCE is an 8 letter word. So, checking the pattern with bold characters we get aeybssdvxgabsstvdtmes Applying Vigenère cipher with PATIENCE as key we get LeftOFBrighTOfrrotTwo which means Left OF B(blue) righT Of r(red) rot-Two(rot-2)

Hence,

Taking the letter's left of Blue and right of Red colored ones, we get - aclrpyjnypi and applying rot-2 gives centralpark

So the rendez-vous will take place at -

Central Park

All thanks to @Gareth, @dcfyj, @Ankoganit and @Sconibulus. Would have costed a lot of brainstorming solving this one.

With that done, I would say, the solution will be too difficult without hints as the puzzle doesn't speak of what to apply where.

probability - One Hundred Lockboxes of Wood and Steel

A bit beyond perceptions reach,
I sometimes believe I see
that life is two locked boxes, each
containing the other’s key.
― Piet Hein

You have one hundred lockboxes, fifty made of steel, fifty made of wood. You only have one copy of each box's key.

A prankster breaks into your house and randomly places the 100 keys in the boxes, one key per box. He then shuts all the boxes, locking the keys inside.

Though the steel boxes are impenetrable, the wooden ones can be broken quite easily. What is the probability that breaking open the wooden boxes will allow you to open all of the steel ones?

Answer

The probability is $1/2$.

We have a permutation that maps each box to the box whose key it contains. Once we open a box, we can open the box it maps to. So, we can open all the boxes exactly if there is no all-steel cycle.

Label the boxes $1$ through $100$. We denote the permutation in cycle format like $(31)(542)(6)$. To make this canonical, write each cycle with the greatest number first, and sort the cycles by increasing first number. Now, we can extract the cycle structure just from the sequence $315426$ because the cycles start at numbers that greater any before them ($3$, $5$, $6$). So, each of the $n!$ such sequences represents one of the $n!$ permutations.

Now, let the steel boxes be $1$ through $50$ and wooden boxes be $51$ through $100$. There's a steel cycle exactly if the first number is at most $50$. If so, the first cycle contains only numbers at most $50$, and if not, every cycle starts with a number at least $51$. Since the cycle representation is a uniformly random ordering, the first number is uniformly random. So, this happens with probability $1/2$, and the remaining $1/2$ the time we can open all the boxes.

More generally with $w$ wooden boxes and $s$ steel boxes, the probability of opening all the boxes equals the fraction of wooden boxes $\frac{w}{w+s}$.

pattern - What is a Regular Word™?

If a word has a certain property, I call it a Regular Word™.

You can use the examples below to find the property:

$$\begin{array}{|c|c|} \hline \bbox[yellow]{\textbf{Regular Words™}}&\bbox[yellow]{\textbf{Not Regular Words™}}\\ \hline \text{NAP}&\text{SLEEP}\\ \hline \text{AXED}&\text{FIRED}\\ \hline \text{QUIT}&\text{EXIT}\\ \hline \text{WHIZ}&\text{GENIUS}\\ \hline \text{DEATH}&\text{EXPIRY}\\ \hline \text{GREEK}&\text{LATIN}\\ \hline \text{BOXING}&\text{FIGHTING}\\ \hline \text{KIDNAP}&\text{ABDUCT}\\ \hline \text{OBJECT}&\text{THING}\\ \hline \text{REVIEW}&\text{PERUSE}\\ \hline \text{SPHINX}&\text{RIDDLER}\\ \hline \text{HEXAGON}&\text{PENTAGON}\\ \hline \text{MOSQUES}&\text{TEMPLES}\\ \hline \text{FEMINISM}&\text{EQUALITY}\\ \hline \text{LUXURIANT}&\text{LAVISH}\\ \hline \text{CONTEXTUAL}&\text{RELATED}\\ \hline \text{EXCAVATION}&\text{DIGGING}\\ \hline \text{JUXTAPOSES}&\text{OVERLAPS}\\ \hline \text{PERPLEXITY}&\text{CONFUSION}\\ \hline \text{INTOXICATES}&\text{INEBRIATES}\\ \hline \end{array} $$

For those without MathJax, or if you want to pop this into a spreadsheet, here is a CSV version:

Regular™,Not Regular™
NAP,SLEEP
AXED,FIRED
QUIT,EXIT

WHIZ,GENIUS
DEATH,EXPIRY
GREEK,LATIN
BOXING,FIGHTING
KIDNAP,ABDUCT
OBJECT,THING
REVIEW,PERUSE
SPHINX,RIDDLER
HEXAGON,PENTAGON
MOSQUES,TEMPLES

FEMINISM,EQUALITY
LUXURIANT,LAVISH
CONTEXTUAL,RELATED
EXCAVATION,DIGGING
JUXTAPOSES,OVERLAPS
PERPLEXITY,CONFUSION
INTOXICATES,INEBRIATES

Answer

A Regular Word™ is a word in which

The length of the word is the same as the distance between the first and last letters in the alphabet.

In other words, the number of letters between the first and last letters of the word is also the number of letters between them in the alphabet.

for example

"MOSQUES" has the same number of letters as "MNOPQRS," and "EXCAVATION" has the same number of letters as "EFGHIJKLMN"

visual - Three-dimensional light bulbs

This is a three-dimensional Akari puzzle (also known as Light Up). The six squares represent the layers of a $6\times6\times6$ cube, top to bottom. The objective is to add light bulbs into any number of cells so that the resulting grid satifies the following rules:¹

• Black cells are walls and cannot contain light bulbs.


• Numbers in black cells indicate how many light bulbs are directly adjacent to that cell (vertically, horizontally or along the Z-axis).


• A light bulb illuminates its own cell as well as every cell visible from it in all six directions (up/down, right/left, and both ways along the Z-axis), continuing until a wall comes in the way.


• Every white square must be illuminated by at least one light bulb.


• No light bulb may be illuminated by another light bulb.

Note: The solution is unique and solvable by logic alone. No guesswork or trial-and-error is necessary here.

_{¹ Paraphrased from the original rules on Nikoli}

Answer

Like that?

Red = light bulb, yellow = light Start from bottom
1 because you can't place light bulb above (in z) "2" and next to "1" can be only one, so it has to be there.
2 because you can't place light bulb above (in z) "2" and there is already light from first light bulb
3 since second light bulb eliminates one solution
4 because 1st and 3rd light bulbs eliminates rest
5 because of 4th and so on…

pattern - State the rule!

State the rule behind the following partition of letters, and fill in the questionmarks. (The puzzle has a clear and justifiable trivia-type solution.)

      A . . D E . G H I . K . . N O . . . S T . . . ? ? ?
   ---------------------------------------------------------
      . B C . . F . . . J . L M . . P Q R . . U V W ? ? ?

Answer

The rule appears to be:

The last letters of the names of U.S. states.

A

Alabama
Alaska
Arizona
California
Florida
Georgia
Indiana
Iowa
Louisiana
Minnesota

Montana
Nebraska
Nevada
North Carolina
North Dakota
Oklahoma
Pennsylvania
South Carolina
South Dakota
Virginia

West Virginia

D
Maryland
Rhode Island

E
Delaware
Maine
New Hampshire
Tennessee

G
Wyoming

H
Utah

I
Hawaii

Mississippi
Missouri

K
New York

N
Michigan
Oregon
Washington
Wisconsin

O
Colorao
Idaho
New Mexico

Ohio

S
Arkansas
Illinois
Kansas
Massachusetts
Texas

T
Connecticut
Vermont

So of the last three letters,

Y (Kentucky, New Jersey) is on the top row, X and Z on the bottom.

construction - Stepladder Game, Introduction and Puzzle 1: Turning

I want to introduce a new game to PSE: the Stepladder Game!^*
_{* You may already know this game by another name.}

The rules for the Stepladder Game:

1. The game is played on a grid.
   
   
   
   • Our grid is rectangular, and in the grid, each square, apart from the edges, has four neighbours.
   
   
   • The grid is "big enough", meaning that there will always be an edge in any direction, but unless the edges are shown, there is always "enough empty space" for any construction.
   
   
   
   
   


2. The squares can be either "white", "black", or "empty".


3. On the grid, there is always a stepladder source. It looks like this:
   


4. There are two players, black and white. They take turns; each turn consists of colouring exactly one empty square with their own colour.


5. After colouring a square, check that every opponent's square has a path, along adjacent same coloured squares, to an empty square. Those squares that don't, are caught and emptied


6. Black wins, if he catches the white squares in the source.


7. White wins, if she catches the inner black squares in the source.


8. White goes first.

---

Examples

Here's an example game with nothing except the source and the eventual edge. Both players are playing optimally. (For the purposes of these puzzles, "optimal play" means that the losing side always chooses the longest resistance, and the winning side always chooses the shortest path to a guaranteed victory.) The first couple of plays are numbered for your convenience.

As you can see, if the grid is empty, black will win; in the above diagram, it's white's turn, but whatever white does, black will win on the next move.

This all changes, if some squares are already coloured in. Here's an example with just one white square added:

This time around, with optimal play from both players, it is white who will win:

In the above diagram, it's black's turn, but white is already adjacent to 3 empty squares. Whatever black plays now, white can use her following two turns to catch the inner black shape and win.

That should just about be enough for the rules, so it's finally time for the first puzzle.

---

Stepladder Puzzle 1: Turning.

As you have no doubt noticed, the stepladder source always sends the stepladder in the northeast direction. Your job is to colour the minimum number of squares in the target area (marked in the diagram) so that with optimal play from both players, the stepladder continues, but is now proceeding towards the northwest.

What constitutes "optimal play" isn't always self-evindent, as Lolgast pointed out. To remove any ambiguities, here's the official definition of "optimal play":

• This is a game of full information, so for any given situation, the winning player can always be figured out, even though it may get complicated at times.


• The losing player will always choose the variation that leads to the most turns before losing.



• The winning player will always choose the variation that leads to a guaranteed win in the fewest possible turns.


• For the purpose of these variation length calculations, any stepladder terminating in an unseen edge will have ”many” steps.


• Given choices that are equal by the above criteria, the players will choose the option that is the most inconvenient for the purposes of solving the puzzle.

Here's a text version of the puzzle for the graphically challenged. X is black, O is white, dots are empty squares, and the target area is marked with underscores.

    . . . . . . . . . . . . . . . . . . . .
    . . . . . . . . . . . . . . . . . . . .
    . . . . . . . . . . . . . . . . . . . .
    . . . . . . . . . . . . . _ _ _ _ _ _ .

    . . . . . . . . . . . . . _ _ _ _ _ _ .
    . . . . . . . . . . . . . _ _ _ _ _ _ .
    . . . . . . . . . . . . . _ _ _ _ _ _ .
    . . . . . . . . . . . . . _ _ _ _ _ _ .
    . . . . . . . . . . . . . _ _ _ _ _ _ .
    . . . . . . . . . . . . . _ _ _ _ _ _ .
    . . . . . . . . . . . . . . . . . . . .
    . . . . . . . . . . . . . . . . . . . .
    . . . . . . . . . . . . . . . . . . . .
    . . . X X X X X . . . . . . . . . . . .

    . . . X O O O O O X . . . . . . . . . .
    . . . X O X X X O X . . . . . . . . . .
    . . . X O X . X O X . . . . . . . . . .
    . . . X O X . X O X . . . . . . . . . .
    . . . X O X X X O X . . . . . . . . . .
    . . . X O O O O O X . . . . . . . . . .
    . . . X X X X X X X . . . . . . . . . .
    . . . . . . . . . . . . . . . . . . . .
    . . . . . . . . . . . . . . . . . . . .

---

Looks like its time for the daily hint then! (2018-01-10)

Jimmy's answer is good. Very good, actually. But there is still room for improvement.

Answer

Here is my solution

With three pieces added

Two for the black player (in gray)
One for the white player (in blue)

After a few laps, Black has a choice : play in A or B

The boxes contain the corresponding play number

If he plays in A, there are two possibilities that lead to the victory of the white player:

White wins because Black can't stop him to have three liberties.

White wins by catching n11 and having one more liberty.

So he will play in B and get to these results:

---

Here is my previous solution which is not valid since, as commented by Bass, b14 should be under n13 to catch n9 and win.

strategy - Sliding Bolt Puzzle - fastest solution (time-wise)

This is a follow-up question for the Sliding Bolt Puzzle. If you have not solved it yet, you might want to head there first, as the extended discussion of its solution in this question will contain spoilers.

In the original version of the Sliding Bolt Puzzle, I asked for the fastest sequence in terms of the maximum number of button presses it takes to definitely open the door, which was given in the accepted answer. However, I was wondering whether there might be a sequence which, while having more button presses, is still faster time-wise when you take into account the probability of the unknown initial position.

To explain a bit more what I mean: As Kendall Frey explained, there are four possible states of the bolts, numbered 1 to 4. (We will neglect state 1 as an initial position, because it is specified that the door is locked at first.) However, these four states are not equally probable as an initial position: The probabilities are

• 8/14 for state 2,


• 4/14 for state 3,


• 2/14 for state 4.

(obtained by simply listing all 16-2=14 possible initial configurations and categorizing them accordingly).

So we see that state 2 is by far the most probable one; the sequence CBCACBC will open it after four touches of a button at the earliest, while the most improbable state 4 will be opened directly.

Hence, my question is: Which is the sequence with the smallest expected value E[t], where t is the number of button pushes until the door opens, taking into account the probability distribution of the initial configuration?

Note that there is also some randomness regarding the next state when you press button A while in state 2 or button B while in state 3 (see the transition state matrix in the accepted answer), which should also be considered.

mathematics - How long can a population last without incest?

A group of $n$ women and $m$ men have gotten themselves stuck on a desert island. Or, they're the last people left alive after the apocalypse. Whichever scenario you prefer. If they start reproducing, and assuming these people are all completely unrelated, how long will their colony last for if they want to avoid any and all incest, no matter how far removed?

More precisely, the rules are:

1. 
   

   A person's generation is the maximum of their parents', plus one. The initial population all have generation zero.

   
   


2. 
   

   You can reproduce with (and only with) anyone of the opposite gender with whom you share no common ancestor in the initial population (and a person is considered to be their own ancestor).

   
   



3. 
   

   Ignore the fact that people have a finite lifespan and that pregnancy takes nine months. You can have as many children as you want.

   
   


4. 
   

   You're trying to find the theoretical highest possible number of generations, so it's fine if your strategy requires that the kids come out with specific genders.

   
   

Try the case with, say, 1 man and 2 women (or vice versa) to whet your appetite, but the point of the puzzle is to work out the general case. Remember that you don't just need some upper bound, you need a tight upper bound - so you'll probably want to show an explicit strategy attaining that bound, along with a proof that that's the best you can do.

I have already solved this, so I can guarantee there is an elegant solution.

Answer

Considering numbers per gender creates a kind of distraction. Let $g(N)$ be the maximum number of 'incest-free generations' that can result from $N$ unrelated individuals, not all of whom are of the same gender. An additional individual unrelated to the $N$ others can mate with any of them who happens to be of the opposite sex. Selecting this mating partner from the most recent generation allows the formation of exactly one additional incest-free generation. So we have: $$g(N+1)=g(N)+1$$ Avoiding incest, a single man and a single woman ($N=2$) can produce no more than one generation of offspring: $$g(2)=1$$ Combining the above two results, it follows that $$g(N)= N-1$$ In this equation one can replace $N$ with $m+n$ to obtain the answer requested.

riddle - Wild Flash Graphics, Bro!

I've had many tricks played on me, but to my credit, I'm rarely so green anymore.

I used to pack quite a punch, things even got spicy with a mom.

I'll post a video greeting if and only if I'm in Wales.

Calling me Jack is sound, but what's my real ID?

Answer

I'm going to lay this one down:

card

I've had many tricks played on me, but to my credit, I'm rarely so green anymore.

Card tricks with playing cards, credit card, and green cards are scarcer with Trump.

I used to pack quite a punch, things even got spicy with a mom.

A pack of cards, a punch (or punched) card used for computer code entry, cardamom is a spice.

I'll post a video greeting if and only if I'm in Wales.

Postcard, video card, greeting card, and card + iff → Cardiff the Welsh capital.

Calling me Jack is sound, but what's my real ID?

A calling card, a jack is one of you, a sound card and an ID card.

Note to report on the suitable number of words the puzzle is carrying:

I count 52 words in this story - a full deck!

Title: Wild Flash Graphics, Bro!

Wild card, flash card, graphics card and courtesy of Chowzen the bro card which you can use occasionally with a good bro.

mathematics - 10 digit number where first n digits are divisible by n

What is the complete set of 10-digit numbers using each value ($0$-$9$) once, such that the first $n$ digits form a number divisible by $n$ (for $n \in \{1,2,...,10\}$)?

E.g. if the number were $1234567890$, $1$ must be divisible by $1$, $12$ must be divisible by $2$, $123$ must be divisible by $3$, etc...

Answer

There's only one answer:

3816547290

If leading zeroes are excluded, then there are 2492 numbers that satisfy the successive divisibility condition. This C program finds them all in a few seconds:

#include 


int main(void)
{
  long i,j,n,c=0;
  for (i=1000000080; i<=9999999990; i+=90) {
    for (n=i,j=10; j>1; n/=10,j--) if (n%j) break;
    if (j==1) {
      c++;
      printf("%ld ",i);
    }
  }

  printf("\nTotal: %ld\n",c);
  return 0;
}

// Output: 1020005640 1020061620 1020068010 ... 9876062430 9876069630 9876545640

However, 3816547290 is the only one of these numbers that contains all the digits 0-9.

---

On a computer, this can be solved quite quickly by restricting the range of numbers that has to be searched.

We know that the last digit must be zero, so the only remaining option for the fifth digit is 5. All the digits in the other even positions must be either 2, 4, 6 or 8. And since there are four even positions to be filled, the numbers in the odd positions (apart from 5) must be 1, 3, 7 or 9 (because there are no even digits left to go around.)

Digit:       1       2       3       4       5       6       7       8       9      10
Value:    1/3/7/9 2/4/6/8 1/3/7/9 2/4/6/8    5    2/4/6/8 1/3/7/9 2/4/6/8 1/3/7/9    0

There are only 4^8 (=65536) numbers that fit this pattern. These can be tested in just a few milliseconds.

mystery - Murder on Stack Exchange!

Disclaimer: no offence is intended to any of the puzzling.SE users mentioned (or not mentioned) here! I was inspired by this question to use them as characters, with no regard for reality.

You are walking through the mansion of the mighty Lord D'alar'cop, Duke of Puzzling in the vast empire of Stackexchange. You see almost nobody apart from the occasional servant. As you pass outside through the great double doors at the entrance to the mansion, you overhear part of a heated conversation between two footmen, Warspyking and Michael.

"His Lordship will hear about this! You'll be out on your ear before the end of the week!"

"Oh, will he indeed? We'll see about that. If you tell him anything about me, your adventures with a certain maid might also come to his attention."

"He'll never hear a word from you again, I can promise you. You..."

As you walk further out along the sweeping drive towards the edge of the estate, the footmen's conversation fades out of your earshot. It is November, and bitterly cold even at noon. The parkland surrounding you is covered in deep snow, but the drive has been kept clear of snow and ice for his Lordship's carriage. The winter panorama is astonishingly beautiful, and you do not even notice the cold.

You reach the Folly, a small tower just off the main drive where his Lordship is normally to be found at this time of day, poring over his books. A gardener in snowshoes, Oblongamous, is adjusting one of the snow sculptures that are dotted about the grounds, and fails to notice you as you walk up to the Folly and enter silently.

When you open the single door to the Folly, you find his Lordship sprawled over his writing desk in a pool of blood, clearly quite dead. His body is stiff and ice-cold; he must have been dead for some time. You wonder why nobody has yet found the body and alerted the authorities. Looking around for clues, you see a bloodstained knife lying on the floor just behind the door. Also the electric light is switched off, the only light coming from the sun shining directly through the open doorway. There are no windows. The killer must have turned off the light as (s)he left.

You leave the Folly, moving northwards across virgin snow towards the mansion. But you have only got a few metres when you hear a piercing scream from behind you. Looking back, you see a maid, Avigrail, running out of the Folly in a terrible state. Oblongamous is still nearby; when he hears her story, he checks quickly inside the Folly and then they both go running to the mansion to raise the alarm.

The bloodstained knife is found to bear the fingerprints of Gilles, a well-known rival of his Lordship, and the blood is of his Lordship's blood group. At Gilles's trial, his defence lawyer makes the case that his Lordship's death wound is much wider and rounder than the knife blade, and the bloodstains on the knife show it could not have entered the body. He argues that the blood was only smeared onto the knife, and finds a cut on his Lordship's arm which could have been used to produce blood for this purpose. "Someone," he thunders in his final summing-up, "has been trying to frame my client!" His eloquence impresses the jury, and Gilles is acquitted.

Avigrail and Oblongamous both appear as witnesses in court, and say they saw nobody entering or leaving the Folly other than his Lordship.

The police have nothing to go on, but you can help!

Who killed his Lordship?

Bonus question: how was he killed?

Answer

As Gilles has told us, I (the protagonist of the OP story, who discovered the body) was standing directly between the Folly and the mansion, but the maid and gardener did not see me, nor even my prints in the snow. Nor did my presence quell the footmen's argument. Furthermore, I do not notice the cold. I must conclude that I am invisible and incorporeal, though I'm rather proud of the supernatural feat of telekinetic concentration by which I opened the Folly door to discover the body. I suspect, in fact, that I am the ghost of His Lordship. This would explain why the author is so certain that I can help with the inquiry. If only I could remember what happened: it's all such a blur. Everything moved so fast, and then it was all dark...

Wait! it's coming back to me: I was killed by

my own hand! I was in despair at the constant bickering and intriguing of my staff and the popular failure of my snow sculpture exhibition. My last act was to frame my hated rival, Gilles, by cutting my arm with one of his knives. Then I plunged an icicle into my own heart, allowing the weapon to melt with my dissipating body heat. Goodbye cold, cruel world.

What I do not understand is how I could have seen Avigrail exiting the Folly. We know that the Folly's only door is on its South side, that there are no windows, and that I was a few metres North of the Folly when she came out. That must mean... my goodness! The walls of the Folly are fading away before my eyes! And the mansion - it's... everything is fading away, everything except that point of light in the distance. I feel myself drawn towards that light...

Warning: since His Lordship seems to have correctly solved the mystery of his own death, the comment thread below also contains spoilers.

Motive (added by OP):

His Lordship D'alar'cop had learned that riddles and challenge questions were about to be summarily banned from Puzzling.SE, and wished to end it all in his misery, starting a new life outside of the "vast empire of Stackexchange" in which his duchy (now in the process of being destroyed) lay. His final act was to attempt to frame his hated enemy Gilles.

Some extra explanation added by the OP:

The above answer already explains how one can deduce that the narrator is his Lordship's ghost. To work out that his Lordship committed suicide, consider the following:

When you open the single door to the Folly, you find his Lordship sprawled over his writing desk in a pool of blood, clearly quite dead. His body is stiff and ice-cold; he must have been dead for some time. You wonder why nobody has yet found the body and alerted the authorities. Looking around for clues, you see a bloodstained knife lying on the floor just behind the door. Also the electric light is switched off, the only light coming from the sun shining directly through the open doorway. There are no windows. The killer must have turned off the light as (s)he left.

How come the knife is behind the door? If a killer dropped it as he was running out, it would have been either within the room, not contained in the sweep of the door, or within the sweep of the door but on the outside, in which case it would have been swept outside by the closing door (the door was closed when the narrator arrived) and not swept behind the door as the narrator opened it.

I hope that's clear; if not, I'll try and make a picture!

mathematics - The Computer Science Department

On a tour of a nearby university, we went into the computer science department. I was fascinated by the courses they offered, of course, but there were weird grids displayed on screens along the hallways. Halfway along the tour, a student challenged me to find out what they represent.

Can you work it out?

Oh, and if it helps, there were a few ads displayed as well:

---

PDF with the 13 pictures in text form

---

(The story is entirely fictional)

---

Hints:
1.

The ads aren't necessary to solve the puzzle, but they will clarify the steps you need to make

Answer

Using Jonathan Allan's results, I suspect the final answer to be

INFORMATICS

How I reached this:

Every keyword Jonathan found should be used as a mask on the corresponding tile, and tiles are ordered according to the numbers which were also decoded by Jonathan. The original color of cells is not needed anymore, only the characters themselves are relevant from now on.

So here's how to decode each tile, but let me skip tile number 0 for a short moment, as it seems to have some anomaly - maybe my understanding is not yet perfect. EDIT: see below for updates.

Tile number 1 has the keyword 'oddness'. This suggests characters with an odd number as their ASCII code should be masked:

Looks like a letter N.

Tile number 2 - 'fourset' - character should me masked if its ASCII code has exactly 4 digits of 1 in its binary representation, or if the character itself is the number 4:

Letter F.

Tile number 3 - 'ispower' - character should me masked if its ASCII code is a power (with exponent greater than one) of a prime number - letter O.

Tile number 4 - 'rbshift' - character should me masked if its ASCII code and one of its neighbouring characters' ASCII code are in a right-bit-shift relation (that's an integer division by 2) - letter M.

Tile number 5 - 'alphnum' - character should me masked if it is alphanumeric - letter A.

Tile number 6 - let me skip this one for a while.

Tile number 7 - 'sixfact' - character should me masked if its ASCII code has exactly 6 different integer divisors - letter I.

Tile number 8 - 'divfour' - character should me masked if its ASCII code is divisible by 4 - letter C.

Tile number 9 - 'sqrfree' - character should me masked if its ASCII code is not divisible with any square number (except 1) - letter S.

So far we have letters: _NFOMA_ICS. Let's get back to tiles 0 and 6.

Tile number 0 - 'collatz' - I think it means character should me masked if its ASCII code and one of its neighbouring characters' ASCII code are consecutive elements of a Collatz-sequence, however the results are a littly bit dirty here:
It's almost a letter I, but the upper right part is missing. There are also some additional black cells on the left side, which might be a result of unintentional match between 68/2=34, or my interpretation is imperfect.

Tile number 6 - 'isprime' - I think it means character should me masked if its ASCII code is a prime number, but this one is dirty too:

It could be a letter T if there weren't those two extra black cells on the left side. A fault in my interpretation or in the puzzle generating process, maybe.

Anyway, I'll go with I and T as best guesses for these two, resulting in the final solution INFOMATICS - which again, seems to be missing an R. But maybe that's a proper english word - I've never heard it, but I'm not a native speaker. Or is this a reference to programming language R, and that it should be skipped because python is superior according to the posters?
EDIT: Tiles number 0 and 6 have been corrected, and now clearly say I and T, respectively. Also, tile number 3 has been updated, and instead of a single letter O, there are two letters now on it: O and R. I wasn't completely right about its masking though, as it seems, it's not only prime number's powers which should be masked, any integer to the power of at least two should be masked in the ASCII codes.

With all these modifications the solution is clearly: INFORMATICS.

riddle - Down the Rabbit Hole

There's a secret hidden in the text below. What mind-blowing truth does it reveal?

Not imaginative? Then Zeke's Clues may assist! Unsure, lad? With our patented Obscurity Clues, begin verifiably outwitting the great VIGENÈRE! My useful helper, Mr. Riddler, envisions heavenly light revealing clever elements in grabbers. For 6 Krones, cousin Kryptos has made plenty new devotees. But taste softly, victim, this advice: GO ACROSTIC. DOMAIN solves everything. Please review wise masters (xenophobic sociopaths) very selectively. Some KEYS percolate behind masks of kingly visages.

---

In re-solving this myself, I noticed a difficulty over which I have no control: At some point, capitalization will matter. The order is lowercase, uppercase, uppercase, lowercase. When see a tiger, you know you've got it right.

---

Hint 1

The capitalized words are obviously clues. The relate to how you perform the first step and, depending on how you look at it, the second step. The DOMAIN that is mentioned is fully qualified. I wanted to fit that into the original text but couldn't find a spot for it.

---

Hint 2

If you don't recognize the clue words, you should Google them. You should at least figure out what kind of puzzle the first step is just from that. The fully qualified domain will help you solve that first step. Don't get too caught up on the text.

Answer

I'll continue where alexmc Left off.

The tiger image can be decrypted (at http://www.mobilefish.com/services/steganography/steganography.php) with the pass code 13312(The question number). That leaves you with a qr code (called "Almost there") that, when decrypted gives the result: Jet fuel can't melt steel beams. This is probably referring to that which is said by 9/11 conspiracy theorists.

Me and my friend Callan worked on this one all day at school today. He was the one who finished the puzzle though, so full credit to him. I'm sorry for my terrible formatting, and I'm sure that I have not followed proper etiquette, but this is my first time on here so I'm a bit of a noob. Thanks for the puzzle! (feel free to edit up my post, it's probably gonna look pretty ugly)

mathematics - Meeting of hands on a clock

We have a regular analog clock with three hands - second, minute and hour. On a given day, how many times do:

(a) the minute and hour hands

(b) the minute and second hands

(c) the hour and second hands

(d) all three hands

meet?

One approach would be to just list out all the times they meet, but I'm looking for a more logical/mathematical approach.

Answer

I think there is a much simpler solution than all provided so far.

Consider the following three facts:

• hour hand will rotate 2 full times in a day


• minute hand will rotate 24 full times in a day


• second hand will rotate 1440 full times in a day

So then:

(a) the minute and hour hands, will meet exactly:

22 times: 24 - 2 (once every 24/22 hours)

(b) the minute and second hands, will meet exactly:

1416 times: 1440 - 24 (once every 24/1416 hours)

(c) the hour and second hands, will meet exactly:

1438 times: 1440 - 2 (once every 24/1438 hours)

(d) all three hands, will meet exactly:

twice: only at exactly 12:00:00 o'clock (noon and midnight)

Simply because:

The faster hand passes the slower hand by the number of laps it makes minus the number of laps the slower hand makes.

With the special case:

One hand lapping another won't necessarily coincide with the third hand being there. Try to find the common multiples of the three fractions 24/22, 24/1416, and 24/1438 and you will see there are only two at 24/1 and 24/2. (ie. 12 hours and 24 hours after the start).

mathematics - A truly amazing way of making the number 2016

Find a mathematical expression that yields the value $2016$ while obeying the following rules:

• Each of the digits $1,2,3,4,5,6,7,8,9$ is used exactly once


• Decimal points are allowed


• You may use brackets "(" and ")" to structure your expression, and to make it well-defined


• The only allowed mathematical operations are addition (+), subtraction (-), multiplication (*) and division (÷)


• The only allowed mathematical functions are square-roots and logarithms. Logarithms must be written in the form $\log[b](x)$ to denote the base-$b$ logarithm of number $x$

Note that in particular the following is not allowed:

• Juxtaposition of digits (as juxtaposing 1 and 3 to get "31")


• other mathematical operations and functions (cube-roots, exponentiation, factorials, absolute values, trigonometric functions, etc)


• matrices and determinants


• integration, differentiation, limits

Answer

$$\frac{9\cdot 8\cdot 7\cdot6\cdot 2}{3}+5-4-1$$

And three more à la Perry

$$1\cdot(2-3+4-5+6)\cdot7\cdot8\cdot 9$$
$$(1+2+3+4+5+6+7)\cdot8\cdot9$$
$$ 1\cdot(2+3+4+5)\cdot 6 \cdot(7+8+9)$$

lateral thinking - A UniCODE Puzzle

NOTE Please do not edit this puzzle or close for off-topic unless you REALLY know what you're doing. The whole thing, except this note, is part of the puzzle. Thanks!

What unicode character am I?

---

Hey guys! I'm not sure if this is the right site to put my question on but I am new to Javascript and I am trying to do a variable swap. When I run it it doesn't even run. Could someone explain my error to me?

var x = "E2";
var y = "9A";
var swap = x;
x = y;
y = swap;
console.log(x);

What is the error in My code? i'm very confused. no one I knew cuuld help me. Feel free to delete this if it'ssnot good ?

---

Hint 1

Find the errors. ALL OF THEM.

Hint 2

Does anyone want to edit my puzzle?

Hint 3

There are 3 parts to this puzzle. Find 2 characters and a message, then you put them together and you will get the answer.

Hint 4

Find the problem with my code first, the secret word second, then another character third.

Hint 5

No hex math operations are necessary.

Hint 6

Order: x y comment

Hint 7

Use Hex UTF-8 when you figure out Hint 6.

Hint 8

Use INCORRECT not CORRECTED (if you figured out this step- it's the one with the U)

Hint 9

Think .?!

---

The different pieces of this puzzle were solved by many people. Many of the answers expanded on previous knowledge to finally come to the solution. I am crediting the first person to find it.

Character 1 (debugging the code):

@athin

Character 2 (Find the 3 pieces):

Third bit discovered by: @wolfram42

Put together by: @nneonneo

Word Puzzle:

@Buildstarted found the majority of it, but a crucial part was missing. @Reinier found the rest of it, which enabled them to find the correct solution.

Put together (solved):

@Reinier

Answer

I think you might be

🇬🇷, or the flag of Greece.

As others have pointed out,

in the code there is a Greek question mark, and also ⚑, so a flag, can be found by putting together the values of x, y and the comment. Furthermore, others have found the mistakes spelling out "Minus". However, also the space and the question mark on the end should be a period, so this would make the message "Minus ?". So the complete result is "Greek question mark flag minus ?", making "Greek flag".

enigmatic puzzle - (9 of 11: Shikaku) What is Pyramid Cult's Favorite Stationeries?

<sub>Dear PSE users and moderators,
I’m new here in PSE, but I really need your help. There was this person who gave me a black envelope consisting 10+1 pages of puzzles, and also a scribble saying: “Find our favorites and you will be accepted to join our ‘pyramid cult’. Feel free to ask for help from your beloved friends on PSE. They will surely guide you into all the truth.” I’m also a newbie on grid puzzles, so, could you please give me any hint to solve these? It’s getting harder and harder later on..
- athin</sub>

_{Jump to the first page: #1 Numberlink | Previous page: #8 Ripple Effect | Next page: #10 Nurikabe}

---

Rules:

1. Divide the grid into triangles or quadrilaterals with the numbers in the cells.


2. Each triangle or quadrilateral is to contain only one number showing the number of cells in the triangle or quadrilateral.

---

^{Special thanks to chaotic_iak for testing this puzzle series!}

Answer

The finished puzzle

Answer: (thanks Deusovi!)

Take the middles of the odd-sized "line" sections to get HOLDERS.

geometry - Fair share of a square watermelon?

One hot day, Stan, Kyle, and Kenny were sitting outside with a square watermelon (actually it was a cube like the picture below).

Stan says "Let's cut the watermelon into 3 equal slices (like the right-side diagram below)."

Kyle says "But then the end slices have more rind."

Can you show them how to cut this square watermelon into 3 equal (congruent) pieces, so that each piece has an equal amount of rind? The correct answer should include a sketch or a picture and uses just a few straight cuts (no tricks).

Answer

This image from http://demonstrations.wolfram.com/ThreePyramidsThatFormACube/

solves the problem - but Stan or Kyle or Kenny would have trouble cutting their cubical watermelon this way since the point of the knife needs to follow the long diagonal of the cube while the edge of the blade follows a face diagonal.

Each individual pyramid has a square base and one vertical edge. It's the most interesting piece in the geoblocks available in many K-2 classrooms - the central one in this image from http://catalog.mathlearningcenter.org/store/product-924015.htm

This dissection is really much more than the answer to a puzzle. It's the essence of the theorem that says that the volume of a pyramid is 1/3 of the height times the area of the base, and of the fact that the integral of $x^2$ is $x^3/3$. The analogue in the plane is cutting a square in half along the diagonal. In four dimensions you can cut a tesseract into four congruent pyramids each of which has a cube for a base.