computer puzzle - MathJax looks kool

^{(This puzzle relies on familiarity with MathJax, the mathematics renderer that is automatically available for use in answers here at Puzzling Stack Exchange. Examples and more information can be found in related puzzles.)}

[Spejlet] fór dem ud af hænderne og styrtede ned mod jorden, hvor det gik i hundrede millioner, billioner og endnu flere stykker, ...
               _{— from Snedronningen (The Snow Queen) by Hans Christian Andersen}

The $\small\texttt{\mirror} \raise2.5ex\strut$ code flew out of their hands and crashed to earth, where it broke into hundreds of millions, billions and even —well, actually just fifteen in all — pieces:

  \mirror   \mirror   \mirror   {   }   %   #1   #1   #1   #2   #2

      $$\require{begingroup}\begingroup   \def   |look|   \endgroup$$

               _{— translated to MathJax}

Please help fit these pieces back together, like a jigsaw puzzle, for a reflective result:

$$|look|kool| \huge\raise-.5ex\strut$$

Spaces and multiple lines are allowed. For a hint and a half, see Davide Cervone’s solution to MathJax reflex. Your browser page might need to be reloaded in order to reset MathJax after errors while testing. A solution should render the result and need not display the assemblage.

Answer

$$\require{begingroup}\begingroup
\def\mirror#1#2
{
#1\mirror#2
%
#1}
\mirror|look|
\endgroup$$

$$\require{begingroup}\begingroup \def\mirror#1#2 { #1\mirror#2 % #1} \mirror|look| \endgroup$$

Trace, with $\, \raise-.4ex{\unicode{8629}} \:$ for line breaks: $$\require{begingroup}\begingroup \def \Type #1#2#3{ \def \Typf % ##1#1##2#1%\Typf##3{\texttt{##1}##3\Typf % ##2#1%\Typf{##3}} \Typf % #3#1%\Typf{#2}#1%\Typf% } \def \RETURN #1 #2\strut{ \texttt{#1} & \kern-1em \Type { } {\kern.3em\raise-.4ex{\unicode{8629}}\kern.2em} { #2} \\[.5ex] \normalsize #1 #2\strut } % \def\mirror#1#2 { & \kern-2em \texttt{\mirror} \, \underline { \texttt {#1} \tiny\strut } \, \underline{ \texttt{#2} \tiny\strut } & \longrightarrow & \RETURN #1\mirror#2 % #1} \small\begin{array}{lrcrl} \mirror|look| \strut \end{array} \endgroup$$

optimization - Odd hours with two timers

Two motorized 24-hour light timers are daisy chained between a power outlet and a light bulb.

For these timers, devise schedules and choose initial times that produce the following repeated 9-hour lighting pattern, with the largest possible whole number $d$ less than 9, beginning when the outlet's power is switched on.

     Light is on for $\:d~$ hours,$~$ off for $\,9\,$–$\:d~~$ hours,
      on for $\:d~$ hours,$~$ off for $\,9\,$–$\:d~~$ hours,

      on for $\:d~$ hours,$~$ off for $\,9\,$–$\:d~~$ hours,
      $~\,\vdots$

If you are unfamiliar with these timers

Each timer repeatedly cycles through its schedule of 24 intervals that last an hour each.
•$~$ A circular dial determines the current point in the schedule
•$~$ A motor rotates the dial to advance through its schedule whenever power is supplied to the timer
•$~$ You may initially set the dial to any minute of any interval
•$~$ You preset each interval to ON or OFF
•$~$ When the dial is in an interval that was set to ON, the timer acts as a direct connection for power to flow to whatever is plugged into the timer
•$~$ When the dial is in an interval that was set to OFF, the timer does not provide a power connection

•$~$ The resulting ON and OFF durations could be fractions of an hour if the timer is set to begin within an interval or if incoming power is interrupted during an interval

The first timer is plugged into the outlet.
•$~$ It runs nonstop once the outlet is switched on
•$~$ It supplies power— but only when its dial is in an ON interval — to the second timer

The second timer has the light bulb plugged into it.
•$~$ It advances through its schedule only when the first timer supplies power
•$~$ It lights the bulb, but only while powered by the first timer and when its dial is in an ON interval

Related puzzles
Halve time with two timers
Day and night of the two timers

Third timer's a charm

^{(This puzzle is directly from an actual botany experiment)}

Answer

Obviously, you can't obtain a 9 hour cycle with just one 24 hour timer, because 24 is no integer multiple of 9. So there have to be hours at which the first timer is turned off. As the first timer runs at a 24 hour cycle, it will be off at certain fixed hours every day.

The greatest common divisor of 24 and 9 is 3, so the 9-hour cycle occurs at all phase shifts that are multiples of three. If the on-time is at least 3 hours, there is no hour the bulb is off every day, which contradicts the observation that there must be such hours. So obviously $d \le 2$.

I will show a solution with $d = 2$, which thus is optimal. Set the first timer to 8 copies of 2 hours on and 1 hour off. The second timer is powered 6 hours per 9 hour interval. As 6 is a divider of 24, it is possible to program the second timer in a way that it has a cycle repeating every 6 hours it is powered, which is 9 hours real time. Set the timer to 4 copies of 2 hours on and 4 hours off, and have both timers start at the beginning of their on phases.

geometry - Tiling rectangles with Heptomino plus rectangle #3

Inspired by Polyomino T hexomino and rectangle packing into rectangle See also series Tiling rectangles with F pentomino plus rectangles and Tiling rectangles with Hexomino plus rectangle #1

Next puzzle in this series: Tiling rectangles with Heptomino plus rectangle #4

...up to 100 to come... I'll post them a few at a time. Why is this first one #3: Numbering is as per my heptomino data file and will skip rectifiable and uninteresting heptominoes. Some of them will be posed as no-computer hand-tiling only puzzles.

The goal is to tile rectangles as small as possible with the given heptomino, in this case number 3 of the 108 heptominoes. We allow the addition of copies of a rectangle. For each rectangle $a\times b$, find the smallest area larger rectangle that copies of $a\times b$ plus at least one of the given heptomino will tile.

Example with the $1\times 1$ you can tile a $2\times 6$ as follows:

Now we don't need to consider $1\times 1$ further as we have found the smallest rectangle tilable with copies of the heptomino plus copies of $1\times 1$.

I found 87 more but lots of them can be found by 'expansion rules'. I considered component rectangles of width 1 through 11 and length to 31 but my search was far from complete.

List of known sizes:

• Width 1: Lengths 1 to 20, 22 to 25, 29 to 30


• Width 2: Lengths 2 to 18, 22 to 24, 29 to 31


• Width 3: Lengths 3 to 8, 14 to 15


• Width 4: Lengths 4 to 25, 27, 29 to 31


• Width 5: Lengths 7 to 8


• Width 7: Length 8



• Width 8: Lengths 9 to 10

Many of them could be tiled by hand fairly easily.

wordplay - Four-words (----||||)

Four-words :
My first is indivisible , but not really.
My second is a cloak worn by men , but in ancient Rome.

My third is a cruel wicked person , but not a human.
My fourth is about 20 quires , but in reverse.

I was not allowed a very short title for this puzzle , but I was able to add a clue to make it cross the 15-letters-minimum limit for titles.

Answer

My first is indivisible, but not really.

The word ATOM means "indivisible" in ancient Greek, but atoms aren't really indivisible since they have protons, neutrons, quarks, etc. inside them.

My second is a cloak worn by men, but in ancient Rome.

A TOGA

is the answer there.

My third is a cruel wicked person, but not a human.

An OGRE.

My fourth is about 20 quires, but in reverse.

20 quires

form a ream, so the answer to this is MAER.

And the full answer is:

A T O M
T O G A
O G R E
M A E R

Lovely puzzle!

story - The SErial Killer (Part 3)

This puzzle is part of a series - part 1 is here and part 2 is here. Any comments I make on the OP are canonical and out of character (OOC). Any comments I post on answers are for flavour only.

Since part 2 turned out to be too easy (or at least, solved too quickly), part 3 is harder. If it turns out to be too hard, I will start hinting, but that won't happen within the first few hours. It's a puzzle - get thinking :-) This version has gone through about 5 revisions and I'm finally happy with it, and where the series is now going (shakes fist at skv).

---

You take a deep breath and settle yourself. In a moment of brilliance you realise that $\overline{XC}CCCLXXXIII$ is the Roman numeral for 90383, and the b on the end forms a hexadecimal colour. You quickly map this to the colour used on the & logo of the English Language & Usage Stack - it's an exact match!

Something's nagging at you in the back of your mind though - Handel seems strangely familiar. You cast your mind around... do you know any German composers? No, that's dumb. Male friends who wear wigs? Well, you know plenty of them but you don't think any of them are called Handel.

Hang on a second... "called Handel"... what if Handel is actually a handle for the real name? As soon as that occurs to you, you realise that a Doorknob is a synonym for "handle". Just like that, in one moment of brilliance you identify "Handel" and absolutely infuriate the narrator of the series, who was hoping to sneak that one under the radar for a little while.

Deciding that since you've already caused so many headaches, you figure your best lead would be to check out Doorknob's profile - it's more likely to have an unintentional clue than whatever fiendish and convoluted puzzle awaits you in EL&U. Besides, you're on the same Stack so it's not like you're going to cause the deaths of any more Stacks, right?

You open Doorknob's profile, and straight away you twig that something's clearly not right.

As you try to make sense of the mess, suddenly pop-up windows start appearing all over the screen. Initially they're white, but then they all start loading different Stacks! "Oh no... no, no, no, not good!" you cry, trying desperately to close them but there are just too many. As a last-gasp effort, you flick the power switch on the wall to turn your computer off. The screen goes black and the fans slowly whir to a stop.

You wait.

You fear starting the computer up again, but know that it's not going to be very helpful in bringing the attacker(s) to justice if you just leave it powered off all day, and it could be disastrous if they're allowed to run loose, unchecked. You steel yourself, take a deep breath and turn the computer back on.

You pull up your browser when the computers finishes booting up, and breathe a sigh of relief when absolutely nothing unusual at all happens. Browsing to Stack Exchange also seems safe enough - nothing explodes, the site looks normal, and for a moment you let yourself think that maybe everything's OK.

Then you look at the list of Stacks. An ominous message greets you, there are even more dead Stacks and critically, English Language & Usage has been killed. Your last lead has gone cold. You slump back in your seat, a sense of defeat beginning to crash down on you. What on earth can you do next?

OOC: Ask Ubuntu should be dead too - I missed that when making the image. I'll fix it for part 4.

Answer

This is all I've got so far but I'm at work and shouldn't be on PSE. :)

Check out Area 51.

Why?

Collectively earth, wind, fire, and water are the elements. Badly annoy Tim: anagram ANNOYTIM into the element antimony, symbol Sb, atomic number 51. "We are not alone" also fits Area 51.

riddle - Immovable object

I have a long history from east to west,
to some I'm the worst and others the best.
With a core made of resin I dance and I flutter,
no one judges me when I speak with a stutter.

I may be an object of legend or myth,

at first you won't find me but maybe in fifth.
For at any point, you summon power of three,
And look to the wind, you may just find me.

My presence strengthens each link in the chain,
certain times that you use me, it may be in vain.
Storms when you're older and peace as a child,
with us two together, adventures are wild.

A limited spectrum between ceiling and floor,
Shot full of holes from twelve until four.
In solving this riddle you're just out Of Time,

you should hear my voice as it is sublime.

It should be clear by now that I'm not a Paid Genius Poet,
"What am I, I ask?," by now you should know it.

(Edit) Rearranged stanzas and hints and got rid of spoilers >!

Part 3

Answer

You are

An ocarina

I have a long history from east to west,
to some I'm the worst and others the best.

"The ocarina belongs to a very old family of instruments, believed to date back over 12,000 years."

With a core made of resin I dance and I flutter,
no one judges me when I speak with a stutter.

They can be played legato or staccato. The ocarina has a resonating chamber (on which resin is a pun); being a wind instrument, it's a relative of the flute (hence: flutter)

I may be an object of legend or myth,
at first you won't find me but maybe in fifth.

It's present in the fifth Legend of Zelda game released, The Ocarina of Time, in which it was part of the game's lore.

For at any point, you summon power of three,
And look to the wind, you may just find me.

OK, we're into video game territory for sure here. The power of three could be the Triforce. It also could have to do with the three stones you have to gather from the Forest, Fire, and Water temples before you get the Ocarina of Time. Wind has to do with the type of instrument it is.

My presence strengthens each link in the chain,
certain times that you use me, it may be in vain.

Sometimes when you use the ocarina in the game, nothing happens. And "link" refers to, well, Link

Storms when you're older and peace as a child,
with us two together, adventures are wild.

Adult Link saves areas from evil - prior to this, the areas are ravaged with fire and storms; afterward, they are calm, as they were when he was a child. Also, the Song of Storms is a song that is taught only to Adult Link.

A limited spectrum between ceiling and floor,
Shot full of holes from twelve until four.

OK, this is back to regular ocarinae: The ocarina doesn't allow for a very wide range of notes, and also: "[v]ariations exist, but a typical ocarina is an enclosed space with four to twelve finger holes and a mouthpiece that projects from the body." - ibid.

In solving this riddle you're just out Of Time,
you should hear my voice as it is sublime.

The Ocarina of Time! Hooray! And "sublime voice" refers to the beautiful sound it makes.

It should be clear by now that I'm not a Paid Genius Poet,
"What am I, I ask?," by now you should know it.

"Paid Genius Poet" is an anagram of Giuseppe Donati, inventor of the ocarina

riddle - A Quick Easy Riley

I've hit a road block with my story riddles and need to refresh my mindset; therefore, I am writing a quick and easy puzzle for the community to solve. Good luck to all of you!

Look for the prefix, in order of real teal; using this combo, I'll help you feel.

It's the suffix you seek, contained in all light; look at the table, 89 on the right.

Echoed the infix, two in a row; after the rain, you'll see quite a bow.

Study this riddle, and take your time; if you think hard, the answer you'll find.

Hint:

You’re stuck on this riddle, analyze each word; flip it, reverse it, unscramble each term.

Hint:

The prefix you need, tis not in a whale; but can be found, in a tree, on a trail.

Answer

Before the answer, big thanks to gabbo1092 for doing the hard work on this puzzle. If you like my answer, go upvote his as well.

The word is

Truth

Explanation

Acrostic

The poem's acrostic spells out LIES, implying that we need to find the opposite or reverse of what the hints point to.

Look for the prefix, in order of real teal; using this combo, I'll help you feel.

Real Teal start with the letters R and T. Flip them and you get TR

Echoed the infix, two in a row; after the rain, you'll see quite a bow.

This is a lie. the letter only happens once. But it is shapped like a rainbow, just upside down: U

It's the suffix you seek, contained in all light; look at the table, 89 on the right.

Element 89 is on the left (not right) of Element 90, which is Thorium TH. The letters TH are found in "all light" with the TH being reversed.

Finally

TRUTH itself is the opposite of LIES and thus fits the theme of the puzzle quite well.

calculation puzzle - "Magic: the Gathering" Challenge #3: Legends Rule

_{Previous Challenge}

_{Next Challenge}

BACKGROUND:

Assume that this is a normal 20-life game of magic with no banned list (excepting cards from the 'Un' sets). If multiple solutions are tied for the fastest, the solution that provides the most damage will be chosen. Turns are counted such that the first player has turn 1, the other has 1.5, the first then has turn 2, the other 2.5 etc. So if you win on turn 5 on the draw, you have a turn 5.5 win. If you are on the draw and win on your opponent's 3rd turn, you have a turn 3 win. Your solution cannot involve opponent's cooperation (e.g. they choose to damage themselves). Assume that your opponent's hand and library consist entirely of basic Mountains and that they will play a Mountain every turn. Should they prove unable to play a Mountain for the turn, they will proceed to discard as normal.

PUZZLE SETUP:
This one is going to be different than the first two, instead of giving you a board state with a specific answer, I pose an open challenge: Perform the fastest win using only legendary cards (not just creatures, all cards used in the solution must be legendary, including lands--this means no instants/sorceries). You may choose to play or draw and assume that you will draw whatever cards are necessary, regardless of shuffling.

The best I have been able to come up with is a turn 3.5 win for 35 damage, but that doesn't mean you can't post a solution that is worse than that but still better than the others posted. A baseline is still useful and your solution may inspire someone to improve upon it and find the optimal solution (which may well be better than mine). Best of luck and be sure to be detailed in your solutions regarding sequencing, keep track of how many cards you have left in hand, and how lands are tapped for mana. Please do not post a solution if a better one has already been posted.

User Ninety-Three has beaten my original solution by a turn, but for those who were curious, see below for my initial solution.

Turn 0: Choose to be on the draw, play Gemstone Caverns, pitching anything (too bad Stormcrow isn't legendary, let's just say Tetsuo Umezawa). 5 cards left in hand
Turn 1.5: Draw, play Karakas, cast Kytheon, Hero of Akros with Karakas and Isamaru, Hound of Konda with Gemstone Caverns. 3 cards left in hand
Turn 2.5: Draw, play Gaea's Cradle, cast Bushi Tenderfoot with Karakas and Zurgo Bellstriker with Caverns. Then tap Cradle for GGGG and cast Surrak the Hunt Caller. Give himself haste and attack with kytheon, isamaru, and surrak for 2(Isamaru)+2(Kytheon)+5(Surrak)=9 damage. Flip Kytheon into Gideon, Battle-Forged. 0 cards left in hand
Turn 3.5: Draw, activate Gideon's 0 ability, turning him into a 4/4 creature. Tap Cradle for GGGGG, Karakas for W, and Caverns for R and cast Atarka, World Render. Give it haste with Surrak. Attack with all, Atarka gives itself double-strike to deal 1(Bushi)+2(Zurgo)+2(Isamaru)+4(Gideon)+5(Surrak)+12(Atarka)=26. Add that the the 9 damage last turn, and your opponent is at -15 life!

Answer

Update (11/17): Based on some comments on my previous solutions and a little more time to think, I'm submitting a single new solution, which I think is a (minor) upgrade.

Notable Features:

1) Kills on the first main phase of turn 2.5

2) Deals infinite damage (technically, infinite life loss)

3) Has no turn 0 plays, and no extra land drops

Solution:

Turn 1.5: Untaidake, the Cloud Keeper, tapped - pass turn (7 cards)

Turn 2.5: Pay 2 life to tap Untaidake for (2), Sword of the Animist; Mox Opal; Tolarian Academy; tap Academy for UU, Umezawa's Jitte (4 cards)

Tap Mox Opal for B, play Mox Opal (sacrificing the first), tap Mox Opal for R, play Grenzo, Dungeon Warden, play Mox Opal, sacrificing the second, tap Mox Opal for B, play Mox Opal, sacrificing the third, tap Mox Opal for B (BB floating, 0 cards)

Pay (BB) for Grenzo, revealing Derevi, Empyrial Tactician. Put her into play, untap Tolarian Academy, tap Academy for UUU. Repeat this three more times, sacrificing previous Derevis. (UUUUUU floating, 0 cards)

Pay (UU) for Grenzo, revealing Emrakul, the Aeons Torn, put it into your graveyard, shuffle your graveyard into your library, adding back in three Derevis and Emrakul (along with the Moxen).

Keep paying (UU) at a time for Grenzo, revealing (Derevi, Derevi, Derevi, Emrakul), untapping Academy with each Derevi before shuffling them all back in. This loop gains one U (-UU+UUU-UU+UUU-UU+UUU-UU), so repeat until you have infinite U. (Infinite U, 0 cards)

Pay (UU), (UU), (UU) for Grenzo, revealing Derevi, Derevi, Derevi - untap/tap Mox Opal for BBB. (BBB, Infinite U, 0 cards)

Pay (UU) for Grenzo, revealing Azami, Lady of Scrolls, put her into play, use Azami to tap Darevi to draw a card. (BBB, Infinite U, 1 card)

Play Maga, Traitor to Mortals for X = infinite, killing your opponent.

Alternatively: for a milling victory, tap the Mox for RB, pay (UU) for Grenzo to reveal Ambassador Laquatus, play Lord of Tresserhorn from your hand, and respond by paying infinite for Laquatus' ability, targeting your opponent.

Old Solutions:

I'll take a crack at this, since I'm enjoying these puzzles quite a bit! I've got full-infinite turns/mana/damage/cards on turn 2.5. I can also deal 20+ damage normally on that turn with minimal difficulty, in case we dislike taking infinite turns. I don't have enough reputation to post links, so you'll have to find some of the cards yourself (sorry!).

Turn 0: Gemstone Caverns, exiling a second Gemstone Caverns (5 cards in hand)

Turn 1.5: Shinka, the Bloodsoaked Keep; Zurgo Bellstriker; Kytheon, Hero of Akros (off Gemstone) (3 cards in hand)

Turn 2.5: Gaea's Cradle; Grenzo, Dungeon Warden for RB (Gemstone for B); tap Gaea's Cradle for GGG floating. Pay (2) for Grenzo, reveal Derevi, Empyrial Tactician - Put her into play, untap Gaea's Cradle. Tap Cradle for GGGG. (GGGGG floating, 2 cards in hand)

Pay (2) for Grenzo, reveal Azami, Lady of Scrolls - put her into play. Pay (2) for Grenzo, reveal Azusa, Lost but Seeking - put her into play. Play Gaea's Cradle #2, sacrificing the first one. Tap Gaea's Cradle for GGGGGG. (GGGGGGG floating, 1 card in hand)

Pay (2) for Grenzo, reveal Prime Speaker Zegana - put her into play with two counters due to Grenzo. Draw three cards. Pay (2) for Grenzo, reveal Endrek Sahr, Master Breeder - put him into play. Pay (2) for Grenzo, reveal Derevi, Empyrial Tactician #2, put her into play, sacrificing the first - untap Gaea's Cradle. Tap Gaea's Cradle for GGGGGGGG. (GGGGGGGGG floating, 4 cards in hand)

Pay (2) for Grenzo, reveal Derevi, Empyrial Tactician #3, put her in play. Sacrifice the one already in play, untap Gaea's Cradle. Tap Gaea's Cradle for GGGGGGGG (GGGGGGGGGGGGGGG floating, 4 cards in hand). Play Emrakul, the Aeons Torn, triggering Endrek Sahr, Master Breeder for 15 Thrull tokens. Sacrifice Endrek as a result. Play Karakas (your third land drop thanks to Azusa). Take your extra turn from playing Emrakul. Do whatever you like, tap Karakas to return Emrakul to your hand, tap Gaea's Cradle to generate >15 mana due to Thrulls, replay Emrakul. Take infinite turns, using a second Emrakul and the legend rule to avoid decking. Draw any cards you like off Azami, Lady of Scrolls. Use Omath, Locus of Mana for infinite mana and infinite power/toughness, because why not.

For the no-extra-turns version of this:

Use your giant pile of mana on turn 2.5 for Heroes' Podium. Attack with Zurgo, Kytheon, and your favorite 2 power creature with haste revealed off Grenzo (I'm partial to Kiki-Jiki, Mirror Breaker) with at least four other legendaries in play to deal 21+. I didn't try to optimize this, but 100+ shouldn't be hard; there are a lot of hasty 1 or 2 power legendaries and it goes up as the square of the number of creatures.

And, as a bonus answer, a second way to start off that infinite turn 2.5 combo:

Turn 0: Gemstone Caverns, exiling a second Gemstone Caverns (5 cards in hand)

Turn 1.5: Mox Opal; Tolarian Academy; tap Academy for U, Sword of the Chosen (3 cards in hand)

Turn 2.5: Tap Academy for UU; Sword of the Animist; tap Mox Opal for B and Gemstone Caverns for R. Play Grenzo, Dungeon Warden. Play a second Tolarian Academy (sacrificing the first), tap for UUU floating, 1 card in hand.

Pay (2) for Grenzo, reveal Azusa, Lost but Seeking. Play Gaea's Cradle as your last card, tap for GG (GGU floating). Pay (2) for Grenzo, reveal Derevi, Empyrial Tactician, untap Gaea's Cradle. Tap for GGG (GGGU floating). Pay (2) for Grenzo, reveal Prime Speaker Zegana, draw 3. Pay (2) for Grenzo, reveal Endrek Sahr (0 mana floating, 3 cards).

Play a second Gaea's Cradle, sacrificing the first. Tap it for GGGGG. Play Titania, Protector of Argaroth, triggering Endrek for 5 Thrulls and returning your first Gaea's Cradle to the battlefield. Sacrifice the tapped one, triggering Titania for a 5/3 elemental token. Tap Gaea's Cradle for GGGGGGGGGGGG. Pay (2) for Grenzo, reveal Derevi, Empyrial Tactician (sacrificing the old one), untap Gaea's Cradle. Tap Cradle for GGGGGGGGGGGG (GGGGGGGGGGGGGGGGGGGGGG floating, 1 card in hand). Play Emrakul, the Aeons Torn triggering Endrek for another 15 Thrulls, then go infinite as before.

situation - The Strangely Clairvoyant Hobo

One day, I was waiting to get on the subway. I was eating the apple I always have for breakfast, and I decided to sit next to a homeless man on the bench. We saw a rather... rotund man walk by us, to which the hobo muttered, "Pig."

I didn't give it a second thought, except how rude the comment was. Another man walked past, this time tall and in a business suit. As he walked past, the hobo said "Human." and I thought nothing of this remark, other than how obviously he was, in fact, human. I soon after boarded the train for work.

The next day, the hobo was still there. Today, as I watched from a distance, he was at it again. Several people walked past him: a skinny woman, a muscular man, and an old lady, to which he respectively muttered "Soup", "Pork", and "Cookie". Odd, I thought, considering none of them was a cookie. Throughout my day, I couldn't stop thinking about this weird hobo in the subway.

Over the next few days, I kept observing him. He continued this odd behavior. "Bread", "Carrot", "Milk", "Rabbit", and other utterances as people walked past. So the next day, I walked by the homeless man again and he muttered, "Apple."

I finally recognized what this hobo was doing... and after thinking on it for a little bit, remembered something and was stricken with horror.

What was this hobo's ability? What had me so disturbed?

Answer

He was listing what people had eaten (at some point in time) and had said "human" at one point. That meant on the first day he had seen a cannibal.

language - The Babel Brotherhood - #2

Inspired by @Racso's Babel Brotherhood #1, I've decided to create one as well. Good luck to you all! Also, @Racso, I would love some feedback on my first attempt so I can make sure my future renditions do your puzzles justice!

---

The Babel Brotherhood is composed of people from all around the world.

---

Two members of the Brotherhood, codenames Typhoon and Washington, where enjoying a nice vacation after their previous missions. They were sitting on the beach drinking Mai Tais and trying to figure out what activities they wanted to do the next day. After half an hour, Washington asked Typhoon what time it was; she responded and Washington began scratching her back. He then asked again, what time is it?

What did Typhoon say, and why did Washington begin scratching her back?

Answer

Perhaps Typhoon said

ichi, the Japanese for "one"

and Washington

thought she had said "itchy".

But

I'm not sure whether you can just say "one" when it's one o'clock,

so another possibility is that Typhoon

said something like "hachi-ji", meaning "eight o'clock"

and Washington

thought she had said "back itchy". Though that seems a bit of a stretch.

mathematics - Can consecutive numbers form a palindrome?

Consider a number formed by concatenating all the natural numbers from $1$ to $n$, for some $n>1$.

(E.g. with $n=13$ this number would be $12345678910111213$.)

Is it possible for such a number to be a palindrome?

Please provide either an example or a proof of impossibility.

_{I found this puzzle in the 1996 All-Russian Mathematical Olympiad.}

Answer

(I'm sure I saw almost exactly this question somewhere in the last week or thereabouts. Maybe my brain's playing tricks. I don't think I read the 1996 All-Russian Mathematical Olympiad recently. Perhaps there was another question about concatenated consecutive numbers.)

The answer is that

it is not possible.

Why?

Suppose $n$ has $k$ digits. By inspection $k>1$. Let $m$ consist of the last $k-1$ digits of $n$; then the end of our monstrous number looks like $k-1$ 9s, then the first digit of $n$, then $k-1$ 0s, then another $km$ digits. Therefore its start looks like the reverse of that: $km$ digits, then $k-1$ 0s, then the first digit of $n$, then $k-1$ 9s.

But

this can't happen. That string of $k-1$ zeros after the first $km$ digits can't include the first digit of any of the constituent numbers making up the monstrous number, so they are in fact the final $k-1$ digits of a $k$-digit number. But then they cannot possibly be followed by a single digit and then $k-1$ 9s; rather, they must be followed by a single digit, $k-2$ 0s, and a single 1. Contradiction.

visual - The untimely death of Dr. Wenzel: What give-away clue is hidden in the crime scene image?

Dr. George Wenzel and his colleague Dr. Martin Hupp are two expert biochemists researching brain-wave-changing drugs. Their work has long been regarded as being worthy of the Nobel Prize, in particular since their recent success with large cats. They showed that they could boost the intelligence of those animals to near-human level. The main challenge they had to overcome was the side-effects on the cats. Being so intelligent made them extraordinarily aggressive. (Maybe they realized their situation in full?)

Anyway, the doctors' latest discovery was generally assumed to be so fundamental, that there wouldn't be a choice but award them the prestigious prize.

So it came as a shock that on the very day a press-conference was announced which would disclose their latest results, Dr. Wenzel was found dead in the animal laboratory. He was ripped to pieces by one of the tigers they were experimenting on. His colleague Dr. Hupp could not explain at all how Dr.Wenzel could have been so careless.

"He normally took all precautions very, very seriously," he said. "And this time it was especially tragic. We had just discovered the proper drug to get all the positive effects and keep the animals calm. I tested it myself, and I've left George all the notes - see there on the table. I really don't understand this. The formulae does work properly. I have even cross-checked it this morning. The only explanation I can possibly think of is that he tried to modify the mixture even a bit more... But why would he do that?"

Inspector Edo Clu, who is leading the investigation, is wondering the same thing. However, it does not take him long to have Dr. Hupp as his prime suspect. Who else would have more profit of the untimely end of Dr. Wenzel than him? The Nobel Prize, after all, can only go to living scientists.

So he gives the surroundings another very close look, then calls Dr. Wenzel's mother to ask her a single question, and based on the answer arrests Dr. Hupp on suspicion of murder.

The puzzle is: What did inspector Edo Clu ask Mrs. Wenzel and what was the answer?

Now the two relevant photos from the crime-scene are shown below. (please use your imagination a bit, I'm not an artist) One shows the desk of Dr. Wenzel on which a note from Dr. Hupp is found together with a couple of post-it blocks and some pencils which have been used all over the lab.

The second shows a group of labelled bottles all standing in a group (no sorting) on the shelf. According to Dr. Hupp, these contain the latest drugs. The bottles are labelled with the post-its and pencils found on the desk (as are other items in the lab on different photos not shown.) There are two distinct handwritings on them.

---

Transcription of the letter:

George, we did it! Yesterday's experiment series went perfectly fine. None of the animals showed any of the side-effects at all! It seems that you were right about adding the Xonophoricum Hauratio to the mixture. It totally neutralized all of the aggression while keeping the brain-stimulating effects. You have to see this for yourself, and then you should prepare for the call from Sweden! Have you ever cuddled with a tiger? Now you can. He actually purred while I stroked him yesterday. Unbelievable. I have left one of the bottles in the prep-room, if you want to see for yourself. As it was a long day, and tomorrow will be even longer with the press interviews and stuff, I’ve gone home already. See you tomorrow. Make sure you come shaved, our pictures will go around the world!

and the five bottles:

Series VII/3/2.1; What do you think,George?; no good

Series VII/5/1; extra aggressive!

Series VII/5/1.1; placebo

Series VII/6/1; new mixture;brain-boost! This is it!; too aggressive; Don’t use!

Series VII/6/1.0; +XoHa

---

---

This is my first 'crime' puzzle and I'm sure it could be better, but I hope it's still good fun. Asking questions in the comments is welcomed and may lead to edits. If there is no solution in a week (which I highly doubt), then I will start giving explicit hints.

Answer

He asked her

Did your son suffer from deuteranopia (red-yellow-green color blindness)?

Deuteranopia is characterized by an inability to distinguish between colours in the red-orange-yellow-green spectrum except by brightness.

The two Series VII/6 boxes are identical save for the labeling. We can reasonably infer that Dr. Wenzel died when Mr. Cuddles (the tiger) mauled him to death after receiving the too-violent version of the serum. This suggests that the means of committing the murder was related to doctoring the labels.

The serum labeled "+XoHa" is the ideal serum, but carries a green-on-yellow label that Dr. Wenzel wouldn't be able to read easily. This wasn't a large problem when there was only one series 6 serum, but when a second series 6 label was accidentally printed and applied to a too-aggressive serum, it became important that he be able to distinguish old from new. Dr. Hupp's green writing barely contrasts against the yellow background and so Dr. Wenzel writes "new mixture" in red, which he's able to see slightly better on the new too-aggressive serum where the "too aggressive" warning isn't visible to him.

Knowing about Dr. Wenzel's colour blindness, Dr. Hupp pens in "brain-boost! this is it!" in bold blue letters on the too-aggressive serum label and invites Dr. Wenzel to cuddle with Mr. Cuddles. Dr. Wenzel, upon seeing the labels, immediately sees the bright blue nicely contrasted and doesn't even bother to check for his red "new mixture" label contraindicating the serum since it's difficult for him to discern and is redundant thanks to the helpful, clear label applied by his colleague.

Cuddling with Mr. Cuddles doesn't go so well for him.

Upon learning that Dr. Wenzel is color-blind and vulnerable to this kind of attack, Inspector Clu arrests Dr. Hupp on suspicion of deliberately confusing the identities of the two serums, resulting in a classic case of tigercide.

mathematics - The grandest bridge in Appelhaken

In the capital of the grand country of Appelhaken there is a plane garden containing four magnificent monuments. So magnificent are they that the country has a Law requiring that there must be a clearly marked carpeted path all the way from each monument in the garden to each other monument, and no two such paths may share parts or intersect.

A fifth monument is installed. In order to comply with the Law, the best architect in all of Appelhaken designs a grand and magnificent bridge in the garden, and one path is laid on top of the bridge while another is laid underneath it, for without this bridge the Law could not be upheld.

Shortly afterwards, two more monuments are installed, but there is no more money for additional bridges because the first bridge was too grand. How can the chief gardener of Appelhaken reroute paths to preserve the Law?

Hint:

The bridge might need to be used for several paths.

Answer

Four monuments

Easy map to draw:

Or, topologically equivalently (and more usefully for our purposes):

Five monuments

Starting from the last map above:

(Green is the bridge, with one path crossing over it.)

Seven monuments

Starting from the five-monument map above:

The two new monuments are in red, and new paths in thin lines.

Left red can be connected directly (no bridge use) to three blues: top left, top right, bottom left.
Right red can be connected directly (no bridge use) to three blues: top right, middle, bottom right.

We draw in these six paths right away with thin black lines, but in two cases (left red to top left blue, right red to middle blue), we must choose which side of the existing mini-path (leading up to the bridge) to put the new path. We make the choice in such a way as to block off further access to the top right blue, which is now connected to everything so we don't need it any more.

The remaining paths (left red to middle and bottom right; right red to top left and bottom left; red to red) must cross the bridge. We draw these paths carefully so that they don't cross each other. I've used thin red lines for paths to the right of the existing path on the bridge, thin blue lines for paths to the left of the existing path on the bridge, and the red-to-red path could be on either side so it doesn't matter and I've just used a thin black line.

physics - A pencil in a beaker of water

You have a beaker containing 100ml of (slightly salty) water, standing on a digital scale. The scale has been calibrated to the beaker so that it shows only the weight of the water, which is exactly 100.0 grams. The beaker is perfectly cylindrical, and the depth of the water in it is exactly 5 cm.

You have a pencil which is half the density of the water. The rear end of the pencil is flat and has a surface area of 0.5 square centimetres. You put the rear end of the pencil into the water until it barely touches the bottom, and hold it here.

What will the scale say?

(Hand holding the pencil not pictured)

Answer

Well it seems to me that

The delta weight will be equal to the force of buoyancy against the pencil. This is Archimedes Principle in that the buoyant force is equal to the weight of the water displaced.

Also, as @Daniel Mathias points out

the water level goes up. So we have initially $5 \pi r^2 = 100$. Afterwards our new height is $x$ so $x \pi r^2 - 0.5 x = 100$. Solving this gives $x = 5.128205$.

Therefore

The weight of the water displaced is .5 sq. cm x 5.13 cm x 1000 kg/ cubic meter (the density of water) x 1/1000000 cubic meter/ cubic cm which comes out to be .002564 Kg or 2.564 grams

So the scale would read very close to

102.564 grams

mathematics - A row of 2015 red and white chips

There is a row of 2015 chips, of which 2014 are white and one is red. You are allowed to make moves of the following type: "Choose one red chip, and flip the colors of its two neighboring chips (from red to white, respectively from white to red)." Your goal is to turn all chips red by a sequence of such moves.

Question: For which positions of the red chip (in the initial situation) can you reach your goal?

Answer

As has been conjectured by others, this is solvable for only the chip in the middle. To show this, let us look at the problem in a new way; instead of considering which chips are red, let us consider which have been chosen an odd number of times.

___*___

___X___
__XX___
__XXX__
_XXXX__
_XXXXX_
_XX_XX_

where a move is legal on the starting square (marked by *) when it has either $0$ or $2$ neighbors which are $X$'s (meaning it remains red) and a move is legal on any other square when it has exactly $1$ neighbor which is an $X$ (meaning it was flipped from white to red). Obviously, we win when a move would be legal on every square, since that means that all squares are red. This is the case in the last line of the solution above.

We will show that all reachable positions have a common form. In particular, any position may be reached by alternatively adding and subtracting intervals from atop each other where each successive interval contains the initial position and is strictly contained in the previous interval. So, for instance, here is a few steps of addition and subtraction showing the form of such an interval:

__________*_______________

_XXXXXXXXXXXXXXXXXXXXXX__
_X___________________XX__
_X_XXXXXXXXXXXXXXXX__XX__
_X_XXX____________X__XX__
_X_XXX_XXXXXXX____X__XX__
_X_XXX_XX___XX____X__XX__
_X_XXX_XX_X_XX____X__XX__

where all the lines are legal positions. Notice that any such position is easily obtainable since one can flip the center when it is strictly contained in an interval (as it would have either $2$ or $0$ neighbors) and then one may expand out the new interval until it hits the boundary of the old interval, since until that point, the change from new interval to old interval would make moves legal at the edge where it went from all X's to all _'s.

Conversely, if we are in a state which is such a nest of intervals, any legal move leads us to a state which is still expressible as a nest of intervals. A convenient way to show this is to note that an equivalent characterization is that "the number of changes from _ to X is the same going out from the initial block to either edge" To show this, one may consider two cases. First, if the move must either be in the original chip's position, meaning that it was either contained wholly in some other interval (meaning both sides gain a change) or was the unique point in some interval (meaning both sides lose a change). Otherwise, the move must have occurred at a point . in the pattern X._ or _.X. Either way, the number of changes does not depend on whether . is X or _ so this move maintains the nested interval property.

This means that this is a complete characterization of the legal moves from a given center. The important thing to note is that there is are precisely two solutions in terms of where the $X$'s go for any starting solution. To prove this, notice that flipping any interior position flips two pieces and that no non-trivial such set of flips yields no change, as the rightmost/leftmost affected piece are affected precisely once. Thus, any change affected by interior pieces may be made in precisely one way. Then, a boundary piece may be flipped, and if so, so must the other one - giving one degree of freedom in whether to flip both boundaries. Moreover, by the argument presented by @DrunkWolf, odd positions are non-solvable. One may check that the following four patterns solve every possible board (as long as the length of the board is of the form $4n-1$ and the * is at an even position). In particular, if * is at a position of the form $4n$ then the following two patterns apply:

...___________*___________...
..._XX__XX__XX_XX__XX__XX_...

or

...___________*___________...
...XX__XX__XX___XX__XX__XX...

and if * is at a position of the form $4n+2$ then the following two patterns apply:

..._____________*_____________...

..._XX__XX__XX__X__XX__XX__XX_...

or

..._____________*_____________...
...XX__XX__XX__XXX__XX__XX__XX...

Merely verifying that the relevant two these are indeed solutions for any setup suffices to show that they are the unique two solutions. Note that all solutions are shown on a row where, without the ellipses, they are valid. All of these change state every two lengths, meaning that, unless they are within one chip of the center, the number of changes from X to _ will not be balanced on each side, meaning that the move will not be possible to reach. Conversely, if * is directly in the center, either solution is symmetric, meaning both sides obviously have the same number of changes and the solution is obtainable (and we can achieve it without flipping the outer pieces, since exactly one of the solutions above will do that and the other will not)

geometry - Building a pyramid from di-spheres

Suppose you want to build a square pyramid made up of 30 spheres. So the bottom layer is a 4x4 square arrangement of touching spheres, and it has 3x3, 2x2 and 1x1 layers on top.

The building blocks you have are 15 di-spheres, i.e. they consist of two spheres stuck together.

At all times while building the pyramid, you want every sphere to be fully supported underneath, either by four spheres in the layer below or by the ground.

Is it ever possible to get stuck, where there is no possible way to complete the pyramid without first removing pieces?

This question arose when someone I know was trying to design a two-player game. It is not a very difficult puzzle, but fun to work out.

Answer

Does this arrangement work?

Fist layer:

AGBB
AEEH
IFFD
CCJD

Second layer:

.G.

I.H
.J.

Same letter denotes same disphere. The dispheres are placed in alphabetical order.

story - Internship Available!* - Figure out what you're being asked to do before you sign up

Sine waves bound in rectangles shaped into prisms. Mass increases, position changes twice, and mass decreases. The prisms become rectangles and are either discarded or stored for later use. If that sounds exciting to you, you may have an exciting future in our field! We need as many able-bodied candidates as possible to help with our project starting ASAP!

Due to lack of funding to actually pay a skilled team, we offer these fantastic incentives instead! How would you like some soggy old gramineaeze? Perhaps acidified coagulation and bio-viscera upon heavily triticumized water? Imagine what you could do with all of that! Sign up today to start this FANTASTIC ADVENTURE!

YOUR FUTURE IS WAITING! CORPORATIONS DON'T UNDERSTAND THAT ALL CAPS IS BAD. JOIN US!

*Company is not responsible for death or injury caused by crushing or falling.

What are you being asked to do and how are they paying you?

Answer

I think that you're asking someone to

help you move.

Reasoning:

The prisms are

cardboard boxes.

The sine waves are the

cardboard's corrugation.

You move them twice: once from

the house to the truck, then back into the new house.

When you're done with

the boxes, you break them down

so they're rectangles again. You can store them for later or throw them out.

You will be paid for your help in

pizza and beer.

Gramineaeze refers to

Poaceae (aka Gramineae), the family of plants that includes the cereals used in the brewing of beer. Through drenching and allowing the cereals to ferment, magical beer is born!

The other item is

Pizza, a combination of: | cheese, a milk product that is acidified (soured) and whose proteins are made to coagulate | Sauce/meat/veg, biological life that has been eviscerated (a term whose dictionary definition means the removal of organs, but whose popular definition often refers to the ripping or cutting apart of a living entity in a very brutal or astonishing way). | Dough, a combination of flour (ground grains of the triticum family) and water.

wordplay - Simple twisting riddle

One of the riddles we used to have painted on the walls of one of our classrooms in secondary school (with the last part added in :+ ):

What runs but doesn't walk,
has a mouth yet still can't talk,
has a head but doesn't think,
has a bed but will not sleep
and twists yet is not a riddle?

Took us 12 year old students a couple of days to figure out back then (despite it being terribly obvious in hindsight of course). So just wondering how many minutes it will take here :P .

Answer

The answer is

A river. As you said, it is pretty obvious. They run, have heads, mouths, and beds, and they often twist.

cipher - Linguistics: Japanese Squares

These are the squares from 1 to 100 in Japanese kanji:

百、六十四、十六、九、八十一、三十六、四、一、二十五、四十九

Which is which?

If you already know Japanese/Chinese numbers, then sorry, this puzzle is not for you. Needless to say, looking it up is cheating, even to check your answer.

Answer

Great puzzle!

I think the numbers are:

1 一
4 四
9 九

16 十六
25 二十五
36 三十六
49 四十九
64 六十四
81 八十一
100 百

Here's how I reasoned.

There are four one-symbol numbers 一, 四, 九, 百. There's only three one-digit squares, so I reasoned that one must be 100. 百 fits this by appearing nowhere else. I guess that the stick 一 is probably 1. Of 4 and 9, there's a square 49, which uses both symbols 四十九. But I don't know if I should read left to right or right to left. But 4 also appears in 64, and 9 doesn't appear elsewhere, so that means 六十四 is 64 and 四 is 4. So, left to right. Only here did I notice that 十 was in the middle of every three-symbol example. I had thought it meant 50 before, but that doesn't make sense. It must be a digit separator, or maybe the symbol ten (like "six tens and four"). The second one works great with 十六, which must be sixteen, since "ten" makes sense for "one ten". And that matches the 6 from 64, great. This leaves only 81, recognized from the 1, and 25, which remains and overlaps no digits.

pattern - The Secret Club

Since I was a kid, nobody wanted me in their super-exclusive clubs... the "House in the Tree Club"... the "Girls Only Club"... the "Whatever you are NOT Club"... but now that I am a skilled (and a bit nerdy) guy, time has come to take my revenge!

I've just discovered that not far from my house in New York City, there's a secret place where some guys seem to meet every week. I don't know what kind of club it is, but I definitely want to be a part of it!

Every night, each member knocks at the door of the Secret Club, and each time, a voice comes from inside, asking some strange questions...

At this very moment, someone is knocking on the door... ( - Let's listen to what they say... - )

"Knock! Knock!"

"Who's knocking?" asks the voice.
"I am a member Sir, let me in!" answers the guy.
"If you're a member, tell me what are the results of these operations: 17 + 5? 24 + 3? 1 + 16?".
"Of course... 22... 27... 17!" is the guy's response.
"Very well! Come on in!".

(- Well, I guess it's not the MENSA Club... - I think...)

A second guy approaches the door, knocks at it, and again the same scene takes place:

"Prove that you're a member! Answer these: 38 - 37? 28 - 10? 14 + 4?"

"Ehm... 1... 18... and again 18!".
"Good boy! Come in and take a seat near the other guys!".

(- OK, it must be some sort of math-nerd association... but better than nothing!...)

One last guy knocks at the door, but this time the question they ask him is very very strange...

"Well, if you are really one of us, tell me... What's our special number... the one with the unique property we all know?..."
The third guy seems slightly surprised... keeps thinking for a while, and then shouts: "Ah ah!.. You're referring to 12! Are you?".
The voice from inside the house starts laughing "Ah ah! For sure you are one of us!", opens the door and lets the guy in...

(- Well... I am a bit disappointed... but I'll give it a try... I'm sure they'll let me in!-)

"Knock knock!... ehm... I'm a member! Let me in!"
"Are you sure? If so, tell me... What's the result of 11 + 1? 10 + 17? 21 + 3?"
"...Ehm... 12... 27..." I start answering...
"OK... and what about the last one?" says the voice
"24!!" I shout... just a second before the guy from inside starts to yell:
"You liar! Go away and never come back to this respectful club!!"

At the end of this shocking experience, I still have a question in my mind:

"WHAT KIND OF CLUB WAS IT THAT I WAS TRYING TO ENTER?"

---

HINT 1:

The club is a real club. Many clubs like this one exist in real life. Probably, many members of the Puzzling Community are members of such a club in real life. Of course, the secrecy about the club is purely "fictional" and for the purpose of the puzzle.

HINT 2:

One of my guesses about the club's members was completely wrong: they ARE NOT "math nerds", and you don't need particular math skills to solve the calculations asked by the man at the door. Anyway, not everybody could answer those questions like the members did.

HINT 3:

Pay attention to the tags: no calculation-puzzle is present...

HINT 4:

It appears that the right answer to the last question of the man at the door should have been "23" instead of "24"...

HINT 5:

In my last edit I added a location for the story. Though it is not strictly necessary, it could be important to solve the puzzle.

HINT 6:

Twelve is "special" for club's members, because is the only number with a specific characteristic; anyway it's not so obvious, that's why the member who was asked about it had to think a while before answering...

Answer

The club is a

Scrabble club

The calculations are

the scrabble values of the letters. For example TWENTYONE = 15 THREE = 8

So the answer for 21+3 is 23.

Twelve is a special number, in that

it has its own scrabble score: T (1) W (4) E (1) L (1) V (4) E (1) gives a total of 12

---

Edit from the OP:

This is the point values of each tile in the American-English version of Scrabble (that's why I added the reference to NYC as a location for the story) : A=1, B=3, C=3,D=2, E=1, F=4, G=2, H=4, I=1, J=8, K=5, L=1, M=3, N=1, O=1, P=3, Q=10, R=1, S=1, T=1, U=1, V=4, W=4, X=8, Y=4, Z=10

According to it

these are the values of each number/word mentioned in the puzzle: ONE=3, THREE=8, FOUR=7, TEN=3, ELEVEN=9, FOURTEEN=11, SIXTEEN=14, SEVENTEEN=12, TWENTYONE=15, TWENTYFOUR=19, TWENTYEIGHT=21, THIRTYSEVEN=20, THIRTYEIGHT=21

How many squares can standard chess pieces control?

How many squares can you control with only one set of chess pieces of the same color? (8 Pawns, 2 Knights, 2 Bishops, 2 Rook, Queen and King)

Edit: Pieces cannot attack (control) the square they are currently occupied.

Answer

I believe this is relatively easy and that many combinations will exist. This seems to be an example of such a position:

To answer the question, the set of pieces for one side can attack all squares (and then some).

no computers - Hand tiling puzzle

Here's a set of polyominoes (sizes 4,5,6,7,8,9,10) that you can print and cut out. You can use the two smallest to make a 3x3. Nice easy one to get you started. Add a piece to make a 3x5. Add another to make an 11x2. Then a 5x6, 3x13 and finally a 7x7.

There's only one way to make each rectangle. You are allowed to flip pieces over.

This should be relatively easy to do by hand in a few minutes. A computer would do them all in a split second... and spoil the fun for you.

Answer

Looks like prog_SAHIL left the job unfinished, so I'll continue.

This is a really pleasant puzzle, and you really should try it yourself. If you're stuck at the beginning, just remember that

at each phase, you can calculate exactly which pieces you should use.

or more specifically,

For each puzzle, you use the pieces in the previous puzzle, plus the smallest possible new one.

Here are the solutions:

3x3

5x3

11x2

6x5

13x3

7x7

As a note to myself (and anyone else wanting to make reasonably clean looking block puzzle images), here's the method I used for the newest version of the 7x7 solution. If you want to play with the pieces, you should create each of them on a new layer. Since I already had a solution in mind, I only wanted the final image, and made everything on a single layer.

1. Create a new image in Gimp. (I also opened OPs image, so I could use the eye dropper to pick the colours)


2. Filters->Render->Pattern->Grid. Choose any grid size you want, but make sure to use 2 as the line width.


3. Solve the puzzle using the paint bucket tool


4. Make the grid disappear (I just used the paint bucket to fill the grid with the background colour. If you used layers, just hide the layer with the grid.)



5. Use the select-by-color tool to select all squares of one block


6. Select->Grow, and grow the selection by 1. (This is why the grid lines needed to be 2 pixels thick. Now the piece shape is nicely selected.)


7. Paint the piece solid (since the selection is active, just paint with a large brush)


8. Change the colour to black, and Edit -> Stroke Selection (line width 2, antialiasing off) to make the outlines.


9. Repeat for each piece.

Apart from the 7x7, the other solutions still look a bit shabby around the edges, but since they are quite serviceable, the effort to redo them seems a bit too much.

family - The Three Coffins

Two mothers and two daughters were in a multi car accident. They were all pronounced dead at the scene. All the bodies have been moved to the local mortuary. There are only three coffins. No one is sharing a coffin and none are left out. How is it possible?

Answer

Two mothers and two daughters but in

three different generations.

One of them is

both a mother and a daughter.

cipher - Portraits and quotes - Clue Twenty Four

_{<<---First clue
<--Previous clue}

---

_{A note from @Mithrandir: I have given explicit permission for @Randal'Thor to post this, and gave him the answer to encode. If you want to post one, ping @Mithrandir in chat and we'll talk.}

---

[redacted], you say, and instantly a panel slides aside in the wall in front of you to reveal a new puzzle.

Feeling buzzed after your quick solution of the last clue, but also worried that the next one will be devilishly hard, you approach the wall and stare at it in bewilderment. All you see are four images:

Warning:

keep your mind out of the gutter.

---

_{Next clue--->}

Answer

The cipher used is

Sharky's implementation of a Vigenere to encode textual Morse

First cipher:

Key: Vera
Vigenere: POINTLINE
Morse: A

Second cipher:

Key: succulent (this is a cryptic: def. delicious, "suck you audibly" = succu, "fast" = Lent)

Vigenere: DOTDASHDOTDOT
Morse: L

Third cipher:

Key: Abdul
Vigenere: THREEHYPHENS
Morse: O

Fourth cipher:

Key: Aioli (another cryptic: saucy = definition; mAId + hOLe + I)
Vigenere: DEEOHTEA (D-O-T)
Morse: E

Now we've got the answer

ALOE (ha! Vera and succulent were even references to this answer)

cipher - The Three Little Pigs are hiding something

As you work through one of your countless puzzle books, you find a sheet of paper inside with a poem on it. Remembering the last time you found something like that, you decide to take a crack at this one, hoping to solve it a bit more quickly...

A Fable

The three little pigs hid a little rhyme,
But it's only a matter of how much time
Until the big bad wolf commits the big bad crime
And blows their whole house down.

“Mark my words, pigs, count on these:
I'll find those hidden rhymes with ease,
I'll find those letters and those ABC's
And split them into threes!”

The big bad wolf said this little rhyme,

And he committed the big bad crime,
And he cooked all three of the pigs with thyme,
Wolfing his bacon down.

Nonsense indeed! But what is the fable hiding?

---

Hint:

Upon closer inspection of the poem, you notice that some words have a faint underline, perhaps hinting at a ciphering method: words, count, rhymes, letters, threes, bacon

Hint 2:

Treat each line as a unit, and ignore the stanza breaks. The number 3 is very important.

Answer

The clue about bacon hints at a Baconian cipher. Zandar suggested that it might be a variant of the Baconian cipher that uses a ternary instead of a binary system. So I started by assigning a pattern to each letter, using the values A, B, C:

AAA  A            BAA  J            CAA  S
AAB  B            BAB  K            CAB  T
AAC  C            BAC  L            CAC  U
ABA  D            BBA  M            CBA  V

ABB  E            BBB  N            CBB  W
ABC  F            BBC  O            CBC  X
ACA  G            BCA  P            CCA  Y
ACB  H            BCB  Q            CCB  Z
ACC  I            BCC  R            CCC  (unassigned)

Then I tried to figure out a way to get a pattern out of the poem. Based on the clue words word, count, and threes, and with the help of the second hint, I counted the words in each line, and assigned each line a letter as follows:

• number of words is divisible by 3: A


• number of words $\div$ 3 leaves a remainder of 1: B



• number of words $\div$ 3 leaves a remainder of 2: C

(as a more mathematical way of looking at it, use 0, 1, 2 instead of A, B, C, and just use (word count $\text{mod}$ 3) to assign a value)

This left me with the following:

          Word                Letter
 Line #   count   Remainder   assigned
 1        8       2           C
 2        9       0           A
 3       10       1           B
 4        6       0           A

 5        7       1           B
 6        7       1           B
 7        7       1           B
 8        5       2           C
 9        8       2           C
10        7       1           B
11       10       1           B
12        4       1           B

This leaves us with four triplets: CAB, ABB, BCC, BBB

If we refer back to our alphabet table, these correspond to the letters

T, E, R, N.

Now if we look at the rhyme scheme of the poem, we get another pattern, based on which words rhyme with each other: AAAB CCCC AAAB.

Splitting that into triplets, we get AAA, BCC, CCA, AAB, which correspond to the letters

A, R, Y, B.

If we do the same thing a third time, this time using the letter count instead of the word count, we end up with the sequence AAAAACBBCBBB.

Splitting that into triplets, we get AAA, AAC, BBC, BBB, which correspond to the letters

A, C, O, N.

Putting these three results together, we get

TERNARYBACON or Ternary bacon, which describes the cipher used in encrypting this.

---

A huge thanks to Zandar who actually solved most of this, and SpiritFryer as well.

logical deduction - 3 guards, 3 objects, 3 monsters and 3 questions

You are standing before 3 objects. You need to give the right one to the right monster or else you'll die.

There are 3 people in front of you for you to question.

One always lies, one always tells the truth, one varies between truth and lies, and you don't know which is which.

You can only ask 3 yes-or-no questions. How do you figure out which object goes to which monster?

EDIT: You are on a different planet, and you don't know the difference between "yes" and "no".

mathematics - Get 20160 with 5 2's and any Mathematical Operator

Here is a mathematical riddle:

Find a way to get 20160 with 5 2's and any mathematical operators you want.

Have fun!

Don't forget that I said any. Feel free to look online for crazy whacky math operators :))

(..and please don't close this as too broad. It's just a fun riddle)

Answer

Uh...I don't think this'll pass as an answer but how about

$\dfrac{(2+2+2+2)!}{2} = 20160$

Lol lucky guess.

EDIT: Apparently here: A new PSE member with square reputation the OP of this question created a bounty for my already correct answer on that question because he wanted "explanation of answer" so here is my explanation:

The first thing you notice is that 20160 is not some ordinary number. It's dvisible by 8, 7, 6, 5, 4, 3, 2, 1. So first thing you do, consider factorials. Then, you realize that 20160 is exactly $8!/2$. Then, you go like whoaaa how do you calculate $8$ using 4 $2$'s and then you stare at that problem for like 5 minutes and then you facepalm and realize that it's $2+2+2+2$...done.

cipher - Cryptograms in invisible ink

Solve the followng encrypted message.

--2b1-21-1b112----b1b---21-b--12--1----b1b---b2-12-1--b22---2b21--211---b--121b-121-b21bb--b2-----211-1-b2b1---1b-12----2b21--211---2112b112-12-111-b2-----21b-12----2b21--211---2112b112-12-111-b2

Be brave and frail, and you'll surely be able to see the answer!

Answer

The message reads:

Fire and ice, and death was coming. But he was steel. He was steel.

(the final line of Chapter 13, Floating like Snow, in The Path of Daggers, book 8 of Wheel of Time; the "he" referred to is Rand al'Thor).

It is encrypted like so:

The message is written in braille. I wouldn't have sen that without dmg's comment, but once he had pointed it out and I had seen the other heavy hints, it seemed to be obvious.

Braille letters are written in a grid of 2×3 dots. The message has 195 letters, which is divisible by three. The four characters encode one pair of horizontal dots: The dash means no dot, the 1 means the first or left dot; the 2 means the second or right dot and the b means both dots.

So the first letter made up of --2 is ⠠, which means the next letter is capital. The next letter b1- is ⠋ or F. The message also makes use of abbreviations like ⠯ for "and". Wikipedia has a detailed description on how Braille is used for English.

logical deduction - Five logicians with one or two hats [7,8,9]

There are five logicians seated around a table. They are blindfolded and hats are placed on their heads. After removing the blindfolds, they are told a true statement:

Each of you has either one hat or two hats on his head. The total number of hats is 7, 8 or 9.

The actual number is 9, but this is not told to them.

They are then told to answer the question "How many hats are on your head?" in clockwise fashion (starting from any of the 5). If it is possible to logically deduce the number of hats on their own heads, they do so, otherwise reply "I don't know." and wait till the question cycles back to them.

They have no means of knowing the number of hats on their own heads (except logically), but can see every other person. They have no other means of communication, and no tricks involved.

At what number should the one-hatted logician be placed so that:

• The game is completed?


• The game goes on forever?

Note

I made a small error while posting the question. Since people have anyway up-voted and answered this one, I've left it as it is and posted the actual question as a second version (click here to view)

Answer

Preamble: The following are all of the possible (unordered) hat combinations that could exist for a total of 7, 8, or 9 hats:
22221: 9 hats
22211: 8 hats
22111: 7 hats

TL;DR

The game only completes when the logician with 1 hat is asked first.

In order for 7 hats to be on the logicians heads, none of the logicians can see three 2-hatted logicians. All of the logicians see at least three 2-hatted logicians, so 7 hats is not possible. This leaves the logicians considering only the 8- and 9-hat scenarios.

Case 12222:

$L_1$ sees 8 hats and can immediately deduce "I have one hat on my head."
From the viewpoint of the other logicians, $L_1$ must have seen zero 1-hatted logicians. Had he seen one, $L_1$ would not have enough information to deduce whether this was an 8-hat or 9-hat scenario.
$\therefore L_{2,3,4,5}$ are all able to deduce that they each wear 2 hats and the game completes.

Case 21222:

$L_1$ sees 7 hats. He cannot determine if he has 1 or 2 hats. The table could be arranged as 11222 or 21222 from the perspective of $L_1$.
It is unknown to the logicians at this point that 7 is not a valid hat count. As such, $L_1$ would also not be able to determine if he saw two 1-hatted logicians.
So at this point, $L_{3,4,5}$ see the table as 21x22, 212x2, 2122x respectively, where x is still unknown.
$L_2$ sees 8 hats and can determine that he has 1 hat.
If $L_2$ saw 1 hat on any of $L_{3,4,5}$, he would be unable to deduce that he is wearing 1 hat. It is at this point that $L_{3,4,5}$ know they are each wearing 2 hats.
If $L_2$ saw 1 hat on $L_1$, he would still be able to make his statement, so $L_1$ is not able to guess and the game does not complete.

Case 22122:

$L_1$ sees 7 hats. He cannot determine if he has 1 or 2 hats. The table could be arranged as 12122 or 22122 from the perspective of $L_1$.
From the perspective of $L_{2,4,5}$, the table could additionally be set up as 2x122, 221x2, 2212x respectively.
$L_2$ having no new information cannot make a decision when seeing only one hat.
$L_3$ of course knows that he's wearing 1 hat because he sees 8.
Again, if $L_3$ saw 1 hat on either of $L_{4,5}$, he would be unable to deduce that he is wearing 1 hat. It is at this point that $L_{4,5}$ know they are each wearing 2 hats.
If $L_3$ saw 1 hat on $L_2$, he would still be able to make his statement because he knows $L_2$ was unable to guess (i.e., he knows $L_2$ did not see 8 hats), so $L_2$ is not able to guess and the game does not complete.

Case 22212:

$L_1$ sees 7 hats. He cannot determine if he has 1 or 2 hats. The table could be arranged as 12122 or 22122 from the perspective of $L_1$.
From the perspective of $L_{2,3,5}$, the table could additionally be set up as 2x212, 22x12, 2221x respectively.
$L_2$ having no new information cannot make a decision when seeing only one hat.
$L_3$ having no new information cannot make a decision when seeing only one hat.
$L_4$ of course knows that he's wearing 1 hat because he sees 8.
Again, if $L_4$ saw 1 hat on any of $L_{1,2,5}$, he would be unable to deduce that he is wearing 1 hat. It is at this point that $L_{1,2,5}$ know they are each wearing 2 hats.
If $L_4$ saw 1 hat on $L_3$, he would still be able to make his statement because he knows $L_3$ was unable to guess (i.e., he knows $L_3$ did not see 8 hats), so $L_3$ is not able to guess and the game does not complete.

Case 22221:

$L_1$ sees 7 hats. He cannot determine if he has 1 or 2 hats. The table could be arranged as 12122 or 22122 from the perspective of $L_1$.
From the perspective of $L_{2,3,5}$, the table could additionally be set up as 2x212, 22x12, 2221x respectively.
$L_2$ having no new information cannot make a decision when seeing only one hat.
$L_3$ having no new information cannot make a decision when seeing only one hat.
$L_4$ having no new information cannot make a decision when seeing only one hat.
$L_5$ of course knows that he's wearing 1 hat because he sees 8.
Again, if $L_5$ saw 1 hat on any of $L_{1,2,3}$, he would be unable to deduce that he is wearing 1 hat. It is at this point that $L_{1,2,3}$ know they are each wearing 2 hats.
If $L_5$ saw 1 hat on $L_4$, he would still be able to make his statement because he knows $L_4$ was unable to guess (i.e., he knows $L_4$ did not see 8 hats), so $L_4$ is not able to guess and the game does not complete.