enigmatic puzzle - The sign unseen

Another puzzle in the spirit of the Density™ puzzle. Enjoy!

Final answer: (7)

Answer

The answer is:

BRAILLE

How to solve? Notice first that there are:

6 different-shaped symbols in the image, each of which can represent a section in a diagram of 2×3 sections that looks like a domino:

If we then:

Group (and order) the symbols by rainbow colour, positioning each symbol as per the domino-style diagram above (i.e. _ | = top left corner, | _ = top right, etc.) we can see that they form symbols in English Braille. Translating them into letters, we get the letters of the word BRAILLE itself - the answer to this puzzle:

At this point, now that we know the encoding, we can see that there was another clue in the puzzle that I inadvertently skipped past altogether! Notice that:

If we split the initial puzzle into 6 equal-sized blocks and replace each symbol (regardless of its shape and colour) with a dot, we have 6 more Braille letter dominos! Translating these as above we see the clue 'RAINBW', suggesting that if we'd got to the Braille conclusion just from its initial appearance we should now focus on the symbols within their rainbow colours, and derive the solution as laid out in the spoiler block above... (However, since I'd solved the previous tic-tac-toe themed puzzle in this series, I was already subliminally programmed to associate these symbols with a grid of some sort, and initially missed this subtle step!)

This solution also explains the title, since:

Braille is a letter system for the blind, made of raised dots that are read by touch (hence, 'unseen').

mathematics - Deducing Two Numbers based on their Difference and Ratio

Yesterday I met the perfect logicians Divvy and Subtra. I told them:

I have chosen two positive integers $x$ and $y$ with $2\le x\lt y\le 100$ and with $x$ a divisor of $y$. I will now whisper the ratio $r=y/x$ into the ear of Divvy and I will whisper the difference $d=y-x$ into the ear of Subtra.

And then I whispered the ratio and the difference into their ears.

Then the following conversation took place:

• Divvy: I don't know the numbers.


• Subtra confirmed: I already knew this.


• Divvy: I already knew that you knew that I don't.


• Subtra: I still can't determine the numbers.


• Divvy: At the very beginning of this conversation, I knew that you did not know the numbers. I forgot to mention this. Does this information change something for you?


• Subtra: Yes, it changes everything! Finally I am able to determine the numbers!

Question: What on earth are these magic numbers $x$ and $y$?

Answer

We are dealing with four integers $x$, $y$, $r$ and $d$ that satisfy $y=rx$ and $d=(r-1)x$. Note that $x

(1) Divvy: I don't know the numbers.

• If $r\ge34$, then $x\ge3$ is impossible (as this would imply $y=rx\ge3\cdot34=102$). Hence this would imply $x=2$ and $y=2r$, so that Divvy knows $x$ and $y$ in contradiction to his statement.


• If $r\le33$, then $x=2$ and $y=2r$ as well as $x=3$ and $y=3r$ are feasible solutions. Divvy is not able to determine $x$ and $y$.

Divvy's first statement is equivalent to $2\le r\le33$.

(2) Subtra confirmed: I already knew this.

As Subtra's difference is $d=(r-1)x$, her statement means that $d$ cannot be written as the product of two factors $r-1$ and $x$, one of which is $\ge33$ and one of which is $\ge2$.

Subtra's statement means that the difference $d$ is either from the range $0\le d\le65$, or that $d$ is odd and from the range $67\le d\le97$. Equivalently, $d\notin\{66,68,70,\ldots,98\}$.

(3) Divvy: I already knew that you knew that I don't.

Divvy's statement means that there does not exist any integer $x$ with $2\le x\le100/r$, so that the corresponding difference $(r-1)x$ is an even integer in the so-called bad range $B=\{66,68,\ldots,98\}$. We make some case distinctions on $2\le r\le33$.

• If $r$ is odd, then $r-1$ is an even integer from the range $2,4,\ldots,32$. As the diameter of the bad range $B$ is $98-66=32\ge r-1$, the bad range contains an (even!) integer multiple $(r-1)x$ of $r-1$. Hence all such values $r$ collide with Divvy's statement.


• If $r=2$, then for any $x$ with $2\le x\le 100/r=50$, the corresponding product $(r-1)x=x\le50$ will not fall into the bad range. This case is possible.



• If $r$ is even with $4\le r\le24$, then $2(r-1)$ is an even integer from the range $6,10,14\ldots,46$. The values $6,10,\ldots,30$ are smaller than the diameter of the bad range, and hence have an integer multiple in $B$. The value $34,38,42,46$ respectively have the multiples $68,76,84,92$ in $B$. For a multiple of the form $2(r-1)k$ in $B$, we may choose $x=2k$. Hence all such values $r$ collide with Divvy's statement.


• If $r\in\{26,28,30,32\}$, then none of the multiples of $r-1\in\{25,27,29,31\}$ fall into $B$. Hence these four cases are possible.

Summarizing, we see that Divvy's second statement is equivalent to $r\in\{2,26,28,30,32\}$.

(4) Subtra: I still can't determine the numbers.

At this point of the conversation, we are left with the following options for the pairs $(x,y)$:

• $r=2$: The pairs $(x,2x)$, where $x\in\{2,3,\ldots,50\}$. Then $d=x$.



• $r=26$: The pairs $(2,52)$ and $(3,78)$. Then $d=50$ or $d=75$.


• $r=28$: The pairs $(2,56)$ and $(3,84)$. Then $d=54$ or $d=81$.


• $r=30$: The pairs $(2,60)$ and $(3,90)$. Then $d=58$ or $d=87$.


• $r=32$: The pairs $(2,64)$ and $(3,96)$. Then $d=62$ or $d=93$.

Since Subtra is not able to determine the numbers $x$ and $y$ from this, her value $d$ must be compatible with two distinct pairs $(x,y)$ in this enumeration. This reduces the possible options for $(x,y)$ to the following short list:

• $r=2$ and $d=50$: The pair $(50,100)$.


• $r=26$ and $d=50$: The pair $(2,52)$.

Subtra's second statement means that $d=50$.

(5) Divvy: At the very beginning of this conversation, I already knew that you did not know the numbers.

We are only left with two possible values $r=2$ and $r=26$ for the ratio.

• If $r=2$, then at the very beginning of the conversation the pair $(x,y)=(2,4)$ with difference $d=2$ would have been a feasible option for Divvy. But then Subtra would have been able to deduce the two numbers right away, as $d=(r-1)x=2$ and $x\ge2$; a contradiction.


• If $r=26$, then at the very beginning of the conversation only the two pairs $(x,y)=(2,52)$ with $d=50$ and $(x,y)=(3,78)$ with $d=75$ would have been feasible options for Divvy. The first difference $d=50=(r-1)x$ is compatible with $(r,x)=(26,2)$, as well as with $(r,x)=(6,10)$. The second difference $d=75=(r-1)x$ is compatible with $(r,x)=(26,3)$, as well as with $(r,x)=(6,15)$. In either case, at the very beginning of the conversation Subtra is not able to deduce the numbers from the difference $d$.

Divvy's third statement implies $r=26$.

(6) Subtra: But now I can determine the numbers.

Subtra now knows that $d=50$ and $r=26$, and deduces the numbers.

Answer: The magic numbers are $x=2$ and $y=52$.

english - Children's riddle

I have a 6-year old niece who is beginning to understand the English language (our native language is Spanish). A few days ago I came up with a short riddle and I want to try it before giving it to her.

What is it like to live in the middle of nowhere?

It's like living here in the first place or there in the second.

It feels like living in Hamburg but not in Germany.

Do you have some advice on how to make this riddle understandable, and enjoyable to her?

Answer

It's pretty obvious to seasoned puzzlers that the answer is

The letter 'h'. In the middle of "nowhere", the first letter in "here", and the second in "there".

Can't speak for what a six-year-old will think of it, but it works.

construction - Building towers with colored blocks

You have blocks in five colors (red, green, yellow, blue and black - or 1 to 5). Each color comes in two varieties: a cube and one as big as two cubes stacked atop each other (sizes 1 and 2).

You are supposed to use them to build two towers next to each other such that two blocks of the same color (in either tower) are separated by the sum of their heights. For example:

Note that this only says that each color must repeat within a given distance, but not that every block of the tower must be of a particular color, so you get white blocks to fill any holes (they don't do anything else, and they aren't subject to any of the rules above or below).

You don't know how high the tower is supposed to be, hence give the shortest repeating pattern that can be used to extend such a tower upwards for any height. Within that pattern, each block of each size and color should repeat an equal numbers of times (ignoring white, of course).

Now if that were all, finding a solution would be simple, e.g. this one:

However, in the pattern of each tower each block size of a color should be opposite a block of each other color in the other tower. The above example doesn't fulfill this: consider the blue 2-block: while it occurs on the left opposite to the black 2-block, but never vice versa, and never opposite to the green 2-block, nor does the blue 1-block occur opposite the black 1-block.

Thus for any pair of two colors (not involving white and not both colors the same) the following patterns must occur (given as example for red and blue):

Note that in 3 and 4 they don't need to be perfectly lined up and in 5-8 the small one doesn't need to be at the same height, but there must be one box in which they are opposite, so e.g. for 3 and 5 these are alternative examples:

riddle - Falcon Circus, Prologue: The Twins

You get home from work, grab the mail, take off your coat, and sit down on the couch with the TV on the news. "Bills, bills, junk..." you sift through your mail, and stop to look at a very colorful looking flyer.

The flyer is bright red and purple- two colors that don't seem to sit well together, but do the job in drawing your attention. The title labels it as some sort of circus ad. The middle of the flyer has a group photo of what you assume to be the curiosities of the circus, but everything is a silhouette. "A strange way to market yourself, but it does get you interested, now doesn't it?" you muse. At the bottom, in large, white letters is the name L. C. Snake. "Snake? That's not a great last name- that seems to be begging for trouble!" you say to yourself jokingly.

$\hskip2.5in$

This flyer seems like some kind of joke, or a very poor attempt at marketing. But... you are free this weekend, and if their advertisement looks like this, you're dying to know what the service quality will be.

As the weekend comes, you make your final decision: you'll go. Flipping the flyer over, you see a hastily scrawled address. "Now I'm not so sure..." the hand-written location isn't a good vibe to give your customers. Nonetheless, you've decided you'll go, and go you shall. You get in the car and set up the GPS.

$\hskip2.5in$

After driving into what seems to be the middle of nowhere, you begin to see the lights and the very top of their red and yellow tent. No wonder it's called Forest Grove- there's trees as far as the eye can see. Now you just have to stay not-murdered! You pull up to the tent and find an empty lot. Getting out of the car, you notice that quite a few people have parked and must be inside the tent. Flyer in hand, you walk inside.

$\hskip0.75in$ $\hskip0.75in$

Two children in sweaters and overalls walk up and greet you. One of them looks incredibly sleepy, and the other seems very energetic. To be seeing such odd kids around here must mean they're part of the circus gang- their faces are painted, to boot. You'll call them Red and Green for short. They block your path, and ask you a question before you can tell them to get out of the way:

Green: "Hey Mister. what's our names?"
Red: "Yeah! You have to guess our names before going further!"
Green: "We'll give you a riddle."
Red: "A riddle! And if you get it, we'll let you meet Snake!"
The one in the green sweater punched his twin.

Green: "No, we won't."
Red: "Fine. But you still have to guess, okay? Okay! Get Ready!"

You weren't prepared for children, and these two seem especially unsettling. Well, whatever. You came here to have an interesting time, and you've been stopped at the door already- this is going to be a great trip. You glance at your flyer. The very front silhouettes seem to mirror each other. Are these two the ones? Before you could ask, they began their riddle.

---

tldr;

Red: "I am to be feared!"
Green: "People want me cleared..."
Red: "I'll howl, I'll bite- I'll give you a fright!"
Green: "We are two, or are we three...?"
Red: "Selene grants me power, my goddess, the lovely!"

Green: "...My first partner... neither plant*, animal, or fungi."
Red: "Stop calling me that! I am not a rat!"
Green: "I'm not a parasite, I'm not a plant... stop calling me that..."
Red: "I'll eat you next month, you'll see!"
Green: "...Friend of Apollo, we live in harmony, with rock or tree."
Green: "Were you counting when we spoke?"

_{*With the definition that a plant must photosynthesize, have a root/stem/leaf system, and have cell walls.}

Luckily no one has entered the tent after you, giving you all the time in the world to solve their riddle. After giving it some thought, you decide to tell them your answer, making sure to note the similarity between their names.

"You- the one in the green... your name is ______. And you, the red one- you're _____."

---

Answer

Red and Green are

Lycan and Lichen

Red clues:

Lycan is short for lycanthrope, or werewolf, hence "feared", "howl", "bite". Selene is the Greek goddess of the moon, which generally triggers the lycanthropic transformation; "eat you next month" also refers to the moon cycle. I think the reference to "rats" is the D&D creature "wererat", which is classified as a lycanthrope.

Green clues:

A lichen is a composite organism consisting of algae and/or cyanobacteria and fungi ("two, or three"). Some algae are technically plants, I think, but many are not; non-plant algae, as well as cyanobacteria, are neither "plant, animal, or fungi". It is not parasitic, but a symbiotic relationship. They are generally found on rocks and trees.

knowledge - Future countries – party like it's 19999!

I recently met a time traveller from a distant hypothetical future, and we had an interesting discussion about the geographies of our respective worlds. It turns out that the world map is pretty similar in the year 19999 (they may be using a different numbering system) to what it is now. There are a few changes, though. Here are the descriptions of some sovereign states from that future.

1. I am a landlocked country with significant Turkish and Arabic speaking minorities. I have a violent history and have been ruled by more countries than I can count.


2. I am surrounded by water on three sides. In my south I have a popular vacation island. For decades I was split through the middle.


3. My mountainous terrain was for years heavily restricted for visitors. Since our independence we have been opening our doors more to outsiders Mountaineering and spiritual tourism are important sources of income.


4. A popular Mediterranean tourist destination, my secession spiked an economic crisis in the kingdom I used to be a part of. Relations are better now, even though we had a long dispute over the possession of an important island chain.

Can you guess the names of the countries?

Answer

Could 1 be

KURDISTAN?

I am a landlocked country with significant Turkish and Arabic speaking minorities.

A landlocked region where Kurdish would be the majority language, but there are plenty of Turks and Arabs in the area too.

I have a violent history and have been ruled by more countries than I can count.

The Ottomans, the Soviets, the Turks, the Saudis, the Iraqis, etc., etc.

Could 2 be (Thanks to @Jaap Scherphuis for this wonderful answer!)

UNIFIED KOREA?

I am surrounded by water on three sides.

It's a peninsula in the East China Sea.

In my south I have a popular vacation island.

Jeju Island, a popular tourist destination off the coast of current South Korea.

For decades I was split through the middle.

Split between the North and South Koreans in the 20th century.

Could 3 be

TIBET?

My mountainous terrain was for years heavily restricted for visitors.

Restricted by the Chinese who currently claim it.

Since our independence we have been opening our doors more to outsiders.

Potential future independence from China.

Mountaineering and spiritual tourism are important sources of income.

Because of the Himalayas and the Dalai Lama and Tibetan monks.

Could 4 be (I just noticed @Eutherpy had the same answer!! Full credit to @Eutherpy!)

CATALONIA?

A popular Mediterranean tourist destination, my secession spiked an economic crisis in the kingdom I used to be a part of.

A popular part of the Kingdom of Spain, although it might not take much to push them into an economic crisis nowadays.

Relations are better now, even though we had a long dispute over the possession of an important island chain.

Relates to a dispute over the Balearic Islands.

riddle - You dirty Mind - What am I?

I may sometimes be rather dry

But you prefer moist and hot.
The flesh is soft, you must apply
Just the right touch, don't get caught.

I must be pierced to bring you joy
My smell is known and often craved
But you should know I'm not a toy,
And there are many I have enslaved.

Now what am I? Oh no, not that.
Have I pulled you into my trap?
I'm not at all related to a cat.

And found nowhere near a lap.

Some think me rare, or maybe not.
Can you rise above the cut?
Again I say, some like it hot.
What am I? Follow your gut.

Answer

The answer is

Meat

I may sometimes be rather dry
But you prefer moist and hot.
The flesh is soft, you must apply
Just the right touch, don't get caught.

Dry meat exists, but it's usually not so tasty. The flesh is synonym of meat, which is not that hard to slice, if you use the knife properly.

I must be pierced to bring you joy
My smell is known and often craved
But you should know I'm not a toy,

And there are many I have enslaved.

Pierced with fork to be eaten. The smell of meat is great, also food isn't a toy, don't waste it! The enslaved may be the animals used to produce meat.

Now what am I? Oh no, not that.
Have I pulled you into my trap?
I'm not at all related to a cat.
And found nowhere near a lap.

The trap might be Vegetarianism. Pussy and cat are synonyms. I don't want to say more...

Some think me rare, or maybe not.
Can you rise above the cut?
Again I say, some like it hot.
What am I? Follow your gut.

Poor people can't afford and consider it rare.

How to choose the best answer in puzzles?

I gave my friends a lateral thinking puzzle

There are six eggs in the basket. Six people each take one of the eggs. How can it be that one egg is left in the basket?

My intended answer was

One of them put the egg again inside the basket.

But one of my friend answered

The last person took the basket with the last egg still inside.

Clearly my friend's answer is more creative then my own (as lateral thinking puzzle demands). Now I am confused if I should take his answer for granted (as I should because I found his answer more creative) or should I say that his answer is wrong just because it was not my intended answer?

The definition given by wikipedia on puzzle is: A puzzle is a game, problem, or toy that tests a person's ingenuity or knowledge.

It got to me thinking if puzzles are just to find the intended answer of the question asker or is it to find the most appropriate and reasonable answer. Which one of them would be better? Or is there something I am missing?

Answer

^{(Not a guess at any specific answer that the question's poser had in mind.)}

Premise

The more the merrier.   Different solutions have different virtues. Eventually every solution will be appreciated by some puzzle lover.

Solutions have a variety of virtues

Far from being exhaustive or mutually exclusive, these qualities are listed alphabetically within two broad categories. Edits welcome.

More for sport:
  •  approval
  •  brevity
  •  creativity / inventiveness / originality
  •  esoterica / sophistication
  •  firstness
  •   humor   humour  mirth
  •  obviousness
  •  simplicity / ease of calculation

  •  subtlety

•  surprise

More for the long term:
  •  clarity
  •  completeness
  •  education / cross reference
  •  generality
  •  resourcefulness

  •  variety, the spice of life

Brief case study of the kind of puzzle in question

Relabeling two 20-sided dice without changing their total

This puzzle received a wonderful solution that was accepted by the poser, who then proceeded to present the intended solution so that it wouldn't be overlooked.

Three solutions have been posted to date, each with its virtues.

1. 
   

   A complete solution that transforms the puzzle into algebra

   
   

   This solution is so complete, educational, general and sophisticated, who could ask for more?

   
   



2. 
   

   An old-fashioned solution

   
   

   This detective-like solution does not require the mathematics or computer employed by the complete solution. Then again, only one of several possible solutions is found and a general solution is only alluded to.

   
   


3. 
   

   The poser's inventive solution

   
   

   Quite possibly the poser all along had subconsciously hoped that nobody else would actually think of this gem of an approach, one that even adds to the understanding of the other solutions.

   
   

Conclusion

The more the merrier.   As in the case study: a puzzle's creator can acknowledge an unintended excellent solution, someone else can provide an incomplete solution for a different audience, and the poser can present their original solution for posterity and still blow our minds.

---

Corollary

Some unexpected solutions simply deserve their own puzzles.   The original puzzle (P) may be restated as two different puzzles tailored differently for the originally intended solution (S) and for the unexpected one (S2).

P 🡒 P1.   There are six eggs  in the basket  on a shelf. Six people each take one of the eggs. How can it be that one egg is left  in the basket  on the shelf?

S 🡒 S1.   One of them put the egg again  inside the basket  on the shelf.
       (Presumably a person cannot take the shelf, or occupy it while holding their egg.)

Chaotic’s answer provides a way to restate part of the puzzle so as to exclude the intended solution.

P 🡒 P2.   There are six eggs in the basket. Six people each take one of the eggs and nobody put it again in the basket. How can it be that one egg is left in the basket?

     S2.   The last person took the basket with the last egg still inside.

Examples of solutions that earned their own puzzles

cipher - The Temple of Puzzling

_{Part of the Community Metapuzzle, although it can be solved on it's own. The password in the puzzle below is also the word needed for the Metapuzzle. Link to the metapuzzle by Deusovi is here.}

---

Welcome Indinana Bones to your first quest. To unlock the next quest, you must take the sacred Golden Statue of Puzzling from the temple.

You must also find a hidden code which will act as a password in the next quest.

If you wish to accept this quest, then click below to start your adventure...

Good luck explorer. You're going to need it.

---

_{This puzzle is inspired by the film Indiana Jones - Raiders of the Lost Ark. Knowledge of the film is required but you only need to watch the first 10 minutes. All the information you need can be read here. Apologies for any mistakes or typos, I was running out of time towards the end so couldn't do a thorough check.}

---

_EDIT:

_{Despite double checking, it seems I made a mistake. 'x' should be 'z'. You'll know where when you get to it. Sorry for the delay and the inconvenience.}

Answer

(Don't worry, I haven't been spoiled for this! I just know the final passcode and that's it.)

After a few story images, we get to this:

There seems to be a key to the cipher hidden in the stone - it's a variant of pigpen. If the numbers are to be trusted, it means that dots come before non-dotted shapes. That means the top left corner DOTTED is A, not the undotted top left as it would be in the standard pigpen cipher. Using the pattern:
[.] [ ] [.][ ]
ABC JKL
DEF MNO ST WX
GHI PQR UV YZ
and reading around the loops clockwise from the north gives us:

 outer loop: REGA[l]R?D

 ----------- EA?C[k]HOB
 ----------- STAC[B]LE?
 ----------- METI[c]?CU
 inner loop: L?OU[H]SLY

The fourth column has helpful arrows. I have indicated this with bracketed letters. The rest spell out the message "REGARD EACH OBSTACLE METICULOUSLY". Unfortunately, the fourth's columns don't seem to lead to any valid stack.imgur URLs in the indicated cases. In chat, however, Rubio found that reading them "upside down" (i.e. as you seem them) makes a valid url: http://i.stack.imgur.com/bGhQP.jpg

---

Now we have a branch!

Each of the links in the previous image has one of the traps from the movie and a pigpen letter hidden inside:

The first room, the pit, has a K.
The second room, the spiders, has an x.
The third room, the boulder, has a c.
The fourth room, the spike gate, has an H.
The fifth room, the pressure plates, has a T.
Putting these all together in movie order (spiders-spikes-pit-plates-boulder) gives http://i.stack.imgur.com/xHKTc.jpg, which... doesn't seem to lead anywhere.

---

BeastlyGerbil has clarified that it was meant to be a z in the spider room, not an x. Putting those together gives along with a few more story images leading ultimately to this image:

That final image contains a code using the same method as the other codes we've seen so far. That code decodes to the password:

BIGFOOT.

cipher - Hidden in plain sight

My good friend Joseph Schmo, who I probably just made up, handed me this mysterious note. And right after I figured out what my other fictitious friend's note said! What is Joseph's favorite pastime?

Hint:

Joseph's favorite number is 7

Answer

Joseph probably likes

Creating and solving puzzles on https://puzzling.stackexchange.com! If we namely group the symbols into groups of 7 and put the different groups (without the circles) below each other in a grid, we get the following:

dissection - Placing 2x1 dominoes on a chessboard with two corners removed

Suppose you have a checkerboard with two opposite corner squares removed, like this:

Is it possible to place 31 dominoes of size 2x1 so as to cover all of these squares?

Answer

This puzzle is known as the mutilated chessboard problem. The other answer correctly explains that such a covering is impossible because it would require an equal number of black and white squares (since each domino must cover one black and one white square), which the corner-cut board does not have.

The link above contains a slightly more general result as well:

The same impossibility proof shows that no domino tiling exists whenever any two white squares are removed from the chessboard. However, if two squares of opposite colors are removed, then it is always possible to tile the remaining board with dominos; this result is called Gomory's theorem, and is named after mathematician Ralph E. Gomory, whose proof was published in 1973. Gomory's theorem can be proven using a Hamiltonian cycle of the grid graph formed by the chessboard squares; the removal of two oppositely-colored squares splits this cycle into two paths with an even number of squares each, both of which are easy to partition into dominos.

riddle - a spell for summoning help

Mutter dos de thou around a rhetorical shape,
then pause with the hero of Card's game.
Then let it sit for far too long

while the celts tell you how many bards came.
Our protagonist makes another appearance.
This sentence exists for poetic adherence.
A short commercial makes it done and unique.
If compelled, you can choose to mark it oblique.

Hint #1:

If you are reading this, chances are very high you've cast this spell

Hint #2:

For this hint, you will need to appeal to Aristotle

Hint #3:

vs vg jnf n fanxr

Hint #4:

A certain Irish singer/songwriter may help you "track" down one of the lines

Hint #5:

No line represents something longer than 3

Answer

Credit is due to other answerers and commenters.
I have pieced together various things to arrive at this answer.

I think what we're looking for is...

That most helpful of sites:

www.rot13.com

Mutter dos de thou around a rhetorical shape,

Dos de thou -> two of you -> double-you -> w
I'm not totally clear on how this clues three Ws for www

(Side note: for so long I thought this line contained an anagram.)

then pause with the hero of Card's game.

As indicated in other answers/comments, this refers to the book Ender's Game by Orson Scott Card. Here, specifically, we want a dot (or period) which commonly is the ender of a sentence.

Then let it sit for far too long

Many things rot if you leave them too long.

while the celts tell you how many bards came.

I haven't tracked down this reference. I'm guessing there were 13.

Our protagonist makes another appearance.

Ender again, for another dot.

This sentence exists for poetic adherence.

Fair enough. Nothing to see here.

A short commercial makes it done and unique.

Short commercial = com

If compelled, you can choose to mark it oblique.

Made clear by other answers that this refers to the forward slash character '/'. If you want to, you can end the URL with a forward slash.

riddle - I am three, but which one am I?

This is a puzzle i came up with earlier, with a few amendments. Work out what the three clues are, and then use the final clue to decide which one is the correct one.

I am three things, but which one am I?

1: The internal temperature North, maybe a half more, and eleven elevens towards apollo's destination. I am almost modest, but not quite there.

2: I am the first of many, the largest and the brightest, but still small in the scale of things; I fit inside a belt.

3: I am plentiful harvest, but the sides of skulls no longer echo my name. Some even use an alternate name.

Of these three, I am one, the one whose period is 1680

I will only accept an answer that fits all of the clues and explains them. Good luck ;)

Answer

Are you:

(Credit goes to @Ivanhoe for the clarification of clue 2)

Ceres, asteroid

The internal temperature North, maybe a half more, and eleven elevens towards apollo's destination. I am almost modest, but not quite there.

Looks like coordinates: 37.5 (or 36.5) north. 121 towards apollo, god of the sun. The sun heads west, so 121 W. At this coordinates there is a city Modesto and there is (not quite there) a nearby city Ceres.

I am the first of many, the largest and the brightest, but still small in the scale of things: I fit inside a belt.

Ceres is the largest object in the asteroid belt that lies between the orbits of Mars and Jupiter

I am plentiful harvest, but the sides of skulls no longer echo my name. Some even use an alternate name.

Demeter or Ceres is the Goddess of the Harvest. Sides of skulls, i.e. temples - Ceres is a Roman goddess and is not worshiped any longer

Of these three, I am one, the one whose period is 1680

It's orbital period is 1680 days

physics - The Twins with a Different Age

Warning: To answer this riddle you need to have a thorough understanding of how time works.

Two twins were born during the same year, on the same month, and the same day, exactly 10 minutes apart. By the time twin A was 100, twin B was younger by more than 10 minutes, fairly slightly. Neither twin died during the 100 years.

How could Twin B be younger than Twin A by more than 10 minutes?

Answer

Due to time dilation, moving faster will make you age slightly slower. This is most commonly mentioned with astronauts and space calculations, but it can also be applied to lower speeds.

Technically, if Twin A normally drives his car at 50mph over the course of his lifetime because he wants to save fuel, whereas Twin B dashes around at 70mph because he's always running late, Twin A will have experienced more time than Twin B, so Twin B would be slightly more than 10 minutes younger after a century.

Thanks to Clint's comment below, there is a similar effect that could happen - gravitational time dilation. If Twin B lives in Washington DC (250ft above sea level) but Twin A moved to Denver (5000ft above sea level) at a young age, time will pass slightly more slowly in Washington because it's closer to the gravitational centre of the earth, again resulting in Twin B having experienced less time.

riddle - Enter the Vermillion

I have a new riddle for everyone here at Puzzling.SE, and I believe it's a safe bet to say it will be difficult to solve.

My suffix is my prefix, defined for both as well;

My infix has one anagram, its colored yellow, and smells.

What am I?

Note: In order for an answer to be accepted, you'll need to supply the correct answer, along with an explanation as to how you found your answer.

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Update: As I believe this is a difficult riddle to solve, and in my honest opinion, unfair with the small amount of detail that was given; I will be adding a few hints to help you with your search. I will post three today, and then one a day for three days. However, don't expect me to go easy on you; the bread and butter of software development (in my opinion), is the ability to overcome all obstacles and use lateral thinking to solve problems both old and new. If you have questions or partial answers, don't be afraid to share.

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Hint #1:

The answer you seek, can be found in due time; the solution you need, is as wide is to fine.

Hint #2:

Attila the Hun, Vlad the Impaler, Ivar the Boneless

Hint #3:

Every mom, makes everything terrific.

Hint #4:

Distilled from tar, used as flavor; don’t say I haven’t, done any favors.

Hint #5:

I run all day, I run all night; one by one, HURRAH! For the queen I fight.

Final Hint:

I have seen, in my short time; a creep of a thing, that feeds at night.
Creeping around, silent in flight; you can't seem to feel, the little guy's bite.
Often it seems, that if you have hair; he shaves it away, leaving you unaware.
A creature of nightmares, or so it would seem; usually leaves little trace at the scene.
Glowing with heat, there's no need to hear; just look for the spot, where food's flowing near.

Up until now, you've had a hard time; not quite as hard, as making this rhyme.
Leaving you hints, has surely been fun; but as this is the best, it is the last one.

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Update: This riddle is definitely a difficult one to solve; there are multiple types of puzzles here:

• Riley Riddle


• Basic Rhyme


• Enigmatic


• Acrostic



• Knowledge Based

The hints are given to help get you thinking in the proper manner; some refer directly to the word you seek, and others point to other things that can relate back to the word. To help everyone out, the answer to Hint 1 is:

Thick and Thin; I will break this down in the Making Of.

To solve from there, you'll need to figure out how each hint relates back to the word. I also want to point out that though the first hint can be related back to the word it is the most useless in helping you find the word. Each hint will get easier to solve, and in succession will reveal more and more vital information about the word you seek.

Other Useful Information:

• Not an animal.



• Not a commonly used word.


• Word has more than seven letters.


• Can be solved without hints (though incredibly difficult).


• Think literal for the riddle; literal and lateral for the hints.

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Note: I will definitely be doing a Making Of for this one after the bounty expires. However, I really want to see someone solve this!

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Update: As this riddle is proving to be difficult, and as such is not receiving many responses; if you are working the riddle and wish for me to not post the answer (as the Making Of) after the bounty expires, please let me know in the comment section. Also, don't forget, help is always given to those who ask.

Answer

I believe the answer is:

Anticoagulant

Some work is done here which is really just my though process so forgive the lack of structure:

Blood is an important part of this. The acrostic in the final hint spells out "I COAGUL...". Combine this with the acrostic where the multiple puzzles are defined: RBEAK is easily arranged to form BREAK. There is also some hints heading towards "Breakfast" "Bread and Butter". Thick and Thin is more about blood. Something that thins blood is an ANTICOAGULANT which begins and ends with "ANT" and contains "ICOAGUL" for the rest. This is probably not the end though, since ICOAGUL is not an anagram. The last hint seems to refer to leeches or ticks as well which inject an anticoagulant. Hint 5 is talking about "Ants" who march 1 by 1 and fight for the queen. Vermillion is a hint at the red color of blood. Anticoagulants are medications for people with heart conditions that are taken at breakfast as well.

Better structure: My suffix is my prefix, defined for both as well;

ANT is both the prefix and the suffix

My infix has one anagram, its colored yellow, and smells.

Not sure but: Coagulant -> Platelet has one anagram: Pallette, platelets in plasma are yellow and smell strongly i believe

Hint 1:

"As wide is to fine" -> Thick is to Thin as explained in the puzzle. Also "Solution" hints at liquids. So a thick liquid to a thin one

Hint 2:

As was said these were all known for drinking blood

Hint 3:

No idea

Hint 4:

Turpentine is a paint thinner (anticoagulant) distilled from tar

Hint 5:

"Ants go marching 1 one by one hurrah" they fight for the queen

Final Hint:

Mostly just the acrostic "ICOAGUL" is the infix. But it also describes animals that have anticoagulants in their saliva: Leeches, in particular are blind and just go after heat, leaving only a mark and you never feel the bite

After the final hint: Explanation of hint one:

Thick and Thin. Coagulated vs not coagulated

How do I solve this sudoku without guessing?

I have filled all the possible singles. Also identified and shortlisted candidates, pointing pairs. How do I move ahead without guessing?

PS: I applied "Nishio", choosing one of the two possible candidate values on row 1, column 1. 7 will come in top left. Using which I can solve the rest of the puzzle. But it is "guessing". Any other way?

Fill the grid (pattern)

Can you guess what comes in place of the question mark?

Here's a text-based version of the table:

+----+----+----+----+
|1536| 48 | 96 | 3  |
+----+----+----+----+
|384 |192 | 24 | 12 |
+----+----+----+----+
|768 | 96 | 48 | 6  |
+----+----+----+----+
|192 | ?  | 12 | 24 |
+----+----+----+----+

Answer

It's:

$384$. Starting from the top-left, go down and do a U'y at the bottom, go up, do a U'y at the top, you get: $1536, 384, 768, 192, ?, 96, 192, 48, 96, \dots$, which is divide by $4$, multiply by $2$, and repeat.

weighing - A balance with three pans, detecting the lightest pan (find the one heavier ball)

This is a continuation of the question A balance with three pans, detecting the lightest pan (find the one lighter ball). It was told to me by a friend, Markus Götz, who put it online here: Deviating Ball Puzzles (pdf).

The three-pan balance

Imagine a balance with not two, but three pans. Weightings using the balance follow these rules:

• If there exists a pan that is lighter than each of the other two pans, then this pan goes up and the other two pans go down to a stop. (Note that one cannot see which of the two heavier pans, if any, is the heaviest.)


• If there is no single lightest pan, then nothing happens. (This includes the case of two equally light pans and one heavier pan.)

Let's call this the "lightest-pan-detection-rule" (LPDR).

You are given n balls, one of which is heavier. What is the largest n, so that the heavier ball can be identified with at most k weightings? (k >= 1)

The "normal" balls are all of the same weight. You are to identify the deviating ball by using the balance a maximum number of weighings stated in the puzzle, weighing only the given balls. You are also to present a method to identify the deviating ball.

Answer

If all three pans have the same number of balls, at most one will be heavier and the other two the same, so the pans will never move. However, if there are $x$ balls on two pans and $x+1$ on the third, then the scale works like a normal 2-pan scale, comparing the two pans with $x$ balls. If the heavier ball is one one of the two pans with $x$ balls, the other one will rise. If it is on the third pan (or anywhere else), then no pan will rise. This matches the behavior of a normal 2-pan scale.

If you have three candidate heavy balls and one or more balls that have been eliminated (I'll call these "ringers"), then you can put one candidate on each pan and also one ringer on the third pan. If one pan goes up, then the other pan with 1 ball holds the heavy one. If neither goes up, then the one with the ringer is the heavy one.

So, with a ringer, you can differentiate up to 3 balls in one weighing. Similarly, if you put $x$ candidate balls on each pan and add a ringer to one pan, then you can determine which pan contains the heavy ball. Iterating, you can differentiate $3^k$ balls in $k$ weighings, if you have a ringer to start with.

• $k=1$: $n=1$

Without a ringer, you can't differentiate anything in 1 weighing.

• $k=2$: $n=7$

Put 2 balls on A and B; put 3 balls on C. After the first weighing you have at most 3 candidates and at least one ringer. The second weighing will suffice. So for k=2, n=7 is the maximum.

• General $k$: $n=3^k-2$.

Put $3^{k-1}-1$ balls on A and B, and $3^{k-1}$ on C. This weighing leaves you with at most $3^{k-1}$ balls and at least one ringer. Iterate through k weighings.

Since the maximum number of balls that you can differentiate in $k-1$ steps is $3^{k-1}$, pan C cannot have more than that number, and the other two pans cannot have more than $3^{k-1}-1$, so the total cannot be more than $3^k-2$.

mathematics - Unfair coins at South Park Elementary

Wendy Testaburger and Sally Turner play a game with two unfair coins. A coin flip with Wendy's coin shows head with probability $\frac {1} {100}$. A coin flip with Sally's coin shows head with probability $p$. Wendy does the first coin flip. Wendy and Sally flip their coins alternately until a coin shows heads. The girl with the coin showing head wins the game. What value of $p$ makes this game a fair game?

Answer

Wendy wins on the first flip with probability $1/100$. Otherwise, the game keeps going and Sally has probability $p$ to win on the next flip, which has overall probability $99/100 \times p$. If not, the game returns to the start, which has no effect on fairness. So, to be fair, these two probabilities should be equal, which happens for $p=1/99$.

word - Hangman- No misses left

**** No programs allowed. You must find the answer using your own brain. ****

Your hangman game allows for 9 misses, but on the 10th miss, you're hanged!

You don't want to be hanged. The letters marked red underneath have already been guessed.

What is the English word?

Answer

I reckon the answer is

Pizzazz
http://www.thefreedictionary.com/pizzazz

combinatorics - Attacking queens

Can you place 3 queens on a 6x6 chess board such that they can attack every square?

Good luck!

Answer

Here's the solution:

It's interesting to note that the domination problem has very few solutions for 3 queens on a 6x6 chessboard as compared to other possibilities. (Neither of those links contains the answer! Just background info.)

Perhaps the key realisation is that

some squares which you'd intuitively expect to be attacked diagonally can be attacked orthogonally. With one queen at $a1$ covering the central white diagonal, it'd seem natural to use two more to cover the black diagonals to either side, including those difficult squares $b3/c2$ and $e6/f5$. But putting the other two queens on the next diagonals leaves not enough space to reach out to the edges ... and we realise there's a nice configuration to cover $b3/c2$ and $e6/f5$ orthogonally.

Before that, starting with

a queen at the corner is a counterintuitive move in which I was inspired by the 5 queens on an 8x8 chessboard problem.

mathematics - Black and White

OK guys, I think this is my best puzzle yet. Hope you enjoy it, the solution is neat and simple.

A boy draws 2015 unit squares on a piece of paper, all oriented the same way. The squares can overlap. Then he colors the resulting picture in black and white chess-wise, such that any area belonging to an even number of squares is painted white and any area belonging to an odd number of squares is painted black.

Prove that there is at least one square unit of black on the paper when he is finished.

Answer

Let's think of the problem as taking XOR's (symmetric differences) of the square areas, with black as 1 and white as 0. The black area is the XOR of all the squares.

Imagine doing this all on graph paper whose cells are sized and oriented like the squares. Take the resulting figure, cut up the occupied cells, and stack them on top of each other without rotating. Now, take the XOR of that stack.

Note that any square on the graph paper gets cut up and rearranged to make exactly the full unit cell. So, the XOR of the stack is the XOR of 2015 fully black squares, which is all black, since 2015 is odd. Then, since each black point in the XOR requires at least one black point the contribute, the total area of black in the stack must be at least 1.

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Needlessly formally:

We are interested in the group $G$ of (well-behaved) finite-area subsets of the plane under symmetric difference (XOR). We quotient out the plane by integer translations to get $(\mathbb R \times \mathbb R) / (\mathbb Z \times \mathbb Z)$, which induces a homomorphism $\phi$ from $G$ to the corresponding group $G'$ on the unit square.

We have $2015$ single-square elements $x_1, ..., x_{2015}$ of $G$ and are interested in their sum $x$. As single-square elements, they all map to the same full-square element $\phi(x_i) = s$ in $G'$, so $\phi(x) = s + \cdots + s$ summed $2015$ times. Since $G'$ has order $2$, this is simply $s$. But the homomorphism $\phi$ cannot increase the size of the subset, so

$$\mathrm{Area}(x) \geq \mathrm{Area}(\phi(x)) = \mathrm{Area}(s) =1 $$

logical deduction - 99 chips into more chips

You are at a casino in Vegas and you have earned 99 chips by playing poker!

While you are checking out a slot machine, someone comes to you and congratulates you that you have a chance to make more chips by using your own chips into a strange four slot machine.

With this machine, you can put as many coins as you want into the four slots available and pull the trigger only once to make more coins but all four slots where you put your coins in the machine behave differently:

• One of them makes your coins four times as many as before!


• Another slot just gives your coins back.


• The last two slots do not give your coins back at all.

But you do not know which slot is which and you can take your coins back after pulling the trigger from somewhere else as a whole.

At most how many coins can you guarantee to have at the end when playing with this machine?

Answer

Quick lower bound:
If you were to

Evenly divide the chips between the four slots (24 in each, with 3 left over), you would get 96 from one, 24 from another, and 0 from the others, for a guaranteed 123 chips in the end.

logical deduction - The Peculiar File

So I was browsing through my computer files earlier and I found a file I'd not seen before. I was wondering if you guys could help me figure out what it's about.

This is where I found it:

This is its contents:

What does it say? And what is it referring to?

Answer

It's a

nonogram.

The filled-in cells say (not perfectly clearly)

THE GREAT ESCAPER. Presumably in homage to the PSE participant. The eye in the window's title bar is (credit to user TrojanByAccident in TSL chat for noticing this) a reference to the eye that appears in many of the grids of TGE's "WITLESS" puzzle that's currently bending the brains of many PSE folks.

Actual filled-in image:

enigmatic puzzle - A simple alien safe

Oh no!! Abducted by aliens and put into a labyrinth for ability testing - again!

There is (obviously) no deeper meaning behind this all, but can you help me open this safe, as it undoubtedly contains some strange device I will need in the next room of the labyrinth just to survive?

That's a picture of the safe in the room:

(Safe)

The only other objects are a couple of plates lying on a glass table next to the safe:

(Plates on a table next to the safe)

Well, that really is all I have. The lid of the box on top of the safe seems to be connected by five (flexible) cables, disappearing seamlessly into the back of the safe. The safe handle can be rotated, but it obviously does not open the safe unless a proper combination is set with the many brass locks on the right. The plates are made of some heavy plastic type material, the box on the safe is some kind of tinted glass with the front-section more transparent.

I'm sending you a few more detail snapshots below. (Clicking the images for full-res version may be advised.)

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(Key-locks detail)

(Box detail)

(Box detail with lid removed)

(Slices arranged, with label, view angle 1)

(Slices arranged, with label, view angle 2)

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You don't have to - all information should be in the images here on SE - but if you fancy it, you can examine the things on an external site with a 3D viewer. Here are the links:

• Safe


• Plates A, B, C, D, E, F, G, H, I, J, K, L, M

Answer

I believe I got it.

Crossing the colors of the wires at the bottom plate, you can see there is a pattern of colors you need to create:

So you have to create paths of the colors in the lid trough the holes of the plates. Also, there is a visible space in the container, where you can see the arrows at the side of each plate.
13 plates, 13 key locks, so you have to put the key locks in that order(there is plates without visible arrow, so I am assuming you leave the key lock in the center).

I did it in the following way(starting from top):
J -> Right
E -> Down
H -> Right
A -> Center
F -> Right
K -> Down
D -> Center

C -> Center
L -> Up
B -> Center
I -> Up
G -> Down
M -> Left

A colorful step-by-step:

And the verified correct solution:

riddle - Anonymous letter from somewhere sometime today

Not many know of me, and even less have touched what is mine.
A lonely fella I am, like being the smallest pebble in the puddle,
Unless you ask the bishop, he owns the title, according to a beaver's idea.
Reached I am by twelve shiny points, across that much and two more...
Almost exposed myself by now, but before you find me, you should know,
That robbing me of my salty gold, is viable no more for some time now.
And as a goodbye, I thank you for this conversation, and may God's will be above all!

P.S. You may want to rob the the Devil instead, just break the king's seal.

I hope you'll find this interesting enough. Don't know what else to say, time will tell.

The answer is a single word, who or what is the author of this letter, which was sent to you?

Hint:

The answer is a geographical place.

Answer

I think this letter describes

Nauru

Not many know of me, and even less have touched what is mine.

Nauru is the least visited country in the world with less than 200 tourist visits in 2011.

A lonely fella I am, like being the smallest pebble in the puddle,

Nauru is a remote island in the South Pacific - it is the world's smallest island (pebble/rock) nation.

Unless you ask the bishop, he owns the title, according to a beaver's idea

Bishop Rock of the Isles of Scilly is the smallest island with a structure. This is from the Guinness World Records, which started as Hugh Beaver's idea. Thanks OP for this!

Reached I am by twelve shiny points, across that much and two more...

On its flag, Nauru is symbolized by a 12-pointed star, each point representing one of the 12 indigenous tribes on the island.
That much and two more = 12 + 2 = 14 refers to the fourteen districts the island is divided into.

Almost exposed myself by now, but before you find me, you should know,

Note the acrostic down the left side the first four lines begin: NAUR

That robbing me of my salty gold, is viable no more for some time now.

The salty gold is the Phosphate (chemically a salt) deposits originating from sea bird droppings.

Nauru had the highest per-capita income of any sovereign state in the world during the late 1960s and early 1970s. The primary phosphate reserves were exhausted, crashing the local economy in the process.

And as a goodbye, I thank you for this conversation, and may God's will be above all!

Nauru's official motto is "God's will first"

P.S. You may want to rob the the Devil instead, just break the king's seal.

The Devil refers to Morocco. Its flag is red and it has a five pointed star, which refers to the pentagram. Red color and pentagram associate to the devil. Morocco has the world's biggest phosphate reserve in the world, and is second in its production (behind China). If you were planning a phosphate heist, this is the place to rob instead. The king's seal refers to the five-pointed star actually representing the Seal of Solomon (King Solomon). Thank you, Vepir, for this!

geography - Fieldwork ahead

_{This puzzle is part 11 of Gladys' journey across the globe. Each part can be solved independently. Nevertheless, if you are new to the series, feel free to start at the beginning: Introducing Gladys.}

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Dear Puzzling,

I can't believe it has taken this long before I visited the first art gallery of this trip. The place I visited today has a large collection of beautiful works by artists whose names I couldn't possibly spell correctly, let alone pronounce!

Wish you were here!
Love, Gladys.

Across
1. Molten rock
2. Poker with four hole cards
3. Van Halen vocalist
4. Caribbean island
5. Playwright Coward

Down
1. Actress with a beach club

2. Herbal liqueur
3. Unclear, imprecise
4. Fictional captain
5. 0.2 g

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_{Gladys will return in "For the price of one".}

Answer

I think Gladys is at

Lauba

Completed grid

Letter locations

Using the rules of minesweeper we can discover where bombs are hidden

mathematics - Automatically a Knight, Knave, and Joker

Let M be a finite positive integer. It's exact value is not known.

Suppose we have three classes of automaton, all of which accept a bit stream as input, produce a bit stream as output (one bit per input bit), and have a non-observable state, S, that is an integer coordinate on an infinite bidirectional 1D line:

• knightbots start with S = M. Whenever a 1 bit is inputted, knights move 2 units to the right (i.e. S is incremented by 2). Whenever a 0 bit is inputted, knights move 1 unit to the left (i.e. S is decremented by 1).


• knavebots start with S = -M. Whenever a 1 bit is inputted, knaves move 1 unit to the right. Whenever a 0 bit is inputted, knaves move 2 units to the left.


• jokerbots start randomly at either S = M or S = -M. On cromulent days, jokers move 1 unit right on a 1 input and 1 unit left on a 0 input, but on non-cromulent days, jokers move 1 unit left on a 1 input and 1 unit right on a 0 input. You have no idea whether or not the current day is cromulent, and there's really no way to find out.

After each input, the automata output either a 1 if they have moved into the state S = 0, or a 0 otherwise.

Simple Problem (for "Correct Answer" Credit)

You are given an automaton that is either a knightbot, a knavebot, or a jokerbot, but you don't know which of the three. The automaton is guaranteed to be in its starting state. You can input bits to the automaton and observe its output, but otherwise you cannot observe its state or inner workings. Also note that you cannot "reset" the automaton.

Your task is to produce an algorithm that is mathematically guaranteed to properly classify the automaton as a knight, knave, or joker using an input sequence of finite length.

Your solution should include a (reasonably) detailed description of your algorithm, as well as a guaranteed (although not necessarily tight) upper bound on required input length as a function of M.

Challenge Problem (for 100 Bonus Rep!)

Instead of the aforementioned jokerbots, consider jesterbots:

• jesterbots start with S start randomly at either S = M or S = -M


• jesterbots have a secondary state, b, which assumes the value of the most-recently-inputted bit. The initial value of b is 0.


• jesterbots have a tertiary state, D, which takes on the values 1 or -1, and may be treated as a direction. The initial value of D is 1.


• on cromulent days, jesterbots move in the direction of D (i.e. increment S by D) iff an input bit matches b (the previous input bit), or else stand still and about face (multiply D by -1)


• on non-cromulent days, jesterbots move in the direction of D iff an input bit does not match b, or else stand still and about face


• after each input, the jesterbot outputs a 1 if it has moved into the state S = 0, or a 0 otherwise

Your task is identical to that for the simple problem, only substituting jester(bot) for joker(bot) in the task description.

I will award 100 bounty rep to the correct, defensible solution to this challenge problem that has the strictest upper input length bound (in terms of $\mathcal{O}\left( M\right)$), or to the earliest solution in the case of a tie.

Good luck automating. :)

Answer

I claim a bound linear in $M$.

First of all, the challenge is really to make the automaton hit 0. Once it hits 0, we can easily determine what type it is using a constant number of moves.

Let $A_1,A_2,...,A_n$ denote different types of automatons. For us, the $A_i$ consist of a knightbot, a knavebot, and then many different automatons for each initial setting of a jester/jokerbot.

We also assume that, given an automaton $A_i$, if we know what state it is in, then we can always move it one space to the right or one space to the left in constant time. This is clearly true for any automaton in the problem.

Now, suppose we have a finite upper bound $B$ on the distance from our automaton to the origin. Then the following series of moves is guaranteed to make the automaton hit 0 in $O(B)$ time:

• Pretend the automaton is $A_1$. Move it $B$ spaces to the right, one at a time. Then move it $2B$ spaces to the left. This takes time $O(B)$.



• If we hit 0 along the lines, great! Otherwise, our automaton is not $A_1$. Pretend the automaton is $A_2$. Then we have spent $O(B)$ moves so far, so there is a number $k=O(B)$ such that our automaton is at most $k$ steps from the origin. So, first calculate the internal state that $A_2$ would be in, given the entire sequence of inputs we have fed it so far. Then, based on this state, move the automaton $k$ steps to the right, one at a time, and then $2k$ steps left, one at a time.


• If we hit 0, great! Otherwise, pretend our automaton is $A_3$. You can see where this is going.... The important thing to note is that, since the number of automatons $n$ is a constant, the total number of steps spent to test all $n$ automatons is still $O(B)$. The constant factor may be large, but whatever, it's still $O(B)$.

Great. Unfortunately, we don't know an upper bound $B$ on the distance from our automaton to the origin. The fix is very simple. The above procedure runs in $O(B)$ moves, so there exists a constant $c$ such that the above procedure never moves any automaton $A_i$ by more than $cB$ steps.

Now, run the above procedure with $B=1,3c,(3c)^2,(3c)^3,(3c)^4,\dots$.

To see that this works, fix $M$. Pick $k$ so that $\frac 12(3c)^k\leq |M| \leq \frac 12(3c)^{k+1}$. Then after the first $k$ iterations of the procedure, by our choice of $c$, the automaton has moved at most $c(1+3c+(3c)^2+\cdots+(3c)^k)=c\frac{(3c)^{k+1}-1}{3c-1}\leq\frac 12(3c)^{k+1}$ spaces. So, it is at most $\frac 12(3c)^{k+1}+|M|\leq(3c)^{k+1}$ distance from the origin. Therefore, it will be found on the next iteration (because on the next iteration we use $B=(3c)^{k+1}$).

Furthermore, because each iteration takes linear time, the entire sequence of iterations takes at most $O(1)+O(3c)+O((3c)^2)+\cdots+O((3c)^{k+1})$ time. This sum is $O((3c)^k)$, and therefore $O(M)$.

Therefore, this procedure is guaranteed to hit 0 in $O(M)$ moves. As noted previously, as soon as we hit 0, we can easily determine which automaton we are dealing with using a constant number of moves. Therefore, this procedure guarantees a linear bound. QED

logical deduction - The seven piece silver chain

You have to stay for 7 days in a hotel outside your city. But you forgot to bring your credit card and are out of cash.

You have a silver chain with 7 links. The rule in this hotel is that residents should pay their rent every morning. The manager of the hotel is well aware of your situation and says that you can pay one link of your silver necklace every morning instead of money.

But he wants the chain in good condition. So he suggests a solution for your every morning payment with only ONE link cut in the chain.

What is his solution?

source: fekraneh.ir

Answer

Answer:

Cut the 3rd link.

1. On day 1 you give him the cut link
2. On day 2 you give him 1-2, he returns the cut link
3. On day 3 you give the cut link, along with the 1-2 he has.
4. On day 4 you swap your 4-5-6-7 for the 1-2 and 3
5. On day 5 you give the cut link (so manager has 4-5-6-7 and the cut link 3)
6. On day 6 you swap 1-2 for cut link 3 (so manager has 1-2 and 4-5-6-7)
7. On day 7 you give the remaining cut link 3.