Extend this infinite series by five more terms:
17, 3, 5, 7, 11, 17, 2, 3, 5, 11, 13, 17, 2, 5, 7, 11, 13, 2
Answer
My answer is
3, 5, 7, 13, 17
When the prime numbers listed are replaced by their ordinal in the prime sequence, this relates to the OEIS sequence A116369 "Day of the week corresponding to Jan 01 of a given year." (n=0 for the year 2000)
The next five terms are 2, 3, 4, 6, 7 which are the ordinals of the prime numbers I gave.
The puzzle title is a hint!
I just got another picture in the mail, and can't for the life of me piece together why I've been getting such unusual letters...
To make things easier for colourblind solvers: the second panel is #FF7F50, the fourth panel is #008080 and the bottom numbers/arrow are #FF0000.
Answer
Each of the rebus works as follows:
A string X is derived from the rebus itself, and a string Y from the overall typesetting (font size, color etc) of the rebus. The answer is then X placed inside Y, since the image literally shows X in (the style) Y. The cute figures below each of the top four images act as definitions for the answer, allowing us to confirm our answers as we progress.
Rebus #1:
The rebus is L IN DF written in BOLD, so the answer is BLINDFOLD.
Rebus #2:
P OR E in CORAL giving CORPOREAL.
Rebus #3:
T IF IC in ARIAL giving ARTIFICIAL.
Rebus #4:
E TO T in TEAL, giving TEETOTAL.
And there's more!
As hinted by the numbers below, we can take the 3rd, 7th, 7th and 4th letters of these answers to form a new rebus:
So the final answer is:
CT IF IE in RED, giving RECTIFIED.
My prefix prefix does exhilarating kicks,
My infix infix uncovered the wizard's tricks,
My suffix suffix is quieter than a tick,
Gone a century since, he made infinite picks.
Answer
I think the answer is
Cantor
My prefix prefix does exhilarating kicks,
My infix infix uncovered the wizard's tricks,
Toto pulls the curtain in The Wizard of Oz.
My suffix suffix is quieter than a tick,
RR? Or possibly RR
Edit: As mentioned by Tom, this instead refers to a Rolls Royce vehicle for which the loudest noise comes from the tick of the electric clock.
Gone a century since, he made infinite picks.
Cantor.
His diagonalisation argument involves making infinite picks, plus he died in 1918.
Edit: As suggested by Bass in the comments, the infinite picks could refer to the process of generating the Cantor set
Reminder:
Everybody knows that we can place 8 queens in a chessboard without threatening each other (see here). Same reasoning can be applied for knights, bishops, rooks and kings. Giving respectively 32 knights, 14 bishops, 8 rooks, and 16 kings.
Problem:
If we assign to each type of piece a value inversely proportional of the number of this we can place. It means Knights = 1/32. Bishop = 1/14. Rook = 1/8. Queen = 1/8 and King = 1/16.
What is the best sum value we can achieve mixing these pieces still with none able to take each other?
Example
In this position we have 1 queen, 4 rooks, 1 knight and 2 kings, so the value would be 1/8 + 4/8 + 0/14 + 1/32 + 2/16 = 25/32 = 0.78125.
Can you beat that score?
Hard Question: Can you prove that your answer is optimal?
Actual High scores
@evargalo 1.2857
@keeta 1.303
@Blcknght 1.3125
@oray 1.3348 (optimal).
source: Gyozo Nagy in IBM Research Ponder this - August 2003 [found in Diophante.fr]
Answer
Here is the most probable optimal solution with some extra modification of previous answers:
The score is 1.3348.
Here is the brute Force Code written by @fireflame241 confirming this is actually optimal.
I make billions (#3)
My products are quite edgy,
But most have been a bust,
I'm only one of many,
Lord knows who we can trust,I'm not a politician,
Yet I work with presidents,
And like a mathematician,
I seek proofs of excellence,I resemble a detective,
Who fills in all the blanks,
And if gold is your objective,
I've more than all the banks,Who / what am I?
Hint for line 3:
Lines 3 and 4 are two sides of the same coin.
Previous riddles:
I make billions #1
I make billions #2
Answer
It’s
The US Mint(s)
My products are quite edgy,
Coins have edges
But most have been a bust,
They have busts of politicians
I'm only one of many,
In the US the coins say “E Pluribus Unum”, meaning “Out of Many, One”. Thanks to @postmortes!
Lord knows who we can trust,
In the US the coins say “In God We Trust”
I'm not a politician,
Mints aren’t politicians
Yet I work with presidents,
Presidents have their likenesses on coins
And like a mathematician,
I seek proofs of excellence,
Per @JonMark Perry, a proof is a non-legal tender coin made pre-release in order to check the die for flaws
I resemble a detective,
Who fills in all the blanks,
A blank is a coin that hasn’t yet been stamped
And if gold is your objective,
I've more than all the banks,
The US Federal Reserve stores its gold bullion reserve in 4 Mint locations around the US including at Fort Knox, a US Mint with nearly 5000 tonnes of gold. Goldfinger famously went after Fort Knox rather than any bank when he was looking to steal gold from the US.
Here's an old puzzle I found in a book once.
A box has some number of locks on it. It is guarded by a Grand Vizier and four slaves, each of which hold keys to the locks.
The Grand Vizier and slaves each hold a different set of keys to the locks, and each key they hold opens a different lock.
The keys they hold are in a certain combination such that as long as the Grand Vizier is with one slave, or if any three slaves are together, they will have all the keys to open the box; but if less than that are together, they cannot. How many locks did this box have at minimum?
Answer
The best solution I've found has 7 locks.
The Grand Vizier has keys 1, 2, 3, 4, 5, 6.
Slave 1 has keys 1, 2, 3, 7.
Slave 2 has keys 1, 4, 5, 7.
Slave 3 has keys 2, 4, 6, 7.
Slave 4 has keys 3, 5, 6, 7.
A couple of things simplify the deduction of this puzzle:
If the Grand Vizier has every key but one, and every slave has the missing key, that reduces the problem to 4 slaves and one less lock.
Every key must be held by at least two slaves.
The rest I just brute forced with a (digital) pencil and paper.
This is a Statue View puzzle, an original invention combining two logic puzzle genres: Statue Park and Canal View.
Rules of Statue View:
Answer
Some initial observations:
The 4 can have at most two cells shaded to the north, so the two cells east of it must be filled in.
Then, since the unshaded cells (including the numbers) must be connected, the cell north of the 4 is also unshaded, and the four cells to the east of the 4 must be the I-tetromino.
Also,
The 5 can have at most two cells to the north and two to the west, so the cell south of it must be shaded.
Then,
The lone unshaded square at the east end of the "I" doesn't have any way of connecting with the other unshaded cells just by going farther east, so the cell north of it must also be unshaded.
Then,
The 5 must also be connected to the other unshaded cells, meaning at least one orthogonally adjacent cell is also unshaded. That means two cells are connected to it going either north or west, and all three cells to the south of the 5 are shaded (this is the exact same logic I applied to the 4 and the 2, but for some reason I failed to apply it to the 5 until after it was pointed out by @Omega_Krypton).
Then,
It's obvious that the only shape that fits into the bottom left is the L-tetromino, which means the cell west of the 5 is adjacent to it and unshaded, and the two cells between the 5 and the 2 must be filled in.
From that,
Again the 2 must be connected to the other unshaded cells, so the square to the west is unshaded, and the only shape that fits below it is the O-tetromino
Finally,
The T- and S/Z-tetrominoes are left to connect to the last two numbers, both 3s (in a symmetrical layout, no less). Now, if either of the shapes were to contribute to both numbers, it would be by sharing the two cells between them, meaning the final shape would have to connect just one cell to one or both of the numbers, which is just not possible in the space given.
Thus, the last two shapes each connect 3 cells to one of the last two numbers.For the T-tetromino, this is only possible on the north vertical of either number. The S/Z can then sit a top the other number, with two cells shaded to its north and one to the inside, touching only the corner of the T.
However, if the T is north of the central 3 and the S is north of the western one, the block of unshaded cells in the northwest corner is disconnected from the rest.Thus, the T must be above the western 3, and the Z is above the 3 in the center:
You have exactly 990 edges. Assemble them into a simple undirected graph with two distinguished vertices A and B, such that the number of different simple paths from A to B is as large as you can make it.
One example. 990 edges is precisely what you need to create a complete graph on 45 vertices. Call one of the 45 vertices A and another B. Then the number of paths from A to B is $$ 1 + 43 + 43\cdot42 + 43\cdot42\cdot41 + \cdots + 43! = \lfloor 43!\cdot e\rfloor \approx 1.64\times 10^{53} $$
Another example. We can do much better than that, though. For example consider this graph:
A---O---O---O---O.....O---O---B
\ / \ / \ / \ / \ / \ /
O O O O O O
made of 330 triangles. Here there are $2^{330}\approx 2.19\times 10^{99}$ simple paths from A to B.
How many paths can you make, and how? I know that solutions with plenty more than $10^{100}$ paths exist, but have no particular reason to believe that what I'm thinking of is optimal.
Most paths win. Answers must state the number of paths in ordinary scientific notation, and must explain clearly how the paths were counted (in addition to explaining how to construct the graph they're counted in, of course). It is acceptable to compute a lower bound instead of an exact count.
(Inspired by this math.SE question, but I don't think anyone will be able to prove a maximum, so it's more a puzzle than an exact question.)
Answer
If I'm not mistaken the following grid must give the optimal result:
where are 990/5 = 198 layers with 2 vertices.
Let's calculate number of paths. I number 2-vertice layers from 0 to 197. CD is 0th, GH is 197th. There are 3 different possibilities at each transition between two layers:
1) straight
/
/ __
2) vertical + straight
__
|\ |
| \ |
3) zigzag
__ __
\/ \
/\ _\
A rule here is that you can make vertical line only at the beginning of the transition. I call type 1 and type 2 open and type 3 - closed. After open transition you can have any other transition, but after closed you can have only type 1. Let's call number of paths, which lead to open layer $i$: $a_i$ and n of paths, which lead to closed layer $i$: $b_i$.
$a_0 = 2$, $b_0 = 0$ (see the vertical-line-rule above).
Open pattern can be followed by 4 open (type 1 and 2) or by 2 closed. Closed patern can be followed only by 2 open (type 1). Thereby: $$ \begin{pmatrix} a_{n} \\ b_{n} \end{pmatrix} = \begin{pmatrix} 4 & 2\\ 2 & 0 \end{pmatrix} \begin{pmatrix} a_{n-1} \\ b_{n-1} \end{pmatrix} $$
Finally, to reach B-point after last layer we can use 2 ways if we ended with open pattern and 1 way with closed. So the total number of paths are:
$$ \begin{pmatrix}2&0\end{pmatrix} \begin{pmatrix} 4 & 2\\ 2 & 0 \end{pmatrix}^{197} \begin{pmatrix} 2 \\ 1 \end{pmatrix} $$
The result is $10^{135.325} = 2.11\cdot10^{135}$.
To be sure I haven't made mistakes, I have compared my calculations for 20 vertices and 45 edges graph with recursive program. They do agree.
The "inductive prove" (guess based) that most probably you can't do better you can find below.
The old answer:
@Falk Hüffner, gave the improved number $6.76 \cdot 10^{130}$ and the fact that you need to use 15 edges per "link" in the "chain", but there is still no prove and no example of the "link", which allows you to achieve this number. So let me do it here and give some extrapolations, which allows to imagine a probable optimal solution.
First, my chain of thoughts:
I got noted that @Henning Makholm (3 edges -> 2 paths), @Roland (5 edges -> 4 paths) and @f'' (9 edges -> 15 paths) can be reached easily from the complete graphs (for 3,4 and 5 vertices correspondingly) by rejecting the most useless edges. In case of 3 vertices there are no edges to reject. In case of 4 and 5 vertices you need to reject the one who connects A to B directly (and adds only 1 path).
Now, if we take complete graph on 6 vertices and will reject edges, which connect A to B via only 1 vertices we will get a graph, which have 10 edges and 20 paths. Which leads to $6.34 \cdot 10^{128}$ and is quite close to result with 5 vertices.
Here comes the example for current max-paths:
But you can improve result if you take 7 vertices and reject edges, which connect A to B via only 1 vertices. You will have 96 paths per 15 edges. And $6.76 \cdot 10^{130}$ paths for 990 edges.
To do so you need to take complete graph on 5 vertices (10 edges) and any 3 of them connect to A and other 2 connect to B.
Complete graph on 5 vertices allows you to reach any point from any-other point in $1+3+3\cdot 2+3\cdot 2+1$ = 16 paths. Plus you have 3 different ways to reach A and 2 ways to reach B, this leads to $16 \cdot 3 \cdot 2$ = 96 paths. Then $96^{990/15} \approx 6.76 \cdot 10^{130}$.
And the extrapolation:
When you try the same trick with 8 vertices, you will get $(65\cdot3\cdot3)^{990/(15+3+3)} \approx 2.83\cdot 10^{130}$ which is already smaller and since 990 is not dividable by 21 the actual result will be even worse.
Let's try now to reject also paths with 2 vertices. This is achieved in the following configuration: 
And the result here is exactly the same as with 7 vertices: 15 edges and 96 paths.
So I suppose to improve this you need to consider 9 vertices and reject all paths via 0,1 and 2 vertices (I still need to figure out how to build and count this, may be it is even to reject all paths, which go through 2 vertices only).
Now, it looks like you need to add 2 vertices and reject 1-vertice longer path each time. Finally, we can guess here that the most optimal result will be a complete graph on X vertices with all paths via 0,1,2,...,(X-5)/2 vertices excluded. So if you stretch it between A and B you should see a chain with width of ~2 vertices.
P.S. Another @Falk Hüffner result $8.16 \cdot 10^{130}$ with 18 edges and 240 paths in "link" is achieved in the following configuration of the link: 
This configuration follows the general sense rule, which I mentioned above: you need to take complete graph and rid of most useless links (those, which contribute to the shortest paths).
Unfortunately it was hard for me to consider all possibilities on paper, but my program confirmed that there are exactly 240 paths.
A week ago, when I was travelling on an airplane to my city London, I found an envelope. The strange thing was that I found my name on it. So, I opened it. It read:
Hi,
Did you miss me? I hope you are alright.
You remember me, right? I am your archenemy.
And your closest person. What is he doing? Does he get any patients now?
I also hope your elder girlfriend Nerie is fine.
I am waiting for you at your home. And yeah, I have the key to the last part of your name.Yours,
The letter was not signed. Can you guess who he was? Or as a matter of fact, who am I? I am quite famous.
Answer
Are you and your archenemy:
Sherlock Holmes and Professor Moriarty?
Did you miss me? I hope you are alright.
"Did you miss me" is what Moriarty says on every screen in the last episode of Sherlock (credit to Lord of dark)
And your closest person. What is he doing? Does he get any patients now?:
Dr. Watson, explaining the 'patients' part
I also hope your elder girlfriend Nerie is fine.:
Irene Adler, an anagram of 'Nerie'
I am waiting for you at your home. And yeah, I have the key to the last part of your name.:
Sherlock, is actually pretty self-explanatory
A man walks out of his house and over to his flower garden. He holds out his right hand and a bee lands in it.
What is in his eye?
Answer
Beauty
Because:
"Beauty is in the eye of the beholder (bee holder)"
These words! --they come falling free;
speech: daft here, deft there,
in that old, linking tongue.
We take these precious signs and tones,
play them on our organs grand,
sing them in our voices strong--
yet spoken in our poor, restricted speech,
we find them dulled again.
So hear ye of this simple speech,
and grasp it phrase for phrase;
key words to link the harmony,
re-form the dictum true:
The first, the object--
abstract; a guide without a muse.
By itself lives unrestricted,
but here we find it,
linked and woven in.
The second, the tongue--
morphing; with an effortless building
of an ethereal wind;
here one moment, gone the next,
sent ever on its merry way.
The third, the tongue itself--
waiting on society lost,
but alas!-- we blinked,
and now it is dead.
The fourth, the death of wind--
pushing; with a quiet passing.
The spread that does not stop,
until it breaks on us,
and ends its blink of life.
The fifth, the deep--
revelation; a formative thought.
Passing without trace,
but leaving the hint of discovery.
So bring ye of these keys and tones,
enmesh them! string them well.
Then bring them to their linking tongue,
to build the dictum true.
And it should matter not from here,
if mind be bent or broke;
for spoken in the linking tongue,
lo, even Madmen earn respect.
A hint, should you need it:
Unless you've encountered the phrase already, or are very familiar with language, you may not be able to easily find the full, correct answer without the aid of a search engine. You should be able to get pretty dang close, though, unless I've messed up difficulty.
Answer
I am not awfully sure, but I think the answer may be:
the popular quote: "Anything said in Latin sound profound", or its Latin equivalent "Quidquid latine dictum sit, altum sonatur". This RationalWiki page contains a variation of this quote.
The first, the object--
abstract; a guide without a muse.
By itself lives unrestricted,
but here we find it,
linked and woven in.
"Anything"- refers to an indefinite abstract object, without explicitly naming it. Usually it refers to any object (unrestricted), but here it is deeply related to the rest of the sentence.
The second, the tongue--
morphing; with an effortless building
of an ethereal wind;
here one moment, gone the next,
sent ever on its merry way.
"said (in ?)" - It means the speech (generated by the tongue), that we almost effortlessly create in the air every now and then, and then it fades away (unless written down or otherwise recorded).
The third, the tongue itself--
waiting on society lost,
but alas!-- we blinked, and now it is dead.
"Latin" - The dead tongue (language).
The fourth, the death of wind--
pushing; with a quiet passing.
The spread that does not stop,
until it breaks on us,
and ends its blink of life.
"Sounds"- sound is created by an agitation in the air, which propagates through the air and finally reaches us to be heard.
The fifth, the deep--
revelation; a formative thought.
Passing without trace,
but leaving the hint of discovery.
"profound"- deep, having (or appearing to have) some sublime significance. Not quite sure about this part.
So bring ye of these keys and tones, them! string them well. Then bring them to their linking tongue, to build the dictum true.
Now joining them together, we get the sentence "Anything said in Latin sound profound", and converting to the 'linking tongue' (Latin), we get its more popular variant "Quidquid latine dictum sit, altum sonatur."
And it should matter not from here, if mind be bent or broke; for spoken in the linking tongue, lo, even Madmen earn respect.
That's what the sentence means.
Interestingly enough,
this sentence appears in the chat profile of the OP.
I'm new to this StackExchange - I'm more active on programming stack exchanges. So here's a computer-themed puzzle:
4K5L3RdR-PASTE
Find the code (6 numbers)
That's it. I'll leave you with one clue - The title.
A man is told to make a circle
He makes this:
Where is the man?
Answer
In Manhattan, because that is what a "circle" (defined to be the set of points of a certain set distance d away from a given point) looks like when using the taxicab (or Manhattan) metric.
Small note on the first image:
I wish romans had a number for 0.
Hint 1 for the squares image:
It is mostly an arithmetic problem.
Hint 2 for the squares image:
Color and relative position matter.
Hint 3 for the squares image (alternative image that should help a little bit):
Hint 4 for the squares image:
This is a Mathematical Rebus, so what about changing the squares by numbers? Which? It's your work to discover.
Answer
Since this remains unsolved, I proceed to answer the question.
Most of it had been solved, so I only put together the already given answers and add the missing clue.
First clue, by Gareth:
is 0 when a=0 and increases by 1 each time a increases by 0.001. So it's floor(1000a) which, given all the Roman numerals, I should probably write as $\lfloor Ma\rfloor$.
Second clue:
The trick is using the colors. White is #FFFFFF and yellow #FFFF00. It was intended to be FFFFFF module FFFF00, or $FFFFFF_{FFFF00}=FFFFFF \mod FFFF00 \equiv FF$
Third clue, also by Gareth:
is probably "voltage, in volts, over resistance, in ohms", yielding "current, in amps" or $I_A$.
All together, the answer is:
Maffia, as actually Gareth already got.
This puzzle is not mine, although I don't have any reference as to where it originated from. I discovered it when I was a student, some fifteen years ago. It has been among my favourite chess puzzles since then. Could be very well known or a duplicate, but I couldn't find it in the database.
The following position is legal, except that the White King (and the White King only) is not displayed. Find it.
Answer
The white king must be on square
c3
Explanation:
First, we determine which side is to move. Unless the white king blocks the a4-d1 diagonal, the black king is in check. The king can't be on c2 since it would attack the black king; on b3, White would be to move but there's no way this double-check could have happened in a real game; there's no way to clear both the b3-d5 diagonal and the b3-b5 file with one move from the black rook or the black bishop. Black simply couldn't have played Bb4-d5 or Rc4-b5. So the black king is in check and White made the last move.
So, what was the last move?
The bishop can't have moved along the d1-a4 diagonal, because it would give check on c2 and d3 as well. So it must be a discovered check, caused by a move from the white king, from b3 to another square.
But ...
doesn't this lead to the same problem as before? On b3, the White king is still in double check. No, because there is a move which simultaneously clears the b3-d5 diagonal and the b3-b5 file, but it requires extra pieces: an en-passant capture from b4 to c3, capturing a pawn on c4. Since we have an extra (half-)move compared to the situation above, we can use that move to get rid of the black pawn.
So,
if White played 1. c2-c4, Black can play 1... b4xc3 e.p., White responds with 2. Kxc3 and we're in the diagram in the question with the White king on c3.
Ciphertext: 1010, 111, 11, 0110, 001, 1, 0, 010, 000
This cryptogram is based off of AE's first one. Enjoy.
The answer to this is a clue to this question which is currently unanswered and has a bounty.
Technical Stuff
No Roman numerals
No abbreviations
No slang or shorthands
Answer is relatively short...
Have fun!
Answer
I think it's
COMPUTERS. If you translate 0 as dot and 1 as dash in Morse code, except the0011needs to be split as001(U) and1(T). There's no Morse code symbol for0011. Edit: The dash character-is apparently encoded by0011so possiblyCOMP-ERS?
William Herschel III, heir to the Herschel Paper Company, is dead. His body was found in fishing gear, discovered near the family estate's lake by his Labrador. Police immediately suspected foul play. CCTV footage proved that his death took place around 3pm the previous day. Eight suspects were called, each had a personal grudge with Mr. Herschel. As the case was coming to a dead end, you tried to scan through Mr. Herschel's family photos and discovered something: he always wore a gold chain that said "WAKE" whenever he would go out fishing. However, the chain was not found at the crime scene. Dedicated house searches by police found that the chain was in the room of Jack, his personal flight attendant. Pressed, Jack finally confessed that he was the son of Mr. Herschel and an unknown maid, and murdered Mr. Herschel in a rage after Mr. Herschel never acknowledged Jack as an heir.
The case seems to end here, but there's another meaning behind this. Can you find it out?
Hint:
Who was dead? Where and how was his body found?
Hint 2:
Hint 3:
Answer
I’ll take the second clue and try to stretch it as far as:
"William Herschel III, heir to the Herschel Paper Company", is...
the original script (hence paper) of the eighth Quentin Tarantino movie, "Hateful Eight". The script was leaked online prior to the start of the filming, causing the movie to be initially "killed" by Tarantino.
"His body was found in fishing gear" —
fishing gear is a "net", reference to the Internet, where the original script was leaked.
Another way to look at it, a murder of William Herschel III is...
A reference to "Kill Bill". There were already 2 parts of the movie and the 3rd one in the works, hence "III".
And the Paper Company is a...
reference to "Pulp Fiction"
"lake" and "Labrador" —
reference to "Reservoir Dogs"
"Eight suspects, each had a grudge" —
reference to "Hateful Eight"
"chain was not found" —
reference to "Django Unchained"
"son of Mr. Herschel and an unknown maid" —
i.e. a bastard, a nod to "Inglourious Basterds"
"CCTV footage proved that his death..."
a reference to "Death Proof"
"Jack, his personal flight attendant" —
a reference to "Jackie Brown", whose heroine was a flight attendant.
"his death took place around 3pm the previous day" —
basically he was lying dead "From Dusk Till Dawn".
The name of the puzzle, "Another Sensational Crime Fiction" —
reference to "Pulp Fiction".
And the name "William Herschel" is...
also a name of a famous astronomer, which by a long shot could be a reference to the upcoming "Star Trek" by Tarantino.
Did I forget something?
You are the president of a secret society of mathematicians with $n$ members, including yourself. No one in the society knows what $n$ is. The dictator of the world, in an effort to erase mathematics from human history, has finally managed to capture all $n$ members of the society, and has placed them in prison.
The prison has a funny design. There are $n$ identical soundproof, windowless prison cells arranged in a circle, each containing a mathematician. The cells all have a light switch the prisoners can use, but the wiring system is screwed up, so the switch controls the light bulb in the clockwise neighboring cell. Furthermore, the switch is only capable of delivering a single flash of light at noon each day. Specifically, if this switch is set to "on" at noon, it will flash the next cell, and do nothing otherwise.
The prison warden worries that the prisoners might try to communicate using the lights, so every night at midnight, he fills all of the cells with knockout gas, cleans the cells so they can't communicate by leaving messages, sets all of the light switches to "off", and rearranges the prisoners in whatever fashion he pleases (still one prisoner per cell).
One day, the warden visits your cell. He confesses that he loves mathematics, and decides to offer your society a game to win their freedom. If any of the prisoners are able to figure out what $n$ is, they may shout out loud "There are $n$ prisoners!" (the cells are monitored with security cameras/mics). If they are correct, all prisoners will go free, and if not, they will all be executed.
The warden allows you to devise a plan for everyone else to follow. He will make $n-1$ copies of this plan, and allow every other prisoner to read it. The warden of course will also read your plan, and will perform the cell rearrangements in such a way to make it fail if he can.
What plan allows the prisoners to guarantee their freedom?
Extra for Experts: Can you find a plan which doesn't require the prisoners to make random decisions?
Notes: There is no lateral thinking required to solve this. Assume the prisoners have perfect, infinite memory, including knowledge of how many days have elapsed.
It is significant that you are one of the prisoners, and that you devise the plan. It means that while all $n-1$ other prisoners must follow the same strategy, you may follow a different strategy.
I will give priority to accepting answers which fulfill the "Extra for Experts" condition. This means there must exist a function $B(n)$ such your plan is guaranteed to stop after $B(n)$ days when there are $n$ prisoners.
This is the hardest puzzle I know, but there is a solution.
Answer
This was a monster of a puzzle. But I think I got it!
The following is a solution that does not use randomness, finishes in bounded time for any given circle size, and includes a (mostly- maybe a detail or two left to the reader =)) full proof of validity. Tell me if it makes sense, or if further clarification is needed.
We first define the subroutine $fill(S,k)$. This takes in a set of people, $S,$ such that everybody knows whether they are in $S$ or not. It also takes in a paramater $k$ related to how long to run for. It works as follows
1) On round 1, everybody in set $S$ turns on their light.
2) On every round from $2$ to $k$, everyone in $S$ and everyone who has seen a light on during $fill(S,k)$ turns on their light.
3) From round $k+1$ to $2^k+k+1$, everybody who saw a light on at the end of stage 2), and in every round since then, turns on their light.
Denote $S'$ to be the set of people who saw lights on round $k$. We first claim that, if S is only the leader, then regardless of $k$, at the end of this subroutine, everyone in the circle either sees a light, or does not see a light.
To see this, note that $S'$ either contains everyone or does not contain everyone. If it contains everyone, it is clear that stage 3) of the subroutine simply involves everyone turning on their light every round, and thus everyone sees a light at the end. If it does not contain everyone, then we note that the set of lights seen in each round of stage 2) at most doubles in size. Thus, at the end of stage 2), at most $2^k$ lights are on. But it's also easy to see that, if $S'$ is not the whole circle, then the set of lights on in stage 3) decreases by at least one each round. Thus, by the end, no lights will be on.
Now, suppose $ U > n$. Then, we claim that $ fill(S, U)$ will result in every light being on if and only if $S$ is not the empty set, and every light being off otherwise. This is easy to see because, if $U > n$, then $ S'$ will have to be the entire circle, as long as $S$ was not the empty set. Otherwise, if $S$ was the empty set, then clearly nobody will ever see a light.
Okay, now we're ready to describe our strategy:
Step 1. Deduce an upper bound $U$ on $n$, as done in many other solutions. We run $fill(L, k)$, where $L$ is the leader, for increasing $k$ starting at $1$. At some point, the result will be positive, with everyone seeing a light. At this point, we can be assured that the circle contains at most $2^k$ people where $k$ was the final value used to $fill$.
Step 2. Initialize by considering the leader to be in set $S_0$, and everyone else to be in set $S_1$. Let $k = 2$ represent the number of sets at any point.
Note that currently, everybody knows which indexed set they are in, and which sets exist. We will maintain this invariant so that everyone can coordinate their strategies.
Step 3. Set $done = false$. While not $done:$
3 A) Set $done = true$. For each subset $ I \subset \{1, 2, \cdots, k\}$, other than the empty set and the entire set:
3 A i) Everybody in the set $ \bigcup_{i \in I} S_i$ turns on their light once. Let $T$ be the set of people who saw lights on this day.
3 A ii) For every set $S_j$:
3 A ii a) Run $fill(S_j \cap T, U)$.
3 A ii b) Run $fill(S_j / T, U)$.
3 A ii c) If both $fills$ return positive results, then increment $k$ by 1, let $ S_j = S_j \cap T$, $S_k = S_j / T$, and set $ done = false$.
Step 4. We will prove that, when we reach this point, everyone in the circle can deduce the circle's size.
First, we prove that we eventually reach Step 4. Note that $done$ is only set to $false$ in 3) if a new set $ S_k$ is created. Furthermore, every new set created is not empty, and no nonempty set becomes empty during the algorithm, because $ S_j$ is only split if both $ S_j \cap T$ and $ S_j / T$ are nonempty. Thus, if there are $n$ people in the circle, we can create at most $n-2$ new sets before there are $n$ total sets, one person to a set, and no further sets can be split. At this point, set $3$ must terminate with $done = true$.
Now, suppose we reach step 4. For every set $S_i$, let $x_i = |S_i|$. For any partition $(A,B)$ of the $x_i$ into two nonempty sets, there was a corresponding round in the final iteration of Step 3 in which every person corresponding to $A$ turned on their light. This then resulted, for each $S_i$, in everyone in $S_i$ seeing a light on, or everyone seeing a light off. Furthermore, at least one person in an $S_i$ represented by $B$ saw a light, or else $A$ would have to be a closed loop. The key here is that we know that the number of people in $A$ who didn't see a light must be equal to the number of people in $B$ who did. This gives us an equation in the variables $x_i$, of the form
\begin{equation} \sum_{i \in X} x_i = \sum_{i \in Y} x_i, \end{equation},
where $ X \subset A$, $ Y \subset B$, and neither $X$ nor $Y$ is empty. We say that such an equation "satisfies" the partition $(A,B)$. More generally, we say that $(A,B)$ is satisfied by any equation of the form
\begin{equation} \sum_{i \in X} r_i x_i = \sum_{i \in Y} s_i x_i, \end{equation}
where all coefficients $r_i,s_i$ are positive, and $X$ and $Y$ are nonempty subsets of $ A$ and $B$. If a set of equations satisfies every partition of a set of variables, we say that that set of variables is fully satisfied. At step 4, we thus have a set of fully satisfied variables. Thus, to prove that we can deduce the size of the circle, we show that, given any set of variables such that $x_i = 1$, along with a satisfiable and fully satisfying set of equations on those variables, we can deduce the value of every variable
We use induction. For $ k = 2$ variables, we have $ x_1 = 1$, and the partition $ (x_1, x_2)$ yields $ k_1x_1 = k_2x_2$, where $ k_1, k_2 \neq 0$, so clearly both variables are determined.
Now, consider a set of $k$ variables $ x_1, x_2, \cdots, x_k$, and assume our hypothesis holds for all sets of $k-1$ variables. We will eleminate $x_k$ from the equations to construct a set of equations which fully satisfies $ x_1, \cdots, x_{k-1}$. For each partition $(A,B)$ of the remaining $k-1$ variables, there are two partitions of the whole set of variables: $ (A \cup c, B)$, and $ (A, B \cup c)$. Both of these partitions have some satisfying equation. If either of the satisfying equations don't include $c$, then we have an equation which satisfies $(A,B)$ on the first $k-1$ variables. Otherwise, if both satsifying equations include $c$ with some weight, then we can linearly combine them to eliminate $c$ and get a new equation, which some basic algebra tells us will also satisfy $(A,B)$ in the first $k-1$ variables. Thus, by doing this for every partition, we can construct a set of fully satisfying equations for $ x_1, \cdots, x_{k-1}$, which by our inductive hypothesis determines all of their values. Finally, we can consider the partition which separates $x_k$ form the other $k-1$ variables. This partition is only satisfied by a linear equation writing $x_k$ in terms of the other variables; as these are all determined, we can also determine $x_k$. Thus all variables are determined, and we are done.
Finally, since every person in the circle knows a fully satisfying set of equations on all of the subsets $S_i$, and knows that $ |S_1| = 1$, everyone in the circle can deduce the size of each subset, and of the entire circle.
Disclaimer: I'll be using the gender-neutral pronoun "xe" for the singular third person.
After escaping from d'alar'cop's cellar, Doorknob♦ was understandably upset. After almost two months of planning, xe gathered some support and the revenge scheme was enacted...
d'alar'cop awoke in a darkened room on a hard surface. Xe was sore from lying there for some unknown time. As soon as xe shifted xyr weight to stand up, lights came on all around the room. Blinking in the sudden brightness, xe took in xyr surroundings. Xe was on a raised daïs in the middle of a round room. A soft glow emanated from underneath and illuminated most of the room. There were 31 other puzzlers in the room, each in their own alcove. The recesses were painted a variety of colors and even the daïs was painted blue. Each puzzler was either staring at d'alar'cop or facing directly away. In front of each puzzler was a set of two lights that was facing the daïs, facing the alcove, or turned off. (AJ Henderson was staring directly into two powerful lights blasting into xyr face. Ow.) Just as d'alar'cop had finished taking this all in, all 31 puzzlers began to chant in unison:
You've been naughty, kidnapper
Taking prisoners, doing wrong
Now it's time for you to learn
If you're mean, you will get burnedYou are trapped here in this room
From that daïs you may not move
You must solve our riddle soon
Or from our pack you will be hewnThe door at south is your escape
But it shan't open lest you say
The secret message known to us
Speaking quickly is a plusPoison daggers now descend
From ceiling to your flesh to pierce
They will slow your very breath
Bringing soon a painful deathIf you belong then you will see
An answer not quite seen by thee
Although mad we are not cruel
We shall grant a single toolYou have a laptop, internet
Seek the answer on the web
Careful, though, where you tread
There are tricks and traps ahead
d'alar'cop's eyes adjusted to the light during the chant and xe could see the laptop. Thankfully, it's already booted up and xyr favorite browser was open to a blank tab. With a quick glance upward at the glistening blades, xe set to work as the puzzlers started up their chant again.
Diagram of the room: (The diagram was made in Excel. Any minor errors in position or orientation are not part of the puzzle.)
Puzzler Positions:
1: Rainbolt
2: Andrea
3: RBerteig
4: zar
5: mau
6: SztupY
7: Hubble07
8: senshin
9: The Wobbuffet
10: Adele C
11: AJ Henderson
12: user88377
13: Doorknob♦
14: gsingh2011
15: bluefeet♦
16: Robert Cartaino♦
17: Raystafarian
18: vivekpoddar
19: glenneroo
20: James Jenkins
21: Young Guilo
22: CaffeinePwrdAl
23: DaG
24: crownjewel82
25: Piero Alberto
26: Michal
27: Danubian Sailor
28: Crispy
29: Gilles
30: strikers
31: igelkott
32: d'alar'cop
Trap Level 1 found by Sebu (+1)
Trap Level 2 not yet found
Trap Level 3 not yet found
Password found by Shawn Holzworth (Tick after another day or two to keep it active so maybe people will try to find the remaining traps)
Answer
Well what do you know, it seems rand al'thor was on to something:
When you take the user id numbers for users 1-31 as ascii, you get a message:
JET Fuel_CAN't-m3l+~S┼Éé| Bèams
Here's what I'm saying
Moving 2 sticks, what is the largest number that you can create?
Answer
How about...
Take the top and bottom sticks of the first zero, and place them upright on the bottom left, making $11^{11000}$?
If that's unrealistic...
... because the base is usually in bigger font, then $11000^{11}$ could work as well.
[This puzzle was previously posted for humans only, but it may have been a little too tough to do manually, so I'm opening it up to computers to solve the remaining unsolved words.]
I am inspired by the story of the boy who came upon an assortment of holiday cookies beautifully arranged on a platter. The boy quietly helped himself to a cookie and artistically rearranged the remaining cookies to make it appear as if none were missing. Then, he helped himself to another and rearranged again. And another. And another. Until all the cookies were gone. (And that's when his mother noticed...)
It's very simple:
For example:
Starting word is ALTERCATIONS
-------------------------------------------
Solution:
ALTERCATIONS
(remove "T" and rearrange)
LACERATIONS
(remove "C" and rearrange)
SENATORIAL
(remove "A" and rearrange)
RELATIONS
(remove "O" and rearrange)
ENTRAILS
(remove "N" and rearrange)
REALIST
(remove "L" and rearrange)
SATIRE
(remove "T" and rearrange)
RAISE
(remove "I" and rearrange)
EARS
(remove "R" and rearrange)
SEA
(remove "E" and rearrange)
AS
(remove "S" and rearrange)
A
(remove "A" and you're done)
You can do as much or as little rearrangement as you like. E.g., you are permitted to remove the "R" from BRAKE and leave BAKE, or even just snap the "-S" off the end of a word and leave all the remaining letters as they are.
As always, I construct my puzzles in such a way that they can be solved using only well-known words, so if you find yourself going down a path of increasingly obscure words, you may be overthinking it.
Multiple solution paths may exist, so don't worry if your path varies slightly from your neighbor's path, especially at the tail ends.
Below are 12 starting words. See if you can reduce each one to nothing through attrition.
1. DISSEMINATORS (solved)
2. ENCRUSTATIONS
3. DECENTRALIZED (solved)
4. IMPERSONATING (solved)
5. SHAREHOLDINGS
6. INVESTIGATORS
7. INDOCTRINATED (solved)
8. REGISTRATIONS
9. DETERIORATING (solved)
10. INDETERMINATE
11. REINSTALLATION
12. RESUSCITATIONS
@hexomino — That was fast work!
Here is a solution to #8:
REGISTRATIONS
REGISTRATION
INTEGRATORS
RESTARTING
RESTATING
ANGRIEST
SEATING
GIANTS
STING
TINS
SIN
IS
I
An alternative solution to #2:
ENCRUSTATIONS
ENCRUSTATION
TRANSECTION
CONTAINERS
REACTIONS
ANCESTOR
COASTER
TRACES
RATES
RATE
EAT
AT
A
An alternative solution to #5:
SHAREHOLDINGS
SHAREHOLDING
HIGHLANDERS
HIGHLANDER
HERALDING
ADHERING
READING
GARDEN
RANGE
NEAR
RAN
AN
A
An alternative solution to #6:
INVESTIGATORS
INVESTIGATOR
INVIGORATES
ORIGINATES
ORIGINATE
RIGATONI
RIOTING
ORIGIN
GROIN
IRON
ION
IN
I
An alternative solution to #10:
INDETERMINATE
INTERMEDIATE
DETERMINATE
TERMINATED
DETRIMENT
REMITTED
MERITED
TIERED
TRIED
DIET
TIE
IT
I
An alternative solution to #12:
RESUSCITATIONS
RESUSCITATION
CREATIONISTS
CREATIONIST
RECITATION
INTRICATE
INTERACT
CERTAIN
RETAIN
INERT
TIRE
TIE
IT
I
Answer
Certainly easier with computers
2.
ENCRUSTATIONS
ENCRUSTATION
TRUNCATIONS
CONSTRAINT
CONSTRAIN
CONTAINS
CONTAIN
CATION
TONIC
COIN
ION
IN
I
5.
SHAREHOLDINGS
SHAREHOLDING
HIGHLANDERS
HIGHLANDER
HERALDING
HARDLINE
HANDLER
HANDLE
ELAND
LAND
LAD
AD
A
6.
INVESTIGATORS
INVESTIGATOR
INVIGORATES
INVIGORATE
ORIGINATE
RIGATONI
ORATING
RATING
GRAIN
RING
GIN
IN
I
10.
INDETERMINATE
INTERMEDIATE
DETERMINATE
TERMINATED
TERMINATE
MARTINET
MINARET
MARINE
RAMIE
MARE
MAR
AM
A
11.
REINSTALLATION
ALLITERATIONS
ALLITERATION
RETALIATION
ALTERATION
RATIONALE
RELATION
ELATION
ENTAIL
INLET
TINE
TIN
IN
I
12.
RESUSCITATIONS
RESUSCITATION
CREATIONISTS
CREATIONIST
RECITATION
ITERATION
INTRICATE
INTERACT
CATTIER
ATTIRE
TREAT
TEAR
TEA
AT
A
Still not sure on 8, possibly involves some more obscure words.
You and your friend Marcus celebrated his birthday at your house, but when you wake up, you find an audio recorder with a white label on it, saying: “sorry buddy”. You play the recorder and practise your scales and pieces of music. Then you put down the instrument and press play on the audio recorder. You make out the words:
“Easily the hardest part of recording this is that I... uhh just y’know, am... erm sorry for what I am doing to ya. Cuts me deep to say this, but I held... sigh this secret close to me. Hard to understand, it is, but I am at... odds with the fact that you trusted me. I’m not the person you thought I was. I might sound a bit jittery but that’s just me, noooooothing else, no deeper meaning or anything. Of course you’re smart, so I know you’ll get this. Involving this, the... ummm question if we should stay united... uhh and, the United Kingdom.. errr I will head... erm away. Next to my friendship with you there is a quarter... umm of me that you can’t see. Centrally, I’m sorry, and goodbye and good luck.”
You sit back, and try to process what happened. This isn’t like Marcus, and the way he was saying it was extremely odd. After an hour of studying this, you find out where he is, and the name of what could’ve taken him.
Answer
The hidden message is:
I am held at the united Kingdom head quarter
You get it by taking every word before he pauses speaking.
He probably got caught by:
Secret Intelligence Service
The following are clues to 18 six-letter words that must be filled into the following three linked word squares.
The clues are in no particular order, but just for fun, I tried to arrange them to tell a little story (which has no relevance to the puzzle's solution).
Overlapping portions of each square will obviously share the same letters.
Answer
The eighteen words are (in the same order as in the question):
PRINCE(the artist later known as) SymbolDEALERBuyer and sellerCREDITAcknowledgeIODINE(the chemical element whose symbol is) IUNRESTDissentSTOLENPlagiarizedTEASEDMockedENDURELastSINFULEvilAGENDAPlanRECIPEInstructionsESTEEMVenerateGROTTOChamberBUGLESMilitary instrumentsLETHALDeathlyENGINEWord with fire or searchERRORSBall-drops (in Baseball, I believe)ESTATELands
This gives the following grid:
B U G L E SU N R E S TG R O T T OL E T H A LE S T A T E A S E DS T O L E N G I N EA G E N D AS I N F U LE N D U R E C I P ED E A L E R R O R SC R E D I TI O D I N EP R I N C EE S T E E M
Have fun with my self-made Sudoku puzzle.
Solve this Sudoku puzzle using the standard Sudoku rules:
Fill any row, column and bold-framed area with the numbers 1 to 9.
Hint: It is not as difficult as it might look!
Answer
Right at the beginning, we can tell that certain sets of cells must all contain the same number. Starting from the extreme bottom right cell, we can find a sequence of equal numbers (say this number is A) by considering each bold-framed area in turn starting from the bottom left and working to the right. This gives us a sequence of A's culminating in a 6, so A=6. Similarly, by starting from the B marked on the top row, we get a sequence of B's culminating in a 3, so B=3. This enables us to fill in all the 3's and 6's:
We can then find another sequence of equal numbers (say C) by starting from the extreme bottom left cell and considering each bold-framed area working up and to the right from there. This gives us a sequence of C's culminating in a 9, so C=9. We can also get a couple more 4's by considering where 4 can be in the second column and in the third bold-framed area from the left. This enables us to fill in all the 9's and 4's:
We can then use the same argument again to fill up the next diagonal and the next and the next:
Now the sixth row is almost full, so its leftmost cell must be 7; then there's only one possibility for where 7 can be in the fifth row, then in the fourth row, and so on. Once this diagonal of 7's is filled in, we can apply the same reasoning to get a diagonal of 5's, then a diagonal of 1's, then a diagonal of 2's:
You are going to a food party on a small deserted island. The host asks that you bring food but first check with them about if it can be brought or not.
Some other people have already gotten accepted for:
To determine their criteria, you asked them a variety of questions:
What is their criteria for what to allow?
Hints:
- 12 brain - Probably Not allowed
- 12 pieces of pizza - Allowed
- 12 pieces of metal - Not allowed
Hint #2:
Re-read the first sentence
Answer
I believe the criteria is:
Items that float. The amount does not matter.
Reasoning:
Lead, potatoes, ripe blueberries, and metal all sink in water. I couldn't find a consensus on whether brains float in water hence the "probably not allowed". Cooked Ravioli, apples, bread, zucchini, grass, Bananas, and pizza should all float. Regarding the second hint, you are going to a deserted island, so you probably want things that can easily float.
Here's my first attempt at a Riley! Good luck!
The thought of my prefix brings many a smile;
Sit down, relax, or sleep for a while.
My infix is, mathematically;
One point six times five point eight, then less three.
My suffix is a speech, mad and long;
Which definitely is not a song.
Answer
Restaurant
Prefix
Rest is a synonym for relax, etc. and people look forward to it wistfully, with smiles on their faces.
Infix
Tau (twice pi) is approximately $6.28$, which is $1.6 * 5.8 - 3$. Clued in by @Omega Krypton.
Suffix
Rant, which is an unhappy speech
Sometimes I am born in silence,
Other times, no.
I am unseen,
But I make my presence known.
In time, I fade without a trace.
I harm no one,
but I am unpopular with all.
What am I?
Answer
You are a
fart
Sometimes I am born in silence,
There can be silent farts.
Other times, no.
There can be noisy farts.
I am unseen,
No one can see a fart.
But I make my presence known.
Farts can smell.
In time, I fade without a trace.
The smell fades away over time.
I harm no one,
Farts are not harmful.
but I am unpopular with all.
No one likes a fart.
19 BBY, the Galactic Republic spots General Grievous in Utapau, the Separatists' Council Base; the Jedi Obi-Wan Kenobi is sent there to deal with him.
After a long search, Obi-Wan comes face to face with the Supreme Commander:
Obi-Wan: Surrender, it's over!
Grievous: Mwhahaha, fool! How can you beat me?!?!
Obi-Wan: With my lightsaber, of course!
Grievous: Mwhahaha, have you ever seen my set of four lightsabers?
Obi-Wan: Do you feel advantaged? May the math be with you! There's no difference between one and four!
Grievous: Can you prove it?
Obi-Wan:
$x=4$
$x(x-1)=4(x-1)$
$x^2-x=4x-4$
$x^2-4x=x-4$
$x(x-4)=x-4$
$x=1$
Grievous: You're trying to use the Force on me, but it won't work!
Is Obi-Wan's math as strong as his Force? Explain it!
If you like problems like this, check A dollar, a penny, there's no difference
Answer
When you multiplied both sides by $(x-1)$, you introduced the new extraneous solution $x=1$ to the equation. Later on when you divided by $(x-4)$, you forgot to case check that $(x-4)$ might equal $0$. If we do so we get $x = 1$ or $4$, as expected.
How do you know if you're addicted to riddles?
1.
I use my head, I move the earth,
So ancient none recall My birth,
One of me spells death and yet,
You'd call me as I am I'll bet,
You know me well, you've got my card,
Who am I? it isn't hard.
2.
A game of fortune,
Dice will roll,
Name and goal the same,
I have two kinds,
And just one chance,
Can you speak my name?
3.
A head of thick, wet, braided hair,
I carry more than can be seen,
And in my wake a brightening glare,
A slick, treacherous sheen,
My story short, but often told,
Me in your hands, what would you hold?
4.
My string tightens as I loose weight,
To remain full, my desired fate,
I can be woven, plastic, skin,
With nothing, coin or fish within,
I travel pressed to many hips,
Surely my name is on your lips?
5.
A random chance will see you fold,
I am not one for the old,
A vibrant grid to break your back,
Colours Prime I do not lack,
More difficult the more that try,
Can you tell me, what am I?
6.
I'm something with an antique feel,
A binding, thing to hold and seal,
Breaking me takes no strength,
Just weakness and time of unknown length,
Made by many, kept by few,
Know my name? Speak if you do.
7.
By clever folk I can be read,
Less clever folk will ask instead,
I stole the likeness of my home,
Else here without direction roam,
Providing clues a familiar game,
So can you tell me, what's my name?
8.
I'm not a doll on strings,
But my steps are not my own,
Just one hole in such things,
Unless we're torn or blown,
The wind blows now I'll help you see,
Now if you can, you should name me.
Answer
FULL ANSWER: CREDIT TO EVERYONE
1. Spade (Credit to Reibello)
2. Yahtzee (Credit to Jason Horner)
3. Mop (Credit to Reibello)
4. A Purse (Credit to Mohirl)
5. Twister
6. An Oath, (Credit to Reibello)
7. A Map
8. Sock (Credit to hagfy)
The answer of each line spells,
SYMPTOMS
I'm pretty sure #5 is
Twister.
A random chance will see you fold,
Refers to the twisting body shapes you make.
I am not one for the old,
It most certainly is not.
A vibrant grid to break your back
The game is set up in a grid.
Colours Prime I do not lack,
The dots are made up of the four basic colors, red, green, yellow, and blue.
More difficult the more that try,
Twister gets much harder with more people.
I think #7 is
A map
By clever folk I can be read,
Because of technology, people don't read maps anymore (which is really a shame, because they can be really useful. Especially when camping.)
Less clever folk will ask instead,
People will ask for directions.
I stole the likeness of my home,
Maps resemble the place they are showing.
Else here without direction roam,
Without a map, you are lost.
Providing clues a familiar game,
Maps are used for scavenger hunts.
With the beginning, I can make you totally emotionless
Add a few letters to it and you can even find a special kind of school
In the middle, a special kind of energy awaits you
Something that flows in every living thing, they say
The ending is quite a short one
But it is actually what you do to this riddle
What am I?
Hint (Usefulness Level : 0)
This word is not commonly used in daily conversations
Hint (Usefulness Level : 1)
This word consists of 13 letters
Hint (Usefulness Level : 1 again)
You need to use mathematics when doing (The word)
Hint (Usefulness Level : 2)
The special school in the second line was created a very long time ago (BC)
Hint (Usefulness Level : 2)
The word is actually a part of a subject at school
Hint (Usefulness Level : 2)
What I meant by few letters in the second line was 3 letters exactly
Note :
Will add more hints if no one gets it after a while
Answer
I think I may be getting somewhere here :)
Stoichiometry
Prefix:
Stoic (emotionless) and stoicism (the school)
Infix (thanks to @Duck and @PerpetualJ):
Chi
Suffix
Try (what you do to this riddle)
All the hints add up as well
13 letters, involves math, is part of a subject (chemistry), three letters to turn stoic in stoicism
A boundary of red and blue squares is given. Can you fill in the interior such that each 5-square pattern consisting of an interior cell plus its four nearest neighbours always contains an even number of red squares?
Example:
Now try this one:
Answer
My answer :
And the number of red cells for each pattern :
every interior pattern has an even number of red cells.
How to find it :
I hoped there were a lot of symmetry so I had only a few cells to choose ! All the yellow cells are deduced from the blue cells by symmetry.
Then a few random tries gave me the answer
This is wholly inspired by Get 6 pawns in file A or H
Is it possible for either white or black to get 6 pawns of the same colour in file A or in file H (you must choose one) without promoting?#
Possible (not necessarily this) end position for positive case (same as in Marius' question):
# I do know the answer
Answer
I believe that it is
Not Possible
Reasoning
To get, say, six white pawns in A2-A7, the pawns will need to capture a total of 15 black pieces, essentially one for every square that's on a diagonal between the pawn's initial condition and the pawn's final position.
In the diagram below, each highlighted square will need to have a black piece on it at some point, and that piece will be captured by a white pawn.
However, for this to happen...
Every black piece, (except the king) needs to end up in one of the red squares. This means that the pawns starting in F7-H7 will need to capture pieces to move to the left. But how many pieces must they capture?
This is admittedly the weakest part of my proof, but so far it seems that black must capture at least 11 white pieces to get these pawns into the highlighted region.
For example, the pawn in H7 can capture three pieces to get to E4, then move normally to E3. The pawns in G7 and F7 do the same to get to C3 and D3, respectively. So far, we're already at 9 white pieces captured. But now the pawn starting in E7 has to capture a piece, because the only highlighted spot in column E is already taken, and then the pawn in D7 also has to capture a piece, because the two slots in column D are already taken. This means that 11 white pieces must be captured. All my attempts at reshuffling this have, so far, failed to produce anything better.
But what does that mean?
While we're allowed to capture as many black pieces as we need (except the king), we don't have nearly as many white pieces that we can capture. Aside from the 6 pawns we need to get to column A, the pawn starting in H2 won't help any of the black pawns move to where they need to be.
That means that there are only 9 white pieces that we can afford to have black capture, and one of those won't help us achieve our end goal, so we really only have 8 pieces that will help us get black pawns into the locations they need to be in.
This picture shows white pieces we can afford to have captured in green, and ones that we can't have captured in red.
So, in conclusion,
The desired layout is not possible because there are not enough unneeded white pieces for the black pawns to capture to get them into the proper position for the white pieces to capture so they all end in the same column. The general argument presented here is symmetric, and should apply to getting either six black or six white pawns to either column A or column H.
What was it?
Answer
Answer:
A coffin/tombstone
Obviously the man who made it doesn't want it! Who would want a [answer]?
The man that bought it didn't use it, if he needed it, he wouldn't have been in a state in which he was capable of buying anything.
The man who received it obviously didn't wish for it, refer back to the man who made it.
The man who used it couldn't have seen it, as he does not have the ability to do so!
Each cryptic clue yields the keyword to the next, terminating in a message to you, gentle puzzler. With the exception of the first, which requires a little lateral thinking, all the clues are standard cryptic fare. The only prerequisite is knowledge of simple ciphers.
A nut for Caesar’s tiger
Pb rpctg ml mbmln' snxgst nl tfmpr agld
Oj tnavf stjbju fh frhuqflc tn sfv aflbjux!
Nabjmwaq?
Adshn lond rmg bcsmcnb hnf
Lmtdn gs jmlb oqg ndmur omttgqy
Jkn rdanjokuj rny, qrsn rejck jrdaqa okjt, weh hpas wsyb nkirdaqas
Ediaj dss lr diturd fsrfcet
E nbpc mpkbp mc n pmkgqgld bppmp
S rpsa jekt ck spguluq acii epcqb, pesp pskaeg
Jfb dfqq em pt. Rml’tw sq tdtfkd.
Hint:
Mind the gap
Answer
Solved by M Oehm.
Solved by Levieux.
As found by Levieux, the decrypted version is
Place vital fluid is during active influx
which is a cryptic clue for
VEIN (the place where blood [vital fluid] is, and found in "active influx").
Decrypt "Nabjmwaq?" using the key found in Clue 3 to get
Deflower?
Thanks to Will, the solution to this is
DAM - a dam is a de-flow-er because it prevents flow.
Decrypt "Adshn lond rmg bcsmcnb hnf" using the key found in Clue 4 to get
Basin mona rch descend ing or Basin monarch descending.
The solution to this is
SINKING - basin+monarch becomes sink+king, and descending = sinking.
Decrypt "Lmtdn gs jmlb oqg ndmur omttgqy" using the key found in Clue 5 to get
Notic ea mong pre cious pottery or Notice among precious pottery.
The solution for this is
SPOT (synonym of notice, found in "precious pottery").
Decrypt "Jkn rdanjokuj rny, qrsn rejck jrdaqa okjt, weh hpas wsyb nkirdaqas" using the key found in Clue 6 to get:
Nor therncoun try, star tingo nthese cond, wil lbea wayf romthesea or Northern country, starting on the second, will be away from the sea.
The solution is
INLAND (Finland is a northern country; start from the second letter to get a word meaning "away from the sea").
Decrypt "Ediaj dss lr diturd fsrfcet" using the key found in Clue 7 to get
Headl ess cr eature isright or Headless creature is right.
The solution is
EAST (creature = beast, remove the first letter to get east, which is to the right on a map).
Decrypt "E nbpc mpkbp mc n pmkgqgld bppmp" using the key found in Clue 8 to get
A perf ormer of p romising error or A performer of promising error.
The solution for this is
SINGER (a performer, and found in "promising error").
Decrypt "S rpsa jekt ck spguluq acii epcqb, pesp pskaeg" using the key found in Clue 9 to get
A frag ment in arduous gibb erish, rear ranged or A fragment in arduous gibberish, rearranged.
The solution is
SHARD (a fragment, and found in the rearranged "gibberish arduous).
Decrypting "Jfb dfqq em pt. Rml’tw sq tdtfkd." with the key found in Clue 10 gives:
Lif eiss ho rt. Don’tw as tetime. or Life is short. Don't waste time.
Ironic, given how much time I've just wasted solving this puzzle! :-P
