geometry - Dissecting a square

You are asked to dissect an $N \times N$ square into polyomino pieces such that each piece shares portion of its boundary with exactly $D$ other pieces, and no piece has area exceeding $N$. This can be achieved for $D \le 5$.

For $D=2, 3, 4$ the smallest such squares are of size $2 \times 2$, $3 \times 3$, $4 \times 4$, respectively:

Find the smallest square for $D=5$.

Credit: inspired by this puzzle.

Answer

I think I've found a solution for an 8x8 square. I do not know if it is the minimum solution or how to prove that:

It was definitely fun to try and find this! Took me a while. Excellent puzzle.

---

Some comments on how I got to the solution (Rather a chronology than a full deduction):

- It has been proven in the other puzzle that at least 12 tiles would be needed.
- It was clear, that each piece has to have at least 2 squares.
- I was intuitively convinced that the solution would have a 4-fold rotational symmetry. (No proof for that whatsoever. More a 'feeling'.) So I set out with trying patterns in this symmetry. If I get 1/4th of the pieces in place, the other 3/4th would be correct automatically.

- I figured the rim would be the difficult part, as long-stretched tiles are needed, and their length is limited by the square-size.
- So I first tried to create a 7x7 square with fitting tiles to the border and leaving room for 5 connections.

- This did not work out (because of the symmetry centre being 'left out'). So I then retried the same with a 8x8 grid, ending up with:

- Trying to extend the colours towards the centre (always using 4-fold rotational symmetry on the extensions) I soon ran into obstacles, which forced me to 'shift' the outer rim to:

- And then it was just a matter of extending inwards and realizing that the number of tiles is not yet enough. Adding 4 colours did the trick.

mathematics - Pythagorean coins

To make payments, the Pythagoreans use coins in no more than three denominations. The three denominations are in whole Oboloi amounts, and the sum of the squares of the two smaller denominations equals the square of the largest denomination. The amount of $1,000$ Oboloi, and any larger whole Oboloi sum, can be obtained using the three distinct types of coins. However, the exact amount of $999$ Oboloi can not be obtained (and hence requires change).

What coin denominations do the Pythagoreans use?

word - Rolling Dice puzzle

The answer is a six-letter word.

---

---

Here's an attempt at a text version, but you're really better off looking at the picture.

Rolling Dice

To get a full tour, you must be a man of many faces and blend into all the right spots.

     |
     v    
O6TAL2
RA2U1Y
DI■ROT
LLHTP3
6FE3PR
MISMO6
^

|

Answer

I'll use a different spoiler for each level of solving the puzzle, such that you can read only as far as you want to be spoiled, if you want to only look at the first few steps for hints. Note that there is some Mathematica code further down, which is not inside a spoiler tag.

The "full tour" indicates that we need to traverse every cell of the grid. It's likely that we need to traverse each cell exactly once, hence asking for a Hamiltonian path. The arrows indicate the start and end of the path.
The "man of many faces" probably just indicates that faces of a (six-sided) die play a role.
"Blend into all the right spots" means that we somehow need to match up the faces with the grid cells.

The puzzle asks us to place a six-sided die on the bottom left cell, and roll it around the grid, such that we visit each cell exactly once. Furthermore, whenever the die is on top of a number, the top face has to show that same number.

To simplify the search for the path, I started at the end, because there are only 4 possible die orientations with a 2 on top.

These constraints result in a unique tour, which I found using the following Mathematica code:

board = Characters /@ StringSplit@"XXXXXXXX
    XO6TALXX

    XRA2U1YX
    XDIXROTX
    XLLHTP3X
    X6FE3PRX
    XMISMO6X
    XXXXXXXX";

moves = {
   (*up*)
   {c = #1 + {-1, 0}, Extract[board, c], <|u -> #3[s], n -> #3[u], w -> #3[w], 

      s -> #3[d], e -> #3[e], d -> #3[n]|>} &,
   (*left*)
   {c = #1 + {0, -1}, Extract[board, c], <|u -> #3[e], n -> #3[n], w -> #3[u], 
      s -> #3[s], e -> #3[d], d -> #3[w]|>} &,
   (*down*)
   {c = #1 + {1, 0}, Extract[board, c], <|u -> #3[n], n -> #3[d], w -> #3[w], 
      s -> #3[u], e -> #3[e], d -> #3[s]|>} &,
   (*right*)
   {c = #1 + {0, 1}, Extract[board, c], <|u -> #3[w], n -> #3[n], w -> #3[d], 
      s -> #3[s], e -> #3[u], d -> #3[e]|>} &

   };
f[path_] := Module[{i, next},
  If[Length@path == 35, Print@Reverse@path; 
   Print@TableForm@Reverse@path; Return[]];
  For[i = 1, i <= 4, ++i,
   next = moves[[i]] @@ path[[-1]];
   If[! (next[[2]] == "X"
       || next[[1]] == {7, 2} && Length@path < 34
       || DigitQ@next[[2]] && next[[2]] != next[[3]][u]),
    (board[[##]] = "X") & @@ next[[1]];

    f[Append[path, next]];
    (board[[##]] = next[[2]]) & @@ next[[1]];
    ]
   ];
  ]
f[{{{2, 7}, "2", <|u -> "2", n -> "3", w -> "1", s -> "4", e -> "6", d -> "5"|>}}];
f[{{{2, 7}, "2", <|u -> "2", n -> "6", w -> "3", s -> "1", e -> "4", d -> "5"|>}}];
f[{{{2, 7}, "2", <|u -> "2", n -> "4", w -> "6", s -> "3", e -> "1", d -> "5"|>}}];
f[{{{2, 7}, "2", <|u -> "2", n -> "1", w -> "4", s -> "6", e -> "3", d -> "5"|>}}];

The only solution is the following: start in the bottom left corner, with 4 facing up and 6 facing south. Then this is the path (red numbers indicate number facing up):

If we now pick out the letters in that path where each number is on the top face (see red letters in the image), we get 6 words. In order from 1 to 6 they are SLUMP, DROOP, OR, MILITARY, FELT, HAT.

"to slump" and "to droop" are synonyms for "to slouch", but "a slouch" can also be a military felt hat. Hence the solution is slouch.

geometry - Solutions for generic polyomino puzzles

Inspired by Mosaic with tetris blocks I was wondering if there were any generic algorithms to solve or show there was a solution to these types of problems (i.e. placing polyominos on a 2D board).

Breaking it down to the simplistic form of dominos, there's the classic puzzle of placing them a checkerboard with either two white or black squares missing. This is known to be impossible and left as an exercise for the reader =). But what if we're given a less uniform board. Is there any known algorithms other than brute-force to show there is a solution or to even solve the puzzle? We know there have to be an even number of square and an equal number of white and black squares, but that doesn't mean there is a solution.

Moving up in complexity, there's another classic problem of filling an $n \times n$ board with L-triominos except for one marked square. Recursive solutions exist here, but I've never seen anything for non-uniform board or allowing non-L blocks.

Finally we hit the Mosaic puzzle. What if we're given a set of tetris pieces and an irregular board? Can we easily determine if there's a solution? Is it easier if we aren't constrained by the number of each piece we can use?

Answer

The problem is a type of "Packing Problem". Regarding polyominoes we are interested in tiling specific regions.

It is known that the problem of answering whether a specific region can be filled with tiles is undecidable. And that finding a solution, in general, is NP-Complete. So we can never do much better than trying all possible configurations - i.e. pick something and exhaustively show that it doesn't work, backtrack and try again.

As to the "simpler" problems like L-trominoes, it is essentially the same problem.

So, to answer directly: there is no easy way to determine whether there is a solution. Just as there is no easy solution for the travelling salesman problem, the tiling problem, or the boolean satisfiability problem (SAT) (on a personal note, I immediately think of this problem, and could fairly easily cast the polyomino problem as a SAT problem).

Related:
More Papers
Be Bruijn's Theorem

knowledge - More important than the destination

_{This puzzle is the 26th and final part of Gladys' journey across the globe. If you are new to the series, feel free to start at the beginning: Introducing Gladys.}

_{Spoiler alert: This puzzle gives away the answers to the previous parts in the series.}

---

Dear Puzzling,

As promised, I'm sending you the correct answers to the puzzles I have sent you during my travels. Thank you for following my journey through to the end. I wish you all the best.

One last question: What do you think I got out of my trip?

Love, Gladys.

1. Heinz Memorial Chapel, Pittsburgh
2. Emerald Necklace, Boston
3. Tredegar Iron Works, Richmond
4. International Church of Cannabis, Denver
5. Eldorado Resort Casino, Reno
6. Sian Ka'an, Tulum
7. Estadio Garcilaso, Cusco

8. African Renaissance Monument, Dakar
9. Oran 1 University, Oran
10. Flaminio Obelisk, Rome
11. Lauba, Zagreb
12. Ile aux Cygnes, Paris
13. Rogier Tower, Brussels
14. Ransomes Sports Pavilion, Ipswich
15. Irish Museum of Modern Art, Dublin
16. Fossil Grove, Glasgow
17. Elbphilharmonie, Hamburg

18. Museum Island, Berlin
19. Matsesta Microdistrict, Sochi
20. Oman Oil and Gas Exhibition Centre, Muscat
21. Mehrangarh, Jodhpur
22. Osh Bazaar, Bishkek
23. Wolong Temple, Xi'an
24. Tjong A Fie Mansion, Medan
25. SM Seaside City Cebu, Cebu

Hint:

The 25 entries can be arranged into a 5×5 table.

Answer

I think Gladys got

A lifetime's worth of memories

Reasoning

Take the first letters of each answer: HETIESEAOFLIRRIFEMMOMOWTS
and take an anagram to produce the answer: A LIFETIMES WORTH OF MEMORIES

logical deduction - A Grave Issue Discovered

In the town of Fairfield (There are 35 Fairfield's in America), there is a graveyard. In 1874, the mayor, Arthur Hither, declared that 25 new graves be created, in a square. Five years later, he was buried in the center. One day, a hurricane comes through and destroys the graves. The people are still buried, but nobody knows where everyone is. You put a hint box in the center of the city, and get these hints:

• My grandfather was East of the mayor 1 spot


• The police chief was 2 spots North of Mrs. Francine Hither


• The mayor's family was in a row of 3 plots going East to West


• The mayor's mom, Mrs. Francine Hither, is just East of one of the old professors


• The first note was by Holly Hither


• There are three old professors. One, Prof. Tasker, is next to the Police Chief


• Mr. Henry Carpenter is South of Officer Smith



• Every doctor has a professor on the far West of him


• The Army Captain is somewhere South of Mr. and Mrs. Williams


• The doctors, Dr. Oak, Dr. Birch, and Dr. Cedar, are all North of Oliver Twist's grave


• The Old Millionaire is South of the Evil Villain


• The Evil Villain is West of Ms. Johnson


• Mr. Mackarel is a few graves South of the mayor's mom


• Prof. Rathburne and Dr. Oak are across from each other


• Dr. Birch is next to Mr. Williams, and Mrs. Williams is West of Dr. Oak


• Ms. Sarah Williams is West of her mom


• Mr. Crabapple is buried South of the Police Chief



• Mr. Crabapple is Southwest of the Fire Chief by one space, and is one space West of Sarah


• The Evil Villain is South of Professor Vocab


• The Army Captain has Arthur Z. on his East side, and Mr. Mackarel two spots West


• I remember someone named Nancy Smith

With these hints, you think you can rebuild the gravestones, and get a huge payment from the city. You have no advanced technology available, and the city says you can't dig up the people to find out.

Answer

The layout of the graves is

In text format (row by row):

Row 1

Prof Tasker......./ Police Chief...../ Fire Chief........../ Mr. Williams......./ Dr. Birch

Row 2

Prof Rathburne./ Mr. Crabapple../ Sarah Williams / Mrs Williams....../ Dr. Oak

Row 3

Prof Vocab......./ Francine Hither / Mayor Hither..../ Holly Hithers GF/ Dr. Cedar

Row 4

Evil Villian......../ Ms Johnson....../ Officer Smith...../ Nancy Smith...../ Oliver Twist

Row 5

Old Millionaire../ Mr. Mackarel..../ Henry Carpenter/ Army Captain.../ Arthur Z

I apologize for the formatting, but I was having problems posting an image so this is the "brute force" way of showing the layout. If anyone can see my image through the link and edit it in, I would greatly appreciate it.

One possible alternate:

If the statement "The Evil Villain is West of Ms. Johnson" means that the Evil Villain must be immediately West of Ms. Johnson, then this is the only solution (as far as I know). If the EV can be anywhere West of Ms. Johnson, then her grave and Nancy Smith's could be switched.

My new cipher with numbers

I have thought up one simple cipher. Hope it didn't exist before. Try to crack it!

Here is encoded one sentence in English consisting of three words.

$(2_ 2,5_ 2,7_ 5,8_ 5,9_ 4,14_ 4),(6_ 2,7_ 1),(3_ 2,5_ 4,30_ 4,70_ 2)$

• Number with index corresponds to one letter

Hint:

Numbers are not numbers

Answer

The answer is:

winter is here

Method:

- spell each integer (excluding index numbers) - two five seven eight etc.
- select letters from each word using index numbers - w i n t etc.

letters - The pound is not a finish

Starting with The pound is not a finish, apply all of the following modifications in the correct order to reveal a popular scientific fact:

• Remove one word.



• Remove one letter from one word.


• Change one letter for another letter.


• Replace one word with an antonym of itself.


• Remove two letters from one word.

Answer

Answer:

The sun is a star

Steps:

The pound is not a finish

Replace one word with an antonym of itself.

The pound is not a start

Remove one letter from one word.

The pound is not a star

Remove one word.

The pound is a star

Change one letter for another letter.

The sound is a star

Remove two letters from one word.

The sun is a star

How I got there:

The antonym is obviously the critical step. If it had to be applied after some of the other steps, the puzzle would have been more difficult. But giving it a shot on the initial sentence, "finish" -> "start" was the obvious candidate, then "scientific fact" suggests "star", making the rest easy.

calculation puzzle - Odds of duplicate birthdays

I found the following problem in an number of places online:

John once bet a fellow gambler that two of the first thirty persons they met and spoke to would prove to have the same birthday. Strong in the thought that he had 365 days running for him, the second hustler was pleased to accept. Suspecting, not unnaturally, a frame-up, he was careful to approach total strangers and chance passers-by, who could not be known to John. He lost the bet on the twenty-eighth question, when a duplicate birthday turned up.

“To tell you the truth,” said John afterward, “on each of the last five guys we spoke to, the odds were already better than even money in my favor. I'll explain the mathematics to you some time.”

Why is this true? At what point did John's chance of filling his bet pass 25%? 50%? 75%? 90%? 99%? (For your calculations, assume there are only 365 unique birthdays.)

N.B. John's statement about "the last five guys" assumes they will talk to all thirty people, regardless of whether the bet condition has been met or not, and the "better than even money" refers to the chance he will win the bet on or before that person.

Answer

Probability two people have the same birthday: $\frac{1}{365}$

Probability that (if the first two of people don't have the same birthday) the third person is a duplicate: $\frac{2}{365}$

So the probability that by the third person, there is a duplicate is: $\frac{1}{365}+(1-\frac{1}{365})(\frac{2}{365})$

The probability that by the nth person they will have found a duplicate:

$P(n) = P(n-1)+(1-P(n-1))*\frac{n-1}{365}=\frac{n-1}{365}+P(n-1)\frac{366-n}{365}$

I don't feel like calculating this exactly by hand so:

$P(15) \approx .255$
$P(23) \approx .509$
$P(32) \approx .754$
$P(41) \approx .903$

$P(47) \approx .955$
$P(57) \approx .99$
$P(366) = 1$

These answers give the probability that he would win the bet by this turn. If, however, he were to say "Given that I have not yet won on turn $n-1$ the probability I will win on turn $n$ is greater than 50%" this does not occur so early. In this case, $P(n-1)=0$ so:

$P(n)=\frac{n-1}{365}$

In order for $P(n)>.5$ then

$n>183$.

chess - Megan mated by a king move

"Been playing some chess with my husband recently", my friend Megan began, as we sat with our drinks in Coffee Monarchy.

"How'd you get on?" I asked.

"Didn't get off to a good start, I'm afraid. In our first game, he managed to beat me in only seven moves. I've still got the moves here, for what it's worth."

Megan put her big shoulder-bag on her lap and brought out a scrap of paper from it, but just as she put it out on the table for me, she knocked her coffee over. Several tissues later we'd cleaned up the spill as well as we could. But that scrap of paper was in a sorry state, and the ink had run. Now the only bit I could make out was the very last move:

K x b7#

Now, given that this was the winner's seventh move, what were the moves that led up to this victory?

---

This problem was invented by Alex Fishbein and published in The Problemist, March 2016. It is reproduced at PDB. The bottom of that page contains links to the PDB site's publication details and privacy policy (in German).

This is an entry in Fortnightly Topic Challenge #40: Retrograde Analysis.

Answer

Oooh, that's a nice one. First of all,

a mating move by a king is always discovered check.

Further more, it's a mating move by

Black, because White's king can get to b7 in 7 moves but there's no move left to set up a 'battery', i.e. a queen on a4 which would give mate after Kc6xb7.

I first tried (in vain) to

create a discovered check over the b8-h2 diagonal; it's easy to get the white king to e5, and the white bishop via a6 to b7 where it can be captured, but there's not enough time to weave a mating net.

So it must be

a major piece on c8 giving the discovered check.

It turns out that

we can just clear up enough pawns to make this work

with the following moves:

1. d4 c5 2. dxc5 Na6 3. Qxd7+ Kxd7 4. Kd2 Kc7+ 5. Kc3 Be6 6. c6 Rc8 7. cxb7 Kxb7#

Final position:

wordplay - Longest sentence with each letter repeated n times

Inspiration:
This question was inspired by Word sets with no repeating letters

Rules:

• Sentences can be of any length.


• Every letter used in this sentence, must be used n times.


• n can be any number.



• Capitals/Punctuation do not count.


• No dialogue!

Objective:
The objective is to find the longest sentence which abides by the rules and clarifications.

Length Measuring:
The way your sentence is measured is by characters, including spaces, and punctuation!

Further Clarifications:

• The sentence cannot consist of mainly the same word repeated, or the same 2, etc.



• You cannot make up new words or names.


• It has to be in English.


• No uncommon names.

Checking your answer:
http://jsfiddle.net/tLcvk2o4/

---

Current Best:

Sentence: I'm the master's nightmarish, gorgonian hatemonger, his moth-eaten gonorrhoea, smothering mightiest heroism, thrashing egomania's (or, to me, ignorant mismanagement's) strong-arm mishmash or staggering high treason.

Length: 215

N: 16

Letters: AEGHIMNORST (11 total)

Current Winner: uri zarfaty

---

The answer format should be the same format as the Current Best section!

If your answer does not beat the current best, but you still wish to post it, feel free, you may still get +1s!

Answer

Sentence

I'm the master's nightmarish, gorgonian hatemonger, his moth-eaten gonorrhoea, smothering mightiest heroism, thrashing egomania's (or, to me, ignorant mismanagement's) strong-arm mishmash or staggering high treason.

Length: 215

N: 16

Letters: AEGHIMNORST (11 total)

Produced by interleaving and manually modifying multiple computer-generated anagram fragments. With more work it should be possible both to improve the readability and extend it further.

Cryptic Crossword Sudoku

_{This puzzle is inspired by Matt’s excellent Cryptic Sucrosswordoku.}

Below is a sudoku as well as some cryptic clues. The clues lack enumeration and are in no particular order. The solutions to the clues can be assembled into a path-connected rotationally symmetric crossword inside the sudoku, with each solution beginning in a square whose number is the same as the clue’s enumeration. Given this information, you should be able to deduce the possible enumerations and the form of the crossword itself. Your job is to assemble the crossword atop the sudoku.

1. Discouraged media forced into action


2. Take over a French square


3. Tree residue after a fire


4. Foolish person captured near canyon


5. Sad strange spots


6. Goes over subpar art and poetry


7. Character at the end finds holes in disorganized site



8. What the despairing cut in impulsive inspiration!


9. Solution lacking wild lions and bird


10. Top dog’s meat placed in empty trophy

Answer

Part 2: The Cryptics

Discouraged media forced into action

DE(PRESS)ED

Take over a French square

SEIZE (French for 16) (thanks, Sp3000!)

Tree residue after a fire

ASH (ddef)

Foolish person captured near canyon

A(BY)SS

Sad strange spots

ADS*

Goes over subpar art and poetry

(TRA*)VERSES (thanks, Sp3000!)

Character at the end finds holes in disorganized site

What the despairing cut in impulsive inspiration!

_VEINS_ (thanks, Sp3000!)

Solution lacking wild lions and bird

EMU(-lsion) (thanks, Sp3000!)

Top dog’s meat placed in empty trophy

C(-u, +HAM)P (thanks, Sp3000!)

Part 3: The Grid

I'm fairly sure this is the only grid possible with exactly ten entries without using bars between squares:

The final grid is:

mathematics - Primes, composites, and an 11-by-11 square

I really enjoyed attempting to find a solution to Filling an 11-by-11 square (and am a fan of Gamow's square puzzles), so I was inspired to make a puzzle in the same spirit.

Is it possible to fill all $121$ entries of an $11\times11$ grid with the numbers $0$, $1$, and any composite number less than or equal to $10$ such that the row sums and column sums contain the first $22$ prime numbers?

Answer

If the row sums and the columns sums contain the first 22 prime numbers, their sum must be equal to the sum of the first 22 prime numbers, which is odd.

However, the sum of the row sums is equal to the sum of the column sums, because they are both equal to the total of all the numbers in the grid. Therefore, their sum must be even.

Therefore, the row and column sums cannot be the first 22 prime numbers.

mathematics - How to make 21?

How do you make $21$ from the numbers $1$, $5$, $6$, and $7$?

• You can use the operations of addition, subtraction, multiplication and division, as well as brackets.


• You must use each number exactly once.


• You cannot juxtapose numbers (i.e., 1 and 5 cannot be used as 15).

Answer

If this is an acceptable question, here is the answer:

$$\frac{6}{1-\frac{5}{7}}=21$$

This is a pretty well known problem that, while the math obviously works, is unintuitive enough that it is sometimes difficult to solve the first time you see it.

strategy - Titanic Tic-Tac-Toe

Xavier and Oliver were playing tic-tac-toe on the Titanic using a wooden board and metal X's and O's. As the stern of the ship sank lower and lower they found, to their annoyance, that the pieces began to slide from left to right. After some thought they decided to turn the situation to their advantage. They took turns holding the board level and added the following rules.

At the start of any player's turn he may allow the board to tilt causing all of the pieces to slide one square to the right. Any pieces leaving the 3x3 grid are removed. The player may then play a normal tic-tac-toe move.

A player may not play two successive moves on the same square.

A player may not allow the board to tilt if doing so will cause one or more of his opponent's pieces to be removed unless one or more of his own pieces will also be removed.

To clarify, a player may tilt the board if a) one or more of his own pieces is in the right hand column or b) if there are no pieces in the right hand column. Tilting only happens in the one direction: left to right. Tilting always takes place during a player's turn before he has placed his piece.

Assuming best play, can either player force a win?

Answer

The first player can always force a win.

I wrote this Python script to analyze the tree of possible game states reachable from an empty starting board. No matter what the second player does, the first player can make a move that eventually leads to a win in ten moves or less.

I took the tree and trimmed it down to only the optimal moves for the first player, plus all possible counter-moves for the second player. Excerpt:

["001", {"120": ["000", {"122": ["011", {"012": ["001", {}], "002": ["001", {}], "001": ["012", {}], "000": ["001", {}], "111": ["022", {}],

Each string represents a possible move one of the players could make. The first character of the string is "1" if the player shifts the board, and "0" otherwise. The second character is the X coordinate of the piece placed, and the third character is the Y coordinate. Each list contains two items: first, the first player's optimal move; second, a dictionary of the second player's possible counter-moves. Each dictionary is keyed by the possible move the second player could take, and each value is the two-item list representing the first player's optimal counter-counter-move to that counter-move. If the dictionary is empty, that means the game is over and the first player won.

I took this data and wrote a simple interactive game that you can play against the unbeatable computer.

---

edit: here is an interactive version with a better interface.

password - Shortest Number Containing the Numbers 1-100?

Exactly what the title says: What is the shortest (fewest digits) number you can find such that somewhere in it, you can find each number 1- 100?

My thinking for this puzzle is you have an electric numeric lock that can have any code from 1-100 and you want to try all possibilities in the shortest time. (Assuming you don't have to push enter or whatever).

For examples:

1-10 can be 1023456789

1-11 can be 11023456789

1-12 can be 110123456789

...

Also, is there some sort of formula that can extend this to other numbers (1-1000, 1-50, ...)

Answer

If you use a decimal De Bruijn sequence of order 2, you get a string of 100 digits that contains every possible substring of length 2 - that is, every possible two-digit number, although you need one last digit to account for the wraparound, bring the total to 101.

Suppose further that the first three digits are "100", we get 100 without adding any additional digits. This must be possible because any valid de Bruijn sequence can be "rotated" so that the second and third digits are identical, and then the digits remapped so that the first digit is 1 and the next two digits are 0.

So the shortest number that fits your criteria is 101 digits long.

How can I swap opposite corners on the same face of a 2x2 cube?

I've been experimenting with using Y-permutations to solve the 2x2 cube as described here. I ended up with the cube solved but for a single face where the opposite corners were swapped and disoriented. I managed to solve the cube, but I wonder if there is a shorter algorithm to accomplish this task.

Answer

If you're using Y-perms, then simply rotate a face containing one of the corners 90 degrees, to position them opposite each other. This is called a setup move. Doing so puts the two corners opposite each other, and a Y-perm will then swap the correct corners.

You can then undo that setup move after completing the Y-perm by rotating the face back in place.

mathematics - Unknown weight of four identical objects

Four identical looking objects weigh 3oz, 5oz, 8oz and 11oz. You do not know which objects weigh what and they are too close in weight to tell by holding them. Using a balance scale, how can you determine the weight of each object in only four weighings?

Answer

1. 
   

   First weigh two of your objects against the other two. Whichever pair is heavier must contain the 11-oz object, since even $11+3>5+8$.

   
   
   


2. 
   

   Now you have two objects of which you know one weighs 11 oz. Weigh them against each other to find out which one it is.

   
   


3. 
   

   Weigh the 11-oz object against two of the remaining three. If the scales balance exactly, those two are 3 and 8 oz. If the 11-oz object is heavier, they are 3 and 5 oz. If the 11-oz object is lighter, they are 5 and 8 oz. Either way, you know the weight of the fourth object.

   
   


4. 
   

   Now you have only two objects of unknown weight, so you can weigh them against each other to find out which is which.

   
   

cipher - The Cursed Breakfast

So yesterday on the bus ride back from school I was sitting beside my friend James talking about food (we were really hungry from all the still life paintings in Art class). Suddenly, his gaze wandered away and he stopped talking. I was getting used to this - lately James had took up a new hobby, cryptography, and sometimes had "brilliant puzzle ideas" he liked to share with me. I sat silently for a minute or so until he got down the bus and said simply "Tell you 'bout it tomorrow". "Sure", I said, "See you tomorrow".

The next day, as I entered the classroom, he handed me a scribbled piece of paper:

Today at breakfast the weirdest thing happened. I was eating something tasty, when all of a sudden a witch popped up right in front of me! She cracked a laugh and said:

"Behold, I am the Breakfast Witch, computer age vegan activist.

For you crimes against nature you shall pay! But fear not, child, your skin I shall not flay.

A curse upon your breakfast is enough - so you may never taste this stuff.

There is even a way to lift it, in fact; there has to be, per the Witch Contract.

Your evil treats I shall transform into twelve five-element magic forms,

And upon that another layer of deception shall further obscure the information.

But in the case you can see beyond the lies, I shall restore your murderous meal to its original guise"

The Breakfast Witch disappeared, and I instantly forgot what I was eating. I couldn't tell for the life of me what I was having for breakfast! Instead, what I saw in my plate, smelled in the air and tasted in my mouth was (don't ask me how!) "218D103061009AC". What was I eating?

I had no clue at all how to solve that. To me it looked like a crazy riddle and an even crazier sequence of letters and numbers! During break I went online and searched for the numbers, but found nothing. James kept asking me every chance he got if I'd "figured it out yet". So annoying.

By the end of the day, I was still stumped. I really want to crack this, if just to put an end to the smug way James keeps smiling at me. Searching online at home I found this site and figured it would be a great place to solve this and finally put an end to at least this one puzzle. So, can you help me find out what he was having for breakfast?

Answer

The answer is:

EGGS AND BACON

Let's start from the end. Combination of digits and letters looks like a hex number. Let's try converting them... If we convert it to binary, we get 60 (which is 12*5) digits (leading zeros appear from digit-by-digit conversion). This looks good, as it suits for "twelve five-element magic forms" and conversion to hex is "another layer of deception"

00100 00110 00110 10001 00000 01100 00011 00001 00000 00010 01101 01100

The Breakfast Witch indicated "another layer of deception", which we remove by the transformation $0\to A$ and $1\to B$:

AABAA AABBA AABBA BAAAB AAAAA ABBAA AAABB AAAAB AAAAA AAABA ABBAB ABBAA

This is the Baconian Cipher, which finally yields

EGGS AND BACON

And that's what James was having for breakfast.

word - Teapot Riddle no.22 (fan made)

Teapot Riddle no.22 (fan made).
Alien.

Rules:

I have one word which has several (2 or more) meanings.
Each of the meanings is a teapot (first, second ...).
You try to figure out the word with my Hints.

• First Hint:
  

  
  

  My first teapot grows, and my second will rest.
  You might see your nose or some weird clouds at best.

  
  
  

  
  


• Second Hint:
  

  
  

  My first teapot shows itself in a commandment.
  My second has chosen the fallen, much absent.

  
  

  
  


• Third Hint:
  

  
  
  

  My first teapot cannot keep promises much;
  It began with a plan of betrayal and such.

  
  

  
  


• Final Hint (to make it easy):
  

  
  

  My second is down and through it you see dreams.
  It can be a noun or verb under the seams.

  
  

  
  

Good luck and have fun!
Last teapot riddle. Big thanks to @Jannis for making them!

Also, the title of this puzzle was my inspiration for this puzzle's title.

Answer

I believe it is the word

lie

My first teapot grows, and my second will rest. You might see your nose or some weird clouds at best.

Lying causes Pinocchio's nose to grow; you lie down to rest, and if you're outdoors you'll see clouds in the sky.

My first teapot shows itself in a commandment. My second has chosen the fallen, much absent.

Thou shalt not lie; if you fall down, you'll end up lying down.

My first teapot cannot keep promises much; It began with a plan of betrayal and such.

If you lie, you would not keep your word. Lies are also often thought out in advance.

My second is down and through it you see dreams. It can be a noun or verb under the seams.

When you lie down, you sleep and see dreams. Sleep can be either a noun or a verb, and you'd sleep under a blanket made of seams!

knowledge - Steve misses _________?

Congrats! Weather Vane for finding the answer of :Steve asks his brother to book tickets

---

$\bullet$Yesterday West Indies won by 7 wickets against Pakistan.

Even though Steve belongs to England, he loves Cricket players of other countries too. In this World Cup, Steve misses one of his favorite players.

EDIT: But the team the player belongs to is still there in this world cup.

From the pictures below, can you find whom does he miss?

HINT

rad$(2\pi)$

Answer

I think the answer is:

AB De Villiers

Because

It is 2pi rad, a circle, which is the 360 degrees that he is famous for, and the second one, you rot13 the letters, which turn out to be abde, which ~ AB de Villiers.

cipher - Lolcat Steganography: Find the message hidden within the transport medium of humorous feline photography

---

• 
  

  The Bad Guys (curse them!) have begun communicating in lolcats, to enable their Internet communications to take place unnoticed.

  
  


• 
  

  This is a job for the NSA/GCHQ crack team of cryptanalysts.

  
  



• 
  

  Unfortunately they're a bit busy today, so you'll have to do it.

  
  


• 
  

  A secret message is concealed in the lolcatstream above.

  
  


• 
  

  It's a pencil-and-paper cipher. You do not need a computer to decipher it.

  
  


• 
  
  

  All the information you need is above the horizontal line. These instructions (everything below the horizontal line) do not contain hidden information.

  
  


• 
  

  From HUMINT assets, you have received the Bad Guys' code book. You know for certain that the message above consists entirely of one of the following code words:

  
  

  YODA (meaning "Your cover has been blown!")

  
  

  LUKE (meaning "Get me Tom Cruise's autograph!")

  
  

  LEIA (meaning "Explode the enormous bomb on Tuesday!")

  
  

  FISHCAKE (meaning "Send more food!)

  
  

  DURANCE (meaning "Give yourself up to the police!")

  
  



• 
  

  Summary (just for clarity). There is a hidden message in the lolcats above. It consists of one of this set of words: YODA, LUKE, LEIA, FISHCAKE, DURANCE. Your task is to discover which of those words the message consists of. Please explain your answer.

  
  

It's been a few days and nobody's cracked it, so here's a clue (if you need it):

Cats like bacon.

Answer

Well, I finally figured it out (thanks to the hint, of course).

The hint is "Cats like bacon", which means that

the message is encoded with a Baconian cipher.

In this case, we need to figure out

the binary distinction in the images upon which the encoding was based.

The characteristic used to encode the message is

the color of the cats: a fully-white cat is B, others are A.

So the encoded message is:

ABABA BAABB ABAAB AABAA

Which translates to

L U K E

mathematics - The Security to the Party [Part 20]

A party is being held at a local mansion. The host is very rich and his success is because of one thing — his famous recipe for Spaghetti!

The only guests that may attend are people who correctly reply to the guard at the door.

Here's where you come in. You and a friend are trying to steal this recipe. You sneak by and listen to the passwords.

The first guest arrives. The guard says, "1", the guest replies "2" and is let in.

The second guest arrives. The guard says, "3", the guest replies "5" and is let in.

The third guest arrives. The guard says, "8", the guest replies "4" and is let in.

Your friend says, "Aha! This is an easy one." He slips away and show up at the gate a minute later. "13", the guard says. Your friend replies "3" and is immediately trampled by an elephant.

Where did your friend go wrong?

Hint:

You decide to hang around a bit longer. The next guest arrives, and the guard says, "2". The guest replies "2" and is let in.

Second Hint:

Another guest arrives, and the guard says "89". The guest replies "8" and get in.

Final hint:

Still scratching your head, you wait for another guest to approach. "Two" you hear the guard start to say, but at that exact moment a passing truck honks somewhere behind you, nearly startling you out from the tree. You barely recover in time to hear the guest reply, "8" and the guard lets her in.

Having read the above hints, now you decide to approach the guard. He says, "34", what do you reply?

As usual please post your answer, an explanation of what process you used to get the answer and any hints used in spoiler tags.

Note: Whoops, flubbed the final hint. It should be correct now.

mathematics - Trapped in my Cellar

I have taken Doorknob hostage in my cellar. He is "perfectly" trapped - solid walls, solid floor, solid roof, no windows, etc. The only way out is a steel door and the only way to unlock the door is by means of two-pan balance scale (accurate to the quacogram).

I have provided 100 weights. Each weight will have its own (non-zero positive integer) weight (in quacograms) accurately written on it. He must put a non-zero number of weights on either side of the scale pans such that they balance perfectly (no trickery on the weights allowed - just the weights). Doorknob has no computer or writing implements - also he is is always able to somehow manoeuvre the weights no matter their weights.

1. How might I select the weights' weights such that he may NEVER escape?


2. How might I minimise the total weight? (having achieved 1.)



3. What is the lowest highest weight that a weight must have? (having achieved 1. and 2.)


4. How much wood must the woodchuck chuck? Just kidding :D

As always explanations/"proofs" are maybe more important than the figures themselves.

Answer

Part 1: What is a possible combination of weights?

Use power of 2 weights from 2^0 to 2^99 (obvious, everyone already said so)

Part 2: What is the lowest total weight?

User12408 already showed that the lowest total weight can be obtained by using power of 2 weights to reach total weight 2^100-1. A quick summary of the proof:

With 100 weights, there are 2^100 possible weight combinations (each weight can either be in a subset or not). Given that "no weights" is one of these 2^100 subsets, the best to hope for is for the subsets to cover weight values 0..2^100-1, which can be achieved by using power of 2 weights.

Part 3: What is the lowest highest weight? This is the interesting part.

Some people said that 2^99 was the answer for this. But that is not true.

Take for example the case of 5 weights. One might think that 1 2 4 8 16 would be the optimal set.

However, a counterexample is the set: 6 9 11 12 13

This set produces the following distinct subset sums: 6 9 11 12 13 15 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 36 38 39 40 42 45 51

(Edit: A four weight counterexample is: 3 5 6 7 by the way)

Something similar can be done for the 100 weight case, but it would be hard to find the optimal solution.

As Jakub Tarnawski pointed out, this is an open problem that even has some kind of reward for a solution (this link comes from Jakub's answer): http://www.openproblemgarden.org/op/sets_with_distinct_subset_sums

Also:

The woodchuck must chuck all it could if the woodchuck could chuck wood.

What's the Password, again?

War has broken out in the kingdom. The king's castle is guarded pretty heavily, by guards with instructions to kill anyone who can't produce the correct password.

Unbeknownst to the guards, the beggars near the main gate are in fact enemy spies. They listen intently as the first knight approaches. "Six!", the guard calls out. "Three," answers the knight. He is allowed through. "Aha," the spies think.

But they needs to be sure. They see another knight approaching and listen in on his exchange with the guard. "Twelve!", "Six!"

One spy decides he's heard enough, so he leaves and returns dressed as a knight. "Eighteen!" The spy answers "nine" and is killed on the spot by the guards.

Another spy, who used to browse Puzzling.SE before he signed up for the Espionage Division feels a moment of deja vu, and decides he now knows the answer. As night falls, he goes off and disguises himself as a knight. He comes back in the morning and walks up to the guard.

"Eighteen!" shouts the guard, "Eight!" the spy replies. The guard cuts him down, and he makes a surprised "gurgle...splot" noise as he dies.

The last spy, still in hiding is confused. As he's thinking, another knights walks up. "Twelve!" is the challenge, "Zero!" comes the reply. The guard steps aside and the knight enters.

Finally, the last spy gets up, goes off and does the whole dress-as-a-knight thing and comes back. "Eighteen!" shouts the guard. What is the correct response?

Update: The answer to TWELVE could have been Zero or Six either time - both were correct. This is one of only two such cases of 2 valid answers.

Edit for anyone who sees this in the future: I'm aware that it's not fully-scoped. It's my first puzzle and I neglected to give any odd-number examples, which is kind of important. It's totally solvable without them, I just should have included at least one.

Answer

After reading Paul's answer:

Start with an analog clock at noon (or midnight). The number the guard starts with is the number of printed digits to advance the minute hand (in other words, the number of 5-minute intervals to advance the time). Take the number of minutes past the hours and divide by 10. That's the answer the knight gives.

This yields an answer of:

three

Expressed as a math formula, where N is the number the guard gives, the solution is:

(N % 12)*5 / 10 (more or less - modulo doesn't actually give the answer x for x%x, but humans who don't understand how modulo works sometimes do).

strategy - How do I choose between letter possibilities for 2- and 3-letter words in a cryptogram?

I've tried to begin solving a cryptogram by making educated guesses from the context: position of punctuation, the presence of single-letter words, or assumptions about the frequency of the letter E, for a few examples. I often find 2-letter and 3-letter words in the puzzle and am looking for a more logical approach to such words than just blind guessing. Any ideas?

Answer

I would suggest compiling a list from this:

http://www.shabanali.com/upload/1000words.pdf

This is a list of the 1000 most common words in the English language.

I have no use for such a list and will, therefore, not make it for you... sorry but I'm selfish.

I would suggest that you note the following: of the first 100 most common words, 5 are two letter words that end in "o" (do to go so no). It may be worth focusing on that possibility first. "e" of course is the most common overall, so should be easier to find. "be" and "he" are somewhat common. Look for "xyz" and "yz" both appearing in longer cryptograms as that can commonly be "the" and "he" respectively. Once these are found, "she", "it", and "on" are easy. You will never see "ne" so that can help differentiate "one" from "the".

Finally, one of the first things to do is look for "xy" and "yx" as these will quickly help find on and no with a relatively high degree of certainty that can easily be checked. "ot" "eb" "eh" and "sa" are fairly uncommon!

riddle - I am shaggy below, Sometimes the beautiful. Who am I?

I am a wonderful help to women,
The hope of something to come. I harm
No citizen except my slayer.
Rooted I stand on a high bed.

I am shaggy below. Sometimes the beautiful
Peasant's daughter, an eager-armed,

Proud woman grabs my body,
Rushes my red skin, holds me hard,

Claims my head. The curly-haired
Woman who catches me fast will feel
Our meeting. Her eye will be wet.

Who am I?

Answer

I guess you are a

Red onion

Because:

You shouldn't look at the lines, but you should read the sentences.

I am a wonderful help to women, the hope of something to come
Food? Food!

I harm no citizen except my slayer
Unless it's a poisoned onion it won't harm anyone except the one who's slaying it

Rooted I stand on a high bed
Here the word bed means garden bed or a flowerbed and onions have roots

I am shaggy below
Yes you are!

Sometimes the beautiful Peasant's daughter, an eager-armed, Proud woman grabs my body, Rushes my red skin, holds me hard, Claims my head
A peasant probably has a bed(or a few) of onions, so his daughter will grab one from there and peel it

The curly-haired Woman who catches me fast will feel Our meeting
I guess if you throw an onion really fast into somebody, he/she will certainly feel it :)

Her eye will be wet

When you cut the onion normally your eyes start crying. Or if you throw an onion reeeeeeally hard into somebody it might hurt enough for him/her to start crying :)

word - Riley Riddle #359

I'm jumping on the bandwagon.

My prefix comes before.
My suffix is a cure.
My infix knows the rules.
Now answer this, you fools!

Answer

The answer is:

Prefix? ( M E T A )

My prefix comes before.

Comes before => Prefix or Pre

My suffix is a cure.

Fix

My infix knows the rules.

The ref knows the rules

strategy - This Dice Blackjack game - is there Nash equilibrium?

This dice Blackjack is game for two - bettor and dealer. It is played with fair six-sided dice.

Bettor: Bettor starts the game. Bettor can roll dice as many times he/she desires while the sum of rolls is lower than 21.

Dealer: Dealer plays after bettor. So dealer knows what sum bettor achieve. Dealer can roll dice as many times he/she desires while the sum of rolls is lower than 17.

Other rules: Winner is the one with lower penalization $ = \begin{cases} 21 - \sum,& \text{if } \sum \leq 21\\ 100, & \text{otherwise} \end{cases}$

where $\sum$ is sum of player rolls. If tie, dealer wins. Goal is to achieve as many wins as possible in huge number of games.

Example game:

Bettor starts:

1. roll: 6, sum=6, no reason to stop


2. roll: 5, sum=11, no reason to stop


3. roll: 5, sum=16, risk to overcome 21 in next roll is acceptable for this guy


4. roll: 4, sum=20, time to stop, risk is too high

Dealer:

1. roll: 5, sum=5, no reason to stop


2. roll: 5, sum=10, no reason to stop


3. roll: 6, sum=16, 16 < 20, so the risk of overcome 21 is not important


4. roll: 2, sum=18, still 18 < 20, but 17 was reached, dealer cannot continue and loose the game

Hypothesis (Nash equilibrium): There is simple ideal strategy for bettor - (for example stop rolling at sum of 18). Dealer can also have simple ideal strategy, if she/he expects that the bettor plays the ideal strategy.

Question: If the dealer plays the ideal strategy according to the hypothesis above, can a bettor do better with a strategy different from the optimal one from the hypothesis above?

In other words - can bettor use his/her strategy to fool dealer and get better outcome of the game?

If anything is not clear, please comment, I will do my best to improve the question.

Answer

PARTIAL ANSWER (Until I understand it deeply)

Since we roll in this game with a dice and no limited and numbered cards included, it is the same game rolling a dice until 6. So the game becomes just roll a dice until you get as close as possible to 6. If you exceed you lose.

It is possible to show this with statistics, but I just wrote a program not to think over much since there is no no-computer tag in the question. but I believe it is also possible to do this with statistics.

For example, let say bettor tend to stop when he gets 1 or over. That means bettor will roll only once whatever so:

• if bettor gets 1, he will lose whatsoever since the dealer will roll at least 1 and gets equality and win. Dealer lost chance is $0$


• if bettor gets 2 on his roll, he will stop again, dealer will lose only if he gets 1 then 6. so the chance is $1/6*1/6=1/36$.


• if bettor gets 3 on his roll, dealer will lose only if he get $1+6, 2+6, 2+5,1+1+6$. That means $1/6*1/6+1/6*2/6+1/6*1/6*1/6$


• if bettor gets $4, 1+6, 2+(5,6),3+(4,5,6), 2+1+(4,5,6), 1+1+1+6, 1+2+(4,5,6),$ etc.

and this goes on like this, at the end you will sum these and find dealer's lost or bettor's win chance, that's why I just put this into a code :)

As a result,

if dealer plays until getting 17 or more and assuming bettor knows this, the the best strategy for bettor will stop after getting 18 of course.

The code is here here you can run it if we wish.

and the best strategy for dealer will of course play until he/she got more than bettor and this will change the bettor stopping point, which becomes 19 but the probability to win for bettor becomes much less than before $33.15\%$. so there is no better strategy for bettor whatever sum dealer will stop. always playing until 19 is the best!

meta knowledge - Explaining Cheryl's birthday

Relevant link : Cheryl's birthday

I understand the solution. But it feels wrong to me, because Albert cannot possibly make the third statement with the information he has at the moment he makes it.

Hear me right : if he does, and if it is true, then we know for sure the answer is what it is. But if I understand correctly, he can't.

The first statement eliminates May and June for everyone. Bernard, had he been told "14th", could still not deduce the correct date. But he was not told the 14th. So he can make the second statement.

Then, the only thing Albert can deduce is "Bernard has been told 15th, 16th or 17th"

These three dates are enough, with the first statement, to fully deduce the date.

Bernard is able to make the second statement because he as a piece of information Albert does not, and the second statement only provides little information to Albert.

I say again, if Albert actually makes the third statement just for the sake of the problem, and if the statement is true, I do understand why the solution is July 16th. But Albert, with the information he has after the second statement, cannot make the third statement.

Let's take an example : let the date be August 15.

Albert can still make the first statement. Bernard can still make the second statement, because "15" (initial information he has) and "not May/June" (which he can deduce from first statement) are enough to deduce August 15 at this point for him. Albert cannot make the third statement.

Now, the date is of very little importance. let the date be July 16th again. Albert can make the first statement. Bernard can still make the second statement, because "16" (initial information) and "not may/june" are enough for him to deduce the date. But the second statement only tells Albert "Bernard was not told 14th".

What would be a good answer to this question ? There are two possible good answer for my question. The first one would be "you're just supposed to accept the clues and not question them". The second would be a detailed explanation of why Albert can make the third statement righteously.

How can Albert, with the information he has after the second statement, make the third one?

enigmatic puzzle - Parallelisation

^{An entry in Fortnightly Topic Challenge #32: Grid Deduction Hybrids}

---

Puzzlers, I am presenting a set of six duo-grids! Of course, I need to tell you how they work - you are given a grid. Looking at it, you are given the two puzzle types embedded in it and the operation. You are also given the values for some of the squares. Because some of the puzzles were... difficult to make, I have modified some rules to existing puzzles. In retrospect, this is pretty much the same as my Adderlink puzzle (see there for an example). Ultimately, the goal is to deduce the two sub-puzzles in each puzzle. So, for example, if I had a 5 in a square and the operation is addition, then I know that I have to have either: 0 and 5, 4 and 1, 3 and 2, etc. in the puzzles (if you still don't understand, just click the adderlink above, the rules are the same except that the puzzles and operations change).

Grid 1: Slitherlink $+$ Sudoku (i.e latin square)

Grid 2: Kakuro $\div$ Skyscraper

Grey squares mean that nothing goes there for the Kakuro ONLY, and also the Kakuro has the additional property that no two squares are the same horizontally, vertically or diagonally including through the center square.

$\color{red}{\text{user39583 proved that this puzzle's ambiguous; see his solution for the intended sol.}}$ (don't forget to +1 his solution!)

Grid 3: Kenken ($+\times \div -$ ^) Haisu

Oh, how careless of me! I forgot to put the regions in. Well, I'll tell you this, the Kenken and Haisu have the same regions. The (large) number appearing in the cell is Kenken and ANY of the operations on the Haisu. So, like, a 1 could be a 6 in the Kenken and a 5 in the Haisu (6-5) or it could be a 1 in the Kenken and anything in the Haisu (1^anything = 1) etc. Same applies for the little numbers, except that the little numbers in the Kenken are again, any operation: $+\times \div -$ ^. So, for example, if the numbers in the Kenken region were 3 and 4, and the little number in the Haisu was, say 10, then the little number in the final grid could be: 22 ($3\times 4+10$) (12 is a valid Kenken clue), 17 ($3+4+10$) (7 is a valid Kenken clue) but not 43 ($3+4\times 10$) (no combination of a valid Kenken clue and Haisu 10 exist).

$\color{red}{\text{user39583 has pointed out that the bottom left 50 should really be a 40}}$

Lastly, the Kenken has numbers 1-6: at most one of any number in a row or column

^{The Haisu was used with permission from TGE}

Grid 4: Nachbarn $\times$ Nawabari

Well, they were the names I found on here (Don't worry, there are English explanations). Two things: The Nachbarn regions do not necessarily have to be the same size. And also, although every black number has to be in its own region, the grey one does not.

Grid 5: Fillomino $+$ Ripple effect

Ripple effect regions are fillomino regions.

Grid 6: ??? $\square$ ???

Weird, huh? I forgot everything about this puzzle - the sub-puzzles used, operations, everything. Well, actually, I did remember something - that these sub-puzzles are different from everything above. Also the operations used ($+\times \div -$ ^) form a latin square - so for example, if the top left number was formed by addition, then no other number in the first row or column was formed by addition.

I'm extremely sorry about the gross underestimation of difficulty in this and the inconvenience therecaused. Sorry to all attempters who attacked the earlier version without this picture.

Any queries or comments, feel free to ask below. I will answer any question about how the puzzles work

_{Hint helpfulness level 0}

There is almost a 1-1 bijection between squares and letters

_{Hint helpfulness level 1}

There is also a hint in the fluffle of text at the beginning

_{Hint helpfulness level 1 (again)}

The puzzles come in pairs; the extraction also deals with them in pairs

_{Hint helpfulness level 2}

Each square corresponds to a letter, much like in a polybius cipher. Then you need to do some anagramming

Answer

The final solution is

Double vision

The solution can be found by

taking the squares in a duo-grid which have the same number in both grids and taking a corresponding letter from a polybius square. Anagramming these letters gives two sets of words, first for anagrams from all individual puzzles, second for anagrams for a fixed number from all the puzzles (e.g. gathering all the letters from all the puzzles that have number 1 in them):
\begin{array}{c|c|c|}&Puzzle&Number\\\hline1&dive&devil\\\hline2&coli&logic\\\hline3&ughs&utes\\\hline4&basil&braid\\\hline5&reload&halo\\\hline6&tend&dens\\\hline\end{array}

Taking from both columns only those letters which are in the same position for both words gives two letters for each word
\begin{array}{|c|c|}\hline1&dv\\\hline2&oi\\\hline3&us\\\hline4&bi\\\hline5&lo\\\hline6&en\\\hline\end{array}

Reading these letters from top to bottom gives double vision.

Partial answer

Here are solutions to all the sub-puzzles, but since it is a metapuzzle there should be some way of using them to get a final answer, which I haven't figured out yet.

First just the solutions for easy finding because this is going to be long:

Slitherlink $+$ Sudoku

Kakuro $\div$ Skyscraper
The solution for this puzzle isn't unique, so here is the solution intended by OP
Kenken ($+\times\div-$ ^) Haisu
I believe there must be a mistake in the small number 50 at the bottom row of the puzzle

Nachbarn $\times$ Nawabari

Fillomino $+$ Ripple effect

The operator Latin square for Grid 6

Seethrough ($+\times\div-$ ^) Shikaku

Slitherlink $+$ Sudoku

3 is the maximum cell value in Slitherlink and 5 in Sudoku so 8 = 3+5 giving also 7 = 3+4, since the 5 in Sudoku is already taken on that row. Because the loop in Slitherlink cannot touch itself, we can also deduce some borders around the 3s.

If the upper right corner had a 5 in Sudoku, it would have a 1 in Slitherlink, which could only go below the cell. This would eventually force the border to form a small loop in the last column, because of the 3 there. Therefore the only possible place for the 5 in the first row is the first column. This also gives the remaining 5s.

Due to the 0 in the first cell, there is only one border that can be filled for the cell below it, and because the corresponding cell can have at most 4 in Sudoku, it must have 1 in Slitherlink. This also gives the borders and numbers for the cell below it.

After this it is fairly straightforward to fill the Sudoku starting with the first column, then filling the 4s, then 3s and then the rest.


Then it is easy to fill the slitherlink starting from the first row and going downwards.

Kakuro $\div$ Skyscraper

Some of the numbers cannot be multiplied in Kakuro, so we know that they are 1 in Skyscraper and must therefore have 5 as the nearest building. After this the second column and second row cannot have 5 as their nearest buildings so they must see at least two buildings, but neither of them can be multiplied by more than two in Kakuro.

The fourth row and fourth column must be multiplied by at least three so we can deduce the last two 5s in Skyscraper. The second column must contain the numbers {4,6,7,8,9} and the last two numbers in third column must be {8,9}, so we can deduce that the last rows multiplier must be 3 and the fourth rows multiplier must be 4 (they cannot be more due to the skyscraper 5s). Also this means that there is no possible way for the third columns 9 to be in the fourth row, which gives the placement of 9 and 8.


The remaining numbers on the fourth row must be either {1,2,3,6} or {1,2,4,5} and the 14 on the third row must be due to diagonal limitations either 8+6 or 5+9. If it were 8+6, the only possible order for the second column would be 8,9,6,4,7, making the last number on last row a 2. Since the fourth row must also contain a 2, it would have to be in the first column, but then the remaining number of that column would have to be a 2 as well. Therefore it follows that the 14 must be 5+9.

The last number on the fourth line can be at most 5, so the last number on the third line must be at least 4 making that rows sum at least 6. If the first rows sum were 20 then the sum of all the Kakuro cells would be 106, 108 or 110. The columns make only 106 possible by having the column sums be 12, 34, 10, 15, 18 and 17. This would mean that in the Skyscraper’s first row, the 4 couldn’t be in the third or fourth column, but this would contradict the assumption that the sum of the first row in Kakuro was 20. Therefore the sum must be 10 or 15.

Now we can see that the first row’s sum is 10 (or 15) if and only if the fourth column’s sum is 15 (or 20) since those are the only row and column affecting whether the whole sum is even or odd. Furthermore, if the first column’s sum were 12, then the third row’s left part would be 10 and third column would be 4. This would mean that in the Skyscraper the third row would be 54321, and the first cell in the third column would be 4. However, since this would make the first row in Kakuro 15 and fourth column 20, the 2 in Skycraper’s fourth column would give a contradiction. Therefore the sum of the first column is 8. Since the Skyscraper’s first column cannot have the 4 in the last two cells and now it cannot have it in the second cell either, it must be in the first cell. This in turn gives the Kakuro sums for the first row and fourth column.

There would be solutions both for the remaining column and row sums being 6 and 8 or them being 8 and 10. Therefore I assume that the rule that the row cannot contain same numbers also applies for the sums, making the solution for the column sum 8 invalid. This logic doesn't work too well however, since the 18s are on the same diagonal.

Edit: The intended solution had the sums being 8 and 10, which would after that lead to a unique solution. The intended solution has been added to the top of this answer.

We can see that the 8 for the second column (having numbers {4,6,7,8,9}) has to go to the second row. Furthermore, since the first cell of the second row can have at most a 2 and the third and fourth cells can have at most a 5 making their sum at most 9, then the fifth cell must have a 9 and the rest need to have their maximum.

Since now the last cell of the last row cannot have 2 it forces the 7 of the second column to be on the first row and since the last cell of the first row cannot be a 1 the first row reads 7,1,2.

Now the second cell of the fourth column cannot be a 4 because of the diagonal so it must be 6, giving also the last row. The 1 of the fourth column must now go on the third row giving the rest of the Kakuro. The solution for the Skyscraper isn’t unique and I have found 7 different solutions. Two of them can be seen below, where the remaining empty cells can be filled by two different ways. Note that this is not the intended solution of OP. The intended solution has been added to the list of solutions at the top.

Kenken ($+\times\div-$ ^) Haisu

We can start by noticing that the 28 can only be 7*4, so the Haisu region containing it must contain at least seven cells. Since there is a long connected region where the litlle numbers are only 19 or 33, and furthermore 19 = 26-7, 33=26+7, we can guess that those form the region, which is entered 7 times in Haisu and for which the sum of cells is 26 in Kenken. There is two possibel ways to draw a path which enters the region 7 times, but only one of those is possible with the given numbers.

The sum of the remaining 3 cells in that region must be 6, which leaves only one possibility. The first row must have 4 cells in the same region to find a solution for the little numbers.

For the last column the solutions are found if there is one cell by itself and the other two in the same region. Then the three remaining cells in the upper right are also together in their region.

Now, I believe there must be a mistake in the last row, since the only logical solution for that 117 would be 120-3. Since it is in the corner, this would require the size of the region to be at least 4, but no matter which cells you would select, there would always be at most 3 possible cells. If the small number 50 is a mistake, and should be for example 5, 40 or 45 then we can find the following solution

Nachbarn $\times$ Nawabari

The 1 must be 1*1. Since no region can have 8 neighbors, the 16s must be 4*4. The one in nachbarn means that some region must fully surround the region with the one, but it cannot be the region with the 12, since that wouldn’t leave enough roome for both the 4s. There fore it must be the 4 on the right which surrounds the one.

Now in the Nawabari the first two cells of the third column must belong to the same rectangle, which implies that there must be a 2*2 square below the 4s. Also the third cell of the first column must be it’s own region in Nachbarn with two neighbors. The lower right corner must have two borders in the Nawabari, so it can have only one neighbor in Nachbarn. The 4 region coming from the upper right cannot be that neighbor, since it would make it neighbor to the middle square in the last row as well, and it would thus have 5 neighbors. Therefore that region must end in the middle row.

There is no way in the Nawabari, which would work if the ? in the first column were 3 and the ? in the last row were 4 at the same time. This means that the ? in the first column of Nachbarn cannot be 2, because there would be no way of making the ? of the last row of Nachbarn a 4. Therefore the ? in the first column is 3 in Nachbarn and 2 in Nawabari. This means that the extra hint (8) has to belong to the 4 region coming from the up, because it has to belong to a group with 4 neighbors and if it belonged to the middle ? then the 4 region from the up would have only 3 neighbors, and if it on the other hand belonged to the bottom ? then the just found 3 region in the left side would have only 2 neighbors. Since the 4 region cannot be connected to any more neighbors after that, we can also connect the 3 region to the bottom ? and extend the middle ? downwards.


Now since the bottom ? has to have at least 3 neighbors, it has to be the one surrounding the corner 1. Now we can also deduce the rest of the Nawabari.

Fillomino $+$ Ripple effect

We can start by putting borders between equal numbers and numbers where one number is at least twice the other minus one, because those cannot belong to the same region. The 2 must be a region of size 1. The upper left 10 must be in a region of at least size 5 so all the four cells in that corner must belong to the same region as the 6 must also belong to a region of multiple cells and there cannot be a region of size 1 next to the previous one. Both 6 and 10 belonging to the region makes the size of the region 5.

The 8 in the lower right corner has to be connected to the 10 above it or else it will end up blocking either that same 10’s region or the 8’s next to it. It will also have to be connected to the 12 or else it would block it’s region since the 12 needs at least 6 cells and cannot contain the ones with 6 or 5. Since the remaining 6 in the first row cannot belong to a region of size 5 and the region with a 12 in it must block the other 10 from either up or down, the only direction for the 10 to go is down and it must be connected to the 8 and 9.


From here it is quite easy to see the final solution for the Fillomino and we can also find the only possible way to fill the remaining Ripple effect cells.

??? $\square$ ???

It is quite easy to deduce the most probable latin square for the operators.

This gives also already some of the numbers as well as long as we assume some reasonable upper limit for them.


We can already notice that the middle cell of the first row must contain a 7 in one of the puzzles. One puzzle where such a large number is possible for the edge is Seethrough (also goes by some other names), where the numbers denotes how many cells can be seen in horizontal and vertical directions before hitting a wall. However, neither puzzle would work for the 7 if the cell isn’t also counted itself like I believe is usually the case in Seethrough. If we count the cell itself also there isn’t a problem at least yet so we shall try solving the first puzzle as Seethrough and put the 7 there. I will erase the known open walls.

The upper right corner cannot see the 7 because it is at most 2 (assuming positive numbers also for the other puzzle). There must be a wall between the 3 and 7 as well. This gives only one possible way for the 7 and solves the neighboring 5 as well.

Since the 3* in the first column can be at most 3, the 6 in the first column must see at least 3 other cells in the same row. Now the cell 24* on the right side of that 6 can see 5-6 cells so it must be a 6 and there are no walls between that row. This gives also the 14+ to be 8 in the Seethrough. The 3/ in that row must then be 6.

The 30* near the middle must be at least 5 and since it can see at most 3 cells in the same column there cannot be a wall on the left side of it. The 20* in the bottom right must be at least a 4 so there cannot be a wall above it nor on the left of it. The 3 on the bottom row cannot have a wall on the right of it but then it must have one on the left.

At this point the right side puzzle starts to look a lot like Shikaku especially when we take into account that 30* is most likely 5*6 and 20* is 4*5. Assuming it’s Shikaku we can already see the shapes of the 4, 5, 6 and 2 regions where we can deduce the second 2 region by noting that it will have to go to the 16^ cell. Filling these in we get

Filling most of the Seethrough walls is now easy. we can also see that the 3* cannot be 3 in the Shikaku so it must be 3 in the Seethrough. The 3 region in Shikaku must be in the 3 middle cells then followed by another 1. This gives the final solution