riddle - This day in history II

This day in history I was correctly answered by PotatoLatte. I will describe an event from a certain number of years ago that happened today (28/07). I would like you to tell me both he event and the amount of years ago it happened.

A glove, these dark people hold
They did something - stupid or bold?
One split to two
Thousands of miles from those who

Were affected by this awful event
Only 2 million minutes would last
Before this disaster was in the past

Answer

Could be

Start of World War I - Started 28.07.1914 (105 years ago) and it lasted 2.2 million minutes.
The Europe split to two: the Triple Entente—consisting of France, Russia, and Britain—and the Triple Alliance of Germany, Austria-Hungary, and Italy.

crosswords - Movies, movies, movies!

Each answer is the title of a well-known movie...

ACROSS
4. Teardrop flows for a broken relationship (8)
6. Spin the hated frog, until brown (3,9)
9. Immediate disaster as Neb crewman’s sly weapon explodes (first part, before 8 down) (10)
10. Unbalanced, like a legendary bird (5)
13. Happy interlingual anonymous fighter (9)
15. Disaster movie provokes small laughs (5)
16. Prideful monarch is audibly recumbent (3,4,4)
17. Selfish idea (now an internet phenomenon) and digit as a keepsake (7)
19. Patents, copyrights and trademarks lost in shipwreck! (5)

20. Taken, awaits rye dip mixture (second part, after 3 down) (4)
22. Mechanised fruit is on time (second part, after 12 down) (6)
23. Subway city-state, or milepost repositioned (10)

DOWN
1. A frog is confused by long-distance travel (5)
2. Flight to the Bahamas (2)
3. Taken, awaits rye dip mixture (first part, before 20 across) (8)
5. Nitrate mother turns from the light to the dark side (3,10)
7. Rick’s White House (10)
8. Immediate disaster as Neb crewman’s sly weapon explodes (second part, after 9 across) (3)
11. Marilyn’s Sugar is naturalised newspaper mogul (7,4)

12. Mechanised fruit is on time (first part, before 22 across) (1,9)
14. Shake saline strangely (6)
18. Threat mix disrupted Blair’s chambers (3,6)
21. Moot-like heist turns farcical (first part, before 25 down) (4,4)
24. Nutty carbohydrate, gangnam-style (6)
25. Moot-like heist turns farcical (second part, after 21 down) (2,3)

Answer

1. 
   

   
   
   

   Fargo (Anagram of "A Frog")

   
   

   
   


2. 
   

   
   

   Up (Two-letter word, "Flight")

   
   

   
   


3. 
   

   
   
   

   Spirited Away (Anagram of "Awaits rye dip")

   
   

   
   


4. 
   

   
   

   Predator (Anagram of "Teardrop")

   
   

   
   


5. 
   

   
   
   

   The Terminator (Anagram of "Nitrate mother")

   
   

   
   


6. 
   

   
   

   The Godfather (Anagram of "The hated frog")

   
   

   
   


7. 
   

   
   
   

   Casablanca (Means "white house")

   
   

   
   


8. 
   

   (See 9)

   
   


9. 
   

   
   

   Apocalypse Now ("Disaster", "Immediate")

   
   

   
   
   


10. 
    

    
    

    Rocky ("Unbalanced", Roc-like)

    
    

    
    


11. 
    

    
    

    Citizen Kane ("Newspaper mogul")

    
    

    
    
    


12. 
    

    
    

    A Clockwork Orange

    
    

    
    


13. 
    

    ???

    
    


14. 
    
    

    
    

    Aliens (Anagram of "Saline")

    
    

    
    


15. 
    

    
    

    Gigli

    
    

    
    


16. 
    
    

    
    

    The Lion King ("Pride" -> Lion, "Monarch" -> King)

    
    

    
    


17. 
    

    
    

    Memento ("Keepsake")

    
    

    
    


18. 
    
    

    ???

    
    


19. 
    

    
    

    Shrek (Subsequence of "shipwreck")

    
    

    
    


20. 
    

    (See 3)

    
    



21. 
    

    ???

    
    


22. 
    

    (See 12)

    
    


23. 
    

    
    

    Metropolis (Anagram of "Or milepost")

    
    

    
    
    


24. 
    

    
    

    Psycho (Psy + CHO)

    
    

    
    


25. 
    

    ??? (See 21)

    
    

Thanks to rand al'thor for (7) and Christian Rau for (15)

logical deduction - Barrel - Part 1

^{An entry in Fortnightly Topic Challenge #35: Restricted Title 1. Title based on the very first xkcd.
Continued in Barrel - Part 2, Barrel - Part 3, Barrel - Part 4, and Barrel - Part 5.}

---

You manage a barrel warehouse. The warehouse is divided into an $8\times 8$ grid of cells. A barrel takes up one cell when upright, and two cells when laid down. You only remember where some of the barrels are in the warehouse, and you must determine where the rest of them are. Luckily, you remember two things that should help you out:

• There are indicators along each row/column representing the number of barrels that occupy at least one cell in that row/column.


• Barrels that are lying down have no room to roll to either side, i.e. there is a barrel or the wall stopping them from rolling on either side.

Here is a diagram of the warehouse, with the row/column indicators and the barrels you know for sure. Please determine where the remaining barrels are, and show your steps.

Answer

Answer

Step by step solution

The start of the puzzle. Several rows and columns are already completely filled, so let's shade them in:

R2C2, R2C4, R6C5, and R8C5 must be occupied by a barrel to prevent their adjacent barrels from rolling now since the other space can't be occupied. They could be in any orientation, but a barrel must exist there.

I counted the number of barrels remaining. The placing of these barrels nearly fills up row 2, 6, and 8, so many spaces can't be placed now:

R2C4 can't extend to R2C5 or it can roll north. Likewise R6C5 can't extend to R5C5 or it can roll east. In particular, R2C5 can't be occupied by another barrel either, since that would exceed row 2's barrel limit.

Therefore, R3C5 is a barrel to block the barrel to the east of it.


Now R5C5 can't be a barrel due to column 5's barrel restriction. R4C2 must be occupied to prevent the barrel west of it from rolling. R2C4 can't extend down due to row 3's barrels.

R4C2 can't extend east or it could slide up. R4C5 must be occupied by a barrel or else the barrel to the west of it could slide, importantly since column 5 is already at the barrel limit that means that R4C5 and R3C5 are part of the same barrel. This means that R3C4 isn't occupied due to row 3, and therefore both R6C5 and R8C5 extend left to satisfy column 4's restriction.

We do another inventory check of barrels remaining. We need 3 more barrels on columns, and 2 more barrels on rows. An interesting property is that upright barrels contribute 1 to both row and column counts, horizontal barrels contribute 1 to row counts, and 2 to column counts, and vertical barrels contribute 1 to column counts, and 1 to row counts. Hence we need at least one more horizontal barrel.

The only place we could place a horizontal barrel is R4C7-8 (R5C2-3 would slide south). The other barrels are therefore upright and we are done.

mathematics - Does this mathy square have any solutions? (And how many?)

Consider a $5\times5$ grid of math operators and numbers that encodes 8 math equations:

A + B = C
= + + + +
D + E = F
+ = = = =
G = H + I

There are 3 horizontal equations, 3 vertical, and 2 diagonal. Specifically:

A + B = C
D + E = F
H + I = G
D + G = A
B + E = H
C + F = I
A + E = I
C + E = G

However, to make the grid less strict, each equation has 3 additional allowed forms:

• 
  

  The + can be replaced with *. e.g. A * B = C. (We'll avoid - / ^ since they aren't commutative.)

  
  


• 
  

  The + and = can be swapped. e.g. A = B + C.

  
  


• 
  
  

  Both of the above. e.g. A = B * C.

  
  

Thus you can massage the equations a bit in order to form a more workable grid. For example, it might look like:

A * B = C
+ = * * +
D = E * F
= = = + =
G * H = I

The question is: Can the letters A through I be replaced with all one-digit numbers 1 through 9 in any order such that all 8 grid equations are satisfied?

Are there lots of solutions or none at all?

(Bonus: What if - or / or ^ is allowed to be an operator (and you make some assumptions about reading direction).)

mathematics - 30 fake coins out of 99 coins

You are given 99 coins which consists of 30 fake ones. You also have a digital balance scale with perfect precision that shows how much difference between weighs you put on. For example, if you put 10 g on the left side and 20 g on the other side, it will show -10, otherwise +10.

You are asked to find a fake coin among given 99 coins:

• You know that all genuine coins have the same weight but you do not know their weights.



• You also know that every fake coin is heavier or lighter by 1 gram than any genuine coin.

EDIT: The intended question was to not allow a mix of heavier and lighter coins. Since all answers were based on this assumption, changing this requirement now would invalidate them all. I'll leave this question as is (and allow a mix) but don't know an optimal solution myself.

So, what is the minimum amount of weighing which guarantees to find any fake coin you are looking for? (The fake coin you are going to find might be heavier or lighter, it does not matter, you just need to find any fake one.)

Note: You may assume weights are positive integers, but it is not supposed to change the result.
You may also weigh one or more coins against nothing to get their total weight (originally asked and answered in comments.)

mathematics - Odd birthdate surprise

Last year, just before my birthday in February, I noticed something unsual. The last two digits of my birth year were the same as my age. How old was I then and was this an unusual occurrence?

Answer

Generally, people born in $1900 + n$ will have this birthday in the year $1900 + 2n$, when they are $n$ years old. The same can be applied to any century. You'll notice this is an even number; the year in the question is odd because you noticed this just before your birthday. That is, the Beddian birthday (see Kate Gregory's link in the quetion) was in 2012, but you noticed the phenomenon in the next year, 2013.

Working backwards, if this year is $c + m$ (where $c$ is a century such as 1900 or 2000 and $m$ is even and less than 100), those with a Beddian birthday this year were born in $c + \frac{m}{2}$ or $(c - 100) + \frac{m + 100}{2} = (c - 100) + \frac{m}{2} + 50 = c + \frac{m}{2} - 50$.

You can obtain the second option by "going back" a century, making $c - 100$ the new century part and $m + 100$ the new "$2n$" part from the first paragraph. Since $m < 100$, $\frac{m}{2} + 50 < 100$, and so the year is valid. You can't go back another century, or else the "$m$" portion will be more than two digits, which wouldn't make any sense.

Thus, for the year 2012: $$ c = 2000, m = 12\\ \text{birth year is } \boldsymbol{6}=\frac{m}{2} \text{or } \boldsymbol{56}=\frac{m}{2} + 50 $$

As Rob Watts notes, you're more likely to have been 56 than 6. Additionally, this happens every even year for those born turning $(year ~mod~ 100)/2$ and the same age plus or minus 50.

mathematics - No broken eggs puzzle

Three men went to buy eggs from an eggs seller:

1. The first ordered half of the seller's stock plus half of an egg.


2. The second ordered half of what's left in the stock plus half of an egg.


3. The third ordered half of what's left in the stock plus half of an egg.

How many eggs did the seller have in his stock knowing that:

1. No eggs were broken in the process.



2. The seller's stock is now empty (no eggs left).

Answer

This can be easily solved starting from the third guy since no egg is left after he took the half of the egg.

Third guy:

Half of the egg has to be equal to half of the stock, because he took first half of the stock then the rest (which is half of an egg), resulting third guy got $1$ egg only.

Second guy:

When second guy bought whatever he says, there are 1 egg left. We know this from above. So $1$ egg+$\tfrac{1}{2}$ egg has to be equal to the half of the stock. So there were $3$ in the stock when second guy decided to buy some eggs, meaning he got $2$ eggs.

First guy:

With the same logic, after the first guy bought the eggs, there are $3$ eggs left. meaning $3$ and and a half should be the other half of the egg, resulting $7$ was the total number of eggs at the very beginning and first guy bought $4$ eggs actually.

optimization - Barrel - Part 3

^{An entry in Fortnightly Topic Challenge #35: Restricted Title 1. Title based on this xkcd.
This is a continuation of Barrel - Part 1 and Barrel - Part 2, but this puzzle is still self-contained.
Continued in Barrel - Part 4 and Barrel - Part 5.}

---

Having excelled at your previous two barrel warehouse management jobs, you have been promoted to the honorable position of Barrel Sorter! I know what you're thinking, that sounds more like a demotion. But just go with it; let's say it has a higher salary. Anyway, it's just like Bucket Sort, but with barrels (not really).

First, some terminology that comes with the new job:

• A barrel is said to face up or down if the opening is on the top or bottom, respectively.


• A nest is a group of barrels positioned such that for any two barrels in the group, one of them is inside the other. All barrels in a nest are either up-facing or down-facing.



• A nest is sorted if all of its barrels are facing up.


• A sorting platform is a platform used for sorting (duh). A platform can hold at most one nest. For your job, you only have two sorting platforms.


• A barrel is available if you can see it from above. That is, it isn't inside a down-facing barrel.


• A barrel is empty if it doesn't contain any other barrels.

To sort barrels, you have two possible operations, which have different effects for up-facing and down-facing barrels:

• Flip: You may take an up-facing available barrel and invert it. Doing so will also invert all barrels inside of the one you flip. Or you may take a down-facing, available, empty barrel and invert it.


• Move: You may take an up-facing available barrel and move it to the other sorting platform. Doing so will also move the barrels inside of it. Or you may take a down-facing available barrel and move it to the other sorting platform. Doing so will not move the barrels inside of it. In either case, when you move barrel(s) to a new platform, you must be able to place them directly onto the other platform from above, without inverting any barrels, and while keeping everything a nest on both platforms.

Here are the six barrels you need to sort. In the end, you should have one sorted nest on one of the platforms. Oh, and please do this in the minimum number of operations. :)

Answer

My best so far is

11 operations: Move the upside down barrels to the other platform, flipping each on the way, and then stack everything up starting with the smallest barrel.

Calling the barrels A-F with A being the largest one:

1: Move A

2: Flip A
3: Move C
4: Flip C
5: Move E
6: Flip E
7: Move F
8: Move E (containing F)
9: Move D (containing EF)
10: Move C (containing DEF)
11: Move B (containing CDEF)

I think I could shave off at least one move if I were allowed to flip a face-down barrel with its contents, but sadly that isn't an option. (EDIT: not that sadly, it turns out. Flipping those would allow a trivial, boring 6-move solution.)

strategy - Guffin factory regulated heist

I work at a guffin factory. I manufacture guffins and have an unlimited access to guffin storage. I can fit up to 20 guffins in my bag, get them home from the factory and nobody will chase me.

You see, the factory wants to encourage brave and risky workers. When you are hired, they give you a random (evenly distributed) secret number N from 1 to 20 inclusively. If you ever steal N (or more) guffins a single day, the next day they will proclaim you fired. So, if I steal 20 I would be fired for sure. But even if I get only one, I can be fired if my luck is too low. The factory has a secure invisible x-ray machine, so they will see you stealing anyways.

Today (not a working day), my boss left me a message.

You' re a total wuss, Thomas. You spend five years working for us, and haven't stolen even one guffin. I'm sick of you and want to fire you. However, I had to warn you. The next 20 work-days (two weeks) are your last days at the factory (however, they can end earlier if you act correspondingly).

My plan is to steal 20 guffins once I get to the factory, and leave. A guffin market value now is quite high, so my family won't starve until I find a new job. However, I can see a nice puzzle here.

Having my information (no info on my secret number, and 20 days to quit the job), what is the best (mean) amount of guffins I can steal off my factory? What is the best strategy here?

Answer

New answer

Denote $E(d, k)$ be the expected gain of optimal actions given d remaining days and a knowledge that the limit is at least $k$. In particular, once we know that $k$ is safe, there is no reason to pick any number below $k$. Additionally, looting $20 \geq l \geq k$ guffins should have a $\frac{20-l}{20-k}$ chance of success, resulting in an expectation of $l+\frac{20-l}{20-k}E(d-1,l)$. Hence $E(d,k)$ is the maximum of these choices, and we therefore wish to compute $E(20, 0)$.
I wrote a program to compute the optimal strategy and it attains $110$ guffins on expectation. It does so by taking $10$ guffins on the first night, which succeeds exactly half of the time (if your secret number is at least 11). Then it keeps taking $10$ guffins on all subsequent nights except the last, where it takes $20$ guffins, in which its total loot becomes $210$ guffins.

Strategy table (row = # of days remaining, column = highest known safe number)

[{}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}],
[{20}, {20}, {20}, {20}, {20}, {20}, {20}, {20}, {20}, {20}, {20}, {20}, {20}, {20}, {20}, {20}, {20}, {20}, {20}, {20}, {}],
[{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20}, {1}, {2}, {3}, {4}, {5}, {6}, {7}, {8}, {9}, {10}, {11}, {12}, {13}, {14}, {15}, {16}, {17}, {18}, {19}, {}],
[{10}, {9, 10}, {9}, {8, 9}, {8}, {8, 7}, {7}, {7}, {8}, {9}, {10}, {11}, {12}, {13}, {14}, {15}, {16}, {17}, {18}, {19}, {}],
[{10}, {10}, {9}, {9}, {9}, {9}, {9}, {8}, {8}, {9}, {10}, {11}, {12}, {13}, {14}, {15}, {16}, {17}, {18}, {19}, {}],
[{10}, {10}, {10}, {9, 10}, {9}, {9}, {9}, {9}, {9}, {9}, {10}, {11}, {12}, {13}, {14}, {15}, {16}, {17}, {18}, {19}, {}],
[{10}, {10}, {10}, {10}, {9, 10}, {9}, {9}, {9}, {9}, {9}, {10}, {11}, {12}, {13}, {14}, {15}, {16}, {17}, {18}, {19}, {}],
[{10}, {10}, {10}, {10}, {10}, {9, 10}, {9}, {9}, {9}, {9}, {10}, {11}, {12}, {13}, {14}, {15}, {16}, {17}, {18}, {19}, {}],

[{10}, {10}, {10}, {10}, {10}, {10}, {9, 10}, {9}, {9}, {9}, {10}, {11}, {12}, {13}, {14}, {15}, {16}, {17}, {18}, {19}, {}],
[{10}, {10}, {10}, {10}, {10}, {10}, {10}, {10}, {9}, {9}, {10}, {11}, {12}, {13}, {14}, {15}, {16}, {17}, {18}, {19}, {}],
[{10}, {10}, {10}, {10}, {10}, {10}, {10}, {10}, {10}, {9}, {10}, {11}, {12}, {13}, {14}, {15}, {16}, {17}, {18}, {19}, {}],
[{10}, {10}, {10}, {10}, {10}, {10}, {10}, {10}, {10}, {9, 10}, {10}, {11}, {12}, {13}, {14}, {15}, {16}, {17}, {18}, {19}, {}],
[{10}, {10}, {10}, {10}, {10}, {10}, {10}, {10}, {10}, {10}, {10}, {11}, {12}, {13}, {14}, {15}, {16}, {17}, {18}, {19}, {}],
[{10}, {10}, {10}, {10}, {10}, {10}, {10}, {10}, {10}, {10}, {10}, {11}, {12}, {13}, {14}, {15}, {16}, {17}, {18}, {19}, {}],
[{10}, {10}, {10}, {10}, {10}, {10}, {10}, {10}, {10}, {10}, {10}, {11}, {12}, {13}, {14}, {15}, {16}, {17}, {18}, {19}, {}],
[{10}, {10}, {10}, {10}, {10}, {10}, {10}, {10}, {10}, {10}, {10}, {11}, {12}, {13}, {14}, {15}, {16}, {17}, {18}, {19}, {}],
[{10}, {10}, {10}, {10}, {10}, {10}, {10}, {10}, {10}, {10}, {10}, {11}, {12}, {13}, {14}, {15}, {16}, {17}, {18}, {19}, {}],
[{10}, {10}, {10}, {10}, {10}, {10}, {10}, {10}, {10}, {10}, {10}, {11}, {12}, {13}, {14}, {15}, {16}, {17}, {18}, {19}, {}],

[{10}, {10}, {10}, {10}, {10}, {10}, {10}, {10}, {10}, {10}, {10}, {11}, {12}, {13}, {14}, {15}, {16}, {17}, {18}, {19}, {}],
[{10}, {10}, {10}, {10}, {10}, {10}, {10}, {10}, {10}, {10}, {10}, {11}, {12}, {13}, {14}, {15}, {16}, {17}, {18}, {19}, {}],
[{10}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}, {}]]

Old answer, which assumed that the guffin limit was not fixed.

If you only had one day to work, clearly the best strategy is to steal the maximum number of 20 guffins since you would be fired anyway.
Now, what if you had two days to work? If you stole $k$ guffins, you will be fired with $k/20$ probability, and in the other $1-k/20$ chance, you would be able to steal 20 guffins tomorrow, so your expected value is $k+20(1-k/20)=20$. Oh, I guess it doesn't matter at all what $k$ you pick - your expected value remains at 20 guffins. This can be repeated inductively to conclude that you can't do better than taking 20 guffins and running. Or if you value the job more than, say, a different job, then you should wait until the very last day and taking the 20 guffins. Or if there's any chance of being rehired, try to do that quickly enough so that you can take your 20 guffins again. The sky's the limit! Or rather, 20 guffins is the limit.

logical deduction - Strategy for solving 0hh1 and Unruly?

Q42's game 0hh1 and Simon Tatham's game Unruly are two almost identical logical deduction games. In each one, we're given a square grid which is to be filled by tiles of two colours, and a certain initial setup of given tiles. We need to fill in the rest of the tiles, subject to the following restrictions:

• No three adjacent tiles of the same colour in any row or column.


• Each row and column is equally divided between the two colours.


• (0hh1 only) No two rows and no two columns have identical patterns of colours.

From any initial setup either of the sites can give us, using these criteria will be enough to derive the final solution logically. But what deductions can we make from these criteria, specifically? What are the patterns we should be looking out for in order to fill in new tiles as fast as possible?

In short, what is the best strategy for solving these games?

Answer

What to look for

At any stage of play in 0hh1 or Tatham's Unruly, here's what you should be looking for, in order, and what you should do in each case if you find it:

$\def\G #1{\color{green}{\textbf{#1}}}$

1. 
   

   $\G{Two}$ $\G{adjacent}$ $\G{tiles}$ $\G{of}$ $\G{the}$ $\G{same}$ $\G{colour}$ $\G{in}$ $\G{the}$ $\G{same}$ $\G{row}$ $\G{or}$ $\G{column.}$

   
   

   If you find this, fill the cells on either side of this pair with the other colour.

   
   

2. 
   

   $\G{An}$ $\G{empty}$ $\G{cell}$ $\G{flanked}$ $\G{by}$ $\G{two}$ $\G{tiles}$ $\G{of}$ $\G{the}$ $\G{same}$ $\G{colour}$ $\G{in}$ $\G{the}$ $\G{same}$ $\G{row}$ $\G{or}$ $\G{column.}$

   
   

   If you find this, fill this empty cell with the other colour.

   
   

3. 
   

   $\G{A}$ $\G{row}$ $\G{or}$ $\G{column}$ $\G{with}$ $\G{all}$ $\G{of}$ $\G{a}$ $\G{single}$ $\G{colour}$ $\G{already}$ $\G{filled}$ $\G{in.}$

   
   
   

   If you find this, fill the rest of that row or column with the other colour.

   
   

4. 
   

   $\G{A}$ $\G{row}$ $\G{or}$ $\G{column}$ $\G{with}$ $\G{all}$ $\G{but}$ $\G{one}$ $\G{of}$ $\G{a}$ $\G{single}$ $\G{colour}$ $\G{already}$ $\G{filled}$ $\G{in,}$ $\G{and}$ $\G{two}$ $\G{empty}$ $\G{cells}$ $\G{adjacent}$ $\G{to}$ $\G{a}$ $\G{tile}$ $\G{of}$ $\G{the}$ $\G{other}$ $\G{colour.}$

   
   

   If you find this, then one of those two cells must contain the final tile of the first colour in that row/column, so fill all but those two empty cells with the second colour.

   
   

The final two configurations to look for are for 0hh1 only, not Tatham's Unruly:

$\def\P #1{\color{purple}{\textbf{#1}}}$

5. 
   

   $\P{A}$ $\P{row}$ $\P{or}$ $\P{column}$ $\P{with}$ $\P{all}$ $\P{but}$ $\P{two}$ $\P{tiles}$ $\P{filled,}$ $\P{which}$ $\P{is}$ $\P{so}$ $\P{far}$ $\P{identical}$ $\P{to}$ $\P{a}$ $\P{full}$ $\P{one.}$

   
   

   If you find this, fill the two empty cells in the opposite way from how they are in the corresponding full row/column.

   
   

6. 
   

   $\P{A}$ $\P{row}$ $\P{or}$ $\P{column}$ $\P{with}$ $\P{all}$ $\P{but}$ $\P{one}$ $\P{of}$ $\P{a}$ $\P{single}$ $\P{colour}$ $\P{already}$ $\P{filled}$ $\P{in,}$ $\P{which}$ $\P{is}$ $\P{so}$ $\P{far}$ $\P{identical}$ $\P{to}$ $\P{a}$ $\P{full}$ $\P{one.}$

   
   

   If you find this, then you know the final tile of that colour must not be in the same position as in the corresponding full row/column, so fill that cell with the other colour.

   
   

Some notes and comparisons

Not all of the above methods, based on possible configurations of tiles to keep an eye out for, are to be applied in the same way.

• 1 and 2 use only the first of the three game restrictions given in the question; 3 uses only the second; 4 uses the first and second together; 5 and 6 use the second and third together.


• 1 and 2 are the easiest configurations to spot. You should always start with these, filling in as many tiles as possible at the beginning until there are no more instances of 1 or 2 on the board. After each application of any of the other techniques, you should then check again to see if that's introduced any more 1 or 2 configurations.


• Many simpler instances of this game can be solved using only 1, 2, and 3, and 5 in the case of 0hh1. 4 and 6 are more advanced techniques, to be used only when nothing else works.

Demonstrations

First let's see a demonstration of how to spot each of the possible configurations listed above and what to do once you've found one. In each case, I've made at least one gif from each of 0hh1 and Tatham's Unruly, except of course for 5 and 6, which are only valid for 0hh1.

1. 
   

   

   
   



2. 
   

   

   
   


3. 
   

   

   
   


4. 
   

   

   
   


5. 
   
   

   

   
   


6. 
   

   

   
   

And finally, a couple of example games played out in full. I've deliberately used relatively small boards here, so that the game and its analysis don't go on for too long. Homework for the reader: go and try out the same techniques on larger boards :-)

• 
  

  A game of 0hh1: solved by a bunch of 1 and 2, 3 in third column, more 1, 3 in second row, another 1, 3 in fourth row, 5 in fifth column w.r.t. second, more 2, 5 in sixth column w.r.t. third, 3 in fifth and sixth rows, 5 in third row w.r.t. second, 3 in first and fourth columns.

  
  
  

  

  
  


• 
  

  A game of Unruly: solved by a bunch of 1 and 2, 3 in second row, more 1 and 2, 3 in second and seventh columns, more 1 and 2, 3 in third row and sixth column, more 1 and 2, 4 in seventh row, another 1, 3 in sixth row, another 1, 4 in fifth row, 3 in first and third and fourth columns.

  
  

  

  
  

Rubik's cube 4x4 parity moves

^{Note: since asking this question I've purchased several other twisty puzzles and now regret taking the "look up algorithms online" approach to solving - in the end I get much more enjoyment from the puzzles where I worked out to solve them myself than the ones where I didn't. If you're stuck at this point I would recommend seeking resources to understand the concept of parity and then try to come up with a solution yourself. You won't end up with a super efficient algorithm, but you will understand it, which is ultimately more satisfying. Or it is for me at least. That said, answers to this question are still welcome.}

I'm solving my 4x4 Rubik's cube (aka Rubik's revenge) for the first time, and I have the "OLL parity" case:

There are plenty of algorithms for this available online, but different people use different notation and nobody ever says which one they're using, so I risk messing up my cube if I guess wrongly.

So, please give me algorithms for solving the OLL parity, using the following notation:

• R: turn just the rightmost face clockwise


• Rw (R wide): turn the rightmost half of the cube clockwise, i.e. the rightmost face and the slice adjacent to it



• r: turn the rightmost slice clockwise, but not the rightmost face


• R2, Rw2, r2: as above but turning it twice


• R', Rw', r': as above but turning it anticlockwise


• L, F, B, U, D: left, front, back, up, down. (All clockwise by default, can be modified as above)


• x: if you use this, please explain what it means, as I have no clue.


• - (hyphen): if you use this, please explain what it means, as I have no clue.


• (, ) (parentheses): if you use these, please explain what they mean, as I have no clue.

If you use anything else at all in your notation, please explain what it means. If you prefer a different notation from the above that's fine, but please explain it.

I'm holding my cube with the unsolved cubies facing me at the top. (i.e. the red face would be the front in the diagram above.) If I should start in some other orientation, please say so.

It would be helpful to have an algorithm that leaves all other pieces alone, as my cube is in exactly the almost-solved state above. But if algorithms that don't do this are significantly shorter and/or easier to remember then that would be fine too. (Please tell me which I can expect in your answer!)

enigmatic puzzle - Piece de Resistance - Four Dots in a Line

Four - Dots in a Line

_{This puzzle is part of the "Piece de Resistance" series. Go back to Part 1 (Ace) for the story.
Ace Two Three Four Five ...}

Looking for the next challenge, you got rid of the coating, and this time it amazed you, as it was a colour image with a beige background instead of the previous monochrome cards! Things are getting fun...

Aha! And you vaguely found five short lines near the bottom of the card, in red ink...

Hint

Well, you expected 4 lines instead of 5, huh? Maybe look it up in a dictionary? Any other spellings? Oh, and just to say, the lines are something like this:
_ _ _ _ _

Answer

First panel:

The top-left character is Loki, but with the two halves switched, so kilo (love that one). Then we have groin $-$ Óin, which leaves gr. The country is Armenia, so am: KILOGRAM

Second panel:

The logo is the Seal of the U.S. Securities and Exchange Commission, so SEC, on D: SECOND.

Third panel:

The MET, a building in Thailand, followed by the logo of Regal Cinemas minus that of the Galway County Council (GalwayCoco), so regal minus gal: RE. METRE

Fourth panel:

A picture of Americium (symbol Am), "per e": AMPERE.

Now

If you look into the list of derived units (credit to @JProblems in the comments for the idea), you see that the numbers from the axis ($-6, -2, 2, 4$) likely correspond to exponents, and since a Volt is $$1V=1m^2\cdot kg\cdot A^{-1}\cdot s^{-3}$$ it follows that the puzzle links to a VOLT SQUARED, with symbol $V^2$. To be noted, there is no reason given in the puzzle to think in volt squared rather than in another equivalent unit, such as WATT OHM ($1 V^2=1W\cdot \Omega$). One of these interpretations might explain what to do with the five red lines at the bottom of the card.

One way to conclude (unlikely):

The five red lines may stand for a five-line staff, and link to music. There is a Spotify list called Volt Squared.

Another way to conclude (much better):

Perhaps the five lines aren't straight, and actually shaped like "$/W$" (credits to @hdsdv in the comments for this idea). In this case the final answer would be OHM (volt squared per watt). Note that it relates to the title of the series, Piece de Resistance, (since Ohm is the unit of resistance), and that it is consistent with the red ink used to write "per e" in the fourth panel.

A third conclusion given the hint that was added:

The scale of the horizontal axis could be $1/2$ (credits to @AHKieran for this idea following the OP's hint in the comments below) in which case the exponents are half what we thought they were, and the answer (in five letters) is VOLTS. It could also be VOLTA, as Volts were named after Alessandro Volta, or something else given the hint of finding another spelling for the four-letter word VOLT.

Edit: the answer is indeed VOLTE as confirmed by the OP in the comments below, who thought that this was a possible spelling of Volt.

mathematics - Escaping a hungry lion you can't outrun

You are at the edge of an enormous circular arena. A hungry lion is eying you from the centre of this area. You are both capable of running at the same maximum speed, but constraint within the arena. The lion has worked out a strategy of always running at maximum speed in an outward direction such that he stays positioned on the line thru you and the center of the arena.

The starting signal sounds and the lion starts moving. You can't outrun the lion. How do you ensure you stay out of the lion's claws?

Clarifications:

• the arena is truly gigantic, and you can think of you and the lion as point objects constrained within a circle of unit radius


• there is no latency in the lion's reactions to your movements: at any moment in time you, the lion, and the center of the arena remain co-linear


• the lion catches you if, and only if, his position coincides with yours

Hint:

Running at maximum speed along any circular path centered on the center of the arena is not going to help you. Starting from the center, the lion will run at the same speed along a circular path of half the radius and catch you as soon as you have completed a quarter of a circle. (It is easy to check that these paths keep you, the lion and the center co-linear.)

Answer

First, run 1/3 of a unit toward the center of the arena. The radius of the arena is 1 unit and the lion runs at the same speed as you, so the lion will not catch you.

Now, we run along a series of straight line paths. The $n$-th path will be orthogonal to the the line segment connecting your position and the lion's (at the start of the path), and have a length of $\frac{1}{2n}$ units.

These paths will never take you outside the arena. To see this, if you are $d$ units from the center of the arena immediately before running along the $n$-th path, your distance from the center after the $n$-th path will be $$ \sqrt{d^2+\left(\frac{1}{2n}\right)^2}. $$ This means your distance from the center after the $n$-th path is $$ \sqrt{\left(\frac{2}{3}\right)^2+\left[\left(\frac{1}{2}\right)^2+\ldots+\left(\frac{1}{2n}\right)^2\right]}, $$ and this quantity is always less than 1.

Since each straight line path heads in a direction orthogonal to the lion, the lion cannot catch you.

Finally, the total distance you are running is $$ \frac{1}{3}+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\ldots\right)=\infty $$ (this is essentially the fact that the harmonic series diverges), so you can continue to run along the chosen paths indefinitely.

In fact, this strategy enables you to evade the lion no matter how the lion decides to pursue you.

word - I was inspired to create a third (and not final) anagram puzzle

Inspired by @QuantumTwinkie's anagram puzzles.

---

First one to answer all of them correctly will get the tick! Good luck!$\;\color{green}{\checkmark}$

Partial answers are also welcome :)

1. 
   

   The _____ felt the paint on his bike. It was much _____ than the day before, so he finally took it for a spin.

   
   


2. 
   

   "And then, right when mum ____ him into bed, a ____ of wind blows through the window and blankets fly everywhere!" she laughed.

   
   


3. 
   

   Who were the characters ____ring in Ratatouille? Well, obviously, a bunch of ____.

   
   


4. 
   
   

   They were crazy, ______ ___, ready to head off for the adventure, before they all scrapped the idea and replied to the email with one word: "_________".

   
   


5. 
   

   But the ____ ___ that had the _______ cheat sheets, was hidden in Joseph's locker... though nobody said anything because they all got A+ in maths for the first time.

   
   


6. 
   

   Sadly, he never made it to his brother's _______ because the ____ ___ out and he couldn't drive anymore, but he still mourned his death.

   
   


7. 
   

   She found him quite ________ when she saw him ________ in the parade.

   
   
   


8. 
   

   That _____, I crept out of bed thinking that everyone was asleep... until I saw some_____: a pair of yellow eyes staring at me from outside my window.

   
   


9. 
   

   She knew I wouldn't like the movie — it was so scary! (I think it was called "The ____".) But what was even scarier, was when she ____ned at me and turned off the lights...

   
   


10. 
    

    I knew he was an expert at surviving out in the woods — he even knew how to build a ____ that could actually float! But what I didn't know was that he would let out a huge ____ every ten minutes. They smelt so bad, I couldn't sleep in the tent anymore.

    
    



11. 
    

    The wooden planks holding the mattress of the bed ____ped when he jumped on it, just like I predicted. His ____ would now have to take place on the couch.

    
    


12. 
    

    My grandpa calls flight engineers "______" and ______ engineers "seamen". But why do the words have to end with "men"? Women work, too.

    
    


13. 
    

    When my sister noticed that the ________ were each not wearing a name-tag, she thought they were ________. That made me wonder whether or not there exists a person who has no name...

    
    


14. 
    
    

    ____ _______ in famous tales and myths kill dragons, unlike David, who killed Goliath. But why? Dragons don't exist! I bet you there is not one dragon-like creature in the entire _____ ______. With such advanced technology nowadays, we should have found one by now!

    
    


15. 
    

    His ______ parents were found dead in the ______. "Slenderman does exist," he muttered. "Now the question is: where are my real parents?"

    
    


16. 
    

    Her ______s pretended to be proud of her when she got the job of a surgeon. Deep down, they weren't at all — they found the job absolutely disgusting, especially after hearing that her first medical procedure would involve ______ning a person's skull!

    
    


17. 
    

    The character in the game wears these special _____ that give you a _____ to make you jump higher, a bit like Barry in Jetpack Joyride, except there is no jetpack.

    
    
    


18. 
    

    When I looked at her new pair of _______s, I got jealous and started to become _______ at myself. Why couldn't I have those? But then I thought that perhaps if I wore them, people would notice my big ears, which is definitely not what I wanted.

    
    


19. 
    

    _ ____ trees, especially the _____ ones. They always provide a cool shade in a hot day, and they make me think of piece — but I would never hug a tree! Ugh.. yuck. Imagine getting sap or ants all over your fingers.

    
    


20. 
    

    He goes to church every Sunday, but he can't stand the sound of the _____. Every time it starts playing, he would just _____ and cover his ears.

    
    

---

Edits do not change any answers (unless requested); they just improve grammar.

Also, @Eutherpy found a solution to 2, that is not the original. Thus, there are two solutions to that. The same applies to how @Kevorobin found a solution to 14.

Answer

1.

rider / drier (@jafe)

2.

urges / surge

3.

star / rats

4.

driven men / nevermind

5.

real bag / algebra (@Alzinos)

6.

funeral / fuel ran

7.

charming / marching

8.

night / thing

9.

Ring/ Grin(ned) (@Shahriar Mahmud Sajid)

10.

raft / fart :D

11.

snap / naps (@jafe)

12.

airmen / marine (@jafe)

13.

salesmen / nameless (@jafe)

14.OMG

most slayers / solar system

15.

foster / forest (@Shahriar Mahmud Sajid)

16.

parent(s)/trepan(ning) (@Hugh Meyers)

17.

boots / boost (@Shahriar Mahmud Sajid)

18.

earring / angrier

19.

I love / olive (@jafe)

20.

organ / groan

calculation puzzle - Find the longest arithmetic expression with value equal to number of letters in its english representation

The expression twenty minus five has 15 letters (we ignore spaces) and also has the arithmetic value of 15.

You can use standard math constants, operators and functions, represented by english words. Spaces and parenthesis do not count towards the expression length.

Find the longest such expression.

EDIT: As correctly states in https://puzzling.stackexchange.com/a/6304/7403, no such exists, so only bonus now remains.

As a bonus, find the most interesting such expression.

Answer

There is no determinate result to this question.

plus twenty minus two: equals 18, 18 letters

plus twenty minus two plus twenty minus two: equals 36, 36 letters

plus twenty minus two plus twenty minus two plus twenty minus two: equals 54, 54 letters

and so on...

mathematics - A clock where the hour and minute hands are the same length

Your buddy Frankie sold you a shoddy clock: it keeps good time, but the minute and hour hands look exactly the same! Both of these hands move continuously, and there is no second hand.

How many times a day is it impossible to tell what time it is?

Answer

When the hour hand has moved $x$ degrees around the clock from the top, the minute hand has moved $y = 12x$ degrees. If the time is still a valid configuration when the hands are switched around then $x = 12y$ as well.

Therefore, we want the values of $x, y$ that satisfy the following two equations:

\begin{align} 12x &\equiv y \pmod {360} \\ 12y &\equiv x \pmod {360} \end{align}

Conveniently, this reduces to $x \equiv 144x \pmod {360}$ or $143x \equiv 0 \pmod {360}$, so whenever the hour hand moves exactly $x/143$ of the way around the clock where $x$ is an integer, it's impossible to tell which hand is which.

---

There's just one problem, though. The above doesn't take into account the times when the hour and minute hands are in the exact same position, in which case it doesn't matter which hand is which. This occurs whenever $x \equiv 12x \pmod {360}$, or $11x \equiv 0 \pmod {360}$. Naturally this is a total of 11 times, so there are $143 - 11 = 132$ times when the time is actually ambiguous in a 12-hour period, making it $264$ times a day.

geometry - Tiling rectangles with U pentomino plus rectangles

Inspired by Polyomino Z pentomino and rectangle packing into rectangle

Also in this series: Tiling rectangles with F pentomino plus rectangles

Tiling rectangles with N pentomino plus rectangles

Tiling rectangles with T pentomino plus rectangles

Tiling rectangles with V pentomino plus rectangles

Tiling rectangles with W pentomino plus rectangles

Tiling rectangles with X pentomino plus rectangles

The goal is to tile rectangles as small as possible with the U pentomino. Of course this is impossible, so we allow the addition of copies of a rectangle. For each rectangle $a\times b$, find the smallest area larger rectangle that copies of $a\times b$ plus at least one U-pentomino will tile. Example shown, with the $1\times 1$, you can tile a $2\times 3$ as follows:

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

There are at least 6 more solutions. I tagged it 'computer-puzzle' but you can certainly work some of these out by hand. The larger ones might be a bit challenging.

Answer

Here is a way to tile a

6x13 = 78

rectangle with U pentominoes and 1x4 rectangles, which is an improvement over @athin's 9x10 solution:

As a bonus, here are two suboptimal solutions, one of which is asymmetric:

link to two 11x8 = 88 solutions

For 1x5:

12x20 = 240

for 1x6:

14x24 = 336

and for 3x4:

19x40 = 760

cipher - Strange women and their strange ways of giving you phone numbers

While Bob was walking down the street one day, he met a wonderful woman that was attractive and smart and a great conversationalist.

After they had been chatting for a while the woman looked at her watch and quickly got up and scribbled something down while saying "I have to go, I'm running very late. Here's my phone number, if you know my name you can call me, otherwise don't bother."

This is what she wrote on the paper:

1.4- 2-2.2- 1.4- 2.3- 2-2.2- 1.4- 1-4. 2-2.2- 3-2. 2-2.2- 1.4-
4-1. 2-2.2- 1.4- 4.1- 2-2.2- 4.1- 2-2.2- 4-1. 2-2.2- 1.4-
2.3- 5- 2-2.2- 1.4- 2-2.2- 1.4- 4.1- 2-2.2- 2-3. 2-2.2- 1.4- 5.
1.4- 4-1. 2-2.2- 4-1. 2-2.2- 5. 2-2.2- 1.4- 3-2. 2-2.2- 1.4- 3-2. 2-2.2- 1.4-
4-1. 2-2.2- 1.4- 4.1- 2-2.2- 4.1- 2-2.2- 4-1. 2-2.2- 1.4-
2.3- 4.1- 2-2.2- 1.4- 3-2. 2-2.2- 1.4- 2-2.2- 2.3- 5.
1.4- 4-1. 2-2.2- 4-1. 2-2.2- 5. 2-2.2- 1.4- 3-2. 2-2.2- 1.4- 3-2. 2-2.2- 1.4-
5. 2-2.2- 3.2- 2-2.2- 3-2. 2-2.2- 1.4- 5.
2.3- 2.3- 2-2.2- 4-1. 2-2.2- 3.2- 2-2.2- 2.3- 5- 2-2.2- 1.4- 5. 2-2.2- 1.4- 3-2.
5. 2-2.2- 3.2- 2-2.2- 3-2. 2-2.2- 1.4- 5.

1.4- 4.1- 2-2.2- 1.4- 5. 2-2.2- 2.3- 2.3- 2-2.2- 5. 2-2.2- 1.4- 3.2- 2-2.2- 2.3- 2-2.2- 5. 2-2.2- 1.4- 3-2.
1-4. 2-2.2- 1.4- 5. 2-2.2- 2.3- 4.1- 2-2.2- 2.3- 5- 2-2.2- 1.4- 3-2. 2-2.2- 1.4- 5. 2-2.2- 2.3- 5-
4-1. 2-2.2- 1.4- 4.1- 2-2.2- 4.1- 2-2.2- 4-1. 2-2.2- 1.4-
1-4. 2-2.2- 1.4- 5. 2-2.2- 2.3- 4.1- 2-2.2- 2.3- 5- 2-2.2- 1.4- 3-2. 2-2.2- 1.4- 5. 2-2.2- 2.3- 5-
5. 2-2.2- 3.2- 2-2.2- 3-2. 2-2.2- 1.4- 5.
2.3- 5- 2-2.2- 1.4- 2-2.2- 1.4- 4.1- 2-2.2- 2-3. 2-2.2- 1.4- 5.
1.4- 3-2. 2-2.2- 1.4- 5. 2-2.2- 1.4- 3.2- 2-2.2- 5. 2-2.2- 1.4- 5.
5. 2-2.2- 3.2- 2-2.2- 3-2. 2-2.2- 1.4- 5.
5. 2-2.2- 3.2- 2-2.2- 3-2. 2-2.2- 1.4- 5.
2.3- 1-4. 2-2.2- 2.3- 1.4- 2-2.2- 1.4- 2.3- 2-2.2- 2.3- 1.4-

5. 2-2.2- 3.2- 2-2.2- 3-2. 2-2.2- 1.4- 5.
5. 2-2.2- 3.2- 2-2.2- 3-2. 2-2.2- 1.4- 5.
1.4- 3-2. 2-2.2- 1.4- 5. 2-2.2- 1.4- 3.2- 2-2.2- 5. 2-2.2- 1.4- 5.
1.4- 5. 2-2.2- 1.4- 4-1. 2-2.2- 3.2- 2-2.2- 1.4- 2-2.2- 1.4- 3-2.
1.4- 4.1- 2-2.2- 1.4- 5. 2-2.2- 2.3- 2.3- 2-2.2- 5. 2-2.2- 1.4- 3.2- 2-2.2- 2.3- 2-2.2- 5. 2-2.2- 1.4- 3-2.
4-1. 2-2.2- 1.4- 4.1- 2-2.2- 4.1- 2-2.2- 4-1. 2-2.2- 1.4-
1.4- 4.1- 2-2.2- 1.4- 5. 2-2.2- 2.3- 2.3- 2-2.2- 5. 2-2.2- 1.4- 3.2- 2-2.2- 2.3- 2-2.2- 5. 2-2.2- 1.4- 3-2.
5. 2-2.2- 3.2- 2-2.2- 3-2. 2-2.2- 1.4- 5.
1.4- 3-2. 2-2.2- 1.4- 5. 2-2.2- 1.4- 3.2- 2-2.2- 5. 2-2.2- 1.4- 5.

For obvious reasons, Bob was rather confused for a while. He realized that in their entire conversation she had never said her name.

All the same, he began trying to figure out the message she had written him. He was beginning to despair but then as he finished working out what she had written him, he immediately knew her name and called her number the first chance he got.

What was her name?

Answer

Her name is

Jenny

Her phone number is

867-5309

Because

The message is in morse code, the numbers represent the number of .s and -s should be in that place. E.g. 1.4- = .----

The translated message is:

.---- --..-- .---- ..--- --..-- .---- -.... --..-- ---.. --..-- .---- --..-- ----. --..-- .---- ....- --..-- ....- --..-- ----. --..-- .---- --..-- ..--- ----- --..-- .---- --..-- .---- ....- --..-- --... --..-- .---- ..... --..-- .---- ----. --..-- ----. --..-- ..... --..-- .---- ---.. --..-- .---- ---.. --..-- .---- --..-- ----. --..-- .---- ....- --..-- ....- --..-- ----. --..-- .---- --..-- ..--- ....- --..-- .---- ---.. --..-- .---- --..-- ..--- ..... --..-- .---- ----. --..-- ----. --..-- ..... --..-- .---- ---.. --..-- .---- ---.. --..-- .---- --..-- ..... --..-- ...-- --..-- ---.. --..-- .---- ..... --..-- ..--- ..--- --..-- ----. --..-- ...-- --..-- ..--- ----- --..-- .---- ..... --..-- .---- ---.. --..-- ..... --..-- ...-- --..-- ---.. --..-- .---- ..... --..-- .---- ....- --..-- .---- ..... --..-- ..--- ..--- --..-- ..... --..-- .---- ...-- --..-- ..--- --..-- ..... --..-- .---- ---.. --..-- -.... --..-- .---- ..... --..-- ..--- ....- --..-- ..--- ----- --..-- .---- ---.. --..-- .---- ..... --..-- ..--- ----- --..-- ----. --..-- .---- ....- --..-- ....- --..-- ----. --..-- .---- --..-- -.... --..-- .---- ..... --..-- ..--- ....- --..-- ..--- ----- --..-- .---- ---.. --..-- .---- ..... --..-- ..--- ----- --..-- ..... --..-- ...-- --..-- ---.. --..-- .---- ..... --..-- ..--- ----- --..-- .---- --..-- .---- ....- --..-- --... --..-- .---- ..... --..-- .---- ---.. --..-- .---- ..... --..-- .---- ...-- --..-- ..... --..-- .---- ..... --..-- ..... --..-- ...-- --..-- ---.. --..-- .---- ..... --..-- ..... --..-- ...-- --..-- ---.. --..-- .---- ..... --..-- ..--- -.... --..-- ..--- .---- --..-- .---- ..--- --..-- ..--- .---- --..-- ..... --..-- ...-- --..-- ---.. --..-- .---- ..... --..-- ..... --..-- ...-- --..-- ---.. --..-- .---- ..... --..-- .---- ---.. --..-- .---- ..... --..-- .---- ...-- --..-- ..... --..-- .---- ..... --..-- .---- ..... --..-- .---- ----. --..-- ...-- --..-- .---- --..-- .---- ---.. --..-- .---- ....- --..-- .---- ..... --..-- ..--- ..--- --..-- ..... --..-- .---- ...-- --..-- ..--- --..-- ..... --..-- .---- ---.. --..-- ----. --..-- .---- ....- --..-- ....- --..-- ----. --..-- .---- --..-- .---- ....- --..-- .---- ..... --..-- ..--- ..--- --..-- ..... --..-- .---- ...-- --..-- ..--- --..-- ..... --..-- .---- ---.. --..-- ..... --..-- ...-- --..-- ---.. --..-- .---- ..... --..-- .---- ---.. --..-- .---- ..... --..-- .---- ...-- --..-- ..... --..-- .---- .....

Which, converted to plain text is:

1,12,16,8,1,9,14,4,9,1,20,1,14,7,15,19,9,5,18,18,1,9,14,4,9,1,24,18,1,25,19,9,5,18,18,1,5,3,8,15,22,9,3,20,15,18,5,3,8,15,14,15,22,5,13,2,5,18,6,15,24,20,18,15,20,9,14,4,9,1,6,15,24,20,18,15,20,5,3,8,15,20,1,14,7,15,18,15,13,5,15,5,3,8,15,5,3,8,15,26,21,12,21,5,3,8,15,5,3,8,15,18,15,13,5,15,15,19,3,1,18,14,15,22,5,13,2,5,18,9,14,4,9,1,14,15,22,5,13,2,5,18,5,3,8,15,18,15,13,5,15

The result above

is the letter number in the English alphabet. (e.g.: 1 = A, 2 = B, etc.)

Translating this, we get:

ALPHA INDIA TANGO SIERRA INDIA XRAY SIERRA ECHO VICTOR ECHO NOVEMBER FOXTROT INDIA FOXTROT ECHO TANGO ROMEO ECHO ECHO ZULU ECHO ECHO ROMEO OSCAR NOVEMBER INDIA NOVEMBER ECHO ROMEO

These are the

Phonetic spellings of the English alphabet.

Which spells out the following:

AIT SIX SEVEN FIFE TREE ZEERO NINER

Which are the

Phonetic spellings of the numbers.

Giving us her phone number:

867-5309

Which, pop-culture tells us belongs to:

Jenny

grid deduction - Double feature: Hot stuff

_{This puzzle is part 4 of the Double feature series (first part here). The series will continue in "Double feature: In cold blood".}

---

Rules of Heyawake¹

• Shade some cells in the grid.


• Inside each room (rectangle with bolded borders), the number indicates how many shaded cells there are inside that room. If there is no number, the number of shaded cells is unknown.


• Shaded cells cannot be connected horizontally or vertically, but they may touch at a corner.


• Unshaded cells must all be orthogonally connected.


• It is not possible draw a straight horizontal or vertical line on unshaded cells so that the line goes inside more than two rooms.

Across
2. Experiment to abruptly reconstruct curved structure (8)
6. Move silently as serpent's tail creeps forward (5)
7. Maxim watched a horror film (3)

8. They're useful documents – the reverse of junk (4)
9. Noblemen not having all internal organs (4)
11. Reboots to uses of silence in music with creative expression (8)
13. Cook a little fish (3)
14. The man standing before a cross reveals a witch (3)

Down
1. Cards for axe-wielding novelist and his portrayer (5)
2. Mountain with extra charge on top (8)
3. Pronounced footprints leading to grasslands (7)
4. "Special Tacos with Coffee" – a chain of cafés (5)

5. Utter exploit of females (4)
9. Shade is essential to recruitment (4)
10. Record a part of Montenegro's territory (6)
12. Broken hand at first leaves annoyance in the flesh (4)

_{¹ Paraphrased from the original rules on Nikoli.}

Solve both puzzles to answer the question: What is 80 degrees Celsius?

Answer

so apparently 80 degrees Celsius is

a SAUNA'S TEMPERATURE.

Oops! I forgot to put in explanations for the cryptic clue solutions. Here they are:

2a. RESE(-t)+ARCH. 6a. SNAKE with E moved earlier. 7a. triple def. 8a. SPAM<. 9a. EARLS minus (-a)L(-l). 11a. REST+ARTS. 13a. double def. 14a. HE+X.
1d. Jack {Torrance, Nicholson}. 2d. RUSH+MORE. 3d. homophone of STEPS. 4d. TACOS*, "Costa Coffee". 5d. homophone of USE. 9d. substring. 10d. substring. 12d. THORN minus H(-and).

(Thanks to Avi for pointing out in comments that I missed a letter.)

knowledge - Piece de Resistance - King Deusovi, Flag this Blo*dy Post!

King Deusovi, Flag this Blo*dy Post!

_{This puzzle is part of the "Piece de Resistance" series. Go back to Part 1 (Ace) for the story.
Ace Two Three Four Five Six Seven Eight Nine Ten Queen King Jack}

Hmmm, what is this?

P.S.: word tag also applies.

Clarification:

The WHITE border of the image is NOT part of the puzzle.

Hint level 1:

ONLY the WHITE border of the image is NOT part of the puzzle. steganography only applies visually.

Answer

Title

From Title "flag this blo*ody post"
The colors blue and red remind me of two flags
- the Flag of Liechtenstein. The crown within could be referenced by the title word "king"...
- The Signal flag for E (ECHO)

Upper half (blue)

There are some hidden words: Ace ice oar Oman over

They are written upside down from right to left
Searching for a pattern, I tried adding letters forming new words:
Adding R gives Race, Rice, Roar, Roman, Rover

Lower half (red) Rebus:

First red square

- ace of spades - short A
- Elephant
- 3/4 of a cent - cent

-> Remove ent from Elephant -> lpha

Alpha

Second red square

- The land on the map is Tibet
- The lower left image is the logo of Texas Instruments - TI
- Ammeter - An instrument to measure the current (in Ampere)
-> Remove TI from Tibet -> bet
-> add A to bet
beta

Fourth red Square

- A 1915 Ford Model T
- A balloon flying with a child... must be filled with Helium, Chemical Symbol is He
- A Taxi
- An image of the chinese president Xi Jinping
Remove Xi from Taxi -> Ta

T He Ta -> Theta

Fifth red Square (Solved by JS1)

$P = V \times I$ (Formula of Electric power) and $I = \frac{V}{R}$
$=> \frac{V^3}{R^2} = \frac{V^2}{R} \times \frac{V}{R} = P \times I = PI$
Pi

Now back to the third square: (Solved by Stiv)

We have
1. $\alpha$ (Alpha, 1. Letter of the Greek Alphabet)
2. $\beta$ (Beta, 2. Letter of the Greek Alphabet)

3. ?
4. $\theta$ (Theta, 8. Letter of the Greek Alphabet)
5. $\pi$ (Pi, 16. Letter of the Greek Alphabet)
Now, it looks like the position of the letter is doubled in each square so the third should be
$\delta$ (Delta, 4. Letter)

Conclusion (thx to M Oehm & Stiv)

E from Echo-flag
R from the hidden words

D from 3. red square Delta
ERD anagrams to RED, fitting the bloody Hint from the title

no computers - If 6 was 9, or 100 was 64, or M was N

         Now if a 6,
         Turned out to be 9,

         I don’ t mind,
         I don’ t mind   – Jimi Hendrix in If 6 was 9

Cool. Suppose a 100 turned out to be 64.

         I don’ t care,
         I don’ t care   – Jimi Hendrix in If 6 was 9

But, but Jimi, what if your computer program is bugged out and can’t tell if its bad self is coming (from octal) or going (to hex)?

          100₈    =  64₁₀
          100₁₀  =  64₁₆
          100₁₆  =  64₄₂

         Let it be,

         It ain’ t me   – Jimi Hendrix in If 6 was 9

You know, 100₈ stands for the digits 100 in base 8, equaling the familiar 64 (in base 10), which itself shows as 64₁₀.

         Dig    – Jimi Hendrix in If 6 was 9

So, that used 6 different digits — 0, 1, 2, 4, 6, 8 — a total of 26 times and bubbles down to...

         K  _W  =  L _X
         K _X   =  L _Y
         K _Y   =  L _Z
         ...where   K = 100,   L = 64,   W = 8,   X = 10,   Y = 16   and   Z = 42.

         I’ m gonna wave my freak flag high! – Jimi Hendrix in If 6 was 9

       Hold tight, hitch-hiker, check out where simpler ingredients can take you.

Just 4 different digits from 0 through 9, taken fewer than 43 times total, can really wig you out.

         M _P   =  N _Q     ( M > N   and   P < Q < R < S < T )
         M _Q  =  N _R
         M _R  =  N _S
         M _S   =  N _T

     Wild.   What M, N, P, Q, R, S and T could do that with the fewest digits total?

         Even a 43-digits-total solution or two would be worth posting.

         Wave on,
         Wave on . . .
         . . .’ Cause everybody knows what I’ m talking about
          – Jimi Hendrix in If 6 was 9_{(vocal track) dailymotion}

Answer

Well, I have a working set, it seems:

1011 (base 1) = 11 (base 2)
1011 (base 2) = 11 (base 10)

1011 (base 10) = 11 (base 1010)
1011 (base 1010) = 11 (base 1030302010)

using 4 digits (0,1,2 and 3) and the total number of times those numbers used is - 49 times (well, it was above the 43 limit...) and

assuming fractional bases are valid, here is another one:

22 (base 2.25) = 11 (base 5.5)
22 (base 5.5) = 11 (base 12)
22 (base 12) = 11 (base 25)
22 (base 25) = 11 (base 51 )

using 3 digits (1, 2 and 5) and the total number of times those numbers used is 33.

Here is another one:

111 (base 2) = 21 (base 3)
111 (base 3) = 21 (base 6)
111 (base 6) = 21 (base 21)
111 (base 21) = 21 (base 231)

using 4 digits (1, 2, 3 and 6) and the total number of times those numbers used is 32. Hope this satisfies @humn's requirements in all manners.

mathematics - Can the Policeman catch the Thief?

The town of Squareshire has six streets: four sides of a square and the lines joining the midpoints of opposite sides.

$\hskip2in$

A policeman is chasing a thief along these streets. If they are ever on the same street, then the policeman can shoot the thief. The policeman runs slightly faster than twice the thief (say 2.0001 times the speed of the thief). How can he shoot the thief?

Source: Moscow mathematical Olympiad, 1978

Clarification:

The squares between the streets contain houses and walls, so the policeman can't see the thief unless they're on the same street. He has no idea about the exact location and strategy used by the thief.

(Thanks to and Milo Brandt and Falco for the suggestion on clarification)

Answer

Yes, the policeman can catch the thief, although it may take a very long time. In particular, the following strategy works: The policeman starts in the center. First, he makes a counterclockwise loop around the top right quadrant. Then, he makes a counterclockwise loop around the top left quadrant. Then, he makes a counterclockwise loop around the bottom left quadrant. He continues making such loops until he catches the thief, each time moving to the next quadrant in counterclockwise order. The strategy is illustrated in the following picture, where he executes the red arrows, then the green arrows, then the blue arrows, then the yellow arrows, and then repeat the whole process ad nauseam. The diagram below labels the four steps in each counterclockwise loop from $1$ to $4$, which will be useful in discussing the strategy.

In the comments, @f'' points out that this strategy may also be described as follows: The policeman walks around the perimeter counterclockwise. Each time he reaches the midpoint of an edge, he walks into the center and then back out, continuing his traversal of the perimeter from where he left off.

To show that this works, let us first notice that the thief can never safely visit the center of Squareshire. This is because, for her to do so, she would have to run from the perimeter to the center and then back in the time it takes the policeman to make one loop. However, this requires her to run past two edges in the time the policeman runs four, which she cannot do.

In particular, this "disconnects" the center in a way, meaning that, if the thief is not in the center, there is a unique point $P$ on the perimeter where she last left the perimeter, and where she must later enter the perimeter again. This will always be a midpoint of one of the outer edges. An important property is that the thief is no better off leaving the perimeter, then coming back, than she is just standing at point $P$. That is, if she leaves the perimeter at a point $P$, and the policeman then sees that point, she will be surely caught.

To show this, consider what happens during the policeman's first loop around the top left quadrant. If $P$ is the midpoint of the top edge, then the policeman sees it after walking two edges. It continues to be visible to the policeman until he reaches the midpoint of the top edge, at which point he sees everywhere the thief could safely be*. If $P$ is the midpoint of the left edge, then the policeman must have already travelled more than one edge before the policeman enters, and the first time the policeman will see $P$ is when he returns to the center. He will also see the whole edge connecting to $P$, so the thief will die regardless of whether she stayed at $P$, or ventured inwards. The case for the midpoint of the bottom or left edges is similar.

This establishes that, if the thief can survive, she can do so while staying on the perimeter. However, the policeman is "sweeping" the perimeter faster than she can move. In particular, let us label every point of the perimeter by a number** with the numbers increasing in the counterclockwise direction such that the difference between two labels is proportional to the distance between them, as walked on the perimeter. We will consider that $1$ and $0$ represent the same point:

During the policeman's first loop, he is able to see the point $0$, then for a while sees all the points in $[-1/8,1/8]$, then sees all the points in $[1/8,3/8]$. On his way back to the center, he sees the point $1/4$. Then, on the next loop around, he sees the same sequence of points, just shifted by $1/4$ forwards, to signify that he is now sweeping the next quadrant. The significance is that the thief can travel at most some constant amount less than $1/4$ around the perimeter in each of these steps, so the policeman's view eventually catches up. A plot of this is provided below, where the thief's position on the perimeter is a yellow line, running as fast as it can away, and the policeman's view is the shaded region bounded by blue and purple lines. The $x$-axis gives the distance walked by the policeman as a fraction of the perimeter.

The important note here is that the thief's line is limited in slope to some constant less than $1/2$, whereas the policeman's view is moving forwards with an average slope of $1/2$. The thief wishes to avoid her line intersecting the policeman's view, but this is, of course, impossible.

(*Note that the thief could have run across the center so that the policeman doesn't see her until he returns to the center, but we already established that running through the center is a fatal mistake, so we assume the thief doesn't do this.)

(**For mathematicians, we're really considering the perimeter to be $\mathbb R/\mathbb Z$, and then lifting both the policeman's view and the thief's position to $\mathbb R$)

word - Will Riley Riddles ever end?

Maybe, maybe not — but the answer won't.

---

My prefix is the first half of what was before.

My suffix curses Scotland in red and no more.

My infix is a hotshot when scrambled (what egg?).

I am twice my definition; I begin with a beg.

What am I?

---

Hint:

The answer to the riddle can be found in front of you,
Including all the clues except the second one through.

The pun is quite a hint — the answer is in every word!
In the riddle, it is seen; in the beginning, it is heard.

Answer

I think you are

Prefix

My prefix is the first half of what was before.

Literally, pre (The word 'prefix' come before in the sentence and half of it is 'pre' and pre = before)

My suffix curses Scotland in red and no more.

Curse of Scotland = 9 of Diamond. 9 in Roman numerals is ix

My infix is a hotshot when scrambled (what egg?).

When you scramble refi, you get 'fire' which is literally hot.

I am twice my definition; I begin with a beg.

Prefix of begin is beg. Hence, begin begins with a beg.

mathematics - Welcome to Oɴᴇderland

                                                    1 1
                                                  1 1 1 1
                                                1 1 1 1
                                1             1 1 1 1
                                1           1 1 1 1
                                1 1 1     1 1 1 1
                                1 1 1 1 1 1 1 1
                                1 1 1 1 1 1 1
                                1 1 1 1 1 1
                                1 1 1 1 1 1 1 1

                                1 1 1 1 1 1 1 1 1 1

         #1.   You are here

            1111111111111111111111111111111
            111                         111
            111   1111111111111111111   111111111111111111111111111
            111   111             111   111                     111
            111   111   1111111   111   111   111111111111111   111

            111   111       111   111   111   111         111   111
            111   1111111   111   111   111   111   111   111   111
            111             111   111   111         111   111   111
            1111111111111111111   111   111111111111111   111   111
                                  111                     111   111
                      111111111111111111111111111111111111111   111
                      111                                       111
                      111   111111111111111111111111111111111111111
                      111   111
                      111   111   111111111111111

                      111   111   111         111
                      111   111   111   111   111
                      111   111         111   111
                      111   111111111111111   111
                      111                     111
                      111111111111111111111111111

         #2.   Through the labyrinth we wind

                           1ooo11ooo1
                       111ooooo1   1oo111o
                       1 oo1111   1      1
                       o1   o1   1        1
                      o1    1   1          1
                       o1   1   1           1
                        1    1   1          1
                         1    1   11ooo1oo1o111oo
                          1ooo11o1     oooo1
                          o1 1111  ooo1111

                         1111    1111111
                         1          1111
                                     1111
                                       o1
                                       o1
                                o1111   o1
                             oo1     o1 o1
                            o1         1111
                           o1      ooo1oo111o
                           1   ooo1      o1

                           1111           1
                            1

         #3.   A rose is a rose is a rose

                                                1
                                               1 1
                                              1o1o1
                                             1 1 1 1

                                    1       1o1o1o1o1       1
                                   1 1     1 1     1 1     1 1
                                  1o1o1   1o1o1   1o1o1   1o1o1
                                 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
                                1o1o1o1o1o1o1o1o1o1o1o1o1o1o1o1o1
                               1 1     1 1             1 1     1 1
                              1o1o1   1o1o1           1o1o1   1o1o1
                             1 1 1 1 1 1 1 1         1 1 1 1 1 1 1 1
                            1o1o1o1o1o1o1o1o1       1o1o1o1o1o1o1o1o1
                           1 1             1 1     1 1     1 1

                          1o1o1           1o1o1   1o1o1   1o1o1
                         1 1 1 1         1 1 1 1 1 1 1 1 1 1 1 1
                        1o1o1o1o1       1o1o1o1o1o1o1o1o1o1o1o1o1
                       1 1     1 1     1 1     1 1
                      1o1o1   1o1o1   1o1o1   1o1o1
                     1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
                    1o1o1o1o1o1o1o1o1o1o1o1o1o1o1o1o1

         #4.   $\sf \small O \scriptsize NE$ders of the ancient world

Obviously (?) the $\sf\scriptsize ZERO$-and$\small/$or-$\sf\scriptsize ONE$ders here represent a pattern of mathematical constructs.

                                            o?1?o?1?
                                           1?      o?
                                                   1?
                                                  o?
                                                1?
                                               o?

                                               1?

         #5.   What picture could be fifth, but not at other #s here?     Why?

                                                               1o111111ooooo
                                                         oooooo1o11111ooooo
                                                   oooooooooooo1o1111ooooo
                                             oooooooooooooooooo1o111ooooo
                                       oooooooooooooooooooooooo1o11ooooo
                                 oooooooooooooooooooooooooooooo1o1ooooo

                           ooooooooooooooooooooooooooooooooooo11o1oooo
                     oooooooooooooooooooooooooooooooooooooooo111o1ooo
               ooooooooooooooooooooooooooooooooooooooooooooo1111o1oo
         oooooooooooooooooooooooooooooooooooooooooooooooooo11111o1o
   ooooooooooooooooooooooooooooooooooooooooooooooooooooooo111111o1

         #6.   One for the road

The answer can be pictured in infinitely many ways. The ?-shaped placeholder presently at #5 is meant to be replaced. Only numbers composed of o zeros and$\small/$or 1 ones are pertinent. Two-dimensional shapes and surrounding words are just gratuitous embellishments. If you’re getting nowhere after considering all this, why not actually go nowhere, to $\sf \small N \scriptsize ONE$derland, for comparison?

Answer

In the $n$th picture, the strings are all

representations of the number $1$ where the base is one of the primitive $n$th roots of unity. (That is, solutions to $z^n=1$ that don't work for any smaller $n$.)

So any picture containing

111111

would fit in slot 5, but not any other.