logical deduction - Infinitely long loaf of Bread

You are visiting your old friend Mike at Infinitely's Baking Shop. Just as you arrived, he was taking out a fresh, infinitely long loaf of bread. Both of its ends extend infinitely long in a straight line. The smell of the breads displayed on the infinitely long shelves feels so good.

But you know that Jesse, your roommate, only eats the crusts of the bread (the ends of the bread). If you ask for the freshly baked bread, you'll grab it by its center, and since it's infinite in both directions, no finite speeds can bring you to either of its ends, and you know you can't travel faster than light anyway.

If you grab one of the loaves of bread from the display, you can grab it by one of its ends and thus cut a single crust piece, but Jesse needs two, or he'll stay hungry.

You might consider buying two loaves of bread from the display to solve this dilemma, but you can afford only one loaf and can't obtain more money for today by any means. Nor can you obtain more loaves by any other means. Mike is also not very giving, to just give you a loaf without receiving the money equaling its total cost.

After standing there shortly, you came up with an idea. At the end of the day, you walked out with two crusts in one hand, and the rest of the infinitely long bread in your infinitely long bag carried in your other hand. How did you manage to do it?

The picture represents a rough estimate of how a loaf of bread would look like if it was finite in length. (Mike only sells infinite breads)

The intended solution, in my opinion, is quite nice. I'm not sure if there are any other tricky means to solve the problem, but go for it.

---

Edit: The accepted answer, in my opinion, is simpler and better than the intended solution, and it can lead to a "problem" regarding infinities.

Intended solution hint:

The "problem" could be represented as a disagreement between yourself and Mike, which can take place some moments after Mike already agreed to your proposal.

Added the "open-ended" tag because it is unknown how many ways the infinities of the loaves can be interpreted.

Answer

You can ask for the freshly-baked bread, grabbing it by its center and

asking that it be packed into the bag with the bread folded at the point you grabbed it, and with that point going into the bag first. The two ends would be within easy reach of the bag's opening, so you just cut them off, hold the bag in one hand and the crusts in the other.

cipher - Michel's cryptic kana

My friend Michel is at it again. Today I found this email in my inbox:

Subject: おいじゃくしゃあむう いみしゃあ りょん ぎをえしゃきょお

てば おんない あきわんのんこお しいしゅう ぞびいけえ るう りゃいじい るう りょん わぐ えきゃ。お うてにょいつ うた ごりゅえや ごんをいれんや ううせえにょうさいじゃ なえ びわびきゃが おえないよお やきょお しゅがやえ きょざよん せ しいしゅう あいが。うにをえきゅう えにゅ おすう ああぬ るう ええこいにをいる るう お うてにょいつ。いいろお せ あえにょおま い あえこえ しいしゃあむう よだ ぴょ いきょお ぴゃ ううぴをいわひをいくう。しゅ えいねづう よだ ばいじゃ おいしをんひゃじょぎい せ えきゃ おえて いえきをぷう びわびきゃが あえねえ がいじゃ えんにゅ あんろんひゃ しゅ おえないよお ぎをねい い えいろんばじゅうなあ。じいかじゅう。

A quick check of Google Translate indicated he was either wishing me and my family a happy new year, or telling me that my nephew is a weak Amu cormorant. Anyway, I'm pretty sure he doesn't speak Japanese, so I'd guess he's using the kana to convey a hidden message.

I emailed back requesting a further sample, and got this:

ぴいる おきゃぎゃえれんや うぴちょおや お おうじょこお げいるえぼきょい うた りわみほきゃ ちをぱあ なあ うた うれんや。お ああくんけえ じゃろお がいじい ええにをんひょぎい しゃあ りょん よぎゃ。ぴいる しゅう しゅ いぎえこお ぎいけえをじえ てやんくう。

おえとん しいしゅう いじゅしょうや お ぎしょお べしゃあそきょい る しゅ ぽしい りょん こいむうべなあ るう お ううる ちをぱあ せ りょん えんじゃりえぎわぐ。お ううか しいる おんない ちみゃ じゃろお しゅ ううしゃあざ おんない ちみゃ りわじわえこお しゅ。お ちをきゅ ちゃきょん しいる しゅ しゅう ううりわじゅよん おえじょすかあ。しゅ おじをえしゅるくう りゃ しいないみい。しゅ しゅう あみょちゃきゅよん うた しいけじょん なえ げりゃ れえりょうちしをえ えいないきょいけえ おえて あみをゆんくう いをい うたが きゃる よすや びじわちをずうくう しゃあ うた ううこうやいごなあ なえ いいしゅう えぱあ ううさいちょん。しゅ しゅう ああぬ うえじわずしょうれい。お ぎいないよお よだ せ ううか いいしゃん おうわぶえこお ちをぱあ しゃあ い しいしをえちょうわぐう じじをえしわま なあ うた あいきょおけえめえないきょお いきょお いぎえこお せ もみゃ うた せえろおくう なえ いをい いいしゅう ぎゃをいのえせえるんけいやじゅう。お おえないよお えいろん い しいないぎきょお せ えきゃ いみしゃあぐ いいしゃん。

おんない おえないよお をが ぴをにをえ あんなあこん てやんくう じゃりわじわえこお じやんよん。るう あえねえ うた いじゅしょうや お おえじをさ しゅ りょんがやえ。

If that's not enough, I've got plenty more where that came from. ;)

-M

What is the encryption method, and what do the messages say?

---

Update:

D: Is a knowledge of the Japanese language needed to solve the puzzle?
M: Grammar and vocab? No.
M: Phonology and spelling conventions? Maaaybe. ^_^

classical mechanics - Every Galilean transformation can be written as the composition of rotation, translation, and uniform motion

Having heard many good things about Arnold's Mathematical Methods of Classical Mechanics, I picked it up and started going through it. While I think I understand all of the definitions he makes, the problems (at least at the beginning) are very mathematical and my proof skills aren't all that great. Can someone show me how to prove this one for instance?

Problem: Show that every Galilean transformation of the space $\Bbb R \times \Bbb R^3$ can be written in a unique way as the composition of a rotation, a translation, and a uniform motion (thus the dimension of the Galilean group is equal to $3+4+3=10$).

Here are some of the relevant definitions:

Definition: Galilean space: An affine space $A^4$ equipped with a Galilean space-time structure.

Definition: Galilean structure: The following properties:

1. parallel displacements of $A^4$ constitute a vector space $\Bbb R^4$.



2. The linear mapping $t:\Bbb R^4 \to \Bbb R$ from the vector space of parallel displacements of $A^4$ to the real axis is given by $t(b-a)$ for $a,b \in A^4$ and is called the time interval. If $t(b-a)=0$ then the events $a$ and $b$ are called simultaneous.


3. The distance between simultaneous events $$d(a,b) = \|a-b\| = \sqrt{(a-b,a-b)} \tag{a,b \in A^3}$$ is given by a scalar product on the space $\Bbb R^3$. This distance makes every space of simultaneous events into a $3$d Euclidean space.

Definition: Galilean Group: The group of all transformations of a Galilean space which preserve its structure (preserves intervals of time and distance between simultaneous events).

Defintion: Galilean transformation: An element of the Galilean group.

---

Two addition questions for others who have read Arnold's book: $1)$ should I try to learn some more abstract math (like maybe abstract algebra) before attempting to read this book? I have already taken all of the math requirements I have as an undergrad -- multivariable calculus, ODEs, PDEs, linear algebra, and a math methods course. $2)$ Is there a solutions manual available for this book?

Answer

Let $\phi:\mathbb{R}^3\times\mathbb{R}\rightarrow\mathbb{R}^3\times\mathbb{R}$ be a Galilean transformation. Let $R\in\mathrm{O}(3)$, $\tau\in\mathbb{R}$ and $\mathbf{v},\mathbf{y}\in\mathbb{R}^3$. The goal of this exercise is to show that $$\phi: \begin{pmatrix} \mathbf{x} \\ t \end{pmatrix}\mapsto \begin{pmatrix} R & \mathbf{v} \\ 0 & 1 \end{pmatrix}\begin{pmatrix} \mathbf{x} \\ t \end{pmatrix}+\begin{pmatrix} \mathbf{y} \\ \tau \end{pmatrix} \tag1 $$ and that this is unique. We first note that the rotation group in $\mathbb{R}^3$ is three dimensional. We then have the boost and the spatial translation, which gives $2\cdot 3=6$. Plus the time translation, we find that the Galilean group is $10$-dimensional.

The most general $\mathbb{R}$-affine map is $$\phi(\mathbf{x},t)=A(\mathbf{x},t)+(\mathbf{y},\tau)$$ Let us write this out in "components": $$A(\mathbf{x},t)=(A_{11}\mathbf{x}+A_{12}t,A_{21}\mathbf{x}+A_{22}t)$$ In order that lengths be preserved, we require $A_{11}$ be an orthogonal matrix. Differences between times of events should also be preserved. Suppose we have two events $x_1$ and $x_2$. This condition means that $$A_{22}(t_2-t_1)+A_{21}(\mathbf{x}_2-\mathbf{x}_1)=t_2-t_1$$ For this to hold at all $\mathbf{x}_1$ and $\mathbf{x}_2$, we require $A_{21}=0$ which implies $A_{22}=1$. We identify the remaining component of $A$ as the linear map $\mathbb{R}\rightarrow\mathbb{R}^3$, which is the velocity.

We have thus shown that the most general Galilean transformation is of the form $(1)$.

For more information and more detailed derivations, see this website and these lecture notes.

riddle - Bored with the Internet

^{An entry in Fortnightly Topic Challenge #35: Restricted Title 1.}

---

You, yeah you! Are you bored with the internet? Well here are some science/math themed jokes for you!

1. 
   

   What is the angriest(not negative) process/operation in math?

   
   


2. 
   

   What bad grade did larry receive in is programming class?

   
   



3. 
   

   What do aliens call a mysterious and grim occurrence in space?

   
   


4. 
   

   What do you call a zany, quantum particle?

   
   

Answer

1. 
   

   
   

   $\times$, because it's cross.

   
   

   
   


2. 
   

   
   

   C++?

   
   

   
   



3. Thanks to @noedne:
   

   
   

   Dark matter?

   
   

   
   


4. 
   

   
   

   A strange quark.

   
   

   
   

mathematics - Peaceable Bishops on a 10x10 grid version 2

Can you place 42 bishops with 6 bishops for each of the 7 colors on a 10x10 grid, such that no two bishops of different colors attack each other?

Here are some similar questions:

Peaceable Bishops on an 8x8 grid

Peaceable Bishops on an 10x10 grid

Answer

This was probably not the most exciting question. Anyway there are many solutions. Here are some examples:

723...5617
2.23456..1
32.......6

.3......65
.4......5.
.5......4.
.6.......4
.1.....7.3
171654...2
71....4327

512...3.65
1.124.7.56

21....6...
.2.....673
.4........
.3......4.
37......24
7.........
6..7342.51
5673..4.15

12.3..4671

21.5346..7
.........6
35......64
.3......4.
..........
467....253
...7..2..5
71.6..5..2
17.4..3521

Interestingly one cannot add a single other bishop of any color.

Is angular momentum conserved if you move off at a Tangent?

Lets imagine a binary system of two astronauts in space connected to one another via light rope.

The rope is taut and they're spinning round and round with their axis of rotation being the the axis perpendicular to the their centre of mass.

Now, my question is this. Lets say they each let go of the rope; they move off at tangents.

Is angular momentum conserved ? And what is their subsequent motion ?

My hypothesis : They will move off at constant velocity at a tangent to their circular motion. I believe that because they now under the effects of no external force by the rotational analogue of newton's second law (dL/dt = T) their angular momentum will be of equal magnitude but opposite sign.

Does this make sense ?

Answer

That's correct; there's no change to the angular momentum here with respect to the center of mass. Remember that $L = r \times p$. If $p$ is constant (as it is once they both let go of the rope), then you can see that $r_\perp$ is constant also, and $r_\parallel \times p$ is guaranteed to be zero.

mathematics - Pirates and gold coins

A group of N pirates has come by a chest containing 200 gold coins. Their rules require that the coins be distributed by the following approach. The pirates are ranked from fiercest to meekest and all pirates know the ranking. The fiercest must propose a division, which is put to majority vote with the fiercest voting and winning ties. If the proposal wins, it is implemented. If not, the fiercest is fed to the sharks and the responsibility falls to the (newly promoted) fiercest. The pirates are selfish and quite rational, with the following priorities:

1. Stay alive


2. Get as much gold as possible, as long as 1 is satisfied


3. If 1 is satisfied and two options are equivalent in terms of gold, feed somebody to the sharks by preference.

Depending on N, what happens? To be complete, your answer should deal with large values of N. Source: I first saw it in a Martin Gardner column.

Answer

Let's put the number of coins at $c$ and see what happens if we increase $n$. We'll number the pirates from meekest ($1$) to fiercest ($n$).

$\begin{array}{rrrll} \begin{array}{c}\text{Nr. of pirates}\end{array} & \begin{array}{c}\text{Pirate} \\ \text{meekest to fiercest}\end{array} & \begin{array}{c}\text{Nr. of coins}\end{array} & \begin{array}{c}\text{Vote}\end{array} & \\ \hline 1 & 1 & c & \text{yes} \\ \\ 2 & 1 & 0 & \text{no} \\ & 2 & c & \text{yes} & \text{breaks tie} \\ \\ 3 & 1 & 1 & \text{yes} \\ & 2 & 0 & \text{no} \\ & 3 & c-1 & \text{yes} \\ \\ 4 & 1 & 0 & \text{no} \\ & 2 & 1 & \text{yes} \\ & 3 & 0 & \text{no} \\ & 4 & c-1 & \text{yes} & \text{breaks tie} \\ \\ 5 & 1 & 1 & \text{yes} \\ & 2 & 0 & \text{no} \\ & 3 & 1 & \text{yes} \\ & 4 & 0 & \text{no} \\ & 5 & c-2 & \text{yes} \\ \\ \text{...} \\ \\ 2c & 1 & 0 & \text{no} \\ & 2 & 1 & \text{yes} \\ & 3 & 0 & \text{no} \\ & \text{...} \\ & n & 1 & \text{yes} & \text{breaks tie} \\ \end{array}$

If we define a solution as stable if nobody gets fed to the sharks, we can say that all solutions so far have been stable. This is because for each $n$, there are more or at least as much (in which case the tie is broken) pirates who would be worse off in the solution for $n-1$.

But now, we're at $n = 2c + 1$. Are we getting ready for carnage? Not quite:

$\begin{array}{rrrll} \begin{array}{c}\text{Nr. of pirates}\end{array} & \begin{array}{c}\text{Pirate} \\ \text{meekest to fiercest}\end{array} & \begin{array}{c}\text{Nr. of coins}\end{array} & \begin{array}{c}\text{Vote}\end{array} & \\ \hline 2c+1 & 1 & 1 & \text{yes} \\ & 2 & 0 & \text{no} \\ & 3 & 1 & \text{yes} \\ & \text{...} \\ & n-1 & 0 & \text{no} \\ & n & 0 & \text{yes} \\ \end{array}$

Here, the fiercest pirate can escape alive, but without gold.

Okay, but now surely the sharks are getting fed? Let's see.

$\begin{array}{rrrll} \begin{array}{c}\text{Nr. of pirates}\end{array} & \begin{array}{c}\text{Pirate} \\ \text{meekest to fiercest}\end{array} & \begin{array}{c}\text{Nr. of coins}\end{array} & \begin{array}{c}\text{Vote}\end{array} & \\ \hline 2c+2 & 1 & 0 & \text{no} \\ & 2 & 1 & \text{yes} \\ & 3 & 0 & \text{no} \\ & \text{...} \\ & n-2 & 1 & \text{yes} \\ & n-1 & 0 & \text{no} \\ & n & 0 & \text{yes} & \text{breaks tie} \\ \end{array}$

The fiercest pirate now broke the tie to stay alive.

How about $n = 2c + 3$ then?

$\begin{array}{rrrll} \begin{array}{c}\text{Nr. of pirates}\end{array} & \begin{array}{c}\text{Pirate} \\ \text{meekest to fiercest}\end{array} & \begin{array}{c}\text{Nr. of coins}\end{array} & \begin{array}{c}\text{Vote}\end{array} & \\ \hline 2c+3 & 1 & 1 & \text{yes} \\ & 2 & 0 & \text{no} \\ & 3 & 1 & \text{yes} \\ & \text{...} \\ & n-3 & 0 & \text{no} \\ & n-2 & 0 & \text{no} \\ & n-1 & 0 & \text{no} \\ & n & 0 & \text{yes} & \text{shark bait} \\ \end{array}$

Here, finally, at $n = 2c + 3$, we reach our first unstable solution. The fiercest pirate can buy $c$ votes, adds his own vote for a total of $c+1$, but the opposition has $c + 2$ votes.
Note that here, pirate $n-3$ is pirate $n-2$ from the solution before it. Voting $\text{no}$ yields a coin. Pirates $n-2$ and $n-1$ get no gold by voting against this proposal, but as per the third rule, vote to feed the sharks.^‡

But does that mean that from now on, all pirates get their feet wet? Not at all.

$\begin{array}{rrrll} \begin{array}{c}\text{Nr. of pirates}\end{array} & \begin{array}{c}\text{Pirate} \\ \text{meekest to fiercest}\end{array} & \begin{array}{c}\text{Nr. of coins}\end{array} & \begin{array}{c}\text{Vote}\end{array} & \\ \hline 2c+4 & 1 & 1 & \text{yes} \\ & 2 & 0 & \text{no} \\ & 3 & 1 & \text{yes} \\ & \text{...} \\ & n-4 & 0 & \text{no} \\ & n-3 & 0 & \text{no} \\ & n-2 & 0 & \text{no} \\ & n-1 & 0 & \text{yes} & \text{potential shark bait} \\ & n & 0 & \text{yes} & \text{breaks tie} \\ \end{array}$

Now both the fiercest pirate and the almost-as-fierce pirate are voting for their lives, since the solution with one pirate less is unstable and will see the almost-as-fierce pirate being fed to the sharks. The fiercest pirate breaks the tie and this solution is stable.

$\begin{array}{rrrll} \begin{array}{c}\text{Nr. of pirates}\end{array} & \begin{array}{c}\text{Pirate} \\ \text{meekest to fiercest}\end{array} & \begin{array}{c}\text{Nr. of coins}\end{array} & \begin{array}{c}\text{Vote}\end{array} & \\ \hline 2c+5 & 1 & 0 & \text{no} \\ & 2 & 1 & \text{yes} \\ & 3 & 0 & \text{no} \\ & \text{...} \\ & n-5 & 1 & \text{yes} \\ & n-4 & 0 & \text{no} \\ & n-3 & 0 & \text{no} \\ & n-2 & 0 & \text{no} \\ & n-1 & 0 & \text{no} \\ & n & 0 & \text{yes} & \text{shark bait} \\ \\ 2c+6 & 1 & 0 & \text{no} \\ & 2 & 1 & \text{yes} \\ & 3 & 0 & \text{no} \\ & \text{...} \\ & n-6 & 1 & \text{yes} \\ & n-5 & 0 & \text{no} \\ & n-4 & 0 & \text{no} \\ & n-3 & 0 & \text{no} \\ & n-2 & 0 & \text{no} \\ & n-1 & 0 & \text{yes} \\ & n & 0 & \text{yes} & \text{shark bait} \\ \\ 2c+7 & 1 & 0 & \text{no} \\ & 2 & 1 & \text{yes} \\ & 3 & 0 & \text{no} \\ & \text{...} \\ & n-7 & 1 & \text{yes} \\ & n-6 & 0 & \text{no} \\ & n-5 & 0 & \text{no} \\ & n-4 & 0 & \text{no} \\ & n-3 & 0 & \text{no} \\ & n-2 & 0 & \text{yes} \\ & n-1 & 0 & \text{yes} \\ & n & 0 & \text{yes} & \text{shark bait} \\ \\ 2c+8 & 1 & 0 & \text{no} \\ & 2 & 1 & \text{yes} \\ & 3 & 0 & \text{no} \\ & \text{...} \\ & n-8 & 1 & \text{yes} \\ & n-7 & 0 & \text{no} \\ & n-6 & 0 & \text{no} \\ & n-5 & 0 & \text{no} \\ & n-4 & 0 & \text{no} \\ & n-3 & 0 & \text{yes} & \text{potential shark bait} \\ & n-2 & 0 & \text{yes} & \text{potential shark bait} \\ & n-1 & 0 & \text{yes} & \text{potential shark bait} \\ & n & 0 & \text{yes} & \text{breaks tie} \\ \end{array}$

And we've found a stable one again.

$\begin{array}{rrrll} \begin{array}{c}\text{Nr. of pirates}\end{array} & \begin{array}{c}\text{Pirate} \\ \text{meekest to fiercest}\end{array} & \begin{array}{c}\text{Nr. of coins}\end{array} & \begin{array}{c}\text{Vote}\end{array} & \\ \hline 2c+9 & 1 & 1 & \text{yes} \\ & 2 & 0 & \text{no} \\ & 3 & 1 & \text{yes} \\ & \text{...} \\ & n-9 & 0 & \text{no} \\ & n-8 & 0 & \text{no} \\ & n-7 & 0 & \text{no} \\ & n-6 & 0 & \text{no} \\ & n-5 & 0 & \text{no} \\ & n-4 & 0 & \text{no} \\ & n-3 & 0 & \text{no} \\ & n-2 & 0 & \text{no} \\ & n-1 & 0 & \text{no} \\ & n & 0 & \text{yes} & \text{shark bait} \\ \end{array}$

And so on.

When $n > 2c$, we'll see more and more large runs of downvoting pirates, until the number of pirates upvoting to save their lives becomes equal to the number of downvoters and we find a stable solution again.

---

This yields the following formula for stable solutions: $$ n \leq 2c \lor n = 2c + 2^z\,|\,z \in \mathbb Z_{\ge 0}$$

The procedure by which to distribute the coins is based on the fact that a pirate can only buy votes if coins are given to the opposite pirates from the next stable solution.
It is as follows:

• For $n \leq 2c$ give a coin to every pirate with the same parity as $n$, with any left over coins going to the fiercest pirate.


• For $2c + 2^{z-1} < n \leq 2c + 2^z\,|\,z \in \mathbb Z_{\ge 0}$ start with the meekest pirate with the opposite parity of $z$ and move up in ferocity.^†

---

^{‡: I prefer to think of the pirates as kind hearted people who care for the sharks and don't want them to go hungry.
†: It can be argued that if a solution is known to be unstable, all other pirates will vote no, since it doesn't matter who gets the coins — they won't get to keep them. But for the stable solution, it is necessary to distribute the coins like this.}