Anmeldelse af “Nej, Alphonse”, af Jesper Goll. Novelle. 2024. Proxima 109.

Skitse: Alphonse har en luftballon, der kan opvarmes med gas. Når han gør sig umulig ved hoffet i Paris, så rejser Madeleine med vestpå. Når ballonen går i stykker, så reparerer hun den, og sukker kun lidt.

Er det science fiction? Ja. Sådan på Niels Klim-måden. Videnskaben er lidt løs i det. Alternate history-agtigt.

Temaer: Alphonse har forelsket sig i, at man kan flyve vestpå, forbi kysten, ud at se hvad der er. Gæt hvem der er skeptisk.

En håndfuld fejl i opsætning og vist et par stavefejl.

Er det godt? Sjovt. 👽👽💀

This week the question is: Where Will the Sorting Hat Put You?

You are waiting in line to be sorted into one of the four houses of Logwarts (a posh wizarding boarding school in the Scottish highlands) by an anthropomorphic sorting hat. The hat is a bit of a snob about the whole matter, and refuses to sort two students in a row into the same house. If a student requests a certain house, but the previously sorted student was already sorted into that same house, then the hat chooses randomly from among the three remaining houses. Otherwise, the hat grants the student’s request.

You are standing 10th in line, and you make plans to request Graphindor house for yourself. As for the other students in line, you can assume that they have random preferences from among the four houses.

The first student steps up, and has a brief, quiet conversation with the hat. After a few moments, the hat proclaims, “Graphindor!”

At this point, what is the probability that you will be sorted into Graphindor?

Highlight to reveal (possibly incorrect) solution:

Spreadsheet.

When a student isn’t sorted into Graphindor, they might end up in Hafflepuff instead. For symmetry reasons, the 3 non Graphindor houses behave the same way, ending up with the same probabilities, so I’m just going to track Graphindor and Hafflepuff.

Student #1 got G.

Student #2 may want G (p=1/4) or something else, like H (p=1/4). If G was requested, H might be granted (p=1/3). The total probability for H is therefore 1/4 + (1/4)*(1/3) = 1/3. The probability for getting G is 0.

This holds in general. A random student wanting something random will end up in a random house, p=1/3, where the previously granted house has p=0.

Student #3 may want H (p=1/4) or something else. The probability this student ends up in G is p(H) from the previous step, or rather p(H) * 3/3. The probability this student ends up in H is (p(G)+…)/3. Therefore p(G) = 1/3 = 3/9 and p(H) = 2/9.

This holds in general. If at a certain step, p(G) = a/c and p(H) = b/c, in the next step, p(G) = 3b/3c and p(H) = (a+2b)/3c.

The general formula for step odd n+1 is p(G) = [3ⁿ/4, rounded down]/3ⁿ and p(H) = [3ⁿ/4, rounded up]/3ⁿ. For even n+1, reverse rounding up and down. Values verified via spreadsheet.

In step 10, I can get G, if G wasn’t granted in the previous step. The probability of this is (6561-1641)/6561 = 4920/6561 = 0.749886… or approximately 1/4.

This week the question is: What’s the Area of a “Pseudo-Square”?

Here’s an image:

Sure enough, the shape (or one very much like it) has four “sides” of equal length, with four right angles. However, two of these sides are curved (in particular, they are arcs of circles), and two of the right angles are exterior, meaning the interior angles measure 270 degrees (rather than the usual 90 degrees).

Let’s call shapes like this one “pseudo-squares.” A pseudo-square has the following properties:

• It is a simple, closed curve.
• It has four sides, all the same length.
• Each side is either a straight line segment or the arc of a circle.
• The four sides are joined at four corners, with each corner having an internal angle of 90 degrees or 270 degrees.

The pseudo-square pictured above has two straight sides, which run radially between arcs of two concentric circles.

Assuming this is a unit pseudo-square (i.e., each side has length 1), what is its area?

Highlight to reveal (possibly incorrect) solution:

Diagram. Calculations.

I’ve built a larger diagram with 2 circles and 3 variables.

• The radius of the small circle is x.
• The circumference of the small circle is 1 + y.
• The radius of the large circle is 1 + x.
• The circumference of the large circle is 1 + z.
• The 2 straight lines in the original drawing lie on top of radii of the large circle.
• Circumference of small circle: 2πx = 1 + y.
• Circumference of large circle: 2π(1+x) = 1 + z.
• Ratio between circumference and smaller arc: y / (1+y) = 1 / (1+z).
• This is a system of 3 equations with 3 unknowns. Calculations tell me, that x = (1-π+√(1+π²))/2π (about 0.184), y = √(1+π²) – π (about 0.155), z = π + √(1+π²) (about 6.439).
• The requested area is small circle + large circle sector – small circle sector. (Avoiding double counting.)
• Small circle area: πx².
• Large circle sector: π(1+x)² * 1/(1+z).
• Small circle sector: πx² * 1/(1+z).
• Further calculations find this area is about 0.684.

This week the question is: Can You Solve the Tricky Mathematical Treat?

It’s Halloween time! While trick-or-treating, you encounter a mysterious house in your neighborhood.

You ring the doorbell, and someone dressed as a mathematician answers. (What does a “mathematician” costume look like? Look in the mirror!) They present you with a giant bag from which to pick candy, and inform you that the bag contains exactly three peanut butter cups (your favorite!) [no, it isn’t!], while the rest are individual kernels of candy corn (not your favorite!).

You have absolutely no idea how much candy corn is in the bag—any whole number of kernels (including zero) seems equally possible in this monstrous bag.

You reach in and pull out a candy at random (that is, each piece of candy is equally likely to be picked, whether it’s a peanut butter cup or a kernel of candy corn). You remove your hand from the bag to find that you’ve picked a peanut butter cup. Huzzah! [Yuck!]

You reach in again and pull a second candy at random. It’s another peanut butter cup! You reach in one last time and pull a third candy at random. It’s the third peanut butter cup! [Triple yuck!]

At this point, whatever is left in the bag is just candy corn. How many candy corn kernels do you expect to be in the bag?

Highlight to reveal (possibly incorrect) solution:

Program. Wikipedia 1 and 2.

First let’s assume there is a finite number of kernels of candy corn (kcc), at most m. What is the probability I got 3 peanut butter cups (pbc) (and yuck)?

• If I have n kcc, that’s 3+n pieces of candy in all. How many different ways can they be sorted? This is a stars and bars question, and the answer is C(3+n n) = (3+n)!/n!3!
• Of these (3+n)!/n!3! permutations, only 1 is any good, because it has the 3 pbc at the top.
• P(3 pbc | n kcc) = 1 / (3+n)!/n!3! = n!3!/(3+n)!
• n has m+1 different possible values.
• P(n kcc) = 1 / (m+1)
• P(3 pbc) = sum of all P(3 pbc | n kcc) * P(n kcc)
• Use Bayes’ theorem.
• P(n kcc | 3 pbc) = P(3 pbc | n kcc) * P(n kcc) / P(3 pbc)
• The expected number (E) of kcc is the sum of all n * P(n kcc | 3 pbc).
• And I made a program to calculate that given m.

Varying m hopefully tells me something about the behavior of E. And yes, it seems to. When m grows, apparently E gets closer to 1. So my guess is the expected number of kernels of candy corn is 1.

Also, I’m still fascinated by the soup bowl last week. Here’s a video!

This week the question is: Can You Make the Biggest Bread Bowl?

I have a large, hemispherical piece of bread with a radius of 1 foot. I make a bread bowl by boring out a cylindrical hole with radius r, centered at the top of the hemisphere and extending all the way to the flat bottom crust.

What should the radius of my borehole be to maximize the volume of soup my bread bowl can hold?

And for extra credit:

Instead of a hemisphere, now suppose my bread is a sphere with a radius of 1 foot.

Again, I make a bowl by boring out a cylindrical shape with radius r, extending all the way to (but not through) the curved bottom crust of the bread. The central axis of the hole must pass through the center of the sphere.

What should the radius of my borehole be to maximize the volume of soup my bread bowl can hold?

Highlight to reveal (possibly incorrect) solution:

Calculations. WolframAlpha assistance.

In my calculations I create a formula for the volume of the cylinder, depending on r. I then find the maximum volume. At r = √(2/3) foot, approximately 0.817, the volume is 2π/3√3, approximately 1.209 foot³.

And for extra credit:

ETA: I interpret “a cylindrical shape” as being the same as a cylinder. Flat top, flat bottom. Others interpret it differently.

Everything is multiplied by 2. The radius ends up being the same.

Anmeldelse af October Daye-serien, af Seanan McGuire. Romanserie. 2009-23, foreløbig. Hugo-finalist, bedste serie.

Skitse: October Daye er ret langt nede. Hendes mor er fae, det var hendes far ikke, så hun kan magi, men ikke ret meget, altså ser fae i høj grad ned på hende, og overfor mennesker kan hun ikke fortælle alt om sin baggrund. Hun var så alligevel ved at få gang i noget, mand, barn, hus, job som privatdetektiv. Men så blev hun forvandlet til en fisk. I 14 år. Bagefter opgiver hun (delvis nødtvunget) sit tidligere liv og prøver at klare sig som ekspedient. Langt fra alt det der magi. Men så …

Er det science fiction? Fantasy fra morgen til aften.

Temaer: Henover de 18 bøger, det er blevet til (der kommer vist 2 til) bliver October (Toby) mere og mere veltilpasset. Hun får det bedre. I starten er hun en ret god privatdetektiv, men hun kaster sig ud i den ene dødsensfarlige situation efter den anden, bl.a. fordi hun er lidt ligeglad. Det går delvist over, i takt med at hun får en ny familie.

I øvrigt en noget utraditionel familie.

Hun kæmper gang på gang mod racisme.

Regelmæssigt viser det sig, at person A i virkeligheden er person B, eller at person C, en ven, faktisk er en fjende, eller omvendt. Og eftersom nogle af de her personer er store kanoner (konger og dronninger), så bliver det hele også mere og mere vigtigt. Som at opdage, at naboen i virkeligheden er Thor, og at kun jeg kan redde hans liv.

Er det godt? Ja. Jeg er nogle gange ved at få pip af stilen, der bruger aaaaalt for mange ord efter min smag. Men jeg har hængt på, mere og mere, fordi det er spændende. Hvem har gjort hvad? Hvordan får Toby løst det den her gang? 👽👽👽

Note: Udover den her serie, der altså var Hugo-finalist, har jeg i den her omgang kun læst en af de andre serier systematisk, som beskrevet i Et vidunderligt lys osv. Glemte i den forbindelse, at Toby også var finalist.

Jeg har løbende lavet mini-anmeldelser af bøgerne. Det har taget mig et års tid at læse dem.

• 1 Rosemary and Rue (2009) Mini (Facebook)
• 2 A Local Habitation (2010) Mini
• 3 An Artificial Night (2010) Mini
• 4 Late Eclipses (2011) Mini
• 5 One Salt Sea (2011) Mini (Mastodon)
• 6 Ashes of Honor (2012) Mini
• 7 Chimes at Midnight (2013) Mini
• 8 The Winter Long (2014) Mini
• 9 A Red-Rose Chain (2015) Mini
• 10 Once Broken Faith (2016) Mini
• 11 The Brightest Fell (2017) Mini
• 12 Night and Silence (2018) Mini
• 13 The Unkindest Tide (2019) Mini
• 14 A Killing Frost (2020) Mini
• 15 When Sorrows Come (2021) Mini
• 16 Be the Serpent (2022) Ikke nogen mini!
• 17 Sleep No More (2023) Mini
• 18 The Innocent Sleep (2023) Mini

Anmeldelse af A City on Mars: Can We Settle Space, Should We Settle Space, and Have We Really Thought This Through?, af Kelly & Zach Weinersmith. Fagbog. 2023. Hugo-finalist, best related.

Skitse: Hvis man kigger på emnet “at bygge en koloni på Mars” og andre nærliggende emner, så er der en masse underemner. Der er noget med teknologi, ja, og noget med medicin. Men der er fx også noget med lovgivning. Lad os tage et grundigt kig på det vigtigste!

Er det science fiction? Nej, det er en fagbog.

Temaer: Formen er sjov. Snurrige formuleringer og deciderede vittigheder. Uærbødige tegninger. Anekdoter. Og det er en god ting. Vi vil jo godt have folk til at læse den her bog. Det var i hvert fald en rigtig god ting for mig. Det er længe siden, jeg kværnede den her type fagbog.

Det hjælper også på, at hovedbudskabet er, nej, vi kan ikke det her endnu, og vi bør ikke prøve. Worst case får Enol Muks sendt en million mennesker til Mars, der prompte dør.

Det er pudsigt, at min sf-radar hele tiden slår ud. Der er rigtig mange historier, der tager fat i nogle af de her emner, mere eller mindre korrekt.

Er det godt? I sidste ende ikke godt nok til sådan en som mig. Der er masser af tankevækkende stof og sjove indslag, men det var stadig svært at komme igennem det hele. 👽👽💀

Note: Det her er den eneste finalist i den her kategori, jeg rigtig har gidet se på i sin helhed. (Omend jeg nok også vil skæve til en af de andre, en samling af anmeldelser.) Den vandt i øvrigt.

Jeg læste bogen som en pdf, der til tider blev konverteret til noget sært.

This week the question is: Will You Top the Leaderboard?

You’re doing a 30-minute workout on your stationary bike. There’s a live leaderboard that tracks your progress, along with the progress of everyone else who is currently riding, measured in units of energy called kilojoules. (For reference, one kilojoule is 1000 Watt-seconds.) Once someone completes their ride, they are removed from the leaderboard.

Suppose many riders are doing the 30-minute workout right now, and that they all begin at random times, with many starting before you and many starting after. Further suppose that they are burning kilojoules at different constant rates (i.e., everyone is riding at constant power) that are uniformly distributed between 0 and 200 Watts.

Halfway through (i.e., 15 minutes into) your workout, you notice that you’re exactly halfway up the leaderboard. How far up the leaderboard can you expect to be as you’re finishing your workout?

As an added bonus problem (though not quite Extra Credit), what’s the highest up the leaderboard you could expect to be 15 minutes into your workout?

And for extra credit:

Again, suppose there are many riders starting their 30-minute workouts at random times, and that their powers are uniformly distributed between 0 and 200 Watts. Now, suppose you decide that you too will be pedaling with a random (but constant) power between 0 and 200 Watts.

If you look down at the leaderboard at a random time during this random workout, how far up the leaderboard can you expect to be, on average?

Highlight to reveal (possibly incorrect) solution:

Plot 1 and 2. Calculations. Program.

How high up you are on the leaderboard is determined by how many kJ you spent, which is in turn determined by how much time you spent and how hard you’re pedaling (power). Time: somewhere between 0 and 30 minutes. Power: somewhere between 0 and 200 W. Furthermore you look at the leaderboard at your personal time of 15 minutes.

Let’s simplify to Energy = Time * Power. E = T * P. Both T and P are evenly distributed. E can be somewhere between 0 kJ and 30 minutes * 200 W = 30 * 60 seconds * 200 W = 360000 Ws = 360 kWs = 360 kJ.

Trying to plot some sections of values for E, we see that a lot of space is taken up by the less than 100 kJ points. (The sections have borders corresponding to 90, 180 and 270 kJ.) While T and P are evenly distributed, E isn’t.

So, what we’re looking for here is some curve, f(x,y) = T * P = C, a constant, so that the area below the curve is half of the available rectangle. All those points below (or to the left of) this curve are the many, many people below you on the leaderboard. Because of the time or power they’ve spent, or maybe both, they’re behind you. The guy with 1 Watt is probably behind you the whole time. The gal beginning 1 minute ago is probably behind you too. They’re both in the lower left corner of the plot.

Some calculations reveal, that C = 67.2 kJ. A new plot sort of confirms this. A program also hits in this neighborhood. ETA: This uses the equation for the area C * (1 + log(360k/C) = E. In this case, C * (1 + log(360k/C) = 180 kJ, half the available area.

In other words, your energy use at 15 minutes is 67.2 kJ. This corresponds to 74.7 W. At the end of the 30 minutes, you will hit double this energy use, 134.4 kJ. Now we do the calculation the other way, T * P = 134.4 kJ. What is the area below this curve? ETA: 134.4k * (1 + log(360k/134.4k) = E = 266.8 kJ. Out of an area of 360 kJ in all, that’s 74.1%. So you can expect to rise above 74% of your competitors, just before you drop out of the leaderboard.

So, I’m not doing any more of this fiddler. I’m not in love with probabilities.

Bonus question:

The highest you can be on the leaderboard implies, that you are going full power, 200 W, ending up with 360 kJ. At 15 minutes you would be at 180 kJ. Again using the equation 180k * (1 + log(360k/180k) = E = 304.8 kJ. Out of the available area of 360 kJ, that means you are above 84.7% of the board.

And for extra credit:

We want the average value of E, because that’s the E you have at a random time. This requires calculating the “volume” of the figure in the coordinate system with axis T, P and E, and then dividing by the area, 360 kJ. A quick integral gives the volume as (1/2 * 360k)². This gives an average E of (1/2)² * 360k. = 90k. What is the area below E = 90 kJ? 90k * (1 + log(360k/90k) = 214.8 kJ. Out of an area of 360 kJ, that’s 59.7%. You will be above 59.7% of the competition.

And I am not too sure about those extra bits. ETA: My program says something else. Sigh.

This week the question is: How Many Dice Can You Roll the Same?

… To get started, you roll all 10 dice—whichever number comes up most frequently becomes your target number. In the event multiple numbers come up most frequently, you can choose your target number from among them. At this point, you put aside all the dice that came up with your target number. …

Now, consider a simplified version of the game in which you begin with three total dice (call it “THREEZI”).

On average, how many dice will you put aside after first rolling all three?

And for extra credit:

Let’s return to the original game of TENZI, which has 10 dice.

On average, how many dice will you put aside after first rolling all 10? (What if, instead of 10 dice, you have N dice?)

Highlight to reveal (possibly incorrect) solution:

Program.

Let’s call the 3 dice rolls a, b and c.

• There are 3 possible cases:
  • 3 dice are put aside. This means a = b = c. This again means a is something with p = 1, and b and c are the same, each with p = 1/6. So p(3) = (1/6)² = 1/36.
  • 2 dice are put aside. This means e.g. a = b != c. This again means a is something with p = 1, b is the same with p = 1/6, and c is different with p = 5/6. Furthermore, I could have chosen the different one in 3 different ways. So p(2) = (1/6) * (5/6) * 3 = 15/36.
  • 1 die is put aside. This means a != b, a != c and b != c. This again means a is something with p = 1, b is something else with p = 5/6, and c is something else again with p = 4/6. So p(1) = (5/6) * (4/6) = 20/36.
• A quick check: 1 + 15 + 20 = 36. The sum of our probabilities is 36/36 = 1. Good.
• The average we’re looking for is 3 * p(3) + 2 * p(2) + 1 * p(1) = 3 * 1/36 + 2 * 15/36 + 1 * 20/36 = (3 + 30 + 20)/36 = 53/36 = 1 17/36, about 1.472.
• A quick check: my program says the same.

And for extra credit:

Program 1 and 2.

I figure out a way to extend the previous program to run with more dice. Mainly by programming the dice as a number in base 6. With 10, the average number of dice set aside is 3.445… In general, with N dice, on average I set aside about √N dice.