Anmeldelse af “Hjemme igen, hjemme igen”, af Cory Doctorow. Novelle. 2024. Så fremmed et sted.
Skitse: Som 10-årig bor Chet i Stærekassen, sammen med alle de andre, som uransagelige rumvæsner har vurderet til at have pip. Der er skole, samtaler med en udenjordisk vejleder og besøg hos en fyr, der bl.a. har et stort akvarium med koraller og fisk.
Er det science fiction? Ja.
Temaer: Chets forældre må altså have pip, mindst en af dem, men de ved ikke hvilken, og det er ikke godt for den mentale sundhed at havne i den her kategori. Ham med akvariet er heller ikke helt jordbunden. Håbet er, at Chet har en bedre fremtid. Der er i øvrigt system i galskaben. Efter en generation vil stærekasserne blive nedlagt igen, fordi der ikke længere er behov for dem.
Sideløbende følger vi den noget ældre Chet, der har været ude i universet, men nu er kommet hjem igen. Hans bedre fremtid indeholdt selv at blive vejleder.
Fyren med akvariet tror, at han er Tesla. Det kommer du ikke til at glemme et sekund under læsningen! 👎🏻
Synspunktet skifter mellem 1. og 3. person. Tja. Skiftene er ret bratte.
Anmeldelse af “Skyggen af moderskibet”, af Cory Doctorow. Novelle. 2024. Så fremmed et sted.
Skitse: En ny kult jager lykken og bygger huse af hærdet skum. Da rumvæserne ankommer, er det kultens leder, der bliver inviteret op på moderskibet og taget med på en lille udflugt. Hvad siger man så, som lederens efterladte søn Maxes?
Er det science fiction? Jeppe-depper.
Temaer: Ikke uventet er livet kaotisk, både for alle Jordens beboere generelt og Maxes helt specifikt. Skrivestilen afspejler det: Store hop i tid, store forandringer.
I was recently a guest at Disney World, which has a new system called “Lightning Lane” for reserving rides in advance—for a fee, of course.
By purchasing “Lightning Lane Multi Pass,” you can reserve three of the many rides in a park, with each ride occurring at a different hourlong time slot. For simplicity, suppose the park you’re visiting (let’s say it’s Magic Kingdom) has 12 such time slots, from 9 a.m. to 9 p.m. So if you have the 3 p.m. time slot for a ride, then you can skip the “standby lane” and board the much shorter “Lightning Lane” at any point between 3 and 4 p.m. Assume you can complete at most one ride within each hourlong time slot.
Once you have completed the first ride on your Multi Pass, you can reserve a fourth ride at any time slot after your third ride. This way, you always have at most three reservations. Similarly, after you have completed your second ride, you can reserve a fifth ride at any time slot after your fourth, and so on, up until you are assigned a ride at the 8 p.m. (to 9 p.m.) time slot. That will be your final ride of the day.
Magic Kingdom happens to be very busy at the moment, and so each ride is randomly assigned a valid time slot when you request it. The first three rides of the day are equally likely to be in any of the 12 time slots, whereas subsequent rides are equally likely to occur in any slot after your currently latest scheduled ride.
On average, how many rides can you expect to “Lightning Lane” your way through today at Magic Kingdom?
And for extra credit:
If you’re a Disney aficionado, then you know that week’s Fiddler is in fact an oversimplification of how Lightning Lane actually works. Let’s make things a little more realistic.
This time around, after you complete the first ride on your Multi Pass, you can reserve a fourth ride at any time slot after your first ride (rather than after your third ride). Similarly, after you have completed your second ride, you can reserve a fifth ride at any time slot after your second, and so on, until there are no available time slots remaining.
As before, the first three rides of the day are equally likely to be in any of the 12 time slots, whereas subsequent rides are equally likely to occur in any remaining available slots for the day.
On average, how many rides can you expect to “Lightning Lane” your way through today at Magic Kingdom?
Highlight to reveal (possibly incorrect) solution:
I actually solved this 3 different ways. The first time around I also got 3 different results. So I kept going until the results lined up.
The spreadsheet is the more mathy. I’m looking at the expected numbers directly. What can we expect if 3 rides are already booked, and after having done the 1st ride, I’m looking to book 1 more?
If the last booked ride is in the 12th hour (E(12)), 0 more rides are expected.
If the last booked ride is in the 11th hour (E(11), 1 more ride + E(12) are expected. This is 1.
If the last booked ride is in the 10th hour (E(10)), 1 more ride + either E(11) or E(12) are expected. E(11) and E(12) are equally likely, as they consist in the next ride chose being in either the 11th or 12th hour. So E(10) = 1 + (E(11) + E(12))/2 = 1 + (1 + 0)/2 = 1 + 1/2 = 1 1/2.
If the last booked ride is in the 9th hour (E(9)), by a similar logic, 1 + (E(10) + E(11) + E(12))/3 rides are expected. This is 1 5/6.
This goes on until E(3).
To calculate the expected number of rides in total, I also need the probabilities for each situation. Given I have booked 3 rides, how many ways can I do this, ending up with the latest being in the nth hour?
If the last booked ride is in the 12th hour, the other 2 rides are chosen randomly among the first 11, 11 choose 2 = 11*10/2 = 55 = P(12)/sum. (Sum: the sum of all these ways.)
If the last booked ride is in the 11th hour, the other 2 rides are chosen randomly among the first 10, 10 choose 2 = 10*9/2 = 45 = P(11)/sum.
All the way until…
If the last booked ride is in the 3rd hour, the other 2 rides are chosen randomly 😉 among the first 2, 2 choose 2 = 2*1/2 = 1 = P(3)/sum.
Adding up all these ways, there are sum = 220 in all.
Now I just take each product, E(n) * P(n), calculate the sum, and hey presto! The result is 1.2699. Oh, and then I have to add the 3 rides already booked, so the result is actually 4.2699.
My first program does a Monte Carlo. My second program goes through all the options, calculating the numbers of rides and averaging out. These 2 programs confirm the result.
And for extra credit:
For this one I define E(a, b) as the expected number of rides added, given that at time a (in the ath hour) there are b booked rides after this time.
E(12, 0) = 0 – I am at the 12th hour, of course there are no rides left after this one, of course I can’t add any rides.
E(11, 0) = 1 + E(12, 0) = 1 – I am at the 11th hour, the 12th slot is empty, I can therefore add that ride and progress to the 12th hour.
E(11, 1) = 0 + E(12, 0) = 0 – I am at the 11th hour, the 12th slot is already booked, I can add no rides and simply progress to the 12th hour.
E(10, 0) = 1 + (E(11, 0) + E(12, 0))/2 = 1 + (1 + 0)/2 = 1 + 1/2 = 1 1/2 – I am at the 10th hour, the 2 slots after this one are empty, either the 11th or 12th will be booked, with equal probability, and I can move forward to that slot.
E(10 , 1) = 1 + E(11, 1) = 1 + 0 = 1 – I am at the 10th hour, 1 of the 2 next slots is already booked, so now the empty is also booked, I can move forward to the 11th hour.
E(10, 2) = E(11, 1) = 0 – I am at the 10th hour, both slots after this one are already booked, I simply move 1 hour forward.
E(9, 0) = 1 + (E(10, 0) + E(11, 0) + E(12, 0))/3 = 1 + (1 1/2 + 1 + 0)/3 = 1 + (5/2)/3 = 1 + 5/6 = 1 5/6 – I am at the 9th hour, the next 3 slots are open, with equal probability one of them is chosen, and I move forward to that hour. (These calculations, where b = 0 are duplicates of calculations from the fiddler, E(9).)
E(9, 1) = 1 + (E(10, 1) + E(10, 1) + E(11, 1))/3 = 1 + (1 + 1 + 0)/3 = 1 + 2/3 = 1 2/3 – With equal probability (*) the last 3 slots are b__, _b_ or __b. (b for booked.) After booking one of them, the situation is, with equal probability (*) bb_, b_b or _bb. 2 of these represent E(10, 1), and 1 of these represent E(11, 1).
E(9, 2) = 1 + E(10 , 2) = 1 + 0 = 1 – There’s an empty slot in front of me, I fill it and move on to the next hour.
E(8, 0) = 25/12 = 2 1/12.
E(8, 1) = 1 + (E(9, 1) * 3 + E(10, 1) * 2 + E(11, 1))/6 = 1 + (1 2/3 * 3 + 1 * 2 + 0)/6 = 1 + 7/6 = 2 1/6 – The new states are bb__, b_b_, b__b, _bb_, _b_b and __bb.
E(8, 2) = 1 + (E(9, 2) * 3 + E(10, 2))/4 = 1 + (1 * 3 + 0)/4 = 1 3/4 – The new states are bbb_, bb_b, b_bb and _bbb.
E(7, 0) = 2 17/60
E(7, 1) = 1 + (E(8, 1) * 4 + E(9, 1) * 3 + E(10, 1) * 2 + E(11, 1))/10 = 1 + (2 1/6 * 4 + 1 2/3 * 3 + 1 * 2 + 0)/10 = 1 + 1 17/30 = 2 17/30 – The new states are bb___, b_b__, b__b_, b___b, _bb__, _b_b_, _b__b, __bb_, __b_b and ___bb. (I smell a formula. Let c = 12-a. This represents the number of slots left. Part of the sum is going from E(a, 1) to E(a+1, 1) * (c-1) + E(a+2, 1) * (c-2) + … + E(a+(c-1), 1) * 1. And dividing by the sum of the integers from 1 to c-1, also called the triangle number.)
E(7, 2) = 1 + (E(8, 2) * 6 + E(9, 2) * 3 + E(10, 2))/10 = 1 + (1 3/4 * 6 + 1 * 3 + 0)/10 = 1 + 1.35 = 2.35. (Again, there’s structure. Let d = 12-a-1. The sum goes from E(a, 2) to E(a+1, 2) * tri(d-1) + E(a+2, 2) * tri(d-2) + … + E(a+(d-1), 2) * tri(1). And then dividing by the sum of the triangle numbers.)
I continue this exercise in the spreadsheet.
(*) I assume.
The next step, after having calculated all of these, going all the way to E(1, 2), is again to compute the expected number, P*E. This gives, that the expected number of rides added, after the first 3 rides already added and the first of these at a, is 3.8096. Adding the 3 first rides, we get 6.8096.
Anmeldelse af “Billy Baileys rebranding”, af Cory Doctorow. Novelle. 2024. Så fremmed et sted.
Skitse: Billy har været noget af en rod siden børnehaven, men hævn over Mitchell kræver en anden profil. Lad forhandlingerne med agenter og sponsorer begynde.
Er det science fiction? Ja, det er det vel egentlig. En fremtid med mere aggressive reklamer.
Temaer: Billy er helt klart intelligent, og selvom der er voksne omkring ham, så træffer han selv beslutningerne.
Anmeldelse af “Verdens største fjols”, af Cory Doctorow. Novelle. 2024. Så fremmed et sted.
Skitse: Verdens klogeste mand har løst en masse opgaver, mens en computer har kigget med. Således kan man lave en chip, der kan gøre en anden mand lige så klog.
Er det science fiction? Ja.
Temaer: Hvad er klogskab egentlig? Kan den kloge kontrolleres af den mindre kloge?
Er det godt? Jeps. Tilfredsstillende afslutning. 👽👽👽
Anmeldelse af “Så fremmed et sted”, af Cory Doctorow. Novelle. 2024. Så fremmed et sted.
Skitse: Som 10-årig flytter vores hovedperson, James, til 1975 fra Utah. Utah i 1898 altså.
Er det science fiction? Ja.
Temaer: 1975 er ikke helt, som det skal være i vores øjne. Futuristiske materialer, transportformer, undervisning osv. Kontakt mellem årstallene bliver overvåget og kontrolleret, men åbenlyst er der stadig ting, der slipper igennem. I øvrigt er James’ far ambassadør.
James har et liv i 1898 og frem, der dog rammer 70’erne af og til. Der er lidt drama. At far en dag ikke kommer hjem fra arbejde. At mor efter noget tid finder en ny mand. At James’ lærer har en spændende baggrund og en interessant tilgang til James’ skolegang.
Der er lidt fnidder her med mangel på kolon før anførselstegn, men hvis jeg husker ret, så har den originale tekst noget lignende. Til gengæld er oversættelsen i sig selv rigtig god, ligesom i resten af bogen.
Anmeldelse af “Møghund”, af Cory Doctorow. Novelle. 2024, dog også tidligere 2015. Så fremmed et sted.
Skitse: Jerry tjener bl.a. penge på at købe genstande billigt på loppemarkeder og sælge dem på auktion. Han er rigtig gode venner med Møghund, et rumvæsen, der primært samler. Indtil de har et sammenstød over en samling cowboyting for børn.
Er det science fiction? Jeps.
Temaer: Der findes jo alle slags personer, så hvorfor dog ikke også en, der støvsuger loppemarkeder for guf?
Og hvorfor skulle sådan en ikke kunne være et rumvæsen? Der normalt handler med væsentlige dyrere produkter, men det andet kan jo være en hobby. Der er et interessant fænomen her, hvor et menneske kommer i øjenhøjde med det fremmede.
Er det godt? Jeps. De ganske få tråde samles tilfredsstillende. 👽👽👽
Note: Jeg har et blødt punkt for Doctorow, så jeg giver selvfølgelig også gode karakterer til hans historier.
You are at the point (0, 0) on the coordinate plane. There is a tree at each point with nonnegative integer coordinates, such as (5, 6), (0, 7), and (9, 1). The trees are very thin, so that they only obstruct trees that are directly behind them. For example, the tree at (2, 2) is not visible, as it is obscured by the tree at (1, 1).
Now, you can’t see infinitely far in this forest. Suppose, for example, that the farthest you can see is 4 units. The diagram below shows the trees you would see and the angles between them:
In truth, you can see much farther than 4 units into the forest. You’re not sure exactly how far you can see, but it’s pretty dang far. To be extra clear about this, the diagram above is just an illustration, and you can in fact see much farther than 4 units.
As you look around, you can make out very narrow angular gaps between the trees. The largest gaps are near the positive x-axis and the positive y-axis (similar to the illustrated case above). After those, the next largest pair of gaps are on either side of the tree at (1, 1), 45 degrees up from the x-axis.
Consider the view between 0 and 45 degrees up from the x-axis. The next largest pair of adjacent gaps in this range are on either side of what angle up from the x-axis? (To be clear, you are not considering the gap just above 0 degrees or the gap just below 45 degrees.)
And for extra credit:
The fifth largest pair of adjacent gaps in this range are on either side of what angle up from the x-axis? (Again, you are not considering the gap just above 0 degrees or the gap just below 45 degrees.)
Highlight to reveal (possibly incorrect) solution:
To get a sense of the problem, I made a spreadsheet and played around with it a little. Then I wrote a program, that could easily handle much bigger distances. The answer seems to be the angle is 26.56505. This is the angle corresponding to (2,1).
(Observation: the big gaps occur around angles with a lot of potential hits. In this case (2,1), (4,2), (6,3) etc. There’s a lot of hits, because the first pair of coordinates have low values (2 and 1).)
And for extra credit:
Same program. The 5th pair of gaps, sorted on size, occurs at the angle 36.86990. This is the angle corresponding to (4,3).
I’ve recently read Doomsday Book, Connie Willis. These notes are about an aspect of the book, I didn’t really like. Spoilers.
Of course it’s drama and conflict. Of course our hero wants something and can’t get it right away. It’s the obstacles I, ahem, stumble over.
You can’t get what you want:
Because you fell ill. Really ill.
Because somebody else fell ill. Really ill.
Because a lot of people fell ill and quarantine rules apply.
Because somebody else has to authorise it, and they are on vacation for 2 weeks and impossible to find.
Because somebody else has to authorise it, and they are arrogant and won’t listen to reason. (*)
Because your question (or the answer) isn’t understood correctly.
Because you talking to the other relevant person Just Isn’t Done, and you can’t figure out a way to do it in secret.
Because your sense of duty says you should do something else.
Because a 5 year old is being 5 and spoiled.
Because a 12 year old is trying to stay away from that creepy guy.
Because young people just don’t listen. (*)
Because this one person is suspicious and schizophrenic and generally angry and negative. (*)
Because this one person is overly protective of her son. (*)
Because this one person thinks cheering up = listening to horrible Bible stories. (*)
The (*) means this rubs me the wrong way. Stupid, irrational people not listening. It’s supposed to be funny? Instead it bores me. And it happens a lot.
You and your opponent are competing in a golf match. On any given hole you play, each of you has a 50 percent chance of winning the hole (and a zero percent chance of tying). That said, scorekeeping in this match is a little different from the norm.
Each hole is worth 1 point. Before starting each hole, either you or your opponent can “throw the hammer.” When the hammer has been thrown, whoever did not throw the hammer must either accept the hammer or reject it. If they accept the hammer, the hole is worth 2 points. If they reject the hammer, they concede the hole and its 1 point to their opponent. Both players can throw the hammer on as many holes as they wish. (Should both players decide to throw the hammer at the exact same time—something that can’t be planned in advance—the hole is worth 2 points.)
The first player to reach 3 points wins the match. Suppose all players always make rational decisions to maximize their own chances of winning.
Good news! You have won the first hole, and now lead 1-0. What is your probability of winning the match?
And for extra credit:
Instead of playing to 3 points, now the first player to 5 points wins the match.
Good news (again)! You have won the first hole, and now lead 1-0. What is your probability of winning the match?
Highlight to reveal (possibly incorrect) solution:
For any given hole, for any player X, p(X wins) = 50%.
Each win is worth 1 point, unless a hammer is thrown and accepted, in which case it’s worth 2 points.
A player can decide to throw a hammer.
A player can decide to accept or reject a thrown hammer. Rejecting a hammer is the same as losing, worth 1 point.
If both players throw hammers, it counts as an accepted hammer.
One player throwing a hammer and the other accepting is equivalent to both players throwing hammers at the same time.
The winner of the whole match is the first to reach 3 points.
Players choose the optimal strategy.
Alice won the first hole, and is in front 1-0.
What is probability that Alice will win?
The notation Px,y is adopted to mean: the probability Alice will win when the match is at x-y. So we’re looking for P1,0.
A strategy consists of these choices: Given x-y, should Alice throw the hammer, or, if the hammer is thrown by Bob, accept or reject?
In a table for Px,y, choosing the optimal strategy for Alice, I can also see Py,x and the optimal strategy for Bob.
Trivially, P3+,y = 1 and Px,3+ = 0.
With this in mind, let’s look at a table for P2,2. Alice has the choices in the left column, Bob has the choices in the top row.
(My tables don’t quite reflect, that accept/reject occurs before the hole is played. I try to capture this nuance in the text instead. It’s important, because a match may be decided right there.)
Bob
P2,2
Alice
accept
reject
hammer
accept
50% * P3,2 + 50% * P2,3 = 50% * 1 + 50% * 0 = 50%
50% * P3,2 + 50% * P2,3 = 50% * 1 + 50% * 0 = 50%
50% * P4,2 + 50% * P2,4 = 50% * 1 + 50% * 0 = 50%
reject
50% * P3,2 + 50% * P2,3 = 50% * 1 + 50% * 0 = 50%
50% * P3,2 + 50% * P2,3 = 50% * 1 + 50% * 0 = 50%
P2,3 = 0
hammer
50% * P4,2 + 50% * P2,4 = 50% * 1 + 50% * 0 = 50%
P3,2 = 1
50% * P4,2 + 50% * P2,4 = 50% * 1 + 50% * 0 = 50%
First, let’s notice, that if player X throws a hammer, player Y should never reject, as that leads to instant defeat. Therefore the middle column and middle row drop out.
Alice has to choose the best row out of (50, 50) and (50, 50). As long as she accepts a thrown hammer, it doesn’t matter what she chooses otherwise.
Bob is trying to choose the optimal (i.e. lowest) column from (75, 75, 50), (75, 75, 1) and (50, 50, 50). He will choose the third one and therefore throw a hammer.
P1,2 = 50%.
Knowing Bob will throw a hammer, it doesn’t really matter what Alice chooses. All her choices give 50%.
The best row for Alice is the third, and she throws a hammer.
P1,0 = 75%.
The best columns for Bob are those where he doesn’t throw a hammer. Given that Alice does throw one, he ends up in the bottom row.
P0,1 = 25%.
Oh! We’ve actually found the value, we were looking for. P1,0 = 75%.
Just to make sure, I also wrote a program to confirm this (and the next) result.
And for extra credit:
Everything is the same, we’re just going for a win of 5 instead of 3.
Trivially, P5+,y = 1 and Px,5+ = 0.
If we read Px,y in the fiddler as “Alice is 3-x away from having 3 points and Bob is 3-y away”, we can translate all of this into Px+2,y+2 for this extra credit puzzle.
#probabilities #random 
Stuff from ommadawn.dk (Lise Andreasen) 