Anmeldelse af Warp Your Own Way, af #RyanNorth. #GraphicNovel. 2024. Hugo-vinder. #StarTrek

Skitse: Ombord på Cerritos vågner Mariner desværre. Hendes sambo skal tidligt på arbejde, mens hun selv kan sove et par timer til. Skidt pyt. Man falder jo bare i søvn igen. Men der bliver ved med at være forstyrrelser, så til sidst opgiver hun, står op, laver noget kaffe og går ud for selv at forstyrre nogle andre.

Er det science fiction? Ja! 🖖🏻🖖🏻🖖🏻

Temaer: Lynhurtigt kommer vi i gang med du-er-helten-delen. Hvilken kaffe skal det være? Og hvem skal forstyrres? Fordi jeg har prøvet det her før, så skrev jeg mine valg ned, så en “blind vej” bare kunne blive rullet tilbage. Og det var der behov for. Rigtig mange af vejene er nemlig blinde. Og der dukker også elementer op, der vist ikke plejer at høre til i den her genre. Et par gange måtte jeg ændre min forståelse af tingene. En mindre doven læser lavede et diagram over bogen og forklarede et par ting, jeg ikke havde fanget.

Det her føles virkelig som en integreret del af Star Trek. Visuelt ligner det 100 %, og det er flot. De forskellige personer opfører sig som de plejer. Plottet er også indenfor skiven, og moralen til sidst er vist en gentagelse fra et af afsnittene, der var det vist bare Rutherford, der viste os, hvordan man bør ordne tingene.

Plottet har en skurk, der på sin egen måde bryder en af “lovene”, og derfor taber. Jeg vil ikke afsløre hvilken lov. Men det kan sammenlignes med, at hvis de var kommet til en paradisisk planet, så ville loven sige, at Tingene Er I Virkeligheden Som Helvede, og at vores helte derfor skulle vælte tyrannen eller ødelægge computeren eller sådan noget. Den slags lov.

Oprindeligt var det her en papirbog, men den virker også fint som pdf på en tablet. Bogstaverne er store nok. Dobbeltsider kan man se ved at blade lidt frem og tilbage. Man kan ikke bare klikke på en boks og hoppe til den side, hvor historien fortsætter (i hvert fald ikke med lige min pdf-læser), men der er en fin indholdsfortegnelse, så skiftet trods alt kan ske hurtigt. Hele bogen tog mig 2-3 timer.

Er det godt? Ja! 🖖🏻🖖🏻🖖🏻 Det er en af de bedste oplevelser, jeg har haft med franchiset i papirform. Og medmindre Paramount får fundet ud af at lave episoder, der kopierer Black Mirrors interaktive film, så kunne det her kun laves som papir- eller computerspil. Og det er charmerende, at de valgte papir. 👽👽👽

This week the #puzzle is: How Much Free Money Can You Win? #probabilities #bets #minimum #maximum

And for extra credit:

How Much Free Money Can You Win?

Highlight to reveal (possibly incorrect) solution:

PDF Image

This time I chose to write out all the possibilities. Big PDF file resulted. The important bit is the part shown in the image. This shows this strategy:

• I first bet 2 x 10 against 25.
  • Either I win $20 and have 3 x 10 left. Bet 10 against 10 twice, resulting in a further $20. $40 in all.
  • Or I win $25 and have 1 x 10 and 1 x 25 left. Bet them against each other and win at least $10. $35 in all.
• Win at least $35.

And for extra credit:

PDF Image Program

I adjust my PDF to look at probabilities instead of minimum. This gives me the strategy:

• 1 first bet 10 against 25.
  • Either I win $10 and have 3 x 10 left. Now bet 2 x 10 against 10.
    • Either I win $20 and have 2 x 10 left. Now bet 10 against 10, win $10, have 10 left, bet that 10 alone, 50% of getting $10. That is in total 50% for $40, 50% for $50.
    • Or I win $10 and have 10 left. Bet that 10 alone, 50% of getting $10. That is in total 50% for $20, 50% for $30.
    • Combined this is 25% for $20, 25% for $30, 25% for $40, 25% for $50.
  • Or I win $25 and have 2 x 10 and 25 left. Now bet 2 x 10 against 25.
    • Either I win $20 and have 2 x 10 left. Now bet 10 against 10, win $10, have 10 left, bet that 10 alone, 50% of getting $10. That is in total 50% for $55, 50% for $65.
    • Or I win $25 and have 25 left. Bet that 25 alone, 50% of getting $25. That is in total 50% for $50, 50% for $75.
    • Combined this is 25% for $50, 25% for $55, 25% for $65, 25% for $75.
  • Combined this is 12.5% for each of $20, $30, $40, $50, $50, $55, $65, $75.
  • The median here is $50. That is therefore my 50% chance result.

I am little surprised at this result. I doodled with adding more bets (if I win my final bet above, I can bet again), but that didn’t change the result. It is close to a theoretical $55, the value of all the vouchers.

Ahem.

There’s a strategy where I begin by betting everything on red. Either I lose, 50% chance, or I win $55 and keep all my vouchers, 50% chance. If I win, I can bet again. This way I get at least $55 with 50% chance. A more complicated strategy has to beat that.

After many thoughts I tried writing a program. This approach uses a little monte carlo. It also uses going through all the options for strategies. A typical output looks like this:

Median:       70.
Array
(
    [10101025] => 7c
    [101025] => 6b
    [101010] => 1
    [1025] => 1
    [1010] => 3b
    [25] => 2
    [10] => 1
    [0] => end
)

The numbers (7c, 6b etc.) refer to the program and to the PDF. Translated this strategy is:

• Bet 10 on red and 3 other vouchers on the rest. (7c)
  • Either win on red, get $10 and keep a 10 voucher. By all means keep playing (1), but it probably won’t get much better. At least $10 in all.
  • Or win on black. Get $45 and keep vouchers for 10, 10 and 25. Bet 2 x 10 on red and 25 on black. (6b)
    • Either win on red and get $20 more and keep 2 x 10 vouchers. Bet 10 on red and 10 on black. (3b) Get $10 more and keep a 10 voucher. Feel free to keep playing (1). At least $75 in all.
    • Or win on black and get $25 more and keep the 25 voucher. Feel free to keep playing (2). At least $70 in all.

Winning on the 1st black guarantees winning at least $70, with 50% chance. And that’s what we were looking for! W = 70!

It’s a rather special feeling, writing a program just to make sure everything is right, and then finding a better solution.

Addendum.

Program

Just for fun I converted the program back to solve the fiddler. And found another solution. And an error in my math in the prodigious, pretty PDF. Hm. This strategy also works:

• Play 10 against 10 (3b). Lose 1 x 10, gain $10. Play 2 x 10 against 25 (6b).
  • Either lose 25 and gain $20. Play 10 against 10 (3b). Lose 1 x 10, gain $10. $40 in total.
  • Or lose 2 x 10 and gain $25. $35 in total.

Minimum:       35.
Array
(
    [10101025] => 3b
    [101025] => 6b
    [101010] => 1
    [1025] => 1
    [1010] => 3b
    [25] => 2
    [10] => 1
    [0] => end
)

This week the #puzzle is: Can You Canoodle at the Coldplay Concert? #probabilities #geometry #integral #estimat #montecarlo

And for extra credit:

Can You Canoodle at the Coldplay Concert?

Highlight to reveal (possibly incorrect) solution:

Program

Information:

• This particular couple canoodles half of the time.
• 1% of couples at the concert will be shown on the jumbotron, randomly chosen.
• Going to n concerts, there’s at most 50% chance that this couple will be on the jumbotron, canoodling. Find n.

My first guess is:

• At any given concert:
  • There’s a 1% chance this couple will be on the jumbotron.
  • There’s a 50% chance they are canoodling.
  • There’s a 1% * 50% = 0.5% chance they are canoodling on the jumbotron.
  • There’s a 100% – 0.5% = 99.5% chance they are not canoodling on the jumbotron.
• At n concerts:
  • There’s a (99.5%)ⁿ chance they are not canoodling on the jumbotron.
  • We want (99.5%)ⁿ > 50%.
  • (99.5%)ⁿ > 50%
  • log(0.995ⁿ) > log(0.5)
  • n * log(0.995) > log(0.5)
  • Divide by negative number on both sides.
  • n < log(0.5)/log(0.995) = 138.282572986
  • n = 138

My second approach was to write a program to confirm this number.

And for extra credit:

Spreadsheet Desmos Program

Information:

• For each couple, each member wants to canoodle t_i of the time, t_i randomly chosen between o and 1.
• The couple then canoodles t = t₁ * t₂ of the time.
• A random canoodling couple will be shown on the jumbotron.
• The chance of being chosen for the jumbotron is proportional to the time spent canoodling.
• Slicing the couples into groups depending on their t, the couples shown most frequently belong to the group that has t = C. Find C.

My first guess didn’t fit the programs I also wrote, so this is my second guess:

• Assume that all the couples are sorted into a grid. With e.g. 10 * 10 couples, there are 10 * 10 cells. In the first row, all the first members of the couples have t₁ = 0.05 +- 0.05, in the next row it’s 0.15 +- 0.05 etc. In the first column, all the second members of the couples have t₂ = 0.05 +- 0.05. This way we get a regular sample of couples.
• For a value a, we can look at all the couples with t = a. Estimating the size of this group, we look at a slice going from a to a + b.
• a = x * y or y = a/x.
• Look at the area A between a/x and (a+b)/x.
• The probability that a couple from this group will be shown is A * a.
• Maximize A * a.
• I use a combination of Desmos and a spreadsheet graph to find this a = 0.368.

My program produced this histogram:

                               * *      *    *   *     *                                            
                     ******************** ****** ***** *   *                                        
                ************************************************ *  *                               
           ******************************************************* *** **                           
        ************************************************************** ***** *                      
      ************************************************************************** *                  
    *********************************************************************************** *           
  **************************************************************************************** *        
 ********************************************************************************************* *    
0---------1---------2---------3---------4---------5---------6---------7---------8---------9---------A
Histogram shows probabilities from 0 through 0.1 and 0.9 to 1 (0, 1, 9, A).
Probability 0.31 appeared 150030 times.
Probability 0.33 appeared 152955 times.
Probability 0.4 appeared 164107 times.
Probability 0.45 appeared 152542 times.
Probability 0.49 appeared 157924 times.
Probability 0.55 appeared 151192 times.
1000 loops, 10000 appearances in a concert, 10000 couples

It confirms the basic shape of the curve, and points towards the same interval of max values, but I couldn’t calculate a with the same precision.

This week the #puzzle is: Can You Squeeze the Squares? #optimal #placements #packing

And for extra credit:

Can You Squeeze the Squares?

***

“Last Week’s Fiddler. Congratulations to the (randomly selected) winner from last week: 🎻 Lise Andreasen 🎻 from Valby, Copenhagen, Denmark.” 😁😁😁

***

Highlight to reveal (possibly incorrect) solution:

Desmos 1 Desmos pictures 1 and 2.

Some assumptions:

• The placement of the 1st square has to be symmetrical in some way. It will be in the middle. If it is not in the middle, some corner or side will leave more room for the 2nd square, and of course my friend doesn’t want that.
• The 1st square will either have sides parallel to the sides of the big square or 45 degrees shifted.

Some experimentation:

• As picture 1 shows, if the 1st, green square is tilted, A is about 2.7. As picture 2 shows, if the 1st, red square isn’t tilted, A is 3. (This picture also demonstrates, that the 2nd square shouldn’t be tilted.) My friend will choose not to tilt the square. A = 3.
• If A was slightly smaller, the ring around the 1st square wouldn’t be wide enough for an extra unit square to fit.

And for extra credit:

Desmos 2

Experimentation:

• A small square in the middle of the big square, aligned with the sides of the big square, dominates all 4 corners of the big square nicely. It’s very hard to come up with something using 2 extra squares, that could dominate a bigger square better. I therefore say as a 1st guess B >= 3.
• A tilted square might dominate 2 corners. 2 other aligned squares in the 2 other corners handle the rest. (Desmos shows 2 green squares, when there’s room for them.) In that case B = 2 + √2 = 3.4142.

This week the #puzzle is: Can You Sprint to the Finish? #probabilities #strategy #optimal

And for extra credit:

Can You Sprint to the Finish?

Highlight to reveal (possibly incorrect) solution:

Formulating what we already know:

• There are 2 competitors, A (me) and B.
• There will be 1 winner.
• legs(X) describes the state of X’s legs, somewhere between 0 and 100 percent.
• For each competitor these 2 steps will be done in order:
  • Determine legs(X).
  • Decide whether to sprint.
• If both sprint, winner will have max(legs(X)).
• If 1 sprint, they will win.
• If 0 sprint, p(A wins) = p(B wins) = 50%
• If legs(B) >= 50%, they will sprint.
• A strategy is a list of values, that will lead to a sprint. For B the strategy is 50-100.

Now lets look at the different situations. Let a = legs(A) and b = legs(B).

Situation 1: a < 50.

In this situation, the strategy of always sprinting dominates. Situation 2, a = 50.

In this situation, all strategies lead to the same results. Finally situation 3, a >= 50.

The dominant strategy in this situation is that A sprints. This leads to an overall strategy of always sprinting. However, that wasn’t the question.

• p(A wins)
• = p(a < 50) * (p(b < 50) * 1 + p(b > 50) * 0) + p(a > 50) * (p(b < 50) * 1 + p(b > 50) * 1/2)
• = 1/2 * (1/2 + 0) + 1/2 * (1/2 + 1/4)
• = 1/4 + 3/8
• = 5/8

My result: 5/8 = 0.625 = 62.5%.

And for extra credit:

Program

I couldn’t quite get the math right in my head, so I monte carloed the problem. My result: 0.307 / 30.7%.

Det er på tide at skæve til Hugo-kategorien Best related. Diverse med andre ord.

“Charting the Cliff: An Investigation into the 2023 Hugo Nomination Statistics” og “The 2023 Hugo Awards: A Report on Censorship and Exclusion” er nyheder. De gav et stort plask, da de var nye, og de har påvirket Worldcon. Specielt den anden.

“The Spectacular Failure of the Star Wars Hotel” er en video på YouTube. Jeg så den faktisk, da den var ny. En yderst grundig analyse. Jeg blev dog lidt irriteret over de jævnlige skift af kostume.

r/Fantasy’s 2024 Bingo Reading Challenge (r/Fantasy on Reddit) er mest af alt et fænomen. Jeg kender ikke så meget til det, men det lyder da interessant.

Jeg har allerede behandlet Speculative Whiteness: Science Fiction and the Alt-Right og Track Changes: Selected Reviews.

Åh, det er svært. Men Hugo-rapporten står mit hjerte nær.

Anmeldelse af Track Changes: Selected Reviews, af #AbigailNussbaum. Bog. 2024. Hugo-finalist.

Skitse: En bog fuld af anmeldelser.

Er det science fiction? Bl.a.

Temaer: Anmeldelser. Ikke min yndlingsgenre. Men hvad jeg lige prøvesmagte virkede velskrevet. Analysen af WandaVision er fx tankevækkende. Også af The City and the City.

Citat: “… what bothered me about Discovery were the obvious limitations in its understanding of the Star Trek franchise, and of what could – and should – be done in it:

Like the reboot movies before it, Discovery seems to think that the most – perhaps the only – interesting question to ask within the Star Trek universe is “should we have a Federation?” […] But the thing is, this is literally the most boring, basic question one can ask about Star Trek. The real challenges posed by a society like the Federation aren’t questions of if, but of how. How do you create a truly just, fair, equal society? How do you balance freedom of conscience and opinion with your core values of tolerance and peace? How do you prevent the exploitation of those who are weaker than you? How do you help people outside your society, and do you have the right to encourage them to be more like you?”

Er det godt? Ja. Men som sagt ikke min yndlingsgenre. 👽👽☠️

Anmeldelse af Speculative Whiteness: Science Fiction and the Alt-Right (gratis), af #JordanSCarroll. Bog. 2024. Hugo-finalist.

Skitse: Den her ser på forskellige måder, hvorpå fascister bruger science fiction. Hvad der bliver skrevet og filmet, læst og set. Hvilke holdninger det kommunikerer og understøtter. Og hvilke videnskabelige fejl, det er nødt til at begå. Her er en flok citater:

Er det godt? Det virker godt researchet. Jeg er jo ikke den rigtige til at vurdere, om det hele er korrekt. Men det virker godt og sammenhængende. Og naturligvis er det her et godt tidspunkt at skrive den her bog på. 👽👽👽