Skitse: Der foregår mærkelige ting i USA, så vores fortæller må rejse over og undersøge det, måske endda stoppe det. I mellemtiden må det tolereres, at udenfor Gallacia forstår man ikke det her med en person, der ligner det ene køn (har bryster), opfører sig som det andet køn (er soldat) og gerne vil bruge et tredje køns stedord (3 køn???).
Er det science fiction? Nej. Fantasy/horror.
Temaer: Mærkelige ting, nede i en kulmine, en person er forsvundet, bla bla bla. Mindelser om Lovecraft. Fortælleren kan ikke lide det her og tog gerne benene på nakken. Og det er så her, jeg falder af og ikke læser historien færdig. Der er alt for meget “uha, hvad sker der nu”.
Det her er vist den tredje i en serie. Jeg kom vist gennem nr. 2, men husker godt, at mine nakkehår blafrede vildt under læsningen. Puha. Hvad værre er, forfatteren vil godt være helt sikker på, at læserne kan huske detaljer fra de foregående historier, så der er store klumper af “jeg huskede med rædsel” osv. Nix. For meget.
“… his Boston accent was so thick that you could stand a spoon up in it.”
“The way he pronounced dark as dahhhk was so pure that I had an involuntary urge to snatch up the teapot and find a harbor to dump it in.”
Skitse: På den anden side af floden bor sommerfolket. (*) (Sommerhusejere? Elvere?) Man skal passe på med at tage derover, man kan blive meget forandret. Og så er der ovenikøbet fred mellem os i øjeblikket. Nåh. Celia har 2 brødre, Argent og Roric. Argent har et stort opgør med deres far og forlader så huset for evigt. Celia elsker ham højt og tager det personligt, at han ikke engang ville sige farvel til hende. Og ups, hun vidste ikke, at hun kunne magi, så hun kommer til at kaste en forbandelse på ham.
(*) 2 historier så tæt på hinanden, der har samme udgangspunkt? Okay så.
Er det science fiction? Nixen bixen. Fantasy.
Temaer: Familien er mægtig nok til, at der er politik involveret i ægteskaber og sådan noget. Da det pludselig viser sig, at Celia kan magi (eller grammatik?), så træder et gammelt løfte i kraft, og hun skal giftes med prinsen. Celia forstår nok af politikken til, at hun accepterer situationen. Hun kan også godt forstå, at hele situationen er mere kompliceret, end man skulle tro. At deres far må træde varsomt for ikke at miste alt igen.
Problemet med Argent var ikke så meget, at han ikke er til piger, men at det blev opdaget. Eller sådan tænker deres far i hvert fald. Men resultatet er i hvert fald, at Argent krydser floden. Efter historierne at dømme, så bliver han en stor helt. Og Celia “opdager” pludselig, at hun har endnu en bror, der også er ensom.
Sommerfolket bedrev krig på en underlig måde. Krigen tog mange år. Den bestod altid i, at sommerfolket krydsede floden, når sommeren begyndte. Så blev der kæmpet lidt hist og her. Om efteråret tænkte de pludselig “guuud, der er noget der hedder efterår, så kan vi jo ikke være her”, og så skyndte de sig hjem. Lidt glemsomme folk. Nogle år glemte de vist også, at der overhovedet var en krig. Skidt pyt. Celias far var god til strategi, så han var med til at afslutte krigen.
Naboerne er i det hele taget lidt underlige. Det er ikke kun hukommelse, der virker anderledes. Deres idéer om ære er mærkelige. Man kan godt have en kamp til døden formiddag og eftermiddag og spise sammen til frokost. Løfter spiller en meget stor rolle.
“My lady, I ask you to aver that you come here with your family’s consent and of your own free will to be married to the prince, as your king has commanded…” Både fri vilje og kongens ordrer? Hm.
“A murmuring swelled through the court, but not of the objections that his people might sensibly have made to Elithyon giving an enemy more chances to kill them. Instead the summerlings all seemed pleased by their prince’s graciousness, even the other knights who were waiting for their chance to fight.”
Er det godt? Egentlig okay. Men den fangede mig ikke helt. Der er igen det her med, at der er regler, men at de er lidt hemmelige. ##-
This week the #puzzle is: The Ant Goes Marching #geometry #minimum #coding
June the ant is on a cylinder. More specifically, she is on the edge of one of the cylinder’s two circular faces. Her dinner is on the edge of the opposite circular face, and all the way around on the other side of that face. The radius of the cylinder is 2 meters and its height is 2 meters.
Your job is to help June find the shortest path along the surface of the cylinder so that she can chow down as quickly as possible. What’s the length of this shortest path?
And for extra credit:
Now, June is on a hollowed-out cylinder, also known as a “cylindrical shell.” The shell’s outer radius is 2 meters and its inner radius is 1 meter. The shell is 2 meters tall. June is on the outer edge of one of the cylinder’s two flat faces. Her dinner is on the opposite face, and all the way around on the other end of that face.
Once again, your job is to help June find the shortest path along the surface of the shell so that she can chow down as quickly as possible. What’s the length of this shortest path?
While I did mess around with various techniques, I ended up using a program to go through a lot of options.
There’s an angle between the beginning point, the center of the top disk and the point, where June begins walking on the curved part of the cylinder. Similarly for the bottom disk. Assumption: I only have to look at cases where these 2 angles sum to 180° or less, and where the shortest possible path is taken on the curved part.
I go through a lot of options where these 2 angles are anything from 0° to 180°.
The length of the path across a disk with radius r (in this case 2) depends on the angle:
For the curved part, I can simply assume, that it has been laid flat. It has height 2, and the width depends on how much is left, after the 2 angles have used some. E.g., if the 2 angles eat up π/2 between them, there’s 3π/2 left. At the most, it has width 2π.
Add these 3 paths parts together, and we have the length of the whole path.
Go through a lot of options, and remember the best one.
I also made a heat map of the options. It tells me, that the shortest paths occur in the corners, when 1 angle is 0 and 1 angle is 180. The program reaches the same conclusion. In this situation the path has length 6: 4 across a disk and 2 going straight down the curved part of the cylinder.
And for extra credit:
Similar to the fiddler, I use a program, with small modifications.
I assume again there are 2 angles. This time I cap both at 90°. My program actually handles 90° slightly wrong, but the solution isn’t close to that situation, so it doesn’t matter.
Across a disk, June travels between the outer circle and the inner. She also travels down the inner cylinder.
For the length traveled across a disk:
This time the solution is close to 26° for both angles. Length 5.36896.
Heat maps. Read 540 on the left as 5.40 and 5368 on the right as 5.368. Further read 0-10 on the right as 20-30.
Skitse: Detektiven Dorothy vågner uventet, i en elevator, og ikke i sin egen krop. (Normalt bliver en backup installeret i en klon af sin egen krop og på et fredeligt sygehus, her ombord på rumskibet, der skal rejse et årtusind med de samme 10.000 passagerer hele vejen. Det er muligt at “ligge på hylden” et stykke tid, og Dorothy havde meldt sig ud af samfundet 2 år, da hun altså pludselig blev flået tilbage.) Altså, Dorothy vågner. Rumvejret er barskt, så skibets computer er høj/skæv/fuld, og Dorothy kan ikke få et ordentligt svar på, hvad der egentlig foregår.
Er det science fiction? Nemlig, sf/krimi.
Temaer: Dorothy vågner i Glorias krop. Hun trisser lidt rundt, opsøger Glorias hjem og arbejdsplads og Dorothys egen nevø. Det er praktisk for en detektiv sådan at ligne en anden! Det viser sig hurtigt, at der er sket en eller flere forbrydelser, inklusive et mord.
Dorothy er en rigtig god detektiv, og det var først her på rumskibet, hun fik det job. Henover 300 år har det vist sig, at hun er dygtig.
Selvom personerne kan tage backup, så bliver Jorden mere og mere tåget for dem. Løsningen er at drikke mindesprut. En god cocktail kan indfange fornemmelsen af at stå på en bakke i regnvejr.
Rumskibet har lidt sine egne regler. Politi, dommere, straf. Økonomi, borgerløn, lån.
Er det godt? Ja. Godt skruet sammen, forbrydelsen bliver langsomt afdækket, og forbryderen giver mening psykologisk. ###
Skitse: Da Ellas far dør (der var gift i teen), ryger Ella (16 år gammel) også selv med. Da huset er magisk, kan det give Ella en slags nyt liv, som spøgelse. Den onde stedmor og de 2 onde stedsøstre finder hurtigt ud af, at spøgelses-Ella stadig kan lave mad og rydde op; glimrende ordning.
Er det science fiction? Nope. Fantasy.
Temaer: Det her er naturligvis en genfortælling af Askepot (Cinderella), og som altid kan en genfortælling i den her længde have mange flere detaljer, mums. Den her gang er hovedpersonen et spøgelse / et hus. Det giver nogle nye aspekter af historien. Såsom: Ella skal være hjemme ved midnat, fordi reglerne for spøgelser er skruet sådan sammen, det har ikke noget med ballet at gøre. Og: Stedsøstrene kan pine Ella via huset, så når Ella ikke vil makke ret, så er det i gang med smadrede vinduer og gennemsavede vindueskarme.
Ella er vist queer. Det får betydning senere, da hun møder prinsen og dennes politisk relevante fremtidige forlovede.
I denne magiske verden har prinsen også nærkontakt med magi, på sin egen pinefulde måde. Både Ella og prinsen kæmper med, at de ikke bliver set ordentligt, og de svælger i at se hinanden. Det har stor betydning for Ella, når hun får en ny ven; det sker så sjældent.
“When she moved, the gauziest layer of her skirts took some time to come dreamily to a halt.”
“This was all she’d wanted: someone who would warmly squeeze her hand and could see her, could hear her, listening to her talk about something she loved.”
Er det godt? Ja! Nogle af de her genfortællinger er forudsigelige, men den her fik mig. ###
This week the #puzzle is: Can You Play Hide-and-Seek? #minimum #maximum #minmax #coding #probabilities #InfiniteSum #integral
I (Nis) am playing hide-and-seek with my nephew. I start at point O, whereas my nephew can hide at point A, B or C. I can walk from O to A in 2 minutes, from O to B in 3 minutes, from O to C in 4 minutes, and from B to C in 5 minutes. To get from A to B or from A to C, I must pass through O.
My goal is to minimize the time it takes to find him, no matter how clever his strategy might be. What is this optimal time?
And for extra credit:
My nephew can no longer hide at C, and is instead limited to A and B. But this time, he has a teleporter that can instantaneously transport him from A to B or from B to A. He can use the teleporter as many times as he wants. However, he can’t react to my approach, and must instead plan out his transport schedule ahead of time. That said, he does know the precise time when the game starts.
My goal is to minimize the average time it takes to find him, no matter how clever his strategy might be. What is this optimal time?
So, I’m looking for a strategy, where the maximum time is minimized. The maximum time is the time required to look in both A, B and C.
A strategy should look in both A, B and C. It can described by the order in which Nis visits these places. Nephew’s strategy can be described by where he is hiding. Let’s look at all the strategies and all the related times. (I did these by hand and confirmed them with a program.) Nis’s strategies going down, Nephew’s strategies going across:
A
B
C
Max
ABC
2
7
12
12
ACB
2
13
8
13
BAC
8
3
14
14
BCA
14
3
8
14
CAB
10
15
4
15
CBA
14
9
4
14
Example of calculation: ACB is actually the path OAOCB, and the time required is 2 + 2 + 4 + 5 = 13.
The min-max occurs with the strategy ABC, and the time is 12 minutes.
Method 1: I modify my program to also look at teleporter strategies. A strategy for Nis can still be described by the order in which he looks in different places, but he might have to look in more than 2 places, and he has the option to stand still in 1 place for a while. For Nephew there’s a sample rate involved. To keep things simple, I say that Nis (when not on the move) looks once every minute. The strategy BBA for Nis would be: go to B and look immediately, wait 1 minute and then look again, and finally go to A. A strategy for Nephew has to describe where he is when Nis is looking. BBA would describe being at B after 1 and 2 minutes and then at A after 3 minutes. Both sets of strategies will stretch to infinity, as the worst case is that Nis never finds Nephew. But I put in a limit in the program, where I simply say, if Nephew hasn’t been found by this time, assume the time will be finite, but large. I also keep track of how many times I invoke this large time.
My program produces something like this:
Winner: AAAAAAAAAAA, time 3.10449, 4 cases out of 4096 of not being found Winner: AAAAAAAAAAB, time 3.10449, 4 cases out of 4096 of not being found
And I go: oooh.
Method 2: Assume AAA… is the best strategy.
1: Can I argue, that this might be the case?
Let’s look at 4 simple classes of strategies for Nis, beginning with AA, AB, BA and BB.
AA uses 2 minutes to get to A and then stays there for 1 minute.
AB uses 2 minutes to get to A and then uses 5 minutes to go to B.
BA uses 3 minutes to get to B and then uses 5 minutes to go to A.
BB uses 3 minutes to get to B and then stays there for 1 minute.
And how are these doing against Nephew’s strategies? * is used as a wild card.
AA finds Nephew with the strategies *A* and **A*.
AB finds Nephew with the strategies *A* and ******B*.
BA finds Nephew with the strategies **B* and ********A*.
BB finds Nephew with the strategies ***B* and ****B*.
AB and BA are horrible. It takes 2-3 minutes to look in 1 place, and then 5 more minutes to look in the next. All that time wasted. AA and BB seems better. Further AA looks better than BB, because it starts looking earlier. So, case argued. Go to A and stay put. Sooner or later Nephew will show up.
2: What is the average time of the strategy AAA…?
If I’m still working with a sample rate, Δ, this goes into my calculation.
After 2 minutes Nis looks for Nephew at A. Half his strategies has him there, and Nis finds him.
After 3 minutes Nis looks for Nephew at A again. Half of his remaining strategies has him there, and Nis finds him.
This represents this average of times:
Or to use my variable Δ:
So far we have the result, that the average time is 2 minutes or more. Makes sense.
If I simply let Δ = 0, the whole average time will be 2 minutes.
HOWEVER. Nis is not playing against all strategies. Nephew knows about this strategy and might simply decide to go to B and stay there. In which case E(t) = ∞. Not good.
Method 1 again: I look at some of the other candidates for strategies. Hm.
Method 3: Assume a best strategy. A best strategy has to include some randomness, otherwise Nephew could just always be the other place. It also has to discard the idea of a sample rate, because Nephew could fight that having a higher teleporter rate. “You didn’t find me at A, because I was at B? Well, now I will stay at B for 5 minutes, knowing you can’t catch me there. Well, 4 minutes, just to make sure.”
Those 5 minutes have to figure into Nis’ strategy.
Randomness. Begin by looking at A or B, but randomly. It can’t pay to look at A early, even though it’s possible, Nephew could simply decide to be at B at this time. So look at either A or B after 3 minutes. (The latter is the strategy “B”. The former is “AA”. Or maybe “OA”, that is, staying at O for 1 minute before going to A. Or maybe simply “-A”, signifying 1 minute of random walking + 2 minutes of walking from O to A.) Choose between “B” and “-A” by the flip of a coin. Both looking at A and looking at B are now equally probable. Nephew doesn’t have any advantage of choosing one over the other. I assume Nephew will also flip a coin to choose either A or B. So after 3 minutes, half of the time Nephew will be found.
The next step is to be at A or B with equal probability some time later. It takes 5 minutes to switch places, so that has to be the next time, 3+5 = 8 minutes. The 4 resulting strategies could be “-A—-A”, “-AB”, “B—-A” and “B—-B”. (Walking between A and B for 5 minutes. Or walking between A and A for 5 minutes.) Again, half of the time Nephew will now be found.
The average time looks like this, upcycling the formula from above.
Overvældende godt felt! Det er svært. Kaijuen skulle måske i virkeligheden kun have 2,5/3. Men så er der stadig 3 kandidater tilbage. Murderbot er jo dejlig. Vegetaren er også stadig charmerende. Så måske er pigen på tredjepladsen. Decisions, decisions! Det bliver Murderbot som nr. 1.
Skitse: Fortælleren har egentlig et solidt liv. Hun blev gift med Charlie, og sammen fik de noget fornuftigt ud af en kedelig grund med en ramponeret lade. Det blev et hjem. De elskede hinanden. Så hvorfor har hun bundet Charlie til sengen, mens han råber en andens navn, og hvorfor sidder hun selv ude på verandaen med et haglgevær?
Er det science fiction? Nix. Fantasy/horror.
Temaer: Det kom snigende ind, at da den andens navn blev afsløret, så kendte jeg jo godt noget af den her historie. Omend det er nyt, at der er overnaturlige kræfter involveret.
Der er en diskussion af begrebet ejerskab. At et ægtepar kan eje et hjem. At en kvinde kan eje sin mand?
Citat: “He bellowed for her out of the belly of the house.”
Er det godt? Ja. Men slutningen virkede brat på mig. ##-
Skitse: Et show kan indeholde alle mulige små numre: den trænede hund, akrobater. Både fortælleren og “Millay”, tidligere Miller, er tryllekunstnere. Og kvinder. Den ene klæder sig ud som mand, den anden optræder fjollet. Fordi kvinder kan jo ikke lave seriøst, dygtigt tryl.
Er det science fiction? Nej. Fantasy?
Temaer: Tryl. Kvinder. Sexisme.
Er det godt? Ja da. Men jeg gættede godt nok næsten slutningen for tidligt. ##-
. Hugo finalists 
Stuff from ommadawn.dk (Lise Andreasen) 