Anmeldelse af “Jamais Vue” (gratis), af Tochi Onyebuchi.

Skitse: En terapeut (?) eller tekniker (?) arbejder med en patients hukommelse. Hvordan går man? Hvordan føles det at gå? Minder den her gåtur om en anden gåtur? Når der sker skade på hjernen, så kan det være indviklet at få det hele gendannet.

Er det science fiction? Ja. Udover det her med at arbejde med andres minder, så er der også fortælleren selv, der kan skrue sine fingre af.

Temaer: Der er en lille gruppe af teknologier, der på forskellig vis påvirker ens hoved. Noget kan huske, hvad man har oplevet. Noget andet kan lægge begrænsninger på ens adfærd. Fortælleren er vant til at arbejde med den slags. Sikken magt, sådan at kunne rode rundt i folks hoveder. Sikken fristelse.

De 2 patienter i historien har noget tilfælles. De repræsenterer hver sin side af en bestemt type situation. Således bliver der en større historie ud af det. En voldsom historie.

Er det godt? Mja. Jeg får en fornemmelse af, at der er noget usikkerhed, noget hemmelighedsfuldhed, der mere er et stilvalg end en nødvendighed for historien. Og fortællerens egen lille udvikling virker heller ikke nødvendig. 👽👽💀

Note: Jeg tager her fat i en gruppe historier, som et magasin stiller gratis til rådighed, fordi Locus har anbefalet dem.

This week the question is: Can You Survive March Madness?

March Madness—the NCAA’s men’s and women’s basketball tournaments—is here!

The single-elimination tournament brackets consists of 64 teams spread across four regions, each with teams seeded 1 through 16. (In recent years, additional teams beyond the 64 have been added, but you needn’t worry about these teams for this week’s puzzle.)

Among the 16 teams in a region, you might wonder which team has the toughest schedule. One way to evaluate a team’s strength of schedule within the region is to compute the geometric mean of strongest opponents a team can face in various rounds.

For example, the 1-seed faces the 16-seed in the first round, then (potentially) the 8-seed in the second round, then (potentially) the 4-seed in the third round, and finally (potentially) the 2-seed in the fourth round. The geometric mean of these opponents is the fourth root (since there are four opponents) of 16 · 8 · 4 · 2, or approximately 5.66. In this computation, we used the 8-seed rather than the 9-seed because 8 is less than 9, we used the 4 seed rather than the 5-seed, 12-seed, or 13-seed because 4 is less than 5, 12, and 13, and so on. The tougher a team’s strength of schedule, the lower this geometric mean.

Of the 16 teams in the region, which two seeds have the toughest strength of schedule?

And for extra credit:

Instead of 16 seeded teams in a region, suppose there are 2^N seeded teams in the region, where N is a very, very large number.

The two seeds with the toughest strengths of schedule have seeds that approach fractional values of 2^N. What are these two fractions?

And here’s my suggested solution. Including my reasoning.

Jeg har lige læst debatindlægget Vi skal lytte godt efter, når vores folkevalgte en gang imellem siger sandheden (paywall). Noget af tesen er, at politikere (i hvert fald dem på Christiansborg) lyver, og det er sådan set bare et vilkår, ja, faktisk er det en god ting.

Jeg synes stort set aldrig, løgn og tavshed er gode ting. Hvis man planlægger noget, der har et godt mål, men ikke kan lykkes, medmindre processens indhold, måske ligefrem processens eksistens, holdes hemmelig? Hvis man deltager i en forhandling, hvor det er virkelig rart at vide, at det kun er det, vi aftaler, der kommer ud senere? Men ellers burde sandhed bare altid være reglen.

Nåh. Lad os se, hvad indlægget snakker om.

Partiets formand:

• En del af fornøjelsen ved at være medlem af et parti er, at partiet udadtil skal være enigt om alting. Deriblandt at formanden er den rigtige.
• Det er selvfølgelig ikke tilfældet i virkeligheden.

Valg og meningsmålinger:

• Påstanden er, at det jo kun er valgresultatet, der tæller.
• Men alle bruger også meningsmålinger.

Integritet:

• Politikere kan ikke bestikkes.
• Men det er dyrt at føre valgkamp …

Politikere har holdninger:

• Holdninger, visioner, fakta.
• Spin, proces, timing osv.

Partiet og de menige medlemmer:

• Det er skønt at møde de menige medlemmer.
• Medlemmerne har reelt ikke så meget indflydelse, det er hårdt arbejde at rejse til møder, man får ikke brugbart input, og en del af øvelsen er at fremstå, som om man ikke lægger afstand til baglandet.

Åh altså! Politik.

Den der med økonomien burde dælendulme fikses. Grænser for bidrag til partierne! For alle partier! Nu til dags burde valgkamp måske være en hjemmeside? Værsgo, I har 10.000 ord til at sige, hvorfor man skal stemme på jer. I kan tilføje undervejs, hvis I ikke bruger alle ordene med det samme. Slut! Stop for valgplakater, annoncer osv. Stop for mediedækning, fordi folk 3 dage før valget “pludselig opdager” noget. Journalister må ikke bruge politikere som kilder i den periode. Sådan. Så bør alle med sindsro kunne gå ud og sige, at de har integritet.

Nu vi er i gang, så har jeg lyst til at forbyde meningsmålinger også. Forbyde at lave dem, deltage i dem, bruge dem. Kan man det? Det ved jeg ikke. Men det kunne være sjovt at prøve. Det ville være skønt, hvis politikerne virkelig gik på job i op til 4 år, med fokus på deres valgløfter og ikke så meget andet.

Så har vi nogle mere komplekse ting. At man kan tro på, at en gruppe udadtil bør fremstå enig, og samtidig være uenig i gruppens kurs. At man kan gøre noget for at fedte for folk. Vi burde kunne rumme i vores små hoveder, at det er sådan, det er. Selvfølgelig er x politikere ikke enige om alting. Selvfølgelig fedter politikere for vælgere, også partimedlemmer. Det er da ikke nogen hemmelighed? Måske er problemet her, at det ikke er svaret, der er dumt, men spørgsmålet. “Hvordan føles det at være hjemme og møde partifællerne?” Hvad tror du? “Bakker du op om jeres formand!” Gæt 3 gange.

Og så den der med spin og timing og alt det der. Hvad søren skal man sige til det? Måske igen noget med mindre presse? Dæk begivenheder i folketingssalen, pressekonferencer osv. Og lad dem så være.

Sådan! Mere sandhed er vejen frem.

This week the question is: Can You Save the Pizza?

[Pizza and snake.] The [pizza] slice … is an equilateral triangle with a side length of 10 inches. I specifically want to request the smallest rectangle of foil (by area) such that I can completely cover the slice above and below. I can’t fold the pizza, but I’m allowed to make as many straight folds in the foil as I want, without tearing it. What is the area of the smallest such rectangle of foil that would do the trick?

And here’s my suggested solution:

Read more: #ThisWeeksFiddler, 20240315

I suggest that the slice can be wrapped in this piece of foil, no less. (That picture has both the top and the bottom of the slice visible. Before cutting the foil, it would actually look more like this.) After a few calculations, I deduce that the area of the foil is 15 * 5√3 = 75√3 = 130, approximately.

ETA: I have since seen a solution with 115. So there…

Reddit has a lot of great puzzles, like this one: MATH PROBLEM

Vehicles are traveling on the same route. First they pass point A, then point B. Vehicles pass point A in equal time steps in the following order: a bus, a motorcycle, and a car. They pass by point B at the same time intervals again transportation: bus, car and motorcycle. Find the speed of the bus if the speed of the car is 90 km/h and the speed of the motorcycle is 60 km/h.

Read more: The velocity of vehicles

Highlight to reveal solution:

So here’s the situation, for B(us), M(otorcycle) and C(ar).

At t0 + 0δ B passes point A.

At t0 + 1δ M passes point A.

At t0 + 2δ C passes point A.

At t1 + 0δ B passes point B.

At t1 + 1δ C passes point B.

At t1 + 2δ M passes point B.

Whatever the distance is between the 2 points, this is how long the vehicles spend crossing.

C: t1 – t0 – δ

M: t1 – t0 + δ

B: t1 – t0

We know that v * t = d, velocity * time = distance. This gives us 3 new equations. For now, I’m not using any units, and t = t₁ – t₀, and we’re looking for v_b.

90 * (t – δ) = d

60 * (t + δ) = d

vb * t = d

In all 3 equations d is the same. Therefore:

90 * (t – δ) = 60 * (t + δ) <=>

90t – 90δ = 60t + 60δ <=>

30t = 150δ <=>

δ = t/5

Returning to the bus, we have:

v_b * t = d <=>

v_b = d/t <=>

v_b = 90 * (t – δ) / t <=>

v_b = (90t – 90δ) / t <=>

v_b = (90t – 90 * t/5) / t <=>

v_b = (90t – 18t) / t <=>

v_b = 90 – 18 <=>

v_b = 72

Adding back the units, the answer is 72 km/h.

Recently I have reread The Dispossessed, Le Guin. Going into it, I had a vague feeling he had short hair on his whole body, but I couldn’t remember anything else about his appearance. So I tried to gather what I could find, and here it is, from knobby baby to tall adult.

• a knobby [baby]
• a lanky eight-year-old with long hands and feet
• His eyes were light, and the light from the window filled them so they seemed clear as water.
• his colorless face, silvered with fine short hair
• He was still lanky, with big hands, protruding ears, and angular joints, but in the perfect health and strength of early manhood he was very beautiful. His dun-colored hair, like the others’, was fine and straight, worn at its full length and kept off the forehead with a band.
• the thin boy with big ears
• light, steady, intelligent eyes
• His light skin had tanned and the fine down that covered his face had bleached to silver
• His hands, work-hard and blackened by frostbite, lay loose on his thighs; his face in relaxation was lined and sad.
• the splendid reticent face, full of life but worn down, worn to the bone
• light, remote eyes
• a tall, frail figure
• the fine shaggy head
• his height, his long hair
• the very fine, soft, short body hair of his race
• He read that he was a towering giant of a man, that he was unshaven and possessed a ‘mane,’ whatever that was, of greying hair
• his long, thin figure and narrow feet
• two legs, two arms, and a head with some kind of brain in it
• count his fingers and toes, add them up to twenty
• his long, rough, dun-grey hair tied back with a piece of string
• You Cetians are all so tall!

And that’s that.

This week the question is: Could You Have Won the Super Bowl?

[Football, football, football.] Every time your team is on offense, suppose there’s a 1-in-3 chance they score a touchdown (which we’ll say is worth a total of 7 points, as we won’t bother with 2-point conversions here), a 1-in-3 chance they score a field goal (worth 3 points), and a 1-in-3 chance they don’t score any points (i.e., they punt or turn the ball over on downs). After any of these three things happens, your team will then be on defense.

Now, here’s how overtime will work: Your team is on offense first. No matter how many points your team does or does not score, the other team then gets a chance at offense. If the game is still tied beyond this point, the teams will continue alternating between offense and defense. Whichever team scores next wins immediately.

Again, your team is on offense first. What is your team’s probability of winning?

And let’s include the extra credit puzzle too:

If your team happens to score a touchdown on its first possession, then it doesn’t make sense for your opponent to then attempt a field goal, since they’d be guaranteed to lose. Instead, they would attempt to score a tying touchdown.

So let’s add the following to our model: When either team is on offense, they now have a choice. They can still opt for a strategy that results in 7 points, 3 points, or 0 points, each with a 1-in-3 chance. Alternatively, they can opt for a more aggressive strategy that results in 7 points or 0 points, each with a 1-in-2 chance.

Your team remains on offense first. Assuming both teams play to maximize their own chances of Super Bowl victory, now what is your team’s probability of winning?

And here’s my suggested solution, highlight to reveal:

Read more: #ThisWeeksFiddler, 20240223

This is the basis of the situation: Every time a team is on offense, p(7 points) = 1/3, p(3 points) = 1/3 and p(0 points) = 1/3. I am on offense once, on defense once, and then it’s sudden death (we take turns, first to score wins, I begin on offense).

Let’s look at just the first 2 plays (I think it’s called?).

Team 1 \ Team 2	7 points	3 points	0 points
7 points	Tie	Win for team 1	Win for team 1
3 points	Loss for team 1	Tie	Win for team 1
0 points	Loss for team 1	Loss for team 1	Tie

Each column and each row has probability 1/3, hence each cell has probability 1/9. Summing the cells, we have probability 1/3 for a win, a loss and more plays for either team.

So what happens in sudden death? The points don’t matter anymore. Every time I’m on offense, I win with probability 2/3, otherwise there’s another play. Every time I’m on defense, I lose with probability 2/3, otherwise there’s another play.

What is the chance I win or lose?

	I win	I lose	Game continues
Play 1 + 2:	1/3	1/3	1/3
Play 3:	2/3	0	1/3
Play 4:	0	2/3	1/3
Play 5:	2/3	0	1/3
Play 6:	0	2/3	1/3

Actually it’s more interesting to look at the “real” probability, not just the “remember that play x only happens because all the preceding plays didn’t settle the score”.

	I win	I lose
Play 1 + 2:	1/3	1/3
Play 3:	1/3 * 2/3	0
Play 4:	0	1/32 * 2/3
Play 5:	1/33 * 2/3	0
Play 6:	0	1/34 * 2/3

Just a quick check, does all of these probabilities add up to 1? The first row is 2/3, and every row after that is 1/3ⁿ * 2/3, from n = 1. Actually, the whole sum can be written as beginning with 1/3ⁿ * 2/3, from n = 0. Sum of terms, first term 1, each next term multiplies previous term with r, infinitely many terms, the sum is 1 / (1 – r). In our case, r = 1/3. So the sum is 1 / 2/3 = 3/2. Sum of 1/3ⁿ * 2/3 (n from 0 to infinity) = 2/3 * sum of 1/3ⁿ = 2/3 * 3/2 = 1. Great.

However, I only want to sum the cases, where I win.

1/3 + 1/3 * 2/3 + 1/3³ * 2/3 … =

1/3 + 1/3 * (2/3 + 1/3² * 2/3 + 1/3⁴ * 2/3 …) =

1/3 + 1/3 * (1/9⁰ * 2/3 + 1/9 * 2/3 + 1/9² * 2/3 …) =

1/3 + 1/3 * 2/3 * (1/9⁰ + 1/9 + 1/9² …) =

1/3 + 1/3 * 2/3 * (1 / (1-1/9)) =

1/3 + 1/3 * 2/3 * (1 / 8/9) =

1/3 + 1/3 * 2/3 * 9/8 =

1/3 + 18/72 =

4/12 + 3/12 =

7/12 =

0.58333…

So the answer is 58.333%.

Just to make sure, I also wrote a program to simulate this thing. It also reports the probability as something like 58.3%.

And for extra credit:

Once we go into sudden death, the optimal strategy is to go for 7/3/0, as that means a 2/3 chance for points (compared to 7/0 giving a 1/2 chance for points). So, nothing changes. Once we reach this stage, it’s the same chance as above, 2/3 * 9/8 = 3/4 chance for me to win.

For my enemy team, if I originally scored 7 points, the optimal strategy is to go for 7/0, as that gives them 1/2 chance of equalizing (compared to 7/3/0 giving only 1/3 chance). If I scored 0 points, the optimal strategy is to go for 7/3/0, as that gives them 2/3 chance of winning (compared to 7/0 giving only 1/2 chance). But what about 3 points? It’s a little more complicated.

Going for 7/3/0 gives them 1/3 chance of win, 1/3 chance of tie and 1/3 chance of loss. Going for 7/0 gives them 1/2 chance of win, 1/2 chance of loss. Multiplying with the chances for winning the sudden death (1 – 3/4 = 1/4) gives them a chance to win in the first case of 1/3 + 1/3 * 1/4 = 4/12 + 1/12 = 5/12, compared to 1/2 in the second case. So the optimal strategy here is also 7/0.

Let’s look at the numbers so far, seeing how often I win.

If team 1 scores	Team 2 chooses	Team 2 scores	Team 2 scores	Team 2 scores	Team 1 wins
		7	3	0
7	7/0	Tie, p = 1/2		Win, p = 1/2	p = 1/2 * 3/4 + 1/2 = 7/8
3	7/0	Loss, p = 1/2		Win, p = 1/2	p = 1/2
0	7/3/0	Loss, p = 1/3	Loss, p = 1/3	Tie, p = 1/3	p = 1/3 * 3/4 = 1/4

And now we’ve worked our way back to my initial choice. If I go for 7/3/0, my chance of winning is 1/3 * 7/8 + 1/3 * 1/2 + 1/3 * 1/4 = 13/24 = 26/48. If I go for 7/0, my chance of winning is 1/2 * 7/8 + 1/2 * 1/4 = 9/16 = 27/48. So my optimal strategy is 7/0. And then my chance of winning is 9/16 = 0.5625 = 56.25%. Ouch. Neat. (I had a math mistake, so at first got less than 50%.)

Jeg har nu (på rette tid for ikke biografen, ikke blueray, men streaming via abonnement) set The Creator. Det her er ikke en helstøbt anmeldelse, men mere et par funderinger, efter også at have set lidt ekstramateriale. Spoilers.

Den ene fundering er meget kortfattet. Vi ser, at robotter og kunstig intelligens lever side om side med mennesker i Asien-ish. Vi ser ikke umiddelbart nogle begrænsninger, omend det kan være svært at se, om en given robot har rigtig kunstig intelligens eller “bare” er programmeret. Der er robot-munke. Robotter kan adoptere (i praksis) menneskebørn.

Så hvad hulen betyder det, når en robot siger “jeg vil være fri”?

Fri for at få en amerikansk bombe i hovedet? Fordi det ville jeg ikke formulere på den måde.

Den anden fundering går på, at vi bliver præsenteret for en konflikt med 3 sider: USA-ish, Asien-ish og AI. AI var vældig nyttigt et stykke tid, men så gik en af dem amok. USA har ganske fornuftigt forbudt alt sådan noget. Asien er ikke lige så fornuftige, men bare rolig, det skal USA nok fikse. Ingen problemer i at komme på besøg og ordne sagerne. Fornylig er der kommet nyt om et AI-supervåben. Det skal selvfølgelig stoppes. Så amerikanerne sadler deres jetjagere, og så afsted.

Bortset fra, at jeg som tilskuer ikke tror på den udlægning i 5 minutter. AI virker som nogle flinke fyre. Supervåbnet skal ikke overvinde USA, men overvinde krig. Faktisk er konflikten mellem USA og hvem de nu end synes er forkert på den. Amerikanerne er de onde! Det er ikke så tit, Hollywood giver os den version.

Underligt nok taler forfatteren/instruktøren om dilemma, nuance og sådan noget. Er det den samme film, vi snakker om? Der er måske 5 mm nuance.

Derudover vil jeg egentlig bare nævne, at filmen er flot. Når der har været behov for at filme rismarker, så har de fundet nogle rismarker. Og så vil jeg også nævne, at de 2 primære skuespillere er gode.

This week the question is: How Many Loops Can You Slither Around?

Nikoli the snake wants to slither along a loop through a four-by-four grid of points. To form a loop, Nikoli can connect any horizontally or vertically adjacent points with a line segment. However, Nikoli has certain standards when it comes to loop construction. In particular:

– The loop can never cross over itself.
– No two corners of the loop can meet at the same point.
– Once Nikoli has crossed the connection between two points, Nikoli can’t cross it again (in either direction).

For example, the following two constructions are valid loops:

Meanwhile, the following three constructions are not valid. The one on the left crosses over itself, the one in the middle has two corners that meet at a single point, and the one on the right requires Nikoli to pass over the same line segment twice.

How many unique loops can Nikoli make on the four-by-four grid? (For any given loop, Nikoli can travel in two directions around it. However, these should still be counted as a single loop.)

And here’s my suggested solution, highlight to reveal:

Read more: #ThisWeeksFiddler, 20240209

Instead of focusing on the loop, I looked at the squares within the loop. E.g. the first valid loop above encloses one small square, the square in the middle of the grid. If I translate the conditions, I’m looking for a combination of squares, combining into one figure, where all squares are connected side to side. No corner connections, please. I then tried to write all the possible figures down. This is easy enough with 1, 2, 3, 8 and 9 squares. With the rest I got some assistance from Wikipedia, with list of tetrominoes, pentominoes etc.

Once I had my list, for each figure I counted, how many ways it could be placed in the grid. If we look again at the first valid loop above, counting all mirrors, rotations and simple moves, there are 9 versions of this loop.

Here are my figures and counts: 1-5 squares, 6-8 squares and 9 squares.

Adding up all the numbers I get 213 possible loops. (Adding mistake. Got 217 the first time.)