Den korte version: Niels Dalgaard og jeg starter nu en ny podcast, Robotter på loftet.
Den lidt længere version: Når man har puslet med noget lidt hemmeligt, så er det skønt endelig at kunne dele det med verden. Så kan man fx sige: Se vores fine logo!
Vi håber, at I har lyst til en podcast, hvor vi snakker science fiction-noveller.
One of the mechanisms in this game is that some of the time the player is rewarded for completing a lot of levels in a row. This is shown by Fennec gradually having more boosters for you. And it looks like this:
Yes, I had some negative stuff to say about Cradle of Empires. But there’s also positive points. Like, would you look at these background. That usually aren’t on the screen very long.
You and a friend each have a standard deck with 52 cards. You thoroughly shuffle your deck, while your friend thoroughly shuffles theirs. Then, you both draw cards one at a time. If the first card you draw is the same as the first card your friend draws, you lose! Otherwise, you draw again. If the next card you draw is the same as the next card your friend draws, you lose! Otherwise … and so on.
If the two of you can make it through your entire decks without ever drawing the same card at the same time, you both win. Otherwise, you both lose.
What is the probability that you and your friend will win this collaborative game?
I wrote a program to simulate the situation, at least with a deck with 2-9 cards. I reduce to a situation, where deck 1 is simply n numbered cards, going from 1 to n. Then I go through all the permutations of deck 2. There seems to be convergence. I guess the answer is 36.79%. I have no idea why.
Skitse: Rustem (et arabisk navn?) bor i en landsby, som er del af en mislykket koloni. De skulle slet ikke have været på den her planet, og de skulle ikke have boet under en ustabil sol. Situationen er deprimerende, og nogle af beboerne har, host, fået fornyet interesse for overtro. Og så en aften ser Rustem, der prøver at holde liv i videnskab, et stærkt lys på himlen.
Er det science fiction? Jeps.
Temaer: Sjov med tid! Og med besøgende, der påstår, de har et lille fartøj, der kan navigere på tidsstrømmene.
Rustem har det godt med orden, så det er hårdt at vænne sig til et stykke tid at have noget andet. Og at vænne sig til en Spørgejørgine.
Skitse: Ava er i starten af en lovende karriere som violinist, eller det var hun i hvert fald, indtil hun mistede noget af den ene arm i et biluheld. Forsikringen ville ikke betale, så nu får hun den gode protese, fordi den tidligere ejers enke af sit hjertes godhed har doneret den.
Er det science fiction? Ja.
Temaer: Musik, selvfølgelig. Den tidligere ejer spillede også violin, og det bliver en kamp for Ava at holde fast i hendes egen stil.
Enken presser på. Med den slags stædighed, jeg forbinder med penge. Der er en plan, og uanset hvordan det skal gå til, så vil planen blive ført ud i livet. Hvad Ava skal spille, og hvordan, og hvornår.
Er det godt? Ja. Jeg blev fanget, nærmest med det samme. 👽👽👽
I’m completing a paint-by-number painting, although this one is a little different from any that I’ve seen before. It’s an infinitely long strip of canvas that is 1 cm wide. It’s broken up into adjacent 1 cm-by-1 cm squares, each of which is numbered zero or one, each with a 50 percent chance. The squares are all numbered independently of each other. Every square with a zero I color red, while every square with a one I color blue.
Once I’m done painting, there will be many “clusters” of contiguous red and blue squares. For example, consider the finite strip of canvas below. It contains 10 total squares and seven clusters, which means the average size of a cluster here is approximately 1.43 squares.
Once I’m done painting, what will be the average size of each red or blue cluster?
Before we get to that though, woohoo!
Congratulations to the (randomly selected) winner from last week: 🎻 Lise Andreasen 🎻 from Valby, Copenhagen, Denmark.
Imagine going through the strip. I can ask: Is this square the beginning of a new cluster? If my color is not the same as the color of the preceding square, the answer is yes. Good! Let’s look at this cluster.
What is the color of the next square? If it’s not the same as mine, my cluster is over, it’s 1 square long. This happens with a probability of 50%.
Otherwise we’re still going. Let’s look at the possible square 3. If it’s not the color same as me, my cluster is over, it’s 2 squares long. This happens with a total probability of 25%. (50% chance square 2 was right, 50% chance square 3 was wrong, 50% * 50% = 25%).
Otherwise we’re still going. Let’s look at the possible square 4. If it’s not the same color as me, my cluster is over, it’s 3 squares long. This happens with a total probability of 12.5%.