Sommerferien 1982

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Af uransagelige årsager dokumenterede jeg sommerferien 1982 med mange detaljer. Da jeg fornylig scannede dit og dat af mine fotos alligevel, så røg den her også gennem maskinen. Sommerferien foregik altså 1982, i juli. Hvert indlæg vil i overskriften have den nøjagtige dato.

#boye26 report

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For some years, a friend of mine has praised the conference Boye Aarhus, as it’s called now. So this year I decided to visit. Just for 1 day. Nothing wild.

Boye 26

I bought an early bird ticket in February. I made arrangements for train and hotel well in advance. As it turned out, the trip then almost collided with the complete transformation of my apartment, because of moth treatments. You know, “we need access to all edges of floors and all kitchen cabinets, in order to spread moth poison”. But what about my stuff? Apparently, the only option for my stuff was to move to a temporary room elsewhere… That project alone was about 1.5 months.

The 1st treatment was Monday morning. Tuesday noon I was on a train, very ready to do something not related to small, flying pests.

I’m pretty sure I have stress. (Why? Moth!) I am easily confused, forgetful, not as detail oriented as I usually am. Also, I’m not in any shape to do a full day, including dinner. (Unrelated to moth.) So. I intended to do as much of Wednesday as possible, but skip the dinner. But maybe I could attend the Tuesday dinner instead? Yes? Great! And then I confusedly didn’t. Sigh. But at least I managed to find the con site (Godsbanen) and a fast route between hotel and con.

Wednesday arrived. I had pretty much made a list of all the User Experience talks. Yeah!

9.00. Welcome to the conference

  • Somebody calls this a digital conference? Huh. Yeah, I guess that’s a good word. Actually helps, to get a word.
  • There are no microphones? Oh, that might be a problem. (Yes, it was, for the usual reasons.)
  • I count about 60 chairs, most of them taken.

  • An athlete describes how retiring from a very demanding career (Olympic silver at some point) is like dying. I don’t know. Maybe? That goes for all kinds of retirement, right?

9.15. Mapping the agentic frontier: What have we learned, and what’s ahead

  • Horror movies are a rest from the horror show that’s reality. 😁 Let’s talk about The Fly, with Jeff Goldblum. 😁
  • What’s your AI attitude, Doomer, Accelerationist or Centrist? (Doomer, in the sense that the bubble can’t burst soon enough.) So. That’s a scale. But it’s not depicted as a scale, with Centrist in the middle. 🙄
  • I get the sense it’s impossible not to talk about AI? 😞
  • I could use a little more depth here. Yes, there’s a known effect, where new technology goes through a phase of disappointment. But there’s also a known effect, where something new generates new words, and there’s a phase of confusion, before the consolidation. The latter seems to be a secret?
  • Goes over time, but keeps talking. (In general I hate that. But this was a small con, short distances etc. It wasn’t actually a real problem.)

10.15. Adapting design systems to constant change

  • There must have been some error here, where something scheduled for the break between items didn’t happen. The 1st 10 minutes: people moving chairs and tables. Loudly. Making the “tell us your name” round even harder to hear. I also think those 10 minutes stay in the program as a delay, until lunch.
  • This is about Figma, a lot. I think that name is new to me. I did some research later. Huh. Neat. But even not knowing exactly what it was, I decide it must be some sort of platform, some underlying part of other programs or maybe a module. Good enough. I can make sense of the rest.
  • Interesting dynamic, where “we need to have a button to change to dark mode, Figma doesn’t support it, we’ll build our own”. And then some day Figma changes, and you have to build that function again, on top of the way Figma does it. Frustrating, but necessary.
  • I sit behind 2 persons who keep doing their own thing. Very distracting. After this item I switch places, so I can’t see it. Works. (The big surprise is actually, that this doesn’t happen a lot. Most people seem to be really paying attention.)

11.00. Who gets left out? Challenging assumptions in design systems

  • Some items begin at 11.15. I get confused. Come to think of it, why can’t I find a table with the programme, to make stuff like that obvious? Am I supposed to photograph the whiteboard? Why not use something like ConClár?

  • A lot of people have disabilities. This influences what they buy. It also influences what 2 of their loved ones buy. That’s a lot of money!
  • Curb cut effect – good for wheelchairs, but also good for so many others.
  • What’s intersectionality? What, I’m the only one who knows? “Well, black women are both black and women.” Kneel before me and my intelligent answers!
  • Endowment effect – I built it, it’s wonderful.
  • Invisible barriers. Solsikkesnoren. A move against invisibility.
  • This item includes 20 minutes of debate in small groups. Brilliant!

12.00. Lunch

  • Networking is so noisy.
  • Surprisingly good lunch.

13.00. What happens when experience, capability and seniority no longer develop at the same speed?

  • Ehm. I don’t know? Halfway through I get the sense the title and description doesn’t quite match the talk. Also, this is UX?
  • Oh. Dark purple letters on a black background.

14.00. Roundtables

  • My head is spinning. Back to the hotel for some rest.

15.15. How the Danish Tax Agency balances design thinking and standardization

  • The Danish state has a department for simplification? Wow. 😲
  • The dreaded “just reading the slides out”.
  • Don’t know about the word tender. I doubt it means soft, in UX Tender Requirements. (Turns out it’s related to “we need a product to do xyz, how do we find the right supplier?”.)
  • Don’t know about the double diamond.

16.15. Small Feature Award

  • A fun item at the end of the day. People are teasing each other a little. Some of them seem to know each other very well.
  • One “small feature” is a program to run a pub quiz. “Which mammal can fly?” Bat. What? Was that a hard question? Ehm. Tremble before my huge brain and its utterances!
  • “Next year the programme will include buttons to build a personal programme.” Or, you know, use ConClár.

And then I was very happy to go back to the hotel.

I think I’ll get a copy of the slides later?

Most of the day was good. I haven’t done something like this in a looong time. It’s inspiring. On the other hand I think I was a very small minority, here on my own dime, not here to further my career in any way. I was also on the older side of the average, though I was very happy to see a guy with white hair present a small feature.

#ThisWeeksFiddler, 20261030

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This week the #puzzle is: Seeking Alpha #counting #average #probabilities

In the figure below, four points are arranged in a square. The top two points can each be labeled with one of eight letters: A, B, C, D, E, F, G, and H. The bottom two points can each be labeled with one of three three letters: J, K, and L. (Hence the numbers “8” and “3” in the figure.)
You are labeling exactly two of the points with letters. The other two points will remain blank.
Once you’ve labeled two points, you must choose one of the colored connectors (yellow, red, green, or blue) between the two points you labeled—note carefully where each color connector is in the diagram. But there are some more rules:
– If the two letters are the same, you cannot use a yellow connector.
– You can only use the blue connector when the bottom points have the same letter.
Finally, two diagrams are equivalent if they are reflections of each other. For example, the following two diagrams are equivalent:
All told, how many distinct (non-equivalent) diagrams can you make?

And for extra credit:

Next, it’s time for a little game. Let’s continue with the list of distinct diagrams you just found for this week’s Fiddler, being sure to choose exactly one representative from each pair of reflected diagrams. And further suppose every diagram in your list actually contains four letters: In addition to the two “visible” letters in the diagram, the remaining two points each have a “secret” letter—8 possible letters (A through H) for a point on top and 3 possible letters (J through L) for a point on the bottom—that are fixed for a given diagram. The top and bottom letters are chosen with replacement, so that a letter can occur more than once.
So here’s the game: In each round, you are randomly generating a new diagram. This time, the top two points are each randomly labeled with one of nine letters: A, B, C, D, E, F, G, H, and I. The bottom two points are each randomly labeled with one of four letters: J, K, L, and M. Letters are chosen independently and with replacement. Once these letters are assigned, you randomly pick two points whose letters are “secret.”
Between the remaining two “visible” letters, you randomly pick from among the four connectors between them shown in the diagram below:
While there are indeed four connectors between any two points, there are five connector types. Here are the point values associated with each type:
– A gray connector is worth 0 points
– A yellow connector is worth 1 point.
– A red connector is worth √2 points.
– A green connector is worth √3 points.
– A blue connector is worth √6 points if it connects identical letters—otherwise, it’s worth 0 points.
Now, you earn points when your randomly generated diagram perfectly matches one of the distinct diagrams from your list, including: the four letters, which letters are visible, their orientation (no reflections allowed this time!), and the color of the connector between the visible letters. If there is indeed a diagram that’s a perfect match, you get the number of points corresponding to its connector. Otherwise, you get no points.
On average, how many rounds of this game can you expect to play for each point that you earn?

Seeking Alpha

Solution, possibly incorrect:

Program

The only method I can think of is to count all the options. Before I start, I label the points.

  • Top left: a
  • Top right: b
  • Bottom left: c
  • Bottom right: d

I also label the letters.

  • A letter is the 1st letter, if the label for the point beneath the letter comes before the letter for the other point with a letter.
  • 1st letter: l1
  • 2nd letter: l2

Now let’s start counting.

  • Points with letters: a, b
    • l1 = l2, connector is G
      • 8 options
      • (8 options for l1, 1 option for l2, 1 option for connector)
    • l1 ≠ l2, connector is Y or G
      • 8 * 7 * 2 / 2 = 56 options
      • (8 options for l1, 7 options for l2, 2 options for connector, remove equivalents (AB is equivalent to BA)
  • Points with letters: a, c
    • l1 ≠ l2, connector is Y or R
      • 8 * 3 * 2 = 48 options
  • Points with letters: a, d
    • l1 ≠ l2, connector is Y or R
      • 8 * 3 * 2 = 48 options
  • Points with letters: b, c (equivalent to a, d)
  • Points with letters: b, d (equivalent to a, c)
  • Points with letters: c, d
    • l1 = l2, connector is G or B
      • 3 * 2 = 6 options
    • l1 ≠ l2, connector is Y or G
      • 3 * 2 * 2 / 2 = 6 options

In all, there are 8 + 56 + 48 + 48 + 6 + 6 = 172 options.

Method 2, I build a program to create all the options and count them. Same result.

And for extra credit:

Let me see, if I can count my way out of this one too.

This time the letters are simply la, lb, lc and ld. The 2 points chosen/visible are p1 and p2. The connector chosen is C.

A diagram is completely described by (la, lb, lc, ld, p1, p2, C).

When does a diagram fit perfectly with 1 from my list? The 1st yellow p is the probability the letters might actually hit something from my list of diagrams. The next is that the hidden letters fit with what I chose. The next is that the connector fits. And then I multiply those 3.

How often does the correct connector occur? The 1st p is the probability from above transferred down here. I multiply the probability with the average point value from the connectors. Then I combine into the 6 cases of point choices. And then I take the average. Finally I calculate the reverse.

Okay, let me see. The average of all of this is 0.009718. This reversed is 102.9005. This means, for every 102.9005 games, I get 1 point. So I guess that’s my answer. I have to play 102.9005 rounds of this game to get 1 point.

Method 2, add monte carlo to my program. Build a list of diagrams. Then create random diagrams and check whether they are in the list.

Number of loops.... 5 * :      3162523
Target deviation........:    0.0001000
Deviation so far........:
................result 1:    0.0000725
======================================
Final result E, 1.......:    0.0112332
                         +/- 0.0000725

Result F 1: average number of games / point, 89.021472

So this gives me about 89, and method 1 gave me about 103. There’s enough of a difference, that I must have made a mistake somewhere.

Method 3. Hm. Instead of generating random diagrams, let me count all of them. WLOG, let’s assume all the diagrams in the good list are the ac and ad reflections.

  • abCAAXY
    • This is a diagram, where the visible points are a and b, the connector is C, the visible letters are A and A (a letter from the A-I group) and the hidden letters are X and Y (letters from the group(s), the hidden letters belong to)
    • This can be done with 4 different connectors, 9 different letters A, 4 different letters X and 4 different letters Y.
    • 4 * 9 * 4 * 4 = 576 options.
    • Of these, 8 are in the list of good diagrams.
  • abCABXY
    • The visible letters are A and B, 2 of the letters from the A-I group.
    • 4 * 9 * 8 * 4 * 4 = 4608 options.
    • Of these, 56 are in the list of good diagrams.
  • acCAJXY
    • The visible letters are A and J, a letter from the A-I group and a letter from the J-M group.
    • 4 * 9 * 4 * 9 * 4 = 5184 options.
    • Of these, 48 are in the list of good diagrams.
  • adCAJXY
    • 5184 options.
    • 48 are good.
  • bcCAJXY
    • 5184 options.
    • 0 are good.
  • bdCAJXY
    • 5184 options.
    • 0 are good.
  • cdCJJXY
    • 4 * 4 * 9 * 9 = 1296 options.
    • 6 are good.
  • cdCJKXY
    • 4 * 4 * 3 * 9 * 9 = 3888 options.
    • 6 are good.

So far, there are 576 + 4608 + 4 * 5184 + 1296 + 3888 = 31104 options, and 172 of them are good. But we also need to count the points.

  • abCAAXY
    • 8 are good. All will have a G connector.
  • abCABXY
    • 56 are good. Half will have a Y connector, half a G.
  • acCAJXY
    • 48 are good. Half will have a Y connector, half a R.
  • adCAJXY
    • 48 are good. Half will have a Y connector, half a R.
  • cdCJJXY
    • 6 are good. Half will have a G connector, half a B.
  • cdCJKXY
    • 6 are good. Half will have a Y connector, half a G.

Connectors of each type:

  • Y: 28 + 24 + 24 + 3 = 79
  • R: 24 + 24 = 48
  • G: 8 + 28 + 3 + 3 = 42
  • B: 3

This gives a sum of points of 79 * 1 + 48 * √2 + 42 * √3 + 3 * √6 = 226.97685414.

So. With 31104 options, I score 226.97685414 points.

31104 / 226.97685414 = 137.03599919. I have to play about 137.04 games to get 1 point.

Hm. I think part of my problem with method 1 was, that I didn’t get the probabilities quite right. Like, the split into the 2 ab choices give 2 probabilities, and I think I just averaged them. But they represent groups of different sizes.

I have no idea what went wrong with method 2. But method 3 was much easier to calculate, so method 2 probably had a mistake in a complicated function.

Oh. I think I made a booboo. It’s important to count options. But I should still have added, that the ab-group and the cd-group are equally likely, even though they don’t have the same size. So. Oops. No time this week to get the correct result.

Method 4. Method 1 done right? About 137? Sheet

Nyt fra sf-land!

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