This week the #puzzle is: Fine, Maybe Tiling Problems Can Be Fun… #tiling

Question row A:

You have an infinite square grid. Each cell in this grid is either red or blue.

Can you create a pattern in which every red cell has exactly:

zero red neighbors, one red neighbor, two red neighbors, ⋮ eight red neighbors?

Neighbors include any cell that is directly adjacent or corner-adjacent. To avoid admitting degenerate solutions, your pattern must have at least some fraction 0<r≤1 of the total board composed of red cells.

Question row B:

Now we will consider patterns in which every red and every blue cell has exactly the same number—nr and nb, respectively—of same colored neighbors. For example, the image above is not a valid pattern since some of the blue cells have 4 blue neighbors while others have 6.

For which pairs (nr, nb) can we construct valid patterns?

Fine, Maybe Tiling Problems Can Be Fun…

Highlight to reveal (possibly incorrect) solution:

Solutions to 1, 2, 3, 4, 5, 5, 6, 7 and 8 neighbors. (Solution for 0 was already given.)

Method: I fooled around with images.

Question row B:

Program

Some more solutions:

nr \ nb	0	1	2	3	4	5	6	7
0	image	image	image	image	image	image	image	(image)
1	image	image	image	image	image	image	(image)	image
2	image	image	image	image	image	image	image	image
3	image	image	image	image	image	image	image	image
4	image	image	(image)	(image)	image	image	image	image
5	image	image	image	image	image	image	image	image
6	(image)	image	(image)	image	image	image	image	image
7	image	image	image	image	image	image	image	image

Thoughts:

• If (n_r, n_b) = (a,b) is possible, (n_r, n_b) = (b,a) is also possible, by reversing the colors.
• If a red has 0 red neighbors, it will have 8 blue neighbors.
  • A blue edge neighbor has at least 4 blue neighbors.
  • A blue corner neighbor has at least 2 blue neighbors.
  • So the cases of (n_r, n_b) = (0, 0), (0,1), (0,2), (0,3) aren’t possible.
  • Similarly for (n_r, n_b) = (1,0), (2,0), (3,0).
• If a red has 1 red neighbor, it will have 7 blue neighbors.
  • A blue edge neighbor has at least 3 blue neighbors.
  • A blue corner neighbor has at least 1 blue neighbor.
  • So the cases of (n_r, n_b) = (1,1), (1,2) aren’t possible.
  • Similarly for (n_r, n_b) = (2,1).
• If a red has 2 red neighbors, it will have 6 blue neighbors.
  • A blue edge neighbor has at least 2 blue neighbors.
  • A blue corner neighbor might have 0 blue neighbors.
  • This does not exclude any new cases.
• Solution with 8 red and x blue can’t work, because there’s no room for the blue. Similarly for 8 blue and x red.
• If a red has 7 red neighbors, it will have 1 blue neighbor.
  • If this is an edge neighbor, it has at most 3 blue neighbors.
  • If it is a corner neighbor, it has at most 5 blue neighbors.
  • So the cases of (n_r, n_b) = (7, 6), (7,7) aren’t possible.
  • Similarly for (n_r, n_b) = (6,7).
• If a red has 6 red neighbors, it will have 2 blue neighbors.
  • A blue edge neighbor has at most 4 blue neighbors.
  • A blue corner neighbor has at most 6 blue neighbors.
  • This does not exclude any new cases.
  • It is however possible to keep building. All the reds have 6 red neighbors. Going through these cases gives a blue cell with at most 5 blue neighbors. (6,6) isn’t possible.
• Method: In some cases I think of a regular pattern, create it and count.
• Further I built a program to create and test patterns. This is how I found the (2, 6) solution.

This week the #puzzle is: How Low (or High) Can You Go? #probabilities #volume #area #random

You’re playing a game of “high-low,” which proceeds as follows:

First, you are presented with a random number, x1, which is between 0 and 1.

A new number, x2, is about to be randomly selected between 0 and 1, independent of the first number. But before it’s selected, you must guess how x2 will compare to x1. If you think x2 will be greater than x1 you guess “high.” If you think x2 will be less than x1, you guess “low.” If you guess correctly, you earn a point and advance to the next round. Otherwise, the game is over.

If you correctly guessed how x2 compared to x1 then another random number, x3, will be selected between 0 and 1. This time, you must compare x3 to x2, guessing whether it will be “high” or “low.” If you guess correctly, you earn a point and advance to the next round. Otherwise, the game is over.

You continue playing as many rounds as you can, as long as you keep guessing correctly.

You quickly realize that the best strategy is to guess “high” whenever the previous number is less than 0.5, and “low” whenever the previous number is greater than 0.5.

With this strategy, what is the probability you will earn at least two points? That is, what are your chances of correctly comparing x2 to x1 and then also correctly comparing x3 to x2?

And for extra credit:

Your friend is playing an epic game of “high-low” and has made it incredibly far, having racked up a huge number of points.

Given this information, and only this information, what is the probability that your friend wins the next round of the game?

How Low (or High) Can You Go?

Highlight to reveal (possibly incorrect) solution:

Desmos

This problem can be modelled as the volume of a shape. Imagine a unit cube. The good parts of the volume are where:

• x was low, x < y, y was low, y < z. E.g. (0.1, 0.2, 0.7).
• And the 3 other variations of x and y being high and low.

See Desmos for a calculation of the volume: 13/24 = 0.54167.

And for extra credit:

Program Spreadsheet

As there is no history in probabilities, I simply need to look at 2 guesses in a row. So this time we’re looking at the good part of the area of a unit square. The first guess can be anything. See Desmos for calculation of the area: 0.75.

Just to make absolutely sure, I wrote a small program to simulate winning a lot of times in a row. Behaves as expected.

Alternate answer: My friend is cheating. The probability is 100%.

I decided to take another look at the data. Wrote a better program. Among other things, it produced data for a spreadsheet. What do you know! The data skews somewhat. There’s a good chance, the latest number drawn was in the middle, was close to 0.5. This has an effect on whether my next guess will be correct. My program says:

Probability for win at length 10: 0.72113
Probability for win at length 20: 0.72152
Probability for win at length 30: 0.71378

So instead of 0.75 (or 1), it appears the probability is 0.71 or 0.72.

This week the #puzzle is: How Far Can You Run Before Sundown? #maximum #strategy #recursion

You’re participating in a trail run that ends at sundown at 7 p.m. There are four loops: 1 mile, 3 miles, 3.5 miles, and 4.5 miles. After completing any given loop, you are randomly assigned another loop to run—this next loop could be the same as the previous one you just ran, or it could be one of the other three. Being assigned your next loop doesn’t take a meaningful amount of time; assume all your time is spent running.

Your “score” in the race is the total distance you run among all completed loops you are assigned. If you’re still out on a loop at 7 p.m., any completed distance on that loop does not count toward your score!

It is now 5:55 p.m. and you have just completed a loop. So far, you’ve been running 10-minute miles the whole way. You’ll maintain that pace until 7 p.m.

On average, what score can you expect to earn between 5:55 p.m. and 7 p.m.?

And for extra credit:

Now let’s add one more wrinkle. At some point during the race, if you’re unhappy with the loop you’ve just been randomly assigned, you’re granted a “mulligan,” allowing you to get another random assignment. (Note that there’s a 25 percent chance you’ll be assigned the same loop again.) You don’t have to use your mulligan, but you can’t use it more than once.

As before, the time is 5:55 p.m. You have just completed a loop, and you haven’t used your mulligan yet.

With an optimal strategy (i.e., using the mulligan at the right moment, if at all), on average, what score can you expect to earn between 5:55 p.m. and 7 p.m.?

How Far Can You Run Before Sundown?

Intermission:

So. I didn’t get the extra credit last week. No shame in that. I had all the points from the previous puzzles of Q3. We were a group of 10 persons able to say that. I am now in the group of 6 persons, who dropped out from that max points group, now counting 4 members.

Still. It made me wonder. What was required to solve the extra credit? How does one find a strategy?

• Make a list of all the strategies you can think of. Remember to include the optimal one. 😉
• Test every item on the list.
• Declare a winner.

This requires good brainstorming skills. Also in this case I couldn’t leap this hurdle:

• A strategy is a list of “given voucher situation A, choose bet option B”.

As it turned out, the strategy also needed to include “is this the 1st bet, yes or no”.

Because I actually had the pieces. I had the $55 strategy for the 1st bet. And I had the $35 strategy for having all vouchers. I could just have combined them to get the correct extra credit result.

Ah well.

Highlight to reveal (possibly incorrect) solution:

Program

Information:

• 4 kinds of loops: 1, 3, 3.5 and 4.5 miles.
• Each loop is randomly assigned.
• The time is 5.55 p.m.
• The run ends at 7.00 p.m.
• Speed: 6 mph. (1 mile = 10 minutes.)
• The score equals the number of miles run.
• A loop not finished when the run ends doesn’t count towards the score.
• Question: What will the expected score be for the remainder of the run?

My solution:

• This could probably be done by hand (as I actually ended up doing below), but I wrote a recursive program.
• Begin with the 65 remaining minutes.
• Go through each option / each of the 4 loops that might be assigned.
  • If there’s time enough for a given loop, add this loop to the score and then go through each option again for the remaining time.
  • Stop when time runs out.
• Going backwards through the options, calculate the average of what happened given the 4 options.
• Finally the average for 65 minutes represents the average score.

Example:

• The program has been running for some time. I am looking at the situation with 30 minutes left.
• Going through the 4 options:
  • Option 1, 1 mile, 10 minutes. Note 1 mile and keep looking at the remaining 20 minutes.
    • Option 1, 1 mile, 10 minutes. Note 1 mile and keep looking at the remaining 10 minutes.
      • Option 1, 1 mile, 10 minutes. Note 1 mile and stop.
      • Option 2, 3 miles. Note 0 miles and stop. (Not time enough to run those 3 miles.)
      • Option 3, 3.5 miles. Note 0 miles and stop.
      • Option 4, 4.5 miles. Note 0 miles and stop.
      • The average for the 4 options is 1/4 miles = 0.25 miles.
    • Option 2, 3 miles. Note 0 miles and stop.
    • Option 3, 3.5 miles. Note 0 miles and stop.
    • Option 4, 4.5 miles. Note 0 miles and stop.
    • The average for the 4 options is (1 + 0.25)/4 miles = 0.3125 miles.
  • Option 2, 3 miles. Note 3 miles and stop.
  • Option 2, 3.5 miles. Note 0 miles and stop.
  • Option 4, 4.5 miles. Note 0 miles and stop.
  • The average for the 4 options is ((1 + 0.3125) + 3) / 4 = 1.078125 miles.

Result: Miles added to score, 4.866.

And for extra credit:

Mindomo Program Image

Information:

• Same as above, plus:
• A mulligan can be invoked once during the run and a new random loop be assigned.
• This leads to an optimal strategy.
• Question: What will the expected score be for the remainder of the run, using this strategy?

My solution. The strategy will look like this:

• Given that there’s A minutes left of the run, and a loop requiring B minutes got assigned, and that the mulligan hasn’t been used, should it be used now, yes or no?
• This can be written in a table with all the combinations of A and B.
• Some of the combinations are very easy. If running this loop means there will be 0 minutes left, keep it, we hit the max possible. If running this loop means the loop won’t be finished before the deadline and a better option exists, use the mulligan.
• In general: If the average score over all 4 loop options without a mulligan is C, and keeping the assigned loop on average adds D to the score, and D < C, use the mulligan.

My table:

\ loop time
time left \	45	35	30	10
5	keep	keep	keep	keep
10	mull	mull	mull	keep
15	mull	mull	mull	keep
20	mull	mull	mull	keep
25	mull	mull	mull	keep
30	mull	mull	keep	keep
35	mull	keep	keep	mull
45	keep	keep	mull	mull
55	keep	mull	mull	keep
65	keep	keep	keep	keep

Let me go through an example, related to the image. (Might contain errors as I later changed the algorithm a little.)

• When there’s 10 minutes left (yellow), from left to right the 4 options (in minutes 45, 35, 30, 10, green) add nothing (-0), nothing, nothing and 10 (*) to the score (calculated in minutes). Furthermore, the first 3 options left no time. The 4th option uses all the time left. Therefore the first 3 options are good for a mulligan, while the 4th is a keeper.
• Mulligan = happy smiley, yes, please use the mulligan.
• Keeper = green flag, you have reached a good option.
• (*) From the yellow to the right most option below, there’s a difference from 10 to 0. This is because the route was through a 10 minute loop. Reaching the green flag, there’s no time left and nothing further added to the score.
• The 4 possible options from the yellow state adds 0, 0, 0 and 10 to the score. This is 10/4 = 2.5 in average.
• Another way to say the green flag is a keeper, is because 2.5 < 10.
• When there’s 20 minutes left (orange), similar construction.
• The 4 possible options from the orange state adds 0, 0, 0 and 10 + 2.5 to the score. Therefore the 3 cyan options are good for a mulligan, while the yellow option is a keeper (purple).
• When there’s 65 minutes left, 1 of the options (orange) will add 45 + 3.13 to the score. As this is less than the average for all 4 options, 48.66, this option is good for a mulligan (red).

I change my program. First I calculate all averages without the mulligan. Then I calculate new averages for the situation with a mulligan. Every time I use a mulligan I drop down to the mulligan free version of the averages. I also fill in a table showing whether I use a mulligan in a given situation.

The result appears to be 5.352 miles.

Just for fun I also implemented a monte carlo program, covering all situations with 0-10 mulligans, assuming the mulligan table is the same all the way up.

Expected score with  0 mulligans: 48.677
Expected score with  1 mulligans: 53.514
Expected score with  2 mulligans: 56.563
Expected score with  3 mulligans: 58.585
Expected score with  4 mulligans: 59.967
Expected score with  5 mulligans: 60.997
Expected score with  6 mulligans: 61.748
Expected score with  7 mulligans: 62.318
Expected score with  8 mulligans: 62.773
Expected score with  9 mulligans: 63.123
Expected score with 10 mulligans: 63.391

It confirms both my results.

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

A casino offers you $55 worth of “free play vouchers.” You specifically receive three $10 vouchers and one $25 voucher.

You can play any or all vouchers on either side of an even-money game (think red vs. black in roulette, without those pesky green pockets) as many times as you want (or can). You keep the vouchers wagered on any winning bet and get a corresponding cash amount equal to the vouchers for the win. But you lose the vouchers wagered on any losing bet, with no cash award. Vouchers cannot be split into smaller amounts, and you can only wager vouchers (not cash).

What is the guaranteed minimum amount of money you can surely win, no matter how bad your luck? And what betting strategy always gets you at least that amount?

Hint: You can play vouchers on both sides of the even money game at the same time!

And for extra credit:

You have the same $55 worth of vouchers from the casino in the same denominations. But this time, you’re not interested in guaranteed winnings. Instead, you set your betting strategy so that you will have at least a 50 percent chance of winning W dollars or more. As before, you cannot split vouchers and cannot wager cash.

What is the maximum possible value of W? In other words, what is the greatest amount of money you can have at least a 50 percent chance of winning from the outset, with an appropriate strategy? And what is that betting strategy?

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

All the many attendees at a particular Coldplay concert are couples. As the CEO of Astrometrics, Inc., you are in attendance with your romantic partner, who is definitely not the head of HR at Astrometrics, Inc. During the concert, the two of you spend half the time canoodling.

The camera operators love to show people on the jumbotron during the concert, but time is limited and there are many attendees. As a result, the camera operators show just 1 percent of couples during the concert. Couples are chosen randomly, but never repeat at any given concert.

You and your partner are shy when it comes to public displays of affection. While you don’t mind being shown on the jumbotron, you don’t want to be shown canoodling on the jumbotron.

How many Coldplay shows can the two of you expect to attend without having more than a 50 percent chance of ever being shown canoodling on the jumbotron?

And for extra credit:

Now, everyone at the concert spends at least some time canoodling. In particular, each member of a couple wants to spend some fraction of the time canoodling, where this fraction is randomly and uniformly selected between 0 and 1. This value is chosen independently for the two members of each couple, and the actual time spent canoodling is the product of these values. For example, if you want to canoodle during half the concert and your partner wants to canoodle during a third of the concert, you will actually canoodle during a sixth of the concert.

Meanwhile, the camera operators love to show canoodling couples. So instead of randomly picking couples to show on the jumbotron, they randomly pick from among the currently canoodling couples. (The time shown on the jumbotron is very short, so a couple’s probability of being selected is proportional to how much time they spend canoodling.)

Looking around the concert, you notice that the kinds of couples who most frequently appear on the jumbotron aren’t constantly canoodling, since there are very few such couples. Indeed, the couples who most frequently appear on the jumbotron spend a particular fraction C of the concert canoodling. What is the value of C?

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

There’s a square board with side length A. Your friend cleverly places a unit square on the board and challenges you to place another unit square on the board—without moving the first one—so that it too is entirely on the board and the squares don’t overlap. (The unit squares can touch each other.)

Alas, it’s impossible for you to do so! But there’s some minimum value of A for which you can always place a second unit square on the board, no matter how cleverly your friend places the first one.

What is this minimum value of A?

And for extra credit:

Now there’s another square board with side length B. This time, your friend cleverly places three unit squares (which can touch but not overlap) and issues a similar challenge, asking you to place one more unit square on the board.

Once again, it’s impossible for you to do so! But there’s some minimum value of B for which you can always place a fourth unit square on the board, no matter how cleverly your friend places the first three squares.

What is this minimum value of B?

Can You Squeeze the Squares?

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“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

… you and a competitor are approaching the finish of a grueling stage of the Tour de Fiddler. One of you will win the stage, the other will come in second. As you approach the finish, each of you will test the feeling of your legs, which will be somewhere between 0 percent (“I can barely go on!”) and 100 percent (“I can do this all day!”). For the purposes of this puzzle, these values are chosen randomly, uniformly, and independently.

Immediately after feeling your legs, you and your opponent each have a decision to make. Do you maintain your current pace, or do you sprint to the finish? Among those who sprint for the finish, whoever’s legs are feeling the best will win the stage. But if no one sprints for the finish, everyone has an equal chance of winning the stage. In the Tour de Fiddler, you must each decide independently whether to sprint for the finish based on your legs—you don’t have time to react to your opponent’s decision.

Normally, teams at the Tour de Fiddler keep their strategy and tactics close to the vest. But earlier today, your opponent’s manager declared on international television that if (and only if) your opponent’s legs were feeling 50 percent or better, they’d sprint for the finish.

As you are about to test your legs for the final sprint and see how they feel, what are your chances of winning the stage, assuming an optimal strategy?

And for extra credit:

Instead of one opponent, now you have two—meaning three riders in total. As luck would have it, the managers for both other riders proclaimed that they’d sprint for the finish if (and only if) their legs were feeling 50 percent or better. Note that your opponents’ feelings are independent of each other.

As the three of you near the finish, your own team manager radios you the following message: “If your legs feel <garbled> percent or better, sprint for the finish!”

You can’t make out what the garbled part of the message is, and you’re too tired to radio back for confirmation. Instead, you somehow muster the energy to randomly, uniformly pick a number between 0 and 100 to fill in the blank from your manager’s message, thereby determining your racing strategy—optimization be damned!

Right before you choose your random strategy and test your legs, what are your chances of winning the stage against both opponents?

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.

A sprints?	b < 50	b => 50
Yes	p(A wins) = 1	p(A wins) = 0
No	p(A wins) = 50%	p(A wins) = 0

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

A sprints?	b < 50	b => 50
Yes	p(A wins) = 1	p(A wins) = p(a > b) = 0
No	p(A wins) = p(a > b) = 1	p(A wins) = 0

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

A sprints?	b < 50	b => 50
Yes	p(A wins) = 1	p(A wins) = p(a > b) = 50%
No	p(A wins) = 50%	p(A wins) = 0

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%.