Advent of Code, day 1-6

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This December I participated in Advent of Code.

Days 1-6 happened without a lot of drama (only 1 question on the forum). This phase also mostly saw me coding in a sandbox (exception: day 3, part 2). Day 7 I permanently changed to my PC. The tasks were rather easy for me. I didn’t really try to come up with a fast answer, in part because the sandbox + my tablet was a slow way to code. My examples below manipulate the test data, but ran on the actual data as well.

DayPartSolution
11Program
2Program
21Program (wrong) Question Program
2Program
31Program
2Program (transplanted back from PC)
41Program
2Program
51Program
2Didn’t save this one!
61Program
2Program

Scroggs, day 6

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A new December and a new bunch of puzzles from mscroggs.co.uk.

n = 99999…999

= 1055 – 1

n3 = (1055 – 1)3

= (1055)3 + 3*(1055)2*(-1) + 3*(1055)*(-1)2 + (-1)3

= 10165 – 3*10110 + 3*1055 – 1

 10000...0000...0000...000
-         300...0000...000
+                300...000
-                        1
==========================
  9999...9700...0299...999

So now we just have to count the 9s and 7s and the 2.

The 1 lives in position 1. The 3 above it has a 3 in position 56. The next 3 is in position 111. Finally the 1 above it is in position 166.

The result of adding and subtracting is 9s from position 1 to 55, a 2 at position 56, then some 0s, a 7 at position 111 and finally 9s from position 112 to 165. 55*9+2+7+54*9 = 990. Which happens to be 55*18.

Just to test my logic here: if n had been 9999 (4 digits), n3 would be 999,700,029,999. 4*9+2+7+3*9 = 72 = 4*18. Checks out.

Scroggs, day 4

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A new December and a new bunch of puzzles from mscroggs.co.uk.

I think the example is also a hint.

A number with 1 factor would be 1. And that’s not 13.

A number with 2 factors would be a prime. Number p, factors 1 and p. The geometric mean would be (1*p)1/2. We can’t get 13 to fit there.

A number with 3 factors has a special property. There’s a square here somewhere. The number, p2, has factors 1, p and p2. This accounts for p2 = 1 * p2 = p * p, the 2 different ways to write p2. Like, 9 = 1*9 = 3*3. The geometric mean works out to be p, isn’t that neat? So my guess is that we’re looking for 132 = 169.

Scroggs, day 3

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A new December and a new bunch of puzzles from mscroggs.co.uk.

My first attempt is further down. But I thought of a better way.

These are equivalent:

  • Writing n as a sum of m odd, positive integers (5 = 1+3+1).
  • Writing n+m as a sum of m even, positive integers (8 = 2+4+2).
  • Writing (n+m)/2 as a sum of m positive integers (4 = 1+2+1).

And how many ways can I write the last sum? Stars and bars. Imagine writing 4 as 1 1 1 1. We want to convert this to 3 positive integers. We do this by adding 2 bars in between, like 1 | 1 1 | 1. (3-1 = 2.) The bars could be added in 3 different positions, between 2 1s. (4-1 = 3.) Only 1 bar in each position. We can do this in (3 c 2) different ways. This is 3*2/2*1 = 3. Anyway, 1 | 1 1 | 1 corresponds to 1 | 1+1 | 1 = 1 | 2 | 1, corresponds to the 4 = 1+2+1.

Going back, writing (n+m)/2 as a sum of m positive integers can be done in ((n+m)/2-1 c m-1) different ways.

Anyway.

  • Write 14 as m odd, positive integers.
  • Write 14+m as m even, positive integers.
  • Write (14+m)/2 as m positive integers.

To write 14 as a sum of m odd, positive integers, m must be at least 2 and at most 14. Also m is even, because the sum of an odd number of odd integers would be odd. Let’s look at these.

m((14+m)/2-1 c m-1)
27
456
6126
8120
1055
1212
141

And when we add up the right column, we get 377.

First attempt:

There’s probably a smarter way to do this. But I needed an intermediary step, before I could answer the primary question. First: How many ways can I write a sum of odd, positive integers, if I don’t care about the order?

SumIntegers
0(nothing)
11*1
22*1 (1)
33*1
1*3 (2)
44*1
1*1+1*3
55*1
2*1+1*3
1*5
66*1
3*1+1*3
1*1+1*5
77*1
4*1+1*3
2*1+1*5
1*1+2*3
1*7
88*1
5*1+1*3
3*1+1*5
2*1+2*3
1*1+1*7
99*1
6*1+1*3
4*1+1*5
3*1+2*3
2*1+1*7
1*1+1*3+1*5
1*9
1010*1
7*1+1*3
5*1+1*5
4*1+2*3
3*1+1*7
2*1+1*3+1*5
1*1+3*3
1*3+1*7
2*5
1*1+1*9
1111*1
8*1+1*3
6*1+1*5
5*1+2*3
4*1+1*7
3*1+1*3+1*5
2*1+3*3
1*1+1*3+1*7
1*1+2*5
2*1+1*9
1*11
1212*1
9*1+1*3
7*1+1*5
6*1+2*3
5*1+1*7
4*1+1*3+1*5
3*1+3*3
2*1+1*3+1*7
2*1+2*5
3*1+1*9
1*1+2*3+1*5
1*5+1*7
1*3+1*9
1*1+1*11
1313*1
10*1+1*3
8*1+1*5
7*1+2*3
6*1+1*7
5*1+1*3+1*5
4*1+3*3
3*1+1*3+1*7
3*1+2*5
4*1+1*9
2*1+2*3+1*5
1*1+1*5+1*7
1*1+1*3+1*9
2*1+1*11
1*1+4*3
2*3+1*7
1*3+2*5
1*13
1414*1
11*1+1*3
9*1+1*5
8*1+2*3
7*1+1*7
6*1+1*3+1*5
5*1+3*3
4*1+1*3+1*7
4*1+2*5
5*1+1*9
3*1+2*3+1*5
2*1+1*5+1*7
2*1+1*3+1*9
3*1+1*11
2*1+4*3
1*1+2*3+1*7
1*1+1*3+2*5
1*1+1*13
1*3+1*11
1*5+1*9
2*7

(1) I find this result for 2 by adding a 1 to the result for 1. 1 = 1*1. 2 = 1*1+1 = 2*1.

(2) I find this result for 3 by first adding a 1 to the result for 2 (getting the first new result), and then adding a 3 to the result for 0 (getting the second new result).

I only actually need the last cell. Let’s look a little closer at it. E.g. 3*1+1*11 is 1+1+1+11 (4 elements), when we don’t care about the order. Let’s look at how many different ways each sum can be written (how many permutations, like 1+1+1+11 = 1+1+11+1 = 1+11+1+1 = 11+1+1+1, 4 different ways).

Sum#Elements#Perms
14*1141
11*1+1*31212 (1)
9*1+1*51010
8*1+2*310(10 c 2) = 45 (2)
7*1+1*788
6*1+1*3+1*588*7 = 56 (3)
5*1+3*38(8 c 3) = 56
4*1+1*3+1*766*5 = 30
4*1+2*56(6 c 2) = 15
5*1+1*966
3*1+2*3+1*566 * (5 c 2) = 6*10 = 60 (4)
2*1+1*5+1*744*3 = 12
2*1+1*3+1*944*3 = 12
3*1+1*1144
2*1+4*36(6 c 2) = 15
1*1+2*3+1*744*3 = 12
1*1+1*3+2*544*3 = 12
1*1+1*1322
1*3+1*1122
1*5+1*922
2*721

(1) Among the 12 positions, I choose 1 for the 3.

(2) Among the 10 positions, I choose 2 for a 3. I can do this in “10 choose 2” ways.

(3) Among the 8 positions, I choose 1 for the 3. I can do this in 8 different ways. Among the remaining 7 positions, I choose 1 for the 5. I can do this in 7 different ways.

(4) Combining methods 2 and 3.

Finally we have to add up the right column. The result is 373. I did some of this manually, I must have missed a step somewhere. Oh yeah, I think I missed 3*3+5, good for 4 further permutations.

Scroggs, day 2

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A new December and a new bunch of puzzles from mscroggs.co.uk.

14 is even, so 2 is a factor. The other factor is 7, a prime, so it can’t be constructed by multiplying lower numbers.

100 d100: One factor is 2, to get an even product the lowest possible way. Then the other factor of the product is a prime p, in this case 101, to make the product impossible. (Anything lower could be reached with 198 * 2 * p.) The product is 202.

Scroggs, day 1

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A new December and a new bunch of puzzles from mscroggs.co.uk.

The only way to reach the sum 16 with 5 different positive integers is 1+2+3+4+6.

The lowest possible sum with 5 different positive integers is 1+2+3+4+5 = 15. In order to reach 16, one of these integers has to be 1 higher. If the integers are still different after this operation, the only candidate is 5.

Alternate proof: Play Killer Sudoku.

The product is 1*2*3*4*6 = 144.

Scroggs, day 0

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A new December and a new bunch of puzzles from mscroggs.co.uk.

Part of the mystery this year is this formula:

  • Santa tells his three-digit number to the first elf.
  • The first elf subtracts her three-digit number then multiplies by her one-digit number. She tells her result to the second elf.
  • The second elf subtracts his three-digit number then multiplies by his one-digit number. He tells his result to the third elf.
  • The third elf subtracts their three-digit number then multiplies by their one-digit number. Their result is a five-digit number that is the code to unlock the warehouse.

It is tempting to try to write this as pure math. I don’t know at this point whether that actually helps.

  • (aaa – bbb)*c = x
  • (x – ddd)*e = y
  • (y – fff)*g = hhhhh

#ThisWeeksFiddler, 20241213

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This week the question is: Can You Survive “The Floor”?

… Let’s consider a slightly simplified version of The Floor with nine contestants on a 3×3 square grid. Each round of the game consists of the following steps:

  • One of the remaining contestants is chosen at random. (Note that each contestant is equally likely to be chosen, regardless of how many squares they currently control.)
  • The set of eligible opponents for this contestant is anyone whose territory shares a common edge with the contestant. One of these eligible opponents is chosen at random. (Again, all eligible opponents are equally likely to be chosen, regardless of how many squares they control or how many edges they have in common with the opponent.)
  • The contestant and their selected opponent have a duel, each with a 50 percent chance of winning. The loser is eliminated, and their territory is added to that of the winner.

These rounds repeat until one contestant remains, and that contestant is the overall winner.

You are a contestant on a new season of this 3×3 version of The Floor. The nine positions on the grid are shown below:

Which position would you choose? That is, which position or positions give you the best chance of being the overall winner?

And for extra credit:

Again, a new season of the 3×3 version of The Floor (as described above) is about to begin.

What is your probability of winning for each of the nine starting positions?

Highlight to reveal (possibly incorrect) solution:

Program. Numbers.

This time I wrote a program to investigate the possibilities. In summation these are the probabilities for a win, rounded:

A, C, G, I (a corner): 13%

B, D, F, H (an edge): 10%

E (the middle): 9%

So I should choose to start in a corner.

Jedier og snotbotter under havet

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Anmeldelse af 7 illustrerede børnebøger, op til omkring 40 sider. 2024.

download (1)

Snotbotterne – snotbotterne ser dagens lys, John Sazaklis.

Skitse: Når man krydser noget radioaktiv stråling med kasseret skrot, så får man snotbotterne. Når snotbotterne spreder for meget snavs, så bekæmper Clean Team dem. 👽👽☠️

Snotbotterne – prutte-bot-invasionen, John Sazaklis.

Skitse: Snotbotterne lader en sværm af prutte-botter invadere byen, men Clean Team er på pletten. 👽👽☠️

Taco og en helt almindelig morgen, Pernilla Gesén.

Skitse: Hos familien Sten har de hund og kat — og rumvæsen. Taco (der egentlig hedder A135) sender rapporter hjem, den her gang om hvad familien laver om morgenen. 👽👽☠️

Taco og monster på gåtur, Pernilla Gesén.

Skitse: Taco går tur med far og hunden Doris. Doris laver hele tiden ulykker. 👽👽☠️

God bedring, Nubs! Christopher Nicholas. Del af Star Wars-universet, ligesom den næste.

Skitse: Nubs er forkølet, så hans venner Kai og Lys, unge jedier, begiver sig ud for at finde en helbredende yara-plante. 👽👽☠️

Jagten på geléfrugterne, Christopher Nicholas.

Skitse: Kai, Lys og deres venner pynter op til geléfrugt-festival, da det viser sig, at alle frugterne er blevet stjålet af pirater. Det må der gøres noget ved. 👽👽☠️

En verdensomsejling under havet, Antonis Papatheodoulou.

Skitse: Et eller andet hærger i både Atlanterhavet og Stillehavet. Professor Aronnax bliver hyret til at undersøge sagen. Efter lang tid finder de endelig synderen: en ubåd! 👽👽👽