Anmeldelse af The Name Ziya (gratis), af #WenYiLee. Langnovelle. 2025. Nebula-nomineret.

Nebula nominees . Hugo finalists .

Skitse: Oprindeligt havde Ziya et navn på 5 stavelser. Men stavelser er magiske. Hver stavelse står for en bestemt magisk evne. Hvis man er fattig, så kan en stavelse sælges. Den tilsvarende “tatovering” bliver skrællet af, og ens navn bliver kortere. Ziya har måttet sige farvel til 3 stavelser, så der blev råd til hendes uddannelse. Men måske er det okay, at magien så kan blive brugt mere generelt i stedet, til at sikre gode afgrøder og sådan noget. I hvert fald, afsted til skolen, hvor Ziya er del af en kvote. Det er helt nyt, at hendes slags også bliver optaget. Med alle den slags problemer, sådan noget giver.

Er det science fiction? Nej. Fantasy.

Temaer: Byen viser sig at være lidt anderledes end forventet. Knapt så ophøjet. Det viser sig bl.a. i, at de andre studerende, der typisk er rige, køber magiske mærker, udelukkende for at pynte sig med dem. Det gør ondt på Ziya at se det.

Undervejs hører vi om venskaber og andet i gruppen.

Er det godt? Ja. Jeg blev meget optaget af historien. Eneste minus er, at vi nærmest ikke ser Ziya eller nogen andre bruge magi på noget tidspunkt. ###

Anmeldelse af “We Begin Where Infinity Ends” (gratis), af #SomtoIhezue. Langnovelle. 2025. Nebula-nomineret.

Nebula nominees . Hugo finalists .

Skitse: 2 drenge er optaget af, at ildfluerne er ved at uddø. Der er et simpelt trick, så gadelys kan lyse på en måde, der ikke skræmmer ildfluerne væk, så drengene lusker rundt om natten for at indføre den slags lys. I al hemmelighed. Men så bliver de opdaget af en pige.

Er det science fiction? Ja.

Temaer: De 3 har den alder, hvor man bliver forelsket. Nogen bliver forelsket i nogen andre. Måske gensidigt.

Det hele bliver meget dramatisk en overgang, og omverdenen hører om ildfluerne og lyset og synes, det er en glimrende idé, der burde praktiseres overalt.

Er det godt? Mja. ##-

This week the #puzzle is: Can You Hop Around the Chessboard? #chess #combinations #coding

A frog is hopping around a chessboard, always from the center of one square to the center of another square. Each square has side length 1, but the board itself is not necessarily 8-by-8. Instead, it’s N-by-N, where N is some large whole number.

Every jump the frog makes must be the same distance, which we’ll call L. The frog wants to make four jumps such that:

– After the fourth jump, the frog has returned to its starting square. – The frog visits a total of four distinct squares along the way, including the square on which it starts (and also stops). – The path the frog takes is not a square loop. – The frog is never on a square that is diagonal (i.e., a bishop’s move away) or horizontal/vertical (i.e., a rook’s move away) from the starting square.

What is the smallest jumping distance L for which this is possible?

And for extra credit:

The frog is jumping around the board with the same minimum distance L you just found.

But this time, the frog also wants to be able to hop to every location on the chessboard. What is the minimum value of N for which this is possible?

Can You Hop Around the Chessboard?

Solution, possibly incorrect:

Program ProofWiki

Method 1. I write a program to go through all the options. Increase L until all requirements are met. Actually, let’s write out those requirements:

1. The frog jumps distance L every time. Each jump goes integer lengths in the x and y directions. E.g. a jump of length √10 could go 3 in the x direction and 1 in the y direction. Describe this jump as (3,1).
2. The frog makes 4 jumps, returning to the starting point, describing some sort of loop.
   • Let’s call these jumps (x₁,y₁), (x₂,y₂), (x₃,y₃) and (x₄,y₄). E.g., the first jump (assume a standard coordinate system) is from (0,0) to the first landing point (x₁,y₁). The next jump will go to (x₁ + x₂,y₁ + y₂). Etc. All 4 jumps will lead back to (0,0).
3. The 4 landing points, (0,0), (x₁,y₁) etc. are different.
4. The loop is not a square.
5. None of the landing points are on a horizontal or vertical line from the starting point.
   1. No (0,y) or (x,0) points.
   2. No (x,x) or (x,-x) points.

Also, let’s look at a few examples below. All of these violate 1 or more requirements. (G is just a doodle, an inspiration for H.)

---

All of these examples of doing it wrong tells us something. But we’ll come back to that.

In my code I focus on finding 2 different ways to combine 2 jumps to reach the same point (x,y). I assume the point has 0 < x, y, WLOG. I assume x < y, again WLOG. I test whether my 4 jumps describe a square (then the jumps would be something like (x₁,y₁), (-y₁,x₁), (-x₁,-y₁) and (y₁,-x₁) like in illustration C above). And I’m done:

( 1, 3) can be reached in 2+ ways. (L^2 = 65.)
		(0,0) -> (-7, 4) -> ( 8,-1)
		(0,0) -> ( 8,-1) -> (-7, 4)
( 1, 5) can be reached in 2+ ways. (L^2 = 65.)
		(0,0) -> (-7, 4) -> ( 8, 1)
		(0,0) -> ( 8, 1) -> (-7, 4)
( 3,15) can be reached in 2+ ways. (L^2 = 65.)
		(0,0) -> (-1, 8) -> ( 4, 7)
		(0,0) -> ( 4, 7) -> (-1, 8)
( 4, 6) can be reached in 2+ ways. (L^2 = 65.)
		(0,0) -> (-4, 7) -> ( 8,-1)
		(0,0) -> ( 8,-1) -> (-4, 7)
( 4, 8) can be reached in 2+ ways. (L^2 = 65.)
		(0,0) -> (-4, 7) -> ( 8, 1)
		(0,0) -> ( 8, 1) -> (-4, 7)
( 6,12) can be reached in 2+ ways. (L^2 = 65.)
		(0,0) -> (-1, 8) -> ( 7, 4)
		(0,0) -> ( 7, 4) -> (-1, 8)
( 5,15) can be reached in 2+ ways. (L^2 = 65.)
		(0,0) -> ( 1, 8) -> ( 4, 7)
		(0,0) -> ( 4, 7) -> ( 1, 8)
( 8,12) can be reached in 2+ ways. (L^2 = 65.)
		(0,0) -> ( 1, 8) -> ( 7, 4)
		(0,0) -> ( 7, 4) -> ( 1, 8)

The illustrations below show each of these solutions.

K and P are the same. L and M are the same. N, O and Q are the same.

These solutions correspond to L = √65.

Method 2: All the illustrations above show loops, where (x₁,y₁), (x₂,y₂), (x₃,y₃) and (x₄,y₄) are strongly related. There are 2 variants:

• G and H are actually impossible. 2 jumps would be equal in every way, (x₁,y₁) = (x₃,y₃). But then at least 1 other jump would be too long.
• The rest have (x₁,y₁), (x₂,y₂), (-x₁,-y₁) and (-x₂,-y₂).

The loop is a parallelogram. There are 2 pairs of parallel sides. Everything reduces to 4 variables: x₁, y₁, x₂, and y₂. We know x₁² + y₁² = x₂² + y₂² = L². Also x₁ != y₁ and x₂ != y₂. So we’re looking for 2 different ways to write a number as the sum of 2 squares. And yes, this has already been investigated. ProofWiki points me towards a list. The first candidate can be thrown out, as one of the sums is 5² + 5². But the next is the same result I have already found.

And for extra credit:

I expand my program to look for “knight’s tour” paths. (Warnsdorff’s Algorithm for the win.) I easily find one for a 15×15 board:

  0  39   2  43   4 159 214  55 196 175 100 115  78  19  30 
135 212 127 204  77  18  29  14  67  74  21 138 143 184 189 
 62  73  22  95 142 133 210 145 136 207 128 205  26  59  28 
131 208 149 162 151 106 103  24 109  84  57  96  33  98  35 
108 193 178 101 114  81  36  31  40 153  44 155 158 215  86 
 13  70  47 140 169 220 187 190 217 192 203 122 119   8  15 
222 171 200 123 198  11  90  63  52  93  50 141 168 219 146 
  5 112  83  56 167 174  99 116 223 160 163  54 197 104  89 
 38   1  42   3 156 213  88 195 180 185 176  79  68  75  20 
181 134 211 126 121  76  17  66  71  60 139 144 137 188 183 
 72  61  94  23 132 209 148 165 172 129 206  25 110  27  58 
117 130 161 150 107 152 105 102 113  82  85  32  97  34  45 
194 179 186 177  80  69  46  37 118  41 154 157 216  87 202 
 65  12  91  48 221 170 201 182 191 218 125 120   7  16   9 
166 173 124 199   6 111  10  53  64  51  92  49 224 147 164 

This makes sense, as 15 = 8 + 7, the 2 largest numbers involved in the jumps. (ETA: Should probably be 8 + 7 + 1 = 16.) I randomly assume, that I begin at (0,0). So N <= 15.

Video:

I modify my program to look at all connected boards. When all cells can be reached from (0,0). (Because then all cells can reach each other.) 14×14 works, 13×13 does not. N = 14. Apparently all cells can be reached in 6 jumps or less.

ETA: The N/O/Q type loop is also possible on this board, as it requires 13 x 9 cells.

ETA: A frog’s tour is easy to find on a 14×14 board, when beginning at (1,1).

Anmeldelse af “Uncertain Sons”, af #ThomasHa. Langnovelle. 2025. Nebula-nomineret.

Nebula nominees . Hugo finalists .

Skitse: Byen har et forsvar på plads, når omgivelserne trænger sig ind i gaderne. Som man nu har, når den vilde natur er svær at holde styr på. Hovedpersonen er tilknyttet en gruppe, der tager hen og fikser problemerne. Lige i dag afviger han fra proceduren, fordi han helt privat også skal have nuppet et par prøver fra udyret.

Er det science fiction? Nej. Horror/fantasy.

Temaer: Den vilde natur er ret præget af tentakler. Der er også tilfælde af, at et forsvundet barn får nyt liv som en masse plante-kopier. Øv.

Vi følger fyren, der har tænkt sig at bruge diverse væsker fra forskellige udyr til at stoppe problemet ved kilden. Han er drevet af mindet om sin far. Han slæber rundt på farmands kranie. Okay så. I starten er det lidt forvirrende, at der foregår en samtale (?) mellem far og søn.

Alt det her med at “holde styr på problemerne” indebærer at slå ihjel, gerne med blod og tarme ud over det hele. Gerne på begge sider. Øv.

Er det godt? Jeg læste den ikke færdig. Alt for mange kropsvæsker. #–

Anmeldelse af “Our Echoes Drifting Through the Marsh” (gratis), af #MarieCroke. Langnovelle. 2025. Nebula-nomineret.

Nebula nominees . Hugo finalists .

Skitse: Når en person dør, så bliver liget bragt ud i mosen, og liget bliver forvandlet til et antal skulpturer af træ, der hver afspiller en lille lydstump, noget af en samtale, afdøde deltog i. Eller sådan var det i hvert fald. Fortællerens datter er ophidset over, at onkel ikke blev begravet på den nye, bedre måde.

Er det science fiction? Nix. Fantasy.

Temaer: Fortællingen er struktureret som en slags afhøring. Datteren, Danai, har en plads i rådet og er blevet udpeget til at finde ud af, hvad der egentlig er sket. Fortælleren, Embri, beretter derfor om forskellige små stumper af sit liv, der peger hen på, hvad der er sket og hvorfor. Hver lille beretning starter med at beskrive, hvordan en hypotetisk dødeskulptur, der passer til beretningen, kunne se ud.

Det helt store emne er: Skal man dvæle ved fortiden? Er det godt? Er det godt at opsøge en afdøds rester for at “have en samtale” med vedkommende? Og hvis man vil gøre opgør med den her praksis, hvor langt må man så gå?

Er det godt? Ja, jo. Velskrevet. Ikke helt oppe at ringe. ##-

This week the #puzzle is: Will You Be Right on Time? #angles #clock #RightAngles #geometry

A standard analog clock includes an hour hand, a minute hand, and 60 minute markers, 12 of which are also hour markers.

At a certain time, both the hour hand and minute hand are both pointing directly at minute markers (either of which could also be an hour marker). The hour hand is 13 markers ahead (i.e., clockwise) of the minute hand.

At what time does this occur?

And for extra credit:

At various times of day, the minute and hour hands form a right angle. But is there a time of day when the hour hand, minute hand, and second hand together form two right angles, as illustrated below? If you can find such a time or times, what are they?

If you cannot find any such times, suppose the measures of the two nearly right angles formed by the three hands measure A and B degrees. What time or times of day minimize the square error function f(A, B) = (A − 90)2 + (B − 90)2?

Either way, your answer should be precise to at least a thousandth of a second.

Will You Be Right on Time?

Solution, possibly incorrect:

We can have any time from 00:00 to 11:59.

The time:

So many minutes have passed since 00:00:

It takes 720 minutes for the hour hand to go all the way around the clock.

When the hour hand points directly at the nth minute marker:

We also have some extra knowledge about the distance between the hands:

There are 2 ways to compute where the hour hand is.

---

Method 1, example:

---

Method 2, example:

---

As these 2 methods don’t result in the same n for the example, this isn’t the time we’re looking for.

---

As we have, from method 1:

n is an integer, so m is a multiple of 12. Further, from method 1:

This leads to MM being a multiple of 12. So:

Then we use method 2 to find n:

MM	00	12	24	36	48
n	13	25	37	49	1

And method 1, backwards, to find m:

MM	00	12	24	36	48
n	13	25	37	49	1
m	156	300	444	588	12

And then find HH:

MM	00	12	24	36	48
m	156	300	444	588	12
HH	2.6	4.8	7	9.2	11.4

Only the time 07:24 gives an integer HH. So this is the solution.

And for extra credit:

Program

I write a program to go through a lot of different options. I fine tune the program to find better and better solutions, with better and better error functions. The results are:

Good time, in seconds..:    13744.07855
 -- Good time..........: 03:49:04.07855
 -- Error function.....: 0.0079550
Good time, in seconds..:    29455.92145
 -- Good time..........: 08:10:55.92145
 -- Error function.....: 0.0079550

So the best possible times are 03:49:04.079 and 08:10:55.921.

I have watched #ProjectHailMary! Boy, did that director send small hails in all directions, primarily to other #movies ? Fun! #references

Rocky has 3 fingers, Grace has 5. Zammis has 3 fingers, Davidge has 5. Enemy Mine.

Sitting on a beach, looking out on the water. Looking to the right. Somebody’s coming! Contact.

A big structure of rods, parallel to each other or perpendicular to each other. Interstellar.

The last one is a nod (I think) to “Story of Your Life”. The story behind Arrival.

“Then Raspberry brought the gourd down between its legs, a crunching sound resulted, and the gourd reemerged minus a bite…”

This week the #puzzle is: Have You Heard the Buzz? #counting #frequency #combinations

I recently introduced my children to a game called “Buzz” (also known as “Fizz buzz”), in which players take turns reciting whole numbers in order. However, in one particular variant of the game, anytime a number is a multiple of 7 or at least one of its digits is a 7, the player must say “buzz” instead of that number.

For example, here is how the first 20 turns of the game should proceed: 1, 2, 3, 4, 5, 6, buzz, 8, 9, 10, 11, 12, 13, buzz, 15, 16, buzz, 18, 19, 20.

How many times should “buzz” be said in the first 100 turns of the game (including those mentioned above in the first 20 turns)?

And for extra credit:

As we just saw, in the first 20 turns of the game, 15 percent of the numbers were “buzzed.” But as the game proceeds, an increasing frequency of numbers get buzzed.

There is a certain minimum number N such that, for the Nth turn in the game and for every turn thereafter, at least half the numbers up to that point have been buzzed. What is this value of N?

Have You Heard the Buzz?

Solution, possibly incorrect:

Program

Method 1: I write a program to produce all the buzzes and count them. There are 30.

Method 2:

• We are looking at the numbers 1-100.
• If the last digit is 7, it’s a buzz: 7, 17, 27, … 97. 10 of these.
• If the next to last digit is 7, it’s a buzz: 70, 71, 72, … 79. 10 of these.
• If the number is a multiple of 7, it’s a buzz: 7, 14, 21, … 98. 14 of these.
• But if we just say 10 + 10 + 14, we’ve counted something twice. 77 will be counted 3 times!
• The first 10, we keep.
• The next 10, we throw 77 away, because we already had it. 10 + 9. (Among the 10 numbers beginning with the digit 7, exactly 1 also ends with a 7, and we already covered that.)
• The next 14, we throw 7, 70 and 77 away, because we already had them. 10 + 9 + 11. (Among the 14 numbers that are multiples of 7, exactly 3 are multiples that “keep” the 7. 1 * 7, 10 * 7, 11 * 7.)
• 10 + 9 + 11 = 30.

And for extra credit:

Desmos

Method 1: I expand the program from above. I look for a frequency of about 0.5 and then zoom in to learn more about this general area. The result is illustrated in the Desmos. Look for the clump at 0.5 and zoom!

Method 2: The graph has a distinctive shape. Every time we run into a whole bunch of numbers, all beginning with 7, the frequency will grow. Afterwards it will go up and down a bit, until the next bunch, but on the whole it will keep growing. For our purposes, once we hit the 700000-799999 bunch, it grows past 0.5 and never dips below again.

I confirm this result via the program, looking at different thresholds. E.g. the 0.45 threshold is 71574. Look for the red stars in the Desmos file.

Anyway. N = 708588.

Method 3: A similar counting argument as above. Left as an exercise to the reader 😉

This week the #puzzle is still unknown to me, as I write this. Because I am busy. But here’s some other content for you.

Some other very fine fiddlers:

• Benjamin Phillabaum (bphillab.github.io)
• Ed Nigma’s Answer Emporium
• Jamshrivelson
• Rohan Lewis (humanrickshaw)
• Sameer Gauria

If you want to compare 2 large numbers, and remember to carry the one(s), this is how you do it.

First row in the table below: Our units. In the middle, there’s a familiar system: 1, 10, 100. This is how our own numbers are constructed. There’s a position for 1s, a position for 10s, etc. To the left and to the right, the units are not the familiar ones, but hopefully we can understand this anyway.

Second row: A sum of a lot of other numbers. The sum hasn’t been normalized to its shortest from yet. We want to normalize and then compare with the last row, another number from the same source.

Normalization example: In the second row, 15 and 15 are bolded. The jump from the first to the second column is to multiply by 2. (1/120 * 2 = 1/60.) So to normalize, we look at the first 15 and divide by 2.

• a = 1/120, b = 1/60
• 2a = b
• 15/2 = 7.5
• 15 = 1 + 7 * 2
  • 15a + 15b
  • = (1 + 7 * 2)a + 15b
  • = a + 7 * 2a + 15b
  • = a + 7b + 15b
  • = a + 22b

So in the next row, instead of 15 and 15, we have 1 and 22.

Move to the next pair of columns and repeat.

1/120

1/60

1/30

1/10

1/5

1

10

100

300

3000

18000

15	15	24	14	19	26	39	11	7	8	5
1	22	24	14	19	26	39	11	7	8	5
1	0	35	14	19	26	39	11	7	8	5
1	0	2	25	19	26	39	11	7	8	5
1	0	2	1	31	26	39	11	7	8	5
1	0	2	1	1	32	39	11	7	8	5
1	0	2	1	1	2	42	11	7	8	5
1	0	2	1	1	2	2	15	7	8	5
1	0	2	1	1	2	2	0	12	8	5
1	0	2	1	1	2	2	0	2	9	5
1	0	2	1	1	2	2	0	2	3	6
1	0	2	1	1	2	2	1	1	3	6

And yes, the end result is, that the 2 numbers aren’t quite the same.

Background: Recently I read this, in From One to Zero, Georges Ifrah :