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joshmaxsilverman.bsky.social @joshmaxsilverman.bsky.social · 05/02/2026

using it, we get a tidy result for the time for the origin to drop to a level L: t ≈ ceil(sqrt(3)/(2πL)), which matches the exact result remarkably well. joshmaxsilverman.github.io/2026-02-04-f...

joshmaxsilverman.github.io

Hexagonal walker

disappearing food

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joshmaxsilverman.bsky.social @joshmaxsilverman.bsky.social · 05/02/2026

we can find them by decomposing the spatial dimensions into fourier modes. on reconstruction, the walk is tightly approximated by a two dimensional gaussian.

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joshmaxsilverman.bsky.social @joshmaxsilverman.bsky.social · 05/02/2026

@xaqwg.bsky.social in my original submission to #thisweeksfiddler, i coded up the recursion eqn and forgot about it. this morning at the vet, i realized it's a nice extension of the basic random walk. the probability distribution is made up of a sum of static patterns on the lattice.

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joshmaxsilverman.bsky.social @joshmaxsilverman.bsky.social · 08/12/2025

in #thisweeksfiddler we descend all levels of the apollonian dartboard, receiving a multiplier bonus on our throw from the dartboard's fractal structure. recursion by hand and by my computer gives a 100M+ circle estimate of 3.71086...

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joshmaxsilverman.bsky.social @joshmaxsilverman.bsky.social · 23/07/2025

the two approximations i make point in opposing directions, so it makes sense that they somewhat cancel, but the result seems too good for coincidence. i suspect it is a glimpse of a deeper calculation that takes into account the time correlation of the excitations.

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joshmaxsilverman.bsky.social @joshmaxsilverman.bsky.social · 23/07/2025

this captures the behavior closely, starting at about 10 friends.

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joshmaxsilverman.bsky.social @joshmaxsilverman.bsky.social · 23/07/2025

i then asked, at what k would we expect such an excitation to spawn once in an hour. this led to a remarkable formula for the excitation δ = (k-μ)

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joshmaxsilverman.bsky.social @joshmaxsilverman.bsky.social · 23/07/2025

on average we'd expect N/4 friends in any 15 min interval. to probe the extremes, I made the approximation that each interval is an independent poisson process. I calculated the probability that such a window would have an excitation to k friends, and found the expected lifetime of such an interval.

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joshmaxsilverman.bsky.social @joshmaxsilverman.bsky.social · 23/07/2025

in #thisweeksfiddler we are asked to find the expected maximum number of friends who are at the mall at the same time, given that they all go at a random 15 min window over an hour. i wondered about the large N behavior.

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joshmaxsilverman.bsky.social @joshmaxsilverman.bsky.social · 26/05/2025

glad to pay you back for your non-vector solution to the tilted plot problem.

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joshmaxsilverman.bsky.social @joshmaxsilverman.bsky.social · 26/05/2025

interestingly, the probability of a river of length ℓ is well approximated by the naive approximation times an extra factor of 2/9

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joshmaxsilverman.bsky.social @joshmaxsilverman.bsky.social · 26/05/2025

with this in hand, we can calculate the probability that the river has length ℓ

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joshmaxsilverman.bsky.social @joshmaxsilverman.bsky.social · 26/05/2025

this oscillates around the long-distance value 2/9 before settling down

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joshmaxsilverman.bsky.social @joshmaxsilverman.bsky.social · 26/05/2025

using generating functions, we can find an exact expression for the chance position ℓ is a space

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joshmaxsilverman.bsky.social @joshmaxsilverman.bsky.social · 26/05/2025

with this in hand, we can calculate the chance of a length ℓ river like

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joshmaxsilverman.bsky.social @joshmaxsilverman.bsky.social · 26/05/2025

the chance position ℓ has a space is the chance position (ℓ-4) was a space times the chance a 3-letter word was used plus the chance position (ℓ-5) was a space times the chance a 4-letter word was used.

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joshmaxsilverman.bsky.social @joshmaxsilverman.bsky.social · 26/05/2025

each line is very long, so we can just find the probability the a given line has a space at position ℓ.

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joshmaxsilverman.bsky.social @joshmaxsilverman.bsky.social · 26/05/2025

in #thisweeksfiddler, we ask how long, on average, are coincidental diagonals of contiguous spaces in books written using 50% 3-letter, 50% 4-letter words with monospaced font.

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joshmaxsilverman.bsky.social @joshmaxsilverman.bsky.social · 19/05/2025

for the puzzle depicted, this is just 2*2*4*4*6*^ = 2,304 and in general, for an L-layer puzzle it is

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joshmaxsilverman.bsky.social @joshmaxsilverman.bsky.social · 19/05/2025

paths from one node to another in a flat layer are unique, so sidestepping doesn't change the multiplicity. a path can enter/exit a layer at any node. since two edges emerge from any given node, the number of paths is just the product of the number of edges emerging at each layer.

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joshmaxsilverman.bsky.social @joshmaxsilverman.bsky.social · 19/05/2025

we are also given a 2D version of the problem, which can be done with the same computational approach but can also be done analytically.

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joshmaxsilverman.bsky.social @joshmaxsilverman.bsky.social · 19/05/2025

putting it all together, we can run Ω(bottom point) which gives 1,093,007,025 for the 7-layer bipyramid.

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joshmaxsilverman.bsky.social @joshmaxsilverman.bsky.social · 19/05/2025

within a layer, pick a corner and call it row = 0, index = 0. all nodes can therefore be labelled by their row, index, and layer. in this scheme, vertical neighbors of node (r,i,l) in the upper half are ((r,i,l-1), (r-1,i,l-1), (r-1,i-1,l-1)) with the row/index shifts reversed in the lower half.

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joshmaxsilverman.bsky.social @joshmaxsilverman.bsky.social · 19/05/2025

finally we can find T(i<-j, l) by searching within the layer. this leaves determining the vertical neighbors for any given node which can be seen by overlaying one layer of the pyramid on the one below.

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joshmaxsilverman.bsky.social @joshmaxsilverman.bsky.social · 19/05/2025

we can find the number of ways to enter a layer at node i as the sum of ways to exit the last layer from one of its neighbors

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joshmaxsilverman.bsky.social @joshmaxsilverman.bsky.social · 19/05/2025

at each layer, a path enters at a node, and moves to exit from any other node. therefore, the number of ways Ω(i) to exit a layer from node i is the number of ways W(j) to enter the layer at node j times the number of ways to move from j to i within their layer, without repeating an edge T(i <- j)

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joshmaxsilverman.bsky.social @joshmaxsilverman.bsky.social · 19/05/2025

In #thisweeksfiddler we're asked to count the paths down a bipyramid, starting at the top and exiting the bottom, without taking any edge twice.

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joshmaxsilverman.bsky.social @joshmaxsilverman.bsky.social · 07/05/2025

the standard credit simplifies to a similar expression, 7/6 + H_N

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joshmaxsilverman.bsky.social @joshmaxsilverman.bsky.social · 05/05/2025

the standard credit is actually a bit more complicated, and is treated in the post. joshmaxsilverman.github.io/2025-05-04-f... @xaqwg.bsky.social

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joshmaxsilverman.bsky.social @joshmaxsilverman.bsky.social · 05/05/2025

recursing down to the the case of 3 time slots, which gives 3 rides by definition, this shows the expected number of rides is (3H_N - 5/2) where H_N is the Nth harmonic number (1 + 1/2 + ... + 1/N). the blue points are the result of a 1E6 round simulation and the gold point are the formula above.

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joshmaxsilverman.bsky.social @joshmaxsilverman.bsky.social · 05/05/2025

this means that the expected number of rides with N time slots is the expected number of rides with (N-1) time slots plus the expected number of times the first time slot is occupied.

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joshmaxsilverman.bsky.social @joshmaxsilverman.bsky.social · 05/05/2025

if so, then it's an itinerary we <could not> have gotten were there one fewer time slot. but, once we take the first rides, we'll draw another time slot and be left with a random itinerary we <could> have gotten if there were one fewer time slot.

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joshmaxsilverman.bsky.social @joshmaxsilverman.bsky.social · 05/05/2025

when we get out initial itinerary, it can either have a ride in the first time slot, like [1,5,9], or not like [2,5,9]. if not, then it's an itinerary we could have gotten if there were one fewer time slot.

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joshmaxsilverman.bsky.social @joshmaxsilverman.bsky.social · 05/05/2025

we can figure this out by thinking about different versions of the game that have different numbers of time slots, with new time slots added at the beginning of the day.

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joshmaxsilverman.bsky.social @joshmaxsilverman.bsky.social · 05/05/2025

in #thisweeksfiddler the question is, on average, how many rides will be taken by holders of the lightning pass, assuming their initial rides are assigned at random from the days hourly time slots and they can hold three reservations at any given time?

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joshmaxsilverman.bsky.social @joshmaxsilverman.bsky.social · 21/04/2025

but, when the target score is even, there's no advantage! that's because the incentives turn it into a run of accepted hammers, thus 2 pt holes. when the distance to the endpoint is even, the advantage of 1 makes no difference as the two players are equidistant from winning. so, the odds even.

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joshmaxsilverman.bsky.social @joshmaxsilverman.bsky.social · 21/04/2025

we can use this to calculate the advantage for whatever player is in the lead. a 1-pt advantage after the first hole translates into a 75% win prob when the target score is 3, and a 68.75% win prob when the target score is 5.

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joshmaxsilverman.bsky.social @joshmaxsilverman.bsky.social · 21/04/2025

with that, we have the following logic for the optimal play of two players (the convention here is that the end state is worth 1 if player A wins, and 0 if player B wins)

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joshmaxsilverman.bsky.social @joshmaxsilverman.bsky.social · 21/04/2025

since the player who doesn't toss the hammer has a choice to accept or reject, each player is incentivized to minimize the maximum expected outcome for the other.

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joshmaxsilverman.bsky.social @joshmaxsilverman.bsky.social · 21/04/2025

in #thisweeksfiddler we analyze the storied subgame of The Hammer in Tomorrow's Golf League. if a player throws the hammer, the opposing hammer has to accept it, doubling the value of winning the next hole, or reject it, automatically ceding 1 pt to the hammer thrower.

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joshmaxsilverman.bsky.social @joshmaxsilverman.bsky.social · 08/04/2025

this is great. i set out hoping to find some way like this but couldn't put it together. blessings.

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joshmaxsilverman.bsky.social @joshmaxsilverman.bsky.social · 07/04/2025

for groups of size 10, it means classroom of size 21. in such a classroom there are binom(21,10) different groups, and any given student is in 10 out of every 21 groups. this means that each student will get 10/21 * binom(21,10) = 167,960 pieces of candy, which is way too much.

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joshmaxsilverman.bsky.social @joshmaxsilverman.bsky.social · 07/04/2025

googling around, this kind of graph is called a kneser graph, and it has hamiltonian cycles whenever the classroom size is bigger than twice the group size. this gives 2x3+1=7 for the case of trios.

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joshmaxsilverman.bsky.social @joshmaxsilverman.bsky.social · 07/04/2025

in other words, it is a bipartite graph where each group is neighbors with every group it shares no members with, and the path we're looking for is one that flip flops between the two sets while visiting every group.

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