joshmaxsilverman.bsky.social
@joshmaxsilverman.bsky.social
this captures the behavior closely, starting at about 10 friends.
July 23, 2025 at 1:10 AM
this captures the behavior closely, starting at about 10 friends.
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)
July 23, 2025 at 1:10 AM
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)
interestingly, the probability of a river of length ℓ is well approximated by the naive approximation times an extra factor of 2/9
May 26, 2025 at 8:52 PM
interestingly, the probability of a river of length ℓ is well approximated by the naive approximation times an extra factor of 2/9
with this in hand, we can calculate the probability that the river has length ℓ
May 26, 2025 at 8:52 PM
with this in hand, we can calculate the probability that the river has length ℓ
this oscillates around the long-distance value 2/9 before settling down
May 26, 2025 at 8:52 PM
this oscillates around the long-distance value 2/9 before settling down
using generating functions, we can find an exact expression for the chance position ℓ is a space
May 26, 2025 at 8:52 PM
using generating functions, we can find an exact expression for the chance position ℓ is a space
with this in hand, we can calculate the chance of a length ℓ river like
May 26, 2025 at 8:52 PM
with this in hand, we can calculate the chance of a length ℓ river like
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.
May 26, 2025 at 8:52 PM
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.
for the puzzle depicted, this is just 2*2*4*4*6*^ = 2,304 and in general, for an L-layer puzzle it is
May 19, 2025 at 2:22 PM
for the puzzle depicted, this is just 2*2*4*4*6*^ = 2,304 and in general, for an L-layer puzzle it is
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.
May 19, 2025 at 2:22 PM
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.
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
May 19, 2025 at 2:22 PM
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
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)
May 19, 2025 at 2:22 PM
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)
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.
May 19, 2025 at 2:22 PM
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.
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.
the blue points are the result of a 1E6 round simulation and the gold point are the formula above.
May 5, 2025 at 12:31 PM
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.
the blue points are the result of a 1E6 round simulation and the gold point are the formula above.
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.
May 5, 2025 at 12:31 PM
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.
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?
May 5, 2025 at 12:31 PM
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?
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)
April 21, 2025 at 3:42 PM
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)
backtracking finds such a sequence for classrooms of 7, the smallest possible size.
April 7, 2025 at 11:05 PM
backtracking finds such a sequence for classrooms of 7, the smallest possible size.
the bracket has a tree structure, so the potential opponents a team has at round k is just the set of all teams in the opposing subtree at level k
March 25, 2025 at 9:31 AM
the bracket has a tree structure, so the potential opponents a team has at round k is just the set of all teams in the opposing subtree at level k
the key insight is that the chance a team makes it to round (k+1) is the product of the chances 1) they made it to round k, 2) their opponent made it to round k 3) they beat their opponent, summed over all possible opponents
March 25, 2025 at 9:30 AM
the key insight is that the chance a team makes it to round (k+1) is the product of the chances 1) they made it to round k, 2) their opponent made it to round k 3) they beat their opponent, summed over all possible opponents
in #thisweeksfiddler, we want to know the chance that a 1-seed makes the final four, given that (2^d-1) inferior teams stand in their way.
March 25, 2025 at 9:30 AM
in #thisweeksfiddler, we want to know the chance that a 1-seed makes the final four, given that (2^d-1) inferior teams stand in their way.
with the pdf in hand, we find the average distance to shore is 1/3-4/(9pi) ~= 0.19186.
but we can better capture the plight of the everyman beachgoer with the median distance. binary searching the cdf, we find the 50th percentile beach distance is approximately 0.17426
@xaqwg.bsky.social
but we can better capture the plight of the everyman beachgoer with the median distance. binary searching the cdf, we find the 50th percentile beach distance is approximately 0.17426
@xaqwg.bsky.social
March 18, 2025 at 5:14 AM
with the pdf in hand, we find the average distance to shore is 1/3-4/(9pi) ~= 0.19186.
but we can better capture the plight of the everyman beachgoer with the median distance. binary searching the cdf, we find the 50th percentile beach distance is approximately 0.17426
@xaqwg.bsky.social
but we can better capture the plight of the everyman beachgoer with the median distance. binary searching the cdf, we find the 50th percentile beach distance is approximately 0.17426
@xaqwg.bsky.social
as the plot shows, both regions have a similar distribution, with the semicircular region slightly more shore-biased.
March 18, 2025 at 5:14 AM
as the plot shows, both regions have a similar distribution, with the semicircular region slightly more shore-biased.
using an unholy mixture of cartesian and polar coordinates, we can find the probability density of the distance to the nearest shore, the deranged expression
March 18, 2025 at 5:14 AM
using an unholy mixture of cartesian and polar coordinates, we can find the probability density of the distance to the nearest shore, the deranged expression
in #thisweeksfiddler, we're kidnapped and wake up in a stupor at some random position on a semi-circular island. what is our distance to the nearest shore, probably?
joshmaxsilverman.github.io/2025-03-16-f...
joshmaxsilverman.github.io/2025-03-16-f...
March 18, 2025 at 5:14 AM
in #thisweeksfiddler, we're kidnapped and wake up in a stupor at some random position on a semi-circular island. what is our distance to the nearest shore, probably?
joshmaxsilverman.github.io/2025-03-16-f...
joshmaxsilverman.github.io/2025-03-16-f...