Tom Keith
@tomkeith.bsky.social
#ThisWeeksFiddler @xaqwg.bsky.social: Anita the Ant
To figure out what was going on, I made an animation. I found that Anita the Ant's path ended up inscribing a 2-3-2 isosceles triangle.
To figure out what was going on, I made an animation. I found that Anita the Ant's path ended up inscribing a 2-3-2 isosceles triangle.
October 7, 2025 at 12:23 PM
#ThisWeeksFiddler @xaqwg.bsky.social: Anita the Ant
To figure out what was going on, I made an animation. I found that Anita the Ant's path ended up inscribing a 2-3-2 isosceles triangle.
To figure out what was going on, I made an animation. I found that Anita the Ant's path ended up inscribing a 2-3-2 isosceles triangle.
For a large-enough number of n participants, the average longest increasing subsequence is roughly proportional to the square root of n.
September 23, 2025 at 1:49 PM
For a large-enough number of n participants, the average longest increasing subsequence is roughly proportional to the square root of n.
#ThisWeeksFiddler @xaqwg.bsky.social
Here are the average lengths of the longest increasing subsequences of all the permutations of n elements.
(The numerators in the table come from oeis.org/A003316.)
Here are the average lengths of the longest increasing subsequences of all the permutations of n elements.
(The numerators in the table come from oeis.org/A003316.)
September 23, 2025 at 1:43 PM
#ThisWeeksFiddler @xaqwg.bsky.social
Here are the average lengths of the longest increasing subsequences of all the permutations of n elements.
(The numerators in the table come from oeis.org/A003316.)
Here are the average lengths of the longest increasing subsequences of all the permutations of n elements.
(The numerators in the table come from oeis.org/A003316.)
#ThisWeeksFiddler @xaqwg.bsky.social
This chart shows how the number of mulligans you have affects your expected score.
This chart shows how the number of mulligans you have affects your expected score.
August 27, 2025 at 2:14 PM
#ThisWeeksFiddler @xaqwg.bsky.social
This chart shows how the number of mulligans you have affects your expected score.
This chart shows how the number of mulligans you have affects your expected score.
#ThisWeeksFiddler @xaqwg.bsky.social:
This chart shows the probability of being able to achieve various dollar amounts of cash if you start with three $10 vouchers and one $25 voucher and play optimally for each amount.
This chart shows the probability of being able to achieve various dollar amounts of cash if you start with three $10 vouchers and one $25 voucher and play optimally for each amount.
August 19, 2025 at 11:51 AM
#ThisWeeksFiddler @xaqwg.bsky.social:
This chart shows the probability of being able to achieve various dollar amounts of cash if you start with three $10 vouchers and one $25 voucher and play optimally for each amount.
This chart shows the probability of being able to achieve various dollar amounts of cash if you start with three $10 vouchers and one $25 voucher and play optimally for each amount.
#ThisWeeksFiddler @xaqwg.bsky.social
I wrote code to emulate playing the game of bowling. I found 420,571 possible states in the game including 25,646 terminal states.
With this list in hand, it was a matter of finding the terminal states that best qualified in each case.
I wrote code to emulate playing the game of bowling. I found 420,571 possible states in the game including 25,646 terminal states.
With this list in hand, it was a matter of finding the terminal states that best qualified in each case.
July 15, 2025 at 11:52 AM
#ThisWeeksFiddler @xaqwg.bsky.social
I wrote code to emulate playing the game of bowling. I found 420,571 possible states in the game including 25,646 terminal states.
With this list in hand, it was a matter of finding the terminal states that best qualified in each case.
I wrote code to emulate playing the game of bowling. I found 420,571 possible states in the game including 25,646 terminal states.
With this list in hand, it was a matter of finding the terminal states that best qualified in each case.
Extra credit: How many different parallelograms in the field of stars on the US flag?
I found 324 different shapes of parallelogram. The smallest appeared at 124 different locations and orientations.
I found 324 different shapes of parallelogram. The smallest appeared at 124 different locations and orientations.
July 8, 2025 at 12:57 PM
Extra credit: How many different parallelograms in the field of stars on the US flag?
I found 324 different shapes of parallelogram. The smallest appeared at 124 different locations and orientations.
I found 324 different shapes of parallelogram. The smallest appeared at 124 different locations and orientations.
#ThisWeeksFiddler @xaqwg.bsky.social
How many different equilateral triangles in the game of Dozo?
I found 11 different sizes of triangle. The smallest appeared at 36 different locations and orientations on the Dozo board. The largest appeared just once.
How many different equilateral triangles in the game of Dozo?
I found 11 different sizes of triangle. The smallest appeared at 36 different locations and orientations on the Dozo board. The largest appeared just once.
July 8, 2025 at 12:55 PM
#ThisWeeksFiddler @xaqwg.bsky.social
How many different equilateral triangles in the game of Dozo?
I found 11 different sizes of triangle. The smallest appeared at 36 different locations and orientations on the Dozo board. The largest appeared just once.
How many different equilateral triangles in the game of Dozo?
I found 11 different sizes of triangle. The smallest appeared at 36 different locations and orientations on the Dozo board. The largest appeared just once.
I was pleased to see there were more than zero but few enough to be listed.
July 2, 2025 at 12:10 PM
I was pleased to see there were more than zero but few enough to be listed.
I think I have a greedier cut for the second borehole. It removes 22% of the original sphere. That's going to affect which are the greediest boreholes after that.
Here are my first three cuts.
Here are my first three cuts.
June 25, 2025 at 1:05 PM
I think I have a greedier cut for the second borehole. It removes 22% of the original sphere. That's going to affect which are the greediest boreholes after that.
Here are my first three cuts.
Here are my first three cuts.
#ThisWeeksFiddler @xaqwg.bsky.social
Here is "a" solution to the extra credit. Very likely not "the" solution because I could only approximately calculate the volume remaining and couldn't figure out a way to find the greediest cylinder at each step.
I obliterated the sphere in 12 steps. YMMV.
Here is "a" solution to the extra credit. Very likely not "the" solution because I could only approximately calculate the volume remaining and couldn't figure out a way to find the greediest cylinder at each step.
I obliterated the sphere in 12 steps. YMMV.
June 24, 2025 at 1:55 PM
#ThisWeeksFiddler @xaqwg.bsky.social
Here is "a" solution to the extra credit. Very likely not "the" solution because I could only approximately calculate the volume remaining and couldn't figure out a way to find the greediest cylinder at each step.
I obliterated the sphere in 12 steps. YMMV.
Here is "a" solution to the extra credit. Very likely not "the" solution because I could only approximately calculate the volume remaining and couldn't figure out a way to find the greediest cylinder at each step.
I obliterated the sphere in 12 steps. YMMV.
#ThisWeeksFiddler @xaqwg.bsky.social
A 10-long river like this one happens only 405 times out of every 1,073,741,824.
A 10-long river like this one happens only 405 times out of every 1,073,741,824.
May 27, 2025 at 1:41 PM
#ThisWeeksFiddler @xaqwg.bsky.social
A 10-long river like this one happens only 405 times out of every 1,073,741,824.
A 10-long river like this one happens only 405 times out of every 1,073,741,824.
After some code optimization and more than 8 hours of computer time, I was able to calculate the 7th bipyramid. It took 463 times as long as the time to calculate the 6th bipyramid.
May 21, 2025 at 12:46 PM
After some code optimization and more than 8 hours of computer time, I was able to calculate the 7th bipyramid. It took 463 times as long as the time to calculate the 6th bipyramid.
Regular: How many distinct paths are there from the top of the rhombus to the bottom if (1) you never visit the same point twice and (2) you move only downward or sideways?
Let's count them up ...
Let's count them up ...
May 19, 2025 at 5:28 PM
Regular: How many distinct paths are there from the top of the rhombus to the bottom if (1) you never visit the same point twice and (2) you move only downward or sideways?
Let's count them up ...
Let's count them up ...
#ThisWeeksFiddler @xaqwg.bsky.social
Extra credit: How many distinct paths are there from the top of a triangular bipyramid to the bottom, if (1) you never visit the same point twice and (2) you move only downward or sideways?
Short answer: A lot, especially with larger bipyramids.
Extra credit: How many distinct paths are there from the top of a triangular bipyramid to the bottom, if (1) you never visit the same point twice and (2) you move only downward or sideways?
Short answer: A lot, especially with larger bipyramids.
May 19, 2025 at 5:25 PM
#ThisWeeksFiddler @xaqwg.bsky.social
Extra credit: How many distinct paths are there from the top of a triangular bipyramid to the bottom, if (1) you never visit the same point twice and (2) you move only downward or sideways?
Short answer: A lot, especially with larger bipyramids.
Extra credit: How many distinct paths are there from the top of a triangular bipyramid to the bottom, if (1) you never visit the same point twice and (2) you move only downward or sideways?
Short answer: A lot, especially with larger bipyramids.
This is the answer I got.
Somehow I was expecting something simpler.
Somehow I was expecting something simpler.
May 12, 2025 at 8:10 PM
This is the answer I got.
Somehow I was expecting something simpler.
Somehow I was expecting something simpler.
Extra credit: Within the interval (a, b), how likely is it that p4 > p7?
Let c = value of p where p7 = p4.
Then the answer is (b - c) / (b - a).
Let c = value of p where p7 = p4.
Then the answer is (b - c) / (b - a).
May 12, 2025 at 8:09 PM
Extra credit: Within the interval (a, b), how likely is it that p4 > p7?
Let c = value of p where p7 = p4.
Then the answer is (b - c) / (b - a).
Let c = value of p where p7 = p4.
Then the answer is (b - c) / (b - a).
#ThisWeeksFiddler @xaqwg.bsky.social
This chart shows how the likelihood of each series result depends on the probability p of the Celtics winning an individual game.
The likelihood of "Celtics in 5" (blue line) is greater than the likelihood of any other result when p is between 0.6 and 0.75.
This chart shows how the likelihood of each series result depends on the probability p of the Celtics winning an individual game.
The likelihood of "Celtics in 5" (blue line) is greater than the likelihood of any other result when p is between 0.6 and 0.75.
May 12, 2025 at 8:07 PM
#ThisWeeksFiddler @xaqwg.bsky.social
This chart shows how the likelihood of each series result depends on the probability p of the Celtics winning an individual game.
The likelihood of "Celtics in 5" (blue line) is greater than the likelihood of any other result when p is between 0.6 and 0.75.
This chart shows how the likelihood of each series result depends on the probability p of the Celtics winning an individual game.
The likelihood of "Celtics in 5" (blue line) is greater than the likelihood of any other result when p is between 0.6 and 0.75.
But what if 2 trees are equally far away, such as (7,4) and (8,1)?
Now the relative size of the gaps depends on the exact visibility. (The tree at (8,1)'s gap is larger 60% of the time.) It doesn't settle one way or the other no matter how large your visibility gets.
Now the relative size of the gaps depends on the exact visibility. (The tree at (8,1)'s gap is larger 60% of the time.) It doesn't settle one way or the other no matter how large your visibility gets.
April 29, 2025 at 11:33 AM
But what if 2 trees are equally far away, such as (7,4) and (8,1)?
Now the relative size of the gaps depends on the exact visibility. (The tree at (8,1)'s gap is larger 60% of the time.) It doesn't settle one way or the other no matter how large your visibility gets.
Now the relative size of the gaps depends on the exact visibility. (The tree at (8,1)'s gap is larger 60% of the time.) It doesn't settle one way or the other no matter how large your visibility gets.
As your visibility grows, the gaps get smaller but their size relative to each other stays the same.
April 29, 2025 at 11:31 AM
As your visibility grows, the gaps get smaller but their size relative to each other stays the same.
#ThisWeeksFidder @xaqwg.bsky.social
The largest gaps seem to lie on either side of the nearest trees. The nearest tree is at (2,1) and it separates the 2 largest gaps. The next nearest trees are at (3,1), (3,2), (4,1), and (4,3).
The illustration shows gaps when visibility is 100.
The largest gaps seem to lie on either side of the nearest trees. The nearest tree is at (2,1) and it separates the 2 largest gaps. The next nearest trees are at (3,1), (3,2), (4,1), and (4,3).
The illustration shows gaps when visibility is 100.
April 29, 2025 at 11:29 AM
#ThisWeeksFidder @xaqwg.bsky.social
The largest gaps seem to lie on either side of the nearest trees. The nearest tree is at (2,1) and it separates the 2 largest gaps. The next nearest trees are at (3,1), (3,2), (4,1), and (4,3).
The illustration shows gaps when visibility is 100.
The largest gaps seem to lie on either side of the nearest trees. The nearest tree is at (2,1) and it separates the 2 largest gaps. The next nearest trees are at (3,1), (3,2), (4,1), and (4,3).
The illustration shows gaps when visibility is 100.
#ThisWeeksFiddler 28 Mar 2025 EC
There seems to be a pattern for the number of teams that can never win a tournament. If the number of teams in the tournament is 2^k, then the number of teams that can never win is 2^(k-1) - k + 1 (in the general case).
There seems to be a pattern for the number of teams that can never win a tournament. If the number of teams in the tournament is 2^k, then the number of teams that can never win is 2^(k-1) - k + 1 (in the general case).
March 31, 2025 at 5:01 PM
#ThisWeeksFiddler 28 Mar 2025 EC
There seems to be a pattern for the number of teams that can never win a tournament. If the number of teams in the tournament is 2^k, then the number of teams that can never win is 2^(k-1) - k + 1 (in the general case).
There seems to be a pattern for the number of teams that can never win a tournament. If the number of teams in the tournament is 2^k, then the number of teams that can never win is 2^(k-1) - k + 1 (in the general case).
#ThisWeeksFiddler 28 Mar 2025
I tried running the tournament with different boost values, namely 0.5, 1.5, 2.5, and 3.5.
Every team won except for the team with power index 3.
I tried running the tournament with different boost values, namely 0.5, 1.5, 2.5, and 3.5.
Every team won except for the team with power index 3.
March 31, 2025 at 4:59 PM
#ThisWeeksFiddler 28 Mar 2025
I tried running the tournament with different boost values, namely 0.5, 1.5, 2.5, and 3.5.
Every team won except for the team with power index 3.
I tried running the tournament with different boost values, namely 0.5, 1.5, 2.5, and 3.5.
Every team won except for the team with power index 3.
I got similar results to yours, with first seed's probability of winning bottoming out at k = 12.
March 25, 2025 at 12:42 PM
I got similar results to yours, with first seed's probability of winning bottoming out at k = 12.