John M. Boyer

752 posts

John M. Boyer

John M. Boyer

@johnboyerphd

Views and comments expressed here are strictly my own.

Katılım Mart 2009
94 Takip Edilen96 Takipçiler
John M. Boyer
John M. Boyer@johnboyerphd·
New arXiv preprint alert. Solved an open problem from the American Mathematical Monthly. Interesting new ways to cover triangles, and particularly pleased to be able to prove that they are optimal. Comments welcome, please see here: arxiv.org/abs/2605.04111
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John M. Boyer
John M. Boyer@johnboyerphd·
@latzplacian Since both methods work but also _almost never_ work, hopefully you and your readers may also like this second Geombinatorics paper that extends the 1st method so that it covers half of all triangles more efficiently than the naïve method: researchgate.net/publication/39…
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John M. Boyer
John M. Boyer@johnboyerphd·
2nd Geombinatorics journal paper on covering triangles. Last one proved theorems whose prior proofs didn't. This is on how to extend the methods from existential proofs. Result: one method that covers half of all triangles better than the naive method. researchgate.net/publication/39…
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John M. Boyer
John M. Boyer@johnboyerphd·
“On the Conway and Soifer Schemas and Methods for Covering Non-Equilateral Triangles” has now appeared in the journal Geombinatorics (get full text here: researchgate.net/publication/39…)
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John M. Boyer
John M. Boyer@johnboyerphd·
I am pleased to announce a major new release of the Edge Additional Planarity Suite Project. This open source project is used by many computer scientists and mathematicians internationally. Release details here: github.com/graph-algorith…
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John M. Boyer
John M. Boyer@johnboyerphd·
@pickover Add right parenthesis to the function compositions for the left 3 fish.
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Cliff Pickover
Cliff Pickover@pickover·
This is a math joke.
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John M. Boyer
John M. Boyer@johnboyerphd·
@ZahlenRMD There is a second valid way to answer this puzzle if one adheres to the order of operations. Label the bottom source node X. The path to the node on the right gives 4X. The clockwise path from the source gives X + 5x3 - 6 = X + 9. Given 4X=X+9, X=3 can also be a valid solution.
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John M. Boyer
John M. Boyer@johnboyerphd·
@ZahlenRMD @BradBMath 4/4 I like how Brad solved it in a ‘goal-directed’ way, by labeling the end, working back to the source, then solving it algebraically. Harder to do mentally though, more like playing mental chess, whereas this arithmetical approach is more like playing mental tic-tac-toe :-)
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John M. Boyer
John M. Boyer@johnboyerphd·
@ZahlenRMD and to @BradBMath 1/4 Yes, there is a way to approach this entirely mentally using only arithmetic, logic, and effort. All circles (vertices) have at least one inward arrow (edge) except the bottom vertex, which is the source of value because it has only outward edges.
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John M. Boyer
John M. Boyer@johnboyerphd·
@ZahlenRMD @BradBMath 3/4 Starting with 1 ends with 4 to the right and 12 going clockwise (difference of 8). Starting with 3, ends with 12 and 18 (difference of 6). Since starting 2 higher decreased the difference by 2, we decrease the difference by 6 more with 3+6=9, which gets 36 in both directions.
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John M. Boyer
John M. Boyer@johnboyerphd·
@ZahlenRMD @BradBMath 2/4 Pick a number, like 0, then do the arithmetic to get a sense of the patterns. To the right, 0x4=0. Going clockwise, 0+5=5, 5x3=15, 15-6=9. Since with an even number (0) produced one even answer (0) and the other odd (9), then we know that we can’t start with any even number.
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John M. Boyer
John M. Boyer@johnboyerphd·
@AGiannas @pickover Since you’re missing 3 of them, I’d guess it’s the 3 that are 2x2 that form in the middle two columns. Yes?
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Cliff Pickover
Cliff Pickover@pickover·
Mathematics. How many squares?
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John M. Boyer
John M. Boyer@johnboyerphd·
@pickover Since pi is irrational, a stairwell with those length and width dimensions can’t be built out of all the atoms in the universe, so the ant could only travel approximately 2 pi distance.
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Cliff Pickover
Cliff Pickover@pickover·
Mathematics. Andy the Ant starts at the green dot at bottom left and carefully crawls to the green dot at top right. What is the distance he has traveled? (Assume this is a 2D problem, right-angle steps, and Andy is not crushed by the humanoid during his journey.)
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John M. Boyer
John M. Boyer@johnboyerphd·
Celebrate Tau Day today by reading The Tau Manifesto! You'll soon realize that Pi Day is only half as important :-) lnkd.in/dQKxAEsG
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John M. Boyer
John M. Boyer@johnboyerphd·
@pickover This is cool. Watched it twice and wondered if it was the same cat or another one just like the first one.
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Cliff Pickover
Cliff Pickover@pickover·
Physics, mathematics, reality, dots. Schrödinger's Cat enters the Matrix. Visualization by Mike Wong, based on charged particles. Source: bit.ly/3lv5rI7
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Cliff Pickover
Cliff Pickover@pickover·
Mathematics. A mathematician emerges from a cave, hands you the the slip of paper below, and asks "Which number do you like better, 2 or 3?" What is your answer?
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Otto Kalliokoski
Otto Kalliokoski@OKalliokoski·
@rtorkar @MaartenvSmeden I want to say that this is basically all that Fisher ever said about why he chose 0.05 as a cut-off heuristic. Basically it was "convenient." From Statistical Methods for Research Workers (1925)
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Maarten van Smeden
Maarten van Smeden@MaartenvSmeden·
Students often ask me: why do so many choose 5% as the threshold for significance (“alpha”)? So let me try to explain the main rationale 1/20
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John M. Boyer
John M. Boyer@johnboyerphd·
I'm happy to see that the new release of the Edge Addition Planarity Suite is making its way into Debian Linux (and related, e.g. Ubuntu and openSUSE): lnkd.in/dCdfWmKq
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John M. Boyer
John M. Boyer@johnboyerphd·
Here's a good paper by Ahmad Biniaz with an elegant proof that the complete graph on 9 vertices cannot be divided into two planar graphs: #graph lnkd.in/gu3ixZGK
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