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July 29

Kulczynski Similarity Coefficient I (K1)

Every similarity metric I've used is bounded, usually [0,1] or [-1,1]. K1 is unbounded [0,inf). Distances and scores are unbounded. Is K1 actually a score and not a similarity? ~2026-41824-87 (talk) 11:56, 29 July 2026 (UTC)

See Qualitative variation#Kulczyński's coefficient. Kulczyński's coefficient varies between 0 and 1. What is this K1, you are talking about? JRSpriggs (talk) 13:20, 29 July 2026 (UTC)

This is the definition I have:

Kulczynski I (K1): The ratio of shared elements to non-shared elements. Counts matches (elements in both sets) against mismatches (elements in exactly one set). Shared absences are ignored, so no universe size is needed.

Formula:

K1(A, B) = a / (b + c)

a = |A ∩ B| (in both: n11)

b = |A only| (in A, not B: n10)

c = |B only| (in B, not A: n01)

Range: [0, inf)

inf = identical sets (no mismatches)

0 = no shared elements ~2026-41824-87 (talk) 13:32, 29 July 2026 (UTC)

Here, the coefficient defined in the Wikipedia article is called "Kulczynski 2".

In the extreme case that A = B, we get indeed K1(A, B) = a / 0.  ​‑‑Lambiam 15:49, 29 July 2026 (UTC)

It considers division by zero to be infinity rather than undefined. Because it is unbounded, I was wondering if it is properly called a "score" instead of a "similarity coefficient". Maybe it is just my limited experience with similarity metrics, but all I've seen are bounded. Score and distance metrics are always unbounded. ~2026-41824-87 (talk) 16:39, 29 July 2026 (UTC)

There is no standard meaning of the term ''coefficient'' that would impose limits on its range. In physics, the coefficient of thermal expansion can even be negative, as it is for fairly pure silicon for temperatures between about 18 and 120 K (−255 and −153 °C; −427 and −244 °F), and the reflection coefficient of a wave vector can become infinite.  ​‑‑Lambiam 12:04, 30 July 2026 (UTC)

- "Maybe it is just my limited experience with similarity metrics, but all I've seen are bounded."

- "There is no standard meaning of the term "coefficient" that would impose limits on its range."

I do not feel that this response fits. Some similarity metrics are coefficients. Some are indexes. ~2026-41824-87 (talk) 11:08, 3 August 2026 (UTC)

AFAIK, the area of similarity measures is a free-for-all, with no agreed conventions.  ​‑‑Lambiam 17:36, 3 August 2026 (UTC)

Not necessarily an answer to your question, but it's worth noting:

1. $b+c=|A-B|+|B-A|=|A\Delta B|$ (the symmetric difference), and $|A\cap B|+|A\Delta B|=|A\cup B|$.

2. ${\frac {x}{x+y}}={\frac {1}{1+{\frac {1}{x/y}}}}$, and ${\frac {1}{1+{\frac {1}{z}}}}$ is bijective and monotone increasing from $[0,\infty )$ to $[0,1)$. Its inverse, ${\frac {1}{{\frac {1}{y}}-1}}$, which takes ${\frac {x}{x+y}}$ to ${\frac {x}{y}}$, is consequently also bijective and monotone increasing from $[0,1)$ to $[0,\infty )$.

3. Since ${\frac {x}{y}}\in [0,\infty )$ for $x,y\in [0,\infty )^{2}$, and all possible values of $x=|A\cap B|$ and $y=|A\Delta B|$ are nonnegative integers, K1 is just a bijective remapping of values of the Jaccard index ${\frac {|A\cap B|}{|A\cup B|}}$ to $[0,\infty )$.

GalacticShoe (talk) 15:28, 29 July 2026 (UTC)

August 9

Pythagorean triple angles

Is it coincidence that the larger acute angle of a 7-24-25 Pythagorean-triple triangle is twice the smaller acute angle of a 3-4-5 triangle? Thanks, cmɢʟee τaʟκ (please add {{ping|cmglee}} to your reply) 00:58, 9 August 2026 (UTC)

No, it is not a coincidence. See trigonometric formulas#Double-angle formulae.

$\tan(2\theta )={\frac {2\tan \theta }{1-\tan ^{2}\theta }}$.

and apply that. JRSpriggs (talk) 01:53, 9 August 2026 (UTC)

In fact, if (a, b, c) and (d, e, f) are Pythagorean triples, then so is (ad-be, ae+bd, cf) (provided ad-be>0). The angle opposite the ae+bd side will then be the sum of the angles opposite the b and e sides in the original triangles. The 7-24-25 is a special case, the "product" of the 4-3-5 triangle with itself. --RDBury (talk) 08:39, 9 August 2026 (UTC)

I naively expected that iterating $(a,b,c)\mapsto (|a^{2}-b^{2}|,2ab,c^{2}),$ starting with $(3,4,5),$ would soon converge to some set of angles. Instead, this keeps jumping all over the place. Then I realized this corresponds to iterating $t\mapsto |2t^{2}-1|,$ where $t=a/c.$ This map has the sole fixpoint ${\tfrac {1}{2}}$ in $(0,1),$ which is not stable, making convergence impossible.  ​‑‑Lambiam 10:30, 9 August 2026 (UTC)

Thanks very much, everyone. cmɢʟee τaʟκ (please add {{ping|cmglee}} to your reply) 11:40, 9 August 2026 (UTC)