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Relation between the number of leaves of a tree and its diameter
Combinatorics · Submitted 27 Apr 2019
abstract compiled · 2026-09-23 · no language-model review
The authors' main claim For example, $B(21,3)=14$ while $L(21,3)=19.$ We prove that for $d\ge 2,$ $ L(n,d)=\left\lceil \frac{2(n-1)}{d}\right\rceil$ if $d$ is even and $L(n,d)=\left\lceil \frac{2(n-2)}{d-1}\right\rceil$ if $d$ is odd.
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The abstract, sentence by sentence
Let $L(n,d)$ denote the minimum possible number of leaves in a tree of order $n$ and diameter $d.$ In 1975 Lesniak gave the lower bound $B(n,d)=\lceil 2(n-1)/d\rceil$ for $L(n,d).$ When $d$ is even, $B(n,d)=L(n,d).$ But when $d$ is odd, $B(n,d)$ is smaller than $L(n,d)$ in general. For example, $B(21,3)=14$ while $L(21,3)=19.$ We prove that for $d\ge 2,$ $ L(n,d)=\left\lceil \frac{2(n-1)}{d}\right\rceil$ if $d$ is even and $L(n,d)=\left\lceil \frac{2(n-2)}{d-1}\right\rceil$ if $d$ is odd. The converse problem is also considered. Let $D(n,f)$ be the minimum possible diameter of a tree of order $n$ with exactly $f$ leaves. We prove that $D(n,f)=2$ if $n=f+1,$ $D(n,f)=2k+1$ if $n=kf+2,$ and $D(n,f)=2k+2$ if $kf+3\le n\le (k+1)f+1.$
background contribution method result interpretation main claim
| # | sentence | role | the authors' own work | states the main conclusion | strength |
|---|---|---|---|---|---|
| s0 | Let $L(n,d)$ denote the minimum possible number of leaves in a tree of order $n$ and diameter $d.$ In 1975 Lesniak gave the lower bound $B(n,d)=\lceil 2(n-1)/d\rceil$ for $L(n,d).$ When $d$ is even, $B(n,d)=L(n,d).$ But when $d$ is odd, $B(n,d)$ is smaller than $L(n,d)$ in general. | background 0.99 | 0.07 | 0.09 | asserted 1.00 |
| s1 | For example, $B(21,3)=14$ while $L(21,3)=19.$ We prove that for $d\ge 2,$ $ L(n,d)=\left\lceil \frac{2(n-1)}{d}\right\rceil$ if $d$ is even and $L(n,d)=\left\lceil \frac{2(n-2)}{d-1}\right\rceil$ if $d$ is odd. | result 0.89 | 0.93 | 0.89 | asserted 1.00 |
| s2 | The converse problem is also considered. | contribution 0.79 | 0.42 | 0.02 | unclear 0.69 |
| s3 | Let $D(n,f)$ be the minimum possible diameter of a tree of order $n$ with exactly $f$ leaves. | background 0.50 | 0.34 | 0.03 | asserted 0.51 |
| s4 | We prove that $D(n,f)=2$ if $n=f+1,$ $D(n,f)=2k+1$ if $n=kf+2,$ and $D(n,f)=2k+2$ if $kf+3\le n\le (k+1)f+1.$ | result 0.99 | 0.94 | 0.21 | asserted 1.00 |
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Original abstract
Let $L(n,d)$ denote the minimum possible number of leaves in a tree of order $n$ and diameter $d.$ In 1975 Lesniak gave the lower bound $B(n,d)=\lceil 2(n-1)/d\rceil$ for $L(n,d).$ When $d$ is even, $B(n,d)=L(n,d).$ But when $d$ is odd, $B(n,d)$ is smaller than $L(n,d)$ in general. For example, $B(21,3)=14$ while $L(21,3)=19.$ We prove that for $d\ge 2,$ $ L(n,d)=\left\lceil \frac{2(n-1)}{d}\right\rceil$ if $d$ is even and $L(n,d)=\left\lceil \frac{2(n-2)}{d-1}\right\rceil$ if $d$ is odd. The converse problem is also considered. Let $D(n,f)$ be the minimum possible diameter of a tree of order $n$ with exactly $f$ leaves. We prove that $D(n,f)=2$ if $n=f+1,$ $D(n,f)=2k+1$ if $n=kf+2,$ and $D(n,f)=2k+2$ if $kf+3\le n\le (k+1)f+1.$
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