Answer by bof for Prove or disprove: $R^{n+1} \supseteq R \cap R^2 \cap...
Let $k_n$ be the least integer $k$ such that, for any digraph $D$ of order $n$ and any vertices $x,y\in D$, if there are $x$-$y$ walks of length $1,\dots,k$, then there are $x$-$y$ walks of all...
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Answer by YCor for Prove or disprove: $R^{n+1} \supseteq R \cap R^2 \cap...
Here's an example of size 9 inspired by domotorp's anwer, where I only consider the primes $2,5$, and where I replace the additional long arrows from the basis of the cycles with arrows from the...
View ArticleAnswer by Ronnie Pavlov for Prove or disprove: $R^{n+1} \supseteq R \cap R^2...
domotorp's lovely solution is by far the best one, but here is an explicit counterexample for $n = 10$, I wonder if it's computationally tractable to figure out the max $n$ for which your statement...
View ArticleAnswer by domotorp for Prove or disprove: $R^{n+1} \supseteq R \cap R^2 \cap...
This is false as shown by the following digraph.From $x$ there is an edge to $v_p$, from $v_p$ there is a cycle of length $p$ to itself, and from $v_p$ there are $p-1$ different paths to $y$, of...
View ArticleProve or disprove: $R^{n+1} \supseteq R \cap R^2 \cap \cdots \cap R^n$ for...
Prove or disprove: $R^{n+1} \supseteq R \cap R^2 \cap \cdots \cap R^n$ for every binary relation $R$ on a set of size $n$.I have verified the statement for $n \leq 4$ with a Mathematica code.I have...
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