AbstractFor n even, a factorization of a complete graph Kn is a partition of the edges into n−1 perfect matchings, called the factors of the factorization. With respect to a factorization, a path is called rainbow if its edges are from distinct factors. A rainbow Hamiltonian path takes exactly one edge from each factor and is called orthogonal to the factorization. It is known that not all factorizations have orthogonal paths. Assisted by a simple edge‐switching algorithm, here we show that for n⩾8, the rotational factorization of Kn, GKn has orthogonal paths. We prove that this algorithm finds a rainbow path with at least (2n+1)/3 vertices in any factorization of Kn (in fact, in any proper coloring of Kn). We also give some problems and conjectures about the properties of the algorithm. © 2010 Wiley Periodicals, Inc. J Combin Designs 18: 167–176, 2010