通过完美的方形对订购1:17

Question

关于R-help有一个有趣的问题：

“将数字1加到17.你能把它们写成一行，这样每一对数字相邻，加上一个正方形数字吗？”

我的解决方案如下，并不是特别特别。我对更优雅和/或更强大的解决方案感到好奇。也许一个解决方案可以采用任意数字串并在可能的情况下对它们进行排序？

sq.test <- function(a, b) {
  ## test for number pairs that sum to squares.
  sqrt(sum(a, b)) == floor(sqrt(sum(a, b)))
}

ok.pairs <- function(n, vec) {
  ## given n as a member of vec,
  ## which other members of vec satisfiy sq.test
  vec <- vec[vec!=n]
  vec[sapply(vec, sq.test, b=n)]
}

grow.seq <- function(y) {
  ## given a starting point (y) and a pairs list (pl)
  ## grow the squaring sequence.
  ly <- length(y)
  if(ly == y[1]) return(y)

  ## this line is the one that breaks down on other number sets...
  y <- c(y, max(pl[[y[ly]]][!pl[[y[ly]]] %in% y]))
  y <- grow.seq(y)

  return(y)
}


## start vector
x <- 1:17

## get list of possible pairs
pl <- lapply(x, ok.pairs, vec=x)

## pick start at max since few combinations there.
y <- max(x)
grow.seq(y)

Answer 1

您可以使用outer来计算允许对。 得到的矩阵是图的邻接矩阵， 你只需要Hamiltonian path就可以了。

# Allowable pairs form a graph
p <- outer(
  1:17, 1:17, 
  function(u,v) round(sqrt(u + v),6) == floor(sqrt(u+v)) ) 
)
rownames(p) <- colnames(p) <- 1:17
image(p, col=c(0,1)) 

# Read the solution on the plot
library(igraph)
g <- graph.adjacency(p, "undirected")
V(g)$label <- V(g)$name 
plot(g, layout=layout.fruchterman.reingold)