## spiral matrix [X-validated]

One recent code-golf challenge was to write the shortest possible code representing the first n² integers in a spiral progression, e.g.,

0 1 2 3 4 15 16 17 18 5 14 23 24 19 6 13 22 21 20 7 12 11 10 9 8

While I did not come close to the best R code (with 67 bytes), this proved an interesting coding exercise, looking for a way to rotate a matrix in R, and then filling one batch at a time. Here is my clumsy if original (?) R output exploiting the vector representation of matrices in R:

r=function(x)apply(t(x),2,rev) #-90⁰ rotation w=which;m=min X=diag(0,n) X[1:n]=n:1 while(!m(X)){ X=r(X) i=m(w(!X)) j=w(!!X) j=m(j[j>i])-1 X[i:j]=-m(-X)+(j-i+1):1} while(X[1]>1)X=r(X)

although I later found a webpage proposing (non-optimised) solutions in most computer languages. Thanks to Robin’s remarks, a tighter version is

r=function(n){ w=which X=diag(0,n) X[n,]=m=n:1 while(!min(X)){ i=w(!X)[1] j=w(!!X[-1:-i]) X[i-1+1:j]=max(X)+j:1 #produces warnings X=t(X)[m,]} #-90⁰ rotation `if`(n%%2,t(t(X)[m,])[m,],X)-1} #180⁰ rotation

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