@*Intro. This program is an iterative implementation of
an interesting recursive algorithm due to
Willard~L. Eastman, {\sl IEEE Trans.\ \bf IT-11} (1965), 263--267:
Given a sequence of nonnegative integers $x=x_0x_1\ldots x_{n-1}$ of
odd length~$n$, where $x$ is not equal to any of its cyclic shifts
$x_k\ldots x_{n-1}x_0\ldots x_{k-1}$ for $1\le k<n$, we output a
cyclic shift $\sigma x$ such that the set of all such $\sigma x$
forms a commafree code of block length~$n$ (over an infinite alphabet).

The integers are given as command-line arguments.

The simplest nontrivial example occurs when $n=3$. If $x=abc$, where
$a$, $b$, and~$c$ aren't all equal, then exactly one of the cyclic shifts
$y_0y_1y_2=abc$, $bca$, $cab$ will satisfy $y_0>y_1\le y_2$, and we
choose that one. It's easy to check that the triples chosen in
this way are commafree.

Similar constructions are possible when $n=5$ or $n=7$. But the case
$n=9$ already gets a bit dicey, and when $n$ is really large it's not
at all clear that commafreeness is possible. Eastman's paper resolved
a conjecture made by Golomb, Gordon, and Welch in their pioneering paper about
comma-free codes (1958).

(Of course, it's not at all clear
that we would want to actually {\it use\/} a commafree code when
$n$ is large; but that's another story, and beside the point. The point
is that Eastman discovered a really interesting algorithm.)

@d maxn 105

@c
#include <stdio.h>
#include <stdlib.h>
int x[maxn+maxn+maxn];
int b[maxn+maxn+maxn];
int bb[maxn];
@<Subroutines@>;
main (int argc,char*argv[]) {
  register int i,j,k,n,p,q,t,tt;
  @<Process the command line@>;
  @<Do Eastman's algorithm@>;
}

@ @<Process the command line@>=
if (argc<4) {
  fprintf(stderr,"Usage: %s x1 x2 ... xn\n",argv[0]);
  exit(-1);
}
n=argc-1;
if ((n&1)==0) {
  fprintf(stderr,"The number of items, n, should be odd, not %d!\n",n);
  exit(-2);
}         
for (j=1;j<argc;j++) {
  if (sscanf(argv[j],"%d",&x[j-1])!=1 || x[j-1]<0) {
    fprintf(stderr,"Argument %d should be a nonnegative integer, not `%s'!\n",
                                     j,argv[j]);
    exit(-3);
  }
}

@*The algorithm.
We think of $x$ as written cyclically, with $x_{n+j}=x_j$ for all~$j\ge0$.
The basic idea in the algorithm below is to also think of $x$ as partitioned
into $t\le n$ subwords by boundary markers $b_j$ where $0\le b_0<b_1<\cdots
<b_{t-1}<n$; then subword $y_j$ is $x_{b_j}x_{b_j+1}\ldots x_{b_{j+1}-1}$,
for $0\le j<t$, where $b_t=b_0$. If $t=1$, there's just one subword,
and it's a cyclic shift of~$x$. The number $t$ of subwords during each phase
will be odd.

Eastman's algorithm essentially begins with $b_j=j$ for $0\le j<n$, so that
$x$ is partitioned into $n$ subwords of length~1. It successively {\it removes\/}
boundary points until only one subword is left; that subword is the answer.
It operates in phases, so that all subwords during the $j$th phase have
length $3^{j-1}$ or more; thus at most $\lfloor\log_3n\rfloor$ phases
are needed. (For example, the case $n=9$ is ``dicey'' because it might
require two phases.)

The algorithm is based on comparison of adjacent subwords $y_{j-1}$ and~$y_j$.
If those subwords have the same length, we use lexicographic comparison;
otherwise we declare that the longer subword is bigger.

(After the first phase, all subwords not only have length $\ge3$, they also
always begin with a nonzero entry; in other words, $x_{b_j}>0$ for every
boundary marker~$b_j$. However, we won't need to use that fact explicitly.)

The algorithm can be described with terminology based on the topography
of Nevada: Say that $i$ is a {\it basin\/} if the subwords satisfy
$y_{i-1}>y_i\le y_{i+1}$. There must be at least one basin; otherwise all
the $y_j$ would be equal, and $x$ would equal one of its cyclic shifts.
We look at consecutive basins, $i$ and~$j$; this means that $i<j$ and that
$i$ and $j$ are basins, and that $i+1$ through $j-1$ are {\it not\/} basins.
If there's only one basin we have $j=i+t$. 
The indices between consecutive basins are called {\it ranges\/}.

Since $t$ is odd, there's an odd number of consecutive basins for which
$j-i$ is odd. Each phase of Eastman's algorithm retains exactly one boundary
point in the range between such basins, and deletes all the others.
The retained point is the smallest $k=i+2l$ such that $y_k>y_{k+1}$.

(For example, suppose $i=2$ and $j=9$ are consecutive basins. Then we have
$y_1>y_2\le y_3\le\cdots\le y_q>y_{q+1}>\cdots>y_9\le y_{10}$, for some
range element $2<q<9$. We choose $k=4$ if $q=3$ or $q=4$,
$k=6$ if $q=5$ or $q=6$, and $k=8$ if $q=7$ or $q=8$.)

@<Do Eastman's algorithm@>=
@<Initialize@>;
for (p=1,t=n;t>1;t=tt,p++)
  @<Do one phase of Eastman's algorithm, putting |tt| boundary points into |bb|@>;

@ @<Initialize@>=
for (j=n;j<n+n+n;j++) x[j]=x[j-n];
for (j=0;j<n+n+n;j++) b[j]=j;
t=n;

@ Here's a basic subroutine that returns 1 if subword $y_{i-1}$ exceeds
subword~$y_i$, otherwise it returns~0.

@<Sub...@>=
int compare(register int i) {
  register int j;
  if (b[i]-b[i-1]==b[i+1]-b[i]) {
    for (j=0;b[i]+j<b[i+1];j++) {
      if (x[b[i-1]+j]==x[b[i]+j]) continue;
      return (x[b[i-1]+j]>x[b[i]+j]);
    }
    return 0; /* $y_{i-1}=y_i$ */
  }
  return (b[i]-b[i-1]>b[i+1]-b[i]);
}

@ @<Do one phase of Eastman's algorithm, putting |tt| boundary points into |bb|@>=
{
  for (tt=0,i=1;i<=t;i++) if (compare(i)) break;
  if (i>t) {
    fprintf(stderr,"The input is cyclic!\n");
    exit(-666);
  }
  for (;compare(i+1);i++) ; /* advance to the first basin */
  for(;i<=t;i=j) {
    for (q=i+1;compare(q+1)==0;q++) ; /* climb the range */
    for (j=q+1;compare(j+1);j++) ; /* advance to the next basin */
    if ((j-i)&1) @<Choose a boundary point to retain@>;
  }
  printf("Phase %d leaves",p);
  for (k=0;k<tt;k++) b[k]=bb[k],printf(" %d",bb[k]);
  printf("\n");
  for (;b[k-tt]<n+n;k++) b[k]=b[k-tt]+n;
}

@ @<Choose a boundary point to retain@>=
{
  if ((q-i)&1) q++;
  if (q<t) bb[tt++]=b[q];
  else {
    for (k=tt++;k>0;k--) bb[k]=bb[k-1];
    bb[0]=b[q-t];
  }
}

@*Index.