\datethis
\everypar{\looseness=-1}

@*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 $y_0\ge y_1<y_2$, 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.)

Note: This program was written after I presented a lecture about
Eastman's algorithm at Stanford on 3~December 2015. While preparing
the lecture I realized that some nice structure was present, and
a day later it occurred to me that the algorithm could therefore
be streamlined. This program significantly simplifies
the method of the previous one, which was called {\mc COMMAFREE-EASTMAN}.
It produces essentially the same outputs, but they are reflected left-to-right.
(More precisely, if the former program gave the codeword $y$ from the
the input sequence $x=x_0\ldots x_{n-1}$, this program gives
the reverse of~$y$ when given the reverse of~$x$.)

@d maxn 105

@c
#include <stdio.h>
#include <stdlib.h>
int x[maxn+maxn];
int b[maxn+maxn];
int bb[maxn];
@<Subroutines@>;
main (int argc,char*argv[]) {
  register int i,i0,j,k,n,p,q,t,tt;
  @<Process the command line@>;
  @<Do Eastman's algorithm@>;
  @<Print the solution@>;
}

@ @<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.

The algorithm is based on an interesting factorization of strings into
substrings that have the form $z=z_1\ldots z_k$ where $k\ge2$ and
$z_1\ge\cdots\ge z_{k-1}<z_k$. Let's call such a substring a ``dip.''
It is not difficult to see that any string $y=y_0y_1\ldots{}$ in
which the condition $y_i<y_{i+1}$ occurs infinitely often can be
factored {\it uniquely\/} as a sequence of dips,
$y=z^{(0)}z^{(1)}\ldots{}\,$. For example,
$3141592653589\ldots{}=
314\,15\,926\,535\,89\,\dots{}\,$.

Furthermore if $y$ is a periodic sequence, its factorization is also
ultimately periodic, although some of its initial factors may not occur in the
period. Consider, for example, the factorizations
$$\displaylines{
1234501234501234501\ldots{}=12\,34\,501\,23\,45\,01\,23\,45\,01\,\ldots{}\,;\cr
1234560123456012345601\ldots{}=
12\,34\,56\,01\,23\,45\,601\,23\,45\,601\,\ldots{}\,.\cr}$$
If the period length is~$t$, and if $i_0$ is the smallest~$i$ such that
$y_{i-3}\ge y_{i-2}<y_{i-1}$, then one of the factors ends at~$i_0$
and all factors are periodic after that point. The value of~$i_0$ is
at most~$t+2$.

Since $t$ is odd, the period contains an odd number of dips of odd length.
Each phase of Eastman's algorithm simply retains the boundary points that
precede those odd dips.

@<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|@>;

@ We might need to refer to |b[n+n-1]|, but not |b[n+n]|.

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

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

@<Sub...@>=
int less(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 (i=1;;i++) if (!less(i)) break;
  /* now |i<=t| and |y[i-1]>=y[i]| */
  for (i+=2;i<=t+2;i++) if (less(i-1)) break;
  if (i>t+2) {
    fprintf(stderr,"The input is cyclic!\n");
    exit(-666);
  }
  /* now |y[i-3]>=y[i-2]<y[i-1]| */
  if (i<t) i0=i;@+ else i=i0=i-t;
  for (tt=0;i<i0+t;i=j) {
    for (j=i+2;;j++) if (less(j-1)) break; /* advance past the next dip */
    if ((j-i)&1) @<Retain |i| as a boundary point@>;
  }
  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;
}

@ If |i>=t| at this point, we have ``wrapped around,'' so we actually
want to retain the boundary point |i-t|. (This case will arise at most
once per phase, because |j>=i+3| and we must have |j=i0+t|. Therefore
|i-t| will be smaller than all of the previously retained points, and
we want to shift them one space to the right.)

@<Retain |i| as a boundary point@>=
{
  if (i<t) bb[tt++]=b[i];
  else {
    for (k=tt++;k>0;k--) bb[k]=bb[k-1];
    bb[0]=b[i-t];
  }
}

@ @<Print the solution@>=
for (j=b[0];j<b[0]+n;j++) printf(" %d",
                                      x[j]);
printf("\n");

@*Index.