@*Intro. This (hastily written) program computes a floorplan that
corresponds to a given Baxter permutation.
See exercises MPR--135 and 7.2.2.1--372 in Volume~4B of {\sl The Art of
Computer Programming\/} for an introduction to the relevant concepts and
terminology.

The input permutation is supposed to satisfy special conditions.
When $k$ is given, let's say that a number~$s$ less than~$k$ is ``small''
and a number~$l$ greater than~$k+1$ is ``large''. Then
$$\displaylines{
\hskip3em\hbox{if $k$ occurs after $k+1$, we don't have
two consecutive elements $sl$ between them;}\hfill(*)\cr
\hskip3em\hbox{if $k+1$ occurs after $k$, we don't have
two consecutive elements $ls$ between them.}\hfill(**)\cr}$$
In other words, if $k+1$ occurs before $k$ in~$P$, any small elements
between them must follow any large ones between them~$(*)$;
otherwise any small elements between them
must {\it precede\/} any large ones between them~$(**)$.
A {\it Baxter permutation\/} is a permutation that satisfies $(*)$ and $(**)$.

Let's call the given Baxter permutation $P=p_1p_2\ldots p_n$.
We'll construct a floorplan whose rooms are the numbers $\{1,2,\ldots,n\}$.
The diagonal order of those rooms will be simply $12\ldots n$; and their
antidiagonal order will be $p_1p_2\ldots p_n$.

Floorplans have an interesting ``four-way'' order, under which 
any two distinct rooms $j$ and $k$ are in exactly one of four relationships to
each other: Either $j$ is left of $k$ (written $j\Rightarrow k$), or
$j$ is above $k$ (written $j\Downarrow k$), or
$j$ is right of $k$ (written $j\Leftarrow k$), or
$j$ is below $k$ (written $j\Uparrow k<$). The diagonal order is the
linear order ``above or left''; the antidiagonal order is the
linear order ``below or left''. 

Therefore we must have the following (nice) situation:
$$\eqalign{
j\Rightarrow k&\iff\hbox{$j<k$ and $j$ precedes $k$ in $P$};\cr
j\Downarrow k&\iff\hbox{$j<k$ and $j$ follows $k$ in $P$};\cr
j\Leftarrow k&\iff\hbox{$j>k$ and $j$ follows $k$ in $P$};\cr
j\Uparrow k&\iff\hbox{$j>k$ and $j$ precedes $k$ in $P$}.\cr
}$$
Furthermore, $j$ precedes $k$ in $P$ if and only $q_j<q_k$, where
$q_1q_2\ldots q_n$ is $P^-$, the inverse of permutation~$P$.

Any permutation $P$ defines a four-way order, according to those
rules. But only a Baxter permutation defines the four-way order
derivable from a floorplan. For example, the ``pi-mutation'' 3142
defines the four-way order with 1 left of 2, 1 above 3, 1 left of 4,
2 above 3, 2 above 4, 3 left of 4; that can happen in a floorplan
only if there's at least one more room. (For instance, we
could put ``room 2.5'' to the right of~1, below~2, above~3,
and left of~4. The Baxter permutation for that floorplan
would be 3 1 2.5 4 2.)

The rooms of a floorplan are delimited by horizontal and vertical
line segments called ``bounds,'' which don't intersect each other.
The number of horizontal bounds in the floorplan we shall output
is two more than the number of {\it descents\/} in~$P$ (that is,
places where $p_k>p_{k+1}$);
and the number of vertical bounds is two more than the number of {\it ascents\/}
(where $p_k<p_{k+1}$). 

By the way, changing $P$ to $P^R$ corresponds to transposing the floorplan
about its main diagonal.
Changing $P$ to $P^C$ corresponds to transposing the floorplan about
its other diagonal. Changing $P$ to $P^-$ corresponds to a top-bottom
reflection of the floorplan.
Thus the eight Baxter permutations obtained from $P$ by
reflection, complementation, and/or inversion correspond to the
eight standard ``isometric'' transformations that can be made to floorplans.

@ The input permutation appears in |stdin|, as the sequence of
numbers $p_1$ $p_2$ \dots~$p_n$ (separated by whitespace).
The output floorplan will be a specification that conforms to the
input conventions of the companion program {\mc FLOORPLAN-TO-TWINTREE},
with the rooms in ascending order.

@d maxn 1024
@d panic(m,k) {@+fprintf(stderr,"%s! (%d)\n",
                   m,k);@+exit(-666);@+}
@d pan(m) {@+fprintf(stderr,"%s!\n",
                     m);@+exit(-66);@+}

@c
#include <stdio.h>
#include <stdlib.h>
@<Global variables@>;
void main(void) {
  register int i,j,k,l,m,n;
  @<Input the permutation@>;
  @<Check for Baxterhood@>;
  @<Compute the floorplan@>;
  @<Output the floorplan@>;
}

@ @<Input the permutation@>=
for (m=n=0;fscanf(stdin,"%d",
                          &inx)==1;n++) {
  if (inx<=0 || inx>maxn) panic("element out of range",inx);
  if (inx>m) m=inx;
  p[n+1]=inx;
}
if (m>n) panic("too few elements",m-n);
if (m<n) panic("too many elements",n-m);
for (k=1;k<=n;k++) q[p[k]]=k; /* compute the inverse */
for (k=1;k<=n;k++) if (q[k]==0) panic("missing element",k);

@ @<Glob...@>=
int inx; /* data input with |fscanf| */
int p[maxn+1],q[maxn+1];

@ The following check might take quadratic time, because I tried to make it as
simple as possible.

If you want to test the Baxter property in linear time, there's a
tricky way to do it: (1)~Feed the permutation~$P$ to {\mc OFFLINE-TREE-INSERTION}.
(2)~Also feed its reflection, $P^R$, to {\mc OFFLINE-TREE-INSERTION}.
(3)~Edit those two outputs to make a twintree
and feed that twintree to {\mc TWINTREE-TO-BAXTER}.
(4)~Compare that result to~$P$. If $P$ is Baxter, you'll get it back again.
[An almost equivalent method was in fact published by Johnson M. Hart,
{\sl International Journal of Computer and Information Sciences\/ \bf9}
(1980), 307--321, and it's instructive to compare the two approaches.]

[If you omit this check, you'll get a floorplan whose anti-diagonal permutation
is~$P$. But if $P$ isn't Baxter, the {\it diagonal\/} permutation of that
floorplan won't be 1~2~\dots~$n$.]

@<Check for Baxterhood@>=
for (k=2;k<n-1;k++) {
  if (q[k]<q[k+1]) {
    for (l=q[k]+1;l<q[k+1]-1;l++) if (p[l]>k && p[l+1]<k)
         panic("not Baxter **",k);
  }@+else {
    for (l=q[k+1]+1;l<q[k]-1;l++) if (p[l]<k && p[l+1]>k)
         panic("not Baxter *",k);
  }
}

@*The key algorithm.
We get to use a particularly nice method here, thanks to the
insights of Eyal Ackerman, Gill Barequet, and Ron Y. Pinter
[{\sl Discrete Applied Mathematics\/ \bf154} (2006), 1674--1684].
The four bounds |lft|, |bot|, |rt|, and |top| of each room can be
filled in systemically as we march through~$P$, taking linear time
because we spend only a small bounded number of steps between
the times when we make a contribution to the final plan.

The algorithm maintains two stacks, |RLmin| and |RLmax|, which record the
current right-to-left minima and maxima in the permutation
read so far. The rooms on |RLmin| are precisely those for
which |lft| and |bot| have been filled, but not yet |top|.
The rooms on |RLmax| are precisely those for which |lft|
and |bot| have been filled, but not yet |rt|.

Values in the |bot| and |top| arrays are indices of horizontal bounds;
values in the |lft| and |rt| arrays are indices of vertical bounds.

At the end, we needn't fill in the missing values of |rt| and |top|,
because they are zero (and that's what we want).

This algorithm is almost too good to be true! It's valid, however,
because it can be seen to create the antidiagonal floorplan, step by step.

@<Compute the floorplan@>=
minptr=maxptr=1,RLmin[0]=RLmax[0]=j=p[1],lft[j]=bot[j]=n;
for (k=1;k<n;k++) {
  i=p[k],j=p[k+1]; /* |i| is at the top of both |RLmin| and |RLmax| */
 if (i<j) @<Create a new vertical bound@>@;
 else @<Create a new horizontal bound@>;
}

@ @<Create a new vertical bound@>=
{
  lft[j]=rt[i]=n-k,maxptr--,RLmin[minptr++]=j;
  while (maxptr && RLmax[maxptr-1]<j) rt[RLmax[--maxptr]]=n-k;
  bot[j]=(maxptr? top[RLmax[maxptr-1]]: n);
  RLmax[maxptr++]=j;
}

@ @<Create a new horizontal bound@>=
{
  bot[j]=top[i]=n-k,minptr--,RLmax[maxptr++]=j;
  while (minptr && RLmin[minptr-1]>j) top[RLmin[--minptr]]=n-k;
  lft[j]=(minptr? rt[RLmin[minptr-1]]: n);
  RLmin[minptr++]=j;
}

@ @<Glob...@>=
int lft[maxn+1],bot[maxn+1],rt[maxn+1],top[maxn+1];
int RLmin[maxn],RLmax[maxn]; /* the stacks */
int minptr,maxptr; /* the current stack sizes */

@ @<Output the floorplan@>=
for (k=1;k<=n;k++)
  printf("%d y%d y%d x%d x%d\n",
            k,n-top[k],n-bot[k],n-lft[k],n-rt[k]);

@*Index.