\datethis
@s mod and
\let\Xmod=\bmod % this is CWEB magic for using "mod" instead of "%"
\def\sqprod{\setbox0=\hbox{\kern-.13em$\times$\kern-.13em}
     \dimen0=\ht0 \advance\dimen0 -.09em \ht0=\dimen0
     \dimen0=\dp0 \advance\dimen0 -.09em \dp0=\dimen0
     \mathbin{\vcenter{\hrule\kern-.4pt
       \hbox{\vrule\kern-.4pt\phantom{$\box0$}\kern-.4pt\vrule}\kern-.4pt
       \hrule}}}

@*Intro. This program finds all of the nonisomorphic graceful labelings
of the graph $K_m\sqprod P_3$. It was inspired by the paper of
B.~M. Smith and J.-F. Puget in {\sl Constraints\/ \bf15} (2010), 64--92,
where Table~5 reports a unique solution for $m=6$. I'm writing it because
I want to gain experience, gracefulnesswise --- and also because Smith and Puget
have unfortunately lost all records of the solution!

The graph $K_m\sqprod P_3$ is ``hardwired'' into the logic of this program.
It has $q=3{m\choose2}+2m$ edges; that's
(7,~15, 26, 40, 57, 77, \dots) for $m=(2$, 3, 4, 5, 6, 7, \dots).
I doubt if I'll be able to reach $m=7$; but I see no reason to exclude that
case, because the algorithm needs very little memory.

Please excuse me for writing this in a rush.

@d m 6 /* the size of the cliques; must be at least 2 and at most 12 */
@d q ((m*(3*m+1))/2) /* number of edges */
@d o mems++ /* count one mem */
@d oo mems+=2 /* count two mems */
@d ooo mems+=3 /* count three mems */
@d delta 10000000000; /* report progress every |delta| or so mems */
@d O "%" /* used for percent signs in format strings */
@d mod % /* used for percent signs denoting remainder in \CEE/ */
@d board(i,j) brd[3*(i)+(j)]
@d leftknown colknown[0]

@c
#include <stdio.h>
#include <stdlib.h>
unsigned long long mems; /* memory accesses */
unsigned long long thresh=delta; /* time for next progress report */
unsigned long long nodes; /* nodes in the search tree */
unsigned long long nulls; /* nodes that need no new vertex placement */
unsigned long long leaves; /* nodes that have no descendants */
int count; /* number of solutions found so far */
int brd[3*m]; /* one-dimensional array accessed via the |board| macro */
int rank; /* how many rows of the board are active? */
int labeled[q+1]; /* what row and column, if any, have a particular label? */
int placed[q+1]; /* has this edge been placed? */
int colknown[3]; /* how many vertices of each clique are labeled? */
int move[q][1024]; /* feasible moves at each level */
int deg[q]; /* number of choices at each level; used in printouts only */
int x[q]; /* indexes of moves made at each level */
int maxl; /* maximum level reached */
int vbose=0; /* can set this nonzero when debugging */
@<Subroutines@>@;
main () {
  register int a,b,i,j,k,l,t,v,aa,bb,ii,row,col,ccol,val,mv,trouble;
  fprintf(stderr,"--- Graceful labelings of K"O"d times P3 ---\n",m);
  @<Initialize the data structures@>;
  @<Backtrack through all solutions@>;
  fprintf(stderr,"Altogether "O"d solution"O"s,",
               count,count==1?"":"s");
  fprintf(stderr," "O"lld mems, "O"lld-"O"lld nodes, "O"lld leaves;",
               mems, nodes, nulls, leaves);
  fprintf(stderr," max level "O"d.\n",maxl);
  if (sanity_checking) fprintf(stderr,"sanity_checking was on!\n");
}

@ The current status of the vertices labeled so far appears in
the |board|, which has three columns and $m$ rows. This is not a
canonical representation: The rows can appear in any order.
When a vertex is unlabeled, the |board| has $-1$. When the vertex
in row~$i$ and column~$j$ receives label~$l$, |labeled[l]| records
the value |(j<<4)+i|; but |labeled[l]| is $-1$ if that label hasn't
been used. If both endpoints of an edge are labeled, and if $d$ is
the difference between those labels, |placed[d]=1|; but
|placed[d]=0| if no edge for difference~|d| is yet known.

The first |rank| rows of the board have been labeled, at least
in part.

@<Initialize the data structures@>=
for (i=0;i<m;i++) for (j=0;j<3;j++) board(i,j)=-1;
rank=0;
for (l=0;l<=q;l++) labeled[l]=-1;
l=0;

@ @<Sub...@>=
void print_board(int rank) {
  register int i,j;
  for (i=0;i<rank;i++) {
    for (j=0;j<3;j++)
      if (board(i,j)>=0) fprintf(stderr,""O"3d",board(i,j));
      else fprintf(stderr,"  ?");
    fprintf(stderr,"\n");
  }
}

@ @<Sub...@>=
void print_placed(void) {
  register int k;
  for (k=1;k<=q;k++) {
    if (placed[k]) {
      if (!placed[k-1]) fprintf(stderr," "O"d",k);
      else if (k==q || !placed[k+1]) fprintf(stderr,".."O"d",k);
    }
  }
  fprintf(stderr,"\n");
}

@ These data structures are somewhat fancy, so I'd better check that
they're self-consistent.

@d sanity_checking 0 /* set this to 1 if you suspect a bug */

@<Sub...@>=
void sanity(void) {
  register int i,j,l,t,v;
  @<Check the rank@>;
  @<Check the labels@>;
  @<Check the placements@>;
}

@ @<Check the rank@>=
for (i=rank;i<m;i++) {
  if (board(i,0)>=0) break;
  if (board(i,1)>=0) break;
  if (board(i,2)>=0) break;
}
if (i<m || rank>m) fprintf(stderr,"rank shouldn't be "O"d!\n",rank);

@ @<Check the labels@>=
for (l=0;l<=q;l++) {
  v=labeled[l];
  if (v>=0 && board(v&0xf,v>>4)!=l)
    fprintf(stderr,"labeled["O"d] not on the board!\n",l);
}
for (i=0;i<rank;i++) for (j=0;j<3;j++) {
  if (board(i,j)>q) fprintf(stderr,"board("O"d,"O"d) out of range!\n",i,j);
  if (board(i,j)>=0 && labeled[board(i,j)]!=(j<<4)+i)
    fprintf(stderr,"label of board("O"d,"O"d) is wrong!\n",i,j);
}

@ @d testedge(i,j,ii,jj) if (board(i,j)>=0 && board(ii,jj)>=0)
    if (t--,!placed[abs(board(i,j)-board(ii,jj))])
      fprintf(stderr,"edge from ("O"d,"O"d) to ("O"d,"O"d) not placed!\n",
                         i,j,ii,jj);

@<Check the placements@>=
for (t=0,l=1;l<=q;l++) t+=placed[l];
for (i=0;i<rank;i++) {
  testedge(i,0,i,1);
  testedge(i,1,i,2);
  for (j=i+1;j<rank;j++) {
    testedge(i,0,j,0);
    testedge(i,1,j,1);
    testedge(i,2,j,2);
  }
}
if (t) fprintf(stderr,"placement count off by "O"d!\n",t);

@ At level |l| of the backtrack procedure I try to place the edge
whose difference is |q-l|, if that edge hasn't already been placed.

Initially there are four symmetries in addition to the $m!$ permutations
of the rows of the board: We can interchange the left and right cliques;
that's called reflection. We can also complement each label, replacing
|l| by |q-l|.

I've set up the levels near the root so that complementation
symmetry is avoided.

Reflection symmetry will disappear as soon as
|leftknown| becomes nonzero. (After that happens, the board implicitly has
$(m-|rank|)!$ symmetries.)

@<Backtrack through all solutions@>=
enter: nodes++;
  if (mems>=thresh) {
    thresh+=delta;
    print_progress(l);
  }
  if (sanity_checking) sanity();
  if (l<=1) @<Make special moves near the root@>;
  if (l>=maxl) {
    maxl=l;
    if (l==q) @<Report a solution and |goto backup|@>;
  }
  if (o,placed[q-l]) @<Record the null move and |goto ready|@>;
  for (t=a=0,b=q-l;b<=q;a++,b++)
    @<Record all possible $(a,b)$ moves in the array |move[l]|@>;
ready: deg[l]=t; /* no |mems| counted for diagnostics */
  if (!t) leaves++;
tryit:@+if (t==0) goto backup;
advance:@+if (vbose) {
    fprintf(stderr,"L"O"d: ",l);
    print_move(move[l][t-1]);
    fprintf(stderr," ("O"d of "O"d)\n",deg[l]-t+1,deg[l]);
  }
  o,x[l]=--t;
  o,mv=move[l][t];
  @<Make |mv|@>;
  if (trouble) {
    if (vbose) fprintf(stderr," -- was bad\n");
    goto unmake;
  }
  l++;
  goto enter;
backup:@+if (--l>=0) {
  o,t=x[l];
unmake: o,mv=move[l][t];
  @<Unmake |mv|@>;
  goto tryit;
}

@ @<Report a solution and |goto backup|@>=
{
  count++;
  printf(""O"d:\n",count);
  for (i=0;i<m;i++)
    printf(""O"3d"O"3d"O"3d\n",board(i,0),board(i,1),board(i,2));
  goto backup;
}

@ @<Sub...@>=
void print_progress(int level) {
  register int l,k,d,c,p;
  register double f,fd;
  fprintf(stderr," after "O"lld mems: "O"d sols,",mems,count);
  for (f=0.0,fd=1.0,l=0;l<level;l++) {
    d=deg[l],k=d-x[l];
    fd*=d,f+=(k-1)/fd; /* choice |l| is |k| of |d| */
    fprintf(stderr," "O"c"O"c",
      k<10? '0'+k: k<36? 'a'+k-10: k<62? 'A'+k-36: '*',
      d<10? '0'+d: d<36? 'a'+d-10: d<62? 'A'+d-36: '*');
  }
  fprintf(stderr," "O".5f\n",f+0.5/fd);
}

@ A ``move'' consists of labeling 0, 1, or 2 vertices and updating
the data structures. A 16-bit packed entry, consisting of
column number (4~bits), row number (4~bits), and label value
(8~bits), specifies what labeling should be done. If two 16-bit
entries are present, the rightmost one is done first.

It turns out that |(row,col,val)| will never be simultaneously zero.
Hence an all-zero move means ``do nothing.''

@d pack(row,col,val) (((col)<<12)+((row)<<8)+(val))

@<Record the null move and |goto ready|@>=
{
  o,move[l][0]=0,t=1,nulls++;
  goto ready;
}

@ @<Sub...@>=
void print_move(int mv) {
  if (!mv)
    fprintf(stderr,"null");
  else if (mv<0x10000)
    fprintf(stderr,""O"d"O"d="O"d",(mv>>8)&0xf,(mv>>12)&0xf,mv&0xff);
  else fprintf(stderr,""O"d"O"d="O"d,"O"d"O"d="O"d",
      (mv>>8)&0xf,(mv>>12)&0xf,mv&0xff,(mv>>24)&0xf,(mv>>28)&0xf,(mv>>16)&0xff);
}
@#
void print_moves(int level) {
  register int i;
  for (i=deg[level]-1;i>=0;i--) { /* we try the moves in decreasing order */
    fprintf(stderr,""O"d:",deg[level]-i);
    print_move(move[level][i]);
    fprintf(stderr,"\n");
  }
}

@ @<Sub...@>=
void print_state(int levels) {
  register int l;
  for (l=0;l<levels;l++) {
    print_move(move[l][x[l]]);
    fprintf(stderr," ("O"d of "O"d)\n",deg[l]-x[l],deg[l]);
  }
}

@ The edge labeled |q| must have endpoints labeled 0 and~|q|. This
can happen in only three essentially different ways: That edge
either belongs to the middle clique, the left clique, or joins
the left and middle cliques. In the latter case, complement
symmetry has been broken. In the former cases, complement symmetry
is avoided by insisting that the edge labeled |q-1| has endpoints
labeled 1 and~|q|.

@<Make special moves near the root@>=
if (l==0) {
  o,move[0][0]=(pack(1,1,0)<<16)+pack(0,1,q);
  o,move[0][1]=(pack(1,0,0)<<16)+pack(0,0,q);
  o,move[0][2]=(pack(0,1,0)<<16)+pack(0,0,q);
  t=3;
  goto ready;
}@+else if (o,x[0]!=2) {
  t=(m==2? 1: 2);
  o,move[1][0]=pack(0,x[0],1);
  if (m>2) o,move[1][1]=pack(2,1-x[0],1);
  goto ready;
}

@ I set |trouble| nonzero if any edge is placed more than once.

@<Make |mv|@>=
for (trouble=0;mv;mv>>=16) {
  val=mv&0xff,row=(mv>>8)&0xf,col=(mv>>12)&0xf;
  o,labeled[val]=(mv>>8)&0xff;
  o,board(row,col)=val;
  oo,colknown[col]++;
  if (col>0) {
    o,v=board(row,col-1);
    if (v>=0) oo,trouble+=placed[abs(val-v)],placed[abs(val-v)]=1;
  }
  if (col<2) {
    o,v=board(row,col+1);
    if (v>=0) oo,trouble+=placed[abs(val-v)],placed[abs(val-v)]=1;
  }
  for (i=0;i<rank;i++) if (i!=row) {
    o,v=board(i,col);
    if (v>=0) oo,trouble+=placed[abs(val-v)],placed[abs(val-v)]=1;
  }
  if (row==rank) rank++;
}

@ @<Unmake |mv|@>=
if (mv>=0x10000) mv=(mv>>16)+((mv&0xffff)<<16); /* undo in opposite order */
for (;mv;mv>>=16) {
  val=mv&0xff,row=(mv>>8)&0xf,col=(mv>>12)&0xf;
  if (row==rank-1 && (o,board(row,(col+1)mod 3)<0) &&
                     (o,board(row,(col+2)mod 3)<0))
    rank=row;
  o,labeled[val]=-1;
  o,board(row,col)=-1;
  oo,colknown[col]--;
  if (col>0) {
    o,v=board(row,col-1);
    if (v>=0) o,placed[abs(val-v)]=0;
  }
  if (col<2) {
    o,v=board(row,col+1);
    if (v>=0) o,placed[abs(val-v)]=0;
  }
  for (i=0;i<rank;i++) if (i!=row) {
    o,v=board(i,col);
    if (v>=0) o,placed[abs(val-v)]=0;
  }
}

@*The nitty gritty. OK, I've put all the infrastructure into place.
It remains to figure out all legal ways to place a new edge
whose endpoints are labeled |a|~and~|b|.
(This is where the graph $K_m\sqprod P_3$ is really ``hardwired.'')

I do this by brute force, while trying to be careful. Sometimes I just
barely avoided a bug, but I hope that I've exterminated them all.

@<Record all possible $(a,b)$ moves in the array |move[l]|@>=
{
  oo,aa=labeled[a],bb=labeled[b];
  if (aa>=0) {
    if (bb>=0) continue; /* |a| and |b| are already on the |board| */
    row=aa&0xf, col=aa>>4;
    @<Record all legal placements of |b| adjacent to |a|@>;
  }@+else if (bb>=0) {
    row=bb&0xf, col=bb>>4;
    @<Record all legal placements of |a| adjacent to |b|@>;
  }
  else @<Record all adjacent placements of |a| and |b|@>;
}

@ @<Record all legal placements of |b| adjacent to |a|@>=
switch (col) {
case 0:@+if ((o,board(row,1)<0) && legal_in_col(b,1) &&
            ((o,board(row,2)<0) || (o,!placed[abs(b-board(row,2))])))
    o,move[l][t++]=pack(row,1,b);
  break;
case 1:@+if ((o,board(row,0)<0) && legal_in_col(b,0))
    o,move[l][t++]=pack(row,0,b);
  if ((o,leftknown) && (o,board(row,2)<0) && legal_in_col(b,2))
    o,move[l][t++]=pack(row,2,b);
  break;
case 2:@+if ((o,board(row,1)<0) && legal_in_col(b,1) &&
            ((o,board(row,0)<0) || (o,!placed[abs(b-board(row,0))])))
    o,move[l][t++]=pack(row,1,b);
  break;
}
if (legal_in_col(b,col)) {
  for (i=0;i<rank;i++) if (o,board(i,col)<0) {
    if (col>0 && (o,board(i,col-1)>=0) &&
         (o,placed[abs(b-board(i,col-1))])) continue;
    if (col<2 && (o,board(i,col+1)>=0) &&
         (o,placed[abs(b-board(i,col+1))])) continue;
    o,move[l][t++]=pack(i,col,b);
  }
  if (rank<m) o,move[l][t++]=pack(rank,col,b);
}

@ @<Sub...@>=
int legal_in_col(val,col) {
  register int i,v;
  if (o,colknown[col]==m) return 0;
  for (i=0;i<rank;i++) {
    o,v=board(i,col);
    if (v>=0 && (o,placed[abs(v-val)])) return 0;
  }
  return 1;
}

@ @<Record all legal placements of |a| adjacent to |b|@>=
switch (col) {
case 0:@+if ((o,board(row,1)<0) && legal_in_col(a,1) &&
            ((o,board(row,2)<0) || (o,!placed[abs(a-board(row,2))])))
    o,move[l][t++]=pack(row,1,a);
  break;
case 1:@+if ((o,board(row,0)<0) && legal_in_col(a,0))
    o,move[l][t++]=pack(row,0,a);
  if ((o,leftknown) && (o,board(row,2)<0) && legal_in_col(a,2))
    o,move[l][t++]=pack(row,2,a);
  break;
case 2:@+if ((o,board(row,1)<0) && legal_in_col(a,1) &&
            ((o,board(row,0)<0) || (o,!placed[abs(a-board(row,0))])))
    o,move[l][t++]=pack(row,1,a);
  break;
}
if (legal_in_col(a,col)) {
  for (i=0;i<rank;i++) if (o,board(i,col)<0) {
    if (col>0 && (o,board(i,col-1)>=0) &&
         (o,placed[abs(a-board(i,col-1))])) continue;
    if (col<2 && (o,board(i,col+1)>=0) &&
         (o,placed[abs(a-board(i,col+1))])) continue;
    o,move[l][t++]=pack(i,col,a);
  }
  if (rank<m) o,move[l][t++]=pack(rank,col,a);
}

@ Finally, the hard case is when a double move is needed.
First I tentatively try all placements of~|a|, actually changing the board.
Then I record the double moves for |b| adjacent to every such placement.
Of course the board has to be restored again.

@<Record all adjacent placements of |a| and |b|@>=
for (o,ccol=(leftknown? 2: 1);ccol>=0;ccol--) if (legal_in_col(a,ccol)) {
  for (ii=0;ii<rank;ii++) if (o,board(ii,ccol)<0) {
    if (ccol>0 && (o,board(ii,ccol-1)>=0) &&
         (o,placed[abs(a-board(ii,ccol-1))])) continue;
    if (ccol<2 && (o,board(ii,ccol+1)>=0) &&
         (o,placed[abs(a-board(ii,ccol+1))])) continue;
    aa=mv=pack(ii,ccol,a);@+@<Make |mv|@>;@+mv=aa;
    if (!trouble) @<Record all double placements of |b| adjacent to |a|@>;
    @<Unmake |mv|@>;
  }
  if (rank<m) {
    aa=mv=pack(rank,ccol,a);@+@<Make |mv|@>;@+mv=aa;
    if (!trouble) @<Record all double placements of |b| adjacent to |a|@>;
    @<Unmake |mv|@>;
  }
}

@ @<Record all double placements of |b| adjacent to |a|@>=
{
  switch (col) {
  case 0:@+if ((o,board(row,1)<0) && legal_in_col(b,1) &&
              ((o,board(row,2)<0) || (o,!placed[abs(b-board(row,2))])))
      o,move[l][t++]=(pack(row,1,b)<<16)+mv;
    break;
  case 1:@+if ((o,board(row,0)<0) && legal_in_col(b,0))
      o,move[l][t++]=(pack(row,0,b)<<16)+mv;
    if ((o,leftknown) && (o,board(row,2)<0) && legal_in_col(b,2))
      o,move[l][t++]=(pack(row,2,b)<<16)+mv;
    break;
  case 2:@+if ((o,board(row,1)<0) && legal_in_col(b,1) &&
              ((o,board(row,0)<0) || (o,!placed[abs(b-board(row,0))])))
      o,move[l][t++]=(pack(row,1,b)<<16)+mv;
    break;
  }
  if (legal_in_col(b,col)) {
    for (i=0;i<rank;i++) if (o,board(i,col)<0) {
      if (col>0 && (o,board(i,col-1)>=0) &&
           (o,placed[abs(b-board(i,col-1))])) continue;
      if (col<2 && (o,board(i,col+1)>=0) &&
           (o,placed[abs(b-board(i,col+1))])) continue;
      o,move[l][t++]=(pack(i,col,b)<<16)+mv;
    }
    if (rank<m) o,move[l][t++]=(pack(rank,col,b)<<16)+mv;
  }
}

@*Index.