#\datethis
#
#@*Intro. This is a modified version of Donald Knuth's program,
#to calculate for values of m,n larger than 16. 
#
#This program finds all of the nonisomorphic graceful labelings
#of the graph $K_{m,n}$. I based it on
#{\mc BACK-GRACEFUL-KMP2}, which handles a more difficult problem.
#
#The graph $K_{m,n}$ is ``hardwired'' into the logic of this program.
#It has $q=mn$ edges.
#Although I won't be able to let $m$ and $n$ get extremely large,
#I'm hoping to see a pattern from which some conjectures will follow.
#(Even some theorems, if I'm lucky.)
#
#Please excuse me for writing this in a rush. A lot of the code has inherited
#awkwardness from its history.
#
#@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 maxmn 32 /* |m| and |n| mustn't exceed this (5-bit limit for |row|) */
#@d maxq 512 /* $mn$ mustn't exceed this (9-bit |val| in |mv| codes) */
#@d board(i,j) brd[(i)+m*(j)]
#
#@c
##include <stdio.h>
##include <stdlib.h>
#int m,n; /* command-line parameters */
#int q; /* the number of edges ($mn$) */
#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 leaves; /* nodes that have no descendants */
#int count; /* number of solutions found so far */
#int nonalf; /* number of non-alpha solutions found so far */
#int brd[maxmn+maxmn]; /* one-dimensional array accessed via the |board| macro */
#int rankl,rankr; /* how many rows of each part are active? */
#int labeled[maxq+1]; /* what row and column, if any, have a particular label? */
#int placed[maxq+1]; /* has this edge been placed? */
#unsigned long long move[maxq][1024]; /* feasible moves at each level */
#int deg[maxq]; /* number of choices at each level; used in printouts only */
#int x[maxq]; /* indexes of moves made at each level */
#int vbose=0; /* can set this nonzero when debugging */
#int left[maxmn],right[maxmn]; /* sorted solution */
#@<Subroutines@>@;
#main (int argc,char*argv[]) {
#  register int a,b,i,j,k,l,t,v,aa,bb,ii,row,col,ccol,val,trouble;
#	unsigned long long mv;
#  @<Process the command line@>;
#  @<Initialize the data structures@>;
#  @<Backtrack through all solutions@>;
#  fprintf(stderr,
#  "Altogether "O"d solution"O"s ("O"d alpha),",
#                count,count==1?"":"s",count-nonalf);
#  fprintf(stderr," "O"lld mems, "O"lld nodes, "O"lld leaves.\n",
#                mems, nodes, leaves);
#}
#
#@ @<Process the command line@>=
#if (argc!=3 || sscanf(argv[1],""O"d",&m)!=1 ||
#               sscanf(argv[2],""O"d",&n)!=1) {
#  fprintf(stderr,"Usage: "O"s m n\n",argv[0]);
#  exit(-1);
#}
#if (m>maxmn || n>maxmn) {
#  fprintf(stderr,"Sorry, m and n mustn't exceed "O"d!\n",maxmn);
#  exit(-2);
#}
#if (m<2 || n<2) {
#  fprintf(stderr,"Sorry, m and n should be 2 or more!\n");
#  exit(-4);
#}
#q=m*n;
#if (q>maxq) {
#  fprintf(stderr,"Sorry, mn mustn't exceed "O"d!\n",maxq);
#  exit(-3);
#}
#printf("--- Graceful labelings of K_{"O"d,"O"d} ---\n",m,n);
#
#@ The current status of the vertices labeled so far appears in
#the |board|, which has two columns, one for each part. This is not a
#canonical representation: The entries of each column can appear in any order.
#The first |rankl| rows of the first column, and the first
#|rankr| rows of the second column, have been labeled.
#When the vertex
#in row~$i$ and column~$j$ receives label~$l$, |labeled[l]| records
#the value |(j<<5)+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.
#
#
#@<Initialize the data structures@>=
#rankl=rankr=0;
#for (l=0;l<=q;l++) labeled[l]=-1;
#l=0;
#
#@ @<Sub...@>=
#void print_board(void) {
#  register int i,j;
#  for (i=0;i<rankl;i++) fprintf(stderr,""O"3d",board(i,0));
#  fprintf(stderr,"  | ");
#  for (j=0;j<rankr;j++) fprintf(stderr,""O"3d",board(j,1));
#  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 a bit fancy, so I'd better check that
#they're self-consistent.
#
#@d sanity_checking 1 /* set this to 1 if you suspect a bug */
#
#@<Sub...@>=
#void sanity(void) {
#  register int i,j,l,t,v;
#  @<Check the ranks@>;
#  @<Check the labels@>;
#  @<Check the placements@>;
#}
#
#@ @<Check the ranks@>=
#if (rankl>m) fprintf(stderr,"rankl shouldn't be "O"d!\n",rankl);
#if (rankr>n) fprintf(stderr,"rankr shouldn't be "O"d!\n",rankr);
#
#@ @<Check the labels@>=
#for (l=0;l<=q;l++) {
#  v=labeled[l];
#  if (v>=0 && board(v&0x1f,v>>5)!=l)
#    fprintf(stderr,"labeled["O"d] not on the board!\n",l);
#}
#for (i=0;i<rankl;i++) {
#  if (board(i,0)>q) fprintf(stderr,"board("O"d,"O"d) out of range!\n",i,0);
#  if (labeled[board(i,0)]!=i)
#    fprintf(stderr,"label of board("O"d,"O"d) is wrong!\n",i,0);
#}
#for (j=0;j<rankr;j++) {
#  if (board(j,1)>q) fprintf(stderr,"board("O"d,"O"d) out of range!\n",j,1);
#  if (labeled[board(j,1)]!=(1<<5)+j)
#    fprintf(stderr,"label of board("O"d,"O"d) is wrong!\n",j,1);
#}
#
#@ @d testedge(i,j) if (!placed[abs(board(i,0)-board(j,1))])
#      fprintf(stderr,"edge from "O"d to "O"d not placed!\n",
#                         i,j);
#
#@<Check the placements@>=
#for (t=0,l=1;l<=q;l++) t+=placed[l];
#if (t!=rankl*rankr)
#  fprintf(stderr,"placement count is off ("O"d not "O"d)",t,rankl*rankr);
#for (i=0;i<rankl;i++) for (j=0;j<rankr;j++)
#  testedge(i,j);
#
#@ 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 two or four symmetries in addition to the $m!\,n!$
#permutations of the two parts of the board:
#We can complement each label, replacing |l| by |q-l|.
#We can also interchange the left and right parts, if $m=n$;
#that's called reflection. 
#
#I've set up the levels near the root so that the label~0 always goes
#into the right part. If $m\ne n$, that avoids complementation.
#If $m=n$, it avoids reflection; and complementation is avoided
#by also forcing~1 into the right part when $m=n$.
#
#@<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==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) 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: ",count);
#  @<Analyze and print the solution@>;
#  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 19-bit packed entry, consisting of
#column number (5~bits), row number (5~bits), and label value
#(9~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)<<14)+((row)<<9)+(val))
#
#@<Record the null move and |goto ready|@>=
#{
#  o,move[l][0]=0,t=1;
#  goto ready;
#}
#
#@ @<Sub...@>=
#void print_move(unsigned long long mv) {
#  if (!mv)
#    fprintf(stderr,"null");
#  else if (mv<0x80000)
#    fprintf(stderr,""O"d"O"d="O"d",(mv>>9)&0x1f,(mv>>14)&0x1f,mv&0x1ff);
#  else fprintf(stderr,""O"d"O"d="O"d,"O"d"O"d="O"d",
#      (mv>>9)&0x1f,(mv>>14)&0x1f,mv&0x1ff,(mv>>28)&0x1f,(mv>>33)&0x1f,(mv>>19)&0x1ff);
#}
#@#
#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|.
#
#@<Make special moves near the root@>=
#{
#  if (l==0) o,move[0][0]=(pack(0,0,q)<<19)+pack(0,1,0),t=1;
#  else if (m==n) o,move[1][0]=pack(1,1,1),t=1;
#  else {
#    o,move[1][0]=pack(1,1,1);
#    o,move[1][1]=pack(1,0,q-1);
#    t=2;
#  }
#  goto ready;
#}
#
#@ I set |trouble| nonzero if any edge is placed more than once.
#
#@<Make |mv|@>=
#for (trouble=0;mv;mv>>=19) {
#  val=mv&0x1ff,row=(mv>>9)&0x1f,col=(mv>>14)&0x1f;
#  o,labeled[val]=(mv>>9)&0x3ff;
#  o,board(row,col)=val;
#  if (col==0) {
#    if (row!=rankl) confusion("left move");
#    rankl++;
#    for (j=0;j<rankr;j++) {
#      o,v=board(j,1);
#      oo,trouble+=placed[abs(val-v)],placed[abs(val-v)]=1;
#    }
#  }@+else {
#    if (row!=rankr) confusion("right move");
#    rankr++;
#    for (i=0;i<rankl;i++) {
#      o,v=board(i,0);
#      oo,trouble+=placed[abs(val-v)],placed[abs(val-v)]=1;
#    }
#  }
#}
#
#@ @<Sub...@>=
#void confusion(char *message) {
#  fprintf(stderr,""O"s!\n",message);
#}
#
#@ @<Unmake |mv|@>=
#if (mv>=0x80000)
#{
# mv=(mv>>19)+((mv&0x7ffff)<<19); /* undo in opposite order */
#}
#for (;mv;mv>>=19) {
#  val=mv&0x1ff,row=(mv>>9)&0x1f,col=(mv>>14)&0x1f;
#  o,labeled[val]=-1;
#  if (col==0) {
#    rankl--;
#    if (row!=rankl) confusion("left unmove");
#    for (j=0;j<rankr;j++) {
#      o,v=board(j,1);
#      o,placed[abs(val-v)]=0;
#    }
#  }@+else {
#    rankr--;
#    if (row!=rankr) confusion("right unmove");
#    for (i=0;i<rankl;i++) {
#      o,v=board(i,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 $K_{m,n}$ 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&0x1f, col=aa>>5;
#    @<Record all legal placements of |b| adjacent to |a|@>;
#  }@+else if (bb>=0) {
#    row=bb&0x1f, col=bb>>5;
#    @<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|@>=
#if (col==0 && right_legal(b)) o,move[l][t++]=pack(rankr,1,b);
#else if (col==1 && left_legal(b)) o,move[l][t++]=pack(rankl,0,b);
#
#@ @<Sub...@>=
#int right_legal(val) {
#  register int i,v;
#  if (rankr==n) return 0;
#  for (i=0;i<rankl;i++) {
#    o,v=board(i,0);
#    if (o,placed[abs(v-val)]) return 0;
#  }
#  return 1;
#}
#
#@ @<Sub...@>=
#int left_legal(val) {
#  register int j,v;
#  if (rankl==m) return 0;
#  for (j=0;j<rankr;j++) {
#    o,v=board(j,1);
#    if (o,placed[abs(v-val)]) return 0;
#  }
#  return 1;
#}
#
#@ @<Record all legal placements of |a| adjacent to |b|@>=
#if (col==0 && right_legal(a)) o,move[l][t++]=pack(rankr,1,a);
#else if (col==1 && left_legal(a)) o,move[l][t++]=pack(rankl,0,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|@>=
#if (left_legal(a)) {
#  aa=mv=pack(rankl,0,a);@+@<Make |mv|@>;@+mv=aa;
#  if (!trouble) @<Record all double placements of |b| adjacent to |a|@>;
#  @<Unmake |mv|@>;
#}
#if (right_legal(a)) {
#  aa=mv=pack(rankr,1,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|@>=
#{
#  if (col==0 && right_legal(b)) o,move[l][t++]=((unsigned long long)(pack(rankr,1,b))<<19)+mv;
#  else if (col==1 && left_legal(b)) o,move[l][t++]=((unsigned long long)(pack(rankl,0,b))<<19)+mv;
#}
#
#@*Looking for patterns. Most of the solutions seem to be derivable in
#a simple way from smaller solutions.
#
#Namely, suppose the left part contains $mn=a_1>\ldots>a_m>0$
#and the right part contains $0=b_1<\ldots<b_n$. It's an ``alpha solution''
#if $b_n<a_m$. (Such bipartite labelings
#were considered by Rosa in his original paper on graceful labelings.)
#Then, if $t$ is any integer $>1$, we can obtain two larger solutions:
#
#\item{1)} Replace each $a_i$ by $ta_i$ and each $b_j$ by
#$tb_j$, $tb_j+1$, \dots, $tb_j+t-1$. This is a graceful alpha solution
#for $K_{m,tn}$.
#
#\item{2)} Replace each $b_j$ by $tb_j$ and each
#$a_i$ by $ta_i$, $ta_i-1$, \dots, $ta_i-t+1$. This is graceful alpha
#solution for $K_{tm,n}$.
#
#I conjecture that all alpha solutions are obtained in this way.
#Equivalently, all alpha solutions can be reduced to a smaller
#alpha solution by inverting (1) or~(2).
#
#The following program marks a non-alpha solution with `\.\#'.
#An alpha solution obtained from a smaller one by~(1) or~(1) is marked
#`\.{L$t$}' or `\.{R$t$}', respectively. An alpha solution not obtained
#by either (1) or~(2) is marked `\.{XXX}'.
#
#@<Analyze and print the solution@>=
#@<Sort the solution into |left| and |right|@>;
#for (i=0;i<m;i++) printf(""O"3d",left[i]);
#printf("  | ");
#for (j=0;j<n;j++) printf(""O"3d",right[j]);
#if (left[m-1]<right[n-1]) nonalf++,printf(" #\n");
#else @<Analyze an alpha solution@>;
#
#@ @<Sort the solution into |left| and |right|@>=
#for (i=0;i<m;i++) {
#  a=board(i,0);
#  for (j=i;j>0;j--) {
#    if (left[j-1]>a) break;  
#    left[j]=left[j-1];
#  }
#  left[j]=a;
#}
#for (j=0;j<n;j++) {
#  b=board(j,1);
#  for (i=j;i>0;i--) {
#    if (right[i-1]<b) break;
#    right[i]=right[i-1];
#  }
#  right[i]=b;
#}
#
#@ @<Analyze an alpha solution@>=
#{
#  for (t=1;;t++) if (t==m || left[t]!=q-t) break;
#  if (t>1) {
#    if (m@q)@>%t) goto irreducible;
#    for (j=0;j<n;j++) if (right[j]@q)@>%t) goto irreducible;
#    for (v=t;v<m;v+=t) {
#      if (left[v]@q)@>%t) goto irreducible;
#      for (i=1;i<t;i++) if (left[v+i]!=left[v]-i) goto irreducible;
#    }
#    printf(" R"O"d\n",t);
#    goto done;
#  }
#  for (t=1;;t++) if (t==n || right[t]!=t) break;
#  if (t>1) {
#    if (n@q)@>%t) goto irreducible;
#    for (i=0;i<m;i++) if (left[i]@q)@>%t) goto irreducible;
#    for (v=t;v<n;v+=t) {
#      if (right[v]@q)@>%t) goto irreducible;
#      for (i=1;i<t;i++) if (right[v+i]!=right[v]+i) goto irreducible;
#    }
#    printf(" L"O"d\n",t);
#    goto done;
#  }
#irreducible: printf(" XXX\n");
#done:;
#}
#
#@*Index.
1 1
2 1
3 1
4 1
5 1
6 1
7 1
8 2
9 2
10 1
11 1
12 5
13 1
14 5
15 1
16 1
17 2
18 4
19 4
20 2
21 1
22 1
23 4
24 2
25 4
26 2
27 4
28 1
29 1
30 8
31 4
32 4
33 4
34 4
35 8
36 1
37 1
38 4
39 3
40 10
41 1
42 10
43 3
44 4
45 1
46 1
47 3
48 4
49 5
50 4
51 4
52 5
53 4
54 3
55 1
56 1
57 14
58 3
59 11
60 3
61 7
62 3
63 11
64 3
65 14
66 1
67 1
68 2
69 5
70 7
71 4
72 5
73 5
74 4
75 7
76 5
77 2
78 1
79 1
80 10
81 2
82 11
83 3
84 16
85 2
86 16
87 3
88 11
89 2
90 10
91 1
92 1
93 20
94 6
95 4
96 5
97 9
98 5
99 5
100 9
101 5
102 4
103 6
104 20
105 1
106 1
107 4
108 3
109 19
110 2
111 15
112 4
113 10
114 4
115 15
116 2
117 19
118 3
119 4
120 1
121 1
122 4
123 4
124 5
125 6
126 4
127 6
128 10
129 10
130 6
131 4
132 6
133 5
134 4
135 4
136 1
137 1
138 39
139 4
140 10
141 3
142 30
143 3
144 17
145 3
146 17
147 3
148 30
149 3
150 10
151 4
152 39
153 1
154 1
155 14
156 6
157 10
158 4
159 5
160 7
161 4
162 10
163 10
164 4
165 7
166 5
167 4
168 10
169 6
170 14
171 1
172 1
173 6
174 2
175 17
176 4
177 14
178 4
179 40
180 3
181 8
182 3
183 40
184 4
185 14
186 4
187 17
188 2
189 6
190 1
191 1
192 68
193 6
194 4
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