\datethis
@*Intro. This program is part of a series of ``SAT-solvers'' that I'm putting
together for my own education as I prepare to write Section 7.2.2.2 of
{\sl The Art of Computer Programming}. My intent is to have a variety of
compatible programs on which I can run experiments to learn how different
approaches work in practice.

Indeed, this is the first of the series --- more precisely the zero-th. I've
tried to write it as a primitive baseline against which I'll be able to measure
various technical improvements that have been discovered in recent years.
This version represents what I think I would have written in the 1960s,
when I knew how to do basic backtracking with classical data structures
(but very little else). I have intentionally written it {\it before\/} having
read {\it any\/} of the literature about modern SAT-solving techniques;
in other words I'm starting with a personal ``tabula rasa.''
My plan is to write new versions as I read the literature, in more-or-less
historical order. The only thing that currently distinguishes me from a
programmer of forty years ago, SAT-solving-wise, is the knowledge that better
methods almost surely do exist.

[{\it Note:}\enspace Actually this is a special version, written
at the end of October 2012. It strips down the old data structures
and uses watched literals instead. I think it represents a nearly
minimal decent {\mc SAT} solver. Algorithm 7.2.2.2B is based on this code.]

Although this is the zero-level program, I'm taking care to adopt conventions
for input and output that will be essentially the same in all of the
fancier versions that are to come.

The input on |stdin| is a series of lines with one clause per line. Each
clause is a sequence of literals separated by spaces. Each literal is
a sequence of one to eight ASCII characters between \.{!} and \.{\}},
inclusive, not beginning with \.{\~},
optionally preceded by \.{\~} (which makes the literal ``negative'').
For example, Rivest's famous clauses on four variables,
found in 6.5--(13) and 7.1.1--(32) of {\sl TAOCP}, can be represented by the
following eight lines of input: 
$$\chardef~=`\~
\vcenter{\halign{\tt#\cr
x2 x3 ~x4\cr
x1 x3 x4\cr
~x1 x2 x4\cr
~x1 ~x2 x3\cr
~x2 ~x3 x4\cr
~x1 ~x3 ~x4\cr
x1 ~x2 ~x4\cr
x1 x2 ~x3\cr}}$$
Input lines that begin with \.{\~\ } are ignored (treated as comments).
The output will be `\.{\~}' if the input clauses are unsatisfiable.
Otherwise it will be a list of noncontradictory literals that cover each
clause, separated by spaces. (``Noncontradictory'' means that we don't
have both a literal and its negation.) The input above would, for example,
yield `\.{\~}'; but if the final clause were omitted, the output would
be `\.{\~x1} \.{\~x2} \.{x3}', in some order, possibly together
with either \.{x4} or \.{\~x4} (but not both). No attempt is made to
find all solutions; at most one solution is given.

The running time in ``mems'' is also reported, together with the approximate
number of bytes needed for data storage. One ``mem'' essentially means a
memory access to a 64-bit word.
(These totals don't include the time or space needed to parse the
input or to format the output.)

@ So here's the structure of the program. (Skip ahead if you are
impatient to see the interesting stuff.)

@d o mems++ /* count one mem */
@d oo mems+=2 /* count two mems */
@d ooo mems+=3 /* count three mems */

@c
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include "gb_flip.h"
typedef unsigned int uint; /* a convenient abbreviation */
typedef unsigned long long ullng; /* ditto */
@<Type definitions@>;
@<Global variables@>;
@<Subroutines@>;
main (int argc, char *argv[]) {
  register uint c,h,i,j,k,l,p,q,r,level,parity;
  @<Process the command line@>;
  @<Initialize everything@>;
  @<Input the clauses@>;
  if (verbose&show_basics)
    @<Report the successful completion of the input phase@>;
  @<Set up the main data structures@>;
  imems=mems, mems=0;
  @<Solve the problem@>;
done:@+if (verbose&show_basics)
    fprintf(stderr,"Altogether %llu+%llu mems, %llu bytes, %llu nodes.\n",
                     imems,mems,bytes,nodes);
}

@ @d show_basics 1 /* |verbose| code for basic stats */
@d show_choices 2 /* |verbose| code for backtrack logging */
@d show_details 4 /* |verbose| code for further commentary */

@<Glob...@>=
int random_seed=0; /* seed for the random words of |gb_rand| */
int verbose=show_basics; /* level of verbosity */
int show_choices_max=1000000; /* above this level, |show_choices| is ignored */
int hbits=8; /* logarithm of the number of the hash lists */
int buf_size=1024; /* must exceed the length of the longest input line */
ullng imems,mems; /* mem counts */
ullng bytes; /* memory used by main data structures */
ullng nodes; /* total number of branch nodes initiated */
ullng thresh=0; /* report when |mems| exceeds this, if |delta!=0| */
ullng delta=0; /* report every |delta| or so mems */
ullng timeout=0x1fffffffffffffff; /* give up after this many mems */

@ On the command line one can say
\smallskip
\item{$\bullet$}
`\.v$\langle\,$integer$\,\rangle$' to enable various levels of verbose
 output on |stderr|;
\item{$\bullet$}
`\.c$\langle\,$positive integer$\,\rangle$' to limit the levels on which
clauses are shown;
\item{$\bullet$}
`\.h$\langle\,$positive integer$\,\rangle$' to adjust the hash table size;
\item{$\bullet$}
`\.b$\langle\,$positive integer$\,\rangle$' to adjust the size of the input
buffer;
\item{$\bullet$}
`\.s$\langle\,$integer$\,\rangle$' to define the seed for any random numbers
that are used; and/or
\item{$\bullet$}
`\.d$\langle\,$integer$\,\rangle$' to set |delta| for periodic state reports.
\item{$\bullet$}
`\.T$\langle\,$integer$\,\rangle$' to set |timeout|: This program will
abruptly terminate, when it discovers that |mems>timeout|.

@<Process the command line@>=
for (j=argc-1,k=0;j;j--) switch (argv[j][0]) {
case 'v': k|=(sscanf(argv[j]+1,"%d",&verbose)-1);@+break;
case 'c': k|=(sscanf(argv[j]+1,"%d",&show_choices_max)-1);@+break;
case 'h': k|=(sscanf(argv[j]+1,"%d",&hbits)-1);@+break;
case 'b': k|=(sscanf(argv[j]+1,"%d",&buf_size)-1);@+break;
case 's': k|=(sscanf(argv[j]+1,"%d",&random_seed)-1);@+break;
case 'd': k|=(sscanf(argv[j]+1,"%lld",&delta)-1);@+thresh=delta;@+break;
case 'T': k|=(sscanf(argv[j]+1,"%lld",&timeout)-1);@+break;
default: k=1; /* unrecognized command-line option */
}
if (k || hbits<0 || hbits>30 || buf_size<=0) {
  fprintf(stderr,
    "Usage: %s [v<n>] [c<n>] [h<n>] [b<n>] [s<n>] [d<n>] [T<n>] < foo.sat\n",argv[0]);
  exit(-1);
}

@*The I/O wrapper. The following routines read the input and absorb it into
temporary data areas from which all of the ``real'' data structures
can readily be initialized. My intent is to incorporate these routines in all
of the SAT-solvers in this series. Therefore I've tried to make the code
short and simple, yet versatile enough so that almost no restrictions are
placed on the sizes of problems that can be handled. These routines are
supposed to work properly unless there are more than
$2^{32}-1=4$,294,967,295 occurrences of literals in clauses,
or more than $2^{31}-1=2$,147,483,647 variables or clauses.

In these temporary tables, each variable is represented by four things:
its unique name; its serial number; the clause number (if any) in which it has
most recently appeared; and a pointer to the previous variable (if any)
with the same hash address. Several variables at a time
are represented sequentially in small chunks of memory called ``vchunks,''
which are allocated as needed (and freed later).

@d vars_per_vchunk 341 /* preferably $(2^k-1)/3$ for some $k$ */

@<Type...@>=
typedef union {
  char ch8[8];
  uint u2[2];
  long long lng;
} octa;
typedef struct tmp_var_struct {
  octa name; /* the name (one to eight ASCII characters) */
  uint serial; /* 0 for the first variable, 1 for the second, etc. */
  int stamp; /* |m| if positively in clause |m|; |-m| if negatively there */
  struct tmp_var_struct *next; /* pointer for hash list */
} tmp_var;
@#
typedef struct vchunk_struct {
  struct vchunk_struct *prev; /* previous chunk allocated (if any) */
  tmp_var var[vars_per_vchunk];
} vchunk;

@ Each clause in the temporary tables is represented by a sequence of
one or more pointers to the |tmp_var| nodes of the literals involved.
A negated literal is indicated by adding~1 to such a pointer.
The first literal of a clause is indicated by adding~2.
Several of these pointers are represented sequentially in chunks
of memory, which are allocated as needed and freed later.

@d cells_per_chunk 511 /* preferably $2^k-1$ for some $k$ */

@<Type...@>=
typedef struct chunk_struct {
  struct chunk_struct *prev; /* previous chunk allocated (if any) */
  tmp_var *cell[cells_per_chunk];
} chunk;

@ @<Glob...@>=
char *buf; /* buffer for reading the lines (clauses) of |stdin| */
tmp_var **hash; /* heads of the hash lists */
uint hash_bits[93][8]; /* random bits for universal hash function */
vchunk *cur_vchunk; /* the vchunk currently being filled */
tmp_var *cur_tmp_var; /* current place to create new |tmp_var| entries */
tmp_var *bad_tmp_var; /* the |cur_tmp_var| when we need a new |vchunk| */
chunk *cur_chunk; /* the chunk currently being filled */
tmp_var **cur_cell; /* current place to create new elements of a clause */
tmp_var **bad_cell; /* the |cur_cell| when we need a new |chunk| */
ullng vars; /* how many distinct variables have we seen? */
ullng clauses; /* how many clauses have we seen? */
ullng nullclauses; /* how many of them were null? */
ullng cells; /* how many occurrences of literals in clauses? */

@ @<Initialize everything@>=
gb_init_rand(random_seed);
buf=(char*)malloc(buf_size*sizeof(char));
if (!buf) {
  fprintf(stderr,"Couldn't allocate the input buffer (buf_size=%d)!\n",
            buf_size);
  exit(-2);
}
hash=(tmp_var**)malloc(sizeof(tmp_var)<<hbits);
if (!hash) {
  fprintf(stderr,"Couldn't allocate %d hash list heads (hbits=%d)!\n",
           1<<hbits,hbits);
  exit(-3);
}
for (h=0;h<1<<hbits;h++) hash[h]=NULL;

@ The hash address of each variable name has $h$ bits, where $h$ is the
value of the adjustable parameter |hbits|.
Thus the average number of variables per hash list is $n/2^h$ when there
are $n$ different variables. A warning is printed if this average number
exceeds 10. (For example, if $h$ has its default value, 8, the program will
suggest that you might want to increase $h$ if your input has 2560
different variables or more.)

All the hashing takes place at the very beginning,
and the hash tables are actually recycled before any SAT-solving takes place;
therefore the setting of this parameter is by no means crucial. But I didn't
want to bother with fancy coding that would determine $h$ automatically.

@<Input the clauses@>=
while (1) {
  if (!fgets(buf,buf_size,stdin)) break;
  clauses++;
  if (buf[strlen(buf)-1]!='\n') {
    fprintf(stderr,
      "The clause on line %lld (%.20s...) is too long for me;\n",clauses,buf);
    fprintf(stderr," my buf_size is only %d!\n",buf_size);
    fprintf(stderr,"Please use the command-line option b<newsize>.\n");
    exit(-4);
  }
  @<Input the clause in |buf|@>;
}
if ((vars>>hbits)>=10) {
  fprintf(stderr,"There are %lld variables but only %d hash tables;\n",
     vars,1<<hbits);
  while ((vars>>hbits)>=10) hbits++;
  fprintf(stderr," maybe you should use command-line option h%d?\n",hbits);
}
clauses-=nullclauses;
if (clauses==0) {
  fprintf(stderr,"No clauses were input!\n");
  exit(-77);
}
if (vars>=0x80000000) {
  fprintf(stderr,"Whoa, the input had %llu variables!\n",vars);
  exit(-664);
}
if (clauses>=0x80000000) {
  fprintf(stderr,"Whoa, the input had %llu clauses!\n",clauses);
  exit(-665);
}
if (cells>=0x100000000) {
  fprintf(stderr,"Whoa, the input had %llu occurrences of literals!\n",cells);
  exit(-666);
}

@ @<Input the clause in |buf|@>=
for (j=k=0;;) {
  while (buf[j]==' ') j++; /* scan to nonblank */
  if (buf[j]=='\n') break;
  if (buf[j]<' ' || buf[j]>'~') {
    fprintf(stderr,"Illegal character (code #%x) in the clause on line %lld!\n",
      buf[j],clauses);
    exit(-5);
  }
  if (buf[j]=='~') i=1,j++;
  else i=0;
  @<Scan and record a variable; negate it if |i==1|@>;
}
if (k==0) {
  fprintf(stderr,"(Empty line %lld is being ignored)\n",clauses);
  nullclauses++; /* strictly speaking it would be unsatisfiable */
}
goto clause_done;
empty_clause: @<Remove all variables of the current clause@>;
clause_done: cells+=k;

@ We need a hack to insert the bit codes 1 and/or 2 into a pointer value.

@d hack_in(q,t) (tmp_var*)(t|(ullng)q)

@<Scan and record a variable; negate it if |i==1|@>=
{
  register tmp_var *p;
  if (cur_tmp_var==bad_tmp_var) @<Install a new |vchunk|@>;
  @<Put the variable name beginning at |buf[j]| in |cur_tmp_var->name|
     and compute its hash code |h|@>;
  @<Find |cur_tmp_var->name| in the hash table at |p|@>;
  if (p->stamp==clauses || p->stamp==-clauses) @<Handle a duplicate literal@>@;
  else {
    p->stamp=(i? -clauses: clauses);
    if (cur_cell==bad_cell) @<Install a new |chunk|@>;
    *cur_cell=p;
    if (i==1) *cur_cell=hack_in(*cur_cell,1);
    if (k==0) *cur_cell=hack_in(*cur_cell,2);
    cur_cell++,k++;
  }
}

@ @<Install a new |vchunk|@>=
{
  register vchunk *new_vchunk;
  new_vchunk=(vchunk*)malloc(sizeof(vchunk));
  if (!new_vchunk) {
    fprintf(stderr,"Can't allocate a new vchunk!\n");
    exit(-6);
  }
  new_vchunk->prev=cur_vchunk, cur_vchunk=new_vchunk;
  cur_tmp_var=&new_vchunk->var[0];
  bad_tmp_var=&new_vchunk->var[vars_per_vchunk];
}  

@ @<Install a new |chunk|@>=
{
  register chunk *new_chunk;
  new_chunk=(chunk*)malloc(sizeof(chunk));
  if (!new_chunk) {
    fprintf(stderr,"Can't allocate a new chunk!\n");
    exit(-7);
  }
  new_chunk->prev=cur_chunk, cur_chunk=new_chunk;
  cur_cell=&new_chunk->cell[0];
  bad_cell=&new_chunk->cell[cells_per_chunk];
}  

@ The hash code is computed via ``universal hashing,'' using the following
precomputed tables of random bits.

@<Initialize everything@>=
for (j=92;j;j--) for (k=0;k<8;k++)
  hash_bits[j][k]=gb_next_rand();

@ @<Put the variable name beginning at |buf[j]| in |cur_tmp_var->name|...@>=
cur_tmp_var->name.lng=0;
for (h=l=0;buf[j+l]>' '&&buf[j+l]<='~';l++) {
  if (l>7) {
    fprintf(stderr,
        "Variable name %.9s... in the clause on line %lld is too long!\n",
        buf+j,clauses);
    exit(-8);
  }
  h^=hash_bits[buf[j+l]-'!'][l];
  cur_tmp_var->name.ch8[l]=buf[j+l];
}
if (l==0) goto empty_clause; /* `\.\~' by itself is like `true' */
j+=l;
h&=(1<<hbits)-1;

@ @<Find |cur_tmp_var->name| in the hash table...@>=
for (p=hash[h];p;p=p->next)
  if (p->name.lng==cur_tmp_var->name.lng) break;
if (!p) { /* new variable found */
  p=cur_tmp_var++;
  p->next=hash[h], hash[h]=p;
  p->serial=vars++;
  p->stamp=0;
}

@ The most interesting aspect of the input phase is probably the ``unwinding''
that we might need to do when encountering a literal more than once
in the same clause.

@<Handle a duplicate literal@>=
{
  if ((p->stamp>0)==(i>0)) goto empty_clause;
}

@ An input line that begins with `\.{\~\ }' is silently treated as a comment.
Otherwise redundant clauses are logged, in case they were unintentional.
(One can, however, intentionally
use redundant clauses to force the order of the variables.)

@<Remove all variables of the current clause@>=
while (k) {
  @<Move |cur_cell| backward to the previous cell@>;
  k--;
}
if ((buf[0]!='~')||(buf[1]!=' '))
  fprintf(stderr,"(The clause on line %lld is always satisfied)\n",clauses);
nullclauses++;

@ @<Move |cur_cell| backward to the previous cell@>=
if (cur_cell>&cur_chunk->cell[0]) cur_cell--;
else {
  register chunk *old_chunk=cur_chunk;
  cur_chunk=old_chunk->prev;@+free(old_chunk);
  bad_cell=&cur_chunk->cell[cells_per_chunk];
  cur_cell=bad_cell-1;
}

@ @<Move |cur_tmp_var| backward to the previous temporary variable@>=
if (cur_tmp_var>&cur_vchunk->var[0]) cur_tmp_var--;
else {
  register vchunk *old_vchunk=cur_vchunk;
  cur_vchunk=old_vchunk->prev;@+free(old_vchunk);
  bad_tmp_var=&cur_vchunk->var[vars_per_vchunk];
  cur_tmp_var=bad_tmp_var-1;
}

@ @<Report the successful completion of the input phase@>=
fprintf(stderr,
   "(%lld variables, %lld clauses, %llu literals successfully read)\n",
                       vars,clauses,cells);

@*SAT solving, version 0. OK, now comes my hypothetical low-overhead
{\mc SAT} solver, with the lazy data structures of Brown and Purdom 1982
grafted back into 1960s ideas.

The algorithm below
essentially tries to solve a satisfiability problem on $n$
variables by first setting $x_1$ to its most plausible value,
then using the same idea recursively on the remaining $(n-1)$-variable
problem. If this doesn't work, we try the other possibility for
$x_1$, and the result will either succeed or fail.

Data structures to support that method should allow us to do the
following things easily:
\smallskip
\item{$\bullet$}Know, for each literal, the clauses in which
that literal is being ``watched.''
\item{$\bullet$}Know, for each clause, the literals that it contains,
and the literal it watches.
\item{$\bullet$}Swap literals within a clause so that the watched literal
is never false.
\smallskip\noindent
The original clause sizes are known in advance. Therefore we can use a
combination of sequential and linked memory to accomplish all of these goals.

@ The basic unit in our data structure is called a cell. There's one
cell for each literal in each clause. This stripped-down version
of {\mc SAT0} doesn't really need a special data type for cells,
but I've kept it anyway.

Each link is a 32-bit integer. (I don't use \CEE/ pointers in the main
data structures, because they occupy 64 bits and clutter up the caches.)
The integer is an index into a monolithic array of cells called |mem|.

@<Type...@>=
typedef struct {
  uint litno; /* literal number */
} cell;

@ Each clause is represented by a pointer to its first cell, which
contains its watched literal. There's also a pointer to another clause
that has the same watched literal.

Clauses appear in reverse order. Thus the cells of clause |c| run
from |cmem[c].start| to |cmem[c-1].start-1|.

The first $2n+2$ ``clauses'' are special; they serve as list heads for
watch lists of each literal.

@<Type...@>=
typedef struct {
  uint start; /* the address in |mem| where the cells for this clause start */
  uint wlink; /* link for the watch list */
} clause;

@ A variable is represented by its name, for purposes of output.
The name appears in a separate array |vmem| of vertex nodes.

@<Type...@>=
typedef struct {
  octa name; /* the variable's symbolic name */
} variable;

@ @<Glob...@>=
cell *mem; /* the master array of cells */
clause *cmem; /* the master array of clauses */
uint nonspec; /* address in |cmem| of the first non-special clause */
variable *vmem; /* the master array of variables */
char *move; /* the stack of choices made so far */

@ Here is a subroutine that prints a clause symbolically. It illustrates
some of the conventions of the data structures that have been explained above.
I use it only for debugging.

Incidentally, the clause numbers reported to the user after the input phase
may differ from the line numbers reported during the input phase,
when |nullclauses>0|.

@<Sub...@>=
void print_clause(uint c) {
  register uint k,l;
  printf("%d:",c); /* show the clause number */
  for (k=cmem[c].start;k<cmem[c-1].start;k++) {
    l=mem[k].litno;
    printf(" %s%.8s", l&1? "~": "", vmem[l>>1].name.ch8); /* $k$th literal */
  }
  printf("\n");
}

@ Similarly we can print out all of the clauses that watch
a particular literal.

@<Sub...@>=
void print_clauses_watching(int l) {
  register uint p;
  for (p=cmem[l].wlink;p;p=cmem[p].wlink)
    print_clause(p);
}

@ In long runs it's helpful to know how far we've gotten.

@<Sub...@>=
void print_state(int l) {
  register int k;
  fprintf(stderr," after %lld mems:",mems);
  for (k=1;k<=l;k++) fprintf(stderr,"%c",move[k]+'0');
  fprintf(stderr,"\n");
  fflush(stderr);
}

@*Initializing the real data structures.
Okay, we're ready now to convert the temporary chunks of data into the
form we want, and to recycle those chunks. The code below is intended to be
a prototype for similar tasks in later programs of this series.

@<Set up the main data structures@>=
@<Allocate the main arrays@>;
@<Copy all the temporary cells to the |mem| and |cmem| arrays
   in proper format@>;
@<Copy all the temporary variable nodes to the |vmem| array in proper format@>;
@<Check consistency@>;

@ The backtracking routine uses a small array called |move| to record
its choices-so-far. We don't count the space for |move| in |bytes|, because
each |variable| entry has a spare byte that could have been used.

@<Allocate the main arrays@>=
free(buf);@+free(hash); /* a tiny gesture to make a little room */
mem=(cell*)malloc(cells*sizeof(cell));
if (!mem) {
  fprintf(stderr,"Oops, I can't allocate the big mem array!\n");
  exit(-10);
}
bytes=cells*sizeof(cell);
nonspec=vars+vars+2;
if (nonspec+clauses>=0x100000000) {
  fprintf(stderr,"Whoa, nonspec+clauses is too big for me!\n");
  exit(-667);
}
cmem=(clause*)malloc((nonspec+clauses)*sizeof(clause));
if (!cmem) {
  fprintf(stderr,"Oops, I can't allocate the cmem array!\n");
  exit(-11);
}
bytes+=(nonspec+clauses)*sizeof(clause);
vmem=(variable*)malloc((vars+1)*sizeof(variable));
if (!vmem) {
  fprintf(stderr,"Oops, I can't allocate the vmem array!\n");
  exit(-12);
}
bytes+=(vars+1)*sizeof(variable);
move=(char*)malloc((vars+1)*sizeof(char));
if (!move) {
  fprintf(stderr,"Oops, I can't allocate the move array!\n");
  exit(-13);
}

@ @<Copy all the temporary cells to the |mem| and |cmem| arrays...@>=
for (j=0;j<nonspec;j++) o,cmem[j].start=cmem[j].wlink=0;
for (c=nonspec+clauses-1,k=0; c>=nonspec; c--) {
  j=0;
  @<Insert the cells for the literals of clause |c|@>;
}
o,cmem[c].start=k;

@ The basic idea is to ``unwind'' the steps that we went through while
building up the chunks.

@d hack_out(q) (((ullng)q)&0x3)
@d hack_clean(q) ((tmp_var*)((ullng)q&-4))

@<Insert the cells for the literals of clause |c|@>=
for (i=0;i<2;j++) {
  @<Move |cur_cell| back...@>;
  i=hack_out(*cur_cell);
  p=hack_clean(*cur_cell)->serial;
  p+=p+(i&1)+2;
  if (j==0)
    ooo,cmem[c].start=k,cmem[c].wlink=cmem[p].wlink,cmem[p].wlink=c,j=1;
  o,mem[k++].litno=p;
}

@ @<Copy all the temporary variable nodes to the |vmem| array...@>=
for (c=vars; c; c--) {
  @<Move |cur_tmp_var| back...@>;
  o,vmem[c].name.lng=cur_tmp_var->name.lng;
}

@ We should now have unwound all the temporary data chunks back to their
beginnings.

@<Check consistency@>=
if (cur_cell!=&cur_chunk->cell[0] ||
     cur_chunk->prev!=NULL ||
     cur_tmp_var!=&cur_vchunk->var[0] ||
     cur_vchunk->prev!=NULL) {
  fprintf(stderr,"This can't happen (consistency check failure)!\n");
  exit(-14);
}
free(cur_chunk);@+free(cur_vchunk);

@*Doing it. Now comes ye olde basic backtrack.

A choice is recorded in the |move| array, as the number 0 if we're
trying first to set the current variable true;
it is 3 if that move failed and we're trying the other alternative.

@<Solve the problem@>=
level=1; /* I used to start at level 0, but Algorithm 7.2.2.2B does this */
newlevel: if (level>vars) goto satisfied;
oo,move[level]=(cmem[level+level+1].wlink!=0 || cmem[level+level].wlink==0);
if ((verbose&show_choices) && level<=show_choices_max) {
  fprintf(stderr,"Level %d, trying %s%.8s",
    level,move[level]?"~":"",vmem[level].name.ch8);
  if (verbose&show_details)
    fprintf(stderr," (%lld mems)",
      mems);
  fprintf(stderr,"\n");
}
nodes++;
if (delta && (mems>=thresh)) thresh+=delta,print_state(level);
if (mems>timeout) {
  fprintf(stderr,"TIMEOUT!\n");
  goto done;
}  
tryit: parity=move[level]&1;
@<Make variable |level| non-watched by the clauses in the non-chosen list;
  |goto try_again| if that would make a clause empty@>;
level++;@+goto newlevel;
try_again:@+if (o,move[level]<2) {
  o,move[level]=3-move[level];
  if ((verbose&show_choices) && level<=show_choices_max) {
    fprintf(stderr,"Level %d, trying again",level);
    if (verbose&show_details) fprintf(stderr," (%lld mems)\n",
                   mems);
    else fprintf(stderr,"\n");
  }
  goto tryit;
}
if (level>1) @<Backtrack to the previous level@>;
if (1) {
  printf("~\n"); /* the formula was unsatisfiable */
  if (verbose&show_basics) fprintf(stderr,"UNSAT\n");
}@+else {
satisfied:@+if (verbose&show_basics) fprintf(stderr,"!SAT!\n");
  @<Print the solution found@>;
}

@ @<Make variable |level|...@>=
for (o,c=cmem[level+level+1-parity].wlink;c;c=q) {
  oo,i=cmem[c].start,q=cmem[c].wlink,j=cmem[c-1].start;
  for (p=i+1;p<j;p++) {
    o,k=mem[p].litno;
    if (k>=level+level || (o,((move[k>>1]^k)&1)==0)) break;
  }
  if (p==j) {
    if (verbose&show_details)
      fprintf(stderr,"(Clause %d contradicted)\n",
                           c);
    o,cmem[level+level+1-parity].wlink=c;
    goto try_again;
  }
  oo,mem[i].litno=k,mem[p].litno=level+level+1-parity;
  ooo,cmem[c].wlink=cmem[k].wlink,cmem[k].wlink=c;
  if (verbose&show_details)
    fprintf(stderr,"(Clause %d now watches %s%.8s)\n",
            c,k&1?"~":"",vmem[k>>1].name.ch8);
}
o,cmem[level+level+1-parity].wlink=0;

@ @<Backtrack to the previous level@>=
{
  level--;
  goto try_again;
}

@ @<Print the solution found@>=
for (k=1;k<level;k++)
  printf(" %s%.8s",move[k]&1?"~":"",vmem[k].name.ch8);
printf("\n");

@*Index.