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puzzles/unfinished/numgame.c
Simon Tatham b3364419da I'm sick and tired of having unfinished puzzle code lying around on 
several different systems in strange directories. So I'm creating an
`unfinished' directory within source control, and centralising all
my half-finished, half-baked or otherwise half-arsed puzzle
implementations into it. Herewith Sokoban (playable but rubbish
generation), Pearl (Masyu - rubbish generation and nothing else),
Path (Number Link - rubbish generation and nothing else) and NumGame
(the Countdown numbers game - currently just a solver and not even a
generator yet).

[originally from svn r6883]
2006-10-29 09:41:02 +00:00

915 lines
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/*
* This program implements a breadth-first search which
* exhaustively solves the Countdown numbers game, and related
* games with slightly different rule sets such as `Flippo'.
*
* Currently it is simply a standalone command-line utility to
* which you provide a set of numbers and it tells you everything
* it can make together with how many different ways it can be
* made. I would like ultimately to turn it into the generator for
* a Puzzles puzzle, but I haven't even started on writing a
* Puzzles user interface yet.
*/
/*
* TODO:
*
* - start thinking about difficulty ratings
* + anything involving associative operations will be flagged
* as many-paths because of the associative options (e.g.
* 2*3*4 can be (2*3)*4 or 2*(3*4), or indeed (2*4)*3). This
* is probably a _good_ thing, since those are unusually
* easy.
* + tree-structured calculations ((a*b)/(c+d)) have multiple
* paths because the independent branches of the tree can be
* evaluated in either order, whereas straight-line
* calculations with no branches will be considered easier.
* Can we do anything about this? It's certainly not clear to
* me that tree-structure calculations are _easier_, although
* I'm also not convinced they're harder.
* + I think for a realistic difficulty assessment we must also
* consider the `obviousness' of the arithmetic operations in
* some heuristic sense, and also (in Countdown) how many
* numbers ended up being used.
* - actually try some generations
* - at this point we're probably ready to start on the Puzzles
* integration.
*/
#include <stdio.h>
#include <limits.h>
#include <assert.h>
#include "puzzles.h"
#include "tree234.h"
/*
* To search for numbers we can make, we employ a breadth-first
* search across the space of sets of input numbers. That is, for
* example, we start with the set (3,6,25,50,75,100); we apply
* moves which involve combining two numbers (e.g. adding the 50
* and the 75 takes us to the set (3,6,25,100,125); and then we see
* if we ever end up with a set containing (say) 952.
*
* If the rules are changed so that all the numbers must be used,
* this is easy to adjust to: we simply see if we end up with a set
* containing _only_ (say) 952.
*
* Obviously, we can vary the rules about permitted arithmetic
* operations simply by altering the set of valid moves in the bfs.
* However, there's one common rule in this sort of puzzle which
* takes a little more thought, and that's _concatenation_. For
* example, if you are given (say) four 4s and required to make 10,
* you are permitted to combine two of the 4s into a 44 to begin
* with, making (44-4)/4 = 10. However, you are generally not
* allowed to concatenate two numbers that _weren't_ both in the
* original input set (you couldn't multiply two 4s to get 16 and
* then concatenate a 4 on to it to make 164), so concatenation is
* not an operation which is valid in all situations.
*
* We could enforce this restriction by storing a flag alongside
* each number indicating whether or not it's an original number;
* the rules being that concatenation of two numbers is only valid
* if they both have the original flag, and that its output _also_
* has the original flag (so that you can concatenate three 4s into
* a 444), but that applying any other arithmetic operation clears
* the original flag on the output. However, we can get marginally
* simpler than that by observing that since concatenation has to
* happen to a number before any other operation, we can simply
* place all the concatenations at the start of the search. In
* other words, we have a global flag on an entire number _set_
* which indicates whether we are still permitted to perform
* concatenations; if so, we can concatenate any of the numbers in
* that set. Performing any other operation clears the flag.
*/
#define SETFLAG_CONCAT 1 /* we can do concatenation */
struct sets;
struct set {
int *numbers; /* rationals stored as n,d pairs */
short nnumbers; /* # of rationals, so half # of ints */
short flags; /* SETFLAG_CONCAT only, at present */
struct set *prev; /* index of ancestor set in set list */
unsigned char pa, pb, po, pr; /* operation that got here from prev */
int npaths; /* number of ways to reach this set */
};
struct output {
int number;
struct set *set;
int index; /* which number in the set is it? */
int npaths; /* number of ways to reach this */
};
#define SETLISTLEN 1024
#define NUMBERLISTLEN 32768
#define OUTPUTLISTLEN 1024
struct operation;
struct sets {
struct set **setlists;
int nsets, nsetlists, setlistsize;
tree234 *settree;
int **numberlists;
int nnumbers, nnumberlists, numberlistsize;
struct output **outputlists;
int noutputs, noutputlists, outputlistsize;
tree234 *outputtree;
const struct operation *const *ops;
};
#define OPFLAG_NEEDS_CONCAT 1
#define OPFLAG_KEEPS_CONCAT 2
struct operation {
/*
* Most operations should be shown in the output working, but
* concatenation should not; we just take the result of the
* concatenation and assume that it's obvious how it was
* derived.
*/
int display;
/*
* Text display of the operator.
*/
char *text;
/*
* Flags dictating when the operator can be applied.
*/
int flags;
/*
* Priority of the operator (for avoiding unnecessary
* parentheses when formatting it into a string).
*/
int priority;
/*
* Associativity of the operator. Bit 0 means we need parens
* when the left operand of one of these operators is another
* instance of it, e.g. (2^3)^4. Bit 1 means we need parens
* when the right operand is another instance of the same
* operator, e.g. 2-(3-4). Thus:
*
* - this field is 0 for a fully associative operator, since
* we never need parens.
* - it's 1 for a right-associative operator.
* - it's 2 for a left-associative operator.
* - it's 3 for a _non_-associative operator (which always
* uses parens just to be sure).
*/
int assoc;
/*
* Whether the operator is commutative. Saves time in the
* search if we don't have to try it both ways round.
*/
int commutes;
/*
* Function which implements the operator. Returns TRUE on
* success, FALSE on failure. Takes two rationals and writes
* out a third.
*/
int (*perform)(int *a, int *b, int *output);
};
struct rules {
const struct operation *const *ops;
int use_all;
};
#define MUL(r, a, b) do { \
(r) = (a) * (b); \
if ((b) && (a) && (r) / (b) != (a)) return FALSE; \
} while (0)
#define ADD(r, a, b) do { \
(r) = (a) + (b); \
if ((a) > 0 && (b) > 0 && (r) < 0) return FALSE; \
if ((a) < 0 && (b) < 0 && (r) > 0) return FALSE; \
} while (0)
#define OUT(output, n, d) do { \
int g = gcd((n),(d)); \
if ((d) < 0) g = -g; \
(output)[0] = (n)/g; \
(output)[1] = (d)/g; \
assert((output)[1] > 0); \
} while (0)
static int gcd(int x, int y)
{
while (x != 0 && y != 0) {
int t = x;
x = y;
y = t % y;
}
return abs(x + y); /* i.e. whichever one isn't zero */
}
static int perform_add(int *a, int *b, int *output)
{
int at, bt, tn, bn;
/*
* a0/a1 + b0/b1 = (a0*b1 + b0*a1) / (a1*b1)
*/
MUL(at, a[0], b[1]);
MUL(bt, b[0], a[1]);
ADD(tn, at, bt);
MUL(bn, a[1], b[1]);
OUT(output, tn, bn);
return TRUE;
}
static int perform_sub(int *a, int *b, int *output)
{
int at, bt, tn, bn;
/*
* a0/a1 - b0/b1 = (a0*b1 - b0*a1) / (a1*b1)
*/
MUL(at, a[0], b[1]);
MUL(bt, b[0], a[1]);
ADD(tn, at, -bt);
MUL(bn, a[1], b[1]);
OUT(output, tn, bn);
return TRUE;
}
static int perform_mul(int *a, int *b, int *output)
{
int tn, bn;
/*
* a0/a1 * b0/b1 = (a0*b0) / (a1*b1)
*/
MUL(tn, a[0], b[0]);
MUL(bn, a[1], b[1]);
OUT(output, tn, bn);
return TRUE;
}
static int perform_div(int *a, int *b, int *output)
{
int tn, bn;
/*
* Division by zero is outlawed.
*/
if (b[0] == 0)
return FALSE;
/*
* a0/a1 / b0/b1 = (a0*b1) / (a1*b0)
*/
MUL(tn, a[0], b[1]);
MUL(bn, a[1], b[0]);
OUT(output, tn, bn);
return TRUE;
}
static int perform_exact_div(int *a, int *b, int *output)
{
int tn, bn;
/*
* Division by zero is outlawed.
*/
if (b[0] == 0)
return FALSE;
/*
* a0/a1 / b0/b1 = (a0*b1) / (a1*b0)
*/
MUL(tn, a[0], b[1]);
MUL(bn, a[1], b[0]);
OUT(output, tn, bn);
/*
* Exact division means we require the result to be an integer.
*/
return (output[1] == 1);
}
static int perform_concat(int *a, int *b, int *output)
{
int t1, t2, p10;
/*
* We can't concatenate anything which isn't an integer.
*/
if (a[1] != 1 || b[1] != 1)
return FALSE;
/*
* For concatenation, we can safely assume leading zeroes
* aren't an issue. It isn't clear whether they `should' be
* allowed, but it turns out not to matter: concatenating a
* leading zero on to a number in order to harmlessly get rid
* of the zero is never necessary because unwanted zeroes can
* be disposed of by adding them to something instead. So we
* disallow them always.
*
* The only other possibility is that you might want to
* concatenate a leading zero on to something and then
* concatenate another non-zero digit on to _that_ (to make,
* for example, 106); but that's also unnecessary, because you
* can make 106 just as easily by concatenating the 0 on to the
* _end_ of the 1 first.
*/
if (a[0] == 0)
return FALSE;
/*
* Find the smallest power of ten strictly greater than b. This
* is the power of ten by which we'll multiply a.
*
* Special case: we must multiply a by at least 10, even if b
* is zero.
*/
p10 = 10;
while (p10 <= (INT_MAX/10) && p10 <= b[0])
p10 *= 10;
if (p10 > INT_MAX/10)
return FALSE; /* integer overflow */
MUL(t1, p10, a[0]);
ADD(t2, t1, b[0]);
OUT(output, t2, 1);
return TRUE;
}
const static struct operation op_add = {
TRUE, "+", 0, 10, 0, TRUE, perform_add
};
const static struct operation op_sub = {
TRUE, "-", 0, 10, 2, FALSE, perform_sub
};
const static struct operation op_mul = {
TRUE, "*", 0, 20, 0, TRUE, perform_mul
};
const static struct operation op_div = {
TRUE, "/", 0, 20, 2, FALSE, perform_div
};
const static struct operation op_xdiv = {
TRUE, "/", 0, 20, 2, FALSE, perform_exact_div
};
const static struct operation op_concat = {
FALSE, "", OPFLAG_NEEDS_CONCAT | OPFLAG_KEEPS_CONCAT,
1000, 0, FALSE, perform_concat
};
/*
* In Countdown, divisions resulting in fractions are disallowed.
* http://www.askoxford.com/wordgames/countdown/rules/
*/
const static struct operation *const ops_countdown[] = {
&op_add, &op_mul, &op_sub, &op_xdiv, NULL
};
const static struct rules rules_countdown = {
ops_countdown, FALSE
};
/*
* A slightly different rule set which handles the reasonably well
* known puzzle of making 24 using two 3s and two 8s. For this we
* need rational rather than integer division.
*/
const static struct operation *const ops_3388[] = {
&op_add, &op_mul, &op_sub, &op_div, NULL
};
const static struct rules rules_3388 = {
ops_3388, TRUE
};
/*
* A still more permissive rule set usable for the four-4s problem
* and similar things. Permits concatenation.
*/
const static struct operation *const ops_four4s[] = {
&op_add, &op_mul, &op_sub, &op_div, &op_concat, NULL
};
const static struct rules rules_four4s = {
ops_four4s, TRUE
};
#define ratcmp(a,op,b) ( (long long)(a)[0] * (b)[1] op \
(long long)(b)[0] * (a)[1] )
static int addtoset(struct set *set, int newnumber[2])
{
int i, j;
/* Find where we want to insert the new number */
for (i = 0; i < set->nnumbers &&
ratcmp(set->numbers+2*i, <, newnumber); i++);
/* Move everything else up */
for (j = set->nnumbers; j > i; j--) {
set->numbers[2*j] = set->numbers[2*j-2];
set->numbers[2*j+1] = set->numbers[2*j-1];
}
/* Insert the new number */
set->numbers[2*i] = newnumber[0];
set->numbers[2*i+1] = newnumber[1];
set->nnumbers++;
return i;
}
#define ensure(array, size, newlen, type) do { \
if ((newlen) > (size)) { \
(size) = (newlen) + 512; \
(array) = sresize((array), (size), type); \
} \
} while (0)
static int setcmp(void *av, void *bv)
{
struct set *a = (struct set *)av;
struct set *b = (struct set *)bv;
int i;
if (a->nnumbers < b->nnumbers)
return -1;
else if (a->nnumbers > b->nnumbers)
return +1;
if (a->flags < b->flags)
return -1;
else if (a->flags > b->flags)
return +1;
for (i = 0; i < a->nnumbers; i++) {
if (ratcmp(a->numbers+2*i, <, b->numbers+2*i))
return -1;
else if (ratcmp(a->numbers+2*i, >, b->numbers+2*i))
return +1;
}
return 0;
}
static int outputcmp(void *av, void *bv)
{
struct output *a = (struct output *)av;
struct output *b = (struct output *)bv;
if (a->number < b->number)
return -1;
else if (a->number > b->number)
return +1;
return 0;
}
static int outputfindcmp(void *av, void *bv)
{
int *a = (int *)av;
struct output *b = (struct output *)bv;
if (*a < b->number)
return -1;
else if (*a > b->number)
return +1;
return 0;
}
static void addset(struct sets *s, struct set *set, struct set *prev)
{
struct set *s2;
int npaths = (prev ? prev->npaths : 1);
assert(set == s->setlists[s->nsets / SETLISTLEN] + s->nsets % SETLISTLEN);
s2 = add234(s->settree, set);
if (s2 == set) {
/*
* New set added to the tree.
*/
set->prev = prev;
set->npaths = npaths;
s->nsets++;
s->nnumbers += 2 * set->nnumbers;
} else {
/*
* Rediscovered an existing set. Update its npaths only.
*/
s2->npaths += npaths;
}
}
static struct set *newset(struct sets *s, int nnumbers, int flags)
{
struct set *sn;
ensure(s->setlists, s->setlistsize, s->nsets/SETLISTLEN+1, struct set *);
while (s->nsetlists <= s->nsets / SETLISTLEN)
s->setlists[s->nsetlists++] = snewn(SETLISTLEN, struct set);
sn = s->setlists[s->nsets / SETLISTLEN] + s->nsets % SETLISTLEN;
if (s->nnumbers + nnumbers * 2 > s->nnumberlists * NUMBERLISTLEN)
s->nnumbers = s->nnumberlists * NUMBERLISTLEN;
ensure(s->numberlists, s->numberlistsize,
s->nnumbers/NUMBERLISTLEN+1, int *);
while (s->nnumberlists <= s->nnumbers / NUMBERLISTLEN)
s->numberlists[s->nnumberlists++] = snewn(NUMBERLISTLEN, int);
sn->numbers = s->numberlists[s->nnumbers / NUMBERLISTLEN] +
s->nnumbers % NUMBERLISTLEN;
/*
* Start the set off empty.
*/
sn->nnumbers = 0;
sn->flags = flags;
return sn;
}
static int addoutput(struct sets *s, struct set *ss, int index, int *n)
{
struct output *o, *o2;
/*
* Target numbers are always integers.
*/
if (ss->numbers[2*index+1] != 1)
return FALSE;
ensure(s->outputlists, s->outputlistsize, s->noutputs/OUTPUTLISTLEN+1,
struct output *);
while (s->noutputlists <= s->noutputs / OUTPUTLISTLEN)
s->outputlists[s->noutputlists++] = snewn(OUTPUTLISTLEN,
struct output);
o = s->outputlists[s->noutputs / OUTPUTLISTLEN] +
s->noutputs % OUTPUTLISTLEN;
o->number = ss->numbers[2*index];
o->set = ss;
o->index = index;
o->npaths = ss->npaths;
o2 = add234(s->outputtree, o);
if (o2 != o) {
o2->npaths += o->npaths;
} else {
s->noutputs++;
}
*n = o->number;
return TRUE;
}
static struct sets *do_search(int ninputs, int *inputs,
const struct rules *rules, int *target)
{
struct sets *s;
struct set *sn;
int qpos, i;
const struct operation *const *ops = rules->ops;
s = snew(struct sets);
s->setlists = NULL;
s->nsets = s->nsetlists = s->setlistsize = 0;
s->numberlists = NULL;
s->nnumbers = s->nnumberlists = s->numberlistsize = 0;
s->outputlists = NULL;
s->noutputs = s->noutputlists = s->outputlistsize = 0;
s->settree = newtree234(setcmp);
s->outputtree = newtree234(outputcmp);
s->ops = ops;
/*
* Start with the input set.
*/
sn = newset(s, ninputs, SETFLAG_CONCAT);
for (i = 0; i < ninputs; i++) {
int newnumber[2];
newnumber[0] = inputs[i];
newnumber[1] = 1;
addtoset(sn, newnumber);
}
addset(s, sn, NULL);
/*
* Now perform the breadth-first search: keep looping over sets
* until we run out of steam.
*/
qpos = 0;
while (qpos < s->nsets) {
struct set *ss = s->setlists[qpos / SETLISTLEN] + qpos % SETLISTLEN;
struct set *sn;
int i, j, k, m;
/*
* Record all the valid output numbers in this state. We
* can always do this if there's only one number in the
* state; otherwise, we can only do it if we aren't
* required to use all the numbers in coming to our answer.
*/
if (ss->nnumbers == 1 || !rules->use_all) {
for (i = 0; i < ss->nnumbers; i++) {
int n;
if (addoutput(s, ss, i, &n) && target && n == *target)
return s;
}
}
/*
* Try every possible operation from this state.
*/
for (k = 0; ops[k] && ops[k]->perform; k++) {
if ((ops[k]->flags & OPFLAG_NEEDS_CONCAT) &&
!(ss->flags & SETFLAG_CONCAT))
continue; /* can't use this operation here */
for (i = 0; i < ss->nnumbers; i++) {
for (j = 0; j < ss->nnumbers; j++) {
int n[2];
if (i == j)
continue; /* can't combine a number with itself */
if (i > j && ops[k]->commutes)
continue; /* no need to do this both ways round */
if (!ops[k]->perform(ss->numbers+2*i, ss->numbers+2*j, n))
continue; /* operation failed */
sn = newset(s, ss->nnumbers-1, ss->flags);
if (!(ops[k]->flags & OPFLAG_KEEPS_CONCAT))
sn->flags &= ~SETFLAG_CONCAT;
for (m = 0; m < ss->nnumbers; m++) {
if (m == i || m == j)
continue;
sn->numbers[2*sn->nnumbers] = ss->numbers[2*m];
sn->numbers[2*sn->nnumbers + 1] = ss->numbers[2*m + 1];
sn->nnumbers++;
}
sn->pa = i;
sn->pb = j;
sn->po = k;
sn->pr = addtoset(sn, n);
addset(s, sn, ss);
}
}
}
qpos++;
}
return s;
}
static void free_sets(struct sets *s)
{
int i;
freetree234(s->settree);
freetree234(s->outputtree);
for (i = 0; i < s->nsetlists; i++)
sfree(s->setlists[i]);
sfree(s->setlists);
for (i = 0; i < s->nnumberlists; i++)
sfree(s->numberlists[i]);
sfree(s->numberlists);
for (i = 0; i < s->noutputlists; i++)
sfree(s->outputlists[i]);
sfree(s->outputlists);
sfree(s);
}
/*
* Construct a text formula for producing a given output.
*/
void mkstring_recurse(char **str, int *len,
struct sets *s, struct set *ss, int index,
int priority, int assoc, int child)
{
if (ss->prev && index != ss->pr) {
int pi;
/*
* This number was passed straight down from this set's
* predecessor. Find its index in the previous set and
* recurse to there.
*/
pi = index;
assert(pi != ss->pr);
if (pi > ss->pr)
pi--;
if (pi >= min(ss->pa, ss->pb)) {
pi++;
if (pi >= max(ss->pa, ss->pb))
pi++;
}
mkstring_recurse(str, len, s, ss->prev, pi, priority, assoc, child);
} else if (ss->prev && index == ss->pr &&
s->ops[ss->po]->display) {
/*
* This number was created by a displayed operator in the
* transition from this set to its predecessor. Hence we
* write an open paren, then recurse into the first
* operand, then write the operator, then the second
* operand, and finally close the paren.
*/
char *op;
int parens, thispri, thisassoc;
/*
* Determine whether we need parentheses.
*/
thispri = s->ops[ss->po]->priority;
thisassoc = s->ops[ss->po]->assoc;
parens = (thispri < priority ||
(thispri == priority && (assoc & child)));
if (parens) {
if (str)
*(*str)++ = '(';
if (len)
(*len)++;
}
mkstring_recurse(str, len, s, ss->prev, ss->pa, thispri, thisassoc, 1);
for (op = s->ops[ss->po]->text; *op; op++) {
if (str)
*(*str)++ = *op;
if (len)
(*len)++;
}
mkstring_recurse(str, len, s, ss->prev, ss->pb, thispri, thisassoc, 2);
if (parens) {
if (str)
*(*str)++ = ')';
if (len)
(*len)++;
}
} else {
/*
* This number is either an original, or something formed
* by a non-displayed operator (concatenation). Either way,
* we display it as is.
*/
char buf[80], *p;
int blen;
blen = sprintf(buf, "%d", ss->numbers[2*index]);
if (ss->numbers[2*index+1] != 1)
blen += sprintf(buf+blen, "/%d", ss->numbers[2*index+1]);
assert(blen < lenof(buf));
for (p = buf; *p; p++) {
if (str)
*(*str)++ = *p;
if (len)
(*len)++;
}
}
}
char *mkstring(struct sets *s, struct output *o)
{
int len;
char *str, *p;
len = 0;
mkstring_recurse(NULL, &len, s, o->set, o->index, 0, 0, 0);
str = snewn(len+1, char);
p = str;
mkstring_recurse(&p, NULL, s, o->set, o->index, 0, 0, 0);
assert(p - str <= len);
*p = '\0';
return str;
}
int main(int argc, char **argv)
{
int doing_opts = TRUE;
const struct rules *rules = NULL;
char *pname = argv[0];
int got_target = FALSE, target = 0;
int numbers[10], nnumbers = 0;
int verbose = FALSE;
int pathcounts = FALSE;
struct output *o;
struct sets *s;
int i, start, limit;
while (--argc) {
char *p = *++argv;
int c;
if (doing_opts && *p == '-') {
p++;
if (!strcmp(p, "-")) {
doing_opts = FALSE;
continue;
} else while (*p) switch (c = *p++) {
case 'C':
rules = &rules_countdown;
break;
case 'B':
rules = &rules_3388;
break;
case 'D':
rules = &rules_four4s;
break;
case 'v':
verbose = TRUE;
break;
case 'p':
pathcounts = TRUE;
break;
case 't':
{
char *v;
if (*p) {
v = p;
p = NULL;
} else if (--argc) {
v = *++argv;
} else {
fprintf(stderr, "%s: option '-%c' expects an"
" argument\n", pname, c);
return 1;
}
switch (c) {
case 't':
got_target = TRUE;
target = atoi(v);
break;
}
}
break;
default:
fprintf(stderr, "%s: option '-%c' not"
" recognised\n", pname, c);
return 1;
}
} else {
if (nnumbers >= lenof(numbers)) {
fprintf(stderr, "%s: internal limit of %d numbers exceeded\n",
pname, lenof(numbers));
return 1;
} else {
numbers[nnumbers++] = atoi(p);
}
}
}
if (!rules) {
fprintf(stderr, "%s: no rule set specified; use -C,-B,-D\n", pname);
return 1;
}
if (!nnumbers) {
fprintf(stderr, "%s: no input numbers specified\n", pname);
return 1;
}
s = do_search(nnumbers, numbers, rules, (got_target ? &target : NULL));
if (got_target) {
o = findrelpos234(s->outputtree, &target, outputfindcmp,
REL234_LE, &start);
if (!o)
start = -1;
o = findrelpos234(s->outputtree, &target, outputfindcmp,
REL234_GE, &limit);
if (!o)
limit = -1;
assert(start != -1 || limit != -1);
if (start == -1)
start = limit;
else if (limit == -1)
limit = start;
limit++;
} else {
start = 0;
limit = count234(s->outputtree);
}
for (i = start; i < limit; i++) {
o = index234(s->outputtree, i);
printf("%d", o->number);
if (pathcounts)
printf(" [%d]", o->npaths);
if (got_target || verbose) {
char *p = mkstring(s, o);
printf(" = %s", p);
sfree(p);
}
printf("\n");
}
free_sets(s);
return 0;
}
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