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]

1 files changed, 914 insertions, 0 deletions

diff --git a/unfinished/numgame.c b/unfinished/numgame.c
new file mode 100644
index 0000000..e6f0985
--- /dev/null
+++ b/unfinished/numgame.c
@@ -0,0 +1,914 @@
+/*
+ * 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;
+}