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puzzles/unfinished/numgame.c
Simon Tatham 4322b90cbb More operations and bug fixes from James H. 
[originally from svn r8107]
2008-07-05 15:40:43 +00:00

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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 <string.h>
#include <limits.h>
#include <assert.h>
#include <math.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 ancestor {
struct set *prev; /* index of ancestor set in set list */
unsigned char pa, pb, po, pr; /* operation that got here from prev */
};
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 */
int npaths; /* number of ways to reach this set */
struct ancestor a; /* primary ancestor */
struct ancestor *as; /* further ancestors, if we care */
int nas, assize;
};
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
#define OPFLAG_UNARY 4
#define OPFLAG_UNARYPREFIX 8
#define OPFLAG_FN 16
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, in expressions and for
* debugging respectively.
*/
char *text, *dbgtext;
/*
* 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 (g < 0) g = -g; \
if ((d) < 0) g = -g; \
if (g == -1 && (n) < -INT_MAX) return FALSE; \
if (g == -1 && (d) < -INT_MAX) return FALSE; \
(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 max_p10(int n, int *p10_r)
{
/*
* Find the smallest power of ten strictly greater than n.
*
* Special case: we must return at least 10, even if n is
* zero. (This is because this function is used for finding
* the power of ten by which to multiply a number being
* concatenated to the front of n, and concatenating 1 to 0
* should yield 10 and not 1.)
*/
int p10 = 10;
while (p10 <= (INT_MAX/10) && p10 <= n)
p10 *= 10;
if (p10 > INT_MAX/10)
return FALSE; /* integer overflow */
*p10_r = p10;
return TRUE;
}
static int perform_concat(int *a, int *b, int *output)
{
int t1, t2, p10;
/*
* We can't concatenate anything which isn't a non-negative
* integer.
*/
if (a[1] != 1 || b[1] != 1 || a[0] < 0 || b[0] < 0)
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;
if (!max_p10(b[0], &p10)) return FALSE;
MUL(t1, p10, a[0]);
ADD(t2, t1, b[0]);
OUT(output, t2, 1);
return TRUE;
}
#define IPOW(ret, x, y) do { \
int ipow_limit = (y); \
if ((x) == 1 || (x) == 0) ipow_limit = 1; \
else if ((x) == -1) ipow_limit &= 1; \
(ret) = 1; \
while (ipow_limit-- > 0) { \
int tmp; \
MUL(tmp, ret, x); \
ret = tmp; \
} \
} while (0)
static int perform_exp(int *a, int *b, int *output)
{
int an, ad, xn, xd;
/*
* Exponentiation is permitted if the result is rational. This
* means that:
*
* - first we see whether we can take the (denominator-of-b)th
* root of a and get a rational; if not, we give up.
*
* - then we do take that root of a
*
* - then we multiply by itself (numerator-of-b) times.
*/
if (b[1] > 1) {
an = (int)(0.5 + pow(a[0], 1.0/b[1]));
ad = (int)(0.5 + pow(a[1], 1.0/b[1]));
IPOW(xn, an, b[1]);
IPOW(xd, ad, b[1]);
if (xn != a[0] || xd != a[1])
return FALSE;
} else {
an = a[0];
ad = a[1];
}
if (b[0] >= 0) {
IPOW(xn, an, b[0]);
IPOW(xd, ad, b[0]);
} else {
IPOW(xd, an, -b[0]);
IPOW(xn, ad, -b[0]);
}
if (xd == 0)
return FALSE;
OUT(output, xn, xd);
return TRUE;
}
static int perform_factorial(int *a, int *b, int *output)
{
int ret, t, i;
/*
* Factorials of non-negative integers are permitted.
*/
if (a[1] != 1 || a[0] < 0)
return FALSE;
/*
* However, a special case: we don't take a factorial of
* anything which would thereby remain the same.
*/
if (a[0] == 1 || a[0] == 2)
return FALSE;
ret = 1;
for (i = 1; i <= a[0]; i++) {
MUL(t, ret, i);
ret = t;
}
OUT(output, ret, 1);
return TRUE;
}
static int perform_decimal(int *a, int *b, int *output)
{
int p10;
/*
* Add a decimal digit to the front of a number;
* fail if it's not an integer.
* So, 1 --> 0.1, 15 --> 0.15,
* or, rather, 1 --> 1/10, 15 --> 15/100,
* x --> x / (smallest power of 10 > than x)
*
*/
if (a[1] != 1) return FALSE;
if (!max_p10(a[0], &p10)) return FALSE;
OUT(output, a[0], p10);
return TRUE;
}
static int perform_recur(int *a, int *b, int *output)
{
int p10, tn, bn;
/*
* This converts a number like .4 to .44444..., or .45 to .45454...
* The input number must be -1 < a < 1.
*
* Calculate the smallest power of 10 that divides the denominator exactly,
* returning if no such power of 10 exists. Then multiply the numerator
* up accordingly, and the new denominator becomes that power of 10 - 1.
*/
if (abs(a[0]) >= abs(a[1])) return FALSE; /* -1 < a < 1 */
p10 = 10;
while (p10 <= (INT_MAX/10)) {
if ((a[1] <= p10) && (p10 % a[1]) == 0) goto found;
p10 *= 10;
}
return FALSE;
found:
tn = a[0] * (p10 / a[1]);
bn = p10 - 1;
OUT(output, tn, bn);
return TRUE;
}
static int perform_root(int *a, int *b, int *output)
{
/*
* A root B is: 1 iff a == 0
* B ^ (1/A) otherwise
*/
int ainv[2], res;
if (a[0] == 0) {
OUT(output, 1, 1);
return TRUE;
}
OUT(ainv, a[1], a[0]);
res = perform_exp(b, ainv, output);
return res;
}
static int perform_perc(int *a, int *b, int *output)
{
if (a[0] == 0) return FALSE; /* 0% = 0, uninteresting. */
if (a[1] > (INT_MAX/100)) return FALSE;
OUT(output, a[0], a[1]*100);
return TRUE;
}
static int perform_gamma(int *a, int *b, int *output)
{
int asub1[2];
/*
* gamma(a) = (a-1)!
*
* special case not caught by perform_fact: gamma(1) is 1 so
* don't bother.
*/
if (a[0] == 1 && a[1] == 1) return FALSE;
OUT(asub1, a[0]-a[1], a[1]);
return perform_factorial(asub1, b, output);
}
static int perform_sqrt(int *a, int *b, int *output)
{
int half[2] = { 1, 2 };
/*
* sqrt(1) == 1: don't perform unary noops.
*/
if (a[0] == 1 && a[1] == 1) return FALSE;
return perform_exp(a, half, output);
}
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, "", "concat", OPFLAG_NEEDS_CONCAT | OPFLAG_KEEPS_CONCAT,
1000, 0, FALSE, perform_concat
};
const static struct operation op_exp = {
TRUE, "^", "^", 0, 30, 1, FALSE, perform_exp
};
const static struct operation op_factorial = {
TRUE, "!", "!", OPFLAG_UNARY, 40, 0, FALSE, perform_factorial
};
const static struct operation op_decimal = {
TRUE, ".", ".", OPFLAG_UNARY | OPFLAG_UNARYPREFIX | OPFLAG_NEEDS_CONCAT | OPFLAG_KEEPS_CONCAT, 50, 0, FALSE, perform_decimal
};
const static struct operation op_recur = {
TRUE, "...", "recur", OPFLAG_UNARY | OPFLAG_NEEDS_CONCAT, 45, 2, FALSE, perform_recur
};
const static struct operation op_root = {
TRUE, "v~", "root", 0, 30, 1, FALSE, perform_root
};
const static struct operation op_perc = {
TRUE, "%", "%", OPFLAG_UNARY | OPFLAG_NEEDS_CONCAT, 45, 1, FALSE, perform_perc
};
const static struct operation op_gamma = {
TRUE, "gamma", "gamma", OPFLAG_UNARY | OPFLAG_UNARYPREFIX | OPFLAG_FN, 1, 3, FALSE, perform_gamma
};
const static struct operation op_sqrt = {
TRUE, "v~", "sqrt", OPFLAG_UNARY | OPFLAG_UNARYPREFIX, 30, 1, FALSE, perform_sqrt
};
/*
* 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
};
/*
* The most permissive ruleset I can think of. Permits
* exponentiation, and also silly unary operators like factorials.
*/
const static struct operation *const ops_anythinggoes[] = {
&op_add, &op_mul, &op_sub, &op_div, &op_concat, &op_exp, &op_factorial,
&op_decimal, &op_recur, &op_root, &op_perc, &op_gamma, &op_sqrt, NULL
};
const static struct rules rules_anythinggoes = {
ops_anythinggoes, 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, int multiple,
struct set *prev, int pa, int po, int pb, int pr)
{
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->a.prev = prev;
set->a.pa = pa;
set->a.po = po;
set->a.pb = pb;
set->a.pr = pr;
set->npaths = npaths;
s->nsets++;
s->nnumbers += 2 * set->nnumbers;
set->as = NULL;
set->nas = set->assize = 0;
} else {
/*
* Rediscovered an existing set. Update its npaths.
*/
s2->npaths += npaths;
/*
* And optionally enter it as an additional ancestor.
*/
if (multiple) {
if (s2->nas >= s2->assize) {
s2->assize = s2->nas * 3 / 2 + 4;
s2->as = sresize(s2->as, s2->assize, struct ancestor);
}
s2->as[s2->nas].prev = prev;
s2->as[s2->nas].pa = pa;
s2->as[s2->nas].po = po;
s2->as[s2->nas].pb = pb;
s2->as[s2->nas].pr = pr;
s2->nas++;
}
}
}
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,
int debug, int multiple)
{
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, multiple, NULL, 0, 0, 0, 0);
/*
* 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;
if (debug) {
int i;
printf("processing set:");
for (i = 0; i < ss->nnumbers; i++) {
printf(" %d", ss->numbers[2*i]);
if (ss->numbers[2*i+1] != 1)
printf("/%d", ss->numbers[2*i+1]);
}
printf("\n");
}
/*
* 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++) {
int jlimit = (ops[k]->flags & OPFLAG_UNARY ? 1 : ss->nnumbers);
for (j = 0; j < jlimit; j++) {
int n[2], newnn = ss->nnumbers;
int pa, po, pb, pr;
if (!(ops[k]->flags & OPFLAG_UNARY)) {
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 */
newnn--;
}
if (!ops[k]->perform(ss->numbers+2*i, ss->numbers+2*j, n))
continue; /* operation failed */
sn = newset(s, newnn, ss->flags);
if (!(ops[k]->flags & OPFLAG_KEEPS_CONCAT))
sn->flags &= ~SETFLAG_CONCAT;
for (m = 0; m < ss->nnumbers; m++) {
if (m == i || (!(ops[k]->flags & OPFLAG_UNARY) &&
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++;
}
pa = i;
if (ops[k]->flags & OPFLAG_UNARY)
pb = sn->nnumbers+10;
else
pb = j;
po = k;
pr = addtoset(sn, n);
addset(s, sn, multiple, ss, pa, po, pb, pr);
if (debug) {
int i;
if (ops[k]->flags & OPFLAG_UNARYPREFIX)
printf(" %s %d ->", ops[po]->dbgtext, pa);
else if (ops[k]->flags & OPFLAG_UNARY)
printf(" %d %s ->", pa, ops[po]->dbgtext);
else
printf(" %d %s %d ->", pa, ops[po]->dbgtext, pb);
for (i = 0; i < sn->nnumbers; i++) {
printf(" %d", sn->numbers[2*i]);
if (sn->numbers[2*i+1] != 1)
printf("/%d", sn->numbers[2*i+1]);
}
printf("\n");
}
}
}
}
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);
}
/*
* Print a text formula for producing a given output.
*/
void print_recurse(struct sets *s, struct set *ss, int pathindex, int index,
int priority, int assoc, int child);
void print_recurse_inner(struct sets *s, struct set *ss,
struct ancestor *a, int pathindex, int index,
int priority, int assoc, int child)
{
if (a->prev && index != a->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 != a->pr);
if (pi > a->pr)
pi--;
if (pi >= min(a->pa, a->pb)) {
pi++;
if (pi >= max(a->pa, a->pb))
pi++;
}
print_recurse(s, a->prev, pathindex, pi, priority, assoc, child);
} else if (a->prev && index == a->pr &&
s->ops[a->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[a->po]->priority;
thisassoc = s->ops[a->po]->assoc;
parens = (thispri < priority ||
(thispri == priority && (assoc & child)));
if (parens)
putchar('(');
if (s->ops[a->po]->flags & OPFLAG_UNARYPREFIX)
for (op = s->ops[a->po]->text; *op; op++)
putchar(*op);
if (s->ops[a->po]->flags & OPFLAG_FN)
putchar('(');
print_recurse(s, a->prev, pathindex, a->pa, thispri, thisassoc, 1);
if (s->ops[a->po]->flags & OPFLAG_FN)
putchar(')');
if (!(s->ops[a->po]->flags & OPFLAG_UNARYPREFIX))
for (op = s->ops[a->po]->text; *op; op++)
putchar(*op);
if (!(s->ops[a->po]->flags & OPFLAG_UNARY))
print_recurse(s, a->prev, pathindex, a->pb, thispri, thisassoc, 2);
if (parens)
putchar(')');
} else {
/*
* This number is either an original, or something formed
* by a non-displayed operator (concatenation). Either way,
* we display it as is.
*/
printf("%d", ss->numbers[2*index]);
if (ss->numbers[2*index+1] != 1)
printf("/%d", ss->numbers[2*index+1]);
}
}
void print_recurse(struct sets *s, struct set *ss, int pathindex, int index,
int priority, int assoc, int child)
{
if (!ss->a.prev || pathindex < ss->a.prev->npaths) {
print_recurse_inner(s, ss, &ss->a, pathindex,
index, priority, assoc, child);
} else {
int i;
pathindex -= ss->a.prev->npaths;
for (i = 0; i < ss->nas; i++) {
if (pathindex < ss->as[i].prev->npaths) {
print_recurse_inner(s, ss, &ss->as[i], pathindex,
index, priority, assoc, child);
break;
}
pathindex -= ss->as[i].prev->npaths;
}
}
}
void print(int pathindex, struct sets *s, struct output *o)
{
print_recurse(s, o->set, pathindex, o->index, 0, 0, 0);
}
/*
* gcc -g -O0 -o numgame numgame.c -I.. ../{malloc,tree234,nullfe}.c -lm
*/
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;
int multiple = FALSE;
int debug_bfs = FALSE;
int got_range = FALSE, rangemin = 0, rangemax = 0;
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 if (*p == '-') {
p++;
if (!strcmp(p, "debug-bfs")) {
debug_bfs = TRUE;
} else {
fprintf(stderr, "%s: option '--%s' not recognised\n",
pname, p);
}
} else while (p && *p) switch (c = *p++) {
case 'C':
rules = &rules_countdown;
break;
case 'B':
rules = &rules_3388;
break;
case 'D':
rules = &rules_four4s;
break;
case 'A':
rules = &rules_anythinggoes;
break;
case 'v':
verbose = TRUE;
break;
case 'p':
pathcounts = TRUE;
break;
case 'm':
multiple = TRUE;
break;
case 't':
case 'r':
{
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;
case 'r':
{
char *sep = strchr(v, '-');
got_range = TRUE;
if (sep) {
rangemin = atoi(v);
rangemax = atoi(sep+1);
} else {
rangemin = 0;
rangemax = 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,-A\n", pname);
return 1;
}
if (!nnumbers) {
fprintf(stderr, "%s: no input numbers specified\n", pname);
return 1;
}
if (got_range) {
if (got_target) {
fprintf(stderr, "%s: only one of -t and -r may be specified\n", pname);
return 1;
}
if (rangemin >= rangemax) {
fprintf(stderr, "%s: range not sensible (%d - %d)\n", pname, rangemin, rangemax);
return 1;
}
}
s = do_search(nnumbers, numbers, rules, (got_target ? &target : NULL),
debug_bfs, multiple);
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 if (got_range) {
if (!findrelpos234(s->outputtree, &rangemin, outputfindcmp,
REL234_GE, &start) ||
!findrelpos234(s->outputtree, &rangemax, outputfindcmp,
REL234_LE, &limit)) {
printf("No solutions available in specified range %d-%d\n", rangemin, rangemax);
return 1;
}
limit++;
} else {
start = 0;
limit = count234(s->outputtree);
}
for (i = start; i < limit; i++) {
char buf[256];
o = index234(s->outputtree, i);
sprintf(buf, "%d", o->number);
if (pathcounts)
sprintf(buf + strlen(buf), " [%d]", o->npaths);
if (got_target || verbose) {
int j, npaths;
if (multiple)
npaths = o->npaths;
else
npaths = 1;
for (j = 0; j < npaths; j++) {
printf("%s = ", buf);
print(j, s, o);
putchar('\n');
}
} else {
printf("%s\n", buf);
}
}
free_sets(s);
return 0;
}
/* vim: set shiftwidth=4 tabstop=8: */
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