James H has helpfully provided yet more silly operators for the -A

mode. I think some user-defined ruleset configuration options are
now required...

[originally from svn r8092]
This commit is contained in:
Simon Tatham
2008-06-24 20:58:35 +00:00
parent 49a077728f
commit 1df94d233a

View File

@ -130,7 +130,7 @@ struct sets {
#define OPFLAG_NEEDS_CONCAT 1
#define OPFLAG_KEEPS_CONCAT 2
#define OPFLAG_UNARY 4
#define OPFLAG_UNARYPFX 8
#define OPFLAG_UNARYPREFIX 8
struct operation {
/*
@ -308,6 +308,26 @@ static int perform_exact_div(int *a, int *b, int *output)
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;
@ -338,18 +358,8 @@ static int perform_concat(int *a, int *b, int *output)
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 */
if (!max_p10(b[0], &p10)) return FALSE;
MUL(t1, p10, a[0]);
ADD(t2, t1, b[0]);
OUT(output, t2, 1);
@ -370,7 +380,7 @@ static int perform_concat(int *a, int *b, int *output)
static int perform_exp(int *a, int *b, int *output)
{
int an, ad, xn, xd, limit, t, i;
int an, ad, xn, xd;
/*
* Exponentiation is permitted if the result is rational. This
@ -384,8 +394,8 @@ static int perform_exp(int *a, int *b, int *output)
* - then we multiply by itself (numerator-of-b) times.
*/
if (b[1] > 1) {
an = 0.5 + pow(a[0], 1.0/b[1]);
ad = 0.5 + pow(a[1], 1.0/b[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])
@ -435,6 +445,72 @@ static int perform_factorial(int *a, int *b, int *output)
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;
}
const static struct operation op_add = {
TRUE, "+", "+", 0, 10, 0, TRUE, perform_add
};
@ -460,6 +536,15 @@ const static struct operation op_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
};
/*
* In Countdown, divisions resulting in fractions are disallowed.
@ -500,7 +585,8 @@ const static struct rules rules_four4s = {
* 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, NULL
&op_add, &op_mul, &op_sub, &op_div, &op_concat, &op_exp, &op_factorial,
&op_decimal, &op_recur, &op_root, NULL
};
const static struct rules rules_anythinggoes = {
ops_anythinggoes, TRUE
@ -744,7 +830,7 @@ static struct sets *do_search(int ninputs, int *inputs,
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("/%d", ss->numbers[2*i+1]);
}
printf("\n");
}
@ -809,11 +895,16 @@ static struct sets *do_search(int ninputs, int *inputs,
addset(s, sn, multiple, ss, pa, po, pb, pr);
if (debug) {
int i;
printf(" %d %s %d ->", pa, ops[po]->dbgtext, pb);
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("/%d", sn->numbers[2*i+1]);
}
printf("\n");
}
@ -895,13 +986,13 @@ void print_recurse_inner(struct sets *s, struct set *ss,
if (parens)
putchar('(');
if (s->ops[a->po]->flags & OPFLAG_UNARYPFX)
if (s->ops[a->po]->flags & OPFLAG_UNARYPREFIX)
for (op = s->ops[a->po]->text; *op; op++)
putchar(*op);
print_recurse(s, a->prev, pathindex, a->pa, thispri, thisassoc, 1);
if (!(s->ops[a->po]->flags & OPFLAG_UNARYPFX))
if (!(s->ops[a->po]->flags & OPFLAG_UNARYPREFIX))
for (op = s->ops[a->po]->text; *op; op++)
putchar(*op);