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Move mul_root3 out into misc.c and generalise it.

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I'm going to want to reuse it for sqrt(5) as well as sqrt(3) soon.
This commit is contained in:
Simon Tatham
2023-07-02 21:22:02 +01:00
parent ad7042db98
commit 6b5142a7a9
3 changed files with 127 additions and 127 deletions
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129
grid.c
View File

@ -3669,131 +3669,6 @@ struct spectrecontext {
tree234 *points;
};
/*
* Calculate the nearest integer to n*sqrt(3), via a bitwise algorithm
* that avoids floating point.
*
* (It would probably be OK in practice to use floating point, but I
* felt like overengineering it for fun. With FP, there's at least a
* theoretical risk of rounding the wrong way, due to the three
* successive roundings involved - rounding sqrt(3), rounding its
* product with n, and then rounding to the nearest integer. This
* approach avoids that: it's exact.)
*/
static int mul_root3(int n_signed)
{
unsigned x, r, m;
int sign = n_signed < 0 ? -1 : +1;
unsigned n = n_signed * sign;
unsigned bitpos;
/*
* Method:
*
* We transform m gradually from zero into n, by multiplying it by
* 2 in each step and optionally adding 1, so that it's always
* floor(n/2^something).
*
* At the start of each step, x is the largest integer less than
* or equal to m*sqrt(3). We transform m to 2m+bit, and therefore
* we must transform x to 2x+something to match. The 'something'
* we add to 2x is at most 3. (Worst case is if m sqrt(3) was
* equal to x + 1-eps for some tiny eps, and then the incoming bit
* of m is 1, so that (2m+1)sqrt(3) = 2x+2+2eps+sqrt(3), i.e.
* about 2x + 3.732...)
*
* To compute this, we also track the residual value r such that
* x^2+r = 3m^2.
*
* The algorithm below is very similar to the usual approach for
* taking the square root of an integer in binary. The wrinkle is
* that we have an integer multiplier, i.e. we're computing
* P*sqrt(Q) (with P=n and Q=3 in this case) rather than just
* sqrt(Q). Of course in principle we could just take sqrt(P^2Q),
* but we'd need an integer twice the width to hold P^2. Pulling
* out P and treating it specially makes overflow less likely.
*/
x = r = m = 0;
for (bitpos = UINT_MAX & ~(UINT_MAX >> 1); bitpos; bitpos >>= 1) {
unsigned a, b = (n & bitpos) ? 1 : 0;
/*
* Check invariants. We expect that x^2 + r = 3m^2 (i.e. our
* residual term is correct), and also that r < 2x+1 (because
* if not, then we could replace x with x+1 and still get a
* value that made r non-negative, i.e. x would not be the
* _largest_ integer less than m sqrt(3)).
*/
assert(x*x + r == 3*m*m);
assert(r < 2*x+1);
/*
* We're going to replace m with 2m+b, and x with 2x+a for
* some a we haven't decided on yet.
*
* The new value of the residual will therefore be
*
* 3 (2m+b)^2 - (2x+a)^2
* = (12m^2 + 12mb + 3b^2) - (4x^2 + 4xa + a^2)
* = 4 (3m^2 - x^2) + 12mb + 3b^2 - 4xa - a^2
* = 4r + 12mb + 3b^2 - 4xa - a^2 (because r = 3m^2 - x^2)
* = 4r + (12m + 3)b - 4xa - a^2 (b is 0 or 1, so b = b^2)
*/
for (a = 0; a < 4; a++) {
/* If we made this routine handle square roots of numbers
* other than 3 then it would be sensible to make this a
* binary search. Here, it hardly seems important. */
unsigned pos = 4*r + b*(12*m + 3);
unsigned neg = 4*a*x + a*a;
if (pos < neg)
break; /* this value of a is too big */
}
/* The above loop will have terminated with a one too big,
* whether that's because we hit the break statement or fell
* off the end with a=4. So now decrementing a will give us
* the right value to add. */
a--;
r = 4*r + b*(12*m + 3) - (4*a*x + a*a);
m = 2*m+b;
x = 2*x+a;
}
/*
* Finally, round to the nearest integer. At present, x is the
* largest integer that is _at most_ m sqrt(3). But we want the
* _nearest_ integer, whether that's rounded up or down. So check
* whether (x + 1/2) is still less than m sqrt(3), i.e. whether
* (x + 1/2)^2 < 3m^2; if it is, then we increment x.
*
* We have 3m^2 - (x + 1/2)^2 = 3m^2 - x^2 - x - 1/4
* = r - x - 1/4
*
* and since r and x are integers, this is greater than 0 if and
* only if r > x.
*
* (There's no need to worry about tie-breaking exact halfway
* rounding cases. sqrt(3) is irrational, so none such exist.)
*/
if (r > x)
x++;
/*
* Put the sign back on, and convert back from unsigned to int.
*/
if (sign == +1) {
return x;
} else {
/* Be a little careful to avoid compilers deciding I've just
* perpetrated signed-integer overflow. This should optimise
* down to no actual code. */
return INT_MIN + (int)(-x - (unsigned)INT_MIN);
}
}
static void grid_spectres_callback(void *vctx, const int *coords)
{
struct spectrecontext *ctx = (struct spectrecontext *)vctx;
@ -3804,9 +3679,9 @@ static void grid_spectres_callback(void *vctx, const int *coords)
grid_dot *d = grid_get_dot(
ctx->g, ctx->points,
(coords[4*i+0] * SPECTRE_UNIT +
mul_root3(coords[4*i+1] * SPECTRE_UNIT)),
n_times_root_k(coords[4*i+1] * SPECTRE_UNIT, 3)),
(coords[4*i+2] * SPECTRE_UNIT +
mul_root3(coords[4*i+3] * SPECTRE_UNIT)));
n_times_root_k(coords[4*i+3] * SPECTRE_UNIT, 3)));
grid_face_set_dot(ctx->g, d, i);
}
}

124
misc.c
View File

@ -536,4 +536,128 @@ char *make_prefs_path(const char *dir, const char *sep,
return path;
}
/*
* Calculate the nearest integer to n*sqrt(k), via a bitwise algorithm
* that avoids floating point.
*
* (It would probably be OK in practice to use floating point, but I
* felt like overengineering it for fun. With FP, there's at least a
* theoretical risk of rounding the wrong way, due to the three
* successive roundings involved - rounding sqrt(k), rounding its
* product with n, and then rounding to the nearest integer. This
* approach avoids that: it's exact.)
*/
int n_times_root_k(int n_signed, int k)
{
unsigned x, r, m;
int sign = n_signed < 0 ? -1 : +1;
unsigned n = n_signed * sign;
unsigned bitpos;
/*
* Method:
*
* We transform m gradually from zero into n, by multiplying it by
* 2 in each step and optionally adding 1, so that it's always
* floor(n/2^something).
*
* At the start of each step, x is the largest integer less than
* or equal to m*sqrt(k). We transform m to 2m+bit, and therefore
* we must transform x to 2x+something to match. The 'something'
* we add to 2x is at most floor(sqrt(k))+2. (Worst case is if m
* sqrt(k) was equal to x + 1-eps for some tiny eps, and then the
* incoming bit of m is 1, so that (2m+1)sqrt(k) =
* 2x+2+sqrt(k)-2eps.)
*
* To compute this, we also track the residual value r such that
* x^2+r = km^2.
*
* The algorithm below is very similar to the usual approach for
* taking the square root of an integer in binary. The wrinkle is
* that we have an integer multiplier, i.e. we're computing
* n*sqrt(k) rather than just sqrt(k). Of course in principle we
* could just take sqrt(n^2k), but we'd need an integer twice the
* width to hold n^2. Pulling out n and treating it specially
* makes overflow less likely.
*/
x = r = m = 0;
for (bitpos = UINT_MAX & ~(UINT_MAX >> 1); bitpos; bitpos >>= 1) {
unsigned a, b = (n & bitpos) ? 1 : 0;
/*
* Check invariants. We expect that x^2 + r = km^2 (i.e. our
* residual term is correct), and also that r < 2x+1 (because
* if not, then we could replace x with x+1 and still get a
* value that made r non-negative, i.e. x would not be the
* _largest_ integer less than m sqrt(k)).
*/
assert(x*x + r == k*m*m);
assert(r < 2*x+1);
/*
* We're going to replace m with 2m+b, and x with 2x+a for
* some a we haven't decided on yet.
*
* The new value of the residual will therefore be
*
* k (2m+b)^2 - (2x+a)^2
* = (4km^2 + 4kmb + kb^2) - (4x^2 + 4xa + a^2)
* = 4 (km^2 - x^2) + 4kmb + kb^2 - 4xa - a^2
* = 4r + 4kmb + kb^2 - 4xa - a^2 (because r = km^2 - x^2)
* = 4r + (4m + 1)kb - 4xa - a^2 (b is 0 or 1, so b = b^2)
*/
for (a = 0;; a++) {
/* If we made this routine handle square roots of numbers
* significantly bigger than 3 or 5 then it would be
* sensible to make this a binary search. Here, it hardly
* seems important. */
unsigned pos = 4*r + k*b*(4*m + 1);
unsigned neg = 4*a*x + a*a;
if (pos < neg)
break; /* this value of a is too big */
}
/* The above loop will have terminated with a one too big. So
* now decrementing a will give us the right value to add. */
a--;
r = 4*r + b*k*(4*m + 1) - (4*a*x + a*a);
m = 2*m+b;
x = 2*x+a;
}
/*
* Finally, round to the nearest integer. At present, x is the
* largest integer that is _at most_ m sqrt(k). But we want the
* _nearest_ integer, whether that's rounded up or down. So check
* whether (x + 1/2) is still less than m sqrt(k), i.e. whether
* (x + 1/2)^2 < km^2; if it is, then we increment x.
*
* We have km^2 - (x + 1/2)^2 = km^2 - x^2 - x - 1/4
* = r - x - 1/4
*
* and since r and x are integers, this is greater than 0 if and
* only if r > x.
*
* (There's no need to worry about tie-breaking exact halfway
* rounding cases. sqrt(k) is irrational, so none such exist.)
*/
if (r > x)
x++;
/*
* Put the sign back on, and convert back from unsigned to int.
*/
if (sign == +1) {
return x;
} else {
/* Be a little careful to avoid compilers deciding I've just
* perpetrated signed-integer overflow. This should optimise
* down to no actual code. */
return INT_MIN + (int)(-x - (unsigned)INT_MIN);
}
}
/* vim: set shiftwidth=4 tabstop=8: */

1
puzzles.h
View File

@ -391,6 +391,7 @@ void obfuscate_bitmap(unsigned char *bmp, int bits, bool decode);
char *fgetline(FILE *fp);
char *make_prefs_path(const char *dir, const char *sep,
const game *game, const char *suffix);
int n_times_root_k(int n, int k);
/* allocates output each time. len is always in bytes of binary data.
* May assert (or just go wrong) if lengths are unchecked. */