day17: code cleaning + results

This commit is contained in:
2021-12-27 11:28:10 +01:00
parent 7b7095b8d7
commit 8748eb0f1f
2 changed files with 29 additions and 92 deletions

View File

@@ -205,3 +205,15 @@ aoc-c : res=923
aoc-c : res=258888628940 aoc-c : res=258888628940
time: 0:00.00 real, 0.00 user, 0.00 sys time: 0:00.00 real, 0.00 user, 0.00 sys
context-switch: 0+1, page-faults: 0+89 context-switch: 0+1, page-faults: 0+89
=========================================
================= day17 =================
=========================================
aoc-c : res=2850
time: 0:00.00 real, 0.00 user, 0.00 sys
context-switch: 0+1, page-faults: 0+96
aoc-c : res=1117
time: 0:00.00 real, 0.00 user, 0.00 sys
context-switch: 1+1, page-faults: 0+95

View File

@@ -60,16 +60,6 @@ static s64 nth(s64 x, s64 n)
static s64 nth_x(s64 x0, s64 n) static s64 nth_x(s64 x0, s64 n)
{ {
return n >= x0? nth(x0, x0): nth(x0, n); return n >= x0? nth(x0, x0): nth(x0, n);
//return ((2 * x0) * -(n - 1)) / 2;
}
/* if we take hypothesis target y is always negative, we can exclude steps
* which are still over 0. We reach back zero after 2 x d1 +1 steps for
* initial d1 positive (0 otherwise).
*/
static s64 nth_yzero(s64 y0)
{
return y0 >= 0? y0 * 2 + 1 : 0;
} }
/* determine max possible initial dx : /* determine max possible initial dx :
@@ -85,25 +75,10 @@ static s64 dx_min(s64 xmin)
bounds.steps_min = 1; bounds.steps_min = 1;
bounds.dx_min = res; bounds.dx_min = res;
/*nmin = (s64) (-(1+2*res) +
sqrt((float)(1 + 2 * res)
* (float)(1 + 2 * res)
- 4 * (2 * res - xmin))) / 2;
*/
//nmin = ((float) 1 + (float)sqrt(1 + 4.0 * xmin)) / (float)2;
//printf("xmin=%ld res=%ld nmin=%ld\n", xmin, res, nmin);
//nmin = ((float) 1 - (float)sqrt((float)1 + 4.0 * xmin)) / (float)2;
//printf("xmin=%ld res=%ld nmin=%ld\n", xmin, res, nmin); //printf("xmin=%ld res=%ld nmin=%ld\n", xmin, res, nmin);
return res; return res;
} }
/*
static s64 max_y_steps(s64 xmax))
{
return (xmax - 1) / 2;
}
*/
/* The highest solution is the solution of: /* The highest solution is the solution of:
* y1 * (y1 +1) / 2 * y1 * (y1 +1) / 2
* (we can ignore x velocity, as we can reach any target x1 with x velocity * (we can ignore x velocity, as we can reach any target x1 with x velocity
@@ -116,7 +91,23 @@ static s64 part1()
static s64 part2() static s64 part2()
{ {
return target.y1 * (target.y1 + 1) / 2; s64 count = 0;
for (s64 dx = bounds.dx_min; dx <= bounds.dx_max ; ++dx) {
for (s64 dy = bounds.dy_min; dy <= bounds.dy_max; ++dy) {
for (s64 step = 1; ; step++) {
s64 newx = nth_x(dx, step);
s64 newy = nth(dy, step);
if (newx > target.x2 || newy < target.y1)
goto nexty;
if (hit(newx, newy)) {
count++;
goto nexty;
}
}
nexty:
}
}
return count;
} }
/* read input /* read input
@@ -158,78 +149,12 @@ int main(int ac, char **av)
return usage(*av); return usage(*av);
read_input(); read_input();
printf("%s : xmin=%ld ymin=%ld xmax=%ld ymax=%ld part1=%ld\n", *av,
target.x1, target.y1, target.x2, target.y2,
part1());
//s64 dxmin = dx_min(target.x1);
//for (s64 i = 0; i < 10; ++i) {
dx_min(target.x1); dx_min(target.x1);
bounds.dx_max = target.x2; bounds.dx_max = target.x2;
bounds.dy_min = target.y1; bounds.dy_min = target.y1;
/* y will comme back at zero with same dy as initial one. next step will be d0+1.
* If d0+1 is > min target y, we will never reach target.
*/
bounds.dy_max = -target.y1; bounds.dy_max = -target.y1;
printf("yzero(%ld) = %ld xmin(%ld) = %ld\n",
target.x2, nth_yzero(target.x2),
target.x1, dx_min(target.x1));
/* loop on x initial acceletation */
s64 count = 0, besty=0;
for (s64 dx = bounds.dx_min; dx <= bounds.dx_max ; ++dx) {
/* maybe here make steps for x only, and find all valid steps.
* need to know if dx becomes zero and too low
*/
for (s64 dy = bounds.dy_min; dy <= bounds.dy_max; ++dy) {
s64 maxy = 0;
for (s64 step = 1; ; step++) {
s64 newx = nth_x(dx, step);
s64 newy = nth(dy, step);
printf("dx=%ld dy=%ld step=%ld x=%ld y=%ld\n", dx, dy, step, newx, newy);
if (newx > target.x2) {
printf("\tnext y1\n");
goto nexty;
}
if (newy < target.y1) {
printf("\t\tnext y2\n");
goto nexty;
}
if (newy > maxy) {
maxy = newy;
printf("new maxy = %ld\n", maxy);
}
/* need to check if x can join minx */
/* need to find max y bound */
if (hit(newx, newy)) {
printf("\tHIT: %ld,%ld\n", newx, newy);
count++;
if (maxy > besty) {
besty = maxy;
printf("new besty = %ld\n", besty);
}
goto nexty;
}
}
nexty:
}
nextx:
}
printf("count=%ld besty=%ld\n", count, besty);
exit (0);
/*
for (s64 dx = 0; dx <=10; ++dx) {
printf("x0=%2ld: ", dx);
for (int i = 0; i < 15; ++i)
printf(" %d=%ld/%ld", i, nth_x(dx, i), nth(dx, i));
printf("\n");
}
*/
printf("%s : res=%ld\n", *av, part == 1? part1(): part2(&target)); printf("%s : res=%ld\n", *av, part == 1? part1(): part2(&target));
exit (0); exit (0);
} }