前言
基本描述
本文主要是为了了解图形基础, 非常仓促; 后期详细学习再补上
参考文章
用C++实现圣诞树http://www.codeceo.com/article/c-christmas-tree.html
用C++心形https://www.zhihu.com/question/27015321/answer/35028446
实现过程
Python实现心形
star3 的 d 生成出现了问题， 有兴趣的可以研究下给我留言
def star1():
for y in [1.5-0.1*i for i in range(30)]:
for x in [0.05*i-1.5 for i in range(60)]:
a = x*x + y*y -1
print '{}'.format('*' if a*a*a - x*x*y*y*y <= 0 else ' '),
print
def star2():
for y in [1.5-0.1*i for i in range(30)]:
for x in [0.05*i-1.5 for i in range(60)]:
a = x*x + y*y -1 #temp value as last example
f = a*a*a - x*x*y*y*y # compared with 0
print '{}'.format('$:-+*#%@'[int(-8*f-1)] if f <= 0 else ' '),
print
def star3():
from math import sqrt
def f(x,y,z):
a = x * x + 9.0/ 4.0* y * y + z * z - 1
return a*a*a - x*x * z*z*z - 9.0/80.0 * y*y * z*z *z
def h(x,z):
for y in [1 - 0.001*i for i in range(1001)]:
if f(x,y,z <= 0):
return y
return 0
for z in [1.5-0.1*i for i in range(30)]:
for x in [0.025*i-1.5 for i in range(120)]:
v = f(x, 0, z)
if(v <= 0):
y0 = h(x, z)
ny = 0.01
nx = h(x + ny, z) - y0
nz = h(x, z + ny) - y0
nd = 1.0/ sqrt(nx * nx + ny * ny + nz * nz)
d = (nx + ny - nz) * nd * 0.5 + 0.5
print '{}'.format('$:-+*#%@'[int(5*d)]),
else:
print ' ',
print
star3()
C++实现心形->存到本地
#ifdef _MSC_VER
#define _CRT_SECURE_NO_WARNINGS
#endif
#include <stdio.h>
#include <math.h>
float f(float x, float y, float z)
{
float a = x * x + 9.0f / 4.0f * y * y + z * z - 1;
return a * a * a - x * x * z * z * z - 9.0f / 80.0f * y * y * z * z * z;
}
float h(float x, float z)
{
for (float y = 1.0f; y >= 0.0f; y -= 0.001f)
if (f(x, y, z) <= 0.0f)
return y;
return 0.0f;
}
int main()
{
FILE* fp = fopen("heart.ppm", "w");
int sw = 512, sh = 512;
fprintf(fp, "P3
%d %d
255
", sw, sh);
for (int sy = 0; sy < sh; sy++)
{
float z = 1.5f - sy * 3.0f / sh;
for (int sx = 0; sx < sw; sx++)
{
float x = sx * 3.0f / sw - 1.5f;
float v = f(x, 1.0f, z); int r = 0;
if (v <= 0.0f) {
float y0 = h(x, z);
float ny = 0.001f;
float nx = h(x + ny, z) - y0;
float nz = h(x, z + ny) - y0;
float nd = 1.0f / sqrtf(nx * nx + ny * ny + nz * nz);
float d = (nx + ny - nz) / sqrtf(3) * nd * 0.5f + 0.5f;
r = (int)(d * 255.0f);
}
fprintf(fp, "%d 0 0 ", r);
}
fputc('
', fp);
}
fclose(fp);
}
心形所用到技术: 分形和简单的数学图形基础,
Ｃ++实现圣诞树的最终源码
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define PI 3.14159265359f
float sx, sy;
typedef float Mat[4][4];
typedef float Vec[4];
void scale(Mat* m, float s) {
Mat temp = { {s,0,0,0}, {0,s,0,0 }, { 0,0,s,0 }, { 0,0,0,1 } };
memcpy(m, &temp, sizeof(Mat));
}
void rotateY(Mat* m, float t) {
float c = cosf(t), s = sinf(t);
Mat temp = { {c,0,s,0}, {0,1,0,0}, {-s,0,c,0}, {0,0,0,1} };
memcpy(m, &temp, sizeof(Mat));
}
void rotateZ(Mat* m, float t) {
float c = cosf(t), s = sinf(t);
Mat temp = { {c,-s,0,0}, {s,c,0,0}, {0,0,1,0}, {0,0,0,1} };
memcpy(m, &temp, sizeof(Mat));
}
void translate(Mat* m, float x, float y, float z) {
Mat temp = { {1,0,0,x}, {0,1,0,y}, {0,0,1,z}, {0,0,0,1} };
memcpy(m, &temp, sizeof(Mat));
}
void mul(Mat* m, Mat a, Mat b) {
Mat temp;
for (int j = 0; j < 4; j++)
for (int i = 0; i < 4; i++) {
temp[j][i] = 0.0f;
for (int k = 0; k < 4; k++)
temp[j][i] += a[j][k] * b[k][i];
}
memcpy(m, &temp, sizeof(Mat));
}
void transformPosition(Vec* r, Mat m, Vec v) {
Vec temp = { 0, 0, 0, 0 };
for (int j = 0; j < 4; j++)
for (int i = 0; i < 4; i++)
temp[j] += m[j][i] * v[i];
memcpy(r, &temp, sizeof(Vec));
}
float transformLength(Mat m, float r) {
return sqrtf(m[0][0] * m[0][0] + m[0][1] * m[0][1] + m[0][2] * m[0][2]) * r;
}
float sphere(Vec c, float r) {
float dx = c[0] - sx, dy = c[1] - sy;
float a = dx * dx + dy * dy;
return a < r * r ? sqrtf(r * r - a) + c[2] : -1.0f;
}
float opUnion(float z1, float z2) {
return z1 > z2 ? z1 : z2;
}
float f(Mat m, int n) {
float z = -1.0f;
for (float r = 0.0f; r < 0.8f; r += 0.02f) {
Vec v = { 0.0f, r, 0.0f, 1.0f };
transformPosition(&v, m, v);
z = opUnion(z, sphere(v, transformLength(m, 0.05f * (0.95f - r))));
}
if (n > 0) {
Mat ry, rz, s, t, m2, m3;
rotateZ(&rz, 1.8f);
for (int p = 0; p < 6; p++) {
rotateY(&ry, p * (2 * PI / 6));
mul(&m2, ry, rz);
float ss = 0.45f;
for (float r = 0.2f; r < 0.8f; r += 0.1f) {
scale(&s, ss);
translate(&t, 0.0f, r, 0.0f);
mul(&m3, s, m2);
mul(&m3, t, m3);
mul(&m3, m, m3);
z = opUnion(z, f(m3, n - 1));
ss *= 0.8f;
}
}
}
return z;
}
float f0(float x, float y, int n) {
sx = x;
sy = y;
Mat m;
scale(&m, 1.0f);
return f(m, n);
}
int main(int argc, char* argv[]) {
int n = argc > 1 ? atoi(argv[1]) : 3;
float zoom = argc > 2 ? atof(argv[2]) : 1.0f;
for (float y = 0.8f; y > -0.0f; y -= 0.02f / zoom, putchar('
'))
for (float x = -0.35f; x < 0.35f; x += 0.01f / zoom) {
float z = f2(x, y, n);
if (z > -1.0f) {
float nz = 0.001f;
float nx = f0(x + nz, y, n) - z;
float ny = f0(x, y + nz, n) - z;
float nd = sqrtf(nx * nx + ny * ny + nz * nz);
float d = (nx - ny + nz) / sqrtf(3) / nd;
d = d > 0.0f ? d : 0.0f;
// d = d < 1.0f ? d : 1.0f;
putchar(".-:=+*#%@@"[(int)(d * 9.0f)]);
}
else
putchar(' ');
}
}说明
只是今天偶尔看到了， 记录下来; 没有特别含义， 以后深入学习了图形学再来深入探讨, 感兴趣的话， 相关内容可以到对应的承接页进行浏览学习
后记
略