图形学基础之画心形和圣诞树

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 ' '),        printdef 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('$:-+*#%@'[int(-8*f-1)] if f <= 0 else ' '),        printdef star3():    from math import sqrt    def f(x,y,z):        a = x * x + 9.0/ 4.0* y * y + z * z - 1        return a*a*a - x*x * z*z*z - 9.0/80.0 * y*y * z*z *z    def h(x,z):        for y in [1 - 0.001*i for i in range(1001)]:            if f(x,y,z <= 0):                return y        return 0    for z in [1.5-0.1*i for i in range(30)]:        for x in [0.025*i-1.5 for i in range(120)]:            v = f(x, 0, z)            if(v <= 0):                y0 = h(x, z)                ny = 0.01                nx = h(x + ny, z) - y0                nz = h(x, z + ny) - y0                nd = 1.0/ sqrt(nx * nx + ny * ny + nz * nz)                d = (nx + ny - nz) * nd * 0.5 + 0.5                print '{}'.format('$:-+*#%@'[int(5*d)]),            else:                print ' ',        printstar3()

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\n%d %d\n255\n", 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('\n', 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 3.14159265359ffloat 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('\n'))        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