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  • 图形学基础之画心形和圣诞树 标签: pythonc语言图形 2017-04-23 16:13 78人阅读 评论(0)

    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

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  • 原文地址:https://www.cnblogs.com/actanble/p/7128678.html
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