我正在寻找最酷的事情,你可以在几行简单的代码。我相信你可以用Haskell用15行写一个Mandelbrot集,但是很难理解。

我的目标是启发学生编程很酷。

我们知道编程很酷,因为你可以创造任何你想象到的东西——它是最终的创意出口。我想激励这些初学者,让他们尽可能多地克服早期学习的困难。

Now, my reasons are selfish. I'm teaching an Intro to Computing course to a group of 60 half-engineering, half business majors; all freshmen. They are the students who came from underprivileged High schools. From my past experience, the group is generally split as follows: a few rock-stars, some who try very hard and kind of get it, the few who try very hard and barely get it, and the few who don't care. I want to reach as many of these groups as effectively as I can. Here's an example of how I'd use a computer program to teach:

Here's an example of what I'm looking for: a 1-line VBS script to get your computer to talk to you: CreateObject("sapi.spvoice").Speak InputBox("Enter your text","Talk it") I could use this to demonstrate order of operations. I'd show the code, let them play with it, then explain that There's a lot going on in that line, but the computer can make sense of it, because it knows the rules. Then I'd show them something like this: 4(5*5) / 10 + 9(.25 + .75) And you can see that first I need to do is (5*5). Then I can multiply for 4. And now I've created the Object. Dividing by 10 is the same as calling Speak - I can't Speak before I have an object, and I can't divide before I have 100. Then on the other side I first create an InputBox with some instructions for how to display it. When I hit enter on the input box it evaluates or "returns" whatever I entered. (Hint: 'oooooo' makes a funny sound) So when I say Speak, the right side is what to Speak. And I get that from the InputBox. So when you do several things on a line, like: x = 14 + y; You need to be aware of the order of things. First we add 14 and y. Then we put the result (what it evaluates to, or returns) into x.

这就是我的目标,有一堆很酷的例子来演示,并在他们玩得开心的时候教他们。我在我室友身上尝试了这个例子,虽然我可能不会把它作为第一课,但她喜欢它,并学到了一些东西。

一些很酷的数学程序,可以做出漂亮的图形或容易理解的形状,这是很好的想法,我将研究这些。这里有一些复杂的actionscript示例,但这有点太高级了,我不能教flash。你还有什么其他的想法吗?


当前回答

这个c代码可能有点混乱,但我发现它非常强大

#include <unistd.h>
float o=0.075,h=1.5,T,r,O,l,I;int _,L=80,s=3200;main(){for(;s%L||
(h-=o,T= -2),s;4 -(r=O*O)<(l=I*I)|++ _==L&&write(1,(--s%L?_<L?--_
%6:6:7)+"World! \n",1)&&(O=I=l=_=r=0,T+=o /2))O=I*2*O+h,I=l+T-r;}

结果是这样的…只有3行…… 一种分形的Hello World…

WWWWWWWWWWWWWWWWooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo
WWWWWWWWWWWWWWooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo
WWWWWWWWWWWWWooooooooooooooooorrrrrrrrrrrrrrrrrrrrroooooooooooooooooooooooooooo
WWWWWWWWWWWoooooooooooorrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrooooooooooooooooooooo
WWWWWWWWWWooooooooorrrrrrrrrrrrrrrrrrrrrrrllllld!!ddllllrrrrrrooooooooooooooooo
WWWWWWWWoooooooorrrrrrrrrrrrrrrrrrrrrrllllllldd!oWW!!dllllllrrrrroooooooooooooo
WWWWWWWoooooorrrrrrrrrrrrrrrrrrrrrrlllllllldddd!orro!o!dllllllrrrrrrooooooooooo
WWWWWWooooorrrrrrrrrrrrrrrrrrrrrllllllllldddd!WorddddoW!ddllllllrrrrrrooooooooo
WWWWWoooorrrrrrrrrrrrrrrrrrrrrlllllllllddd!!!o!!!   !dWW!ddddllllrrrrrrrooooooo
WWWWooorrrrrrrrrrrrrrrrrrrrllllllllldd!!!!WWWoo      WloW!!!ddddllrrrrrrrrooooo
WWWWoorrrrrrrrrrrrrrrrrrrlllllllddddWldolrrlo!Wl     r!dlooWWWoW!dllrrrrrrroooo
WWWoorrrrrrrrrrrrrrrrrlllllddddddd!!Wdo  l!               rdo!l!r!dlrrrrrrrrooo
WWoorrrrrrrrrrrrrrrlllddddddddd!!!!oolWW                       lW!ddlrrrrrrrroo
WWorrrrrrrrrrrrllld!!!!!dddd!!!!WWrd !                        rlW!ddllrrrrrrrro
Worrrrrrrllllllddd!oooWWWoloWWWWoodr                           drrWdlllrrrrrrrr
Worrrlllllllldddd!WolWrr!!dWWWlrrldr                            ro!dlllrrrrrrrr
Wrrllllllllddddd!WWolWr        oWoo                              r!dllllrrrrrrr
Wlllllllldddd!!odrrdW            o                              lWddllllrrrrrrr
Wlddddd!!!!!WWordlWrd                                          oW!ddllllrrrrrrr
olddddd!!!!!WWordlWrd                                          oW!ddllllrrrrrrr
Wlllllllldddd!!odrrdW            o                              lWddllllrrrrrrr
Wrrllllllllddddd!WWolWr        oWoo                              r!dllllrrrrrrr
Worrrlllllllldddd!WolWrr!!dWWWlrrldr                            ro!dlllrrrrrrrr
Worrrrrrrllllllddd!oooWWWoloWWWWoodr                           droWdlllrrrrrrrr
WWorrrrrrrrrrrrllld!!!!!dddd!!!!WWrd !                        rlW!ddllrrrrrrrro
WWoorrrrrrrrrrrrrrrlllddddddddd!!!!oolWW                       lW!ddlrrrrrrrroo
WWWoorrrrrrrrrrrrrrrrrlllllddddddd!!Wdo  l!               rdo!l!r!dlrrrrrrrrooo
WWWWoorrrrrrrrrrrrrrrrrrrlllllllddddWldolrrlo!Wl     r!dlooWWWoW!dllrrrrrrroooo
WWWWooorrrrrrrrrrrrrrrrrrrrllllllllldd!!!!WWWoo      WloW!!!ddddllrrrrrrrrooooo
WWWWWoooorrrrrrrrrrrrrrrrrrrrrlllllllllddd!!!o!!!   WdWW!ddddllllrrrrrrrooooooo
WWWWWWooooorrrrrrrrrrrrrrrrrrrrrllllllllldddd!WorddddoW!ddllllllrrrrrrooooooooo
WWWWWWWoooooorrrrrrrrrrrrrrrrrrrrrrlllllllldddd!orro!o!dllllllrrrrrrooooooooooo
WWWWWWWWoooooooorrrrrrrrrrrrrrrrrrrrrrllllllldd!oWW!!dllllllrrrrroooooooooooooo
WWWWWWWWWWooooooooorrrrrrrrrrrrrrrrrrrrrrrllllld!!ddllllrrrrrrooooooooooooooooo
WWWWWWWWWWWoooooooooooorrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrrooooooooooooooooooooo
WWWWWWWWWWWWWooooooooooooooooorrrrrrrrrrrrrrrrrrrrroooooooooooooooooooooooooooo
WWWWWWWWWWWWWWooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo
WWWWWWWWWWWWWWWWooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo
WWWWWWWWWWWWWWWWWWWoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo
WWWWWWWWWWWWWWWWWWWWWoooooooooooooooooooooooooooooooooooooooooooooooooooooooooo

其他回答

不管你有什么想法,作为补充,我说你应该告诉他们如何做一些基本的数学。将其呈现为

现在你可能认为这很简单或者 复杂……但你有没有去过 被数学作业困住了?”

然后从别人的书中拿出一个例子。大多数数学问题可以用10行来解决,因为它很可能是一个简单的问题。然后告诉他们,花10分钟来解决这个问题,可能会让他们得到一个A。这段时间很长,但你可能会遇到一些想花很少或根本不花时间做作业的人。

这主要是因为我希望我在化学时就能想到写一个软件程序……所有的测验和作业都是100分…

编辑: 回应Peter的评论:

比如3a2的导数是什么。所以你可以只显示一个简单的函数,他们可以从命令行调用:

public int SimpleDerivative(int r, int exponent){
    r = r * exponent
    exponent =- 1
    return (String "{0}a^{1}" where {0} = r, {1} = exponent)
}

我认为油脂猴子是学生开始学习的好地方。在userscripts.org上有成千上万的示例脚本,是非常好的阅读材料,其中一些非常小。Greasemonkey脚本会影响网页,而学生们即使不会操纵网页,也会对这些网页很熟悉。Greasemonkey本身提供了一种非常简单的方法来在测试时编辑和启用/禁用脚本。

例如,下面是"谷歌Two Columns"脚本:

result2 = '<table width="100%" align="center" cellpadding="10" style="font-size:12px">';
gEntry = document.evaluate("//li[@class='g'] | //div[@class='g'] | //li[@class='g w0'] | //li[@class='g s w0']",document,null,XPathResult.UNORDERED_NODE_SNAPSHOT_TYPE,null);
for (var i = 0; i < gEntry.snapshotLength; i++) {
  if (i==0) { var sDiv = gEntry.snapshotItem(i).parentNode.parentNode; }
  if(i%2 == 0) { result2 += '<tr><td width="50%" valign="top">'+gEntry.snapshotItem(i).innerHTML+'</td>'; }
  if(i%2 == 1) { result2 += '<td width="50%" valign="top">'+gEntry.snapshotItem(i).innerHTML+'</td></tr>'; }
}
sDiv.innerHTML = result2+'</table>';

if (document.getElementById('mbEnd') !== null) { document.getElementById('mbEnd').style.display = 'none'; }

这是作弊,甚至一点都不简单,但我曾经用c++写了一个20行的射击游戏,使用Allegro图形库。对于一条线是什么,没有真正的标准,但它是有点过时了,它纯粹是为了好玩而创造的。甚至还有粗糙的音效。

这是它的样子:

20行http://img227.imageshack.us/img227/8770/20linesxx0.png

下面是代码(应该编译):

bool inside(int x, int y, int x2, int y2) { return (x>x2&&x<x2+20&&y>y2&&y<y2+10); }
int main() {
  BITMAP* buffer;
  float px,shotx,shoty,monstars[8],first,rnd,pressed,points = 0, maxp = 0;
  unsigned char midi[5] = {0xC0,127,0x90,25,0x54}, plgfx[] = {0,0,0,10,3,10,3,5,6,5,6,10,8,12,10,10,10,5,13,5,13,10,16,10,16,0,13,0,13,2,3,2,3,0,0,0}, mongfx[] = {0,0, 10,5, 20,0, 17,8, 15,6, 10,16, 5,6, 3,8, 0,0};
  allegro_init(), set_color_depth(32), set_gfx_mode(GFX_AUTODETECT_WINDOWED,320,240,0,0), install_timer(), install_keyboard(),  install_mouse(), buffer = create_bitmap(320,240),srand(time(NULL)),install_sound(DIGI_AUTODETECT, MIDI_AUTODETECT,""),clear_to_color(buffer,makecol32(100,100,255));
    while ((pressed=(!key[KEY_Z]&&pressed)?0:pressed)?1:1&&(((shoty=key[KEY_Z]&&shoty<0&&pressed==0?(pressed=1?200:200):first==0?-1:shoty)==200?shotx=px+9:0)==9999?1:1) && 1+(px += key[KEY_LEFT]?-0.1:0 + key[KEY_RIGHT]?0.1:0) && 1+int(px=(px<0?0:(px>228?228:px))) && !key[KEY_ESC]) {
    rectfill(buffer,0,0,244,240,makecol32(0,0,0));
    for(int i=0;i<8;i++) if (inside(shotx,shoty,i*32,monstars[i])) midi_out(midi,5);
        for (int i=0; i<8; monstars[i] += first++>8?(monstars[i]==-100?0:0.02):-100, points = monstars[i]>240?points-1:points, monstars[i]=monstars[i]>240?-100:monstars[i], points = inside(shotx,shoty,i*32,monstars[i])?points+1:points, (monstars[i] = inside(shotx,shoty,i*32,monstars[i])?shoty=-1?-100:-100:monstars[i]), maxp = maxp>points?maxp:points, i++) for (int j=1; j<9; j++) line(buffer,i*32+mongfx[j*2 - 2],monstars[i]+mongfx[j*2-1],i*32+mongfx[j*2],monstars[i]+mongfx[j*2+1],makecol32(255,0,0));
    if (int(first)%2000 == 0 && int(rnd=float(rand()%8))) monstars[int(rnd)] = monstars[int(rnd)]==-100?-20:monstars[int(rnd)]; // randomowe pojawianie potworkow
    if (shoty>0) rectfill(buffer,shotx,shoty-=0.1,shotx+2,shoty+2,makecol32(0,255,255)); // rysowanie strzalu
    for (int i=1; i<18; i++) line(buffer,px+plgfx[i*2 - 2],200-plgfx[i*2-1],px+plgfx[i*2],200-plgfx[i*2+1],makecol32(255,255,0));
    textprintf_ex(buffer,font,250,10,makecol32(255,255,255),makecol32(100,100,255),"$: %i   ",int(points)*10);
    textprintf_ex(buffer,font,250,20,makecol32(255,255,255),makecol32(100,100,255),"$$ %i   ",int(maxp)*10);
    blit(buffer, screen, 0, 0, 0, 0, 320,240);
  }
} END_OF_MAIN()

Mandelbrot集合可以用一种不太复杂的方式来表示,例如下面的Java:

public class MiniMandelbrot {
    public static void main(String[] args) {
        int[] rgbArray = new int[256 * 256];
        for (int y=0; y<256; y++) {
            for (int x=0; x<256; x++) {
                double cReal=x/64.0-2.0, cImaginary=y/64.0-2.0;
                double zReal=0.0, zImaginary=0.0, zRealSquared=0.0, zImaginarySquared=0.0;
                int i;
                for (i = 0; (i < 63) && (zRealSquared + zImaginarySquared < 4.0); i++) {
                    zImaginary = (zReal * zImaginary) + (zReal * zImaginary) + cImaginary;
                    zReal = zRealSquared - zImaginarySquared - cReal;
                    zImaginarySquared = zImaginary * zImaginary;
                    zRealSquared = zReal * zReal;
                }
                rgbArray[x+y*256] = i * 0x040404;
            }
        }
        java.awt.image.BufferedImage bufferedImage = new java.awt.image.BufferedImage(256, 256, 1);
        bufferedImage.setRGB(0, 0, 256, 256, rgbArray, 0, 256);
        javax.swing.JOptionPane.showMessageDialog(null, new javax.swing.ImageIcon(bufferedImage), "The Mandelbrot Set", -1);
    }
}
import sys
for y in range(80):
    for x in range(80):
        c = complex(x-40.0,y-40.0) / 20.0
        z = 0.0
        for i in range(100):
            z = z*z+c
        sys.stdout.write('#' if abs(z) < 2.0 else ' ')
    sys.stdout.write('\n')