问题：平面内一个物体Object的初始坐标(0,0)，此后，每秒进行一个移动，每次移动1，方向是上下左右。即，newLocation = oldLocation + (x, y)，其中(x,y) = (0,1) or (0, -1) or (1, 0) or (-1, 0)。那么，N秒后，Object离初始坐标的距离，会不会随着N的增加而增加？

分析：问题的本质是求N秒后的期望距离。这里有一点要注意，N秒后的期望距离和N秒后object最有可能出现的位置是两个该问题！搞清了这一点之后，我们就知道E(dist) = sigma(dist[i] * p[i])，其中dist[i]是离(0,0)的距离，p[i]是此距离的概率。由此，我们可以得到一个直观的概念，期望距离会随着N的增加而增加。（实际上，用数学的概率分布可以证明）

以下，用python来仿真这一过程，证明这一结论

代码

#!/usr/bin/python                                                                                                            import pylabimport randomimport mathdef distance(location):    return math.sqrt( location[0]**2 + location[1]**2 )def simulate(count):    location = [0, 0]    countLeft = 0    countRight = 0    countUp = 0    countDown = 0    for i in range(0, count):        direction = random.choice(["left", "right", "up", "down"])#        print "direction = ", direction                                                                                             if (direction == "left"):            location[0] -= 1            countLeft += 1        if (direction == "right"):            location[0] += 1            countRight += 1        if (direction == "up"):            location[1] += 1            countUp += 1        if (direction == "down"):            location[1] -= 1            countDown += 1#       print "location = ", location               dist = distance(location)        result.append(dist)    print countLeft, countRight, countDown, countUp    returnresult = []simulate(10000)pylab.title('random walk simulation')pylab.xlabel('time/s')pylab.ylabel('distance/m')pylab.plot(result)pylab.show()

运行结果：

 改进：

我们看到，一次试验只能给人一个直观映像，即，随着step的增加，与原点的距离也在增加。但是，一次试验的随机性比较高，这导致结果的振动幅度比较大，无法很好的看出dist和step的关系。如何改进这一点呢？

用取平均的方法来消除随机误差。这里，我们对500次试验取平均。

以下是改进后的试验结果：

以下是改进后的代码：

#!/usr/bin/pythonimport pylabimport randomimport mathdef distance(location):    return math.sqrt( location[0]**2 + location[1]**2 )def simulate(count):        location = [0, 0]    countLeft = 0    countRight = 0    countUp = 0    countDown = 0    result = []        for i in range(0, count):        direction = random.choice(["left", "right", "up", "down"])#        print "direction = ", direction        if (direction == "left"):            location[0] -= 1            countLeft += 1        if (direction == "right"):            location[0] += 1            countRight += 1        if (direction == "up"):            location[1] += 1            countUp += 1        if (direction == "down"):            location[1] -= 1            countDown += 1#       print "location = ", location        dist = distance(location)        result.append(dist)    return resultdef performTrial(steps, tries):    distLists = []    for i in range(0, tries):        result = simulate(steps)        distLists.append(result)    return distListsdef main():    steps = 3    tries = 500    distLists = performTrial(steps, tries)    averageList = [0]    for i in range(0, steps):        total = 0        for j in range(0, tries):            total += distLists[j][i]        average = total/tries        averageList.append(average)#        print 'total=', total#        print 'average=', average#        print averageList    pylab.plot(averageList)    pylab.title('random walk simulation')    pylab.xlabel('time/s')    pylab.ylabel('distance/m')    pylab.show()    returnif __name__ == '__main__':    main()

最后用简单的例子来证明正确性：

steps = 0  dist = 0

steps = 1  dist = 1

steps = 2  dist = (2+0+sqrt(2)+sqrt(2))/4 = 1.21

step3 = 3  dist = (3+1+sqrt(5)+sqrt(5)+...+...)/16 = 1.59

如下是steps=3, tries=500的结果图示，和以上数学分析结果一致。