1. Give a single intensity transformation function for spreading the intensities of an image so the lowest intensity is 0 and the highest is L -1. [为了扩展一幅图像的灰度，使其最低灰度为0、最高灰度为L -1，请给出一个单调的变换函数。]
Answer: Let f denote the original image. First subtract the minimum value of f denoted f min from f to yield a function whose minimum value is 0:
1min g f f =-
Next divide g 1 by its maximum value to yield a function in the range [0,1] and multiply the result by L 一1 to yield a function with values in the range [0, L 一1]
1min 1min
11()max()max()L L g g f f g f f --==--
Keep in mind that f min is a scalar and f is an image.
[让f 表示原始图像。首先从图像函数f 中减掉f 的最小值f min , 然后生成一个新的函数g 1，它的最小值为0：
1min g f f =-
接下来让g 1的最大值除g 1得到另一新的函数，它的值域在[0,1]区间，然后再乘上L 一1，得到值域为[0, L 一1]的新函数。请注意f min 是一个标量，而f 是一个图像。
2.Explain why the discrete histogram equalization technique does not,in general,yield a flat histogram. [请解释为什么离散直方图均衡化技术一般不能得到平坦的直方图。]
Answer: All that histogram equalization does is remap histogram components on the intensity scale. To obtain a uniform (flat) histogram would require in general that pixel intensities actually be redistributed so that there are L groups of n/L pixels with the same intensity, where L is the number of allowed discrete intensity levels and n=MN is the total number of pixels in the input image. The histogram equalization method has no provisions for this type of (artificial) intensity redistribution
process.
[直方图均衡化就是把直方图重新映射到灰度尺度上。为了得到平坦的直方图，通常需要图像中像素的灰度值重新分配，这样会产生L个组，每个组中包含相同灰度值的像素的数量是n/L，其中L是允许的离散灰度值的个数，n=MN是输入图像中总的像素的个数。而离散直方图的均衡化没有提供这样一个重新分配的过程。]
3.In a given application an averaging mask is applied to input image to reduce noise, and then a Laplacian mask is applied to enhance small details. Whould the result be the same if the order of these operations were reserved? [在给定的应用中，一个均值模板被用于输入图像以减少噪声，然后再用一个拉普拉斯模板来增强细节。如果交换一下两个操作步骤的顺序，能否得到相同的结果？]
Answer: The student should realize that both the Laplacian and the averaging process are linear operations, so it makes no difference which one is applied first.
[学生应该认识到拉普拉斯和平均化的处理过程都是线性运算过程，所以无论先应用哪一个模板对结果是没有影响的。]
4. You saw in Fig. 2.18 that the Laplacian with a -8 in the center yields sharper results than the one with a -4 in the center. Explain the reason in detail. [在图2.18中所看到的中心系数为-8 的拉普拉斯模板所得到的结果，要比中心系数为-4的模板所得到的结果清晰一些。请详细说明原因。] Answer: The Laplacian mask with a -4 in the center performs an operation proportional to differentiation in the horizontal and vertical directions. Consider for a moment a 3 x 3 "Laplacian" mask with a -2 in the center and 1 above and below the center. All other elements are 0. This mask will perform differentiation in only one direction, and will ignore intensity transitions in the orthogonal direction. An image processed with such a mask will exhibit sharpening in only one
