Đáp án đề kiểm tra học kỳ I năm học 2019-2020 môn Xử lý ảnh công nghiệp - ĐH Sư phạm Kỹ thuật

RU'CSNG DAI HOC SU' PHAM KY THUAT
THANH PHO HO CHI MINH
KHOA CO KHI CHE TAO MAY
BO MON C O DIENTtT

DAP AN CUOI KY HK I NAM HOC 2019-2020
Mon: Xff LY ANH CONG NGHlEP
MS mon hoc: IIPR422529
Be so: 01 Be thi co 02 trang
Ng^ythi: 16/12/2019
Thai gian: 75 phut.
Buoc phep su dung tai lieu giay

Cau 1: (2d)
Xay dung va cai dat thuat toan lam tang do net cua anh bang mat na Laplace. (1.5d)
Tai sao mat na Laplace lai lam tang do net cua anh? (0.5d)
Dap an:
void Sharpen2(Mat imgin, Mat imgout)
{
Mat temp = Mat(imgin.size(),CV_32FCl);
Mat w = (Mat_<float>(3,3) « 1,1,1,1,-8,1,1,1,1);
filter2D(imgin,temp,CV_32FC 1,w);
int M - imgin.size().height;
int N = imgin.size().width;
int x, y;
float r;
for (x=0; xfor (y=0; yr = imgin.at<uchar>(x,y) - temp.at<float>(x,y);
if (r < 0)
r = 0;

if (r > L-l)
r = L-l;
imgout.at<uchar>(x,y) = (uchar)r;
}
return;

Cau 2: (3d)
Xay dung va cai dat thuat toan loai bo cac hat gao nho hon 90% hat gao lan nhat, tuc la
trong anh chi con lai nhirng hat gao lan.

Trang 1/5


Dap an:
void RemoveSmallRice(Mat imgin, Mat imgout)

{

// B1
Mat wl = getStructuringElement(MORPH_ELLIPSE, Size(81, 81));
morphologyEx(imgin, imgout, MORPH_TOPHAT, w l);
// B2
threshold(imgout, imgout, 50, 255, THRESH_BINARY | THRESH_OTSU);
medianBlur(imgout, imgout, 3);
// B3
int M = imgout.size().height;
int N = imgout.sizeQ. width;
int x, y, dem = 0;
int r;
int color = 150;

for (x = 0; x < M; x++)
for (y = 0; y < N; y++)

{
r = imgout.at<uchar>(x, y);
if (r == L - 1)

{

i/

floodFill(imgout, Point(y, x), CV_RGB(color, color, color));
dem-H-;
color++;

}

// B4
int a[L];
for (r = 0; r < L; r++)
a[r] = 0;
for (x = 0; x < M; x++)
for (y = 0 ;y < N ;y -H -)

{
r = imgout.at<uchar>(x, y);
if (r > 0)
a[r]++;

}

int max = 0;
for (r=0; rif (a[r] > max)
max = afr];
max = (int)(max*0.9);

// B5
for (r = 0; rif (a[r] < max)
a[r] = 0;
dem = 0;
for (r = 0; r < L; r++)

Trang 2/5


if (a[r] > 0)
dem++;
for (x = 0; x < M; x++)
for (y = 0; y < N; y++)

{
r = imgout.at<uchar>(x, y);
if (a[r] = 0)
imgout.at<uchar>(x, y) = 0;
else
imgout.at<uchar>(x, y) = L - 1;

}
char s[5];

sprintf(s, "%d", dem);
putText(imgout, s, Point(0, 30), CV_FONT HERSHEY SIMPLEX, 0.8, CV RGB(255, 255,
255));
return;

Cau 3: (3d)

Ta dinh nghTa mang na-ron chap dung de nhan dang 10 chu so viet tay co kfch thuoc 28x28
nhtr sau:
def build(input_shape, classes):
model = Sequential()
# CONV => RELU =» POOL
model.add(Conv2D(20, kerne l_size=5, padding=”same",
input_shape=input_shape))
model.add(Activation("relu"))
model.add(MaxPooling2D(pool_size=(2, 2), strides=(2, 2)))
# CONV => RELU => POOL
model.add(Conv2D(50, kernel_size=5, padding="same"))
model.add(Activation("relu"))
model.add(MaxPooling2D(pool_size:=(2, 2), strides=(2, 2)))
# Flatten => RELU layers
model. add(Flatten())
model.add(Dense(500))
model.add(Activation("relu"))
# a softmax classifier
model.add(Dense(classes))
model.add(Activation("softmax"))
a. Hay ve so do khoi cua mang nof-ron chap tren.
b. Cho biet so luong tham so cua cac bo loc trong cac lop chap (co tmh nut bias).
c. Cho biet so luong trong so cua lap ket noi day du (co tinh nut bias)

Dap an:

Trang 3/5


activation_l (Activation)

(None, 20, 28, 28)

0

uax_pooling2d_l (MaxPooling2 (None, 20, 14, 14)

0

conv2d_2 (Conv2D)

(None, 50, 14, 14)

25050

activation_2 (Activation)

(None, 50, 14, 14)

0

nax_pooling2d_2 (MaxPooling2 (None, 50, 7, 7)

0

flatten_l (Flatten)

(None, 2450)

0

dense_l (Dense)

(None, 500)

1225500

activation_3 (Activation)

(None, 500)

0

dense_2 (Dense)

(None, 10)

activation_4 (Activation)

(None, 10)

Total params: 1,256,080
Cau 4: (2d)
Given the input image and the filter shown in the following figure:

4
7
3
3
1

2
5
8
4
5
5

3
6
9
9
8
7

0
4
5
2
9
4

1
9

0
6
7
6

5
3
1
5
9
0

2

1

-2

0

0

-1

1

0

1

Trang 4/5


a. Determine the size of the output image if zero-padding is 0 and stride is 1. (0.5
points)
b. Compute the output image as the convolution of the input image and the filter. (1.5
points)
Dap an:
a. Wout =
is 4x4
b.
8
4
6
-8

(W-F+2P)/S + 1 = (6-3+2*0)/l + 1 = 4 , therefore the size of output image

16
9
27
13

4
13
32
14

-6
17

21
-5

------------------------ HET-------------------Ghi chu: Can bo coi thi khong giai thick de thi.

Ngay 23 thang 12 nam 2019
Thong qua bQ mon
(ky va ghi ro ho ten)

GV lam dap an
(ky va ghi ro ho ten)

T

I

Trang 5/5

/
O