**複雜度o(m*m*t*logm),m為影象的畫素個數，m為patch size ,t為迭代次數
import numpy as np
from pil import image
import time
def cal_distance(a, b, a_padding, b, p_size):
p = p_size // 2
patch_a = a_padding[a[0]:a[0]+p_size, a[1]:a[1]+p_size, :]
patch_b = b[b[0]-p:b[0]+p+1, b[1]-p:b[1]+p+1, :]
temp = patch_b - patch_a
num = np.sum(1 - np.int32(np.isnan(temp)))
dist = np.sum(np.square(np.nan_to_num(temp))) / num
return dist
def reconstruction(f, a, b):
a_h = np.size(a, 0)
a_w = np.size(a, 1)
temp = np.zeros_like(a)
for i in range(a_h):
for j in range(a_w):
temp[i, j, :] = b[f[i, j][0], f[i, j][1], :]
image.fromarray(temp).show()
def initialization(a, b, p_size):
a_h = np.size(a, 0)
a_w = np.size(a, 1)
b_h = np.size(b, 0)
b_w = np.size(b, 1)
p = p_size // 2
random_b_r = np.random.randint(p, b_h-p, [a_h, a_w])
random_b_c = np.random.randint(p, b_w-p, [a_h, a_w])
a_padding = np.ones([a_h+p*2, a_w+p*2, 3]) * np.nan
a_padding[p:a_h+p, p:a_w+p, :] = a
f = np.zeros([a_h, a_w], dtype=object)
dist = np.zeros([a_h, a_w])
for i in range(a_h):
for j in range(a_w):
a = np.array([i, j])
b = np.array([random_b_r[i, j], random_b_c[i, j]], dtype=np.int32)
f[i, j] = b
dist[i, j] = cal_distance(a, b, a_padding, b, p_size)
return f, dist, a_padding
def propagation(f, a, dist, a_padding, b, p_size, is_odd):
a_h = np.size(a_padding, 0) - p_size + 1
a_w = np.size(a_padding, 1) - p_size + 1
x = a[0]
y = a[1]
if is_odd:#向左上方傳播
d_left = dist[max(x-1, 0), y]
d_up = dist[x, max(y-1, 0)]
d_current = dist[x, y]
idx = np.argmin(np.array([d_current, d_left, d_up]))
if idx == 1:
f[x, y] = f[max(x - 1, 0), y]
dist[x, y] = cal_distance(a, f[x, y], a_padding, b, p_size)
if idx == 2:
f[x, y] = f[x, max(y - 1, 0)]
dist[x, y] = cal_distance(a, f[x, y], a_padding, b, p_size)
else:#向右下方傳播
d_right = dist[min(x + 1, a_h-1), y]
d_down = dist[x, min(y + 1, a_w-1)]
d_current = dist[x, y]
idx = np.argmin(np.array([d_current, d_right, d_down]))
if idx == 1:
f[x, y] = f[min(x + 1, a_h-1), y]
dist[x, y] = cal_distance(a, f[x, y], a_padding, b, p_size)
if idx == 2:
f[x, y] = f[x, min(y + 1, a_w-1)]
dist[x, y] = cal_distance(a, f[x, y], a_padding, b, p_size)
def random_search(f, a, dist, a_padding, b, p_size, alpha=0.5):
x = a[0]
y = a[1]
b_h = np.size(b, 0)
b_w = np.size(b, 1)
p = p_size // 2
i = 4
search_h = b_h * alpha ** i
search_w = b_w * alpha ** i
b_x = f[x, y][0]
b_y = f[x, y][1]
while search_h > 1 and search_w > 1:
search_min_r = max(b_x - search_h, p)
search_max_r = min(b_x + search_h, b_h-p)#不能越界
random_b_x = np.random.randint(search_min_r, search_max_r)#row
search_min_c = max(b_y - search_w, p)
search_max_c = min(b_y + search_w, b_w - p)
random_b_y = np.random.randint(search_min_c, search_max_c)#col
search_h = b_h * alpha ** i#用類似於二分法的方式不斷縮小搜尋範圍
search_w = b_w * alpha ** i
b = np.array([random_b_x, random_b_y])
d = cal_distance(a, b, a_padding, b, p_size)
if d < dist[x, y]:#如果新的隨機距離小於已有的距離，那麼就更新
dist[x, y] = d
f[x, y] = b
i += 1
def nns(img, ref, p_size, itr):
a_h = np.size(img, 0)
a_w = np.size(img, 1)
f, dist, img_padding = initialization(img, ref, p_size)#隨機初始化a到b的匹配
for itr in range(1, itr+1):
if itr % 2 == 0:
for i in range(a_h - 1, -1, -1):
for j in range(a_w - 1, -1, -1):
a = np.array([i, j])
propagation(f, a, dist, img_padding, ref, p_size, false)#先根據周圍畫素的匹配情況來進行本畫素的匹配
random_search(f, a, dist, img_padding, ref, p_size)#繼續進行大範圍的搜尋
else:
for i in range(a_h):
for j in range(a_w):
a = np.array([i, j])
propagation(f, a, dist, img_padding, ref, p_size, true)
random_search(f, a, dist, img_padding, ref, p_size)
print("iteration: %d"%(itr))
return f
if __name__ == "__main__":
img = np.array(image.open("001_0.png"))
ref = np.array(image.open("001_5.png"))
p_size = 3#patch的長寬
itr = 5#迭代次數
start = time.time()
f = nns(img, ref, p_size, itr)
end = time.time()
print(end - start)
reconstruction(f, img, ref)
DUKPT演算法解析
dukpt derived unique key per transaction 1 是什麼？是一種非常安全的金鑰管理技術，主要應用於對稱金鑰加密mac,pin等安全資料方面 2 主要思想 保證每一次交易流程使用唯一的金鑰，採用一種不可逆的金鑰轉換演算法，使得無法從當前交易資料資訊破解上一次交易金鑰...
KMP演算法解析
日期 2013年5月 1日 字串匹配是計算機的基本任務之一。舉例來說，有乙個字串 bbc abcdab abcdabcdabde 我想知道，裡面是否包含另乙個字串 abcdabd 許多演算法可以完成這個任務，knuth morris pratt演算法 簡稱kmp 是最常用的之一。它以三個發明者命名，...
highestOneBit 演算法解析
integer類有個highestonebit 方法，作用是返回具有單個 1 位的 int 值,在指定值中最高位的 1 位的位置，比如5 00000101 返回4 00000100 highestonebit 的演算法如下 public static inthighestonebit int i 為...