Something interesting

不管哪个领域,都可以在上升期做科研,在平稳期做业务,在饱和期做教育,显然Andrew Ng是个明白人。

No matter what field you are in, you can do research in the growing period, dedicate into the industry in the steady period, and develop education in the saturation period. Obviously, Andrew Ng is a sensible person.





To be honest, we have a decent job, house, car at the age of 25, should not complain more. However, cars/houses/good jobs, all of them, are general commodities, others may have them of ten times or even of hundred times than of what I have. But the ten years of youth, everyone has only one time.

So, please remember this thing, when you are 26 years old and decide whether you want to be a 30-year-old Doctor.

Machine learning, can?


Actually, we have been aware of the trending of massive scale parameters in machine learning. The number of CPUs and the access of data determines the final performance, but not the person who researches machine learning algorithms.

Mitchell Approximation

A method of computer multiplication and division is proposed which uses binary logarithms. The logarithm of a binary number may be determined approximately from the number itself by simple shifting and counting. A simple add or subtract and shift operation is all that is required to multiply or divide.


int main() {
    float a = 12.3f;
    float b = 4.56f;
    int c = *(int*)&a + *(int*)&b - 0x3f800000;
    printf("Approximate result:%f\n", *(float*)&c);
    printf("Accurate result:%f\n", a * b);
    return 0;


import numpy as np
import networkx as nx
from functools import reduce
import matplotlib.pyplot as plt

connect_graph = np.array([[0, 1, 0, 0, 0], 
                          [0, 0, 0, 1, 0], 
                          [0, 0, 0, 1, 0], 
                          [0, 0, 0, 0, 1], 
                          [0, 0, 1, 0, 0]])

def ring_add(a, b):
    return a or b

def ring_multi(a, b):
    return a and b

def dot_product(i, j):
    row = connect_graph[i]
    column = connect_graph[:,j]
    return reduce(ring_add, [ring_multi(a, b) for a, b in zip(row, column)])

def next_generation(connect_graph):
    candidate_number = connect_graph.shape[0]

    new_connect_graph = np.zeros((candidate_number, candidate_number))

    for i in range(candidate_number):
        for j in range(candidate_number):
            new_connect_graph[i][j] = dot_product(i,j)
    return new_connect_graph

new_connect_graph = next_generation(connect_graph)

def draw_graph(connect_graph):
    G = nx.DiGraph()
    candidate_number = connect_graph.shape[0]
    node_name = list(range(candidate_number))
    for i in range(candidate_number):
        for j in range(candidate_number):
            if connect_graph[i][j]:
                G.add_edge(i, j)

    nx.draw(G, with_labels=True)


Heap sort

class Test():
    def heap_sort(self, nums):
        i, l = 0, len(nums)
        self.nums = nums
        # 构造大顶堆,从非叶子节点开始倒序遍历,因此是l//2 -1 就是最后一个非叶子节点
        for i in range(l//2-1, -1, -1): 
            self.build_heap(i, l-1)
        # 上面的循环完成了大顶堆的构造,那么就开始把根节点跟末尾节点交换,然后重新调整大顶堆  
        for j in range(l-1, -1, -1):
            nums[0], nums[j] = nums[j], nums[0]
            self.build_heap(0, j-1)

        return nums
    def build_heap(self, i, l): 
        nums = self.nums
        left, right = 2*i+1, 2*i+2 ## 左右子节点的下标
        large_index = i 
        if left <= l and nums[i] < nums[left]:
            large_index = left

        if right <= l and nums[left] < nums[right]:
            large_index = right
        # 通过上面跟左右节点比较后,得出三个元素之间较大的下标,如果较大下表不是父节点的下标,说明交换后需要重新调整大顶堆
        if large_index != i:
            nums[i], nums[large_index] = nums[large_index], nums[i]
            self.build_heap(large_index, l)