# Algorithm diagram

## Binary search

1. For ordered sets, you can use

2. Search from the middle

def binary_search(list,item): low = 0 high = len(list) - 1 while low <= high: mid = (low+high)//2 guess = list[mid] if guess == item: return mid if guess > item: high = mid-1 else: low = mid+1 return None print(binary_search([1,3,4,5,6,7,8,9],4))

3. Only suitable for ordered sets

## Large O representation

1. The large o notation refers to a speed that is not in seconds. The big O notation allows you to compare operands. It indicates the growth rate of algorithm running time.

2. Common large O indicates time:

2.1 O(log n), also known as log time, this algorithm includes binary search.

2.2 O(n), also known as linear time, such an algorithm includes simple search.

2.3 O(n * log n), such an algorithm includes the quick sort introduced in Chapter 4 - a faster sort algorithm.

2.4 O(n**2), such an algorithm includes the selective sorting introduced in Chapter 2 - a slower sorting algorithm.

2.5 O(n!)， Such an algorithm includes the solution to the traveling salesman problem that will be introduced next - a very slow algorithm

## The difference between array and linked list

Array: in memory, elements need to be connected. Once the added elements exceed the obtained memory, they need to be expanded and then reallocated. If a lot of space is prepared in advance, it will become a waste of space. Therefore, the array is not suitable for adding element O(n). Array also has great advantages. When looking for elements, you can find them directly, and the complexity is O(1)

Linked list: it will not be a waste of space. As long as there is a location, it can be stored. One element needs to mark the location O(1) of the next element. However, when searching, you can only view it step by step from the beginning. The time is O(n)

## Select sort

1. Large o representation: O(n**2)

1. If an unordered array is given, sort the array. First, cycle the array n times to find the smallest one, then put it into a new array, and then continue to cycle n times to find the second smallest one, and continue to cycle until the order is taken, so the time complexity is O(n**2)

def findSmallest(arr): smallest = arr[0] smallest_index = 0 for i in range(1,len(arr)): if arr[i] < smallest: smallest = arr[i] smallest_index = i return smallest_index def selectionSort(arr): newArr = [] for i in range(len(arr)): smallest = findSmallest(arr) newArr.append(arr.pop(smallest)) return newArr print(selectionSort([5,3,6,2,10]))

## recursion

### Stack

The difference between queue and queue is that queue is FIFO and stack is FIFO. It's like washing dishes. The first dish to be washed is placed at the bottom. When you use it, you take it from the top, that is, the last plate you put in.

Function is called call stack.

### Recursive call stack

For example, to find the factorial of a number, the simplest recursive method is:

def fact(x): if x == 1: return x else: return x * fact(x-1)

### Quick sort (quick sort)

Fast sorting is mainly to divide an array into two groups greater than this number and less than this number according to a random number, and then use recursion to sort

Quick steps:

(1) Select the reference value.

(2) Divide the array into two subarrays: elements less than the reference value and elements greater than the reference value.

(3) Quickly sort the two subarrays.

For example, to sort a queue:

def quicksort(array): #The baseline condition must be set, otherwise there will be endless calls if len(array) < 2: return array else: pivot = array[0] less = [i for i in array[1:] if i <= pivot] greater = [i for i in array[1:] if i>= pivot] return quicksort(less) + [pivot] + quicksort(greater)

## Hash table

1. It is called dictionary in python and hash in ruby, for example: {'name': 'Lisi'}

### Hash function

1. Hash function "map input to number".

2. Hash function is not irregular. On the contrary, hash function must meet certain requirements before it can be called a qualified hash function. Simple requirements:

it must be consistent. For example, suppose you get 4 when you enter apple, then you get 4 every time you enter apple

Must be 4. If not, the hash table will be useless.

it should map different inputs to different numbers. For example, if a hash function returns 1 regardless of the input, it is not a good hash function. Ideally, map different inputs to different numbers.

## Breadth first search

Brief introduction:

If you're looking for a fruit seller, you're going to look among your friends.

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First, find your first level of friends. If you don't find your friends again. The first layer of friends becomes a one-time relationship, and then it accumulates, such as two-layer relationship, three-layer relationship and so on.

The relationship weight of one layer is the highest, which is searched first, and then searched layer by layer.

This is called breadth first search.

Simply implement this case:

#Create a friend relationship friend = {} friend["you"] = ["alice", "bob", "claire"] friend["bob"] = ["anuj", "peggy"] friend["alice"] = ["peggy"] friend["claire"] = ["thom", "jonny"] friend["anuj"] = [] friend["peggy"] = [] friend["thom"] = [] friend["jonny"] = [] #Create a queue from collections import deque def search_shuiguo(): search_queue = deque() search_queue += friend["you"] while search_queue: #Throw the value to the left of the queue person = search_queue.popleft() if person_is_shuiguo(person): print(person + "It's a fruit seller") return True else: search_queue += friend[person] return False def person_is_shuiguo(name): return name[-1] == "m"

Queue is first in first out, which is very suitable for breadth first search.

## Dixtra algorithm

Brief introduction:

Dixtra algorithm is simply breadth first search plus weight.

For example:

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As shown in the figure, you have to calculate the distance from the twin peaks to the Golden Gate Bridge. Breadth first search can be used.

First calculate the first step away from yourself, two steps away from yourself and three steps away. The nearest distance can be calculated. However, this does not take into account that the length of each distance is different, the tools are different, and the time will be different. At this time, dixtra algorithm can be used.

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The dixtra algorithm consists of four steps:

(1) Find the "Cheapest" node, that is, the node that can arrive in the shortest time.

(2) The cost of updating the neighbors of the node will be described later.

(3) Repeat this process until this is done for each node in the graph.

(4) Calculate the final path.

For example, find the shortest formation from seven points to the end point.

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Create three hash tables

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#Create a distance hash from a neighbor graph = {} graph["start"] = {} graph["start"]["a"] = 6 graph["start"]["b"] = 2 graph["a"] = {} graph["a"]["fin"] = 1 graph["b"] = {} graph["b"]["a"] = 3 graph["b"]["fin"] = 5 graph["fin"] = {} #Create an expense table from the starting point to each location #Represents infinity infinity = float("inf") costs = {} costs["a"] = 6 costs["b"] = 2 #Because I'm not my neighbor, I don't know how far it will be, so I set infinity costs["fin"] = infinity #Stores the hash of the parent node parents = {} parents["a"] = "start" parents["b"] = "start" parents["fin"] = None #Create an array to record the processed nodes processed = [] #Find the node with the least overhead def find_lowest_cost_node(costs): lowest_cost = float("inf") lowest_cost_node = None for node in costs: cost = costs[node] if cost < lowest_cost and node not in processed: lowest_cost = cost lowest_cost_node = node return lowest_cost_node #Find the node with the least overhead among the unprocessed nodes node = find_lowest_cost_node(costs) while node is not None: cost = costs[node] neighbors = graph[node] for n in neighbors.keys(): new_cost = cost + neighbors[n] if costs[n] > new_cost: parents[n] = node processed.append(node) node = find_lowest_cost_node(costs)

## greedy algorithm

Greedy algorithm is a very simple problem-solving strategy. It may not be the optimal solution. But it is also a solution. It is a convenient solution, whether it is a fast solution or not

Suppose you run a radio program that should be heard by listeners in 50 states across the United States. To do this, you need to decide which stations to broadcast on. You have to pay for broadcasting on every radio station, so you try to broadcast on as few radio stations as possible.

Greedy algorithm solution:

(1) Select a radio station that covers the most uncovered states. Even if this radio station covers some already covered

It doesn't matter.

(2) Repeat the first step until all States are covered.

def find_states(): # State to be covered states_needed = set(["mt", "wa", "or", "id", "nv", "ut", "ca", "az"]) # List of broadcasting stations stations = {} stations["kone"] = set(["id", "nv", "ut"]) stations["ktwo"] = set(["wa", "id", "mt"]) stations["kthree"] = set(["or", "nv", "ca"]) stations["kfour"] = set(["nv", "ut"]) stations["kfive"] = set(["ca", "az"]) final_stations = set() while states_needed: best_station = None states_covered = set() for station,states in stations.items(): covered = states & states_needed if len(covered) > len(states_covered): best_station = station states_covered = covered states_needed -= states_covered final_stations.add(best_station) return final_stations print(find_states())