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Hash / (understanding, implementation and application)

2022-07-07 11:07:00 【Exy-】

unordered The reason why the series of associative containers are more efficient , Because it uses a hash structure at the bottom

Hash concept

  Sequence structure and balance tree , There is no corresponding relationship between the element key and its storage location , So in When looking for an element , Must go through Multiple comparisons of clearance key codes . The time complexity of sequential search is O(N), Balance the height of the tree in the tree , namely O( log2
N), The efficiency of the search depends on the number of element comparisons in the search process .
Ideal search method ： Without any comparison , Get the elements to search directly from the table at a time .
If you construct a storage structure , adopt Some kind of function (hashFunc) A one-to-one mapping relationship can be established between the storage location of an element and its key , Then when searching, use this letter You can quickly find this element by counting .
When in this structure ：
Insert elements
According to the key of the element to be inserted , This function calculates the storage location of the element and stores it according to this location
Search element
Do the same calculation for the key of the element , Take the function value as the storage location of the element , Take the element ratio according to this position in the structure a , If the keys are equal , Then the search is successful
This is called hash ( hash ) Method , The conversion function used in the hash method is called hash ( hash ) function , The constructed structure is called a hash table (Hash Table)( Or a hash table ) 

Common hash functions

direct addressing

Take a linear function of keyword as hash address ：Hash（Key）= A*Key + B
advantage ： Simple 、 uniform
shortcoming ： It needs to be done in advance Know the distribution of keywords
Use scenarios ： It is suitable for finding relatively small and continuous cases

Division and remainder

Set the number of addresses allowed in the hash table to m, Take one not greater than m, But closest to or equal to m The prime number of p As divisor , According to the hash function Count ：Hash(key) = key% p(p<=m), Convert key to hash address
Hash Collisions : Different keywords pass through the phase The same hash function calculates the same hash address , This phenomenon is called hash collision

So how to solve hash conflicts ?

Closed hash and open hash

Closed hash

It's also called open addressing , When a hash conflict occurs , If the hash table is not full , It means that there must be a space in the hash table , that You can put key Stored in a conflict location “ next ” Go to an empty place .

So how to find this location ?

  • Linear detection
  • Second detection

Linear detection : The positions mapped by hash have conflicts , Then you need to find an empty space to store data in the future  

Hash(key)=key%len + i(i=0,1,2,3...)

Second detection : The positions mapped by hash have conflicts , Then we need to find a space to store data in the later power 

Hash(key)=key%len + i^2(i=0,1,2,3...)

Closed hash implementation 

namespace CLOSE_HASH
{
	enum State{EMPTY,EXITS,DELETE};
	template<class K,class V>
	struct HashDate
	{
		pair<K,V> _kv;
		State _state=EMPTY;
	};
	//  specialized 
	template<class K>
	struct Hash
	{
		size_t operator()(const K& key)
		{
			return key;
		}
	};
	template<>
	struct Hash<string>
	{
		// "int"  "insert" 
		//  The string is converted to a corresponding integer value , Because of shaping, we can get the modular mapping position 
		//  expect -> String is different , The shaping value transferred out shall be as different as possible 
		// "abcd" "bcad"
		// "abbb" "abca"
		size_t operator()(const string& s)
		{
			// BKDR Hash
			size_t value = 0;
			for (auto ch : s)
			{
				value += ch;
				value *= 131;
			}

			return value;
		}
	};

	template<class K,class V,class KHashFunc=Hash<string>>
	class HashTable
	{
	public:
		bool Insert(const pair<K, V>& kv)
		{
			HashDate<K, V>* ret = find(kv.first);
			if (_table.size() == 0)
			{
				_table.resize(10);
			}
			else if(_size*10 / _table.size() > 7)
			{
				HashTable<K, V, KHashFunc> newHT;
				newHT._table.resize(_table.size() * 2);

				for (auto& e : _table)
				{
					newHT.Insert(e._kv);
				}
				_table.swap(newHT._table);
			}
			KHashFunc hf;
			size_t start = hf(kv.first) % _table.size();
			size_t index = start;
			size_t i = 1;
			while (_table[index]._state == EXITS)
			{
				index = start + i;
				index %= _table.size();
				++i;
				//index += i ^ 2;
			}
			_table[index]._kv = kv;
			_table[index]._state = EXITS;
			++_size;
			return true;
		}
		HashDate<K,V>* find(const K& key)
		{
			KHashFunc hf;
			if (_table.size() == 0)
			{
				return nullptr;
			}
			size_t i = 1;
			size_t start = hf(key) % _table.size();
			size_t index = start;
			while (_table[index]._state != EMPTY)
			{
				if (_table[index]._kv.first==key&&_table[index]._state==EXITS)
				{
					return &_table[index];
				}
				index = start + i;
				index %= _table.size();
				++i;
			}
			return nullptr;
		}
		bool Erase(const K& key)
		{
			HashDate<K, V>* ret = find(key);
			if (ret == nullptr)
			{
				return false;
			}
			else
			{
				ret->_state = DELETE;
				return true;
			}
		}
	private:
		vector<HashDate<K,V>> _table;
		size_t _size=0;// Number of valid data stored 
	};
}

  Hash

Hash method is also called chain address method ( Open chain method ), First, we use hash function to calculate hash address for key set , Keys with the same address Belong to the same subset , Each subset is called a bucket , The elements in each bucket are linked by a single chain table , The head knot of each linked list Points are stored in a hash table , It's also called the Hashi bucket

 

namespace OpenHash
{
	template<class K>
	struct Hash
	{
		size_t operator()(const K& key)
		{
			return key;
		}
	};
	//  specialized 
	template<>
	struct Hash < string >
	{
		// "int"  "insert" 
		//  The string is converted to a corresponding integer value , Because of shaping, we can get the modular mapping position 
		//  expect -> String is different , The shaping value transferred out shall be as different as possible 
		// "abcd" "bcad"
		// "abbb" "abca"
		size_t operator()(const string& s)
		{
			// BKDR Hash
			size_t value = 0;
			for (auto ch : s)
			{
				value += ch;
				value *= 131;
			}

			return value;
		}
	};

	template<class T>
	struct HashNode
	{
		HashNode<T>* _next;
		T _data;

		HashNode(const T& data)
			:_next(nullptr)
			, _data(data)
		{}
	};

	//  Forward declaration 
	template<class K, class T, class KeyOfT, class HashFunc>
	class HashTable;

	//  iterator 
	template<class K, class T, class KeyOfT, class HashFunc = Hash<K>>
	struct __HTIterator
	{
		typedef HashNode<T> Node;
		typedef __HTIterator<K, T, KeyOfT, HashFunc> Self;
		typedef HashTable<K, T, KeyOfT, HashFunc> HT;
		Node* _node;
		HT* _pht;

		__HTIterator(Node* node, HT* pht)
			:_node(node)
			, _pht(pht)
		{}

		Self& operator++()
		{
			// 1、 There is still data in the current bucket , Then go back in the current bucket 
			if (_node->_next)
			{
				_node = _node->_next;
			}
			// 2、 The current bucket is finished , Need to go to the next bucket .
			else
			{
				//size_t index = HashFunc()(KeyOfT()(_node->_data)) % _pht->_table.size();
				KeyOfT kot;
				HashFunc hf;
				size_t index = hf(kot(_node->_data)) % _pht->_table.size();

				++index;
				while (index < _pht->_table.size())
				{
					if (_pht->_table[index])
					{
						_node = _pht->_table[index];
						return *this;
					}
					else
					{
						++index;
					}
				}
				_node = nullptr;
			}

			return *this;
		}
		T& operator*()
		{
			return _node->_data;
		}
		T* operator->()
		{
			return &_node->_data;
		}
		bool operator != (const Self& s) const
		{
			return _node != s._node;
		}
		bool operator == (const Self& s) const
		{
			return _node == s.node;
		}
	};

	template<class K, class T, class KeyOfT, class HashFunc = Hash<K>>
	class HashTable
	{
		typedef HashNode<T> Node;
		template<class K, class T, class KeyOfT, class HashFunc>
		friend struct __HTIterator;
	public:
		typedef __HTIterator<K, T, KeyOfT, HashFunc> iterator;

		HashTable() = default; //  Displays the default construction for the specified build 

		HashTable(const HashTable& ht)
		{
			_n = ht._n;
			_table.resize(ht._table.size());
			for (size_t i = 0; i < ht._table.size(); i++)
			{
				Node* cur = ht._table[i];
				while (cur)
				{
					Node* copy = new Node(cur->_data);
					//  Insert the head into the new table 
					copy->_next = _table[i];
					_table[i] = copy;

					cur = cur->_next;
				}
			}
		}

		HashTable& operator=(HashTable ht)
		{
			_table.swap(ht._table);
			swap(_n, ht._n);

			return *this;
		}

		~HashTable()
		{
			for (size_t i = 0; i < _table.size(); ++i)
			{
				Node* cur = _table[i];
				while (cur)
				{
					Node* next = cur->_next;
					delete cur;
					cur = next;
				}
				_table[i] = nullptr;
			}
		}
		iterator begin()
		{
			size_t i = 0;
			while (i < _table.size())
			{
				if (_table[i])
				{
					return iterator(_table[i], this);
				}
				++i;
			}
			return end();
		}
		iterator end()
		{
			return iterator(nullptr, this);
		}
		size_t GetNextPrime(size_t prime)
		{
			const int PRIMECOUNT = 28;
			static const size_t primeList[PRIMECOUNT] =
			{
				53ul, 97ul, 193ul, 389ul, 769ul,
				1543ul, 3079ul, 6151ul, 12289ul, 24593ul,
				49157ul, 98317ul, 196613ul, 393241ul, 786433ul,
				1572869ul, 3145739ul, 6291469ul, 12582917ul, 25165843ul,
				50331653ul, 100663319ul, 201326611ul, 402653189ul, 805306457ul,
				1610612741ul, 3221225473ul, 4294967291ul
			};

			size_t i = 0;
			for (; i < PRIMECOUNT; ++i)
			{
				if (primeList[i] > prime)
					return primeList[i];
			}

			return primeList[i];
		}

		pair<iterator, bool> Insert(const T& data)
		{
			KeyOfT kot;
			//  eureka 
			auto ret = Find(kot(data));
			if (ret != end())
				return make_pair(ret, false);

			HashFunc hf;
			//  Load factor to 1 when , Increase capacity 
			if (_n == _table.size())
			{
				vector<Node*> newtable;
				//size_t newSize = _table.size() == 0 ? 8 : _table.size() * 2;
				//newtable.resize(newSize, nullptr);
				newtable.resize(GetNextPrime(_table.size()));

				//  Traverse and get the nodes in the old table , Recalculate the position mapped to the new table , Hang in a new table 
				for (size_t i = 0; i < _table.size(); ++i)
				{
					if (_table[i])
					{
						Node* cur = _table[i];
						while (cur)
						{
							Node* next = cur->_next;
							size_t index = hf(kot(cur->_data)) % newtable.size();
							//  Head insertion 
							cur->_next = newtable[index];
							newtable[index] = cur;

							cur = next;
						}
						_table[i] = nullptr;
					}
				}

				_table.swap(newtable);
			}

			size_t index = hf(kot(data)) % _table.size();
			Node* newnode = new Node(data);

			//  Head insertion 
			newnode->_next = _table[index];
			_table[index] = newnode;
			++_n;

			return make_pair(iterator(newnode, this), true);
		}

		iterator Find(const K& key)
		{
			if (_table.size() == 0)
			{
				return end();
			}
			KeyOfT kot;
			HashFunc hf;
			size_t index = hf(key) % _table.size();
			Node* cur = _table[index];
			while (cur)
			{
				if (kot(cur->_data) == key)
				{
					return iterator(cur, this);
				}
				else
				{
					cur = cur->_next;
				}
			}

			return end();
		}

		bool Erase(const K& key)
		{
			size_t index = hf(key) % _table.size();
			Node* prev = nullptr;
			Node* cur = _table[index];
			while (cur)
			{
				if (kot(cur->_data) == key)
				{
					if (_table[index] == cur)
					{
						_table[index] = cur->_next;
					}
					else
					{
						prev->_next = cur->_next;
					}

					--_n;
					delete cur;
					return true;
				}
				prev = cur;
				cur = cur->_next;
			}
			return false;
		}
	private:
		vector<Node*> _table;
		size_t _n = 0;         //  Number of valid data 
	};
}

  encapsulation unordered_map

namespace ljx
{
	template<class K, class V>
	class unordered_map
	{
		struct MapKeyOfT
		{
			const K& operator()(const pair<K, V>& kv)
			{
				return kv.first;
			}
		};
	public:
		typedef typename OpenHash::HashTable<K, pair<K, V>, MapKeyOfT>::iterator iterator;
		iterator begin()
		{
			return _ht.begin();
		}
		iterator end()
		{
			return _ht.end();
		}
		pair<iterator, bool> insert(const pair<K, V>& kv)
		{
			return _ht.Insert(kv);
		}
		V& operator[](const K& key)
		{
			pair<iterator, bool> ret = _ht.Insert(make_pair(key, V()));
			return ret.first->second;
		}
	private:
		OpenHash::HashTable<K, pair<K, V>, MapKeyOfT> _ht;
	};
}

  encapsulation unordered_set

namespace ljx
{
	template<class K>
	class unordered_set
	{
		struct SetKeyOfT
		{
			const K& operator()(const K& k)
			{
				return k;
			}
		};
	public:
		typedef typename OpenHash::HashTable<K, K, SetKeyOfT >::iterator iterator;
		iterator begin()
		{
			return _ht.begin();
		}
		iterator end()
		{
			return _ht.end();
		}
		pair<iterator, bool> insert(const K k)
		{
			return _ht.Insert(k);
		}
	private:
		OpenHash::HashTable<K, K, SetKeyOfT> _ht;
	};
}

The application of hash

Bitmap

So called bitmap , Is to use each bit to store a certain state , For massive data , There is no duplicate data . It is usually used to judge whether a data store does not exist .
Realization 
namespace Y
{
	template<size_t N>
	class BitSet
	{
	public:
		BitSet()
		{
			_bits.resize(N / 32 + 1, 0);
		}

		//  hold x The mapped bits are marked as 1
		void Set(size_t x)
		{
			assert(x < N);

			//  Work out x The mapped bit is at i It's an integer 
			//  Work out x The mapped bit is in the... Of this integer j bits 
			size_t i = x / 32;
			size_t j = x % 32;

			// _bits[i]  Of the j The bit is marked as 1, And it doesn't affect his other positions 
			_bits[i] |= (1 << j);
		}

		void Reset(size_t x)
		{
			assert(x < N);

			size_t i = x / 32;
			size_t j = x % 32;

			// _bits[i]  Of the j The bit is marked as 0, And it doesn't affect his other positions 
			_bits[i] &= (~(1 << j));
		}

		bool Test(size_t x)
		{
			assert(x < N);

			size_t i = x / 32;
			size_t j = x % 32;

			//  If the first j Is it 1, The result is right and wrong 0, Not 0 It's true 
			//  If the first j be for the purpose of 0, The result is 0,0 It's fake 
			return _bits[i] & (1 << j);
		}
	private:
		vector<int> _bits;
	};
}

Apply to finding intersection , Quickly find out whether a number is in a set .

advantage : Save a space , Fast                         shortcoming : Can only handle shaping

The bloon filter

The bloom filter is made of bloom （Burton Howard Bloom） stay 1970 Put forward in A compact 、 More ingenious Probabilistic data structure structure , Characteristic is Insert and query efficiently , It can be used to tell you “ Something must not exist or may exist ”, It uses multiple hash functions Count , Map a data into a bitmap structure . This method can not only improve the query efficiency , It can also save a lot of memory space

The idea of Bloom filter is to map an element into a bitmap with multiple hash functions , Therefore, the bit of the mapped position must be 1. So you can find it in the following way ： Calculate whether the stored bit position corresponding to each hash value is zero , As long as one is zero , Indicates that the element must not be in the hash table , Otherwise, it may be in the hash table .

Judgment is not accurate , There is a miscalculation , Judgment is not there , accuracy ; For more strings

Generally, deletion is not supported

There is a way to delete , Use multiple bits as counters , When mapping multiple values ,++ Count , deleted ,-- Count ;

struct HashBKDR
{
	// "int"  "insert" 
	//  The string is converted to a corresponding integer value , Because of shaping, we can get the modular mapping position 
	//  expect -> String is different , The shaping value transferred out shall be as different as possible 
	// "abcd" "bcad"
	// "abbb" "abca"
	size_t operator()(const std::string& s)
	{
		// BKDR Hash
		size_t value = 0;
		for (auto ch : s)
		{
			value += ch;
			value *= 131;
		}

		return value;
	}
};

struct HashAP
{
	// "int"  "insert" 
	//  The string is converted to a corresponding integer value , Because of shaping, we can get the modular mapping position 
	//  expect -> String is different , The shaping value transferred out shall be as different as possible 
	// "abcd" "bcad"
	// "abbb" "abca"
	size_t operator()(const std::string& s)
	{
		// AP Hash
		register size_t hash = 0;
		size_t ch;
		for (long i = 0; i < s.size(); i++)
		{
			ch = s[i];
			if ((i & 1) == 0)
			{
				hash ^= ((hash << 7) ^ ch ^ (hash >> 3));
			}
			else
			{
				hash ^= (~((hash << 11) ^ ch ^ (hash >> 5)));
			}
		}
		return hash;
	}
};

struct HashDJB
{
	// "int"  "insert" 
	//  The string is converted to a corresponding integer value , Because of shaping, we can get the modular mapping position 
	//  expect -> String is different , The shaping value transferred out shall be as different as possible 
	// "abcd" "bcad"
	// "abbb" "abca"
	size_t operator()(const std::string& s)
	{
		// BKDR Hash
		register size_t hash = 5381;
		for (auto ch : s)
		{
			hash += (hash << 5) + ch;
		}

		return hash;
	}
};

template<size_t N, class K = std::string,
class Hash1 = HashBKDR,
class Hash2 = HashAP,
class Hash3 = HashDJB>
class BloomFilter
{
public:
	void Set(const K& key)
	{
		//Hash1 hf1;
		//size_t i1 = hf1(key);
		size_t i1 = Hash1()(key) % N;
		size_t i2 = Hash2()(key) % N;
		size_t i3 = Hash3()(key) % N;

		cout << i1 << " " << i2 << " " << i3 << endl;

		_bitset.Set(i1);
		_bitset.Set(i2);
		_bitset.Set(i3);
	}

	bool Test(const K& key)
	{
		size_t i1 = Hash1()(key) % N;
		if (_bitset.Test(i1) == false)
		{
			return false;
		}

		size_t i2 = Hash2()(key) % N;
		if (_bitset.Test(i2) == false)
		{
			return false;
		}

		size_t i3 = Hash3()(key) % N;
		if (_bitset.Test(i3) == false)
		{
			return false;
		}

		//  here 3 Everyone is , It could be something else key Account for the , It is not accurate in , There is a miscalculation 
		//  No, it's accurate 
		return true; 
	}

private:
	bit::BitSet<N> _bitset;
	bit::vector<char> _bitset;

};

void TestBloomFilter()
{
	/*BloomFilter<100> bf;
	bf.Set(" Zhang San ");
	bf.Set(" Li Si ");
	bf.Set(" But the king bull ");
	bf.Set(" Red boy ");

	cout << bf.Test(" Zhang San ") << endl;
	cout << bf.Test(" Li Si ") << endl;
	cout << bf.Test(" But the king bull ") << endl;
	cout << bf.Test(" Red boy ") << endl;
	cout << bf.Test(" The Monkey King ") << endl;*/

	BloomFilter<600> bf;

	size_t N = 100;
	std::vector<std::string> v1;
	for (size_t i = 0; i < N; ++i)
	{
		std::string url = "https://www.cnblogs.com/-clq/archive/2012/05/31/2528153.html";
		url += std::to_string(1234 + i);
		v1.push_back(url);
	}

	for (auto& str : v1)
	{
		bf.Set(str);
	}

	for (auto& str : v1)
	{
		cout << bf.Test(str) << endl;
	}
	cout << endl << endl;

	std::vector<std::string> v2;
	for (size_t i = 0; i < N; ++i)
	{
		std::string url = "https://www.cnblogs.com/-clq/archive/2012/05/31/2528153.html";
		url += std::to_string(6789 + i);
		v2.push_back(url);
	}

	size_t n2 = 0;
	for (auto& str : v2)
	{
		if (bf.Test(str))
		{
			++n2;
		}
	}
	cout << " Similar string misjudgment rate :" << (double)n2 / (double)N << endl;

	std::vector<std::string> v3;
	for (size_t i = 0; i < N; ++i)
	{
		std::string url = "https://zhuanlan.zhihu.com/p/43263751";
		url += std::to_string(6789 + i);
		v3.push_back(url);
	}

	size_t n3 = 0;
	for (auto& str : v3)
	{
		if (bf.Test(str))
		{
			++n3;
		}
	}
	cout << " Misjudgment rate of dissimilar strings :" << (double)n3 / (double)N << endl;

}

Hash cut

Hash cutting is an idea of hash cutting , By cutting into uneven parts , Convenient operation

Put the same elements into the same subscript file through segmentation , And then compare

such as Give one more than 100G The size of log fifile, log There is IP Address , Design algorithm to find the most frequent IP Address ？

Method :

  1. Suppose to generate A0~A99 100 Small file , Read... In turn ip, Calculate each ip Mapped file number ,i=HashBKDR()(ip)%100 This ip Just go in Ai Small file No , same ip It must have entered a small file , So we can directly count the times in the small file
  2. To deal with A0~A99, Read Ai file , If the file is larger than 2G, You can slice it again , If it is less than 2G, Then use a map<string,int> Count the times
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