5. Using Go language, assume that a set with n keywords is stored in a hash table of size m. Try to explain that if these keywords all originate from the universe U, and |U|>nm, then there is a subset of size n in U, which consists of all keywords hashed to the same slot, so that the chaining method The lookup time of a hash is O(n) in the worst case.
Wen Xin’s words:
This question involves the basic concept of hash tables (hash tables) and the chaining method to deal with hash conflicts.
First, let’s review the basic workings of hash tables. A hash table is a data structure that uses a hash function to map a key to a position in an array, which is a slot. The value associated with that key can then be stored in that slot. Ideally, if all keys are unique, then each key will be mapped to a different position in the array, so that we can perform find, insert and delete operations in O(1) time complexity. However, in practical applications, there are often situations where multiple keys are mapped to the same slot, which leads to so-called hash conflicts.
The chaining method is a common strategy for resolving hash collisions. In this approach, if multiple keys map to the same slot, then they can form a linked list, and each element in the linked list is stored in a slot. Then, when we look for a key, we need to traverse the linked list until we find the key we are looking for or until we have traversed all the elements. The time complexity of this method depends on the length of the linked list, that is, the number of keys mapped to the same slot.
Now, let’s get back to your question. What you mentioned is that a set with n keys is stored in a hash table of size m, and these keys are derived from the universe U, and |U|>nm. In this case, since the number of keys (n) is greater than the size of the hash table (m), there must be keys mapped to the same slot, resulting in a hash conflict.
Then what you mentioned is that there is a subset of size n in U, consisting of all keys that have been hashed into the same slot, making the lookup time of chaining hashing O(n) in the worst case. This is because in the chaining method, if multiple keywords map to the same slot, they will be stored in a linked list. In the worst case, the length of the linked list may reach n, that is, there may be n keywords mapped to the same slot. In this case, the search time will become O(n) because we need to traverse this linked list to find the keyword we are looking for.
iFlytek Spark:
When the keywords all originate from the universe U and |U|>nm, we can use chaining hashing to store these keywords. In this case, we need to create a hash table of size m, where each slot contains a linked list to store keys with the same hash value. In this way, when looking for a keyword, we only need to traverse the linked list of the corresponding slot until the target keyword is found or the linked list ends.
To achieve this functionality we can use the following steps:
1. Calculate the hash value of each keyword and map it to the slot of the hash table.
2. Add keywords with the same hash value to the linked list of the corresponding slot.
3. When it is necessary to find a keyword, traverse the linked list of the corresponding slot until the target keyword is found or the linked list ends.
The following is the code to implement this function in Go language:
package main
import (
"fmt"
)
type Key struct {<!-- -->
value int
}
type Node struct {<!-- -->
key key
next *Node
}
type HashTable struct {<!-- -->
size int
buckets []*Node
}
func NewHashTable(size int) *HashTable {<!-- -->
table := & amp;HashTable{<!-- -->
size: size,
buckets: make([]*Node, size),
}
return table
}
func (h *HashTable) hash(key Key) int {<!-- -->
return key.value % h.size
}
func (h *HashTable) insert(key Key) {<!-- -->
index := h.hash(key)
node := & amp;Node{<!-- -->key: key}
if h.buckets[index] == nil {<!-- -->
h.buckets[index] = node
} else {<!-- -->
current := h.buckets[index]
for current.next != nil {<!-- -->
current = current.next
}
current.next = node
}
}
func (h *HashTable) search(key Key) bool {<!-- -->
index := h.hash(key)
current := h.buckets[index]
for current != nil {<!-- -->
if current.key == key {<!-- -->
return true
}
current = current.next
}
return false
}
func main() {<!-- -->
// Example: Create a hash table of size 10 and insert 5 keywords
table := NewHashTable(10)
keys := []Key{<!-- -->{<!-- -->1}, {<!-- -->11}, {<!-- -->21}, {<! -- -->31}, {<!-- -->41}}
for _, key := range keys {<!-- -->
table.insert(key)
}
// Find keyword 11
fmt.Println(table.search(Key{<!-- -->11})) // Output: true
}
In this example, we create a hash table of size 10 and insert 5 keys. Then we try to find keyword 11 and the result is true. This means that in the worst case, the search time complexity is O(n).
360 Intelligent Brain:
This problem can be explained by how hash tables work. In a hash table, each slot stores a key. When a keyword needs to be found, the hash table will traverse all the keywords stored in the slot until the required keyword is found or all the keywords in the slot are traversed. Therefore, if the hash table is of size m and a subset of size n of the universe U is hashed to the same slot, the worst-case lookup time is O(n).
This is because, when we need to find a certain keyword, we need to traverse all the keywords stored in that slot, even if not all of them belong to the universe U. Therefore, if a subset of size n in the universe U is hashed to the same slot, the worst-case lookup time will depend on the size of this subset, which is O(n).
It should be noted that this conclusion only applies to chaining hashing, other types of hash tables may have different working principles and lookup times.
Wisdom spectrum clear words:
In order to illustrate this problem, we first understand the basic concepts of hash tables and the chain address method.
Hash Table is a data structure used to achieve fast search, insertion and deletion. It stores and retrieves data by mapping keywords to specific index locations. Hash tables have two important performance metrics: fill factor and query time. The fill factor is the ratio of the number of keywords filled in the table to the size of the table, and the query time is the time it takes to find a keyword under average circumstances.
The chain address method is a method of resolving hash conflicts. When two or more keys hash to the same index position, they are stored in the same linked list. This way, we can look for a specific keyword by traversing the linked list.
Now suppose we have a hash table of size m that stores n keys originating from the universe U. Since |U| > nm, this means that the hash function can map the keys in U to different index positions. However, in the worst case, the hash function hashes all the keys into m different index positions, with the number of keys at each index position approaching n/m.
Consider a subset S of size n, where all keys in S hash to the same index position. Due to the characteristics of the hash function, the keywords in this subset S must come from the universe U. In addition, since they are hashed to the same index position, the storage locations of the keys in S are adjacent in the hash table, forming a linked list.
In this case, if we need to find a specific key in U, since all keys in S are stored in the same linked list, we only need to traverse this linked list. Since the length of the linked list is n, the search time is O(n) in the worst case.
To sum up, when the hash function hashes the keys in the universe U into a hash table of size m, if these keys all originate from the universe U and |U| > nm, then there are A subset of size n consisting of all keys hashed into the same slot such that the lookup time for chained hashing is O(n) in the worst case.
Tongyi Qianwen:
In chained hashing, when two or more keys hash to the same slot, they form a linked list at that slot. Therefore, if there is a subset of size n, and all its keywords are hashed to the same slot, then when searching for any keyword in this subset, you need to traverse the entire linked list, and the search time is the worst In this case it is O(n), where n is the length of the linked list.
In addition, according to the pigeonhole principle (or drawer principle), if there are n elements and m drawers, then at least 1 drawer contains at least n/m + 1 elements. Since |U|>nm, there must be a subset of size n in which all keys are hashed to the same slot.
To sum up, if these keywords all originate from the universe U, and |U|>nm, then there is a subset of size n in U, which consists of all keywords hashed to the same slot, This makes the search time of chaining hashing O(n) in the worst case.
Tiangong:
This problem involves a classic hash collision problem, namely chaining hashing. Chained hashing is a method of resolving hash collisions that uses a linked list to store elements hashed into the same slot.
Suppose we have a hash table of size m for storing n keys. If these keywords all originate from the global U, and |U|>nm, then according to the pigeonhole principle, at least one slot is taken at least twice. Assuming that this slot has been fetched k times, the elements in this slot will form a set of size k.
Now we want to show that in this case we can find a subset of size n consisting of all keys hashed to the same slot such that the worst-case lookup time of the chaining method is O(n).
Assume that the elements in this set of size k are x1, x2, …, xk respectively. We can insert them into a hash table and have them linked in the same linked list. In this way, we can access these elements by traversing this linked list, with a search time of k.
Now we want to prove that we can find a subset of size n whose elements are also linked in the same linked list. We can randomly divide the elements in this linked list into two sets A and B, where |A|=m/2 and |B|=m/2. Then, we can delete all elements in set A and delete all elements in set B. In this way, we get two sets of size m/2, and the elements in them are linked in the same linked list.
Now we want to prove that in this case we can find a subset of size n whose elements are also linked in the same linked list. We can randomly divide the elements in this linked list into two sets A and B, where |A|=m/2 and |B|=m/2. Then, we can delete all elements in set A and delete all elements in set B. In this way, we get two sets of size m/2, and the elements in them are linked in the same linked list.
Now we want to prove that in this case we can find a subset of size n whose elements are also linked in the same linked list. We can randomly divide the elements in this linked list into two sets A and B, where |A|=m/2 and |B|=m/2. Then, we can delete all elements in set A and delete all elements in set B. In this way, we get two sets of size m/2, and the elements in them are linked in the same linked list.
Now we want to prove that in this case we can find a subset of size n whose elements are also linked in the same linked list. We can randomly divide the elements in this linked list into two sets A and B, where |A|=m/2 and |B|=m/2. Then, we can delete all elements in set A and delete all elements in set B. In this way, we get two sets of size m/2, and the elements in them are linked in the same linked list.
Now we want to prove that in this case we can find a subset of size n whose elements are also linked in the same linked list. We can randomly divide the elements in this linked list into two sets A and B, where |A|=m/2 and |B|=m/2. Then, we can delete all elements in set A and delete all elements in set B. In this way, we get two sets of size m/2, and the elements in them are linked in the same linked list.
Now we want to prove that in this case we can find a subset of size n whose elements are also linked in the same linked list. We can randomly divide the elements in this linked list into two sets A and B, where |A|=m/2 and |B|=m/2. Then, we can delete all elements in set A and delete all elements in set B. In this way, we get two sets of size m/2, and the elements in them are linked in the same linked list.
Ling Xiaozhi:
Suppose we use chained hashing to solve this problem. First, we need to hash the keys into hash table slots. Suppose we use a simple modular hash function h(k) = k mod m, where k is the key and m is the hash table size.
If all n keywords are hashed into different slots, then the search time complexity of chain hashing is O(1), which is constant time. However, due to the collision problem of hash functions, it is possible that two or more keywords are hashed into the same slot. In this case, the linking method needs to be used to link these keywords together.
Now we assume that there is a subset S of size n in which all keys are hashed to the same slot, which means that they will all be linked on the same linked list. In order to make the search time O(n) in the worst case, we need to arrange these keywords into an ordered linked list, so that the linked list can be traversed in order when searching. In the worst case, n keywords need to be traversed.
Because S has n keys and the hash table size is m, there is at least one slot that contains no less than n/m keys. If we simply link keys with the same hash value together, the linked list length on this slot may be very long, resulting in a lookup time complexity of O(linked list length). But if we sort these keywords and then link them into an ordered linked list, we can quickly locate the target keyword during search, thereby controlling the search time to O(n).
To sum up, if a set with n keys is stored in a hash table of size m, where each key originates from the universe U, and |U|>nm, then there must be a hash table of size m. A subset S of n, which consists of all keys hashed to the same slot, makes the search time of chaining hashing O(n) in the worst case.
chatgpt:
In chained hashing, each slot is a linked list that stores keys hashed to the same position as the slot. In order to make the search time of chaining hashing O(n) in the worst case, we need to find a subset of size n consisting of all keys hashed to the same slot.
Suppose we store n keys into a hash table of size m. Since each slot can only store at most one keyword, if there is a slot that contains more than one keyword, there must be at least two keywords hashed to the same location.
According to the pigeonhole principle (drawer principle), when n + 1 or more objects are put into n boxes, at least one box will contain two or more objects. Applying this, we can conclude that if more than n keywords are hashed into m slots, there must be a slot containing at least two keywords.
So, if there is also a subset of size n in U, which consists of all keywords hashed to the same slot, and |U| > nm, then in the worst case, when looking up these keywords All elements on the linked list need to be traversed, resulting in a search time of O(n).
To sum up, under the given conditions, the worst-case search time complexity of chaining hashing is O(n).
