6. Using Go language, assume that n keywords are stored in a hash table of size m and conflicts are resolved through the chaining method. At the same time, the length of each chain is known, including the length L of the longest chain. Please describe the process of selecting an element uniformly and randomly from all keys of a hash table and returning the key in the expected time of O(L·(1 + 1/a)).
Wen Xin said, the code cannot run:
In this problem, we need to select an element uniformly and randomly from n keys in a hash table of size m with conflict resolution by chaining method using Go language. In order to achieve the expected time complexity of O(L·(1 + 1/a)), we need to consider the following steps:
First, we need to calculate the load factor of the hash table. The load factor is the ratio of the number of elements n in the hash table to the hash table size m, or n/m.
We then need to use linear probing to resolve conflicts. Linear probing is a method of resolving hash table conflicts by sequentially looking at the next position when a conflict occurs.
Next, we need to generate a random number in the range 0 to n-1. We can use the math/rand package of Go language to generate random numbers.
Finally, we can select a random element by finding the element at this random index position in the linked list. Since the length of the linked list is known, we can directly calculate the random index position.
Here is a possible implementation:
package main
import (
"fmt"
"math/rand"
"time"
)
type ListNode struct {<!-- -->
Val int
Next *ListNode
}
type HashTable struct {<!-- -->
Size int
Buckets[]*ListNode
}
func NewHashTable(size int) *HashTable {<!-- -->
return & amp;HashTable{<!-- -->Size: size, Buckets: make([]*ListNode, size)}
}
func (h *HashTable) Hash(key int) int {<!-- -->
return key % h.Size
}
func (h *HashTable) Insert(key int) {<!-- -->
index := h.Hash(key)
node := NewListNode(key)
h.Buckets[index] = node
}
func (h *HashTable) SelectRandom() (int, *ListNode) {<!-- -->
rand.Seed(time.Now().UnixNano())
index := rand.Intn(h.Size)
node := h.Buckets[index]
return node.Val, node
}
func main() {<!-- -->
hashTable := NewHashTable(10)
for i := 0; i < 100; i + + {<!-- -->
hashTable.Insert(i)
}
for i := 0; i < 10; i + + {<!-- -->
value, _ := hashTable.SelectRandom()
fmt.Println(value)
}
}
This code implements a simple hash table and uses linear probing to resolve collisions. It then uses the SelectRandom function to select a random element from the hash table. This function first generates a random index and then selects an element from that index. Since each element is selected with equal probability, this process meets the expected time complexity of O(L·(1 + 1/a)) required by the question.
Iflytek Spark, the code cannot run:
In order to select an element uniformly and randomly from all keys of the hash table in the desired time of O(L·(1 + 1/a)), we can use the following steps:
1. First, we need to calculate the probability distribution of the length of each chain. This can be achieved by dividing the length of each chain by the total length of all chains. For example, if the length of the longest chain is L and the lengths of the other chains are l1, l2, …, ln, then the probability distribution is P(L) = 1/n, P(l1) = 1/n, …, P (ln) = 1/n.
2. Then, we can generate a random number r in the range [0, 1].
3. Next, we traverse the probability distribution and accumulate the probability value of each chain. When the accumulated probability value is greater than or equal to r, we have found the corresponding keyword.
The following is the code to implement this process in Go language:
package main
import (
"fmt"
"math/rand"
"time"
)
func main() {<!-- -->
n := 10 //Number of keywords
m := 5 // Hash table size
L := 3 //The longest chain length
a := 2 // Parameters for chaining method to resolve conflicts
//Initialize hash table
hashTable := make([][]int, m)
for i := range hashTable {<!-- -->
hashTable[i] = make([]int, L)
}
// Assume that n keywords have been stored in the hash table
for i := 0; i < n; i + + {<!-- -->
hashValue := i % m
chainIndex := i % L
hashTable[hashValue][chainIndex] = i
}
// Calculate probability distribution
probabilities := make([]float64, L)
totalLength := float64(L) * float64(n)
for i := 0; i < L; i + + {<!-- -->
probabilities[i] = totalLength / float64(n)
}
// Generate random numbers
rand.Seed(time.Now().UnixNano())
r := rand.Float64()
// Find the corresponding keyword according to the probability distribution
cumulativeProbability := 0.0
for i := 0; i < L; i + + {<!-- -->
cumulativeProbability + = probabilities[i]
if r <= cumulativeProbability {<!-- -->
selectedKeyword := hashTable[i%m][i%L]
fmt.Printf("Selected keyword: %d
", selectedKeyword)
break
}
}
}
This code first initializes a hash table of size m with conflict resolution via chaining, and then stores n keys into the hash table. Next, it calculates the probability distribution of the length of each chain and generates a random number. Finally, it finds the corresponding keyword based on the probability distribution.
360 Intelligent Brain:
First, we need to define a hash table structure, which contains information such as keywords, linked list, and linked list length. Then, we can use the rand.nextInt() function in the Go language to generate a random number between 0 and m-1 as the subscript of the linked list, thereby obtaining an element in the hash table.
Next, we need to consider how to return this keyword in the desired time of O(L·(1 + 1/a)). Since each node in the linked list has a chain of the same length, we can find the corresponding linked list by comparing the lengths of the linked lists. Specifically, we can first traverse the entire hash table and find the length L of the longest chain in the linked list. Next, we traverse the entire hash table, and for each linked list, calculate its length as n-1, and record the longest chain length L. Finally, we traverse the entire hash table again, find the chain in the linked list with length equal to L, and return the element in the linked list.
In the above process, we need to pay attention to the following points:
1. It is necessary to ensure that the order of traversing the hash table is uniform, otherwise the time complexity may be high.
2. When calculating the length of the longest chain, it is necessary to record the length of each linked list and compare it in subsequent traversals, otherwise the problem of high time complexity may occur.
3. When traversing the hash table, you need to pay attention to the head and tail of the linked list to avoid the problem of high time complexity.
Wisdom is clear, and panic is thrown away:
In the Go language, we can use the following methods to achieve this requirement:
1. First, we need to create a hash table of size m and resolve conflicts through the chaining method. We can use arrays to store keywords and corresponding linked lists.
2. Next, we need to implement a method of selecting keywords uniformly and randomly. We can use the Linear Congruential Generator (LCG) algorithm to generate a random number between [0, m-1]. The LCG algorithm is as follows:
x := (a * x + c) % m
Among them, a, c and m are constants that can be adjusted to change the distribution of random numbers. Here we can set a to 1 and c to 1 to achieve a uniform distribution.
3. After generating a random number, we need to find the keyword in the corresponding linked list. We can use binary search or traverse the head and tail of the linked list. The average time complexity of the binary search method is O(L), and the average time complexity of traversing from the head and tail of the linked list is also O(L).
4. To return the keyword in the desired time of O(L·(1 + 1/a)), we can combine a random generator and a lookup method. Specifically, we can continuously generate n random numbers, each random number corresponding to a keyword. We can then use binary search or traverse the list at the head and tail until all n keywords are found.
5. After all keywords are found, we can use a priority queue (such as a small root heap) to maintain the found keywords and their positions in the linked list. In this way, we can return keywords in order of priority, thus achieving uniform random selection.
6. The following is a simple Go implementation:
package main
import (
"fmt"
"math/rand"
"sort"
)
type Node struct {<!-- -->
Key int
Next *Node
}
type HashTable struct {<!-- -->
Size int
Nodes[]*Node
LCG[2]int
Found bool
}
func NewHashTable(m, n int) *HashTable {<!-- -->
ht := & amp;HashTable{<!-- -->
Size: m,
Nodes: make([]*Node, m),
LCG: [2]int{<!-- -->1, 1},
}
rand.Seed(time.Now().UnixNano())
for i := 0; i < n; i + + {<!-- -->
k := rand.Intn(m)
ht.Nodes[k] = & amp;Node{<!-- -->Key: k}
}
return ht
}
func (ht *HashTable) FindRandomKey() (int, bool) {<!-- -->
x := ht.generateLCG()
for _, node := range ht.Nodes {<!-- -->
for node.Next != nil {<!-- -->
if node.Key == x {<!-- -->
return x, true
}
node = node.Next
}
}
return 0, false
}
func (ht *HashTable) generateLCG() int {<!-- -->
a, b := ht.LCG[0], ht.LCG[1]
ht.LCG[0] = (a*ht.LCG[0] + b) % ht.Size
ht.LCG[1] = (a*ht.LCG[1] + b) % ht.Size
return ht.LCG[0]
}
func main() {<!-- -->
ht := NewHashTable(10, 5)
foundCount := 0
for i := 0; i < 10; i + + {<!-- -->
k, found := ht.FindRandomKey()
if found {<!-- -->
fmt.Printf("Found key: %d\\
", k)
foundCount++
}
}
fmt.Printf("Found %d/10 keys.\\
", foundCount)
}
This implementation creates a hash table of size 10 and uniformly randomly selects 5 keywords. The running results will show the found keywords and their positions.
Tongyi Qianwen:
In the given problem, we can use probabilistic analysis methods to implement this process.
First, we need to maintain an array of pointers, where each element points to a node in the linked list. Then we do a iteration over all the nodes and put them into an array of pointers. In this way, we have an array containing all the linked list nodes.
Next, we treat all linked lists as a whole and calculate the probability of each node being selected based on its total length. Assuming that the total length of the linked list is T, then the probability of each node being selected is 1/T.
Then, we generate a random number R between 0 and T-1 and find the R-th element in the pointer array. If the linked list where this element is located is empty, skip it and continue looking for the next linked list. Repeat this process until a non-empty linked list is found.
Finally, we randomly select an element in the current linked list in a probabilistic manner. Specifically, we can generate a random number K between 0 and L-1, and start from the first node of the current linked list and move K steps each time until a valid node is found.
The above is the general idea of the entire process. The specific implementation may need to be appropriately adjusted according to actual needs and conditions. Overall, this method can return a key in the expected time of O(L·(1 + 1/a)), where a is the probability distribution parameter used to generate the random numbers.
Tiangong:
In a hash table of size m that resolves conflicts by chaining, each key is stored in a chain. The length of the chain can be arbitrary, but the length of the longest chain is known to be L. Now to select an element uniformly and randomly from a hash table and return it in the expected time of O(L·(1 + 1/a)), you can use the following algorithm:
1. First calculate the average length of each chain, that is, the average number of times each keyword is visited in the chain, recorded as a.
2. Select an element from any chain as the starting point, delete the element from the chain, and insert it into a new chain.
3. Repeat step 2 until the number of elements in the new chain reaches m.
4. Randomly select an element from the new chain and return it.
5. Calculate the number of times each keyword is visited, that is, the number of elements in the new chain, and divide it by m to get the access probability of each keyword.
6. Select the keywords to be returned based on the access probability of each keyword.
The time complexity of this algorithm is O(L·(1 + 1/a)), where L is the length of the longest chain and a is the average number of times each keyword is visited in the chain. Specifically, the time complexity of steps 2 and 6 is O(L), and the time complexity of steps 3 and 5 is O(1), so the total time complexity is O(L·(1 + 1/a )).
It should be noted that this algorithm does not guarantee that each keyword has an equal probability of being selected, because the number of times each keyword is visited may be different. However, this algorithm can guarantee to return any keyword within the expected time of O(L·(1 + 1/a)), and has a certain degree of randomness. If you need to ensure that each keyword has an equal probability of being selected, you can use other more complex algorithms, such as the Reservoir Sampling algorithm.
Ling Xiaozhi:
Assume that the size of the hash table is m, the lengths of the chains are L0, L1,..., Ln-1 respectively, where the maximum length of the chain is L, a is the hash table loading factor, then the average length of the chain is L_avg = (L0 + L1 + ... + Ln-1) / n.
Select an element uniformly and randomly from all keywords, that is, select a position i (0 <= i < m), and then randomly select an element in chain i. Assuming that the length of chain i is Li, the time complexity of randomly selecting an element in chain i is O(Li).
Since the hash table uses the linking method to resolve conflicts, the number of elements at each position i obeys a Poisson distribution with mean a. Then the expected length of chain i is E[Li] = a, so the average length of all chains is L_avg = a. Therefore, the expected time complexity of selecting an element uniformly and randomly from all keywords is:
E[T] = Σi=0~m-1 (P(i) * O(Li))
= Σi=0~m-1 (1/m * O(Li))
= (1/m) * Σi=0~m-1 (O(Li))
= (1/m) * n * L_avg
=L_avg/a
Therefore, select an element uniformly and randomly from all keys of the hash table and return that key in the expected time of O(L·(1 + 1/a)).
chatgpt:
To achieve uniform random selection of an element from a hash table and return the key in the expected time of O(L·(1 + 1/a)), you can follow the following steps:
1. Randomly select an index from a linked list.
2. According to the index of the selected linked list, obtain the length L of the linked list.
3. Generate a random number r on the linked list, in the range [0, L-1].
4. Traverse the selected linked list until the r-th element is found and return it.
According to the chaining method to resolve conflicts, we can calculate the average search length a by the length of each chain in the hash table. Then, in step 3, the generated random numbers r and a are used to adjust the search starting point to ensure that a certain element is selected uniformly and randomly.
Overall, this procedure returns the keyword in the expected time of O(L·(1 + 1/a)).
