hash-collision

问题 This question already has answers here : Why does Java's hashCode() in String use 31 as a multiplier? (13 answers) Closed 2 years ago . I'm new to Java, and I've recently learned about hashCode() . On the wikipedia article about Java hashCode(), there is the following example of a hashCode() method: public class Employee { int employeeId; String name; Department dept; // other methods would be in here @Override public int hashCode() { int hash = 1; hash = hash * 17 + employeeId; hash = hash *

问题 This question already has answers here : Why does Java's hashCode() in String use 31 as a multiplier? (13 answers) Closed 2 years ago . I'm new to Java, and I've recently learned about hashCode() . On the wikipedia article about Java hashCode(), there is the following example of a hashCode() method: public class Employee { int employeeId; String name; Department dept; // other methods would be in here @Override public int hashCode() { int hash = 1; hash = hash * 17 + employeeId; hash = hash *

问题 How to identify whether or not the keys in a std::unordered_map have experienced hash collisions? That is, how to identify if any collision chaining is present? 回答1: You can use the bucket interface and its bucket_size method. std::unordered_map<int, int> map; bool has_collision = false; for(size_t bucket = 0; bucket < map.bucket_count(); bucket++) { if(map.bucket_size(bucket) > 1) { has_collision = true; break; } } 来源： https://stackoverflow.com/questions/46137811/how-to-identify-whether-or

问题 How to identify whether or not the keys in a std::unordered_map have experienced hash collisions? That is, how to identify if any collision chaining is present? 回答1: You can use the bucket interface and its bucket_size method. std::unordered_map<int, int> map; bool has_collision = false; for(size_t bucket = 0; bucket < map.bucket_count(); bucket++) { if(map.bucket_size(bucket) > 1) { has_collision = true; break; } } 来源： https://stackoverflow.com/questions/46137811/how-to-identify-whether-or

问题 If counting from 1 to X, where X is the first number to have an md5 collision with a previous number, what number is X? I want to know if I'm using md5 for serial numbers, how many units I can expect to be able to enumerate before I get a collision. 回答1: Theoretically, you can expect collisions for X around 2 64 . For a hash function with an output of n bits, first collisions appear when you have accumulated about 2 n/2 outputs (it does not matter how you choose the inputs; sequential integer

问题 If counting from 1 to X, where X is the first number to have an md5 collision with a previous number, what number is X? I want to know if I'm using md5 for serial numbers, how many units I can expect to be able to enumerate before I get a collision. 回答1: Theoretically, you can expect collisions for X around 2 64 . For a hash function with an output of n bits, first collisions appear when you have accumulated about 2 n/2 outputs (it does not matter how you choose the inputs; sequential integer

问题 I've done quite a bit of checking up on other questions and I'm still uncertain on the issue. Here's my usage case: I have an online shopping cart. Occassionaly, certain clients find the ordering process either too tedious, or there are some clients where an online order will not cut it, and they need an actual PDF estimate (quote) in order to purchase a product. So I coded in a module that takes the shopping cart contents, and lays out neatly as a PDF estimate. Now because this process only

问题 How does google's sparse hash table handle collisions? i.e. when 2 elements map to the same bucket, how does it decide where to put the new (colliding) element? I'm reading What is the main implementation idea behind sparse hash table? but that answer doesn't cover the collision idea. 回答1: Your question is answered in the documentation here, specifically: 2c) If t.sparsetable[i % 32] is assigned, but to a value other than foo, look at t.sparsetable[(i+1) % 32] . If that also fails, try t

问题 Given two different messages, A and B (maybe 20-80 characters of text, if size matters at all), what is the probability that the MD5 digest of A is the same as the MD5 digest of B and the SHA1 digest of A is the same as the SHA1 digest of B? That is: (MD5(A) == MD5(B)) && (SHA1(A) == SHA1(B)) Assume no malicious intent, i.e., that the messages are not selected with an aim of finding a clash. I just want to know the odds of this happening naturally. I'm thinking the chances are "astronomically

问题 Consider the following: HashMap<String, Object> hm = new HashMap<>(); final String prefix = "My objects "; int counter = 0; void put(Object value) { hm.put(prefix+(counter++), value); } Given that the key of each entry starts with the same string and only differs by a number appended to it, is this likely to lead to more collisions? I'm trying to decide whether this way of creating unique keys is a good idea from a performance perspective. 回答1: No it will not. And that is not necessarily