hash-collision | 易学教程

Does a HashMap collision cause a resize?

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问题 When there is a collision during a put in a HashMap is the map resized or is the entry added to a list in that particular bucket? 回答1: When you say 'collision', do you mean the same hashcode? The hashcode is used to determine what bucket in a HashMap is to be used, and the bucket is made up of a linked list of all the entries with the same hashcode. The entries are then compared for equality (using .equals()) before being returned or booted (get/put). Note that this is the HashMap

How to handle hash collisions for Dictionaries in Swift

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问题 TLDR My custom structure implements the Hashable Protocol. However, when hash collisions occur while inserting keys in a Dictionary , they are not automatically handled. How do I overcome this problem? Background I had previously asked this question How to implement the Hashable Protocol in Swift for an Int array (a custom string struct). Later I added my own answer, which seemed to be working. However, recently I have detected a subtle problem with hashValue collisions when using a

Unexpected collision with std::hash

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问题 I know hashing infinite number of string into 32b int must generate collision, but I expect from hashing function some nice distribution. Isn't it weird that these 2 strings have the same hash? size_t hash0 = std::hash<std::string>()("generated_id_0"); size_t hash1 = std::hash<std::string>()("generated_id_1"); //hash0 == hash1 I know I can use boost::hash<std::string> or others, but I want to know what is wrong with std::hash . Am I using it wrong? Shouldn't I somehow "seed" it? 回答1: There's

Is this an appropriate use of python's built-in hash function?

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I need to compare large chunks of data for equality, and I need to compare many per second, fast . Every object is guaranteed to be the same size, and it is possible/likely they may only be slightly different (in unknown positions). I have seen, from the interactive session below, using == operator for byte strings can be slower if the differences are towards the end of the string, and it can be very fast if there is a difference near the start. I thought there might be some way to speed things up using some sort of hash, of course computing the md5 hash and comparing is a fair whack slower,

What's the shortest pair of strings that causes an MD5 collision?

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Up to what string length is it possible to use MD5 as a hash without having to worry about the possibility of a collision? This would presumably be calculated by generating an MD5 hash for every possible string in a particular character set, in increasing length, until a hash appears for a second time (a collision). The maximum possible length of a string without a collision would then be one character less than the longest of the colliding pair. Has this already been tested for MD5, SHA1, etc? intgr Update Ironically, a few weeks after I posted the previous answer, two Chinese researchers,

Examples of Hash-Collisions?

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For demonstration-purposes, what are a couple examples of strings that collide when hashed? MD5 is a relatively standard hashing-option, so this will be sufficient. This page provides these examples of 128 byte values hashing to the same value: d131dd02c5e6eec4693d9a0698aff95c 2fcab58712467eab4004583eb8fb7f89 55ad340609f4b30283e488832571415a 085125e8f7cdc99fd91dbdf280373c5b d8823e3156348f5bae6dacd436c919c6 dd53e2b487da03fd02396306d248cda0 e99f33420f577ee8ce54b67080a80d1e c69821bcb6a8839396f9652b6ff72a70 and d131dd02c5e6eec4693d9a0698aff95c 2fcab50712467eab4004583eb8fb7f89

How does Dictionary use the Equatable protocol in Swift?

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问题 In order to solve this question, I have been playing around with a custom struct that implements the Hashable Protocol. I'm trying to see how many times the equivalency operator overload ( == ) gets called depending on if there is a hash collision or not when populating a Dictionary . Update @matt wrote a much cleaner example of a custom struct that implements the Hashable protocol and shows how often hashValue and == get called. I am copying his code below. To see my original example, check

What is the clash rate for md5? [closed]

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What's the probability for the clash for the md5 algorithm? I believe it is extremely low. You need to hash about 2^64 values to get a single collision among them, on average, if you don't try to deliberately create collisions. Hash collisions are very similar to the Birthday problem . If you look at two arbitrary values, the collision probability is only 2 -128 . The problem with md5 is that it's relatively easy to craft two different texts that hash to the same value. But this requires a deliberate attack, and doesn't happen accidentally. And even with a deliberate attack it's currently not

What's the shortest pair of strings that causes an MD5 collision?

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问题 Up to what string length is it possible to use MD5 as a hash without having to worry about the possibility of a collision? This would presumably be calculated by generating an MD5 hash for every possible string in a particular character set, in increasing length, until a hash appears for a second time (a collision). The maximum possible length of a string without a collision would then be one character less than the longest of the colliding pair. Has this already been tested for MD5, SHA1,

Examples of Hash-Collisions?

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问题 For demonstration-purposes, what are a couple examples of strings that collide when hashed? MD5 is a relatively standard hashing-option, so this will be sufficient. 回答1: This page provides these examples of 128 byte values hashing to the same value: d131dd02c5e6eec4693d9a0698aff95c 2fcab58712467eab4004583eb8fb7f89 55ad340609f4b30283e488832571415a 085125e8f7cdc99fd91dbdf280373c5b d8823e3156348f5bae6dacd436c919c6 dd53e2b487da03fd02396306d248cda0 e99f33420f577ee8ce54b67080a80d1e