快速可哈希协议散列函数是否需要返回唯一值？ [英] Do swift hashable protocol hash functions need to return unique values?

问题描述

我正在研究一个iOS快速的Tetris教程*，并且它已经完成并且可以工作。
但是我对一个特定方面感到困惑-Hashable协议。
函数：

I am working through an iOS swift Tetris tutorial* and have it completed and working. But I am puzzled about one particular aspect - the Hashable protocol. The function:

class Block: Hashable, Printable {
    [...]    
    var hashValue: Int { return self.column ^ self.row }

行转到0..9 ，以及第0..20列。这些函数的注释说：我们返回行和列属性的异或，以为每个块生成一个唯一的整数。但是我的理解是0 ^ 1与1 ^ 0相同，依此类推...我想知道如果Hash函数不是像这样唯一的问题，或者冲突通常可以吗？就像我说的那样，该应用程序似乎可以正常工作...

Rows go 0..9, and Columns 0..20. The notes says of this function "We return the exclusive-or of our row and column properties to generate a unique integer for each Block.". But my understanding is that 0^1 would be the same as 1^0, etc... I would like to know if it is a problem if the Hash function is not unique like this, or are collisions are generally OK? As I say, the application appears to work fine...

* https://www.bloc.io/tutorials/swiftris-build-your-first-ios-game-with- swift＃！/ chapters / 681

推荐答案

冲突不是 通常可以 。基本假设是，当且仅当<$ c $时， x 的哈希值是 y 的哈希值。 c> x == y 。如果您认为第2列第1行与第1列第2行相同，则可以。但是我不认为你愿意！该应用程序似乎可以正常运行，但是想必您尚未执行任何需要哈希处理的操作-

Collisions are not "generally OK". The underlying assumption is that the hash value of x is the hash value of y if and only if x == y. If you consider column 2, row 1 the same as column 1, row 2, then fine. But I don't think you do! The application may appear to work, but presumably you have done nothing that requires hashability - yet.

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