Scala Memoization:这个Scala备忘录是如何工作的？

Question

以下代码来自Pathikrit的动态编程存储库.我的美丽和独特使我感到困惑.

def subsetSum(s: List[Int], t: Int) = {
  type DP = Memo[(List[Int], Int), (Int, Int), Seq[Seq[Int]]]
  implicit def encode(key: (List[Int], Int)) = (key._1.length, key._2)

  lazy val f: DP = Memo {
    case (Nil, 0) => Seq(Nil)
    case (Nil, _) => Nil
    case (a :: as, x) => (f(as, x - a) map {_ :+ a}) ++ f(as, x)
  }

  f(s, t)
}

该类型Memo在另一个文件中实现:

case class Memo[I <% K, K, O](f: I => O) extends (I => O) {
  import collection.mutable.{Map => Dict}
  val cache = Dict.empty[K, O]
  override def apply(x: I) = cache getOrElseUpdate (x, f(x))
}

我的问题是:

1. 为什么type K声明为(Int, Int)subsetSum？

2. 什么是int在(Int, Int)分别代表？

3.如何(List[Int], Int)隐式转换为(Int, Int)？
我看不到implicit def foo(x:(List[Int],Int)) = (x._1.toInt,x._2).(甚至在Implicits.scala它导入的文件中也没有.

*编辑:好吧,我想念这个:

implicit def encode(key: (List[Int], Int)) = (key._1.length, key._2)

我非常喜欢Pathikrit的图书馆scalgos.里面有很多Scala珍珠.请帮助我,这样我就能体会到Pathikrit的机智.谢谢.(:

Answer 1

我是上述代码的作者.

/**
 * Generic way to create memoized functions (even recursive and multiple-arg ones)
 *
 * @param f the function to memoize
 * @tparam I input to f
 * @tparam K the keys we should use in cache instead of I
 * @tparam O output of f
 */
case class Memo[I <% K, K, O](f: I => O) extends (I => O) {
  import collection.mutable.{Map => Dict}
  type Input = I
  type Key = K
  type Output = O
  val cache = Dict.empty[K, O]
  override def apply(x: I) = cache getOrElseUpdate (x, f(x))
}

object Memo {
  /**
   * Type of a simple memoized function e.g. when I = K
   */
  type ==>[I, O] = Memo[I, I, O]
}

在Memo[I <% K, K, O]:

I: input
K: key to lookup in cache
O: output

该行I <% K意味着K可以从中查看(即隐式转换)I.

在大多数情况下,I应该是,K例如,如果您正在编写fibonacci类型函数,则可以单独Int => Int缓存Int.

但是,有时当你编写memoization时,你不想总是通过输入本身进行memoize或cache(I)而是输入()的函数,K例如当你编写subsetSum具有类型输入的算法时(List[Int], Int),你不想要使用List[Int]作为缓存的关键,而是你想使用List[Int].size作为缓存的关键组成部分.

所以,这是一个具体案例:

/**
 * Subset sum algorithm - can we achieve sum t using elements from s?
 * O(s.map(abs).sum * s.length)
 *
 * @param s set of integers
 * @param t target
 * @return true iff there exists a subset of s that sums to t
 */
 def isSubsetSumAchievable(s: List[Int], t: Int): Boolean = {
    type I = (List[Int], Int)     // input type
    type K = (Int, Int)           // cache key i.e. (list.size, int)
    type O = Boolean              // output type      

    type DP = Memo[I, K, O]

    // encode the input as a key in the cache i.e. make K implicitly convertible from I
    implicit def encode(input: DP#Input): DP#Key = (input._1.length, input._2)   

    lazy val f: DP = Memo {
      case (Nil, x) => x == 0      // an empty sequence can only achieve a sum of zero
      case (a :: as, x) => f(as, x - a) || f(as, x)      // try with/without a.head
    }

    f(s, t)
 }

您可以将所有这些缩短为一行: type DP = Memo[(List[Int], Int), (Int, Int), Boolean]

对于常见情况(何时I = K),您可以简单地执行此操作:type ==>[I, O] = Memo[I, I, O] 并使用它来计算递归memoization 的二项式系数:

  /**
   * http://mathworld.wolfram.com/Combination.html
   * @return memoized function to calculate C(n,r)
   */
  val c: (Int, Int) ==> BigInt = Memo {
    case (_, 0) => 1
    case (n, r) if r > n/2 => c(n, n - r)
    case (n, r) => c(n - 1, r - 1) + c(n - 1, r)
  }

要查看上述语法的详细信息,请参阅此问题.

这是一个完整的例子,它通过将输入的参数编码为:来计算editDistance:(Seq, Seq)(Seq.length, Seq.length)

 /**
   * Calculate edit distance between 2 sequences
   * O(s1.length * s2.length)
   *
   * @return Minimum cost to convert s1 into s2 using delete, insert and replace operations
   */
  def editDistance[A](s1: Seq[A], s2: Seq[A]) = {

    type DP = Memo[(Seq[A], Seq[A]), (Int, Int), Int]
    implicit def encode(key: DP#Input): DP#Key = (key._1.length, key._2.length)

    lazy val f: DP = Memo {
      case (a, Nil) => a.length
      case (Nil, b) => b.length
      case (a :: as, b :: bs) if a == b => f(as, bs)
      case (a, b) => 1 + (f(a, b.tail) min f(a.tail, b) min f(a.tail, b.tail))
    }

    f(s1, s2)
  }

最后,规范的斐波那契例子:

lazy val fib: Int ==> BigInt = Memo {
  case 0 => 0
  case 1 => 1
  case n if n > 1 => fib(n-1) + fib(n-2)
}

println(fib(100))