Algorithm goal

A most-recently used cache (MRU) stores a set of key-value pairs in order of their most recent access but evicts the most-recently used one when space is needed. 'MRU' refers to the eviction policy in this case.

Implement an immutable MRU cache which returns the most recently touched key (candidate for eviction), get a value (and 'touches' it), and puts values ('touching' and possibly causing an eviction).

Test cases in Scala

assert(
  MRU.init(1)("X" -> 1).get("X").map(_._1).contains(1),
  "First put element in the cache remains there"
)
assert(
  MRU.init(1)("X" -> 1).maybeNewest.contains("X"),
  "First put element in the cache remains is the head"
)
assert(
  MRU.init(1)("X" -> 1).put("Y", 1).get("X").isEmpty,
  "Cache becomes full, so X is no longer there after putting Y"
)
assert(
  MRU
    .init(2)("X" -> 1, "Y" -> 2)
    .get("Y")
    .map(_._2.put("Z", 3))
    .flatMap(_.get("X"))
    .isEmpty,
  "For a cache of size 1, the least recently used element X gets evicted"
)
assert(
  MRU
    .init(2)("X" -> 1, "Y" -> 2)
    .get("Y")
    .map(_._2.put("Z", 3))
    .flatMap(_.get("Y"))
    .map(_._1)
    .contains(2),
  "For a cache of size 2, the most recently used element Y is not evicted"
)
assert(
  MRU
    .init(2)("X" -> 1, "Y" -> 2)
    .get("Y")
    .map(_._2.put("Z", 3))
    .flatMap(_.get("Y"))
    .flatMap(_._2.maybeNewest)
    .contains("Z"),
  "For a cache of size 2, the oldest element is Z, since X was evicted and the last access was Y"
)

Algorithm in Scala

59 lines of Scala (compatible versions 2.13 & 3.0).

Explanation

MRU caches are typically mutable, implemented using mutable linked lists, however the approach similar to that in MakeQueueUsingMap can be used in order to create a linked list using immutable structures. While it is not typical to use an MRU that is immutable, this is extremely useful in immutable state machines, which otherwise would be more difficult and less performant to implement without immutable underlying structures.

In this approach, we create an explicit EmptyMRU class to make the implementation simpler (otherwise we would have to deal with `oldestKey` that is Optional - complicating computations. This class allows for a simple creation of a non-empty MRU. (this is © from www.scala-algorithms.com)

A Map is used to represent a linked list: we define the start and the end with 'newestKey' and 'oldestKey', and then the chain via 'newer' and 'older' Maps. Note that immutable maps in Scala are highly optimized for immutability.

Scala concepts & Hints

1. foldLeft and foldRight

   A 'fold' allows you to perform the equivalent of a for-loop, but with a lot less code.

   def foldMutable[I, O](initialState: O)(items: List[I])(f: (O, I) => O): O =
     items.foldLeft(initialState)(f)
   

   ▶ Run code in the IDE

2. Option Type

   The 'Option' type is used to describe a computation that either has a result or does not. In Scala, you can 'chain' Option processing, combine with lists and other data structures. For example, you can also turn a pattern-match into a function that return an Option, and vice-versa!

   assert(Option(1).flatMap(x => Option(x + 2)) == Option(3))
   
   assert(Option(1).flatMap(x => None) == None)
   

   ▶ Run code in the IDE

3. Pattern Matching

   Pattern matching in Scala lets you quickly identify what you are looking for in a data, and also extract it.

   assert("Hello World".collect {
     case character if Character.isUpperCase(character) => character.toLower
   } == "hw")
   

   ▶ Run code in the IDE

4. Stack Safety

   Stack safety is present where a function cannot crash due to overflowing the limit of number of recursive calls.

   This function will work for n = 5, but will not work for n = 2000 (crash with java.lang.StackOverflowError) - however there is a way to fix it :-)

   In Scala Algorithms, we try to write the algorithms in a stack-safe way, where possible, so that when you use the algorithms, they will not crash on large inputs. However, stack-safe implementations are often more complex, and in some cases, overly complex, for the task at hand.

   def sum(from: Int, until: Int): Int =
     if (from == until) until else from + sum(from + 1, until)
   
   def thisWillSucceed: Int = sum(1, 5)
   
   def thisWillFail: Int = sum(1, 300)
   

   ▶ Run code in the IDE