X = [ "Alice", "Bob", "Carol", "Dave" ]
    

      print(X[2])  // ->  "Carol"
    

      X[2] = "Eve"
    

      print(X[2])  // -> "Eve"
    

      X = [ Alice, Bob, Carol, Dave ]
    

    X[2] = "Eve"
        

     print(X[2])
        

Mutable Data Structures

• The programmer's intended ordering is unclear.
• Atomicity/Correctness requires locking.
• Versioning requires copying the data structure.
• Cache coherency is expensive.

Can these problems be avoided?

      X = [ "Alice", "Bob", "Carol", "Dave" ]
    

      print(X[2])  // ->  "Carol"
    

      X[2] = "Eve"
    

Don't allow writes!

But what if we need to update the structure?

Idea 1: Copy

Slooooooooooooooooooooooow!

Idea 2: Break it Down

Data is always added, not replaced!

Immutable Data Structures

(aka 'Functional' or 'Persistent' Data Structures)

• Once an object is created it never changes.
• The object persists until all pointers to it go away, at which point it is garbage collected.
• Only the "root" pointer is ever allowed to change, to point to a new version.

Class Exercise 2

Implement a set with:

         set init()
         boolean is_member(set, elem)
         set insert(set, elem)
    

Putting off Work

x = "expensive()"

print(x)

print(x)

Class Exercise 3

Make it better!

Putting off Work

      concatenate(a, b) { 
        a', front = pop(a)
        if a' is empty {
          return (front, b)
        } else {
          return (front, "concatenate(a', b)")
        }
      }
    

What is the time complexity of this concatenate?

What happens to reads?

Lazy Evaluation

Save work for later...

• ... and avoid work that is never requred.
• ... to spread out work over multiple calls.
• ... for better "amortized" costs.

Example: Amortized Queues

Example: Amortized Queues

      queue enqueue(queue, item) {
        return {
          current : queue.current, 
          todo : push(queue.todo, item)
        )
      }
    

What is the cost?

Example: Amortized Queues

    queue dequeue(queue) {
      if(queue.current != NULL){

        return { current: pop(queue.current), todo: queue.todo }

      } else if(queue.todo != NULL) {

        return { current: reverse(queue.todo), todo: NULL }

      } else { 

        return { current: NULL, todo: NULL }

      }
    }
    

What is the cost?

Example: Amortized Analysis

enqueue(): Push onto todo stack

$O(1) + \text{create } 1 \text{ credit}$

dequeue(): Pop current OR Reverse todo

Either:

• Pop current queue: $O(1)$
• Reverse stack: $O(1) + \text{consume } N \text{ credits}$

Critical requirement of amortized analysis: Must ensure that every credit consumed was once created.