10.3 Implementing pointers and objects
10.3-1
Draw a picture of the sequence $\langle 13, 4, 8, 19, 5, 11 \rangle$ stored as a doubly linked list using the multiple-array representation. Do the same for the single-array representation.
-
A multiple array version could be $L = 2$,
$$ \begin{array}{ccccccc} / & 3 & 4 & 5 & 6 & 7 & / \\ & 12 & 4 & 8 & 19 & 5 & 11 \\ & & 2 & 3 & 4 & 5 & 6 \end{array} $$
-
A single array version could be $L = 4$,
$$ \begin{array}{cccccccccccccccccc} 12 & 7 & / & 4 & 10 & 4 & 8 & 13 & 7 & 19 & 16 & 10 & 5 & 19 & 13 & 11 & / & 16 \end{array} $$
10.3-2
Write the procedures $\text{ALLOCATE-OBJECT}$ and $\text{FREE-OBJECT}$ for a homogeneous collection of objects implemented by the single-array representation.
1 2 3 4 5 6 | ALLOCATE-OBJECT() if free == NIL error "out of space" else x = free free = A[x + 1] return x |
1 2 3 | FREE-OBJECT(x) A[x + 1] = free free = x |
10.3-3
Why don't we need to set or reset the $prev$ attributes of objects in the implementation of the $\text{ALLOCATE-OBJECT}$ and $\text{FREE-OBJECT}$ procedures?
We implement $\text{ALLOCATE-OBJECT}$ and $\text{FREE-OBJECT}$ in the hope of managing the storage of currently non-used object in the free list so that one can be allocated for reusing. As the free list acts like a stack, to maintain this stack-like collection, we merely remember its first pointer and set the $next$ attribute of objects. There is no need to worry the $prev$ attribute, for it hardly has any impact on the resulting free list.
10.3-4
It is often desirable to keep all elements of a doubly linked list compact in storage, using, for example, the first $m$ index locations in the multiple-array representation. (This is the case in a paged, virtual-memory computing environment.) Explain how to implement the procedures $\text{ALLOCATE-OBJECT}$ and $\text{FREE-OBJECT}$ so that the representation is compact. Assume that there are no pointers to elements of the linked list outside the list itself. ($\textit{Hint:}$ Use the array implementation of a stack.)
1 2 3 4 5 | ALLOCATE-OBJECT() if STACK-EMPTY(F) error "out of space" else x = POP(F) return x |
1 2 3 4 5 6 7 8 | FREE-OBJECT(x) p = F.top - 1 p.prev.next = x p.next.prev = x x.key = p.key x.prev = p.prev x.next = p.next PUSH(F, p) |
10.3-5
Let $L$ be a doubly linked list of length $n$ stored in arrays $key$, $prev$, and $next$ of length $m$. Suppose that these arrays are managed by $\text{ALLOCATE-OBJECT}$ and $\text{FREE-OBJECT}$ procedures that keep a doubly linked free list $F$. Suppose further that of the $m$ items, exactly $n$ are on list $L$ and $m - n$ are on the free list. Write a procedure $\text{COMPACTIFY-LIST}(L, F)$ that, given the list $L$ and the free list $F$, moves the items in $L$ so that they occupy array positions $1, 2, \ldots, n$ and adjusts the free list $F$ so that it remains correct, occupying array positions $n + 1, n + 2, \ldots, m$. The running time of your procedure should be $\Theta(n)$, and it should use only a constant amount of extra space. Argue that your procedure is correct.
We represent the combination of arrays $key$, $prev$, and $next$ by a multible-array $A$. Each object of $A$'s is either in list $L$ or in the free list $F$, but not in both. The procedure $\text{COMPACTIFY-LIST}$ transposes the first object in $L$ with the first object in $A$, the second objects until the list $L$ is exhausted.
1 2 3 4 5 6 7 8 9 10 11 12 13 | COMPACTIFY-LIST(L, F) TRANSPOSE(A[L.head], A[1]) if F.head == 1 F.head = L.head L.head = 1 l = A[L.head].next i = 2 while l != NIL TRANSPOSE(A[l], A[i]) if F == i F = l l = A[l].next i = i + 1 |
1 2 3 4 5 | TRANSPOSE(a, b) SWAP(a.prev.next, b.prev.next) SWAP(a.prev, b.prev) SWAP(a.next.prev, b.next.prev) SWAP(a.next, b.next) |