ADS_Lec2_Linked_Allocation

Advanced Data
Structures:
Linked Allocation
Sub code: 12CSE102 for Mtech CSE
Hemanth Kumar G,
Assistant Professor,
Department of CSE, NMAMIT, Nitte
http://hemanthglabs.wordpress.in
http://veda-vijnana.blogspot.in

Linked vs Sequential Allocation
• Linked allocation
o Additional mem space for links. May be a dominating factor!
o Info in a node doesn’t take up a whole word.
• So already there is a space for link field.
o An app may combine several items into one node
• So only one link for several items of info.
o Implicit gain in storage as tables can overlap, sharing common parts.
• Sequential allocation
o If large no. of mem locs are left vacant, then only it is useful.

o Easy to delete an item from within a linked list.
o Easy to insert an item into the midst of a list.
o References to random parts of the linked list is slow.
o To gain access to the kth item in the list, we need k iterations to march down
to the right place.
o Deletion requires moving a large part of the list up into different locations.
o Inserting an item into the midst will be extremely time consuming in a long
sequential table.
o References to random parts of the list is faster in the sequential case.
o To gain access to the kth item in the list, takes a fixed time.

o Easier to join two lists together, or to break one apart into two that will grow
independently.
o Lends itself immediately to more intricate structures like
• variable-size lists
• Any nod of the list may be a starting point for another list
• Nodes may simultaneously linked together in several orders corresponding
to different lists, etc.,
o If the elements of a linked list belong to different pages in a bulk memory, the
memory accesses might take significantly longer.
o Simple operations, like proceeding sequentially through a list, are slightly faster.

Theory of Linked Lists
• Node has one word & broken into 2 fields:
• There must be some mechanism to find empty
space available for new node:
o “list of available space” ---- “AVAIL list (stack)”
o Avail refers to the top element of this list.
• If we want to set link variable X to the address of a
new node, and to reserve that node for future use:
o X ← AVAIL, AVAIL ← LINK (AVAIL) (4)

Operations on Linked List
• When a node is deleted and no longer needed, process
(4) can be reversed:
Link (X)  AVAIL, AVAIL  X. (5)
• This operation puts the node addressed by X back onto
the list of raw material; we denote (5) by “AVAIL <= X”.
• Setting up AVAIL stack:
a) linking together all nodes that are to be used for linked mem,
b) setting AVAIL to the address of the first of these nodes, and
c) Making the last node link to Λ
• The set of all nodes that can be allocated is called the
storage pool.

Handling AVAIL Stack (Technique-1)
• Check (4) if all available mem space has been
taken:
o If AVAIL = Λ, then OVERFLOW;
otherwise X ← AVAIL, AVAIL ← LINK (AVAIL) (6)
• Check always for OVERFLOW:
o terminate with regrets or
o Go to Garbage collection routine that attempts to find more
available space.

Handling AVAIL Stack (Technique-2)
• We often do not know in advance how much mem
space should be used for the storage pool.
• There may be a sequential table of var-size that wants to
coexist in mem with linked tables;
o In such a case we do not want the linked mem area to take any more space
than is absolutely necessary.
• L0 = beginning of ascending mem locs
• SEQMIN = current lower bound of the sequential table.
• If we want to place the linked mem area in Li < SEQMIN,
then we can proceed as follows, using a new var
POOLMAX.

Special recovery procedure for Overflow
Address Sequential Table Pointers
L0 + 5c SEQ Node 3
L0 + 4c SEQ Node 2
L0 + 3c SEQ Node 1  SEQMIN
L0 + 2c AVAIL Node 3  POOLMAX
L0 + c AVAIL Node 2
L0 AVAIL Node 1

Special recovery procedure for Overflow
• Keeps storage pool as small as possible.
• Can be applied when all lists occupy the storage
pool area.
• It avoids time consuming operation of initially linking
all available cells together.
• Facilitates debugging.
• You can put sequential list on the bottom & the
pool on the top:
o POOLMIN & SEQMAX

Common list operations
Insertion Operation – (8)
Deletion Operation – (9)
17 cycles, but
12 cycles in
sequential
table
(overflow is
longer)

Cyclic permutation in list operations
• In the insertion operation let P be the value of AVAIL before
the insertion; if P ≠ Λ, we find that after operation

Insertion & Deletion in Lists
• Insertion of 21/2 would have been done by using (8) with
T=LINK(LINK(FIRST))
• Linked allocation also applies to Queues.

Conditions in List design
• List empty condition
• Programming Error:
1. Failure to handle empty lists properly
Soln:- Examine the boundary conditions carefully.
2. Forget about changing some of the links when a structure
is being manipulated.
Soln:- Draw “before & after” diagrams & compare them, to
see which links must change.

Conditions applied on Q design
Before Insertion:-
Insert @ R:-
After Insertion:-
Boundary situation when the Q is empty: Fig (15) shows
situation after insertion. Situation before insertion is yet
to be determined

Queue Deletion Operation
• Check boundary conditions properly
• Basic operations – one way to do the things, there
will be multiple solutions to a problem.
• Abstract discussions  Practical examples.

Case Study
• Network problems. PERT Charts, linguistics.
• Problems involving ”Partial Ordering”.
• A partial ordering of a set S is a relation between
the objects of S denoted by , satisfying the
following properties for any objects x, y, and z in S:

GettinG Bored..? Let’s Have
An Activity On Relations..
• I want 9 students to volunteer on the podium and pick the announced card.
• The card(s) you have, will be your relative(s).
• When I ask that anybody have relative- ‘k’ do the following:
• Lift your hand if you have that relative’s card.
• Go & increment the COUNT node on the blackboard to prove your presence in the
class.
• If you have a relative, go & create nodes on the board for each relative, linking from
your node.
• Fill your relative’s number on the SUC field on the board.
• If nobody has a relative card, he/she will become independent, fills the SUC filed if
he/she will have any relative & walks out of the group to create new party ;)
• Based on the entries on the board rearrange yourselves based on SUC, NEXT &
COUNT fields.

Case Study: Topological Sorting
• S is a finite set & we work inside a computer.
• (II) => there are no closed loops in the diagram.
• In Fig.6, 4  1 violates partial ordering

Case Study: Topological Sorting
• Embed the partial order in a linear order:
o Linear sequence: a1a2…. an : whenever
o Boxes are to be rearranged into a line so that all arrows go towards right.
o May not be possible in all the cases. E.g. No reordering In case of loops.
• Algorithm gives useful operation & proves it is
possible for every partial ordering.

Applications of Topological Sorting
1. Large glossary of technical terms.
• if the definition of the word w1 depends
(in)directly on that of word w2.
o A partial ordering if there are no “circular” definitions.
• Goal: Find a way to arrange the words in the glossary so
that no term is used before it has been defined.
2. Writing programs to process the declarations in certain
assembly & compiler languages.
3. User’s manual describing a computer language
4. Writing textbook about Information Structures.

Topological Sorting
• Simple Algorithm:
1. Take an obj that is not preceded by any other obj in the ordering.
2. This obj may be placed first in the output.
3. Remove this obj from the set S;
• Resulting set is again partially ordered.
4. Repeat the process until the whole set has been sorted.
• Failure:- If there were a nonempty partially ordered set in
which every element was preceded by another.
• If every element is preceded by another, we could construct
an arbitrarily long sequence b1, b2, b3….. in which
• Since S is finite, we must have bj = bk for some j < k; =>
Contradicting (ii).

Implementation of Topological Sorting
• Ready to perform the actions like locate & remove the objects
from the set.
• Influenced by desired I/O characteristics:
o Alphabetic names for gigantic objects exceeding memory limit..??
• Let the objects to be sorted are numbered from 1 to n in any
order.
• The input of the program will be on tape unit 1:
o Each tape record contains 50 pairs of numbers
o Pair (j, k) = object j precedes object k.
o First pair is (0, n), n = # objs.
o Pair (0, 0) terminates the input.
o We shall assume that n + # relation pairs will fit comfortably in memory.
o Assume no input validation.
o O/P:- # objs in sorted order, followed by the number 0, on tape unit 2.

Hand trace of Topological data
• E.g.,
• Don’t give unnecessary more pairs.
o can be deduced from
o i.e., Give only the pairs corresponding to arrows on a diagram (fig 6.)
• Algorithm uses a sequential table X[1], X[2], …, X[n], and
each node X[k] has the form:
• Count[k] = # direct predecessors of object k
• Top[k] = Link to the beginning of the list of direct
successors of object k.
• Latter list contains entries in the format:
• SUC = direct successor of k, NEXT = next item of the list.

Computer Representation corresponding to relations (18)
• O/p the fields whose COUNT = 0. Then COUNT[Successors]--
• Trick:- Avoids searching for nodes with COUNT = 0.
• How? By a Queue containing those nodes. Links for this queue are kept in
COUNT field. Reusability!!
• Use QLINK[k] for COUNT[k] when that field is no longer being used for counting

Algorithm T (Topological Sort)
• I/p: Pairs of relations with 1 ≤ j,k ≤ n.
• O/p: Set of n objects embedded in linear order.
• Internal tables:
o QLINK[0], COUNT[1] = QLINK[1], COUNT[2] = QLINK[2], …, COUNT[N] = QLINK[n];
o TOP[1], TOP[2], …, TOP[n];
o A storage pool with one node for each I/p relation and with SUC & NEXT fields;
o P, a link variable used to refer to the nodes in the storage pool;
o F & R, integer-valued variables of a queue whose links are in QLINK table;
o N, a variable that counts how many objects have yet to be output.

Algorithm T (Topological Sort)

Characteristics of Algorithm T
• Nice interplay b/n Seql mem & linked mem techniques.
• Seql mem: X[1], …, X[n]; with COUNT[k] & TOP[k]
o References to “random” parts of this table in step T3.
• Linked mem: tables of “immediate successors”.
o No particular order in the I/p for these table entries.
• The Q of nodes waiting to be O/p is kept in the midst of the seql
table by linking the nodes together in O/p order.
o This linking is done by table index instead of by address;
• When front of the Q is X[k], we have F=k instead of F = Loc(X[k])

Analysis of Algorithm T
• By Kirchoff’s law Execn time ≈ c1m + c2n
o m = # I/p relations
o n = # Objects
o c1 & c2 = Constants
• Hard to imagine faster algorithm for this algorithm!
• Total running time: (32m + 24n + 7b + 2c + 16)u.
o a = # objects with no predecessor
o b = # tape records in I/p =
o c = # tape records in O/p =

Exercise
Item 5Item 4Item 3Item 2Λ Item 1 First

Programming Practices
• Develop the functions & a test driver for the following:
o Accepts a general list, traverses it, and returns the key of the node with the
minimum key value.
o Traverses a general list and deletes all nodes whose keys are negative.
o Traverses a general list and deletes all nodes that are after a node with a
negative key.
o Traverses a general list and deletes all nodes that are before a node with a
negative key.
o Search list for a given key & return the node number if it is present.
o Returns a pointer to the last node in a general list.
o Appends two general lists together.
o Appends a general list to itself.
o Inverts a general list.

Reference
“The Art of Computer Programming” – Volume 1, on
Fundamental Algorithms by Donald E. Knuth, Stanford
University, published by Pearson Education © 1997.

ADS_Lec2_Linked_Allocation

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