ADS_Lec2_Linked_Allocation

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Linked allocation vs. Sequential allocation, Operations on Linked Lists realized in terms of Stacks & Queues with boundary conditions, programming errors & rectifications, Cyclic permutation in Stack push & pop, Partial Ordering, Group Activity, Case Study of Topological Sorting; Characterization, explanation, analysis & applications of Topological Sorting Algorithm, Exercise on Inverting a Linked list, Programs for practice.

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ADS_Lec2_Linked_Allocation

  1. 1. 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
  2. 2. Linked Allocation
  3. 3. 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.
  4. 4. Linked vs Sequential Allocation • Linked allocation 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. • Sequential allocation 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.
  5. 5. Linked vs Sequential Allocation • Linked allocation 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. • Sequential allocation o Simple operations, like proceeding sequentially through a list, are slightly faster.
  6. 6. 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)
  7. 7. 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.
  8. 8. 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.
  9. 9. 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.
  10. 10. 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
  11. 11. 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
  12. 12. Common list operations Insertion Operation – (8) Deletion Operation – (9) 17 cycles, but 12 cycles in sequential table (overflow is longer)
  13. 13. 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
  14. 14. 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.
  15. 15. 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.
  16. 16. 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
  17. 17. 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.
  18. 18. 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:
  19. 19. Case Study Exercise
  20. 20. 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.
  21. 21. 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
  22. 22. 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.
  23. 23. 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.
  24. 24. 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).
  25. 25. 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.
  26. 26. 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.
  27. 27. 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
  28. 28. Topological Sorting Flowchart
  29. 29. 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.
  30. 30. Algorithm T (Topological Sort)
  31. 31. Algorithm T (Topological Sort)
  32. 32. 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])
  33. 33. 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 =
  34. 34. Exercise Item 5Item 4Item 3Item 2Λ Item 1 First
  35. 35. 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.
  36. 36. Reference “The Art of Computer Programming” – Volume 1, on Fundamental Algorithms by Donald E. Knuth, Stanford University, published by Pearson Education © 1997.

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