1. Advanced Higher Computing Science Stacks Queues and linked lists

2. Why do we need data structures? How we represent data when solving a problem should mirror the real world as closely as possible Data structures should be what you think about when first approaching a problem

3. Are all complex data structures arrays? An array is a set of contiguous blocks of memory. Resizing an array is a complex and memory intensive operation Storing data in non contiguous blocks makes the data structure more flexible but more complex

4. Real world examples: Stacks Browser history Interrupts Blog posts

5. Real world examples: Stacks Undo function To do lists

6. The Stack A stack is a 1-dimensional array or list where data elements can be added or retrieved only from the top of the stack. The terms push and pop refer to adding and removing items respectively. In a stack, a pointer is required to identify the location of the top of the stack A stack is often referred to as a "Last in First Out" structure

7. The stack

8. The interrupt stack: Run programA Interrupt event (interrupt1) Store programA register contents in memory. Push programA memory location on to stack Run interrupt1 routine Interrupt event (Interrupt2) Store interrupt1 register contents in memory. Push interrupt 1 memory location on to stack Run interrupt2 routine Interrupt2 complete Pop interrupt1 memory location from stack Retrieve interrupt1 status and register contents Continue interrupt1 routine interrupt1 complete Pop programA memory location from stack Retrieve programA status and register contents Continue programA

9. Implementing a stack stackPointer is used to keep track of the top of the stack max is the maximum number of items which the stack can hold stack[] is an array to hold the data

10. Implementing a stack CLASS Stack IS {INTEGER stackPointer, INTEGER max, INTEGER stack[]} METHODS CONSTRUCTOR Stack(INTEGER size) DECLARE THIS.stack INITIALLY []*size DECLARE THIS.stackPointer INITIALLY 0 DECLARE THIS.size INITIALLY size END CONSTRUCTOR PROCEDURE Push(INTEGER value) IF THIS.stackPointer=THIS.size THEN ELSE SET THIS.stack[THIS.stackPointer] TO value SET THIS.stackPointer TO THIS.stackPointer+1 END IF END PROCEDURE

11. Implementing a stack FUNCTION pop() RETURNS INTEGER IF THIS.stackPointer=0 THEN ELSE SET THIS.stackPointer TO THIS.stackPointer -1 RETURN THIS.stack[THIS.stackPointer] END IF END FUNCTION END CLASS

12. Real world examples: Queues Tech Support helplines

13. Real world examples: Queues Scheduled tasks Printer queues

14. The Queue A queue is a 1-dimensional array or list, but data elements are inserted and retrieved at different ends. In a queue two pointers are required: one to point to the head of the queue and one to point to the end. A queue is a "First in First Out" structure.

15. The Circular queue

16. When the last space in the array is reached the rear pointer is moved to the start of the array. What would this queue look like after the following operations? Leave 6 Join 77

17. The Circular queue What would this queue look like after the following operations? Leave 6 Join 77

18. Implementing a Circular queue start is used to keep track of the front of the queue rear is used to keep track of the back of the queue maxSize is the maximum number of items which the queue can hold currentSize is the current number of items which the queue is holding queue[] is an array to hold the data

19. Implementing a circular queue CLASS Queue IS {INTEGER start, INTEGER rear, INTEGER currentSize, INTEGER maxSize, ARRAY OF INTEGER queue} METHODS CONSTRUCTOR Queue(INTEGER size) DECLARE THIS.start INITIALLY 0; DECLARE THIS.rear INITIALLY 0; DECLARE THIS.currentSize INITIALLY 0 DECLARE THIS.maxSize INITIALLY size; DECLARE THIS.queue INITIALLY INTEGER[maxSize] END CONSTRUCTOR

20. Implementing a circular queue PROCEDURE join(INTEGER data) IF THIS.currentSize = THIS.MaxSize THEN SEND "Queue Overflow" TO DISPLAY ELSE SET THIS.queue[THIS.rear] TO data SET THIS.rear TO THIS.rear + 1 SET THIS.currentSize TO THIS.currentSize + 1 END IF IF THIS.rear > THIS.maxSize THEN SET THIS.rear TO 1 END IF END PROCEDURE

21. Implementing a circular queue FUNCTION leave() RETURNS INTEGER IF THIS.currentSize = 0 THEN SEND "Queue Underflow" TO DISPLAY RETURN 0 ELSE RETURN THIS.queue[THIS.start] SET THIS.queue[THIS.start] TO 0 SET THIS.currentSize TO THIS.currentSize - 1 SET THIS.start TO THIS.start + 1 END IF IF THIS.start > THIS.maxSize THEN SET THIS.start TO 1 END IF END FUNCTION END CLASS

22. A queue with one pointer Join 56

23. A queue with one pointer queuePointer keeps track of the front of the queue size is the maximum number of items which the queue can hold queue[] is an array to hold the data

24. A queue with one pointer CLASS Queue IS {ARRAY OF INTEGER queue, INTEGER queuePointer, INTEGER size} METHODS CONSTRUCTOR Queue(INTEGER queueSize) DECLARE THIS. size INITIALLY queueSize DECLARE THIS.queue INITIALLY [size] DECLARE THIS.queuePointer INITIALLY 0 END CONSTRUCTOR

25. A queue with one pointer PROCEDURE join(INTEGER value) IF THIS.queuePointer=THIS.size THEN ELSE SET THIS.queue[THIS.queuePointer] TO value SET THIS.queuePointer TO THIS.queuePointer+1 END IF END PROCEDURE

26. A queue with one pointer FUNCTION leave() RETURNS INTEGER DECLARE result INITIALLY IF THIS.queuePointer=0 THEN ELSE SET result TO THIS.queue[0] FOR i FROM 0 TO THIS.queuePointer-2 DO SET THIS.queue[i] TO THIS.queue[i+1] END FOR SET THIS.queuePointer TO THIS.queuePointer-1 END IF RETURN result END FUNCTION END CLASS

27. Real world examples: Linked lists Road train File blocks on disk

28. Linked Lists A linked list is a dynamic data structure, which means that its size is not fixed Each item in a linked list consists of the data and a link to the memory location of the next item. A linked list is very flexible as the order of items in a linked list can be changed without actually moving any data around, just the links between them.

29. File blocks on disk Why does this data need to be stored as a dynamic structure? What causes disk fragmentation and why is it a problem? What is done during disk defragmentation?

30. Linked Lists it is not possible to identify a specific item directly using its index in the way that you can when storing data in an array. To identify the position of an item in a linked list you have to walk through the list from beginning to end. Linked lists can be used to implement queues and stacks as well as arrays, with the advantage that they do not need to have a fixed size, avoiding overflow errors.

31. Implementing a linked list PROCEDURE setupList() DECLARE data INTEGER INITIALLY 0 DECLARE head POINTER INITIALLY NUL # POINTER is a link to a memory location# END PROCEDURE PROCEDURE newNode(INTEGER data, POINTER next) IF head = NUL THEN next = NUL ELSE next = NUL END IF END PROCEDURE

32. Implementing a linked list PROCEDURE deleteNode(INTEGER data) IF head = NUL THEN SEND "List is empty" TO DISPLAY ELSE IF THEN END IF DECLARE found INITIALLY FALSE REPEAT IF THEN SET found TO true END IF UNTIL current item pointer = NUL IF found = false THEN SEND "not found" TO DISPLAY END IF END PROCEDURE

33. Using a linked list to implement a stack or a queue Since a linked list is a dynamic structure, a stack can grow and shrink as items are pushed and popped. The stack pointer effectively becomes the last item in the list If a linked list is used to implement a queue, since it is a dynamic structure, the front and back of the queue are the last and first items in the list. Both structures will require a variable to store the maximum number of items allowed in the linked list.

34. Specimen Paper Q 3a A computerised version of a card game, based on various animals native to Scotland, is being developed for a website. During game play, players can take a card from or place a card on a pile of cards. A stack data structure will represent this pile of cards. The stack is held in a 1-D array and the last item placed in the stack was the Golden Eagle. The 1-D array in which the stack is held is shown below:

35. Specimen Paper Q 3a (i) An item is added to the stack by “pushing” and removed by “popping”. Draw the final state of the stack after the following five operations: 1. Pop 2. Push Loch Ness Monster 3. Pop 4. Pop 5. Push Grouse IndexCharacter 0Ptarmigan 1Otter 2Golden Eagle 3 4

36. Specimen Paper Q 3a (i) An item is added to the stack by “pushing” and removed by “popping”. Draw the final state of the stack after the following five operations: 1. Pop 2. Push Loch Ness Monster 3. Pop 4. Pop 5. Push Grouse IndexCharacter 0Ptarmigan 1Grouse 2 3 4

37. Specimen Paper Q 3 (ii & iii) (ii) Apart from the 1-D array, describe another item of data required to implement a stack. (iii) When a stack is implemented using a 1-D array, adding a valid item can cause an execution error. Explain why an execution error can occur in this situation.

38. Specimen Paper Q 3 (ii & iii) (ii) Apart from the 1-D array, describe another item of data required to implement a stack. A stack pointer (INTEGER) is required to store the index position of the top of the stack (iii) When a stack is implemented using a 1-D array, adding a valid item can cause an execution error. Explain why an execution error can occur in this situation.

39. Specimen Paper Q 3 (ii & iii) (ii) Apart from the 1-D array, describe another item of data required to implement a stack. A stack pointer (INTEGER) is required to store the index position of the top of the stack (iii) When a stack is implemented using a 1-D array, adding a valid item can cause an execution error. Explain why an execution error can occur in this situation. Stack Overflow occurs when the array is full and an attempt is made to access an index position which does not exist.

40. Specimen Paper Q 3b (i) A linked list could have been used rather than a stack to represent the pile of cards. (i) The animals Ptarmigan, Otter and Golden Eagle are entered, in the order given, into a linked list. Draw this linked list.

41. Specimen Paper Q 3b (i) A linked list could have been used rather than a stack to represent the pile of cards. (i) The animals Ptarmigan, Otter and Golden Eagle are entered, in the order given, into a linked list. Draw this linked list.

42. Specimen Paper Q 3b (ii) (ii) Explain why the execution error in part (a) (iii) would not occur if a linked list is used rather than a stack.

43. Specimen Paper Q 3b (ii) (ii) Explain why the execution error in part (a) (iii) would not occur if a linked list is used rather than a stack. A linked list is a dynamic structure (items are not stored contiguously in memory) so the number of items is not fixed and items can be easily added

44. 2014 Q4 In the early days of hand-held calculators, ‘Postfix’ notation was used to reduce memory access during a calculation and made use of the stack to evaluate expressions. Postfix notation avoids the use of brackets. The arithmetical expression 6 * (4 + 3) is written in Postfix notation as 6 4 3 + * and is then processed, left to right, using a stack.

45. 2014 Q4

46. 0123456 6 Stack pointer 0 0123456 64 1 At the beginning of the process the stack is empty. The first element input is a 6 and this is pushed onto the stack. After the number “4” has been pushed on to the stack, the state of the stack is: The arithmetical expression 6 * (4 + 3) is written in Postfix notation as 6 4 3 + * (a) Explain why the “stack pointer” is needed during stack operations.

47. 2014 Q4 (b) (i) Following the algorithm above, show the state of the stack after the 3 has been pushed on to the stack. (ii) Show the contents of the stack, and the value of the stack pointer, after the algorithm has run to completion.

48. 2014 Q4 (i) Following the algorithm above, show the state of the stack after the 3 has been pushed on to the stack. ii) Show the contents of the stack, and the value of the stack pointer, after the algorithm has run to completion.

49. 2014 Q4 (i) Following the algorithm above, show the state of the stack after the 3 has been pushed on to the stack. 0123456 643 Stack pointer 2 ii) Show the contents of the stack, and the value of the stack pointer, after the algorithm has run to completion.

50. 2014 Q4 (i) Following the algorithm above, show the state of the stack after the 3 has been pushed on to the stack. 0123456 643 Stack pointer 2 ii) Show the contents of the stack, and the value of the stack pointer, after the algorithm has run to completion. 0123456 67 Stack pointer 1

51. 2014 Q4 (i) Following the algorithm above, show the state of the stack after the 3 has been pushed on to the stack. 0123456 643 Stack pointer 2 ii) Show the contents of the stack, and the value of the stack pointer, after the algorithm has run to completion. 0123456 67 Stack pointer 1 0123456 42 Stack pointer 0

52. 2014 Q4 (c) The algorithm above makes use of the “pop” operation. Use pseudocode to describe the operation to pop a value from the stack. (d) Describe one problem that should be checked for when pushing a value onto a stack.

53. 2014 Q4 IF THEN SEND "Stack underflow" TO DISPLAY ELSE SET outputValue TO stack[stackPointer] SET stackPointer TO stackPointer -1 END IF Stack c) d) Stack overflow

54. 2012 4b A 1-D array with six elements is used to hold the contents of a queue and the variables front and rear are used to hold the positions of the item at the front and rear of the queue. The following lines of code are executed: MyQ.addtoback(15) MyQ.addtoback(29) MyQ.remove MyQ.addtoback(8) The following diagram shows that 29 and 8 are in the queue. The number 15 is still present in the 1-D array but is not in the queue. Index012345 15298 front 1 rear 2

55. 2012 4b (i)State what happens to the variables front and rear when a number is removed from the queue. ii) The state of MyQ shown above is changed by executing these additional lines of code: MyQ.addtoback(11) MyQ.remove MyQ.addtoback(9) Draw a diagram that shows the new state of MyQ

56. 2012 4b (i)State what happens to the variables front and rear when a number is removed from the queue. SET front TO front + 1 ii) The state of MyQ shown above is changed by executing these additional lines of code: MyQ.addtoback(11) MyQ.remove MyQ.addtoback(9) Draw a diagram that shows the new state of MyQ

57. 2012 4b (i)State what happens to the variables front and rear when a number is removed from the queue. SET front TO front + 1 ii) The state of MyQ shown above is changed by executing these additional lines of code: MyQ.addtoback(11) MyQ.remove MyQ.addtoback(9) Draw a diagram that shows the new state of MyQ

58. 2012c (i) Describe the problem that will arise as items continue to be added and removed from MyQ. (ii) Describe how the problem in (c)(i) could be solved.

59. 2012c (i) Describe the problem that will arise as items continue to be added and removed from MyQ. The end of the array will be reached but there is still room in the array for items to be placed in the queue (ii) Describe how the problem in (c)(i) could be solved.

60. 2012c (i) Describe the problem that will arise as items continue to be added and removed from MyQ. The end of the array will be reached but there is still room in the array for items to be placed in the queue (ii) Describe how the problem in (c)(i) could be solved. Either use a circular queue where items joining the queue will be added to the start of the array and the pointer rear will become 1 Or Reset the array every time an item is removed by shuffling items one place up

61. 2010 Q2

62. (a)Explain why the characters L, E and D have not been moved to the locations identified by indices 0, 1 and 2 after characters C and A have been processed. (b) State the values stored in the variables front and rear after the characters O and N have been added and a further three characters removed from the queue.

63. 2010 Q2 (a)Explain why the characters L, E and D have not been moved to the locations identified by indices 0, 1 and 2 after characters C and A have been processed. It would be very inefficient to move every character one place up every time a key is pressed (b) State the values stored in the variables front and rear after the characters O and N have been added and a further three characters removed from the queue.

64. 2010 Q2 (a)Explain why the characters L, E and D have not been moved to the locations identified by indices 0, 1 and 2 after characters C and A have been processed. It would be very inefficient to move every character one place up every time a key is pressed (b) State the values stored in the variables front and rear after the characters O and N have been added and a further three characters removed from the queue. Front = 5 Back = 6

65. 2010 Q2 c (c) Characters will continue to be added and removed from the queue held in the 1-D array. (i)State the problem encountered as characters continue to be added and removed. (ii) Describe how to solve the problem encountered in (i).

66. 2010 Q2c (c) Characters will continue to be added and removed from the queue held in the 1-D array. (i)State the problem encountered as characters continue to be added and removed. Rear will reach 9 but the queue is not full as there are spaces at the front of the array. (ii) Describe how to solve the problem encountered in (i).

67. 2010 Q2c (c) Characters will continue to be added and removed from the queue held in the 1-D array. (i)State the problem encountered as characters continue to be added and removed. Rear will reach 9 but the queue is not full as there are spaces at the front of the array. (ii) Describe how to solve the problem encountered in (i). The solution would be to wraparound where items are added at the start of the array.

68. 2010 Q2d Another data structure used in programming is a stack. (i)Explain how the operation of a stack data structure differs from a queue. (ii) Explain why a keyboard buffer does not use a stack data structure.

69. 2010 Q2d Another data structure used in programming is a stack. (i)Explain how the operation of a stack data structure differs from a queue. In a stack items are popped and pushed at the same end. In a queue items leave at one end and join at the other (ii) Explain why a keyboard buffer does not use a stack data structure.

70. 2010 Q2d Another data structure used in programming is a stack. (i)Explain how the operation of a stack data structure differs from a queue. In a stack items are popped and pushed at the same end. In a queue items leave at one end and join at the other (ii) Explain why a keyboard buffer does not use a stack data structure. Characters would be processed in the wrong order ( last key pressed processed before earlier characters )