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Arada [10]
2 years ago
15

1. Write a set of routines for implementing several stacks and queues within a single array. Hint: Look at the lecture material

on the hybrid implementation.
Computers and Technology
1 answer:
JulsSmile [24]2 years ago
4 0

Answer:

Out of the two methods The 2nd method is a better option for implementing several stacks and arrays using the single array system this is because Method 2 makes efficient use of the available space

Explanation:

A) Method 1 ( Divide the given single array in the size of n/k.)

  • How to use method 1 to implement several stacks involves :

i)  To implement several stacks through an array (x) is by dividing the array in n/k parts.

ii) The k represents the slots in which the different stacks will be placed and the n represents the size of the array x.

iii) If you need to implement at least two stacks place the first stack in the slot of a [0] to a [n/k - 1], and another stack in the slot of a[n/k] to a[2n/k-1].

note: The  disadvantage of this method is that the use of the space of the array is not much efficient. therefore method 2 is better

  • How to use method 1 to implement queues involves :

i) queues can also be implemented through an array. applying the same method above

ii) Divide the array in slots and place the queues in that slots.

This method has a problem with  the efficient utilization of the space.therefore method 2 is preferred

B) Method 2 ( uses the space efficiently ) uses two more arrays to implement stacks which are : Top_array and Next_array

  • how to use method 2 to implement several stacks

i) Store the indexes of the next item that will also be stored in all stacks in this initial stack

ii)The initial actual array is x[] and this will store the stacks.

iii)Simultaneously with several stacks, the stack which contain the free slots in the array x[] will also be maintained.

iv)  The entries of the Top_array[] will be initialized to -1. This implies that all the stacks are empty.

v)  Firstly the entries of the array Next_array[i] will be initialized to i+1, since all the slots initially are free and are pointed to the next slot.

vi) Initialize The top of the free stack that is maintaining the free slots  as 0.

vii)  The complexity of push () (method to insert an element) and pop () (method to delete an element) operations by using this method is O (1).

  • How to use method 2 to implement several queues

The same method applicable to implementing several stacks is used here but with a difference is the presence of three extra arrays which are :

Front_array[] = indicates the number of queues. This array stores the indexes of the front elements of the stacks.

Rear_array[] = determines the sizeof the array k . This array stores the indexes of the last elements of the stacks.

Next_array[] = The array n indicates the size of the single array say x. This array stores the indexes of the next items that is being pushed.

i) The initial actual array is a[] which will store the queues. The free slots will also be maintained.

ii)The entries of the Front_array[] will be initialized to -1. This means  that all the queues are empty initially

ii) Initially the entries of the array Next_array[i] will be initialized to i+1, since all the slots initially are free and are pointed to the next slot.

iv) Apply The complexity of enqueue ()  and dequeue ()  by using this method is O (1).

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#Write a function called string_finder. string_finder should #take two parameters: a target string and a search string. #The fun
adoni [48]

Answer:

I am writing a Python program:

def string_finder(target,search): #function that takes two parameters i.e. target string and a search string

   position=(target.find(search))# returns lowest index of search if it is found in target string

   if position==0: # if value of position is 0 means lowers index

       return "Beginning" #the search string in the beginning of target string

   elif position== len(target) - len(search): #if position is equal to the difference between lengths of the target and search strings

       return "End" # returns end

   elif position > 0 and position < len(target) -1: #if value of position is greater than 0 and it is less than length of target -1

       return "Middle" #returns middle        

   else: #if none of above conditions is true return not found

       return "not found"

#you can add an elif condition instead of else for not found condition as:

#elif position==-1    

#returns "not found"

#tests the data for the following cases      

print(string_finder("Georgia Tech", "Georgia"))

print(string_finder("Georgia Tech", "gia"))

print(string_finder("Georgia Tech", "Tech"))

print(string_finder("Georgia Tech", "Idaho"))

Explanation:

The program can also be written in by using string methods.

def string_finder(target,search):  #method definition that takes target string and string to be searched

       if target.startswith(search):  #startswith() method scans the target string and checks if the (substring) search is present at the start of target string

           return "Beginning"  #if above condition it true return Beginning

       elif target.endswith(search):  #endswith() method scans the target string and checks if the (substring) search is present at the end of target string

           return "End" #if above elif condition it true return End

       elif target.find(search) != -1:  #find method returns -1 if the search string is not in target string so if find method does not return -1 then it means that the search string is within the target string and two above conditions evaluated to false so search string must be in the middle of target string

           return "Middle"  #if above elif condition it true return End

       else:  #if none of the above conditions is true then returns Not Found

           return "Not Found"

6 0
1 year ago
Remember that ""state space"" refers to the space of all potential possibilities. Which dichotomous questions below will success
egoroff_w [7]

Answer:

Please check options are not given. Please check explanation for corrected version.

Explanation:

Options are not mentioned. Please post the complete question.

However, if "below" is removed, the question makes sense. I am taking it that way.

Dichotomous question means those questions which has two outcomes: true or false.

For the given condition, this is possible only if:

  • coin shows head and dice shows 1
  • coin shows head and dice shows 2
  • head, 3
  • head, 4
  • head, 5
  • head, 6
  • tails, and all above cases

However, each time, each mentioned condition should be strictly followed.

And fewer outcome than 1 is virtually or realistically impossible, as both coin and dice will roll out one outcome in any condition certainly.

7 0
2 years ago
Use the loop invariant (I) to show that the code below correctly computes the product of all elements in an array A of n integer
NeTakaya

Answer:

Given Loop Variant P = a[0], a[1] ... a[i]

It is product of n terms in array

Explanation:

The Basic Step: i = 0, loop invariant p=a[0], it is true because of 'p' initialized as a[0].

Induction Step: Assume that for i = n - 3, loop invariant p is product of a[0], a[1], a[2] .... a[n - 3].

So, after that multiply a[n - 2] with p, i.e P = a[0], a[1], a[2] .... a[n - 3].a[n - 2].

After execution of while loop, loop variant p becomes: P = a[0], a[1], a[2] .... a[n -3].a[n -2].

for the case i = n-2, invariant p is product of a[0], a[1], a[2] .... a[n-3].a[n-2]. It is the product of (n-1) terms. In while loop, incrementing the value of i, so i=n-1

And multiply a[n-1] with p, i.e P = a[0].a[1].a[2].... a[n-2].

a[n-1]. i.e. P=P.a[n-1]

By the assumption for i=n-3 loop invariant is true, therefore for i=n-2 also it is true.

By induction method proved that for all n > = 1 Code will return product of n array elements.

While loop check the condition i < n - 1. therefore the conditional statement is n - i > 1

If i = n , n - i = 0 , it will violate condition of while loop, so, the while loop will terminate at i = n at this time loop invariant P = a[0].a[1].a[2]....a[n-2].a[n-1]

6 0
2 years ago
The microprogram counter (MPC) contains the address of the next microcode statement for the Mic1 emulator to execute. The MPC va
Ray Of Light [21]

Answer:

MAR bit

Explanation:

The MPC computers area set of software and hardware standards that was developed by the consortium of computer firms which is led by Microsoft. It contains the address of a next microcode for the Mic1 emulator foe execution.

The part of the executing microcode instruction that determines the value which is placed in the MOC is the MAR bit. The MAR is the memory address register in the CPU which stores the memory address or such addresses to which some data will be sent and also stored.

6 0
1 year ago
Which statement below correctly differentiates between frames and bits? Frames have more information in them than bits. Frames a
9966 [12]
<span>Frames have more information in them than bits.
</span>
<span>Frames are made up of bits but not vice versa.


A bit (BInary digiT) is the basic unit of digital. It can be 0 (logical false, off) or 1 (not logical false - true, on). Four bits are in a nybble, which can have a value of 0 - 15, eight bits are in a byte which can have a value of 0 - 255. Words vary in size, they consist of multiple bytes and are generally correlated with the system's data bus/processor data width (a 64 bit system has an 8 byte word).
</span>
5 0
2 years ago
Read 2 more answers
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