Convert the following infix notation to its postfix notation. You must show the stack contents in your simulation. r*2 <= s != [{(x == 3+ 5) && (3!= 2 / 3)} == 4 * r] Priority: */,% (high) +,- <,<=,>= =, != && (low)
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- Convert the following infix notation to its postfix notation. You must show the stack contents in your simulation. (5+ x!=d)==|{(3!=e*6) && (1==5/5} != 4 -51 Priority: */,% (high) t, = I= && (low)Given a stack, switch_pairs function takes a stack as a parameter and thatswitches successive pairs of numbers starting at the bottom of the stack.For example, if the stack initially stores these values:bottom [3, 8, 17, 9, 1, 10] topYour function should switch the first pair (3, 8),the second pair (17, 9), ...:bottom [8, 3, 9, 17, 10, 1] topif there are an odd number of values in the stack, the value at the top of thestack is not moved: For example:bottom [3, 8, 17, 9, 1] topIt would again switch pairs of values, but the value at thetop of the stack (1)would not be movedbottom [8, 3, 9, 17, 1] topNote: There are 2 solutions:first_switch_pairs: it uses a single stack as auxiliary storagesecond_switch_pairs: it uses a single queue as auxiliary storage""".Write a Java public static general function which doesn't belong to the Stack class, isGreaterThan that determines if one stack is greater than another stack. If the summation of every other item(starting at the top) from one stack is greater than the summation of every other item(starting at the top) from another, return true, otherwise, return false. After you call this method, both stacks should stay the same as they are called before. Receives: a Stack type argument S1 and another Stack type argument S2 Postcondition: If the summation of every other item (starting at the top) from S1 is greater than the summation of every other item (starting at the top) from S2, return true, otherwise, return false. Both S1 and S2 should stay the same as they are called before. public static boolean isGreaterThan(Stack S1, Stack S2)
- You may operate on a stack WORK and a temporary stack TEMP (if necessary) to enable its ADT operations of PUSH (S,X), POP (S, X), and EMPTYSTACK (S), where X represents an element/variable to be pushed into or popped out of the stack and S represents a stack. You may also utilise one variable to carry out the procedures if necessary. I Given n different random integers to be pushed into WORK, how can you determine the maximum element pushed into it while ensuring that the items placed in WORK are in decreasing order, with the maximum element beginning at the bottom of the stack? You may utilise a single variable and the temporary stack TEMP. ii) Given an array A[1: n] of different random numbers, how can you extract the array's sorted list using only stacks?Imagine a (literal) stack of plates. If the stack gets too high, it might topple.Therefore, in real life, we would likely start a new stack when the previous stack exceeds somethreshold. Implement a data structure SetOfStacks that mimics this. SetOfStacks should becomposed of several stacks and should create a new stack once the previous one exceeds capacity.SetOfStacks. push() and SetOfStacks. pop() should behave identically to a single stack(that is, pop () should return the same values as it would if there were just a single stack).FOLLOW UPImplement a function popAt(int index) which performs a pop operation on a specific substack.Passing Arguments to a FunctionThe sample program in Example 5.4 shows how the stack frame is used to passarguments to a function.The code simply creates some local stack variables, fillsthem with values, and passes them to a function called callex().The callex() function takes the supplied arguments and prints them to the screen.Example 5.4 Stack and Passing Parameters to a Function/* stack2.c */#include <stdlib.h>#include <stdio.h>int callex(char *buffer, int int1, int int2){/* This prints the input variables to the screen:*/printf("%s %d %d\n",buffer,int1, int2);return 1;}int main(int argc, char **argv){Stack Overflows • Chapter 5 173 char buffer[15]="Hello Buffer"; /* a 15-byte character bufferwith12 characters filled/*int int1=1, int2=2; /* two four-byte integers */callex(buffer,int1,int2); /*call our function*/return 1; /*leaves the main function*/}You need to compile this example in MSVC in a console application inRelease mode or in GCC without optimizations. Example…
- A counter automaton is a pushdown automaton that only uses one stack symbol (in additionto the bottom of the stack marker). This means that the stack acts as a simple counter: the push operationincreases the counter, the pop operation decreases it, and you can check if the counter is currently set to 0by looking for the bottom of the stack. In particular, you can not store a string over an arbitrary alphabeton the stack. For example, the automaton for anbn we have seen in class is a counter automaton.Let L = {w ∈{a, b}∗|w contains the same number of symbols a and b}.(2.a) Build a counter automaton accepting the language L. As in Question 1, describe the algorithmused by your automaton, and draw a transition diagram.(2.b) Write down an accepting computation of your automaton on the string abbbabaa.A SpecialStackTM is a stack modified to support the following two operations: PUSHCLEAR(v) successively pops consecutive items from the top of the stack that are less than u, then pushes v onto the stack. POP() deletes the item at the top of the stack. Assume these operations are implemented using a singly-linked list. PUSHCLEAR(v) iter- ates through the linked list to pop the applicable items and then adds v to the beginning. POP() deletes the first item. (a) Describe the ordering of values on the stack. (b) Explain how a single call to PUSHCLEAR(V) could take (n) time. (c) If we assume that every operation takes linear time, we get a naive bound of O(n²) on the total runtime. But this doesn't consider the fact that these expensive operations happen infrequently so let's analyze it more closely. Show that any sequence of n operations takes O(n) time. -Write a program for Stack (Array-based or linked list-based) in Python. Test the scenario below with the implementation and with the reasoning of the answer. Make comments with a short description of what is implemented. Include source codes and screen-captured outputs. Stack: A letter means doing a push operation and an asterisk means doing a pop operation in the below sequence. Give the sequence of letters that are returned by the pop operations when this sequence of operations is performed on an initially empty stack. A*BCE**F*GH***I*
- A stack of integer elements is implemented as an array. The index of the topelement is kept in position 0 in the array, and the stack elements are stored in stack[1], …,stack[stack[0]].1. How does this implementation fare when assessed against the idea of an array as ahomogeneous collection of data elements?2. How would this implementation change the stack specification?3. How would it change the implementation of the functions?Imagine a stack of dishes, literally. The stack may collapse if it becomes too tall.Therefore, when the prior stack above a certain threshold in real life, we probably start a new stack. Create a data structure called SetOfStacks to resemble this. When a stack's capacity is reached, a new stack should be created and added to the SetOfStacks. SetOfStacks. SetOfStacks with push(). In other words, pop() should return the same values as it would if there were only one stack. It should operate exactly like a single stack.TAKE ACTIONCreate a function called popAt(int index) that does a pop operation on a certain substack.S and Q are integer stacks and priority queues, respectively. C mod 3 is the priority of an element C joining the priority queue Q. In other words, the elements' priority numbers are either 0 or 1 or 2. What is the output of the following code if A, B, and C are integer variables? However, the queue procedures have been modified to work on a priority queue. 1.A = 102.B = 113.C = A+B4.while (C < 110) do5. if (C mod 3) = 0 then PUSH (S,C)6. else ENQUEUE (Q,C)7. A = B8. B = C9. C = A + B10.end11.while not EMPTY_STACK (S) do12. POP (S,C)13. PRINT (C)14.end15.while not EMPTY_QUEUE (Q) do16. DEQUEUE (Q, C)17. PRINT (C)18.end