Convert the given NFA to a DFA.
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- Suppose you have a finite state machine that accepts bit strings divisible by 11. The FSM has 11 states R0, R1, ... , R10, each corresponding to the remainder when dividing by 11. Input is fed to the FSM one bit at a time from left to right, the same way we did in our in-class examples. If the machine is currently in state R7, an input of 1 should transition it to which state? Type the subscript only of the new state. Your answer should be an integer between 0 and 10, inclusive.Please answer this Python in the simplest way. A1input.txt B001,book,Patriot Games, 15.95B002,book,Origin, 19.95C001,cloth,Armani Suit, 800.00B003,book,Animal Farm, 9.97B004,book,Grant, 22.50E001,elect,EyePhone 10,795.00E002,elect,First Alert Smoke Alarm, 29.95F001,food,Moose Drool Ale 6-pack, 9.95C002,cloth,Pants, 39.95B005,book,Prairie Fires, 18.95E003,elect,Sony Prtable Radio, 15.00C003,cloth,Vasque Hiking Boots, 109.00C004,cloth,Wool Hat, 14.00F002,food,Jumbo shrimp, 12.50E004,elect,HP Laptop, 350.00C005,cloth,Wrangler Jeans, 24.50B005,book,Ragtime, 17.25F003,food,Fusili - 16 oz., 2.95C006,cloth,Nike T-shirt, 19.00C007,cloth,Gore-Tex Gloves, 39.00C008,cloth,North Face Fleece Jacket, 89.95C009,cloth,Nationals Logo Sweatshirt, 49.00E005,elect,LinkSys Router, 49.95F004,food,Lamb Chops, 21.95C010,cloth,New Balance Trail Runners,69.95E006,elect,Altec Lansing Speakers, 195.95B006,book,Future Shock, 8.95E007,elect,LG 55 UDTV,350.00E008,elect,Dell All-in-One PC, 495.00E009,elect,Brother…2. Q3) The questions refer to the example of a Turing machine deciding the language {a^nb^nc^n | n \in N} a) On input aabbbc, how many transitions take place before the state "no" is reached? (Include the one leading to "no"). b) On input aabbbc, what is the content of the tape (excluding bottom) after 10 transitions? c) On input aabbcc, what is the current tape position in the first configuration we reach having state 4? Remember that we start counting tape positions at 0.
- Using a 5-bit cell to store integers in 2's complement, what is the largest integer that can be stored? Question 74 options: A. 15 B. 31 C. 3 D. 8 E. 7 Consider the following program segment, where a is an integer constant and b is an integer variable that holds a positive value.int r = 0, n = b;while (n != 0){ r += 2*a; n--; }Which of the following is a loop invariant for the while loop? r = 2*a*(b-1); r = a*b; r = 2*a*b ; r = a*(b+1);Modiflow y the beprogram given to include response time program;FCFS CPU SCHEDULING ALGORITHM #include<stdio.h>#include<conio.h>main(){int bt[20], wt[20], tat[20], i, n; float wtavg, tatavg;clrscr();printf("\nEnter the number of processes -- "); scanf("%d", &n); for(i=0;i<n;i++){printf("\nEnter Burst Time for Process %d -- ", i); scanf("%d", &bt[i]);}wt[0] = wtavg = 0; tat[0] = tatavg = bt[0]; for(i=1;i<n;i++){wt[i] = wt[i-1] +bt[i-1];tat[i] = tat[i-1] +bt[i]; wtavg = wtavg + wt[i]; tatavg = tatavg + tat[i];}printf("\t PROCESS \tBURST TIME \t WAITING TIME\t TURNAROUND TIME\n");for(i=0;i<n;i++){printf("\n\t P%d \t\t %d \t\t %d \t\t %d", i, bt[i], wt[i], tat[i]);}printf("\nAverage Waiting Time -- %f", wtavg/n);printf("\nAverage Turnaround Time -- %f", tatavg/n); getch();}Modiflow y the beprogram given to include response time program;FCFS CPU SCHEDULING ALGORITHM #include<stdio.h>#include<conio.h>main(){int bt[20], wt[20], tat[20], i, n; float wtavg, tatavg;clrscr();printf("\nEnter the number of processes -- "); scanf("%d", &n);for(i=0;i<n;i++){printf("\nEnter Burst Time for Process %d -- ", i); scanf("%d", &bt[i]);}wt[0] = wtavg = 0; tat[0] = tatavg = bt[0];for(i=1;i<n;i++){wt[i] = wt[i-1] +bt[i-1];tat[i] = tat[i-1] +bt[i]; wtavg = wtavg + wt[i]; tatavg = tatavg + tat[i];}printf("\t PROCESS \tBURST TIME \t WAITING TIME\t TURNAROUND TIME\n");for(i=0;i<n;i++){printf("\n\t P%d \t\t %d \t\t %d \t\t %d", i, bt[i], wt[i], tat[i]);}printf("\nAverage Waiting Time -- %f", wtavg/n);printf("\nAverage Turnaround Time -- %f", tatavg/n); getch();}
- Suppose the state machine shown below is in State A and receives the string 10010100. What is its output? Choose the correct answer and show all your intermediate steps to receive full points. A) 00 10 11 10 00 01 11 10 11B) 10 11 10 00 01 11 10 11 10C) 00 10 11 00 01 11 01 11 01D) 10 11 10 10 11 10 11 10 10Question. What is the algorithm to solve the following problem? a. Given two DNA sequences from different individuals, what is the longest shared sequence of nucleotides? b. Given a product ID and a sorted array of all in-stock products, is the product in stock and what is the product's price? c. Given a user's current location and desired location, what is the fastest route to walk to the destination.Present a coding technnique that accepts 2 numbers x,y that will result into a transformation L: R^2 -> R^2 Example: L(x,y) = (x-y,x+y) .
- By assuming that X is the last digit of your student number and 3X is a two digitnumber, consider memory storage of a 64-bit word stored at memory word 3X ina byte-addressable memory(a) What is the byte address of memory word 3X?(b) What are the byte addresses that memory word 3X spans?(c) Draw the number 0xF1234567890ABCDE stored at word 3X in both big-endianand little-endian machines. Clearly label the byte address corresponding to eachdata byte value.Suppose that you are given the following logical statement:¬ P ∧ Q → ¬ R ∨ SBased on our discussion of order of operations, which is the correct way to interpretthis?(a) ¬(P ∧ ((Q → (¬R)) ∨ S))(b) (¬(P ∧ (Q → (¬R)))) ∨ S(c) ((¬P) ∧ (Q → (¬R))) ∨ S(d) (¬P) ∧ ((Q → (¬R)) ∨ S)(e) (¬(P ∧ Q)) → (¬(R ∨ S))(f) ((¬P) ∧ Q) → ((¬R) ∨ S)(g) ((¬P) ∧ (Q → (¬R))) ∨ S(h) ((¬P) ∧ (Q → (¬R))) ∨ S(i) (¬P) ∧ ((Q → (¬R)) ∨ S)(j) ¬((P ∧ Q) → (¬(R ∨ S)))Convert this code to bash script to simulate producer-consumer problem using semaphores. #include<stdio.h>int mutex=1,full=0,empty=3,x=0;main(){ int n; void producer(); void consumer(); int wait(int); int signal(int); printf(“\n 1.Producer \n 2.Consumer \n 3.Exit”); while(1) { printf(“\n Enter your choice:”); scanf(“%d”,&n); switch(n) { case 1: if((mutex==1)&&(empty!=0)) producer(); else printf(“Buffer is full”); break; case 2: if((mutex==1)&&(full!=0)) consumer(); else printf(“Buffer is empty”); break; case 3: exit(0); break; } }}int wait(int s){ return (--s);}int signal(int s){ return(++s);}void producer(){ mutex=wait(mutex); full=signal(full); empty=wait(empty); x++; printf(“\n Producer produces the item %d”,x); mutex=signal(mutex);}void consumer(){…