{0,1}* | w starts with 1 and ends with 0}. Its complement L’={w∈{0,1}* | ________ } w starts with 0 and ends with 1 w has length 0 or 1 or starts with 0 or ends with 1 w starts with 1 or en
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L={w∈{0,1}* | w starts with 1 and ends with 0}. Its complement L’={w∈{0,1}* | ________ }
w starts with 0 and ends with 1 |
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w has length 0 or 1 or starts with 0 or ends with 1 |
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w starts with 1 or ends with 0 |
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w has length 0 and 1 and starts with 1 and ends with 0 |
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- 3.37.1: String access operations. Given string inputText, output "Match found" if the first character of inputText is 'w'. Otherwise, output "Match not found". End with a newline. Ex: If the input is: warmth then the output is: Match found Note: Assume the length of string inputText is greater than or equal to 1. 1 2 3 4 5 6 7 8 9 10 11 12 13 import java.util.Scanner; public class CharMatch { publicstaticvoidmain(String[] args) { Scannerscnr=newScanner(System.in); StringinputText; inputText=scnr.nextLine(); /* Your code goes here */ } } import java.util.Scanner; public class CharMatch { public static void main(String[] args) { Scanner scnr = new Scanner(System.in); String inputText; inputText = scnr.nextLine(); /* Your code goes here */ }}Crypto Columns The columnar encryption scheme scrambles the letters in a message (or plaintext) using a keyword as illustrated in the following example: Suppose BATBOY is the keyword and our message is MEET ME BY THE OLD OAK TREE. Since the keyword has 6 letters, we write the message (ignoring spacing and punctuation) in a grid with 6 columns, padding with random extra letters as needed: MEETME BYTHEO LDOAKT REENTH Here, we've padded the message with NTH. Now the message is printed out by columns, but the columns are printed in the order determined by the letters in the keyword. Since A is the letter of the keyword that comes first in the alphabet, column 2 is printed first. The next letter, B, occurs twice. In the case of a tie like this we print the columns leftmost first, so we print column 1, then column 4. This continues, printing the remaining columns in order 5, 3 and finally 6. So, the order the columns of the grid are printed would be 2, 1, 4, 5, 3, 6, in this case. This…Correct answer will be upvoted else Multiple Downvoted. Don't submit random answer. Computer science. You are given a string s of even length n. String s is twofold, at the end of the day, comprises just of 0's and 1's. String s has precisely n2 zeroes and n2 ones (n is even). In one activity you can invert any substring of s. A substring of a string is an adjoining aftereffect of that string. What is the base number of tasks you really want to make string s rotating? A string is rotating if si≠si+1 for all I. There are two kinds of exchanging strings overall: 01010101... or on the other hand 10101010... Input The principal line contains a solitary integer t (1≤t≤1000) — the number of experiments. The principal line of each experiment contains a solitary integer n (2≤n≤105; n is even) — the length of string s. The second line of each experiment contains a parallel string s of length n (si∈ {0, 1}). String s has precisely n2 zeroes and n2 ones. It's surefire that…
- python: def dna_slice(strand, first_nucleotides, last_nucleotides): """ Question 3 - Regex You are working with a long strand of DNA. Your task is to find and return the length of the first substrand that begins with the two-letter sequence specified in first_nucleotides and ends with the sequence specified in last_nucleotides. THIS MUST BE DONE IN ONE LINE. Args: strand (str) first_nucleotides (str) last_nucleotides (str) Returns: int strand_1 = 'TATGGGTCGAGCATGT' >>> dna_slice(strand_1, 'AT', 'CG') 8 >>> dna_slice(strand_1, 'GT', 'TG') 10 """ strand_1 = 'TATGGGTCGAGCATGT' pprint(dna_slice(strand_1, 'AT', 'CG')) pprint(dna_slice(strand_1, 'GT', 'TG'))extern "C" int f(int *,int,int); int a[2][2] = {{11,12},{21,22}}; void setup(){ Serial.begin(115200); while(!Serial); delay(500); int *arr = (int*)a; Serial.println(f(arr,2,2)); } void loop(){ } .global f f: ldr r3,[r0] // get first element mov r2,#0add r2,r3 mov r1, #3 lp: add r0,#4 // add next elementldr r3,[r0]add r2,r3sub r1,#1 bgt lp mov r0,r2 bx lr Describe what operation is being performed in this codeFind the body for:def find_first_occurrence(msg: str, string_to_find: str) -> int: """Return the index of the first occurence of string_to_find in msg, or -1 if it does not occur, ignoring case. >>> find_first_occurrence('October holidays: Halloween and Thanksgiving', 'H') 8 >>> find_first_occurrence('Because DEC 25 == OCT 31', 'H') -1 >>> find_first_occurrence('hip hip hooray', 'ip') 1 """
- Caesars Cypher in C programming language How do you program an encryption and decryption for a Caesars Cypher that covers all the possible ASCII characters? In this there should be a shift over K = 5 but the example below is K = 3. This should result in actual character and not the control ASCII represenations. For example: >>> $ encrypt(I have a key)N%mf{j%f%pj~>>> $ encrypt(see me at 3)xjj%rj%fy%8>>> $ decrpyt(|jfw%ns%gqzj)wear in blue Assume that the text parsing method has been succesful and it's the encryption/decryption of the char array that needs to be factored. Formula Encryption -> C = E(k, P) = (P + k) (mod 26) Decryption -> P = D(k, C) = (C - k) (mod 26) Where k = 5Q1.nq. Given a 2d grid map of '1's (land) and '0's (water),count the number of islands.An island is surrounded by water and is formed byconnecting adjacent lands horizontally or vertically.You may assume all four edges of the grid are all surrounded by water. Example 1: 11110110101100000000Answer: 1 Example 2: 11000110000010000011Answer: 3""" def num_islands(grid): count = 0 for i in range(len(grid)): for j, col in enumerate(grid[i]): if col == 1: dfs(grid, i, j) count += 1 Please code it. .* Assignment: Word operations * * Description: * This assignment asks you to implement common word operations that are * not available in the Scala programming language. The intent is to practice * your skills at working with bits. * */ package wordOps /* * Task 1: Population count (number of 1-bits in a word) * * Complete the following function that takes as parameter * a 32-bit word, w, and returns __the number of 1-bits__ in w. * */ def popCount(w: Int): Int = ??? /* * Task 2: Reverse bit positions * * Complete the following function that takes as parameter * a 16-bit word, w, and returns a 16-bit word, r, such that * for every j=0,1,...,15, * the value of the bit at position j in r is equal to * the value of the bit at position 15-j in w. * */ def reverse(w: Short): Short = ??? /* * Task 3: Left rotation * * Complete the following function that takes two parameters * * 1) a 64-bit word, w, and *…
- /* segvhunt.cFind and eliminate all code that generates Segmentation Fault*/#include <stdio.h>int main() {char **s;char foo[] = "Hello World";*s = foo;printf("s is %s\n",s);s[0] = foo;printf("s[0] is %s\n",s[0]);return(0);}Correct answer will be upvoted else Multiple Downvoted. Don't submit random answer. Computer science. Ridbit begins with an integer n. In one action, he can perform one of the accompanying tasks: partition n by one of its appropriate divisors, or take away 1 from n in case n is more prominent than 1. An appropriate divisor is a divisor of a number, barring itself. For instance, 1, 2, 4, 5, and 10 are appropriate divisors of 20, however 20 itself isn't. What is the base number of moves Ridbit is needed to make to decrease n to 1? Input The principal line contains a solitary integer t (1≤t≤1000) — the number of experiments. The main line of each experiment contains a solitary integer n (1≤n≤109). Output For each experiment, output the base number of moves needed to lessen n to 1.Artificial Intelligence - Local Search Starting from a randomly generated state of the 15-puzzle game (https://en.wikipedia.org/wiki/15_puzzle), steepest-ascent hill-climbing (the vanilla version of hill-climbing search) gets stuck 76% of the time, i.e., solving only 24% of problem instances. But it works very quickly, i.e., it takes just 6 steps on average when it succeeds and 5 steps when it gets stuck. In contrast, if sideways moves are allowed, this raises the percentage of problem instances solved by hill-climbing from 24% to 81%, with the success at a cost: the algorithm averages roughly 7 steps for each successful instance and 32 steps for each failure. Now suppose that we are implementing random-restart hill climbing (i.e., if a search fails, it keeps to try, and try, until it gets a success) by the following two versions: one uses vanilla steepest-ascent hill climbing, and the other one uses hill climbing with sideways moves. Can you please tell which version of…