[] [] dir_euler Write a function "dir_euler(digraph)" that takes in a connected, directed graph as its input, and then determines whether or not a directed Eulerian circuit exists. After compiling the above cell, you should be able to compile the following cell and obtain the desired outputs. print (dir_euler({"A": ["A", "B", "C"], "B" : ["A"], "C": ["A"]}), dir_euler({"A" : ["B", "D"], "B" : ["A", "D"], "C": ["C", "D"], "D": []})) This should return True False Python Python

= = 2×2 and 6 = (a) A composite number is a positive integer that has at least one divisor other than 1 and itself. For example, 2 1×2 is not a composite number but 4 2 × 3 are composite numbers. A logic circuit has four binary input variables, A, B, C and D. The output Z of the logic circuit is 1 if the unsigned integer represented by the binary number ABCD is a composite number. Using variables A and B for the select inputs S1 and S0 of a 4-to-1 multiplexer, implement the logic function Z(A, B, C, D) using this multiplexor and other logic gates.

(Kruskal’s algorithm) The text introduced Prim’s algorithm for finding a minimum spanning tree. Kruskal’s algorithm is another well-known algorithm for finding a minimum spanning tree. The algorithm repeatedly finds a minimumweight edge and adds it to the tree if it does not cause a cycle. The process ends when all vertices are in the tree. Design and implement an algorithm for finding an MST using Kruskal’s algorithm.This is a java code question

In c++ please help me answer this question I will give you a good rating :) implement the three versions of the addupto20() function: iterative, recursive, and divide-and-conquer approach test these functions with a few inputs from your main() Iterative solution * check if there exsits two numbers from vector data that add up to 20 e.g., if data=[2,5,3,15], the function returns true, as data [0] +data [3]==20 e.g., if data=[3,4,0,8], the function return false precondition: vector data has been initialized postcondition: if there are two numbers from list add up to 20, return true; otherwise, return false */ bool AddupTo20 (const vector‹int> & data){ } Come up with a recursive solution to the problem, following the hints given below: * check if there exsits two numbers from vector data[first...right] add up to 20 e.g., if data=[2,5,3,15], first=0, last=3, the function returns true, as data [0] +data [3]==20 e.g., if data=[2,5,3,15], first=2, last=3, the function returns…