{{m, x)| m is a turing machine that runs on input x for an even let even number of steps}. of course, a turing machine running for infinitely many steps neither runs for an odd nor an even number of steps. we want to show that even is undecidable by a proof from scratch (i. e., by diagonalization not by a reduction) first we assume even is decidable, i. e., there is a tm h that accepts (m, x) if m is a turing machine that runs on input x for an even number of steps, and rejects otherwise. we want to do the diagonalization in two stages. (a) first, describe a tm d (based on the supposedly existing tm h) accepting (m) if and only if m is a turing machine that runs on input (m) for an even number of steps. otherwise d rejects note that you may assume the correctness of the church-turing thesis, i. e., instead of describing a turing machine in detail, you can just describe an algorithm in pseudo-code. (b) now define a tm d from d that runs on (m) for an odd number of steps if d accepts (m). otherwise d runs for an even number of steps (c) how does the existence of this produce a contradiction? (d) what does the contradiction prove?

This program should be a short piece of code that prints all of the positive integers from 1 to 100 as described more fully below. the program may contain multiple methods, and if using an oo language, should be contained within a single class or object. the program should be designed so that it begins execution when invoked through whichever mechanism is most common for the implementation language. â–ş print out all positive integers from 1 to 100, inclusive and in order. â–ş print messages to standard output, matching the sample output below. â–ş in the output, state whether the each integer is 'odd' or 'even' in the output. â–ş if the number is divisible by three, instead of stating that the number is odd or even, state that the number is 'divisible by three'. â–ş if the number is divisible by both two and three, instead of saying that the number is odd, even or divisible by three; state that the number is 'divisible by two and three'. â–ş design the logic of the loop to be as efficient as possible, using the minimal number of operations to perform the required logic. sample output the number '1' is odd. the number '2' is even. the number '3' is divisible by three. the number '6' is divisible by two and three.