I am counting 6 multiplications still when you take each x to each of the terms in each of the the parenthesis. The computer would have to treat each of those as independent multiplication operations, wouldn't it?
Isn't it still 3 multiplications since you have to multiple each factor in the parenthesis by the x outside of them? One on each side of the addition makes 2 added to the one on the right factor in the parenthesis.
I watched whole video to learn about something about algorithm but I didn't get the main point. Not even close that what you are talking about and it help in chess. A recommendation for you just give direct message and give some example.
Dr.He, I normally can not watch Tuesdays lectures until Thursday. You had said in Tuesdays video that we would finish module two with problem 5 and E.3. Are we missing this Video? Thank you
I ran into the problem of getting the servlet context before I saw your fix at ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-3WFaVSAgQNA.html . According to stackoverflow.com/questions/35837285/different-ways-to-get-servlet-context , you can also use request.getServletContext(); in the doGet() method (or any service() invoked method).
