Prove or disprove: (a)∃ sets A,B ∀ functions f: A→B, f is a bijection. (b) if f : A → B is a function, C ⊆ A, and D ⊆ B, then f : C → D is always a function. [Note 1: one is true, the other is false.] [Note 2: feel free to draw bubble-arrow pictures here, but make sure you explain what’s going on very carefully.]

Let's revisit our first problem, where we want to set up a series of chess matches so we can rank six players in our class. As we did before, we will assume that everyone keeps their chess rating a private secret; however, when two players have a chess match, the person with the higher rating wins 100% of the time. But this time, we are only interested in identifying the BEST of these six players and the WORST of these six players. (We don't care about the relative ordering or ranking of the middle four players.) Your goal is to devise a comparison-based algorithm that is guaranteed to identify the player with the highest rating and the player with the lowest rating. Because you are very strong at Algorithm Design, you know how to do this in the most efficient way. Here are five statements. A. There exists an algorithm to solve this problem using 6 matches, but there does not exist an algorithm using only 5 matches. B. There exists an algorithm to solve this problem using 7 matches,…

My apologies for misunderstanding. Here's the translation: "A function, f:f(x,y)=x2+2y2−12x+xy is defined on the domain: Ω = {(x, y): -1 ≤ x ≤ 1, -1 ≤ y ≤ x}. Create a contour plot of the function and plot the boundary of Ω with a dashed line: In Python

I'm having a difficult time with this problem. Could you lend me a hand in understanding it? I'd appreciate a step-by-step explanation, along with insights into how you arrived at the solution. here is the question I need help with:4.3 Let ALLDFA = {<A> | A is a DFA and L(A) = Σ∗ }. Show that ALLDFA is decidable