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- (V, E) be a connected, undirected graph. Let A = V, B = V, and f(u) = neighbours of u. Select all that are true. Let G = a) f: AB is not a function Ob) f: A B is a function but we cannot always apply the Pigeonhole Principle with this A, B Odf: A B is a function but we cannot always apply the extended Pigeonhole Principle with this A, B d) none of the abovearrow_forwardFor all a and b in the domain of a function f, the function is injective iff f(a) #f(b) a=Db f(a)=f(b) a=b f(a)=f(b) atb o fly)=x iff f(x)=y A Moving to another question will save this response. 5arrow_forwardH.W2 Minimize the following function using K-Maps: F (A, B, C, D) = Σ m (1, 5, 6, 12, 13, 14) + d (2, 4).arrow_forward
- 5. Simplify the following functions using a K-map:(a) F(X,Y)¼m2+ m3arrow_forwardCreate K-maps and then simplify for the following functions. 3) F(x, y, z) = xyz + xy’z’ + x’yz + x’y’zarrow_forwardQ. Let A = {a, b, c, d, e} and B = {1, 2, 3, 4, 5, 6, 7, 8}. How many functions f : A → B(a) ... are injective?(b) ... are not injective?(c) ... are such that f(a) = f(b) = f(c)?(d) ... are such that exactly three elements of A have 8 as an image?(e) ... are surjective?arrow_forward
- Draw the graph of the function f(x)=⌊x/2⌋ from R to R.arrow_forwardThe function of four variables G (A, B, C, D) = Z (2, 3, 8, 9, 11, 14, 15) + F (6, 10) is given. Here F values are "Doesn't matter" (Indeterminate) values that can take any value. Write this function in a simplified form using the Karnaugh Map.arrow_forward8. X. Let f: RR defined by f(x) = x³ -arrow_forward
- 2. Definition: If f(x) is a function, then we say that a value u is a fixed point of f(x) if and only if f (u) = u. Suppose F(x) is a given continuous function and a # 0 is a given real number. Show that u is a zero of F(x) if and only if u is a fixed point of f (x) f (x) = x + a F (x). b. Suppose F '(x) is continuous, u is a zero of F, and F'(u) ± 0. Define f (x) = x + aF(x). Prove there are values of a + 0 and ɛ > 0 so that if uo E (u – E, u + ɛ) and un+1 = f (Un) for n = Hint: Jun+1 – u| = \f (un) – f (u)]. Use the definition of f (x) and the mean а. = 0,1,2, ... then un → u as n → ∞. | value theorem.arrow_forward3. (a) Consider the following algorithm. Input: Integers n and a such that n 20 and a > 1. (1) If 0arrow_forwardDetermine whether each of the following functions f : {a,b,c,d} -> {a,b,c,d} is one-to-one and/or onto. (a) f(a) = b, f(b) = a, f(c) = b, f(d) = c (b) f(a) = b, f(b) = b, f(c) = d, f(d) = c (c) f(a) = b, f(b) = a, f(c) = c, f(d) = d (d) f(a) = d, f(b) = a, f(c) = c, f(d) = b (e) f(a) = c, f(b) = d, f(c) = aarrow_forwardarrow_back_iosSEE MORE QUESTIONSarrow_forward_ios
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