Computers and Technology, 27.07.2020 01:01, lizdeleon248

(HOF. scala) Define the following two functions: // given f and g, compose returns the composition of f with g. def compose(f:Int=>Int, g: Int=>Int): Int=>Int = p // given f and a positive integer n, repeated returns a function that applies f // to its input n times. i. e. f(f(...(f(x))...)) def repeated(f:Int=>Int, n:Int): Int=>Int = p For example, compose(x=>xx, x=>x+1)(6) = 49 and repeated(x=>3x,4)(8) = 648. Submit just the definitions of the two functions in HOF. scala. We will call these with our test cases.

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Choice 2.the algorithm constantly checks to see if the value in the position before is greater than the value in the current position. if there is even one instance, it will return false and stop looping.using an if-statement without } and { means that the if-statement true condition only applies for the next lineconsider $\texttt{int[] nums = \{3,6,12\}}$ . the condition $\texttt{(nums[i - 1]\textgreater= nums[i])}$ checks if previous position is greater than or equal to current position. if it is, it will make method return false. for that example, it won't.consider $\texttt{int[] nums = \{3,3,12\}}$ , it would return falseconsider $\texttt{int[] nums = \{6,2,12\}}$ . it would return false

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(HOF. scala) Define the following two functions: // given f and g, compose returns the composition o...