Haskell - 通过创建更高级的函数来计算函数的总和

Question

Without the usage of imported code (i can use head, tail , init , last , filter , map , fold , . , generally basic haskell functions ) I want to create a higher-order function int ,type hof :: [Integer->Integer]->(Integer->Integer) , with hof s = \\n ->.... (using lambda) that implements the following math formula

(sorry, it is hand-written)

Need some basic guidence for starting , i thought about implementing a recoursive call each time of the hof function , which hof itself should implement a changed math formula(which i given) suitable for recursive tail calls.

some results that should be produced :

Main> map (hof [(+1)]) [1..10]
[2,3,4,5,6,7,8,9,10,11]

Main> map (hof [(+1),(+2)]) [1..10]
[3,4,6,7,9,10,12,13,15,16]

Main> map (hof [(2^),(2^),(2^),(2^),(2^)]) [5..12]
[42,85,170,341,682,1364,2728,5456]

Main> map (hof [(*2),(+100),(^3),negate,(mod 100)]) [24..40]
[2768,3151,3567,4020,4509,5038,5609,6221,6876,7575,8322,9117,9960,10855,11805,
12806,13863]

Main> map (hof [(mod 100),negate,(+100),(^3),(*2)]) [24..40]
[1181,1351,1562,1767,1989,2230,2490,2771,3071,3395,3774,4145,4539,4958,5402,5873,6369]

Main> map (hof [(‘mod‘ i) | i<-[50..100] ]) [1000..1030]
[23,24,27,28,30,32,35,36,39,40,37,38,40,41,44,45,47,48,52,53,55,31,33,35,37,39,42,
45,46,48,47]

Thanks for reading !

Answer 1

Let's say you have a list of functions:

fs = [f1, f2, f3, f4, f5] -- or more

Now, for each value i from 0 to infinity, we want to apply a function from fs to a corresponding n - i , then divide that value by 2 ⁱ , and we want to sum the resulting floors.

hof fs = \n -> sum [ fi (n-i) `div` (2^i) | (fi, i) <- zip fs [0..]]

Even though [0..] is an infinite list, the result of zip fs [0..] is finite, limited by the length of fs .

To avoid repeated multiplications for each successive power of 2, use the iterate function to compute a sequence of them to zip along with fs and [0..]

hof fs = \n -> sum [fi (n-i) `div` k | (fi, i, k) <- zip3 fs [0..] (iterate (*2) 1)]