Time: Wed 9/13/17 2 pm

**IMPORTANT** In this question, assume that all of `f`

, `g`

, and `h`

are functions that take one non-negative integer argument and return a non-negative integer. You do not need to consider negative or fractional numbers.

Implement the higher-order function

`decompose1`

, which takes two functions`f`

and`h`

as arguments. It returns a function`g`

that relates`f`

to`h`

in the following way: For any non-negative integer`x`

,`h(x)`

equals`f(g(x))`

. Assume that`decompose1`

will be called only on arguments for which such a function`g`

exists. Furthermore, assume that there is no recursion depth limit in Python.`def decompose1(f, h): """ Return g such that h(x) equals f(g(x)) for any non-negative integer x. >>> add_one = lambda x: x + 1 >>> square_then_add_one = lambda x: x * x + 1 >>> g = decompose1(add_one, square_then_add_one) >>> g(5) 25 >>> g(10) 100 """ def g(x): def r(y): if : return else: return return r(0)`

Write a number in the blank so that the final expression below evaluates to

`2015`

. Assume`decompose1`

is implemented correctly.`def make_adder(a): def add_to_a(b): return a + b return add_to_a def compose1(f, g): def h(x): return f(g(x)) return h e, square = make_adder(1), lambda x: x*x decompose1(e, compose1(square, e))(3) +`

Suppose we have a sequence of quantities that we want to multiply together, but can only multiply two at a time. We can express the various ways of doing so by counting the number of different ways to parenthesize the sequence. For example, here are the possibilities for products of 1, 2, 3 and 4 elements:

- 1 element of 1 possibility

`a`

- 2 elements of 1 possibility

`ab`

- 3 elements of 2 possibilities

`(ab)c`

`a(bc)`

- 4 elements of 5 possibilities

`a(b(cd))`

`a((bc)d)`

`(ab)(cd)`

`(a(bc))d`

`((ab)c)d`

Assume, as in the table above, that we don’t want to reorder elements.

Define a function `count_groupings`

that takes a positive integer `n`

and returns the number of ways of parenthesizing the product of `n`

numbers. (You might not need to use all lines.)

```
def count_groupings(n):
"""
For N >= 1, the number of distinct parenthesizations of a product of N items.
>>> count_groupings(1)
1
>>> count_groupings(2)
1
>>> count_groupings(3)
2
>>> count_groupings(4)
5
>>> count_groupings(5)
14
"""
if n == 1:
return
i =
while :
i += 1
return
```

**IMPORTANT DEFINITION** Each digit in a non-negative integer `n`

has a digit position. Digit positions begin at `0`

and count from the right-most digit of `n`

. For example, in `568789`

, the digit `9`

is at position `0`

and digit `7`

is at position `2`

. The digit `8`

appears at both positions `1`

and `3`

.

Implement the find_digit function, which takes a non-negative integer

`n`

and a digit`d`

greater than`0`

and less than`10`

. It returns the largest (left-most) position in`n`

at which digit`d`

appears. If`d`

does not appear in`n`

, then`find_digit`

returns`False`

. You may not use recursive calls.`def find_digit(n, d): """ Return the largest digit position in n for which d is the digit. >>> find_digit(567, 7) 0 >>> find_digit(567, 5) 2 >>> find_digit(567, 9) False >>> find_digit(568789, 8) # 8 appears at positions 1 and 3 3 """ i, k = 0, while n: n, last = n // 10, n % 10 if last == : i = i + 1 return`

Find all values of

`y`

between 1 and 9 (inclusive) for which the final expression below evaluates to`True`

. Assume that`find_digit`

is implemented correctly.`def compose1(f, g): def h(x): return f(g(x)) return h f = lambda x: find_digit(234567, x) compose1(f, f)(y) == y`