Part Four $-$ Implementing the Acorn Generator

Create a class called Acorn.

Instance Variables:

one or more instance variables representing the state $($think about how many you need$)$

don't store extraneous values

Methods:

$\_\_$init$\_\_($self$,$ seed: list$[$int$],$ M: int$)$: constructor. Although

is often considered the seed and

are often considered the initial values, for programming ease, we'll lump them into a single list for the parameter $($where the first element is

and so on$).$ Ensure that seed is a list containing integers. If not, raise a TypeError. Verify that

; if not, raise a ValueError.

$\_\_$iter$\_\_($self$)$: method that returns the iterator

$\_\_$next$\_\_($self$)$: method that returns next numbers in sequence. This should be a list of integers including

$($even though it will be the same$).$ Raises StopIteration exception when seuqence is repeating.

get$\_$state$($self$)$: function that returns the current state. The current state should be represented as a dictionary. This dictionary should have the following keys:

'vals' storing the list of integers representing the previous values

$\text{'}$M$\text{'}$ storing the integer$-$valued modulus

Note: you don't need this dictionary to contain anything that would be beneficial for determining the period $($if we reload the state, we'll assume it's starting from that point with no previous values$)$

set$\_$state$($self$,$ state$)$: function that takes in dictionary representing the state in the same format returned by getState and sets appropriate instance variables. As mentioned above, calling this should reset the random number generator as if it has not seen previous values.

A sample sequence with

$,$ and a seed and values of $[7,6,8,11,5]$:

$[7,6,8,11,5]->[7,(6+7)\%23,(8+13)\%23,(11+21)\%23,(5+9)\%23]=[7,13,21,9,14]$

$[7,13,21,9,14]->[7,(13+7)\%23,(21+20)\%23,(9+18)\%23,(14+4)\%23]=[7,20,18,4,18]$

$[7,20,18,4,18]->[7,(20+7)\%23,(18+4)23,(4+22)\%23,(18+3)\%23]=[7,4,22,3,21]$

$[7,4,22,3,21]->[7,(4+7)\%23,(22+11)\%23,(3+10)\%23,(21+13)\%23]=[7,11,10,13,11]$

$[7,11,10,13,11]->[7,(11+7)\%23,(10+18)\%23,(13+5)\%23,(11+18)\%23]=[7,18,$