## Consider the following algorithm for sorting a list:

1. Starting from the beginning of the list, check each pair of adjacent elements in turn.

2. If the elements are out of order:

a. Move the smallest element of the pair at the beginning of the list.

b. Restart the process from step 1.

3. If all pairs are in order, stop.

For example, the list $\{\,4\,1\,3\,2\,\}$ is sorted as follows:

$\underline{4\,1}\,3\,2$ ($4$ and $1$ are out of order so move $1$ to the front of the list)

$1\,\underline{4\,3}\,2$ ($4$ and $3$ are out of order so move $3$ to the front of the list)

$\underline{3\,1}\,4\,2$ ($3$ and $1$ are out of order so move $1$ to the front of the list)

$1\,3\,\underline{4\,2}$ ($4$ and $2$ are out of order so move $2$ to the front of the list)

$\underline{2\,1}\,3\,4$ ($2$ and $1$ are out of order so move $1$ to the front of the list)

$1\,2\,3\,4$ (The list is now sorted)

Let $F(L)$ be the number of times step 2a is executed to sort list $L$. For example, $F(\{\,4\,1\,3\,2\,\}) = 5$.

Let $E(n)$ be the expected value of $F(P)$ over all permutations $P$ of the integers $\{1, 2, \dots, n\}$.

You are given $E(4) = 3.25$ and $E(10) = 115.725$.

Find $E(30)$. Give your answer rounded to two digits after the decimal point.

### To solve this problem, we need to find a way to simulate the pouring process and calculate the minimal number of pourings needed to get one litre for each pair of prime numbers $p$ and $q$. Finally, we will sum up the results modulo $1\,000\,000\,007$.

We can approach this problem using the Breadth-First Search (BFS) algorithm. In each step of the BFS, we will generate all possible states of the buckets after pouring water and enqueue them for further exploration.

Let’s start by defining a function to pour water between the buckets. We can represent the state of the buckets as a tuple of three values – (S, M, L), where S represents the water level in the small bucket, M represents the water level in the medium bucket, and L represents the water level in the large bucket.

“`python

def pour_water(state, src_bucket, dest_bucket, capacity):

src_water = state[src_bucket]

dest_water = state[dest_bucket]

space_left = capacity – dest_water

poured_water = min(src_water, space_left)

state_updated = list(state)

state_updated[src_bucket] -= poured_water

state_updated[dest_bucket] += poured_water

return tuple(state_updated)

“`

Next, we define the BFS algorithm to explore all possible states and calculate the minimal number of pourings needed to get one litre.

“`python

from collections import deque

def bfs(a, b):

capacity = a + b

target_state = (1, 0, 0) # We want one litre in the small bucket

queue = deque()

visited = set()

initial_state = (a, b, 0) # Start with full small and medium buckets, empty large bucket

queue.append((initial_state, 0)) # Add the initial state to the queue

visited.add(initial_state)

while queue:

state, num_pourings = queue.popleft()

if state == target_state:

return num_pourings

for src in range(3):

for dest in range(3):

if src != dest:

new_state = pour_water(state, src, dest, capacity)

if new_state not in visited:

queue.append((new_state, num_pourings + 1))

visited.add(new_state)

return None

“`

Finally, we can iterate through all pairs of prime numbers $p$ and $q$ such that $p < q < 1000$, calculate the minimal number of pourings using the `bfs` function, and sum up the results modulo $1\,000\,000\,007$. ```python def sum_of_pourings(): prime_nums = sieve_of_eratosthenes(1000) sum_pourings = 0 for i in range(len(prime_nums)): for j in range(i+1, len(prime_nums)): p = prime_nums[i] q = prime_nums[j] a = 2**(p**5) - 1 b = 2**(q**5) - 1 pourings = bfs(a, b) sum_pourings = (sum_pourings + pourings) % 1000000007 return sum_pourings ``` Note: The `sieve_of_eratosthenes` function can be implemented to generate a list of prime numbers up to a given limit using the Sieve of Eratosthenes algorithm. You can find various implementations of this function online. Now, you can call the `sum_of_pourings` function to get the sum of pourings modulo $1\,000\,000\,007$ for all pairs of prime numbers $p$ and $q$ such that $p < q < 1000$. ```python result = sum_of_pourings() print(result) ``` Note: This code will take some time to execute since it needs to iterate through all possible pairs of prime numbers. You can optimize the code by using a more efficient prime number generation algorithm or using memoization to store previously calculated results.

##### More Answers:

SimbersSmallest Prime Factor

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