Buckets of Water

There are 3 buckets labelled $S$ (small) of 3 litres, $M$ (medium) of 5 litres and $L$ (large) of 8 litres.
Initially $S$ and $M$ are full of water and $L$ is empty.
By pouring water between the buckets exactly one litre of water can be measured.
Since there is no other way to measure, once a pouring starts it cannot stop until either the source bucket is empty or the destination bucket is full.
At least four pourings are needed to get one litre:

$(3,5,0)\xrightarrow{M\to L} (3,0,5) \xrightarrow{S\to M} (0,3,5) \xrightarrow{L\to S}(3,3,2)
\xrightarrow{S\to M}(1,5,2)$

After these operations, there is exactly one litre in bucket $S$.

In general the sizes of the buckets $S, M, L$ are $a$, $b$, $a + b$ litres, respectively. Initially $S$ and $M$ are full and $L$ is empty. If the above rule of pouring still applies and $a$ and $b$ are two coprime positive integers with $a\leq b$ then it is always possible to measure one litre in finitely many steps.

Let $P(a,b)$ be the minimal number of pourings needed to get one litre. Thus $P(3,5)=4$.
Also, $P(7, 31)=20$ and $P(1234, 4321)=2780$.

Find the sum of $P(2^{p^5}-1, 2^{q^5}-1)$ for all pairs of prime numbers $p,q$ such that $p < q < 1000$. Give your answer modulo $1\,000\,000\,007$.

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.

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