Chef Showdown

A group of chefs (numbered #$1$, #$2$, etc) participate in a turn-based strategic cooking competition. On each chef’s turn, he/she cooks up a dish to the best of his/her ability and gives it to a separate panel of judges for taste-testing. Let $S(k)$ represent chef #$k$’s skill level (which is publicly known). More specifically, $S(k)$ is the probability that chef #$k$’s dish will be assessed favorably by the judges (on any/all turns). If the dish receives a favorable rating, then the chef must choose one other chef to be eliminated from the competition. The last chef remaining in the competition is the winner.
The game always begins with chef #$1$, with the turn order iterating sequentially over the rest of the chefs still in play. Then the cycle repeats from the lowest-numbered chef. All chefs aim to optimize their chances of winning within the rules as stated, assuming that the other chefs behave in the same manner. In the event that a chef has more than one equally-optimal elimination choice, assume that the chosen chef is always the one with the next-closest turn.
Define $W_n(k)$ as the probability that chef #$k$ wins in a competition with $n$ chefs. If we have $S(1) = 0.25$, $S(2) = 0.5$, and $S(3) = 1$, then $W_3(1) = 0.29375$.
Going forward, we assign $S(k) = F_k/F_{n+1}$ over all $1 \le k \le n$, where $F_k$ is a Fibonacci number: $F_k = F_{k-1} + F_{k-2}$ with base cases $F_1 = F_2 = 1$. Then, for example, when considering a competition with $n = 7$ chefs, we have $W_7(1) = 0.08965042$, $W_7(2) = 0.20775702$, $W_7(3) = 0.15291406$, $W_7(4) = 0.14554098$, $W_7(5) = 0.15905291$, $W_7(6) = 0.10261412$, and $W_7(7) = 0.14247050$, rounded to $8$ decimal places each.
Let $E(n)$ represent the expected number of dishes cooked in a competition with $n$ chefs. For instance, $E(7) = 42.28176050$.
Find $E(14)$ rounded to $8$ decimal places.

To find the value of $P(3141592653589793)$, we can use a Python program to calculate it.

First, let’s define a function to check if two numbers are relatively prime:

import math

def is_relatively_prime(x, y):
return math.gcd(x, y) == 1

Next, we’ll define a function to count the number of primitive Pythagorean triples up to a given limit:

def count_primitive_triplets(limit):
count = 0
for a in range(2, limit + 1):
for b in range(a + 1, limit + 1):
c = math.sqrt(a**2 + b**2)
if c.is_integer() and c <= limit and is_relatively_prime(a, b): count += 1 return count ``` Finally, we can calculate the value of $P(3141592653589793)$ using the `count_primitive_triplets` function: ```python limit = 3141592653589793 result = count_primitive_triplets(limit) print(result) ``` When we run this program, it will output the value of $P(3141592653589793)$. Note: Running this program will take a considerable amount of time since the limit is very large. You may want to consider using a more efficient algorithm or optimizing the program further if you're working with such large inputs.

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