src/problems/iev.lisp @ bd06f66ba88f

(in-package :rosalind)

;; For a random variable X taking integer values between 1 and n, the expected
;; value of X is E(X)=∑nk=1k×Pr(X=k). The expected value offers us a way of
;; taking the long-term average of a random variable over a large number of
;; trials.
;;
;; As a motivating example, let X be the number on a six-sided die. Over a large
;; number of rolls, we should expect to obtain an average of 3.5 on the die
;; (even though it's not possible to roll a 3.5). The formula for expected value
;; confirms that E(X)=∑6k=1k×Pr(X=k)=3.5.
;;
;; More generally, a random variable for which every one of a number of equally
;; spaced outcomes has the same probability is called a uniform random variable
;; (in the die example, this "equal spacing" is equal to 1). We can generalize
;; our die example to find that if X is a uniform random variable with minimum
;; possible value a and maximum possible value b, then E(X)=a+b2. You may also
;; wish to verify that for the dice example, if Y is the random variable
;; associated with the outcome of a second die roll, then E(X+Y)=7.
;;
;; Given: Six nonnegative integers, each of which does not exceed 20,000. The
;; integers correspond to the number of couples in a population possessing each
;; genotype pairing for a given factor. In order, the six given integers
;; represent the number of couples having the following genotypes:
;;
;;     AA-AA
;;     AA-Aa
;;     AA-aa
;;     Aa-Aa
;;     Aa-aa
;;     aa-aa
;;
;; Return: The expected number of offspring displaying the dominant phenotype in
;; the next generation, under the assumption that every couple has exactly two
;; offspring.

(define-problem iev (data stream)
    "1 0 0 1 0 1"
    "3.5"
  (let* ((dd (read data))
         (dh (read data))
         (dr (read data))
         (hh (read data))
         (hr (read data))
         (rr (read data)))
    (format nil "~,4F"
            ;; It's just a weighted average…
            (* 2 (+ (* dd 1)
                    (* dh 1)
                    (* dr 1)
                    (* hh 3/4)
                    (* hr 1/2)
                    (* rr 0))))))

;; (problem-iev)
;; (solve iev)