src/problems/iprb.lisp @ bd06f66ba88f

(in-package :rosalind)

;; Probability is the mathematical study of randomly occurring phenomena. We
;; will model such a phenomenon with a random variable, which is simply
;; a variable that can take a number of different distinct outcomes depending on
;; the result of an underlying random process.
;;
;; For example, say that we have a bag containing 3 red balls and 2 blue balls.
;; If we let X represent the random variable corresponding to the color of
;; a drawn ball, then the probability of each of the two outcomes is given by
;; Pr(X=red)=35 and Pr(X=blue)=25
;;
;; Random variables can be combined to yield new random variables. Returning to
;; the ball example, let Y model the color of a second ball drawn from the bag
;; (without replacing the first ball). The probability of Y being red depends on
;; whether the first ball was red or blue. To represent all outcomes of X and Y,
;; we therefore use a probability tree diagram. This branching diagram
;; represents all possible individual probabilities for X and Y, with outcomes
;; at the endpoints ("leaves") of the tree. The probability of any outcome is
;; given by the product of probabilities along the path from the beginning of
;; the tree; see Figure 2 for an illustrative example.
;;
;; An event is simply a collection of outcomes. Because outcomes are distinct,
;; the probability of an event can be written as the sum of the probabilities of
;; its constituent outcomes. For our colored ball example, let A be the event "Y
;; is blue." Pr(A) is equal to the sum of the probabilities of two different
;; outcomes: Pr(X=blue and Y=blue)+Pr(X=red and Y=blue), or 310+110=25 (see
;; Figure 2 above).
;;
;; Given: Three positive integers k, m, and n, representing a population
;; containing k+m+n organisms: k individuals are homozygous dominant for
;; a factor, m are heterozygous, and n are homozygous recessive.
;;
;; Return: The probability that two randomly selected mating organisms will
;; produce an individual possessing a dominant allele (and thus displaying the
;; dominant phenotype). Assume that any two organisms can mate.

(define-problem iprb (data stream)
    "2 2 2"
    "0.78333"
  (let* ((d (read data))
         (h (read data))
         (r (read data))
         (n (+ d h r)))
    ;; We could expand this all into a giant equation and cancel it out, but
    ;; let's just let the computer do the busywork.
    (flet ((p-same (x also-x)
             (declare (ignore also-x))
             ;; P({X X}) = X/N * (X-1)/(N-1)
             ;;          = X(X-1) / N(N-1)
             ;;          = X²-X / N²-N
             (/ (- (* x x) x)
                (- (* n n) n)))
           (p-diff (x y)
             ;; P({X Y}) = P(X, Y)     + P(Y, X)
             ;;          = X/N * Y/N-1 + Y/N * X/N-1
             ;;          = XY/N(N-1)   + YX/N(N-1)
             ;;          = 2XY/N(N-1)
             ;;          = 2XY/N²-N
             (/ (* 2 x y)
                (- (* n n) n))))
      (format nil "~,5F"
              (+ (* (p-same d d) 1)      ;; AA AA
                 (* (p-diff d h) 1)      ;; AA Aa
                 (* (p-diff d r) 1)      ;; AA aa
                 (* (p-same h h) 3/4)    ;; Aa Aa
                 (* (p-diff h r) 1/2)    ;; Aa aa
                 (* (p-same r r) 0)))))) ;; aa aa

;; (problem-iprb)
;; (solve iprb)