练习3-5

原文

Exercise 3.5. Monte Carlo integration is a method of estimating definite integrals by means of Monte Carlo simulation. Consider computing the area of a region of space described by a predicate P(x, y) that is true for points (x, y) in the region and false for points not in the region. For example, the region contained within a circle of radius 3 centered at (5, 7) is described by the predicate that tests whether (x - 5)2 + (y - 7)2< 32. To estimate the area of the region described by such a predicate, begin by choosing a rectangle that contains the region. For example, a rectangle with diagonally opposite corners at (2, 4) and (8, 10) contains the circle above. The desired integral is the area of that portion of the rectangle that lies in the region. We can estimate the integral by picking, at random, points (x,y) that lie in the rectangle, and testing P(x, y) for each point to determine whether the point lies in the region. If we try this with many points, then the fraction of points that fall in the region should give an estimate of the proportion of the rectangle that lies in the region. Hence, multiplying this fraction by the area of the entire rectangle should produce an estimate of the integral.

Implement Monte Carlo integration as a procedure estimate-integral that takes as arguments a predicate P, upper and lower bounds x1, x2, y1, and y2 for the rectangle, and the number of trials to perform in order to produce the estimate. Your procedure should use the same monte-carlo procedure that was used above to estimate . Use your estimate-integral to produce an estimate of by measuring the area of a unit circle.

You will find it useful to have a procedure that returns a number chosen at random from a given range. The following random-in-range procedure implements this in terms of the random procedure used in section 1.2.6, which returns a nonnegative number less than its input.8

(define (random-in-range low high) 
 (let ((range (- high low)))   
  (+ low (random range))))

分析

蒙特卡罗法（Monte Carlo Method）求圆周率的原理：在长为1单位，面积为1平方单位的正方形中，以正方形的一个顶点作为圆心取1/4圆，面积为pi/4。在正方形内均匀的放入n个点，落在扇形（1/4圆）内的概率=扇形面积/正方形面积=pi/4。所以只要随机产生N个坐标（x,y），其中满足x^2+y^2<1的数据才有效。落在扇形中的次数乘以N再乘以4的数值在理论上接近于圆周率Pi。

然后我们再借用书中第155页底部的monte-carlo函数和题目中给出的random-in-range函数即可。（由于最终结果为浮点型，因此将random-in-range稍作修改更好）。

代码

(define (monte-carlo-pi trials)

  (define (monte-carlo trials experiment)
    (define (iter trials-remaining trials-passed)
      (cond ((= trials-remaining 0) (/ trials-passed trials))
        ((experiment) (iter (- trials-remaining 1)(+ trials-passed 1)))
        (else (iter (- trials-ramaining 1) trials-passed))))
    (iter trials 0))

  (define (random-in-range low high)
    (let ((range (- high low)))
      (+ low 
     (random (exact->inexact range)))))

  (define (estimate-integral p? x1 y1 x2 y2 trials)
    (* 4
       (monte-carlo trials
            (lambda () (p? (random-in-range x1 x2) (random-in-range y1 y2))))))

  (exact->inexact
   (estimate-integral (lambda (x y)
            (< (+ (square x) (square y)) 1.0))
              -1.0
              -1.0
              1.0
              1.0
              trials)))

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