sicp-ex-1.20

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The process generated using the normal-order evaluation is the following. It performs 18 remainder operations: 14 when evaluating the condition and 4 in the final reduction phase.

  
 (gcd 206 40) 
  
 (if (= 40 0) ...) 
  
 (gcd 40 (remainder 206 40)) 
  
 (if (= (remainder 206 40) 0) ...) 
  
 (if (= 6 0) ...) 
  
 (gcd (remainder 206 40) (remainder 40 (remainder 206 40))) 
  
 (if (= (remainder 40 (remainder 206 40)) 0) ...) 
  
 (if (= 4 0) ...) 
  
 (gcd (remainder 40 (remainder 206 40)) (remainder (remainder 206 40) (remainder 40 (remainder 206 40)))) 
  
 (if (= (remainder (remainder 206 40) (remainder 40 (remainder 206 40))) 0) ...) 
  
 (if (= 2 0) ...) 
  
 (gcd (remainder (remainder 206 40) (remainder 40 (remainder 206 40))) (remainder (remainder 40 (remainder 206 40)) (remainder (remainder 206 40) (remainder 40 (remainder 206 40))))) 
  
 (if (= (remainder (remainder 40 (remainder 206 40)) (remainder (remainder 206 40) (remainder 40 (remainder 206 40)))) 0) ...) 
  
 (if (= 0 0) ...) 
 (remainder (remainder 206 40) (remainder 40 (remainder 206 40))) 
  

The number 'R' of remainder operations executed by the normal-order evaluation process is given by the formula

         R = SUM(i from 1 to n, fib(i) + fib(i - 1)) - 1 

where n is the number of gcd invocations required to compute the (gcd a b). The first invocation does not need to compute remainder thus the -1

R is given by the following function:

 (define (count-remainders n) 
     (define (loop n sum) 
         (if (= 0 n) (- sum 1) 
             (loop (- n 1) (+ sum (fib n) (fib (- n 1)))))) 
     (loop n 0)) 

In this particular case that is:

 (count-remainders 5) 
 => 18 

The process generated using the applicative-order evaluation is the following. It performs 4 remainder operations.

  
 (gcd 206 40) 
  
 (gcd 40 (remainder 206 40)) 
  
 (gcd 40 6) 
  
 (gcd 6 (remainder 40 6)) 
  
 (gcd 6 4) 
  
 (gcd 4 (remainder 6 4)) 
  
 (gcd 4 2) 
  
 (gcd 2 (remainder 4 2)) 
  
 (gcd 2 0) 
  
 2 
  

It seems that the formula

         R = SUM(i from 1 to n, fib(i) + fib(i - 1)) - 1 

is incorrect. One can easy check this by letting n=2. The correct count should be 1 while the formula gives 2.

Let b_i be the value of b at the n'th invocation of gcd(a,b). Let b(i) be the count of remainder procedure needed to calculate b_i. Similarly, define a_i and a(i). It is easy to check that

1. a_i=b_(i-1) and thus a(i)=b(i-1).

2. b(i+1) = a(i) + b(i) + 1 = b(i-1) + b(i) + 1 because b_(i+1) = a_i mod b_i

Based on my own derivation, R should be

b(1)+b(2)+...+b(n)+b(n-1) with b(1)=0, b(2)=1 and b(n)=b(n-1)+b(n-2)+1, which is not equivalent with the above R formula

The correct formula is

R(n) = SUM(i from 1 to n - 1, fib(i)) + fib(n - 2) - 1

Where n is the number of applications.

R(2) = 1

R(3) = 4

R(5) = 17

The above formulas are wrong. The truly correct one should be:

R(n) = SUM(i from 1 to n, fib(i+1)- 1) + fib(n) -1

where n is the number of times gcd(a, b) is called.

We have,

R(1) = 0

R(2) = 1

R(3) = 4

R(4) = 9

R(5) = 18

Below is the derivation:

Denote n as the number of times gcd() is called, R(n) the number of times remainder() is invoked.

For (gcd a(k) b(k)), let num_a(k) be the number of remainder() in a(k), num_b(k) be the number of remainder() in b(k), where k=1,2,...n.

Then we have the updating process:

a(k+1) = b(k)

b(k+1) = remainder(a(k), b(k))

Thus,

num_a(1) = 0

num_b(1) = 0

num_a(k+1) = num_b(k)

R(k+1) = num_a(k) + num_b(k) + 1 = num_b(k+1)

By substituting num_a for num_b, we get the following for {num_b(k)}, k=1,2,...n,

num_b(1) = 0

num_b(2) = 1

num_b(k+1) = num_b(k) + num_b(k-1) + 1

Obviously we could get

num_b(k) = fib(k+1) - 1

(hello Lily.x, the above step is not obvious for me, can you explain?)

(It isn't obvious to me either until I looked through the results for k = 1..10. The conjecture is then easily proven by induction.)

|  k  | num_b(k) | fib(k) |
|-----|----------|--------|
|  0  |    -     |   0    |
|  1  |    0     |   1    |
|  2  |    1     |   1    |
|  3  |    2     |   2    |
|  4  |    4     |   3    |
|  5  |    7     |   5    |
|  6  |   12     |   8    |
|  7  |   20     |   13   |
|  8  |   33     |   21   |
|  9  |   54     |   34   |
| 10  |   88     |   55   |

Since

R(n) = SUM(i from 1 to n, num_b(i)) + num_a(n)

which is quite obvious following the normal-order-evaluation rule for if statement, we could get

R(n) = SUM(i from 1 to n, fib(i+1)- 1) + fib(n) -1

:) Lily.X

The correct formula above can also be written as

R(n) = 2*fib(n+2)-n-3