fun t(0)= 0
| t(n)= 2+t(n-1);
This function simply doubles its input. To see this by example consider
the call t(100).
t(100) = 2 + t(99)
= 2 + (2 + t(98))
= 2 + (2 + (2 + t(97)))
= ...
= 2 + 2 + 2 + ... + 2 + 0
= 200
There will be 100 2's in all
fun d(0)= "de"
| d(n)= "do"^d(n-1)^"da";
Each recursion puts a "do" at the front and a "da" at the end - the base
case puts a "de" in the middle. We get a string consisting of n "do"'s
a single "de" followed by n "da"'s.
fun h(0)= 1
| h(n)= h(n-1)+h(n-1);
This function returns 2n. Each call effectively doubles the
previous value. The function is very slow as the doubling is performed
using addition instead of multiplication. In all 2n-1 additions
are performed.
fun m(a,0) = 0
| m(a,b) = a+m(a,b-1);
This is the multiplation function. e.g. m(3,5) = 15. The function causes
b copies of a to be added to 0.
e.g.
m(5,3) = 5 + m(5,2)
= 5 + (5 + m(5,1))
= 5 + (5 + (5 + m(5,0)))
= 5 + (5 + (5 + 0))
= 15
fun f(0)= true
| f(n)= not(f(n-1));
This function returns true if the input is even and false if it is odd.
Follow the logic...
|
f(0) = true | Zero is an even number |
|
f(n) = not(f(n-1)) | The "even-ness" of n is the opposite
of the "even-ness" of n-1. That is n is even if n-1 is odd and vis-versa.
|
fun g(0)= nil
| g(n)= 0::g(n-1);
g returns a list of n 0's. Each call to g conses another 0 onto the list.
fun l(0)= nil
| l(n)= n mod 10 :: l(n div 10);
l returns the decimal digits of its input in reverse order. The value
n mod 10 gives the least significant decimal digit of n,
The value n div 10 returns the number consisting of the decimal
digits of n without the last one. e.g.
1962 mod 10 gives 2. 1962 div 10 gives 196.
l(1962) = 2 :: l(196)
= 2 :: (6::l(96))
= 2 :: (6 :: (9 :: l(1)))
= 2 :: (6 :: (9 :: (1 :: l(0) )))
= 2 :: (6 :: (9 :: (1 :: nil )))
= [2,6,9,1]
fun j(0)= 0
| j(n)= (n mod 10) + j(n div 10);
This gives the sum of the decimal digits of n.