1!
1!
CS1110 7 October 2008!
More on Recursion"
My first job was working in an orange juice factory, "
but I got canned: couldn't concentrate.!
Then I worked in the woods as a lumberjack, but I "
just couldn't hack it, so they gave me the axe.!
After that I tried to be a tailor, but I just wasn't suited "
for it. Mainly because it was a so-so job.!
Next I tried working in a muffler factory but that was
exhausting.!
I worked as a pilot but eventually got grounded for "
taking off too much.!
Then I tried teaching but I couldn't make the grade.!
We derive recursive functions. !
Study Sect 15.1, p. 415. Watch activity 15-2.1 on the CD. In
DrJava, write and test as many of the self-review exercises as
you can (disregard those that deal with arrays).!
Get more of
these from the
course website!
2!
Geometry test!
3!
Recursive functions!
/** = a copy of s in which s[0..1] are swapped, s[2..3] are
swapped, s[3..4] are swapped, etc. */!
public static String swapAdjacent(String s)!
/** = b c. Precondition: c ā„ 0*/"
public static int exp(int b, int c)!
"Properties:"
(1)āÆb c = b * b c-1 "
(2)āÆFor c even"
" "b c = (b*b) c/2"
e.g 3*3*3*3*3*3*3*3!
= (3*3)*(3*3)*(3*3)*(3*3)!
4!
Recursive functions!
/** = b c. Precondition: c ā„ 0*/"
public static int exp(int b, int c) {!
if (c = 0)!
return 1;!
if (c is odd)!
return b * exp(b, cā1);!
// c is even and > 0!
return exp(b*b, c / 2);!
}!
c number of calls!
0 1!
1 2!
2 2!
4 3!
8 4!
16āÆ 5!
32 6!
2n n + 1!
32768 is 215"
so b32768 needs only 16 calls!!
5!
Binary arithmetic!
Decimal Binary ! Dec ! !Binary Oct!
00 ! 00 ! ! 20 = 1 ! 1 ! ! !0!
01 ! 01 ! ! 21 = 2 ! !10 !!!1!!
02 ! 10 ! ! 22 = 4 ! !100 !!!2!
03 ! 11 ! ! 23 = 8 ! !1000!
04 ! 100 ! ! 24 = 16 ! !10000!
05 ! 101 ! ! 25 = 32 ! !100000!
06 ! 110 ! ! 26 = 64 ! !1000000!
07 ! 111 ! ! 215 = 32768 !1000000000000000 !7!
08 ! 1000 !!!!!!!10!
09 ! 1001 !!!!!!!11 !
10 ! 1010 !!!!!!!12!
Test c odd: Test last bit = 1!
Divide c by 2: Delete the last bit!
Subtract 1 when odd: Change last bit from 1 to 0.!
Exponentiation algorithm processes binary rep. of the exponent.!
6!
Hilbertās space-filling curve!
Hilbert(1):!
Hilbert(2):!
Hilbert(n):!
H(n-1)"
left!
As the size of each
line gets smaller and
smaller, in the limit,
this algorithm fills
every point in space.
Lines never overlap.!
H(n-1)"
dwn!
H(n-1)"
dwn!
H(n-1)"
right!
