Concrete mathematics : a foundation for computer science

74 INTEGER FUNCTIONS
By the way, there’s a mnemonic for remembering which case uses floors
and which uses ceilings: Half-open intervals that include the left endpoint
but not the right (such as 0 < 8 < 1) are slightly more common than those
that include the right endpoint but not the left; and floors are slightly more Just like we can recommon
than ceilings. So by Murphy’s Law, the correct rule is the opposite member the date of
of what we’d expect -ceilings for [OL . . p) and floors for (01. . 01.
Columbus’s depart
ure by singing, “In
Similar analyses show that the closed interval [o(. . fi] contains exactly fourteen hundred
Ll3J - [a] +1 integers and that the open interval (01.. @) contains [fi] - LX]- 1; ;o~u~~~-$;~;{~e
but we place the additional restriction a # fl on the latter so that the formula deep b,ue sea ,,
won’t ever embarrass us by claiming that an empty interval (a. . a) contains
a total of -1 integers. To summarize, we’ve deduced the following facts:
interval integers contained restrictions
[a.. 81 1B.l - Toil+1 a6 B,
[a.. I31 Ml - bl a6 B, (3.12)
(a.. Bl LPJ - 14 a< 6,
(a..B) TPl - 14 -1 a< p.
Now here’s a problem we can’t refuse. The Concrete Math Club has a
casino (open only to purchasers of this book) in which there’s a roulette wheel
with one thousand slots, numbered 1 to 1000. If the number n that comes up
on a spin is divisible by the floor of its cube root, that is, if
then it’s a winner and the house pays us $5; otherwise it’s a loser and we
must pay $1. (The notation a\b, read “a divides b,” means that b is an exact
multiple of a; Chapter 4 investigates this relation carefully.) Can we expect
to make money if we play this game?
We can compute the average winnings-that is, the amount we’ll win
(or lose) per play-by first counting the number W of winners and the number
L = 1000 - W of losers. If each number comes up once during 1000 plays,
we win 5W dollars and lose L dollars, so the average winnings will be
5w-L 5w-(looo-w) 6W- 1000
~ =
1000 ;ooo = 1000 .
[A poll of the class
at this point showed
that 28 students
thought it was a
bad idea to play,
13 wanted to gamble,
and the rest
were too confused
to answer.)
(So we hit them
with the Concrete
If there are 167 or more winners, we have the advantage; otherwise the ad- Math aub.1
vantage is with the house.
How can we count the number of winners among 1 through 1 OOO? It’s
not hard to spot a pattern. The numbers from 1 through 23 - 1 = 7 are all
winners because [fi] = 1 for each. Among the numbers 23 = 8 through
33 - 1 = 26, only the even numbers are winners. And among 33 = 27 through
43 - 1 = 63, only those divisible by 3 are. And so on.