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Concrete mathematics : a foundation for computer science

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A ANSWERS TO EXERCISES 537<br />

and we have xk=, qk+k = 0 by taking the limit as n + 03. Hence Q(x,y) is<br />

a multiple of U(x,y), say A(x,y)U(x,y). But U(F,+,,F,) = (-1)” and n is<br />

even, so Pc(x,y) = P(x,y) - (U(x,y) - l)A(x,y) is another polynomial such<br />

that Po(F,+, , F,) = 0. The total degree of PO is less than t, so PO is a multiple<br />

of U - 1 by induction on t.<br />

Similarly, P(x,y) is divisible by U(x,y) + 1 if P(F,+, ,F,) = 0 <strong>for</strong> infinitely<br />

many odd values of n. A combination of these two facts gives the<br />

desired necessary and sufficient condition: P(x, y) is divisible by U(x, y)’ - 1.<br />

6.78 First add the digits without carrying, getting digits 0, 1, and 2. Then<br />

use the two carry rules<br />

O(d+l)(e+l) + lde,<br />

O(df2)Oe --t ldO(e+l),<br />

always applying the leftmost applicable carry. This process terminates because<br />

the binary value obtained by reading (b, . . . bZ)F as (b, . . . b2)2 increases<br />

whenever a carry is per<strong>for</strong>med. But a carry might propagate to the<br />

right of the “Fibonacci point”; <strong>for</strong> example, (1 )~+(l )F becomes (10.01)~. Such<br />

rightward propagation extends at most two positions; and those two digit positions<br />

can be zeroed again by using the text’s “add 1" algorithm if necessary.<br />

Incidentally, there’s a corresponding “multiplication” operation on<br />

nonnegative integers: If m = Fj, +. . . + Fjq and n = Fk, +. 3 . + Fk, in the Fibonacci<br />

number system, let m o n = xz=, EL=, Fjb+k,, by analogy with mul-<br />

Exercise: m 0 n = tiplication of binary numbers. (This definition implies that m o n z x/!? mn<br />

mn+ when m and n are large, although 1 o n M +*n.) Fibonacci addition leads to<br />

l(m+l)/@Jn+ a proof of the associative law 1 o (m o n) = (L o m) o n,)<br />

ml(n+l )/@I .<br />

6.79 Yes; <strong>for</strong> example, we can take<br />

A0 = 331635635998274737472200656430763;<br />

A, = 1510028911088401971189590305498785 .<br />

The resulting sequence has the property that A,, is divisible by (but unequal<br />

to) pk when n mod mk = rk, where the numbers (pk, mk,rk) have the following<br />

18 respective values:<br />

(3,4,1) (2,3,2) (5,5>1)<br />

(7,833) (17,9,4) (11,10,2)<br />

(47,16,71 (19,18,10) (61,15,3)<br />

(2207,32,15) (53,27,16) (31,30,24)<br />

(1087,64,31) (109,27,7) (41,20,10)<br />

(4481,64,63) (5779,54,52) (2521,60,60)

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