Computer Science B Homework #2 Due Monday February 9, 2004, 13:30

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1 Computer Science B Homework #2 Due Monday February 9, 2004, 13:30 In all the problems, all calculations are done mod M where M= in order to limit the size o the integers involved. This way we avoid problems due to overlow. As we have seen in class, the Fibonacci numbers are deined as ollows: n = 1 i n=1,2 n-1 + n-2 i n>2 Thus the most natural way to write a program to compute one uses the next algorithm: ibrec(n) I (n<3) Then return 1 Else return ibrec (n-1)+ ibrec (n-2) [15%] 1) Write a Java program that implements this algorithm. Find the time needed to compute 1, 2, 3, 4, 5, and plot a graph o your results. What is the largest n or which you can compute n within 1 second? The running time should be exponential. Indeed, the running time is Ω( n ) as explained in class beore.

2 An iterative way o computing the same values only keeps track o the last two values and use them to compute the new ones : ibit(n) :=1; g:=1 For k:=2 to n do := +g g:= -g Return [15%] 2 Write a Java program that implements this algorithm. Find the time needed to compute 1, 2, 3, 4, 5, and plot a graph o your results. What is the largest n or which you can compute n within 1 second? The running time should be linear, i.e. O(n). An alternate deinition we have NOT seen in class or the Fibonacci numbers is: 1 i n=1,2 n = 2 2 n n 12 i n>2, n odd 2 2 n 2 +1 n2 1 i n>2, n even

3 [15%] 3a) Prove by mathematical induction that this new deinition o n is correct. basis case: n=1,2: 1 = 2 =1 induction step: Let n>2 and assume the ormulas are correct or all k, 0<k<n. Then notation: F= 2 n odd, (n-1 even, n-2 odd) n even, (n-1 odd, n-2 even) n = n-1 + n-2 = * [F (n-1)/2+1 -F (n-1)/2-1 ]+[F (n-1)/2 +F (n-3)/2 ] = F (n+1)/2 -F (n-3)/2 +F (n-1)/2 +F (n-3)/2 = F (n+1)/2 +F (n-1)/2 NOTE : = * indicates use o the induction hypothesis. All other steps are simple algebraic manipulations. n = n-1 + n-2 = * [F n/2 +F (n-2)/2 ]+[F (n-2)/2+1 -F (n-2)/2-1 ] = F n/2 +F n/2-1 +F n/2 -F n/2-2 = * 2F n/2 +[ n/2 - n/2-2 ] 2 -F n/2-2 = 2F n/2 +[F n/2-2 n/2 n/2-2 +F n/2-2 ]-F n/2-2 = F n/2 +2F n/2-2 n/2 n/2-2 = * n/2 [ n/2 +2 n/2-2 n/2-2 ] = n/2 [ n/2 +2 n/2-1 ] = [ n/2+1 - n/2-1 ] [ n/2+1 + n/2-1 ] = * F n/2+1 -F n/2-1 [15%] 3b) Write a Java program that implements this algorithm. Find the time needed to compute 1, 2, 3, 4, 5, and plot a graph o your results. What is the largest n or which you can compute n within 1 second? The running time should be linear, i.e. O(n). [15%] 4a) Prove the ollowing by mathematical induction or all n> n = n 1 n n n+1

4 basis case: n=1,2 : = , = induction step: Let n>2 and assume the ormula is correct or n-1. Then 0 1 n = n = * n 2 n 1 = n 1 + n 2 n 1 = n 1 n n n+1 n 1 n n + n 1 n A natural way o computing this exponentiation is to use a method similar to the one we used in Homework #1 or numbers (below I stands or the 2 by 2 identity matrix): ExpMOD(A,b) I (b=0) Then return I Else I (b mod 2)=0 Then return ExpMOD( A 2, b/2 ) Else return A( ExpMOD(A 2, (b-1)/2 ) )

5 [15%] 4b) Write a Java program that implements this algorithm. Find the time needed to compute 1, 2, 3, 4, 5, and plot a graph o your results. What is the largest n or which you can compute n within 1 second (give an estimate i n is too big )? The running time should be logarithmic, i.e. O(log n). [10%] 5) I M is rather small (say M<1000) and you wish to compute the value n mod M; ind a very ast algorithm to compute this value, even or very large values o n. hint: Show that there exists an integer R, 0<R<M 2 such that n mod M = n mod R mod M. Since the value o n mod M is completely deined by n-1 mod M and n-2 mod M it is clear that whenever n-1 = n+r-1 mod M and n-2 = n+r-2 mod M then n = n+r mod M. Thereore i we ind a value R such that R+1 = 1 (= 1 ) mod M and R+2 = 1 (= 2 ) mod M then or any positive n we have n = n+r mod M and thus n mod M = n mod R mod M. All is let to prove is that there exists an integer R, 0<R<M 2 such that R+1 = 1 mod M and R+2 = 1 mod M. I we consider the sequence 1 mod M, 2 mod M, 3 mod M,, M 2 mod M, M 2 +1 mod M, M 2 +2 mod M we enumerate M 2 +1 pairs ( i mod M, i+1 mod M) o consecutive values o the Fibonacci sequence. Now since only M 2 such pairs (x,y) exist, there must exist an i and a j such that ( i mod M, i+1 mod M) = ( j mod M, j+1 mod M). Let R=i-j<M 2. Notice that i we start with ( i mod M, i+1 mod M) = ( j mod M, j+1 mod M) then i-1 mod M = j-1 mod M as well because n-1 = n+1 n. We obtain ( i mod M, i+1 mod M) = ( j mod M, j+1 mod M) which means that ( i mod M, i+1 mod M) = ( i+r mod M, i+r+1 mod M) which implies that ( i-1 mod M, i mod M) = ( i+r-1 mod M, i+r mod M) which implies that ( i-2 mod M, i-1 mod M) = ( i+r-2 mod M, i+r-1 mod M) which implies by induction that ( 1 mod M, 2 mod M) = ( R+1 mod M, R+2 mod M).

6 Pre-processing: FindR(M) :=1; g:=2; R:=1 While NOT (=1 and g=1) do := +g mod M g:= -g mod M R:=R+1 Return R Given constant R: FastFib(n) Return(Fibit(n mod R))