Question

2. Consider the following functions. For each of them, determine how many times is ‘hey’ printed in terms of the input n. You should first write down a recurrence and then solve it using the recursion tree method. That means you should write down the first few levels of the recursion tree, specify the pattern, and then solve. (a) def fun(n) { if (n > 1) { print( ‘hi’ ‘hi’ ‘hi’ ) fun(n/4) fun(n/4) fun(n/4) }}

(b) def fun(n) { if (n > 1) { for i=1 to n { print( ‘hi’ ‘hi’ ) } fun(n/4) fun(n/4) fun(n/4) }}

2. Consider the following functions. For each of them, determine how many times is 'hey? printed in terms of the input n. You should first write down a recurrence and then solve it using the recursion tree method. That means you should write down the first few levels of the recursion tree, specify the pattern, and then solve. (a) def fun(n) { if (n > 1) { print( 'hi' 'hi' 'hi') fun (n/4) fun(n/4) fun(n/4) }}

Answer 1

I have attached four images for the question .First two are for part (a) and last two are for part(b) .I would suggest you to please comment your review for the answer.

Q2.

(a) Answer to 1st Part Of Question

2. (a) Resuvunce Relation is an follows: Tim / 3T12/4) + 1 m l otherwise Ruous Tree T(nu) in18) ng Tinig fin86) 7/1/16) (1116) 0 Tinit) Tallo) Tin/16) T1/64) Tim164) T12164) "level 3 - Exar i summing and at each level, the number of one (marked o) Tn1=1+3+ 3² + 3² + 34.3k & k dendes the cast level The last level I will be when m/ uk = 1 = m = 4k - x = logem or logiz (m) 2 .. Tln) = 1 + 3 + 3² + 3² + 34 3k "Geometric Progression" common ratio (8)=3, First term (a)= 1 Total terms (nl=k+1 Sn = alim_1) (0-1) -> T() = (34) = (3x3731_)

Using lega(b) = log{ b) TM- 4( 3x log(a) - 1) . -* 413x (myy ) - ) T Usring log (6) = 1 oba) in <3 (m) "???. 1) TM) - L12007-) Exact Number of Times "Hi" us punted is :- 13x13 logu (m/ 12
(b) Answer to 2nd Part Of Question
- 4m (1 - 13 og 4 (n)+1 T/m) f 31/014) + m inal ml [./16) Tin 116) TINI) Tim116) T116) 116) Tin (16) TH164) T(1/64) zact Tim) - + 3(4) + 3% (42) + 3(%) () = m (1 +18)+($)+(2).(*)") Termination of hee accur at when mal a m =) m = 4 * » k = logy (n) common Ratio (8) = 3 @ First Term o(a) all Tim) = m/M (131) 9-)) - um (1-1914) 4

T(M 1= 40 (1.–134) 934[/+ Tlm) - um (1 – 636) (19logu mi) +(n)= 4m(! – (%) ( a ) ) Oraing boy.com - see Tim) = 4m (1-e 39 cm pituisang = HM (1-2 m ) ay = 40 ( =3 in ) 7,019)) = Um (1-2 (n)1020) T!n) = 4m (1-2 / 1m 0.20 ) log (n)+1 TM) = um 3m 0.80 =) o (m) Escact No. of Times "Hi" is printed = 4m (1-1310