Question

3. Generate random numbers as indicated and comment on the results. (a) generate 10 random numbers using the middle-square method using 20 = 1009. (b) generate 10 random numbers using linear congruence with a = 5, b = 1 and c= 8.

Answer 1

An Sher- (a) In middle square method We have to start with a 4- digit number called seed. Then of every step we have to square our Seed and then next number of the Sequence will be middle four digit of square númber (must have 8-digits) 20=1009 (20) = (1009) .010180B1 => 4,-0180 (20)C0180)? - 0003240 » 22 +0:324 (23)? : (032472-00104976 -> 23 = 1049 (33) = (1049)2 = 01100401 -> 2451004 (ay)2 = (1004) - 200006400> 250064 Gor : (0080)2 - 00004096 => X6-0040 lor. C0064)2 =00000256 => 27-0002 Co2 = C0040)2 = 00001600 => 29-1.0016 (= (0016) · 00000256 -) dq=0002 (2932 = (0002)? = 0000004 => 210-000 O And How all ani nalo xn-00007 with seed 1009 our sequence converges too fast after as and it get its limit at ra Scanned with die foene)

In lineal Congrence We have to start wih!! I seed then next team anumber Wo called Seed is calculated by the 2941(0Xn+b) (noodc) ulated by the fomula. Here to is not given let xo=1. a=5, b=1,0 = 8 ( :(s) +) dB): GCd82 -> - 6 az = (56+) modo = 31 (mod 8) => 22 23 = (567)+1) mod(8) = (36)(mod!) => x3 = 24 = (5(w+1) mod (8) = 2i mods) => 2425 25 = (515)+1) moda = 26 (mods) :) X5 ac - (5(2)+1) mod(8) = 1 (mods) =) 26 27 - (513)+1) modos = 16 (mods) =) 27 = 28 = (560)+1) mod (9) = 1 (modz) -) 12 = 1 aq - C5 (1)+1) mod(s) 6 (modo) =) 2q = 6 As we start repeated here 50 40=22 = 7 211-23 = 4 and soon Here we took (mod 8) so our sequency started repeatation after 8 steps: