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2nd qtn ka answer kya 61 hoga sir? Plz rply kal xm hai..

Sir is there any difference between left to right and right to left sir.. because i understand left to ryt...even i think ryt to left is also same for me If any difference..pls let me know sir

thanks a lot sir your teaching way is so good ❤❤

Excellent explanation

Sir mazaa aa gya smajh ke 😊

are you ignou teacher ???

Sir. Sir Jo apne last me question show Kara h uska answer 29 aa raha h. Sir 29 answer is right or wrong.

Bahut ache se smjaya sir 👍pura pura smj aya

🙂👍 0:51

Excellent explanation

excellent explanation

Nice bro

Thanks alot

Ek hi chapter aur chapter kab tak aaye ga

Sir aapki tabiyat kharab ho rhi h. 2-4 din ka rest lelo.....

👌👌👌

Thanks a lot! It is so clear and easy to understand.

Horner's Rule for Polynomial Evaluation --------------------------------------------------------------- Input: An array of A[0...n] of coefficient of a polynomial of degree n and a point x. Output: The value of polynomial at given point x. Evaluate_Horner(A, n, x) { p = 0; for(i = n; i >= 0; i--) p = p * x + A[i]; return p; } Q1) P(x) = x^2 + 3x + 2 [ans: 12 at x=2] n = 2 (degree of P(x)) array A[n+1] = {2,3,1} x = 2 Evaluate_Horner(A, n, x): P = 0 iteration 1: i = 2 P = P * x + A[2] = 0 * 2 + 1 = 1 iteration 2: i = 1 P = P * x + A[1] = 1 * 2 + 3 = 5 iteration 3: i = 0 P = P * x + A[0] = 5 * 2 + 2 = 12 return P Q2) P(x) = 2x^3 + 3x^2 + 2x + 10 [ans: 42 at x=2] n = 3 (degree of P(x)) array A[n+1] = {10,2,3,2} x = 2 Evaluate_Horner(A, n, x): P = 0 iteration 1: i = 3 P = P * x + A[3] = 0 * 2 + 2 = 2 iteration 2: i = 2 P = P * x + A[2] = 2 * 2 + 3 = 7 iteration 3: i = 1 P = P * x + A[1] = 7 * 2 + 2 = 16 iteration 4: i = 0 P = P * x + A[0] = 16 * 2 + 10 = 42 return P

Left to Right Binary Exponentiation -------------------------------------------------------- Algorithm to compute a^n by Left to Right Binary Exponentiation Methods. Input: a^n, and Binary string of length S for exponent n as array A[S]. Output: a^n Algo: 1. Result = a 2. for ( i = S-2 to 0 ) 3. Result *= Result 4. if A[i] == 1, then 5. Result = Result * a 6. return Result (i.e. a^n) Pseudo: left_to_right(a, A[S], S) { result = a; for (i = S-2; i >= 0; i++) { result = result * result; if(A[i] == 1) result = result * a; } return result; } Q) a^17 4 3 2 1 0 Binary of 17: 1 0 0 0 1 a = 17 S = 5 A[S] = {1,0,0,0,1} result = a iteration 1: i = 3 result = a * a = a^2 A[3] =/= 1 iteration 2: i = 2 result = a^2 * a^2 = a^4 A[2] =/= 1 iteration 3: i = 1 result = a^4 * a^4 = a^8 A[1] =/= 1 iteration 4: i = 0 result = a^8 * a^8 = a^16 A[0] == 1: result = a^16 * a = a^17 loop ends return result (a^17)

Nice ❤️❤️

bro how to solve it when the value of x is not given ??

Voice is not clear.. Pls can do better...

Bro can you please share this notes with us

Algorithm to compute a^n by left to right binary exponentiation method is as follows: Input: a^n and binary string of length s for exponent n as an array A[s] Output: Final value of a^n . 1 result=a 2 for i=s-2 to0 3 result = result *result 4 if A[i]= 1then 5 result= result *a 6 return result (i.e a n ) Let us take an example to illustrate the above algorithm to compute a 17 In this exponent n=17 which is equivalent to binary string 10001 Step by step illustration of the left to right binary exponentiation algorithm for a 17 : s=5 result=a Iteration 1: i=3 result=a *a= a 2 A[3] ≠ 1 Iteration 2: i=2 result= a 2 * a 2 = a 4 A[2] ≠ 1 Iteration 3: i=1 result= a 4 * a 4 = a 8 A[1] ≠ 1 Iteration 4: i=0 result= a 8 * a 8 = a 16 A[0] = 1 result = a 16 * a = a 17 return a 17

Best video

Excellent explanation

The way of teach is awesome