Def multiplyusingshiftandadd a b initialize result result 0...

Name: def multiplyusingshiftandadd a b initialize result result 0 while the multiplier b is not zero while b 0 if the least significant bit of b is 1 if b 1 result a add the multiplicand a to 83102
Uploaded: 2023-01-24T01:07:27-08:00
Duration: 7 min 37 s
Channel: Craig West
Description: def multiplyusingshiftandadd a b initialize result result 0 while the multiplier b is not zero while b 0 if the least significant bit of b is 1 if b 1 result a add the multiplicand a to 83102

def multiply_using_shift_and_add( A, B): # Initialize result result = 0 # While the multiplier (B) is not zero while B > 0: # If the least significant bit of B is 1 if B & 1: result += A # Add the multiplicand (A) to the result # Shift the multiplicand (A) left (multiply by 2) A = A << 1 # Shift the multiplier (B) right (divide by 2) B = B >> 1 return result

Transcript

00:03 So in this video, we're told that one way to do multiplication is through repeated addition.

00:09 And so they give us an algorithm to work through here.

00:13 And so we're given two positive numbers a and b.

00:16 So we're given two positive numbers a and b.

00:20 And we want to come up with an algorithm here.

00:22 So we wanted a number c, which is that contains the product of a times b.

00:30 Okay, that's what we're shooting for, some number c here.

00:33 And so in the algorithm, it says step one, right, set the value of c equal to zero.

00:40 Okay, so c is equal to zero.

00:45 Okay, so c is actually going to contain our product, and we want to start it at zero.

00:50 Because if we're going to do repeated addition, we want to start at zero.

00:54 So step two, set the value of i.

00:59 Let's set the value of i equal to a.

01:03 So set the value of i equal to a.

01:06 So a is how many times we're going to do the repeated addition.

01:11 I could also set i equal to b that would work the same way.

01:16 So three, repeat steps blank and blank until the value of i equals blank.

01:24 So we want to repeat steps blank and blank until i equals something.

01:42 Okay.

01:43 Step 4, we're going to set the value of c to be equal to something.

01:50 In step 5, we're going to set the value of, we're going to subtract 1 from i.

01:56 So i'm going to say i equals i minus 1.

02:01 Okay, that's how we subtract 1 from i.

02:04 Step 6 is to print out the final answer to c and step 7 stop.

02:09 So i'm not going to write those two things.

02:11 Let's try to figure out what's going on here.

02:13 Right so if i let c equals zero to start with and the idea here is that i want to repeat adding a certain amount um i want to add a number a certain amount of times so if i'm doing a times b right i want to add b a times all right so i'm going to add b to c so down here in step five i'm going to step four i'm to say let's see equal c plus b and then remember because i already set a to i and so then i'm going to say i equals i minus one so i is going to come become one less right so what i basically want to do is i want to repeat these two steps right here so i want to repeat step four and step five and i'm going to add in i'm going to do this until i get all the way down to i equals zero, okay, because i'll be subtracting one each time, so i equals zero.

03:26 All right, so let's work, let's do this with two numbers here so that we can see how the flow goes.

03:33 Okay.

03:34 So here, let's start with a equals three and b equals five, right? so three times five, we know is 15.

03:46 So let's go through our algorithm here.

03:48 Okay, step one here, right? step one, we have c equals zero.

03:56 Step two, we have i equals my a value is three...