00:01
Let's first find the value of yi.
00:04
So for that we need to map it.
00:10
1, 2, 3, 4.
00:15
Here we can write the value of a i b i here s 1 s o that is 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 0 0 1 1 1 1 1 1 1 here also we have 1 so how can we map it we can map like this first and here we have a 1 1 also we have of this big one.
00:48
Four ones are together.
00:50
So we can write our output y i is equal to s1 bar s0 plus s0 into a i bar plus s1 s0 bar b.
01:04
Now how can we implement this? so now let's implement using using ai.
01:11
Let's first draw like this.
01:13
So we have a node gate here.
01:19
Now again bi with a node gate.
01:29
Now we have s1 with a node gate.
01:36
So this will be ai bar.
01:39
Node gate means bi bar s1 bar...