## EXPERT ANSWER

The following word problem can be converted into a logic working and circuit diagram for the resultant Boolean function:

Light bulb (L) is ON if switch A (A) is OFF and switch B (B) is ON, or switch C (C) is ON.

The circuit diagram will have two inputs, two outputs, and four states. The two inputs are switches A and B, which are both off when the light bulb is off. The two outputs are on when both switches are on—either switch A is on or switch B is on. The four states of this circuit diagram are:

1) Switch A is both off or on and either switch B is off or on;

2) Switch A and B both off;

3) Switch B only being off; and

4) Switch C only being on.

We can start by thinking about the fact that if switch A (A) is OFF, then the light bulb (L) will be OFF. This means that we don’t need to worry about whether switch B (B) is ON or OFF, because if it’s ON then the light bulb will be ON, and if it’s OFF then the light bulb won’t be on.

We can also think about the fact that if switch B (B) is ON, then the light bulb will be ON. This means that we don’t need to worry about whether switch C (C) is ON or OFF, because if it’s ON then the light bulb will be on.

Now we’ve converted this into a logic working: L = Ω(A(ON)) + Ω(B(ON)) + Ω(C(ON)).

If we are to turn off the light, then the switches must both be OFF. The resulting Boolean function has two possible inputs, and therefore can have only two possible outputs: ON or OFF.

To solve this problem, we will use an AND function: the light bulb will be ON if both switches are OFF or if either switch is turned on.

We can represent this logic with an AND gate. The input of the gate is represented by the first switch (A), and the output is represented by the second switch (B). To toggle A and B, we need to apply a voltage between them at least 5V higher than their minimum voltage levels.

b)i) Compute the weight of the new genome in a decimal scheme.

The weight of the new genome is calculated as follows:

1. The first ten digits of the number must be greater than or equal to 1 and less than 10. The sum of these digits is 175.

2. The next two digits must be greater than or equal to 2 and less than 9, with a sum of 3. This gives a total of 177 (17 + 7 + 7 = 22).

3. The last digit must be greater than or equal to 3 and less than 0, with a sum of 5, giving us 178 (17 + 8 + 8 = 24).

4. Formula for conversion from decimal numbers to hexadecimal numbers:

The weight of a genome in a decimal scheme is equal to the sum of the digits of its two-digit number, divided by 10. For example, if we have a four-digit number with values 1 through 9, then we would say that its weight is 4×9/10=3.

Hexadecimal number = Decimal number × 16 ÷ 10,000 (remembering that 0x is pronounced “zero”): 178 × 16 ÷ 10,000 = 172880 Hexadecimal number = 172880 × 16 ÷ 10,000 = 178,128 Hexadecimal number = 19, the weight of a genome is equal to its product (number+sum) divided by 16. If we have a four-digit number with values 0 through F, then its weight will be (4+0)×16/16=175.

**Explanation:**

Compute the weight of the new genome in decimal scheme.

The new genome will have a weight of 175. The ones (unit) digit of the number exceeds the tens (10) digit by a 3 and the product of the number and its sum exceeds 175 by a 3. Therefore, we can use the following formula to compute the weight:

w = 2^n + 2^m+ 2^(n+m)

The weight of a genome in decimal scheme is equal to the sum of the digits of its two-digit number, divided by 10. For example, if we have a four-digit number with values 1 through 9, then we would say that its weight is 4×9/10=3.

In Hexadecimal scheme, the weight of a genome is equal to its product (number+sum) divided by 16. If we have a four-digit number with values 0 through F, then its weight will be (4+0)×16/16=175.