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- 1 Resume
- 2 Current Parts
- 3 Planned Parts
Resume[edit | edit source]This is the plan of the current calculator I've built :
Of course, all compilations are made with binary code. This is why we have many different decoders.
Current Parts[edit | edit source]
I will try to go in a somewhat logic order to describe the parts.
Control Panel (room)[edit | edit source]
The control panel is the room from which you set the inputs and decide of the operation
Numbers Input Panel[edit | edit source]
Here, the user will decide what numbers he wants to use. Right now, I am using a lever-based binary input system, so the user has to decompose the numbers he wants to use into powers of two.
Operation Panel[edit | edit source]
From this panel, the user chooses between the operations he is going to use : add (+), subtract (-), multiply (*) and divide (/). Again, right now, I am using the lever system.
Input Wires (white and orange)[edit | edit source]
These wires link the input panel and the operation panel to the different logic units. Try to rearrange them in a manner where the same values go together. So, your wires should look like, from left to right: A1; B1; A2; B2; A4; B4; ...
Logic Units[edit | edit source]
These are the machines that perform the operations.
Adder/Subtractor (yellow and red)[edit | edit source]
I've built my own version of an adder/subtractor. Its construction is simple because it is modulated (made of many same parts). That means that if you use more bits, you can just add more parts on the side, but you'll have to change some links.
In this machine, your inputs (in binary code) goes into the bottom (yellow) full adders. Each adder needs the two inputs (A and B) with the same values. Also, the least significant bit has to be on the left, so they should all be connected by their carries. Basically, your inputs look like the wires in the #Input Wires (white and orange) section. Use basic bridges to pass wires over the others without connecting them. Your A inputs (left) are the minuend (X in X-Y=Z), and go straight in the adders. The B inputs, your subtrahend (Y in X-Y=Z), have to pass through a multiplexer, made out of a modified version of a XOR gate which gives to the adder an inverted signal in case of a subtraction. The multiplexer is controlled by a switch (on the picture, that switch is on the left). The sums go into another multiplexer, which, again, gives an inverted input in case of a subtraction. This is controlled by an IMPLIES gate (in the top right) which gives a true output if the switch is on "Subtraction" AND if the last carry is true. This is required because on a subtraction, that last carry actually means the "-" (minus) sign.
The white machines are half-adders, that use, as inputs, the carry of the last adder and the sum of their respective full adder. We need this because, if the answer is negative, it uses the equation " -A = !A (inverted A)+ 1 ", as explained here. The final outputs are all of the top-most wires you can observe, plus the wire on the right (the carry from the last full adder) and the carry that goes in the first (left) half-adder, as the negation sign.
Multiplier (light blue)[edit | edit source]
Let's start with the basics here, because this is the most complicated part. For our purposes, multiplication is a repeated addition. That means that, once again, we will use adders here.
Before them, you actually have to set up an AND gate (not including the control one). Its use is simple : in binary multiplication, because we only use 0's and 1's, the only way that we can have an output is by multiplying 1 by 1.
I will go in order from the least to the most significant bits.
Least significant bits : 1*1 = 1. That means that the output of the second AND gate (the control one) goes straight to the output collective wires.
Second to last: 1*2 = 2 and 2*1 = 2. Those two outputs meet in a full adder. The sum goes to the output, and the carry goes to the next bit.
Next: 1*4 = 4; 2*2 = 4; and 4*1 = 4 The carry from the last bit goes in the first adder as the carry input. The two normal inputs are 2 of the 3 AND gates. The sum of this goes in a second adder, where the second input is the third AND gate. Both carry outs go to the next stage, and the sum goes to the output.
You continue like this until you run out of AND gates, or equations.
Divider (pink)[edit | edit source]
This one is an easy one compared to the multiplication. Again, we will be using the full adders. Basically, for each A input, set up n adders where n=number of B inputs. Also, this time, you have to "reverse them". Now the most significant bit should pass its carry downwards.
Output wires[edit | edit source]
These have to get every output from every machine and redirect them to the next part.
Binary-to-decimal decoder[edit | edit source]
This transforms your binary code into a decimal output. The size of it will be (Binary inputs*2)*(Decimal output)
*Quick note* It uses a "programmable" XOR gate cane tend to a not gate. This activates a line of preset redstone torches to output the correct answer.