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The circuit for the Millionaires' Problem is a digital comparator circuit (which is a chain of full adders working as a subtractor and outputting the carry flag). A full adder circuit can be implemented using only one AND gate and some XOR gates. This means the total number of AND gates for the circuit of the Millionaires' Problem is equal to ...
The sum-output from the second half adder is the final sum output of the full adder and the output from the OR gate is the final carry output (). The critical path of a full adder runs through both XOR gates and ends at the sum bit . Assumed that an XOR gate takes 1 delays to complete, the delay imposed by the critical path of a full adder is ...
An example of a 4-bit Kogge–Stone adder is shown in the diagram. Each vertical stage produces a "propagate" and a "generate" bit, as shown. The culminating generate bits (the carries) are produced in the last stage (vertically), and these bits are XOR'd with the initial propagate after the input (the red boxes) to produce the sum bits. E.g., the first (least-significant) sum bit is ...
To convolutionally encode data, start with k memory registers, each holding one input bit.Unless otherwise specified, all memory registers start with a value of 0. The encoder has n modulo-2 adders (a modulo 2 adder can be implemented with a single Boolean XOR gate, where the logic is: 0+0 = 0, 0+1 = 1, 1+0 = 1, 1+1 = 0), and n generator polynomials — one for each adder (see figure below).
A carry-skip adder [nb 1] (also known as a carry-bypass adder) is an adder implementation that improves on the delay of a ripple-carry adder with little effort compared to other adders. The improvement of the worst-case delay is achieved by using several carry-skip adders to form a block-carry-skip adder.
A conditional sum adder [3] is a recursive structure based on the carry-select adder. In the conditional sum adder, the MUX level chooses between two n/2-bit inputs that are themselves built as conditional-sum adder. The bottom level of the tree consists of pairs of 2-bit adders (1 half adder and 3 full adders) plus 2 single-bit multiplexers.
The first input to the XOR gate is the actual input bit; The second input for each XOR gate is the control input D; This produces the same truth table for the bit arriving at the adder as the multiplexer solution does since the XOR gate output will be what the input bit is when D = 0 and the inverted input bit when D = 1.
A majority gate returns true if and only if more than 50% of its inputs are true. For instance, in a full adder, the carry output is found by applying a majority function to the three inputs, although frequently this part of the adder is broken down into several simpler logical gates.