This circuit is on Tinkercad: https://www.tinkercad.com/things/2B4XSzGDpWY-port-3-3-bit-multiplier/editel?sharecode=lX8hli3wYCjahW9KmGLha8723_c6o1aGP8h2mI43gts

This circuit began as a 2- bit multiplier however I found it a bit too simple, so I changed it to a 3 bit, which ended up with a fairly large circuit. This circuit is pretty self explanatory, it takes inputs of two 3 digit binary numbers, then multiplies them and displays them on the leds to the right. This is done through and gates, xor gates and an or gate only. Each combination of inputs had to be and-ed together then filtered through half adders and full adders depending on how close to the msb the digit was. This circuit was definitely bigger than I initially expected when I decided to upscale but it ended up working correctly. This was only three bit inputs so its only for regular positive binary digits as opposed to 2's compliment with a normal adder or subtractor. Essentially the inputs are anded together in order to get the multiplication product then fed through a series of adders in order to replicate addition on paper with two digits as well as a carry depending on the digit's place. For example, since there is always an empty space under the first product, the first sum does not require addition, then starting with A2xB1 and B2xA1 it is necessary to add, but does not require a carry as sum 1 will not carry.

This is a more in depth binary arithmetic circuit using multiplication rather than addition or subtraction, as well as a general combinational logic circuit.

Above is a simpler view of the circuit on circuitverse and below are the marked half adders and full adders the way I built them in tinkerCad. Each input combination is sent through an and gate, then I had two sets of three half adders and full adders as shown above. Since the lower sums did not need a carry in, a half adder was used, however with Sum three to six, I needed to utilize all three inputs which necessitated full adders. Sum 1 was the result of an and gate, sum two was the result of a half adder, and sum three required a full adder. Sum four required a half adder to add a carry out and sum of the previous full adders, then sum 5 was the sum of my final adder, and sum 6 was the same adder's carry out.