Turing Tumble Community

2.5 bit multiplier


A times B is equal to C
A is 3 bit, B is 2 bit, and C is 5 bit.

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Without explanation it was quite a challenge to understand, how it works. But now I think, I’ve got, how to use it

  1. the non-operand bits need to point left before start
  2. the register A is set as usual, bits pointing left are 0, bits pointing right are 1, the top most is the smallest bit. On the picture, A is set to 7
  3. B is made of two ramps, the right ramp is the smaller bit. If ramp points right, it is 1, if pointing left it is 0. On the picture, B is set to 3
  4. the resulting C has the same logic as A: smallest bits at the top, left is 0, right is 1. Set it to 0 before starting.
  5. start by pulling the left lever

Tested with 4x3 → 12, 1x3 → 3

But multiplying by other values, than 3, e.g. 2, 1 seem not to work. Pointing both ramps to the left (multiply by 0) even leads to a endless loop of red balls. Or how the B and the non-operand bits (bits outside of A and C) are supposed to work?

It should intercept eventually, I dont know why it would loop forever.

The endless Loop is because the ramp-left-positions are connected together and in the end are both triggering a new red ball.
Question: Why is the down left non-operand-bit in the picture pointed right? Shouldn’t it be pointed left?

Yeah I set it up wrong for the picture. Sorry.

The Problem is that if the first ramp is pointed left, it does two 2-turns per left count and if it’s the second ramp, it has two 1-turns per count