How does one actually add with this?

▲ curtisf 6 hours ago | parent | next [-]

It's basically using the "-" embedded in the definition of the eml operator.

Table 4 shows the "size" of the operators when fully expanded to "eml" applications, which is quite large for +, -, ×, and /.

Here's one approach which agrees with the minimum sizes they present:

        eml(x, y             ) = exp(x) − ln(y) # 1 + x + y
        eml(x, 1             ) = exp(x)         # 2 + x
        eml(1, y             ) = e - ln(y)      # 2 + y
        eml(1, exp(e - ln(y))) = ln(y)          # 6 + y; construction from eq (5)
                         ln(1) = 0              # 7

After you have ln and exp, you can invert their applications in the eml function

              eml(ln x, exp y) = x - y          # 9 + x + y

Using a subtraction-of-subtraction to get addition leads to the cost of "27" in Table 4; I'm not sure what formula leads to 19 but I'm guessing it avoids the expensive construction of 0 by using something simpler that cancels:

                   x - (0 - y) = x + y          # 25 + {x} + {y}

	▲	5 hours ago \| parent [-]
		[deleted]

▲ bzax 7 hours ago | parent | prev | next [-]

Well, once you've derived unary exp and ln you can get subtraction, which then gets you unary negation and you have addition.

	▲	freehorse 5 hours ago \| parent [-]
		And then by using the fact that the exponential turns addition into multiplication, you get multiplication (and subtraction gives division).

▲ nick238 6 hours ago | parent | prev [-]

Don't know adding, but multiplication has diagram on the last page of the PDF.

xy = eml(eml(1, eml(eml(eml(eml(1, eml(eml(1, eml(1, x)), 1)), eml(1, eml(eml(1, eml(y, 1)), 1))), 1), 1)), 1)

From Table 4, I think addition is slightly more complicated?

▲ simplesighman 6 hours ago | parent | next [-]

Thanks for posting that. You had a transcribing typo which was corrected in the ECMAScript below. Here's the calculation for 5 x 7:

    const eml = (x,y) => Math.exp(x) - Math.log(y);
    const mul = (x,y) => eml(eml(1,eml(eml(eml(1,eml(eml(1,eml(1,x)),1)),eml(1,eml(eml(1,eml(y,1)),1))),1)),1);
    console.log(mul(5,7));

> 35.00000000000001

For larger or negative inputs you get a NaN because ECMAScript has limited precision and doesn't handle imaginary numbers.

▲ Charon77 6 hours ago | parent | prev [-]

x+y = ln(exp(x) * exp(y))

exp(a) = eml(a, 1) ln(a)=eml(1,eml(eml(1,a),1))

Plugging those in is an excercise to the reader

	▲	jcgrillo 6 hours ago \| parent [-]
		might need to turn the paper sideways