Wrong behavior around multiplication

[odashi]

The current implementation converts a * b into $ab$. This only works if the variables have only 1 character and doesn't generate a correct syntax for other cases:

• Long names: abc * def --> $abcdef$, which is hard to distinguish from other combination of variables.
• Unary operators: x * -y --> $x - y$, which is definitely wrong.

These cases require explicit multiply operator ( $\cdot$ or $\times$ ) between the operands.

I think the Mult(left, right) subtree should be converted to $left \cdot right$ by default, and try to avoid the operator only in the following cases:

• The left-hand side operand is a single character or a bracket (with or without unary operators and/or super/subscripts)
• The right-hand side operand is a single character or a bracket (with or without super/subscripts, without unary operators)

This requires fine-graind name processing, and may be applied after #82.

[odashi]

Anyway, I will handle this issue today.

[odashi]

Anyway some simple heuristic can be applied without waiting for #82

[odashi]

Re-open this issue since it is actually not resolved yet.

[ZibingZhang]

Mind if I take a look into this?

[odashi]

Maybe it's good if we get a list of possible LHS-RHS pairs with/without the multiplication symbol, e.g.,

left	right	tex
3	x	$3 x$
x	y	$x \cdot y$

[ZibingZhang]

I can't think of a case where we would want to remove the \cdot other than the case of constant * name.

left	right	tex	why
3	x	$3x$	basic multiplication with constant (should work with floats and complex)
x	3	$x \cdot 3$	should not reorder for user, if they want lack of \cdot, can change order themselves
two	names	$two \cdot names$	two variables should always have a \cdot
3	x * 4 * y	$3x \cdot 4y$	should work in more complicated expression as well

[odashi]

Some complete examples:

	numeric	alphabet	math	word	function	bracket
numeric	$2 \cdot 3$	$\color{red} 2 y$	$\color{red} 2 \beta$	$\color{red} 2 \mathrm{bar}$	$\color{red} 2 g(y)$	$\color{red} 2 (u + v)$
alphabet	$x \cdot 3$	$\color{red} x y$	$\color{red} x \beta$	$x \cdot \mathrm{bar}$	$x \cdot g(y)$	$x \cdot (u + v)$
math	$\alpha \cdot 3$	$\color{red} \alpha y$	$\color{red} \alpha \beta$	$\alpha \cdot \mathrm{bar}$	$\alpha \cdot g(y)$	$\alpha \cdot (u + v)$
word	$\mathrm{foo} \cdot 3$	$\mathrm{foo} \cdot y$	$\mathrm{foo} \cdot \beta$	$\mathrm{foo} \cdot \mathrm{bar}$	$\mathrm{foo} \cdot g(y)$	$\mathrm{foo} \cdot (u + v)$
function	$f(x) \cdot 3$	$f(x) \cdot y$	$f(x) \cdot \beta$	$f(x) \cdot \mathrm{bar}$	$f(x) \cdot g(y)$	$f(x) \cdot (u + v)$
bracket	$(s + t) \cdot 3$	$\color{red} (s + t) y$	$\color{red} (s + t) \beta$	$\color{red} (s + t) \mathrm{bar}$	$\color{red} (s + t) g(y)$	$\color{red} (s + t) (u + v)$

In cases {alphabet,math,word}-{bracket} it would make some confusion between them and function calls if we removed \cdot.

EDIT(2023-10-16)

Added some consistent rules:

1. Numeric should always keep the preceding \cdot.
2. Functions should always keep the following \cdot due to some ambiguity (e.g., sin(x) * y -> $\sin xy$)
3. Numeric and brackets can remove the following \cdot without ambiguity, except the next operand is numeric.

[ZibingZhang]

I think alphabet-alphabet would also cause confusion, because you could have a (arguably dumb) case like

def (x, y, xy):
    return xy

In general though when user sees $xy$ there's no easy way of telling if it's $x \cdot y$ or a single variable $xy$, so I think it we should keep the \cdot.

edit: I'm realizing that word is stylized differently than alphabet, so it should be clear that $\mathrm{xy}$ is different than $xy$.

[odashi]

I'm realizing that word is stylized differently than alphabet

Yes this is intended in #139, but maybe the result is somewhat confusing for some users.

[ZibingZhang]

Looking through the code this seems like we'll have to utilize information from expr codegen to figure out what nodes are wrapped in parens , as well as information from identifier replacer to differentiate between math vs. alphabet vs. word.

[ZibingZhang]

Another situation, if we have a * b @ x @ y, this should result in $ab \cdot xy$.

[odashi]

Another situation, if we have a * b @ x @ y, this should result in .

It looks $abxy$ is acceptable. $ab \cdot xy$ looks to represent (a*b)@(x@y), but:

• Matrix multiplication is an associative operation: position of parenthesis is meaningless
• We get the AST (((a*b)@x)@y) for this expression.

[odashi]

I think we eventually need more informative object than the raw string for expression codegen to implement this feature (and other stuff that rely on the LaTeX syntax rules). let me implement it later