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2-color your Dynkin diagram, black and white. Let $w$ be the product of the white simple reflections, $b$ the product of the black. Note that $w$ and $b$ are well-defined, as the reflections you're multiplying to make them, commute. You'll have to pick the order if you want an actual word, in what follows.

If $G$ is not $A_{even}$: the affinization of the diagram is also 2-colorable, so you can choose the affine vertex to be white. Let $\chi = w b$, a Coxeter element. The Coxeter number $h$ is even, and $\chi^{h/2} = w_0$. So you get a reduced word $wbwbwb\ldots wb$, where the total number of letters is $h$ (and each letter is a product of commuting reflections).

If $G$ is, unfortunately, $A_{even}$: you have to pick $w$ vs. $b$, and the diagram automorphism shows that the choice is unavoidable. The Coxeter number is odd. But you still get a reduced word, $wbwb\ldots bw$, again with $h$ letters.

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2-color your Dynkin diagram, black and white. Let $w$ be the product of the white simple reflections, $b$ the product of the black. Note that $w$ and $b$ are well-defined, as the reflections you're multiplying to make them, commute. You'll have to pick the order if you want an actual word, in what follows.

If $G$ is not $A_{even}$: the affinization of the diagram is also 2-colorable, so you can choose the affine vertex to be white. Let $\chi = w b$, a Coxeter element. The Coxeter number $h$ is even, and $\chi^{h/2} = w_0$. So you get a reduced word $wbwbwb\ldots wb$, where the total number of letters is $h$ (and each letter is a product of commuting reflections).

If $G$ is, unfortunately, $A_{even}$: you have to pick $w$ vs. $b$, and the diagram automorphism shows that the choice is unavoidable. The Coxeter number is odd. But you still get a reduced word, $wbwb\ldots bw$, with $h$ letters.