Can a chlorinated or brominated compound be reduced by HI thus forming iodine monochloride or iodine monobromide?
Or the other way round, can  iodinated compound be reduced by HBr or HCl?

According to this chapter on reduction, from Bernthsen and Suborough's Textbook of Organic Chemistry, reduction of a chloro substituent by hydriotic acid is possible; at least in one instance:

https://www.thevespiary.org/rhodium/Rhodium/hive/hiveboard/picproxie_docs/000505616-bernsthensudborough-reduction.djvu" target="_blank" title="Download this file">

Wow!
The article is a treasure case for reductions - highly recommended!

Thanks lugh!  

ORG

I'd bet my boots on it.

Step I -- finkelstein swap R-Cl --> R-I
Step II-- HI reduction of iodo compound R-I + HI --> R-H + I2

Since I guess it doesn't matter what the source of ions is for the finkelstein swap as long as they're there, probably HI could sub for NaI quite nicely. Or whatever. A one pot process it would bee, very likely.

The thread on replacing red P with formic acid talked a lot about this, IIRC.

It's really too bad the HI reduction isn't looked at much outside the reduction of ephedrine. It seems to bee very versatile and useful.

Since I guess it doesn't matter what the source of ions is for the finkelstein swap as long as they're there, probably HI could sub for NaI quite nicely. Or whatever.

Ehm... And what is the driving force for the finkelstein reaction? Why is acetone chosen as the solvent?

I'd guess you're hinting that solubility differences will make the formed NaCl drop out of solution, pushing the reaction forward.

In the proposed reaction, an equilibrium between alkyl chloride and iodide is fine, as once the iodide reacts with HI and is reduced to alkane, there's no going back.

Both methods have their "irreversable" step.

If only the alkyl iodide is reduced reasonably fast by HI to the alkane, and the halide equilibrium reaction is slow, the reaction might take a very long time. In that case it would be better to perform the halide swap first.

But if the mechanism of the halide swap is the same, and the reduction is not the rate-limiting step, then they should bee equally fast. How would preforming the iodide improve this?

I think in the formic acid thread it was hinted that replacement of HO- in ephedrine with I- was also the slow step.

Probably it's worth a test.

But if the mechanism of the halide swap is the same, and the reduction is not the rate-limiting step, then they should bee equally fast.

That conclusion cannot be drawn - we are using different solvents (acetone vs. water). The SN2 halide swap is way faster in polar aprotic solvents like acetone than in water.

How would preforming the iodide improve this?

By performing the SN2 reaction in a polar aprotic solvent and the radical/ionic reduction in water, each step is performed in its medium of choice.

One can debate how much of a difference it makes in practice, this is however the theoretical considerations.

Sounds like it's worth some research. A good question.

I put my belief in that it would work in one pot. Maybe it would work better as two.

How versatile that iodine is.

It would be very funny if some clandestine chemists elaborated the reaction in all kinds of situations, and it became a popular method in academic research again.