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Hi,

ok, now I see your point. The issue is indeed a bit more subtle.

Pak algorithm is guaranteed to find a mapping between two equivalent sets of propagators, but
it does not make any statements on how this mapping will be realized via momentum shifts. This
is also not surprising, since while ordering the Feynman parameters $x_i$, the algorithm works with
a momentum free representation.

Now for practical purposes one naturally wants a shift that only involves changing loop momenta.
By selecting the very first ordering of propagators found by Pak one assumes that it will be the most
useful one. Heuristically, mostly it happens to be true, but there are also more pathological cases (like yo…

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@adisurtya
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@vsht
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vsht Jun 21, 2026
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Answer selected by adisurtya
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