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I'm currently reading introduction to algorithms and came by Johnson’s algorithm that depends on making sure that all paths are positive.

the algo depends on finding a new weight function (w') that is positive for all edges and keeps the correctness of the shortest paths relations.

It does so by calculating h(s), h(d) values to be added to the w original value.

My question is, why not just find the smallest w in the graph and add it to all edges ? this will satisfy both conditions and will require less calculation.

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    $\begingroup$ Have you tried proving your claim or finding a counter-example? Hint: your intuition is wrong. (Community, I'm pretty sure this is a duplicate. Can you find it?) $\endgroup$ – Raphael Jan 20 '16 at 19:19
  • $\begingroup$ @Raphael I'm pretty sure it's a dupe, too, but I figured it'd be faster to answer it than find the dupe. $\endgroup$ – David Richerby Jan 20 '16 at 21:30
  • $\begingroup$ @Raphael I'm sorry since I couldn't phrase my question in a specific format I couldn't search for it. $\endgroup$ – Mr.Me Jan 20 '16 at 22:33
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    $\begingroup$ We do have a question that already explains this, but it got marked as a dup of another question that is pretty confusing and hard to understand, and that mixes multiple different questions together. Therefore, I think this question has value over what we had before. If you wanted, I suppose we could re-target the dups (close them as a dup of this instead off what they currently point to). $\endgroup$ – D.W. Jan 21 '16 at 11:20
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Adding a weight to every edge adds more weight to long paths than short paths. (Long in the sense of having many edges.)

For example, suppose the lowest-cost edge has weight $-2$ and there are two paths from $a$ to $b$: a single edge of weight $3$ and a path with two edges, each of weight $1$. The two-edge path has the lowest weight. However, if you add $2$ to every edge, the one-edge path has weight $5$ but the three-edge path now has weight $6$, so you get the wrong answer.

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Increasing every edge weight by the same amount does not necessarily increase every path by the same amount of distance. Rather, the increase to the paths are often disproportional which depends on how many edges the path has.

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    $\begingroup$ This effect is already mentioned in the other answer. $\endgroup$ – Yuval Filmus Sep 20 '17 at 8:47
  • $\begingroup$ I just rephrased it to the point of confusion. $\endgroup$ – Pendechosen Sep 20 '17 at 14:38

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