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I came across this problem:

Find the mistake of the following inductive proof:

Theorem: all algorithms have the same time complexity.

Proof: (By induction on the number of algorithms.)
The statement is trivial for one algorithm.
Let the statement holds for some K. Consider (K+1) algorithms. Remove any and algorithm, then the remaining algorithms have the same time complexity according to the assumption. Return the algorithm back and remove another one, then the remaining algorithms have the same time complexity again. It means all (K + 1) have the same time complexity.
Q.E.D.



My attempt: Consider the case of 2 algorithms (K+1) = 2, each with different time complexity. By removing any of them, we have only one algorithm where the statement holds. However, this is not two if each one has different time complexity.


Is this the correct way to solve the problem?
Thanks.

Salma
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    Yes, the argument fails for $K=2$. In particular, implicit in the proof is that removing one algorithm out of $K$ leaves at least $2$ algorithms. – Clement C. May 15 '16 at 21:53
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    Yes, and you can find some extensive discussion of essentially the same question here. – Brian M. Scott May 15 '16 at 22:01
  • @ClementC. sorry, but I did not get your point. Do you mean we need k≥2 in our base case? or the disprove ? – Salma May 15 '16 at 22:15
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    I am saying that the inductive step only works if $k\geq 2$ to begin with, so one would need a establish a base case for some $k\geq 2$. But, of course, this is not going to happen. – Clement C. May 15 '16 at 22:18
  • This is exactly the same "proof" that all human beings have the very same age. It fails for $;k=2;$ and only because of that we all don't have the same age...:) – DonAntonio May 15 '16 at 22:21

1 Answers1

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There are two mistakes here.

Mistake 1) To prove all members of some given set $A$ have some property, it is not valid to argue by induction on the number of elements of $A$: "By induction on the number of (all) algorithms ..." is not a valid method of proof.

Mistake 2) The induction scheme in the proposed proof is valid if what we were trying to prove is: if $X$ is any finite set of algorithms, then all algorithms in $X$ have the same time complexity. Or in other words, writing $t(A)$ for the time complexity of an algorithm $A$, that $t$ is constant on any finite set of algorithms $X$. However, the inductive step fails in this proof: you can't infer that $t$ is constant on the two-element set $\{A, B\}$ from the (obvious fact) that it is constant on the singleton sets $\{A\}$ and $\{B\}$. (The rhetoric in the question about moving algorithms in and out of the set sounds convincing but does not work when $K = 2$. I think this is what you had in mind in your attempt at the question.)

Rob Arthan
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