Induction

1 PECS

Phases of argument by induction

Preparation -> most important
Execution -> becomes routine if prep is good
Checking -> second most important
Satisfaction

1.1 Preparation

isolate the property that you are trying to verify and the parameter, n, associated with is
- e.g., min possible size of set of rank k is 2^n
Confirm by hand that for small values of the parameter, the property is true
Use previous cases as assumptions
Pause and reflect
If you understand what's going on -> proceed to execution

1.2 Execution

Technical and prescribed (once you're an expert you can take some liberties)

Four parts

statement
verificatio of base case
inductive step
conclusion

e.g.,

we will prove that, for every non-negative integer n, insert property here
For n = 0, The property is true because explicit verification of this case
for any n > 0, assuming the property is true for n-1 (or, for all k < n), the property is true at n because explain why we can take a step up
Therefore, by induction, the property is true for all n.

1.3 Checking

Basically debugging without a compiler to find errors

have you forgotten anything? e.g., the base case
Does the inductive step work fro 0 to 1? or are they irregular
Make sure that you are only assuming the result for things less than n
ideally show someone and try to convince them (dont let them be polite)
if necessary go back to execution or preparation

1.4 Satisfaction

Commence satisfaction. Confidence +100. 😆

2 Examples

2.1 Union Find - min size for set of rank k

Initially every element is its own representative and every element has rank 0;
when we do a union operation, the the two reps have different ranks, the ranks stay the same
when we do a union operation, if the two reps have the same rank, then the rank increases

minimum (and only) size of a rank 0 rep is 1

to get a rank 1 representative, we form a union of either a rank 0 and a rank 1 set or two rank 0 sets for the minimum possible size, it must be the second case, and the two rank 0 sets must be each of minimum size 1, so this gives minimum size for a rank 1 set of 2

To get a rank 2 rep, we form a union of either rank 2 and rank 0 or 1 set, or two rank 1 sets For the minimum possible size, it must be the second cae, and the two rank 1 sets must each be of minimum size 2, so this gives minimum size for a rank 2 set of 4

To get a rank n rep, we form a union of either rank n and rank k set for some k<n or two rank n-1 sets. For the minimum possible size, it must be the second cae, and the two rank n-1 sets must each be of minimum size, which we are assuming 2^(n-1), so this gives minimum size for a rank n set of

2^{n-1} + 2^{n-1} = 2\times2^{n-1} = 2^n

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