quartz/content/out/notes/induction.md

title	sr-due	sr-interval	sr-ease
Induction	2022-04-23	29	272

tags: #review

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Induction

PECS

Phases of argument by induction

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

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

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.

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

Satisfaction

Commence satisfaction. Confidence +100. 😆

Examples

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