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78 lines
2.8 KiB
Markdown
78 lines
2.8 KiB
Markdown
---
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title: Induction
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sr-due: 2022-04-23
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sr-interval: 29
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sr-ease: 272
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---
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tags: #review
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---
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# Induction
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## PECS
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Phases of argument by induction
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- Preparation -> most important
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- Execution -> becomes routine if prep is good
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- Checking -> second most important
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- Satisfaction
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### Preparation
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- isolate the property that you are trying to verify and the parameter, n, associated with is
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- e.g., min possible size of set of rank k is $2^n$
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- Confirm by hand that for small values of the parameter, the property is true
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- Use previous cases as assumptions
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- Pause and reflect
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- If you understand what's going on -> proceed to execution
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### Execution
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Technical and prescribed (once you're an expert you can take some liberties)
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Four parts
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- statement
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- verificatio of base case
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- inductive step
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- conclusion
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e.g.,
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- we will prove that, for every non-negative integer $n$, *insert property here*
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- For $n = 0$, *The property* is true because *explicit verification of this case*
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- 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*
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- Therefore, by induction, *the property* is true for all n.
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### Checking
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Basically debugging without a compiler to find errors
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- have you forgotten anything? e.g., the base case
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- Does the inductive step work fro 0 to 1? or are they irregular
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- Make sure that you are only assuming the result for things less than $n$
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- ideally show someone and try to convince them (dont let them be polite)
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- if necessary go back to execution or preparation
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### Satisfaction
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Commence satisfaction.
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Confidence +100. 😆
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## Examples
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### Union Find - min size for set of rank k
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- Initially every element is its own representative and every element has rank 0;
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- when we do a union operation, the the two reps have different ranks, the ranks stay the same
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- when we do a union operation, if the two reps have the same rank, then the rank increases
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minimum (and only) size of a rank 0 rep is 1
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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
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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
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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
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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
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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.
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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
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> $2^{n-1} + 2^{n-1} = 2\times2^{n-1} = 2^n$
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