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86 lines
2.9 KiB
Markdown
86 lines
2.9 KiB
Markdown
# Recursion
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## Reductions
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Quote: *Reduction* is the single most common technique used in designing algorithms. Reduce one problem $X$ to another problem $Y$.
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The running time of the algorithm can be affected by $Y$, but $Y$ does not affect the correctness of the algorithm. So it is often useful to not be concerned with the inner workings of $Y$.
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## Simplify and Delegate
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Quote: *Recursion* is a particularly powerful kind of reduction, which can be described loosely as follows:
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- If the given instance of the problem can be solved directly, then do so.
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- Otherwise, reduce it to one or more **simpler instances of the same problem.**
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The book likes to call the delegation of simpler tasks "giving it to the recursion fairy."
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Your own task as an algorithm writer is to simplify the original problem and solve base cases. The recursion fairy will handle the rest.
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Tying this to mathematics, this is known as the **Induction Hypothesis**.
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The only caveat is that simplifying the tasks must eventually lead to the **base case** otherwise the algorithm might run infinitely!
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#### Example: Tower of Hanoi
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Assuming you know how the game works, we will describe how to solve it.
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Quote: We can't move it all at the beginning, because all the other disks are in the way. So first we have to move those $n - 1$ smaller disks to the spare peg. Once that's done we can move the largest disk directly to its destination. Finally to finish the puzzle, we have to move the $n -1$ disks from the spare peg to the destination.
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**That's it.**
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Since the problem was reduced to a base case and a $(n - 1)$ problem, we're good. The book has a funny quote "Our job is finished. If we didn't trust the junior monks, we wouldn't have hired them; let them do their job in peace."
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```
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Hanoi(n, src, dst, tmp):
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if n > 0
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Hanoi(n - 1, src, tmp, dst)
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move disk n from src to dst
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Hanoi(n - 1, tmp, dst, src)
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```
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## Sorting Algorithms
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## Merge Sort
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```
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MergeSort(A[1..n]):
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if n > 1
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m = floor(n / 2)
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MergeSort(A[1..m])
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MergeSort(A[m + 1..n])
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Merge(A[1..n], m)
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```
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```
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Merge(A[1..n], m):
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i = 1
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j = m + 1
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for k = 1 to n
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if j > n
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B[k] = A[i]
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i++
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else if i > m
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B[k] = A[j]
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j++
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else if A[i] < A[j]
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B[k] = A[i]
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i++
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else
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B[k] = A[j]
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j++
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Copy B to A
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```
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I think an important part to recall here is that the algorithm will break it down to the lowest level. An array with one element and slowly work its way up.
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That means we can always assume that each subarray that we're merging is already sorted! Which is why the merge algorithm is written the way it is.
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## The Pattern
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This section is quoted verbatim.
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1. **Divide** the given instance of the problem into several *independent smaller* instances of *exactly* the same problem.
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2. **Delegate** each smaller instance to the Recursion Fairy.
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3. **Combine** the solutions for the smaller instances into the final solution for the given instance.
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