106 lines
4.4 KiB
Markdown
106 lines
4.4 KiB
Markdown
---
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layout: post
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title: "Mathematical Induction for proving tiling methods"
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date: 2017-10-19
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permalink: mathematical-induction-proving-tiling-methods/
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categories: math
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math: true
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---
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On my way towards self-taught data science, I've stumbled upon the need to be
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proficient with mathematical proofs, so I picked up the amazing [How To Prove
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It: A Structured
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Approach](https://www.amazon.com/How-Prove-It-Structured-Approach/dp/0521675995)
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by Daniel J. Velleman; and I've been fascinated by mathematical proofs since
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then.
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One of the uses for [Mathematical
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Induction](https://en.wikipedia.org/wiki/Mathematical_induction) which I've
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found to be pretty cool is proving methods for tiling shapes.
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Here is an example:
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Suppose $n$ is a positive integer. An equilateral triangle is cut into $4^n$
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congruent equilateral triangles, and one corner is removed. Show that the
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remaining area can be covered by trapezoidal tiles like this: <img
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src='/img/tiling/trapezoidal.jpg' class='inline' width='30'/>
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<canvas id='tiling-triangle' width='200' height='200' class='centered'></canvas>
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{% include caption.html text='An example of n = 2. The dark tile is removed.' %}
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Just to be clear, by _cover_ we mean covering without any overlaps, so you can't
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have two overlapping trapezoidal tiles.
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As with any mathematical induction solution, we start with the base case, which
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is $n = 1$, in that case the triangle looks like this:
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<canvas id='base-case' width='100' height='100' class='centered'></canvas>
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In this case, it's obvious that we can cover the leftover area using trapezoidal
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tiles consisting of three equilateral triangles (the second row), so:
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----
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Base case: $n = 1$ then the triangle has $4^1 = 4$ tiles, and by removing the
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top-most tile, the second row can be covered using a single trapezoidal tile.
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----
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Now for the induction step, we have to somehow show that after adding 1 to $n$,
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that is, by multiplying the tiles by $4$, we can still cover the triangle by
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trapezoidal tiles. To show this, we start by assuming we have a triangle with
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$4^n$ tiles, which we know can be covered by trapezoidal tiles, then we add the
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new tiles and show that they, too, can be covered by trapezoidal tiles.
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----
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Induction Step: Suppose we have a triangle split into $4^n$ tiles, and we know
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by induction hypothesis that it can be covered by trapezoidal tiles.
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Now suppose we have another triangle with $4^{n+1}$ tiles, that means, $4$ times
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as many triangles as our original triangle. We can then group the new bigger
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triangle into 4 congruent triangles, one of which we know can be split into
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trapezoidal tiles by removing one of it's tiles.
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For the three left triangles, we can find a neighbouring corner and assume the
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tile on that corner to be removed, and then cover the rest by trapezoidal tiles.
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Afterwards, since we had three such corners, and they are neighbouring corners,
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we can cover these three corners with one trapezoidal tile, thus completing the
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triangle.
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----
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Now that's a mouthful, in simpler terms, we are following these steps (for $n = 2$):
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<canvas id='n-2' width='200' height='200' class='centered'></canvas>
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First, we split the whole triangle into 4 smaller groups:
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<canvas id='n-2-grouped' width='200' height='200' class='centered'></canvas>
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Then, we know that one of the triangles can be covered by trapezoidal tiles if
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we remove one of it's tiles, that's the induction hypothesis (the case for $4^n$
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which is proved in the base case):
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<canvas id='n-2-grouped-removed' width='200' height='200' class='centered'></canvas>
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Leaving the top triangle behind, we now find neighbouring corners among the
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three left triangles, and we assume the tiles in those corners to be removed
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(they are not actually removed as we are constrained to remove only one tile):
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<canvas id='n-2-grouped-neighbours' width='200' height='200' class='centered'></canvas>
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Now we can cover the rest of these triangles by a single trapezoidal tile,
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similar to the case of $n = 1$.
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Afterwards, we see that the three neighbouring tiles form a trapezoidal tile,
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therefore we can now put the last piece there to complete the tiles.
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<canvas id='final' width='200' height='200' class='centered'></canvas>
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This procedure can be applied recursively on larger values of $n$ as well, so
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this concludes a proof of tiling an equilateral triangle divided into $4^n$
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equilateral triangles using trapezoidal tiles after removing a single piece.
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<script src="/js/tiling-shapes.js"></script>
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