Add some more to the generalization sections.

content/blog/modulo_patterns/index.md

@@ -277,37 +277,136 @@ factor with 9; 3, 6, 9, 12, and so on. However, those can't all serve as cycle l
 cycles can't get longer than 9. This leaves us with 3, 6, and 9 as _possible_ cycle lengths,
 none of which are divisible by 4. We've eliminated the possibility of spirals!
 
-{{< todo >}}
-This doesn't get to the bottom of it all.
-{{< /todo >}}
-
 ### Generalizing to Arbitrary Divisors
 The trick was easily executable on paper because there's an easy way to compute the remainder of a number
 when dividing by 9 (adding up the digits). However, we have a computer, and we don't need to fall back on such
 cool-but-complicated techniques. To replicate our original behavior, we can just write:
 
-```
+```Ruby
 def sum_digits(n)
   x = n % 9
   x == 0 ? 9 : x
 end
 ```
 
-But now, we can change the `9` to something else. Any number we pick, so long as it isn't
-{{< sidenote "right" "div-4-note" "divisible by 4," >}}
-"Wait", you might be thinking, "I thought you said that 4 can't have a common factor with the divisor,
-and that means any even numbers are out, too."<br>
-<br>
-Good observation. Although the path-not-divisible-by-four condition is certainly sufficient, it is not
-<em>necessary</em>. There seems to be another, less restrictive, condition at play here: even numbers work fine. I haven't
-figured out what it is, but we might as well make use of it.
-{{< /sidenote >}} will work. I'll pick primes for good measure. Here are a few good ones from using 13
+But now, we can change the `9` to something else. There are some numbers we'd like to avoid - specifically,
+we want to avoid those numbers that would allow for cycles of length 4 (or of a length divisible by 4).
+If we didn't avoid them, we might run into infinite loops, where our pencil might end up moving
+further and further from the center.
+
+Actually, let's revisit that. When we were playing with paths of length \\(k\\) while dividing by 9,
+we noted that the only _possible_ values of \\(k\\) are those that share a common factor with 9,
+specifically 3, 6 and 9. But that's not quite as strong as it could be: try as you might, but
+you will not find a cycle of length 6 when dividing by 9. The same is true if we pick 6 instead of 9,
+and try to find a cycle of length 4. Even though 4 _does_ have a common factor with 6, and thus
+is not ruled out as a valid cycle by our previous condition, we don't find any cycles of length 4.
+
+So what is it that _really_ determines if there can be cycles or not?
+
+Let's do some more playing around. What _are_ the actual cycle lengths when we divide by 9?
+For all but two numbers, the cycle lengths are 9. The two special numbers are 6 and 3, and they end up
+with a cycle length of 3. From this, we can say that the cycle length seems to depend on whether or
+nor our \\(n\\) has any common factors with the divisor.
+
+Let's explore this some more with a different divisor, say 12. We fill find that 8 has a cycle length
+of 3, 7 has a cycle length of 12, 9 has a cycle length of 4. What's
+happening here? To see, let's divide 12 __by these cycle lengths__. For 8, we get (12/3) = 4.
+For 7, this works out to 1. For 9, it works out to 3. These new numbers, 4, 1, and 3, are actually
+the __greatest common factors__ of 8, 3, and 7 with 12, respectively. The greatest common factor
+of two numbers is the largest number that divides them both. We thus write down our guess
+for the length of a cycle:
+
+{{< latex >}}
+k = \frac{d}{\text{gcd}(d,n)}
+{{< /latex >}}
+
+Where \\(d\\) is our divisor, which has been 9 until just recently. Indeed, this equation is in agreement
+with our experiment for \\(d = 9\\), too. Why might this be? Let's start once again with
+our equation for paths of length \\(k\\):
+
+{{< latex >}}
+kn \equiv 0\ (\text{mod}\ d)
+{{< /latex >}}
+
+Here we've replaced 9 with \\(d\\), since we're trying to make it work for _any_ divisor, not just 9.
+Now, suppose the greatest common divisor of \\(n\\) and \\(d\\) is some number \\(f\\). Then,
+since this number divides \\(n\\) and \\(d\\), we can write \\(n=fm\\) for some \\(m\\), and
+\\(d=fg\\) for some \\(g\\). We can rewrite our equation as follows:
+
+{{< latex >}}
+kfm \equiv 0\ (\text{mod}\ fg)
+{{< /latex >}}
+
+We can simplify this a little bit. Recall that what this equation _really_ means is that the
+difference of \\(kfm\\) and \\(0\\), which is just \\(kfm\\) is divisible by \\(fg\\):
+
+{{< latex >}}
+fg|kfm
+{{< /latex >}}
+
+But if \\(fg\\) divides \\(kfm\\), it must be that \\(g\\) divides \\(km\\)! This, in turn, means
+we can write:
+
+{{< latex >}}
+g|km
+{{< /latex >}}
+
+We also know that \\(g\\) and \\(m\\) have no common factors -- they were all divided out from \\(d\\)
+and \\(n\\) when we divided by \\(f\\)! We can thus further simplify our claim:
+
+{{< latex >}}
+g|k
+{{< /latex >}}
+
+This says that \\(k\\) must be divisible by \\(g\\). Recall that we got \\(g\\) by dividing
+\\(d\\) by \\(f\\), which is our largest common factor -- aka \\(\\text{gcd}(d,n)\\). We can thus
+write:
+
+{{< latex >}}
+\frac{d}{\text{gcd}(d,n)}|k
+{{< /latex >}}
+
+All that's left is to pick the smallest \\(k\\) that fits this description (that would be the first
+point at which our sequence of multiples of \\(n\\) will loop). Zero is divisible by anything, but
+alas, our sequence numbers start at 1 -- zero's out. The next best thing is \\(d/\\text{gcd}(d,n)\\).
+And there we have it: the first point at which our sequence loops, just like we guessed.
+
+Lastly, recall that a cycle occurs whenever a \\(k\\) is a multiple of 4. Now that we know what
+\\(k\\) is, we can restate this as "\\(d/\\text{gcd}(d,n)\\) is divisible by 4". But if we pick
+\\(n=d-1\\), the
+{{< sidenote "right" "coprime-note" "greatest common factor has to be \(1\)," >}}
+Wait, why is <em>this</em> true? Well, suppose some number \(f\) divides both \(d\) and \(d-1\).
+In that case, we can write \(d=af\), and \((d-1)=bf\). Subtracting one equation from the other:
+{{< latex >}}
+1 = (a-b)f
+{{< /latex >}}
+But this means that 1 is divisible by \(f\)! That's only possible if \(f=1\). Thus, the only
+number that divides \(x\) and \(x-1\) is 1; that's our greatest common factor.
+{{< /sidenote >}} which means that our condition further simplifies to
+"\\(d\\) is divisible by 4".
+Thus, we can state simply that any divisor divisible by 4 is off-limits, as it will induce loops.
+For example, pick \\(d=4\\). Running our algorithm for \\(n=d-1=3\\), we indeed find an infinite
+spiral:
+
+{{< figure src="pattern_3_4.svg" caption="Spiral generated by the number 3 with divisor 4." class="tiny" alt="Spiral generated by the number 3 by summing digits." >}}
+
+Let's try again. Pick \\(d=8\\); then, for \\(n=d-1=7\\), we also get a spiral:
+
+{{< figure src="pattern_7_8.svg" caption="Spiral generated by the number 7 with divisor 8." class="tiny" alt="Spiral generated by the number 7 by summing digits." >}}
+
+A poem comes to mind:
+> Turning and turning in the wydening gyre
+>
+> The falcon cannot hear the falconner;
+
+Fortunately, there are plenty of numbers that are not divisible by four, and we can pick
+any of them! I'll pick primes for good measure. Here are a few good ones from using 13
 (which corresponds to summing digits of base-14 numbers):
 
 {{< figure src="pattern_8_13.svg" caption="Pattern generated by the number 8 in base 14." class="tiny" alt="Pattern generated by the number 8 by summing digits." >}}
 {{< figure src="pattern_4_13.svg" caption="Pattern generated by the number 4 in base 14." class="tiny" alt="Pattern generated by the number 4 by summing digits." >}}
 
-Here are a few from dividing by 17 (base-18 numbers).
+Here's one from dividing by 17 (base-18 numbers).
 
 {{< figure src="pattern_5_17.svg" caption="Pattern generated by the number 5 in base 18." class="tiny" alt="Pattern generated by the number 5 by summing digits." >}}
 
@@ -316,3 +415,86 @@ Finally, base-30:
 {{< figure src="pattern_2_29.svg" caption="Pattern generated by the number 2 in base 30." class="tiny" alt="Pattern generated by the number 2 by summing digits." >}}
 
 {{< figure src="pattern_6_29.svg" caption="Pattern generated by the number 6 in base 30." class="tiny" alt="Pattern generated by the number 6 by summing digits." >}}
+
+### Generalizing to Arbitrary Numbers of Directions
+What if we didn't turn 90 degrees each time? What, if, instead, we turned 120 degrees (so that
+turning 3 times, not 4, would leave you facing the same direction you started)? We can pretty easily
+do that, too. Let's call this number of turns \\(c\\). Up until now, we had \\(c=4\\).
+
+First, let's update our condition. Before, we had "\\(d\\) cannot be divisible by 4". Now,
+we aren't constraining ourselves to only 4, but rather using a generic variable \\(c\\).
+We then end up with "\\(d\\) cannot be divisible by \\(c\\)". For instance, suppose we kept
+our divisor as 9 for the time being, but started turning 3 times instead of 4. This
+violates are divisibility condtion, and we once again end up with a spiral:
+
+{{< figure src="pattern_8_9_t3.svg" caption="Pattern generated by the number 8 in base 10 while turning 3 times." class="tiny" alt="Pattern generated by the number 3 by summing digits and turning 120 degrees." >}}
+
+If, on the other hand, we pick \\(d=8\\) and \\(c=3\\), we get patterns for all numbers just like we hoped.
+Here's one such pattern:
+
+{{< figure src="pattern_7_8_t3.svg" caption="Pattern generated by the number 7 in base 9 while turning 3 times." class="tiny" alt="Pattern generated by the number 7 by summing digits in base 9 and turning 120 degrees." >}}
+
+Hold on a moment; it's actully not so obvious why our condition _still_ works. When we just turned
+on a grid, things were simple. As long as we didn't end up facing the same way we started, we will
+eventually perform the exact same motions in reverse. The same is not true when turning 120 degrees, like
+we suggested. Here's a circle with the turn angles labeled:
+
+{{< figure src="turn_3_1.png" caption="Orientations when turning 120 degrees" class="small" alt="Possible orientations when turning 120 degrees." >}}
+
+We never quite do the exact _opposite_ of any one of our movements. So then, will we come back to the
+origin anyway? Well, let's start simple. Suppose we always turn by exactly one 120-degree increment
+(we might end up turning more or less, just like we may end up turning left, right, or back in the
+90 degree case). Now,
+
+1. Suppose that having performed one complete cycle, we end up away from the center
+by \\(dx\\) on the \\(x\\)-axis, and \\(dy\\) on the \\(y\\)-axis (we do this without loss
+of generality).
+2. We are now turned around by 120 degrees, so once we perform the cycle again, we end up offset
+by \\(dx(\\cos 120)-dy(\\sin 120)\\) on the \\(x\\)-axis, and \\(dx(\\sin 120)+dy(\\cos 120)\\) on
+the \\(y\\)-axis (I got this from the [rotation matrx](https://en.wikipedia.org/wiki/Rotation_matrix)
+page on Wikipedia).
+3. After one more step, we end up with having rotated a total of 240 degrees. As we perform the cycle
+again, we end up having moved by an additional \\(dx(\\cos 240)-dy(\\sin 240)\\) and \\(dx(\\sin 240)+dy(\\cos 240)\\).
+
+Summing up all of these changes, we get:
+
+{{< latex >}}
+    dx(\cos0+\cos120+\cos240) + dy(\sin0+\sin120+\sin240)
+{{< /latex >}}
+
+Why don't we start trying to write this in terms of variables already? For some number of turns
+\\(c\\), a single turn is \\(360/c\\) degrees. We start having turned 0 degrees, then progress
+to having turned \\(360/c\\) degrees, then \\(2\times360/c\\), and so on until \\((c-1)360/c\\).
+We can write this using _summation notation_ (and radians instead of degrees) as follows:
+
+{{< latex >}}
+    \begin{aligned}
+        x &= dx\left[\sum_{i=0}^{c-1} \cos\left(i\frac{2\pi}{c}\right)\right] -
+        dy\left[\sum_{i=0}^{c-1} \sin\left(i\frac{2\pi}{c}\right)\right] \\
+        y &= dx\left[\sum_{i=0}^{c-1} \sin\left(i\frac{2\pi}{c}\right)\right] +
+        dy\left[\sum_{i=0}^{c-1} \cos\left(i\frac{2\pi}{c}\right)\right] \\
+    \end{aligned}
+{{< /latex >}}
+
+For reasons beyond the scope of this article, sums like those between the square brackets
+in the above equations _always_ equal zero. This means that after all the turns have been made,
+we get \\(x=0\\) and \\(y=0\\) -- back at the origin, where we started!
+
+{{< todo >}}Maybe we can prove the sin/cos thing? {{< /todo >}}
+
+What if we turn by 240 degrees at a time: 2 turns instead of 1? Even though we first turn
+a whole 240 degrees, the second time we turn we "overshoot" our initial bearing, and end up at 120 degrees
+compared to it. As soon as we turn 240 more degrees (turning the third time), we end up back at 0.
+In short, even though we "visited" each bearing in a different order, we visited them all.
+
+{{< figure src="turn_3_2.png" caption="Orientations when turning 120 degrees, twice at a time" class="small" alt="Possible orientations when turning 120 degrees, twice at a time." >}}
+
+Let's try put some mathematical backing to this "visited them all" idea. 
+
+{{< todo >}}Remainders, visited them all, etc.{{< /todo >}}
+
+But let's not be so boring. We can branch out some, of course.
+
+{{< figure src="pattern_1_7_t5.svg" caption="Pattern generated by the number 1 in base 8 while turning 5 times." class="tiny" alt="Pattern generated by the number 1 by summing digits in base 8 and turning 72 degrees." >}}
+
+{{< figure src="pattern_3_11_t6.svg" caption="Pattern generated by the number 3 in base 12 while turning 6 times." class="tiny" alt="Pattern generated by the number 3 by summing digits in base 12 and turning 60 degrees." >}}

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