From a573a0b765a202acf5510ce0487bf3a96d889478 Mon Sep 17 00:00:00 2001 From: Danila Fedorin Date: Sat, 8 Jan 2022 17:42:33 -0800 Subject: [PATCH] Update and publish digit sum patterns article --- content/blog/modulo_patterns/index.md | 71 ++++++++++++++++++--------- 1 file changed, 47 insertions(+), 24 deletions(-) diff --git a/content/blog/modulo_patterns/index.md b/content/blog/modulo_patterns/index.md index 4cd9d31..fd30276 100644 --- a/content/blog/modulo_patterns/index.md +++ b/content/blog/modulo_patterns/index.md @@ -2,7 +2,7 @@ title: Digit Sum Patterns and Modular Arithmetic date: 2021-12-30T15:42:40-08:00 tags: ["Ruby", "Mathematics"] -draft: true +description: "In this article, we explore the patterns created by remainders from division." --- When I was in elementary school, our class was briefly visited by our school's headmaster. @@ -79,39 +79,39 @@ divisor (what a mouthful). In mathematics, we typically use \\(a|b\\) to say \\(a\\) divides \\(b\\), or, as we have seen above, that the remainder of dividing \\(b\\) by \\(a\\) is zero. -Working with remainders actually comes up pretty commonly in discrete math. A well-known +Working with remainders actually comes up pretty frequently in discrete math. A well-known example I'm aware of is the [RSA algorithm](https://en.wikipedia.org/wiki/RSA_(cryptosystem)), which works with remainders resulting from dividing by a product of two large prime numbers. But what's a good way to write, in numbers and symbols, the claim that "\\(a\\) divides \\(b\\) with remainder \\(r\\)"? Well, we know that dividing yields a quotient (possibly zero) and a remainder -(also possibly zero). Let's call the quotient \\(k\\). +(also possibly zero). Let's call the quotient \\(q\\). {{< sidenote "right" "r-less-note" "Then, we know that when dividing \(b\) by \(a\) we have:" >}} It's important to point out that for the equation in question to represent division -with quotient \(k\) and remainder \(r\), it must be that \(r\) is less than \(a\). +with quotient \(q\) and remainder \(r\), it must be that \(r\) is less than \(a\). Otherwise, you could write \(r = s + a\) for some \(s\), and end up with {{< latex >}} \begin{aligned} - & b = ka + r \\ - \Rightarrow\ & b = ka + (s + a) \\ - \Rightarrow\ & b = (k+1)a + s + & b = qa + r \\ + \Rightarrow\ & b = qa + (s + a) \\ + \Rightarrow\ & b = (q+1)a + s \end{aligned} {{< /latex >}} In plain English, if \(r\) is bigger than \(a\) after you've divided, you haven't taken out "as much \(a\) from your dividend as you could", and the actual quotient is -larger than \(k\). +larger than \(q\). {{< /sidenote >}} {{< latex >}} \begin{aligned} - & b = ka + r \\ - \Rightarrow\ & b-r = ka \\ + & b = qa + r \\ + \Rightarrow\ & b-r = qa \\ \end{aligned} {{< /latex >}} We only really care about the remainder here, not the quotient, since it's the remainder that determines if something is divisible or not. From the form of the second equation, we can -deduce that \\(b-r\\) is divisible by \\(a\\) (it's literally equal to \\(a\\) times \\(k\\), +deduce that \\(b-r\\) is divisible by \\(a\\) (it's literally equal to \\(a\\) times \\(q\\), so it must be divisible). Thus, we can write: {{< latex >}} @@ -332,7 +332,7 @@ Let's explore this some more with a different divisor, say 12. We fill find that 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 +the __greatest common factors__ of 8, 7, and 3 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: @@ -410,7 +410,18 @@ Lastly, recall that our patterns would spiral away from the center whenever a \\ \\(k\\) is, we can restate this as "\\(d/\\text{gcd}(d,n)\\) is divisible by 4". But if we pick \\(n=d-1\\), the greatest common factor has to be \\(1\\) (see [this section below](#divisors-of-n-and-n-1)), so we can even further simplify this "\\(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 +For example, pick \\(d=4\\). Running our algorithm +{{< sidenote "right" "constructive-note" "for \(n=d-1=3\)," >}} +Did you catch that? From our work above, we didn't just find a condition that would prevent spirals; +we also found the precise number that would result in a spiral if this condition were violated! +This is because our proof is constructive: instead of just claiming the existence +of a thing, it also shows how to get that thing. Our proof in the earlier section (which +claimed that the divisor 9 would never create spirals) went by contradiction, which was +not constructive. Repeating that proof for a general \(d\) wouldn't have told us +the specific numbers that would spiral.
+
+This is the reason that direct proofs tend to be preferred over proofs by contradiction. +{{< /sidenote >}} 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." >}} @@ -420,7 +431,7 @@ 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 +> Turning and turning in the widening gyre > > The falcon cannot hear the falconner; @@ -450,7 +461,7 @@ First, let's update our condition. Before, we had "\\(d\\) cannot be divisible b 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: +violates our 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." >}} @@ -483,8 +494,8 @@ a triangle: As long as you rotate by the same amount each time (and you will, since the cycle length determines how many times you turn, and the cycle length never changes), you can do so for any number -of directions. For instance, here's a similar visualization for a five-turn system, where -each turn is 72 degrees: +of directions. For instance, here's a similar visualization in which +there are 5 possible directions, and where each turn is consequently 72 degrees: {{< figure src="turn_5_anim.gif" caption="Pentagon formed by five 72-degree turns." class="small" alt="Pentagon formed by five 72-degree turns." >}} @@ -537,14 +548,14 @@ after \\(360/\\text{gcd}(360, t)\\) turns, we'll end up facing the way that we s So far, so good. What I don't like about this, though, is that we have all of these numbers of degrees all over our equations: 72 degrees, 144 degrees, and so forth. However, -something like 73 degrees (in a 5-turn system) is just not a valid bearing, +something like 73 degrees (if there are five possible directions) is just not a valid bearing, and nor is 71. We have so many possible degrees (360 of them, to be exact), but we're only -using a handful! That's wasteful. Instead, observe that for a system with \\(c\\) turns, -the smallest possible turn is \\(360/c\\). Let's call this turn \\(\\theta\\) (theta). +using a handful! That's wasteful. Instead, observe that for \\(c\\) possible turns, +the smallest possible turn angle is \\(360/c\\). Let's call this angle \\(\\theta\\) (theta). Now, notice that we always turn in multiples of \\(\\theta\\): a single turn moves us \\(\\theta\\) degrees, two turns move us \\(2\\theta\\) degrees, and so on. If we define \\(r\\) to be -the number of turns after a single cycle, we have \\(t=r\\theta\\), and our turning -equation can be written as to: +the number of turns that we find ourselves rotated by after a single cycle, +we have \\(t=r\\theta\\), and our turning equation can be written as: {{< latex >}} kr\theta \equiv 0\ (\text{mod}\ c\theta) @@ -567,7 +578,7 @@ And finally, observing that \\(kr=kr-0\\), we have: This equivalence says the same thing as our earlier one; however, instead of being in terms of degrees, it's in terms of the number of turns \\(c\\) and the turns-per-cycle \\(r\\). -Now, recall once again that for this equivalence to hold in \\(k>0\\) steps, we need +Now, recall once again that the smallest number of steps \\(k>0\\) for which this equivalence holds is \\(k = c/\\text{gcd}(c,r)\\). We're close now: we have a sequence of \\(k\\) steps that will lead us back to the beginning. @@ -623,6 +634,18 @@ is one more than the _divisor_ (we started with numbers in the usual base 10, bu {{< figure src="pattern_2_11_t7.svg" caption="Pattern generated by the number 2 in base 12 while turning 7 times." class="tiny" alt="Pattern generated by the number 2 by summing digits in base 12 and turning 51 degrees." >}} +### Conclusion +Today we peeked under the hood of a neat mathematical trick that was shown to me by my headmaster +over 10 years ago now. Studying what it was that made this trick work led us to play with +the underlying mathematics some more, and extend the trick to more situations (and prettier +patterns). I hope you found this as interesting as I did! + +By the way, the kind of math that we did in this article is most closely categorized as +_number theory_. Check it out if you're interested! + +All that remains are some proofs I omitted from the original article since they were taking +up a lot of space (and were interrupting the flow of the explanation). They are listed below. + ### Referenced Proofs #### Adding Two Congruences @@ -676,7 +699,7 @@ if \\(a=fc\\) and \\(b=fd\\), then \\(c\\) and \\(d\\) have no common factors ot \\(\\text{gcd}(c,d)=1\\)). __Proof__: Suppose that \\(c\\) and \\(d\\) do have sommon factor, \\(e\\neq1\\). In that case, we have -\\(c=ei\\) and \\(d=ej\\) for some \\(i\\) and \\(j\\). then, we have \\(a=fei\\), and \\(b=fej\\). +\\(c=ei\\) and \\(d=ej\\) for some \\(i\\) and \\(j\\). Then, we have \\(a=fei\\), and \\(b=fej\\). From this, it's clear that both \\(a\\) and \\(b\\) are divisible by \\(fe\\). Since \\(e\\) is greater than \\(1\\), \\(fe\\) is greater than \\(f\\). But our assumptions state that \\(f\\) is the greatest common divisor of \\(a\\) and \\(b\\)! We have arrived at a contradiction.