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Give initial stabs at Arthur's suggestions

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Danila Fedorin 2022-01-01 14:45:11 -08:00
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content/blog/modulo_patterns/index.md
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@ -12,7 +12,7 @@ _"Pick a number"_, he said, _"And I'll teach you how to draw a pattern from it."
The procedure was rather simple: The procedure was rather simple:
1. Pick a number between 2 and 8 (inclusive). 1. Pick a number between 2 and 8 (inclusive).
2. Start generating multiples of this number. If you picked 8, 2. Start generating positive multiples of this number. If you picked 8,
your multiples would be 8, 16, 24, and so on. your multiples would be 8, 16, 24, and so on.
3. If a multiple is more than one digit long, sum its digits. For instance, for 16, write 1+6=7. 3. If a multiple is more than one digit long, sum its digits. For instance, for 16, write 1+6=7.
If the digits add up to a number that's still more than 1 digit long, add up the digits of _that_ If the digits add up to a number that's still more than 1 digit long, add up the digits of _that_
@ -37,7 +37,7 @@ characters (`"17"` becomes `["1", "7"]`), turning each of these character back i
We may now encode the "drawing" logic. At any point, there's a "direction" we're going - which We may now encode the "drawing" logic. At any point, there's a "direction" we're going - which
I'll denote by the Ruby symbols `:top`, `:bottom`, `:left`, and `:right`. Each step, we take I'll denote by the Ruby symbols `:top`, `:bottom`, `:left`, and `:right`. Each step, we take
the current `x`/`y` coordinates (our position on the grid), and shift them by `n` in a particular the current `x`,`y` coordinates (our position on the grid), and shift them by `n` in a particular
direction `dir`. We also return the new direction alongside the new coordinates. direction `dir`. We also return the new direction alongside the new coordinates.
{{< codelines "Ruby" "patterns/patterns.rb" 10 21 >}} {{< codelines "Ruby" "patterns/patterns.rb" 10 21 >}}
@ -109,7 +109,7 @@ larger than \(k\).
\end{aligned} \end{aligned}
{{< /latex >}} {{< /latex >}}
We only _really_ care about the remainder here, not the quotient, since it's the remainder 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 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 \\(k\\),
so it must be divisible). Thus, we can write: so it must be divisible). Thus, we can write:
@ -138,6 +138,9 @@ Multiplying both sides by the same number (call it \\(n\\)) also works:
\textbf{if}\ a \equiv b\ (\text{mod}\ k),\ \textbf{then}\ na \equiv nb\ (\text{mod}\ k). \textbf{if}\ a \equiv b\ (\text{mod}\ k),\ \textbf{then}\ na \equiv nb\ (\text{mod}\ k).
{{< /latex >}} {{< /latex >}}
To see why this works, try rewriting the equivalences back into statements of divisibility,
then into actual equations (just like our very first \\(b-r=ka\\)).
Ok, that's a lot of notation and other _stuff_. Let's talk specifics. Of particular interest Ok, that's a lot of notation and other _stuff_. Let's talk specifics. Of particular interest
is the number 10, since our number system is _base ten_ (the value of a digit is multiplied by 10 is the number 10, since our number system is _base ten_ (the value of a digit is multiplied by 10
for every place it moves to the left). The remainder of 10 when dividing by 3 is 1. Thus, for every place it moves to the left). The remainder of 10 when dividing by 3 is 1. Thus,
@ -248,8 +251,16 @@ and since turning four times leaves us facing "up" again, we'll end up getting _
From this, we can devise a simple condition to prevent spiraling -- the _length_ of the sequence before From this, we can devise a simple condition to prevent spiraling -- the _length_ of the sequence before
it repeats _cannot be a multiple of 4_. This way, whenever the cycle restarts, it will do so in a it repeats _cannot be a multiple of 4_. This way, whenever the cycle restarts, it will do so in a
different direction: backwards, turned once to the left, or turned once to the right. Clearly repeating different direction: backwards, turned once to the left, or turned once to the right. Clearly repeating
the sequence backwards is guaranteed to take us back to the start. The same is true for the left and right-turn sequences, the sequence backwards is guaranteed to take us back to the start. The same is true for the left and right-turn sequences, though it's less obvious. If drawing our sequence once left us turned to the right,
since after two iterations they will _also_ leave us facing backwards. drawing our sequence twice will leave us turned more to the right. On a grid, two right turns are
the same as turning around. The third repetition will then undo the effects of the first one
(since we're facing backwards now), and the fourth will undo the effects of the second.
There is an exception to this
multiple-of-4 rule: a sequence makes it back to the origin right before it starts over.
In that case, even if it's facing the very same direction it started with, all is well -- things
are just like when it first started, and the cycle repeats. I haven't found a sequence that does this,
so for our purposes, we'll stick with multiples of 4.
Okay, so we want to avoid cycles with lengths divisible by four. What does it mean for a cycle to be of length _k_? It effectively means the following: Okay, so we want to avoid cycles with lengths divisible by four. What does it mean for a cycle to be of length _k_? It effectively means the following:
@ -303,10 +314,10 @@ is not ruled out as a valid cycle by our previous condition, we don't find any c
So what is it that _really_ determines if there can be cycles or not? 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? 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 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 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. not 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 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 of 3, 7 has a cycle length of 12, 9 has a cycle length of 4. What's
@ -321,8 +332,8 @@ k = \frac{d}{\text{gcd}(d,n)}
{{< /latex >}} {{< /latex >}}
Where \\(d\\) is our divisor, which has been 9 until just recently. Indeed, this equation is in agreement 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 with our experiment for \\(d = 9\\), too. Why might this be? Recall that sequences with
our equation for paths of length \\(k\\): period \\(k\\) imply the following congruence:
{{< latex >}} {{< latex >}}
kn \equiv 0\ (\text{mod}\ d) kn \equiv 0\ (\text{mod}\ d)
@ -337,7 +348,7 @@ since this number divides \\(n\\) and \\(d\\), we can write \\(n=fm\\) for some
kfm \equiv 0\ (\text{mod}\ fg) kfm \equiv 0\ (\text{mod}\ fg)
{{< /latex >}} {{< /latex >}}
We can simplify this a little bit. Recall that what this equation _really_ means is that the 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\\): difference of \\(kfm\\) and \\(0\\), which is just \\(kfm\\) is divisible by \\(fg\\):
{{< latex >}} {{< latex >}}
@ -351,14 +362,21 @@ we can write:
g|km g|km
{{< /latex >}} {{< /latex >}}
We also know that \\(g\\) and \\(m\\) have no common factors -- they were all divided out from \\(d\\) Can we distill this statement even further? It turns out that we can. Remember that we got \\(g\\)
and \\(n\\) when we divided by \\(f\\)! We can thus further simplify our claim: and \\(m\\) by dividing \\(d\\) and \\(n\\) by their greatest common factor, \\(f\\). This, in
turn, means that \\(g\\) and \\(m\\) have no more common factors (that aren't equal to 1). If they
did have such a common factor, say \\(h\\), we would be able to write \\(m=hi\\) and \\(g=hj\\).
Then, we would _also_ be able to write \\(n=fhi\\) and \\(d=fhj\\). But now, \\(fh\\) is a common
factor of both \\(n\\) and \\(d\\), and it's bigger than \\(f\\) (since \\(h\\) is not 1)! This
violates our assumption that \\(f\\) is the greatest common factor. So there can't be another shared
fadctor between \\(g\\) and \\(m\\). From this, in turn, we can deduce that \\(m\\) is not
relevant to \\(g\\) dividing \\(km\\), and we get:
{{< latex >}} {{< latex >}}
g|k g|k
{{< /latex >}} {{< /latex >}}
This says that \\(k\\) must be divisible by \\(g\\). Recall that we got \\(g\\) by dividing That is, we get 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 \\(d\\) by \\(f\\), which is our largest common factor -- aka \\(\\text{gcd}(d,n)\\). We can thus
write: write:
@ -366,12 +384,23 @@ write:
\frac{d}{\text{gcd}(d,n)}|k \frac{d}{\text{gcd}(d,n)}|k
{{< /latex >}} {{< /latex >}}
All that's left is to pick the smallest \\(k\\) that fits this description (that would be the first Let's stop and appreciate this result. We have found a condition that is required for a sequnce
point at which our sequence of multiples of \\(n\\) will loop). Zero is divisible by anything, but of remainders from dividing by \\(d\\) (which was 9 in the original problem) to repeat after \\(k\\)
alas, our sequence numbers start at 1 -- zero's out. The next best thing is \\(d/\\text{gcd}(d,n)\\). numbers. Furthermore, all of our steps can be performed in reverse, which means that if a \\(k\\)
And there we have it: the first point at which our sequence loops, just like we guessed. matches this conditon, we can work backwards and determine that a sequence of numbers has
to repeat after \\(k\\) steps.
Lastly, recall that a cycle occurs whenever a \\(k\\) is a multiple of 4. Now that we know what Multiple \\(k\\)s will this condition, and that's not surprising. If a sequence repeats after 5 steps,
it also repeats after 10, 15, and so on. We're interested in the first time our sequences repeat after
taking any steps, which means we have to pick the smallest possible non-zero value of \\(k\\). The smallest
number divisible by \\(d/\\text{gcd}(d,n)\\) is \\(d/\\text{gcd}(d,n)\\) itself. We thus confirm
our hypothesis:
{{< latex >}}
k = \frac{d}{\text{gcd}(d,n)}
{{< /latex >}}
Lastly, recall that our patterns would spiral away from the center 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 \\(k\\) is, we can restate this as "\\(d/\\text{gcd}(d,n)\\) is divisible by 4". But if we pick
\\(n=d-1\\), the \\(n=d-1\\), the
{{< sidenote "right" "coprime-note" "greatest common factor has to be \(1\)," >}} {{< sidenote "right" "coprime-note" "greatest common factor has to be \(1\)," >}}