Forum - Other - How many possible friendship bracelets are there?

rhall7464

Professional

3 years, 3 months ago by rhall7464

I'm not too sure how many people are interested in this, but I've been looking at some really big numbers recently and it got me thinking about how many possible friendship bracelets there are given certain parameters. This inspired me to work out a function. I am very prone to making stupid mistakes so if anyone spots one please let me know. I'm also aware that the r/2 section means it sometimes only works for an even number of rows but I don't have the patience to fix that so if anyone wants to then feel free 😂 😂 . I might do that later. And I'm not going to include alphas because they work completely differently, or shaped bracelets because that's just a huge hassle sorry.

Here is my thought process:
Define a function Fb(r, s, c), where r is the number of rows, s is the number of strings and c is the number of colours. This function will spit out the number of possible normal bracelets with these conditions met. (I also decided not every colour has to be used, so for example if there were 3 colours, bracelets with 2 or 1 would also be accepted.)

First off, how many knots are there?
We have s strings. The number of knots changes every other row: half the rows have s/2 knots while the other half have s/2 - 1 knots. Look at any normal pattern on the site and you will know what I mean.
So the half with s/2 knots are represented by: r/2 x s/2.
The half with s/2 - 1 knots are represented by: r/2 x (s/2 - 1).
Then we can add these together and simplify:
r/2 x s/2 + r/2 x (s/2 - 1)
= r/2 x s/2 + r/2 x s/2 - r/2
= r/2 x (s/2 + s/2 - 1)
= r/2 x (s - 1)
Therefore the total number of knots in a bracelet with r rows and s strings is r/2 x (s - 1) (for even r as previously mentioned)

Next, how many knot patterns are possible?
Well, each knot can be one of four: f, b, fb and bf. Disregarding no knot because I'm disregarding shaped bracelets.
As far as I'm aware these don't depend on each other, there are always four possible knot types for every knot.
So the number of knot patterns is:
4 x 4 x 4 x ............ x 4 x 4 x4 where the number of 4s is the number of knots in the bracelet.
In other words, 4 ^ (r/2 x (s - 1))

The final step is to work out how many starting configurations of strings there are, keeping in mind that not every colour has to be used.
We have c colours and s strings. Each string then has c possible options for what colour it can be. This is really similar to the previous step, they don't depend on each other and every string always has c possible colours.
Then we get that the number of string configurations is:
c x c x c x ............... x c x c x c where the number of cs is the number of strings.
In other words, c ^ s.

To get the formula for the function, we multiply the number of possible knot patterns by the number of possible starting string configurations.
This gives Fb(r, s, c) = c ^ s x 4 ^ (r/2 x (s - 1)).

As you can probably guess this function grows relatively quickly. It would probably be possible to graph it with 3d input and 1d output but I'm not sure how to go about this, maybe using colours for the output?
For example, I entered Fb(10, 4, 2) into a calculator as a random test and got an output of about 1.72 x 10^10, which is a decent sized number. To give you an idea of how big this is, the number of bracelets with only 10 rows, 4 strings and 2 colours (which most people on here would see as an incredibly simple bracelet to make) is over 17 billion. That's more than the age of the universe in years. If you made 1 of these bracelets every hour, it would take you over a million years to finish all of them! And even working non stop at one every minute, it would still take almost 20,000 years.

TL;DR because I won't blame people for skipping over the boring maths bit:
The number of possible normal bracelets with r rows, s strings and up to c colours is c ^ s x 4 ^ (r/2 x (s - 1)).
This means there are a REALLY huge number of possible bracelets even for small values of r, s and c.

I hope at least some people found this interesting ^_^ Tell me your thoughts below. Have a nice day!
- Ryan ❤️

KrazyKnotz

Super Moderator

3 years, 3 months ago by KrazyKnotz

WOW! That is AMAZING! 😲

rhall7464

Professional

3 years, 3 months ago by rhall7464

AHHHH thank you so much!!!!! it means a lot 🙂

amazingme

Bracelet King

3 years, 3 months ago by amazingme

...
My dear person, you are a genius. I'm over here struggling with precalc, and you just made an entire formula 😂 you're built different and that formula is amazing!!!

rhall7464

Professional

3 years, 3 months ago by rhall7464

thanks hahaha!! to be honest i just really love maths >_>

rhall7464 Professional	3 years, 3 months ago by rhall7464 I'm not too sure how many people are interested in this, but I've been looking at some really big numbers recently and it got me thinking about how many possible friendship bracelets there are given certain parameters. This inspired me to work out a function. I am very prone to making stupid mistakes so if anyone spots one please let me know. I'm also aware that the r/2 section means it sometimes only works for an even number of rows but I don't have the patience to fix that so if anyone wants to then feel free 😂 😂 . I might do that later. And I'm not going to include alphas because they work completely differently, or shaped bracelets because that's just a huge hassle sorry. Here is my thought process: Define a function Fb(r, s, c), where r is the number of rows, s is the number of strings and c is the number of colours. This function will spit out the number of possible normal bracelets with these conditions met. (I also decided not every colour has to be used, so for example if there were 3 colours, bracelets with 2 or 1 would also be accepted.) First off, how many knots are there? We have s strings. The number of knots changes every other row: half the rows have s/2 knots while the other half have s/2 - 1 knots. Look at any normal pattern on the site and you will know what I mean. So the half with s/2 knots are represented by: r/2 x s/2. The half with s/2 - 1 knots are represented by: r/2 x (s/2 - 1). Then we can add these together and simplify: r/2 x s/2 + r/2 x (s/2 - 1) = r/2 x s/2 + r/2 x s/2 - r/2 = r/2 x (s/2 + s/2 - 1) = r/2 x (s - 1) Therefore the total number of knots in a bracelet with r rows and s strings is r/2 x (s - 1) (for even r as previously mentioned) Next, how many knot patterns are possible? Well, each knot can be one of four: f, b, fb and bf. Disregarding no knot because I'm disregarding shaped bracelets. As far as I'm aware these don't depend on each other, there are always four possible knot types for every knot. So the number of knot patterns is: 4 x 4 x 4 x ............ x 4 x 4 x4 where the number of 4s is the number of knots in the bracelet. In other words, 4 ^ (r/2 x (s - 1)) The final step is to work out how many starting configurations of strings there are, keeping in mind that not every colour has to be used. We have c colours and s strings. Each string then has c possible options for what colour it can be. This is really similar to the previous step, they don't depend on each other and every string always has c possible colours. Then we get that the number of string configurations is: c x c x c x ............... x c x c x c where the number of cs is the number of strings. In other words, c ^ s. To get the formula for the function, we multiply the number of possible knot patterns by the number of possible starting string configurations. This gives Fb(r, s, c) = c ^ s x 4 ^ (r/2 x (s - 1)). As you can probably guess this function grows relatively quickly. It would probably be possible to graph it with 3d input and 1d output but I'm not sure how to go about this, maybe using colours for the output? For example, I entered Fb(10, 4, 2) into a calculator as a random test and got an output of about 1.72 x 10^10, which is a decent sized number. To give you an idea of how big this is, the number of bracelets with only 10 rows, 4 strings and 2 colours (which most people on here would see as an incredibly simple bracelet to make) is over 17 billion. That's more than the age of the universe in years. If you made 1 of these bracelets every hour, it would take you over a million years to finish all of them! And even working non stop at one every minute, it would still take almost 20,000 years. TL;DR because I won't blame people for skipping over the boring maths bit: The number of possible normal bracelets with r rows, s strings and up to c colours is c ^ s x 4 ^ (r/2 x (s - 1)). This means there are a REALLY huge number of possible bracelets even for small values of r, s and c. I hope at least some people found this interesting ^_^ Tell me your thoughts below. Have a nice day! - Ryan ❤️
KrazyKnotz Super Moderator	3 years, 3 months ago by KrazyKnotz WOW! That is AMAZING! 😲
rhall7464 Professional	3 years, 3 months ago by rhall7464 AHHHH thank you so much!!!!! it means a lot 🙂
amazingme Bracelet King	3 years, 3 months ago by amazingme ... My dear person, you are a genius. I'm over here struggling with precalc, and you just made an entire formula 😂 you're built different and that formula is amazing!!!
rhall7464 Professional	3 years, 3 months ago by rhall7464 thanks hahaha!! to be honest i just really love maths >_>

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How many possible friendship bracelets are there?