A recent “Ask the Expert” question was, “How do we actually KNOW that no two snowflakes are alike”?
Is every snowflake unique? Unfortunately, the best answers disagree:
- Yes, by extrapolation from a limited sample.
- We don’t know but it is simply probable.
- Depending on exactly what we mean by ‘unique,’ some have been found that are alike.
After reviewing the standard answers, let’s build a math framework to estimate an answer for this “unique snowflake” question.
Review the Best Answers about Unique Snowflakes
By reviewing the best typical answers, we will also learn some background about snowflakes.
Extrapolating Uniqueness from a Sample of Snowflakes
Chris V. Thangham reported that Wilson Alwyn Bentley was the first person to state he had not found any two snowflakes alike. Since then, almost everyone who looked for a counter-example failed.
Probably No Two Snowflakes are Alike
As the pictures of lacey dendrite snowflakes show, their patterns can be incredibly complex. Whether or not any physical snowflake is a true fractal, the “Koch snowflake” in the GIF image shows how repeated branching leads to a very complicated form.
Since snowflakes can branch differently down to individual water molecules, the number of possibilities is extremely large. Without a restrictive theory that constrains snowflakes to a limited number of shapes, it seems probable that no two snowflakes are alike.
Some Snowflakes Are Alike, Depending on our Definitions
Some snowflakes look more like six-sided plates than lace doilies; other snowflakes are hexagonal prisms. The smallest and simplest plates or prisms may easily have identical twins, since their criteria are nearly limited to the diameter and thickness measured to a reasonable precision.
Thangham also reported that Nancy Knight did indeed find “identical snowflakes of the hollow column type.” Thangham thus supplies the counter-example that disproves his thesis that “no two snowflakes are alike” if we include the simplest shapes.
Approaching Identical Snowflakes with Birthday Mathematics
You may remember the “birthday party” puzzle. “If you want a 50% chance that two people attending a party share the same birthday, how many guests do you randomly invite?”
Each person has a 1/365 chance of sharing a birthday with a random stranger. (Readers are encouraged to do the math including leap years).
Eric W. Weisstein provides an estimate that at least two members in a set of size ‘n’ share the same value if there are ‘d’ possible values.
P = 1 – ( 1 – ( n/(2*d) ) )^(n-1), approximately, where…
- ‘P’ is the probability;
- ‘n’ is the number of members in the set, or “guests at the party”;
- ‘d’ is the number of possible values, with d=365 for the birthday problem;
- ‘^’ is the symbol for exponentiation.
It turns out that “n=23″ gives a probability of about 50% that any two guests at a party share a birthday.
If we knew how many snowflakes are being compared (‘n’), and how many “values” a snowflake might have (‘d’), this formula would estimate the probability that at least two snowflakes were alike.Mike DeHaan, All rights Reserved. Written For: