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.
What is the Value of a Snowflake?
Let’s sketch out an approach that snow-ologists might refine, by digitizing a value for a snowflake.
Thangham notes that snowflakes have “six-fold symmetry.” Let’s concentrate on one-sixth of a snowflake, isolating the area between two major arms. That should be an equilateral triangle, since the angle at the centre is 360/6=60 degrees and the sides from that vertex have the same length.
The argument is that any one triangular plate, or lace pattern, is repeated for the whole snowflake. Therefore we only need to examine one triangle per snowflake.
Charles Q. Choi states that the largest snowflake’s diameter is about 50.8mm, or two inches. So the length of one side of this equilateral triangle is 25.4mm. The area of any triangle is one-half the base times the height, where the height is the distance of a line from the opposite vertex that meets the base at a right angle.
However, the area of any equilateral triangle is (3^(1/2)) * (s^2) /4, after using the Pythagorean theorem to determine the height from the side ‘s’.
By rounding down the length of the side to 25mm and approximating the square root of three as 1.73, the area of this triangle is approximately 270.3 square millimetres.
Resolving a Snowflake’s Value
Orange Coast College introduces microscopes by saying that the “resolution limit of the human eye is about 100 micrometers” or 0.1mm. As well, the “compound light microscope has a resolution limit of about 0.2 micrometers under ideal conditions.” So a microscope can provide up to 500 times greater resolution.
Let’s use the naked eye to resolve each snowflake. If our limit is 0.1mm, then there are about 250 pixels along the side of the snowflake triangle, and 270.3*10*10 = 27030 pixels inside the area of that triangle.
We can digitize the snowflake’s triangle by saying that each pixel is one bit of information: either it has a grain of ice, or not. Let ‘one’ represent the grain of ice, and ‘zero’ its absence.
We can create a unique number for the snowflake by weaving from the central point, whose value is almost certainly ‘1’. The next bit is on the left arm beside the centre; then use the next bits heading toward the other arm. On either arm, step one bit away from the centre and then record the values of the bits heading to the other arm.
The total information is 2^27030 = 10^821.7846 approximately. (See How to Convert the Base of an Exponent with Logarithms for this useful conversion technique). Yes, we need an 822-digit base-10 number to describe a perfectly flat snowflake’s lacey pattern. Let’s round up to 10^822 different possible snowflakes; that’s the ‘d’ in Weisstein’s birthday problem.
How Many Snowflakes Fall per Year?
Choi quoted expert Jon Nelson’s estimates for the mass of one snowflake (“1 millionth of a gram” = 10^-9 Kg) and the total snowfall per year (“1 million billion” = 10^15 Kg). So 10^24 snowflakes fall per year. Let’s invite them all to Weisstein’s party.
The Probability of Identical Digitized Snowflakes
Now we have ‘n’ = 10^24 snowflakes per year, to be compared on ‘d’ = 10^822 values. The equation for the probability that at least two do match is “P = 1 – ( 1 – ( n/(2*d) ) )^(n-1)”.
By substitution, “P = 1 – ( 1 – ( (10^24)/(2*(10^822)) ) )^((10^24)-1)”.
Let’s examine some of the terms.
The “n/(2*d)” term, by itself, is “(10^24)/(2*(10^822))” = “1/(2*(10^(792)” = 0.00…005 with 791 “zeroes” between the decimal point and the ‘5’. Subtracting that from ‘1’ leaves 0.99…95 with 791 “nines”.
To a reasonable approximation, taking “0.99….99^((10^24 )- 1)” removes the 24 least-important “nines”. We now have “0.99…99” with only about 767 “nines”.
Finally, we have the leading “1 – 0.99…99” = 0.0…01 with over seven hundred “zeroes” after the decimal point.
Assigning these values to this mathematical model makes it extremely improbable that any two snowflakes are identical.
Objections to this Snowflake Theory
Readers are welcome to raise objections, and then submit all the rest of their arithmetic in their comments. We suggest starting with the following concerns.
- This number is too high. No snowflake could have a grain of ice surrounded by air; connections are needed; and the apparent self-symmetry reduces the possibilities.
- This number is too low. We totally ignore how thick the snowflake is, which may vary from point to point on the triangle.
- We should examine the snowflake under a microscope, so the resolution is 50 times higher, with 2500 times more bits. This makes it harder to find identical snowflakes.
- We should consider all the snowflakes that ever fell on Earth; so let’s multiply the annual snowfall by 4,000,000,000 years.
- We need to sample different types of snowflakes, and use different mathematical models for each. Duplicates might be more likely given an abundance of “plate” snowflakes.
This article suggests a mathematical framework for approaching this question, rather than claiming to have the definitive solution.
To conclude: at least two visually identical snowflakes have been found, but they had a very simple shape. It is highly improbable that there are two identical snowflakes among the complex dendrites.
Introduction to the Microscope. Orange Coast College. (2012). PDF accessed December 24, 2012.
Choi, Charles Q. Two snowflakes may actually be alike. (2007). Accessed December 24, 2012.
Thangham, Chris V. No two snowflakes are alike. Digital Journal. (2008). Accessed December 24, 2012.
Weisstein, Eric W. Birthday Problem. MathWorld–A Wolfram Web Resource. Accessed December 24, 2012.