# Annual DIMACS REU graph puzzle

Every year we create a network with all the REU students and their mentors. Discovering the rule used to create the specific network configuration is given to the REU students as a puzzle to solve on picture day. The winner receives a small prize and a mention on this webpage!

Solution:
?

## 2023 - 6th annual DIMACS REU graph puzzle

Solution:
Call z the distance in the alphabet between the first and last letter in a student's first name.
Connect two students with a thick line iff their values of z are the same.
Connect two students with a thin line iff their values of z differ by 1 and their first letter is in the same half of the alphabet.
Mentors are simply connected to the students they supervised.

## 2022 - 5th annual DIMACS REU graph puzzle

Solution:
For any letter #, let f(#) be the position of # in the alphabet (ie f('a') = 1, f('b') = 2, etc). If a word w has an odd number of letters with middle letter #, let g(w) = f(#). If w has an even number of letters, with middle letters #_1 and #_2, let g(w) = [f(#_1) + f(#_2)]/2.
Embed a name into 2-dimensional Euclidean space as follows:
x-coordinate = g(first name)
y-coordinate = g(last name)
Any two points within distance 4 are then connected in the graph.
Mentors are simply connected to the students they supervised.

## 2019 - 4th annual DIMACS REU graph puzzle

Solution:
Map each node to a point (x,y) in a Euclidean plane where
x=Distance in the alphabet of the first two letters in student's first name
y=Distance in the alphabet of the last two letters in student's first name
Any two points within distance 4 are then connected in the graph.
Mentors are simply connected to the students they supervised.

## 2018 - 3rd annual DIMACS REU graph puzzle

Solution:
Each student is assigned two variables, r and θ, which correspond to the polar coordinates on a plane:
r = Number of letters in student's last name
θ = (Position in the alphabet of the first letter of the first name)*360/26
[i.e. A=0 degrees, B=1*360/26=13.84 degrees, ... , Z=25*360/26=346.15 degrees.]
Connect two students if their Euclidean distance is smaller than 4.
Mentors are simply connected to the students they supervised.

## 2017 - 2nd annual DIMACS REU graph puzzle

Solution:
Call Lname the distance in the alphabet between the two first letters in a student's first name
(for example, Aaron has Lname=0 and Edgar has Lname=1).
Connect two students with a thick line if they have the same Lname value
and connect them with a thin line if the difference of their Lname values is 1.
Mentors are simply connected to the students they supervised.

## 2016 - 1st annual DIMACS REU graph puzzle

Solution:
Map each node to a point (x,y) in a Euclidean plane where
x=number of letters in student's first name
y=number of letters in student's last name
Any two points within distance 1 are then connected in the graph.
Mentors are simply connected to the students they supervised.