# Annual DIMACS REU graph puzzle

Every year we create a network with all the REU students and their mentors. 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!

## Winners!

### 2018: Scott Harman

2017: Jakub Pekarek

2016: Thomas Murrills

## 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 *L*_{name} the distance in the alphabet between the two first letters in a student's first name

(for example, Aaron has *L*_{name}=0 and Edgar has *L*_{name}=1).

Connect two students with a thick line if they have the same *L*_{name} value

and connect them with a thin line if the difference of their *L*_{name} 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.