/DOS_proj1

Primary LanguageElixir

DOS Project 1 - Finding perfect squares that are sums of consecutive squares on distributed Elixir nodes

Group info

Siddhesh Gupte(UFID 2975-1118) and Sagar Prabhu(UFID 8915-4951)

Instructions

  • To run the program on a single machine use the command- mix run proj1.exs arg1 arg2, where arg1 is n and arg2 is k. For example, mix run proj1.exs 40 24

  • To run the program on multiple machines:

    • First start the iex sessions on individual machines using ip address, name, and cookie. For example: iex --name machine1@192.168.0.7 --cookie somecookie -S mix

    • Then establish a connection using the function Node.connect/1 : Node.connect :"machine2@192.168.0.5"

    • Finally to run the program in an iex session use: Proj1.driver(n,k)

No. of machines

The program has been tested on 3 machines.

iex(machine1@192.168.0.5)7> Proj1.driver(10000000, 50) 7, 245.0, machine3@192.168.0.6 28, 385.0, machine3@192.168.0.6 44, 495.0, machine3@192.168.0.6 67, 655.0, machine3@192.168.0.6 87, 795.0, machine3@192.168.0.6 124, 1055.0, machine3@192.168.0.6 168, 1365.0, machine3@192.168.0.6 287, 2205.0, machine3@192.168.0.6 379, 2855.0, machine3@192.168.0.6 1001139, 7079295.0, machine3@192.168.0.6 512, 3795.0, machine3@192.168.0.6 628, 4615.0, machine3@192.168.0.6 2714784, 19196595.0, machine3@192.168.0.6 843, 6135.0, machine3@192.168.0.6 1099, 7945.0, machine3@192.168.0.6 1792, 12845.0, machine3@192.168.0.6 2328, 16635.0, machine3@192.168.0.6 3103, 22115.0, machine3@192.168.0.6 3779, 26895.0, machine3@192.168.0.6 5032, 35755.0, machine3@192.168.0.6 6524, 46305.0, machine3@192.168.0.6 10563, 74865.0, machine3@192.168.0.6 13687, 96955.0, machine3@192.168.0.6 2096248, 14822885.0, machine3@192.168.0.6 18204, 128895.0, machine3@192.168.0.6 5835184, 41261155.0, machine2@192.168.0.7 22144, 156755.0, machine3@192.168.0.6 29447, 208395.0, machine3@192.168.0.6 38143, 269885.0, machine3@192.168.0.6 61684, 436345.0, machine3@192.168.0.6 1296547, 9168145.0, machine3@192.168.0.6 79892, 565095.0, machine3@192.168.0.6 106219, 751255.0, machine3@192.168.0.6 129183, 913635.0, machine3@192.168.0.6 4389227, 31036695.0, machine2@192.168.0.7 iex(machine1@192.168.0.5)8> 3609127, 25520555.0, machine2@192.168.0.7 iex(machine1@192.168.0.5)8> 171748, 1214615.0, machine3@192.168.0.6 iex(machine1@192.168.0.5)8> 222432, 1573005.0, machine3@192.168.0.6 iex(machine1@192.168.0.5)8> 359639, 2543205.0, machine3@192.168.0.6 iex(machine1@192.168.0.5)8> 465763, 3293615.0, machine3@192.168.0.6 iex(machine1@192.168.0.5)8> 7556948, 53435865.0, machine1@192.168.0.5 iex(machine1@192.168.0.5)8> 619208, 4378635.0, machine3@192.168.0.6 iex(machine1@192.168.0.5)8> 753052, 5325055.0, machine3@192.168.0.6

where the output is in the format start_number, square_number, Node.self

Performance

  • To determine the size of the work unit, we first computed a list of integers which perfectly divides the input n. Then we checked the ratio-CPU time/Real time at various indices of the list, and for different inputs. We found that the ratio is generally highest at 3/4th position in the list. For n= 1000000, the 3/4th position in the list and hence the size of work unit is 15625.

  • There are no sequence of numbers for input n=1000000 and k=4 which satisfies the requirement. Hence on running the command mix run proj1.exs 1000000 4, there is no output.

  • time mix run proj1.exs 1000000 4 Compiling 1 file (.ex) [ ok: #PID<0.130.0>, ok: #PID<0.131.0>, ok: #PID<0.132.0>, ok: #PID<0.133.0>, ok: #PID<0.134.0>, ok: #PID<0.135.0>, ok: #PID<0.136.0>, ok: #PID<0.137.0>, ok: #PID<0.138.0>, ok: #PID<0.139.0>, ... ] real 0m1.067s user 0m2.527s sys 0m0.128s

    Total CPU time / Real time = 2.527s/1.067s = 2.368

  • The largest problem we managed to solve is for n=10000000 and k=50 .

Specific requirements