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Code Retreat @ Bento Miso

Welcome to Code Retreat Toronto! 

The event is for software developers from different paradigms to learn new techniques, languages, and practices from each other by pairing to solve a pre-determined problem. Participants must choose a new partner for each session and new programatic constraints will be introduced with each session.

The format is as follows:

6:00pm - 6:15pm: Introductions

6:15pm - 7:00pm: Session 1

7:00pm - 7:15pm: Session 1 Review

7:15pm - 8:00pm: Session 2

8:00pm - 8:15pm: Session 2 Review

8:15pm - 9:00pm: Session 3

9:00pm - 9:15pm: Session 3 Review


Make sure to bring a laptop and have your development environment ready for the type of coding you'd like to be doing.



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  • dann

    That was definitely atypical in terms of problem difficulty. Next month we'll be back to our regularly scheduled programming -- probably with a computing competition problem, or we'll trot out that old chestnut The Game of Life. Hope you all had fun. See you at the next one!

    April 19, 2013

  • A former member
    A former member

    That was a really tough problem, maybe suggest an easier question next time so we can spend more time learning a new language?

    April 19, 2013

  • Peter B.

    A lot of fun. Too tired to think.

    April 18, 2013

  • Andrey P.

    That was a really hard problem :)

    April 18, 2013

  • Mohammad Z.

    I won't be able to attend today's meetup. Something popped up in the last minute. Sorry.

    April 18, 2013

  • Mehran B.

    Just joined. Hopefully next one...

    April 18, 2013

  • dann

    We've done a lot of parsing problems recently, so this month we're going to change things up with a counting problem. Counting is fun!

    We'll start off by counting the number of ways a square of size N can be dissected into smaller squares. So, for example, a 2x2 square can only be decomposed into either 1 2x2 square or 4 1x1 squares, so square-count(2) == 2

    A 3x3 square is more interesting: it can be (1 3x3 square) or (9 1x1 squares) or (1 2x2 and 5 1x1s) -- but you can do that last dissection in four different ways, so square-count(3) == 6

    This sequence is (note the image links).

    If that proves too easy we'll then try counting the same thing, but ignoring dissections that are reflections or rotations of previously counted dissections. That sequence is and if anyone gets this far maybe we can extend it! More pics:

    April 15, 2013

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  • Bento Miso

    Bento Miso provides meetup space in their amazing facility in Queen West

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