Jake Scruggs

Recently my team needed to find out if a point was inside a given polygon and, as usual, we found some code on the internet that did what we wanted. And, as usual, there wasn't much explanation as to what it was doing. Here's the code (after we converted it to Ruby): 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 def contains_point? (point) c = false i = -1 j = self .size - 1 while (i += 1 ) < self .size if (( self [i].y <= point.y && point.y < self [j].y) || ( self [j].y <= point.y && point.y < self [i].y)) if (point.x < ( self [j].x - self [i].x) * (point.y - self [i].y) / ( self [j].y - self [i].y) + self [i].x) c = !c end end j = i end return c end end So, you know, it works and I'm grateful to somebody for posting it but we had no idea how it works. So we wrapped 2 tests around it (one where the point is not in the polygon and another where it is...

In a previous post I discussed generating a unique id number for an order in our database. Read the original post for the full story, but to sum up, I decided that playing around with rand(100_000_000) was good enough. Martin Ankerl posted a comment pointing out that I was vulnerable to the Birthday Problem . So after a quick trip to wikipedia, I found out that if you have 23 people in a room you might think you have a 23/365 chance of having 2 people with the same birthdays, but no, your chances are closer to 50%! This is because in a group of 23 people there are 253 pairs each of which have a 1/365ish chance to have the same birthday. The formula to calculate how many are needed to give you a 50% chance of having a birthday collision is: .5 + sqrt(.25 + 2 * 365 * ln(2) ) = 22.9999 Which means the 50% chance of collision mark for 100,000,000 is: .5 + sqrt(.25 + 2 * 100,000,000 * ln(2) ) = 11774.600235771 Holy crap! I'm screwed if I can only have 12,000 orders before I get a...

Weird requirement at work today: We needed a number that was alpha-numeric, exactly 11 digits, unique, and non-sequential. At first we thought of using a hash, but MD5 and Sha1 give you way too many digits. We could truncate to 11, but not knowing much about hashing, that made me pretty nervous that we'd have collisions. After a bunch of discussion, we decided on this: MAXIMUM_FOR_10_DIGITS = 0xffffffffff (MAXIMUM_FOR_10_DIGITS + cart_id + Time.now.to_i + rand(100_000_000)).to_s(16) I was thinking about using hex, but I was worried about what would happen if the number got too big and went to 12 digits. So I did this calculation: irb(main):001:0> 0xfffffffffff - 0x10000000000 => 16492674416639 Turns out there's all sorts of room between the lowest and highest 11 digit hex number. Paul and Schubert did some quick calculations assuming lots of carts per day and figured out that we have about 250,000 years before we run out of space. Now since we use a random numb...