Extending Watir

My current project is my first project that is solely a web application. I was looking for a way to automate testing for acceptance and regression tests. In reviewing the available frameworks I chose watir. Watir is built using Ruby which fit well with my project as I am using Rake for my build system.

The app makes extensive use of divs and spans in the page templates. Watir allows divs and spans to be tested but it required a lot of repetitive coding. To test that a div exists, and if so that it meets some criteria you have to do something like this.

# validate the login error message is not displayed
#
divFound = false
@browser.divs.each{ |d|
if d.id == 'ValidationSummary'
 assert( false ==
         d.text().include?('Enter a valid e-mail address.'),
         "Expected the error message: 'Enter a valid e-mail " +
         "address'. Received error message: #{d.text()}\n"
 )
 divFound = true
 break
end
}
assert( divFound, "Did not find the ValidationSummary div\n")

Not a big deal but it gets tedious if just about everything is in a div or span. To make life a bit easier I wrote routines that encapsulate the repetitive code. They will test if the div or span exists and if so call an optional block. Since all objects in Ruby are open I coded these as an extension of watir.

module Watir
module Container

   # Search the list of divs for the one specified by id.
   # If the div is not found return false. If a block is
   # provide return its result. Otherwise return true.
   #
   # id - The id of the div to test for
   # block - The optional code block
   #
   def div?(id, &block )
       self.divs.each{ |d|
           if d.id == id
               return (block) ? yield(d) : true
           end
       }
       false
   end

   # Search the list of spans for the one specified by id.
   # If the span is not found return false. If a block is
   # provide return its result. Otherwise return true.
   #
   # id - The id of the span to test for
   # block - The optional code block
   #
   def span?(id, &block )
       self.spans.each{ |d|
           if d.id == id
               return (block) ? yield(d) : true
           end
       }
       false
   end
end
end

We can then change the first example to:

# validate the login error message is not displayed
#
divFound = @browser.divs?( 'ValidationSummary') { |d|
assert( false ==
            d.text().include?('Enter a valid e-mail address.'),
            "Expected the error message: 'Enter a valid e-mail " +
            "address'. Received error message: #{d.text()}\n")
    true }
assert( divFound, "Did not find the ValidationSummary1 div\n")

If you just wanted to test for the existence of the div you would do this.

assert( @browser.divs?( 'ValidationSummary'), "Did not find....")

I'm working on a patch submission to add these to the watir project.

Fortune Cookie Says

You have much skill in expressing yourself to be effective.

Project Euler Problems 18 and 67

Problem 18 and problem 67 are the same problems. The difference being that problem 67 cannot be brute forced. These problems are given a triangle on numbers starting at the top and moving to adjacent numbers on the row below, find the path with the maximum sum.

My first thought was building a tree and using a greedy algorithm to find the path. However, a little thought, and sketching out trees, will show that you can easily build a tree where a greedy algorithm will fail to find maximum path. So what we need to no is traverse the tree and calculate the maximum value for each node. We can do the going down and then scan for the largest value form the nodes that form the bottom of the triangle, or go from the bottom up which causes the root node to have the largest value.

For my solution the triangle represented as an array of arrays. Where each element represents a row of the triangle, and each row is one element longer than the previous. Also each element e[r,i] has to be compared to the elements e[r+1,i] and e[r+1, i+1] where r is index of the row and i the index of the element in that row. The code is as follows.

# receives the table generated from readTable and calculate the largest
# value. The calculation starts with the second to the last table row and
# modifies every element e[r,i] = e[r,i] + max( e[r+1,i], e[r+1, i+1] ).
#
def calcTable( table )
  (table.length - 2).downto(0){ |r|
      table[r].each_with_index{ |val, idx|
          table[r][idx] =
              table[r][idx] + max(
                 table[r+1][idx],
                 table[r+1][idx+1])     
      }        
 }    
table[0][0]
end

Fortune Cookie Says

Your success in life must be earned with earnest efforts.

Project Euler problems 11 - 15

Here are problems 11 - 16 for project Euler.

Problem 11: What is the greatest product of four adjacent numbers in any direction (up, down, left, right, or diagonally) in the 2020 grid? I used an array of arrays to represent the grid.

def euler11(grid)
   maxVal = 0
   (0..19).each{ |i|
       (0..19).each{ |j|
           # calculate horizontal
           if j < 15 then
               val = grid[i][j] * grid[i][j+1]
                    * grid[i][j+2] *grid[i][j+3]
               maxVal = val if maxVal < val

               # calculate diag down right
               if i < 15 then                   
                  val = grid[i][j] * grid[i+1][j+1]
                       * grid[i+2][j+2] * grid[i+3][j+3]
                   maxVal = val if maxVal < val
               end

               # calculate diag up right
               if i > 3 then
                   val = grid[i][j] * grid[i-1][j+1]
                        * grid[i-2][j+2] * grid[i-3][j+3]
                   maxVal = val if maxVal < val
               end
           end
  
           # calculate vertical
           if i < 15 then
               val = grid[i][j] * grid[i+1][j]
                    * grid[i+2][j] * grid[i+3][j]
               maxVal = val if maxVal < val
           end
       }
   }
   maxVal
end

Problem 12: What is the value of the first triangle number to have over five hundred divisors?

A triangle number is calculated as Tn = (n**2 + n) / 2.

def euler12()
   maxInt = 1 << 64
   (500..maxInt).each{|val|
       numT = (val**2 + val)/2
       limit = Math.sqrt(numT)
       stepVal = (0 == numT % 2) ? 1 : 2
       divs = 0
       (1..limit).step(stepVal){ |x|
           if 0 == numT % x then
               divs += 2
           end
       }                                               
       return numT if divs > 500                      
   }                                                  
end                                                    

Problem 13: Find the first 10 digits (most significant) of the sum of 100 50-digit numbers.

def euler13
numArr = [...]
sum = numArr.inject(0){|s,n| n+s}
val=sum.to_s.slice(0..9)
end

Problem 14: Find the starting number, under 1,000,000, that produces the longest chain for the Collatz problem.

The Collatz conjecture, or problem, generates a chain of numbers that terminates at 1. The chain is generated as Ni+1 = { Ni / 2 if Ni is even, 3 Ni + 1 if Ni is odd. At first read this was an "egads!" problem. But after looking at it was depressing easy to solve. The immediate solution that came to mind is a recursive algorithm using memoization. As follows, this runs in just under 59 seconds on my computer.

def findLen(start, cache)
   len = cache[start]
   if len
       return start, len
   else
       nextVal = (0 == start%2) ? start / 2 : 3 * start + 1
       nextVal, len = findLen(nextVal, cache)
       cache[start] = len + 1
       return start, len +1
   end
end

def euler14()
   cache  = { 1 => 1 }
   maxVal = 1
   maxLen = 1

   (2..1000000).each{ |start|
       val, len = findLen(start, cache)

       if maxLen < len
           maxLen = len
           maxVal = val
       end
   }

   return maxVal, maxLen
end

I came up with an iterative approach as well. This is the same algorithm as the recursive solution, but it pushes values with unknown chain length onto a stack, instead of pushing them onto the call stack. This runs in just over 10 seconds which give a good indication on how expensive function calls can be.

def euler14_b()
   cache = {1 => 1}
   maxVal = 1
   maxLen = 1
   stack = []

   (2..1000000).each{ |start|

       val = start
       len = cache[val]
       while !len
           stack.push(val)
           val = (0 == val%2) ? val / 2 : 3 * val + 1
           len = cache[val]
       end

       val = stack.pop
       while val
           len += 1
           cache[val] = len
           val = stack.pop                              
       end

       if maxLen < len
           maxLen = len
           maxVal = start
       end
   }

   return maxVal, maxLen

end

Problem 15: How many routes are there through a 2020 grid?

This was the most interesting problem out of this set. My first idea was to start at the end and generate valid paths back to the start. My second idea was, hey this is a tree, I should be able to calculate the number of paths directly. I thought that the number of paths should be proportional to the height of the tree. A good initial thought, but a little time at the white board showed that it was incorrect. A Google search turned up a blog entries by James Horsley and Real Ultimate Programming, after reading them and cracking open my dusty old discrete math books the answer is 40! / (20! * 20!) or C(40, 20).

To describe the answer in a bit more detail. One must always make 40 moves to get from the start to the finish. 20 moves down and 20 moves right. This gives 40! possibilities (20 down + 20 right). But since moves are indistinguishable, unordered, and without replacement, the total is reduced by 20! ways to select down moves, and 20! ways to select right moves. Or 40! / (20! * 20!).

Visitor at Work

This needs a more clever title than this, but its been a long day.

Coming into work this morning I saw this guy at the revolving door and had to take his picture.

Fortune Cookie Says

You will have a long and wealthy life.