Exams... what are they good for... really?

I realize that the academic world uses the exam process as a measure for how much information was squeezed into your brain in the last X period of time.

So they sit you down, give you a piece of paper that you quietly download your cranium onto in the hopes that in brief period of time that you are given a grade that means you know or you don't.

The problem is that.

Lets say you're like me, that sometimes a subject is interesting and that you find it invigorating to learn about it, and hence your grey matter soaks it up like a sponge to a puddle of water.

But - the person next to me finds it totally the opposite, dreary and mind numbingly aweful, and hence their brain refuses.. no matter how much they try.. to accept that this is going to be useful, and therefor prehaps you should learn it.

So - exam comes.. me.. i do really well, thats great.. person next to me does... ok - still manages to pass.

10 years pass by... (watch the tumble weeds go by)...

We both go for the same job, ohh look says mr employer... i see that you sat a course on molecular biophysics, just what we need, can you comment on ....

And here is my problem, i haven't used molecular biophysics in 10 years, i've been a truck driver... but yet - it says on my paper.. i sat an exam - and i got an X for it. At this point, i feel my knowledge is about the same as that person who was next to me....

I know - the argument here is that i was going for a job that needed that.. blah blah blah, but that aside, you see my point.

I bet for most of us, 6 months down the track - the information retained in the membrane isn't at the same level that it was during the 3 hrs that you were hurriedly scratching at a piece of paper for.

So while i can see that an exam is a measure of performance, it just sucks... esp when we get to the real world, and work in fricken TEAMS!

oh well...

## Monday, 6 June 2016

## Thursday, 2 June 2016

### My fav little algorithm

So, just doing a bit more study and i thought i would share my favourite Algorithm so far, count sort.

I like count sort cos it's so easy and it just does it.

Count sort is non-comparison (which means it doesn't compare this number vs this number), and it's

So here is how it all goes down, lets work on a list that looks like ..... alist =

It has two stages:

To find the key positions in our list:

Find the maximum number in our list, in our example this would be.. 5

Now create a list the of [

So we go thru the list and when we hit a number eg 5 we increment that index value by 1 ie...

So when we have done that, the final array should look like

a new c array to keep track of our sums (see what happens with that in a minute)

I like count sort cos it's so easy and it just does it.

Count sort is non-comparison (which means it doesn't compare this number vs this number), and it's

__stable.__This means that things are left in the order in which the appeared in the original list.So here is how it all goes down, lets work on a list that looks like ..... alist =

**[2,5,3,0,2,3,0,3]**It has two stages:

__Stage 1__: Finding the key positions, count the appearance of that number in the list__Stage 2__: Using key positions, we place the elements from the input array in the right order in the output arrayTo find the key positions in our list:

Find the maximum number in our list, in our example this would be.. 5

Now create a list the of [

**max_num+1**] and fill it with zero's... so, in our case that would look like:**index 0, 1, 2, 3, 4, 5, 6****alist = [2, 5 , 3, 0, 2, 3, 0, 3]****kp= [0, 0, 0, 0, 0, 0, 0]**So we go thru the list and when we hit a number eg 5 we increment that index value by 1 ie...

**index 0, 1, 2, 3, 4, 5, 6, 7****alist = [2, 5 , 3, 0, 2, 3, 0, 3]****kp = [0, 0, 0, 0, 0, 1, 0]**So when we have done that, the final array should look like

**index 0, 1, 2, 3, 4, 5, 6, 7****alist = [2, 5 , 3, 0, 2, 3, 0, 3]****kp = [2, 0, 2, 3, 0, 1, 0]****Now we have how often the numbers repeat, we need to use a running sum to organize the array to give us the true key positions we do this by... initializing our sum = 0 and our index = 0**

a new c array to keep track of our sums (see what happens with that in a minute)

**index 0, 1, 2, 3, 4, 5, 6****kp [2, 0, 2, 3, 0, 1, 0]****sum = 0**

**index = 0**

**idxValue = 2**

Put our sum in the array at position index position, each step here we will be incrementing index by 1

how it goes, is we get the value @ index i, switch that value for our current sum, then add the switched value to our sum value. I have tried to show this in one line, reading from left to right

how it goes, is we get the value @ index i, switch that value for our current sum, then add the switched value to our sum value. I have tried to show this in one line, reading from left to right

index = 0 idxValue = 2 (kp[index] = 2) sum = 0 kp = [

**0****, 0, 2, 3, 0, 1, 0****]**new sum = 0+2
index = 1 idxValue = 0 (kp[index] = 0) sum = 2 kp =

index = 2 idxValue = 2 (kp[index] = 2) sum = 2 kp =

index = 3 idxValue = 3 (kp[index] = 3) sum = 4 kp =

index = 4 idxValue = 0 (kp[index] = 0) sum = 7 kp =

index = 5 idxValue = 1 (kp[index] = 1) sum = 7 kp =

We don't go around for the last time in the array

So that gives us the key positions that the numbers should be in

Now we just lookup the value as an index and increment the value in that index

alist[0] = 2, kp[2] = 2 (kp = kp[2] + 1 = 3) output_list

alist[1] = 5, kp[5] = 7 (kp = kp[5] + 1 = 8) output_list

alist[2] = 3, kp[4] = 4 (kp = kp[4] + 1 = 5) output_list

alist[3] = 0, kp[0] = 0 (kp = kp[0] + 1 = 1) output_list

alist[4] = 2, kp[2] =

alist[5] = 3, kp[3] = 5 (kp = kp[3] + 1 = 6) output_list

alist[6] = 0, kp[0] =

alist[7] = 3, kp[3] = 6 (kp = kp[3] + 1 = 7) output_list

output_list

Notice in this that in red i have highlighted the two indexes that have changed, this is because we already placed a number in the position that was there, so our next number is to the right of that one.

Some code for you to try out..

**[0****,****2****,****2****, 3, 0, 1, 0****]**new sum = 2+0index = 2 idxValue = 2 (kp[index] = 2) sum = 2 kp =

**[0****,****2****,****2****, 3, 0, 1****, 0****]****new sum = 2+2**index = 3 idxValue = 3 (kp[index] = 3) sum = 4 kp =

**[0****, 2, 2, 4, 0, 1****, 0****]****new sum = 4+3**index = 4 idxValue = 0 (kp[index] = 0) sum = 7 kp =

**[0****, 2, 2, 4, 0, 1****, 0****]**new sum = 7+0index = 5 idxValue = 1 (kp[index] = 1) sum = 7 kp =

**[****2, 0, 2, 4, 7, 7****, 0****]**new sum = 7+1We don't go around for the last time in the array

So that gives us the key positions that the numbers should be in

Now we just lookup the value as an index and increment the value in that index

**index 0, 1, 2, 3, 4, 5, 6, 7**alist[0] = 2, kp[2] = 2 (kp = kp[2] + 1 = 3) output_list

**= [_, _, 2, _, _, _, _, _ ]**alist[1] = 5, kp[5] = 7 (kp = kp[5] + 1 = 8) output_list

**= [_, _, _, _, _, _, _, 5 ]**alist[2] = 3, kp[4] = 4 (kp = kp[4] + 1 = 5) output_list

**= [_, _, _, _, _, 3, _, _ ]**alist[3] = 0, kp[0] = 0 (kp = kp[0] + 1 = 1) output_list

**= [0, _, _, _, _, _, _, _ ]**alist[4] = 2, kp[2] =

**(kp = kp[2] + 1 = 4) output_list**__3__**= [_, _, _, 2, _, _, _, _ ]**alist[5] = 3, kp[3] = 5 (kp = kp[3] + 1 = 6) output_list

**= [_, _, _, _, _, _, 3, _ ]**alist[6] = 0, kp[0] =

**(kp = kp[0] + 1 = 2) output_list**__1__**= [_, 0, _, _, _, _, _, _ ]**alist[7] = 3, kp[3] = 6 (kp = kp[3] + 1 = 7) output_list

**= [_, _, _, _, 3, _, _, _ ]**output_list

**= [0, 0, 2, 2, 3, 3, 3, 5 ]**Notice in this that in red i have highlighted the two indexes that have changed, this is because we already placed a number in the position that was there, so our next number is to the right of that one.

Some code for you to try out..

def key_positions(seq, key): array_items = [] for items in seq: array_items.append(key(items)) # find the max in the array max_value = max(array_items) # initialise with 0's count = [0] * (max_value + 1) # count occurances of integers for value in array_items: count[value] += 1 # make the initial sum zero sum = 0 print("size of count array = " + str(max_value + 1)) # iterate over the length of our counting array for idx in range(0, len(count)): # save the current value we have in the array cur_val = count[idx] # put the current sum to the index pos in array count[idx] = sum print("index = {0}: sum = {1} kp = {2} current_value = {3} sum-> {1}+{3}".format(idx, sum, count, cur_val)) # add the saved value to the sum, and make that the new sum sum = sum + cur_val print("index = {0}: sum = {1} kp = {2} current_value = {3} sum-> {1}+{3}".format(idx, sum, count, cur_val)) return count print(key_positions([2,5,3,0,2,3,0,3], lambda x: x))

## Wednesday, 1 June 2016

### Quick Hull in Python

So my assignment for Algorithms was on creating an implementation for convex hulls using the Giftwrap Algorithm, the Graham-Scan Algorithm, and also another of our choosing.

Many chose to monotone hull as their third, i thought i would give another a go, searched around a bit and came up with an implementation called Quick hull which is based around the Quicksort algorithm for those who have come across it, where a part point is formed and sorted items go on one side and the part point is incremented as it continues through the items.

I found a few posts on the Internet about how it sort of works, but thought i would post my own commentary about it and hopefully provide some help to some poor person that comes across this.

So first up Quick hull uses recursion... if you aren't a fan of recursion, or don't enjoy it... you should go here

The algorithm needs a part line to split the points in your point cloud. So we choose the minimum x value and then the maximum x value. This essentially gives us a line through which to split the points left and right on.

Once we have found that line, we now need all the points on the left hand side of the line. At this point in time we don't care about the right hand side points. So we check each point min, max, pt to see if it does a counter clockwise turn (CCW), if it does then it's on the left side of the line, so we include it in our points list.

When that is finished, we now find in that points list, the point that has the furthest distance from the line, lets call that point ptC. Every point that is now in the triangle formed by the points min,max,ptC can be ignored since they will never be part of the convex hull.

Call our quick hull function again with the new points list and change our line end points to that of min, ptc. This will then again find the points on the left and furthest point etc, eventually leaving us with just 2 sets of points, the minimum and the maximum, of which we want the maximum because that was the last furthest point from the line. Return the max and append to the convex hull points.

Now - that does the left hand side, all we have to do is to call the quick hull function again, passing in our points list and reverse the max, min points, and append that to the previous hull list.

And we're done.

Time for some Pseudocode i think, should see what i described a bit clearer:

Ok - now we have the Psuedocode, we need some real code.

The language i implemented this in is python (version 3), but with the Psuedoocode above you should be able to translate it into which ever code you wish.

Many chose to monotone hull as their third, i thought i would give another a go, searched around a bit and came up with an implementation called Quick hull which is based around the Quicksort algorithm for those who have come across it, where a part point is formed and sorted items go on one side and the part point is incremented as it continues through the items.

I found a few posts on the Internet about how it sort of works, but thought i would post my own commentary about it and hopefully provide some help to some poor person that comes across this.

So first up Quick hull uses recursion... if you aren't a fan of recursion, or don't enjoy it... you should go here

The algorithm needs a part line to split the points in your point cloud. So we choose the minimum x value and then the maximum x value. This essentially gives us a line through which to split the points left and right on.

Once we have found that line, we now need all the points on the left hand side of the line. At this point in time we don't care about the right hand side points. So we check each point min, max, pt to see if it does a counter clockwise turn (CCW), if it does then it's on the left side of the line, so we include it in our points list.

When that is finished, we now find in that points list, the point that has the furthest distance from the line, lets call that point ptC. Every point that is now in the triangle formed by the points min,max,ptC can be ignored since they will never be part of the convex hull.

Call our quick hull function again with the new points list and change our line end points to that of min, ptc. This will then again find the points on the left and furthest point etc, eventually leaving us with just 2 sets of points, the minimum and the maximum, of which we want the maximum because that was the last furthest point from the line. Return the max and append to the convex hull points.

Now - that does the left hand side, all we have to do is to call the quick hull function again, passing in our points list and reverse the max, min points, and append that to the previous hull list.

And we're done.

Time for some Pseudocode i think, should see what i described a bit clearer:

```
Def Get Hull Points (point list):
Find the min x, max in list of points
Hull pts = Quick Hull (point list, min, max)
Hull pts = Hull pts + Quick Hull (point list, max, min)
Return Hull pts
Def Quick Hull (point list, min, max)
New pt list = Find all the points left of the line (min, max)
ptC = A point on the left side with the greatest distance from the line
if a ptC was not found
then return the max point
Hull points = Quick Hull (New pt list, min, ptC)
Hull points = Hull points + Quick Hull (New pt list, max, ptC)
Return Hull points
```

Ok - now we have the Psuedocode, we need some real code.

The language i implemented this in is python (version 3), but with the Psuedoocode above you should be able to translate it into which ever code you wish.

```
'''
Quick Hull in Python
```

` Date: 01/06/2016`

` Written by: Grant McEwan`

` License: Beer license... you should buy me one if you make anything off this :) `

` This program produces a set of hull points using a similar method to the quick`

```
Sort Algorithm, so runs in O(n log n) time
'''
```

```
```

```
```

```
'''
When called returns a list of points that forms a convex hull around
```

```
the listPts Given
'''
```

def get_hull_points(listPts): # get the min, and max from the list of points min, max = get_min_max_x(listPts) hullpts = quickhull(listPts, min, max) hullpts = hullpts + quickhull(listPts, max, min) return hullpts ''' Does the sorting for the quick hull sorting algorithm ''' def quickhull(listPts, min, max): left_of_line_pts = get_points_left_of_line(min, max, listPts) ptC = point_max_from_line(min, max, left_of_line_pts) if len(ptC) < 1: return [max] hullPts = quickhull(left_of_line_pts, min, ptC) hullPts = hullPts + quickhull(left_of_line_pts, ptC, max) return hullPts ''' Reterns all points that a LEFT of a line start->end ''' def get_points_left_of_line(start, end, listPts): pts = [] for pt in listPts: if isCCW(start, end, pt): pts.append(pt) return pts ''' Returns the maximum point from a line start->end ''' def point_max_from_line(start, end, points): max_dist = 0 max_point = [] for point in points: if point != start and point != end: dist = distance(start, end, point) if dist > max_dist: max_dist = dist max_point = point return max_point def get_min_max_x(list_pts): min_x = float('inf') max_x = 0 min_y = 0 max_y = 0 for x,y in list_pts: if x < min_x: min_x = x min_y = y if x > max_x: max_x = x max_y = y return [min_x,min_y], [max_x,max_y] ''' Given a line of start->end, will return the distance that point, pt, is from the line. ''' def distance(start, end, pt): # pt is the point x1, y1 = start x2, y2 = end x0, y0 = pt nom = abs((y2 - y1) * x0 - (x2 - x1) * y0 + x2 * y1 - y2 * x1) denom = ((y2 - y1)**2 + (x2 - x1) ** 2) ** 0.5 result = nom / denom return result

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