scala> :paste
// Entering paste mode (ctrl-D to finish)
trait Environment {
val mode: String
override val toString = "Running in mode '" + mode + "'"
}
// Exiting paste mode, now interpreting.
defined trait Environment
scala> val x = new Environment { override val mode = "production" }
x: java.lang.Object with Environment = Running in mode 'null'
//As you can see the mode is still null so instantiating Traits this way is a bad idea.
//You can however use the construct below looking like an anonymous class definition
scala> class Acceptance extends {val mode = "Acceptance"} with Environment
defined class Acceptance
scala> val y = new Acceptance
y: Acceptance = Running in mode 'Acceptance'
//or you can even skip constructing a class and just create a new Object directly
scala> val dev = new {val mode = "Development"} with Environment
dev: java.lang.Object with Environment = Running in mode 'Development'
Friday, June 15, 2012
Using early member definitions with Traits
Monday, June 4, 2012
Micro Scala benchmark
I wanted to check how well Scala performs doing the same test as in Python on my laptop by zipping 2 identical vectors and calculating the sum. It didn't really surprise me that Scala outperformed Python but the outcome really gives me another WTF moment. I probably made a stupid mistake somewhere but just in case here is the test executed on scala. I did a few tests with smaller vectors and those gave the same results with Python and Scala. The strange outcome is due to the result not fitting into an integer !! Changing generic type [Int] to [Long] will result in ouf-of-memory exception unfortunately. For completeness a link to the simple Profiler used.
import Profiler._
object MicroBenchMark extends App {
val v1 = 1 to 9999999 //scala includes .to(x) while Python excludes .to(x)
val v2 = 1 to 9999999
def vector_op(vector1: Iterable[Int], vector2: Iterable[Int], op: (Int, Int) => Int): Iterable[Int] = {
vector1.zip(vector2).map(op.tupled)
}
profile(println("sum is " + vector_op(v1, v2, (x,y) => x + y).sum))
//python prints 99999990000000 instead of 266447232
}
sum is 266447232 Execution took 8 seconds
Friday, June 1, 2012
Python: Creating dictionary of adjacent elements of circular iterable
In many problems you will need to deal with circular datastructures which are most of the time presented in a simple iterable. If you need to quickly find for a specific element what the left and right adjacent elements are, you can use this generic function presented below.
We can even make it more generic if we want to specify what keys and values to use from the iterables.
#we got a list of persons sitting next to each other in a circle.
#This means Robby is sitting next to Alice (left) and Valerie (right) for below example.
#We want to create a quick lookup dict for who sits next to each other
def getCircularLookupDictionary(iter):
#generic function which returns a dict of items with format (key, (leftadjacent, rightadjacent))
l = len(iter)
"""
return dict(
(iter[i], (iter[l - 1], iter[i+1]) ) if i == 0 else
(iter[i], (iter[i - 1], iter[0]) ) if i == l -1 else
(iter[i], (iter[i - 1], iter[i+1]) ) for i in range(l)
)
"""
#nice oneliner provided by my colleague Ivan Lagunov !!
return dict((iter[i], (iter[i - 1], iter[i + 1 - l])) for i in range(l))
circle = ['Robby', 'Valerie', 'Lindsey', 'Audrey', 'Alice']
neighbours = getCircularLookupDictionary(circle)
print neighbours['Robby']
print neighbours['Lindsey']
print neighbours['Alice']
assert neighbours['Robby'] == ('Alice', 'Valerie')
assert neighbours['Lindsey'] == ('Valerie', 'Audrey')
assert neighbours['Alice'] == ('Audrey', 'Robby')
('Alice', 'Valerie')
('Valerie', 'Audrey')
('Audrey', 'Robby')
We can even make it more generic if we want to specify what keys and values to use from the iterables.
#we got a list of persons sitting next to each other in a circle.
#This means Robby is sitting next to Alice (left) and Valerie (right) for below example.
#We want to create a quick lookup dict for who sits next to each other
def getCircularLookupDictionary(iter, keyf, valuef):
#generic function which returns a dict of items with format (key, (leftadjacent, rightadjacent))
l = len(iter)
"""
return dict(
(keyf(iter[i]), (valuef(iter[l - 1]), valuef(iter[i+1])) ) if i == 0 else
(keyf(iter[i]), (valuef(iter[i - 1]), valuef(iter[0])) ) if i == l -1 else
(keyf(iter[i]), (valuef(iter[i - 1]), valuef(iter[i+1])) ) for i in range(l)
)
"""
#nice oneliner provided by my colleague Ivan Lagunov
return dict((keyf(iter[i]), (valuef(iter[i - 1]), valuef(iter[i + 1 -l])) ) for i in range(l))
#circle contains tuples of persons (name, age)
circle = [('Robby', 35), ('Valerie', 5), ('Lindsey', 9), ('Audrey', 40), ('Alice', 59)]
neighbours = getCircularLookupDictionary(circle, lambda (name,age): name, lambda (name,age): age)
print neighbours['Robby']
print neighbours['Lindsey']
print neighbours['Alice']
assert neighbours['Robby'] == (59, 5) #ages of Alice, Valerie
assert neighbours['Lindsey'] == (5, 40) #ages of Valerie, Audrey
assert neighbours['Alice'] == (40, 35) #ages of Audrey, Robby
(59, 5) (5, 40) (40, 35)
Python: Using attributes with functions
Python offers the possibility to store attributes on a function. This can be very useful in a number of situations. Below an example that shows usage.
import inspect
def viewCount(page):
#returns a function which keeps track of how many times the page got viewed
def f():
f.numberOfViews += 1
return f.numberOfViews
f.__name__ = "ViewCount for page " + page
f.numberOfViews = 0
return f
viewCountOne = viewCount('index.html')
viewCountTwo = viewCount('faq.html')
print viewCountOne()
print viewCountOne()
print viewCountTwo()
print viewCountOne()
print viewCountTwo()
print viewCountOne.__name__
print viewCountTwo.__name__
print hasattr(viewCountOne, 'numberOfViews')
print inspect.isfunction(viewCountOne)
print viewCountOne.__name__.find('index') != -1
1 2 1 3 2 ViewCount for page index.html ViewCount for page faq.html True True True
Thursday, May 31, 2012
Python performance pitfalls (?) or issues (?)
I just want to warn you that Python has its glitches performance wise. Functional programming buzzwords are immutability and lazy evaluation (reduced memory footprint). But Python is doing something wrong here one would say. Check for yourself and I would love Python gurus to explain why my attempts to go with the lazy flow resulted in killing the performance.
Let's not return an array but a generator expression.
What is really surprising is that it even takes more time now. So what are we doing still wrong?
Let's also use xrange which is lazy evaluated and see how far that gets us.
Will izip do us anygood in this case?
What about not using a lambda expression?
Ok... last hope?! Let's inline the addition instead of calling operator function.
That seemed to help a lot, so beware of using too much abstraction when performance really matters. Not making the extra number of function calls reduces the execution time a lot. So now just for fun let us not return a generator expression but an array again.
What the f@ck... The generator expression in combination with the sum function seems like total killing performance. Storing the result from the zip call in an array and next calling sum boosted performance.
This answer from stackoverflow is very useful to understand the outcome of this test. I conclude this article that it always makes good sense to profile your app and decide for yourself what is the best approach.
import cProfile
def vector_op(v1,v2,op):
return [op(x,y) for x,y in zip(v1,v2)]
v1 = range(1, 10000000)
v2 = range(1, 10000000)
cProfile.run('print sum(vector_op(v1,v2, lambda x,y: x + y))')
99999990000000
10000046 function calls (10000040 primitive calls) in 59.064 seconds
ncalls tottime percall cumtime percall filename:lineno(function)
9999999 25.622 0.000 25.622 0.000 :1()
1 31.126 31.126 58.165 58.165 PythonApplication1.py:21(vector_op)
1 0.837 0.837 0.837 0.837 {sum}
1 1.417 1.417 1.417 1.417 {zip}
Let's not return an array but a generator expression.
import cProfile
def vector_op_lazy(v1,v2,op):
return (op(x,y) for x,y in zip(v1,v2))
v1 = range(1, 10000000)
v2 = range(1, 10000000)
cProfile.run('print sum(vector_op_lazy(v1,v2, lambda x,y: x + y))')
What is really surprising is that it even takes more time now. So what are we doing still wrong?
99999990000000
20000046 function calls (20000040 primitive calls) in 91.824 seconds
Ordered by: standard name
ncalls tottime percall cumtime percall filename:lineno(function)
9999999 26.132 0.000 26.132 0.000 :1()
10000000 46.346 0.000 72.478 0.000 PythonApplication1.py:25()
1 17.966 17.966 90.444 90.444 {sum}
1 1.378 1.378 1.378 1.378 {zip}
Let's also use xrange which is lazy evaluated and see how far that gets us.
import cProfile
def vector_op_lazy(v1,v2,op):
return (op(x,y) for x,y in zip(v1,v2))
v1 = xrange(1, 10000000)
v2 = xrange(1, 10000000)
cProfile.run('print sum(vector_op_lazy(v1,v2, lambda x,y: x + y))')
99999990000000
20000046 function calls (20000040 primitive calls) in 96.393 seconds
ncalls tottime percall cumtime percall filename:lineno(function)
9999999 27.329 0.000 27.329 0.000 :1()
10000000 48.976 0.000 76.305 0.000 PythonApplication1.py:27()
1 19.059 19.059 95.364 95.364 {sum}
1 1.028 1.028 1.028 1.028 {zip}
Will izip do us anygood in this case?
import cProfile
from itertools import izip
def vector_op_lazy(v1,v2,op):
return (op(x,y) for x,y in izip(v1,v2))
v1 = xrange(1, 10000000)
v2 = xrange(1, 10000000)
cProfile.run('print sum(vector_op_lazy(v1,v2, lambda x,y: x + y))')
99999990000000
20000045 function calls (20000039 primitive calls) in 91.346 seconds
ncalls tottime percall cumtime percall filename:lineno(function)
9999999 26.304 0.000 26.304 0.000 :1()
10000000 46.759 0.000 73.063 0.000 PythonApplication1.py:27()
1 18.280 18.280 91.343 91.343 {sum}
What about not using a lambda expression?
import cProfile
from itertools import izip
def vector_op_lazy(v1,v2,op):
return (op(x,y) for x,y in izip(v1,v2))
def op_add(x,y):
return x + y
v1 = xrange(1, 10000000)
v2 = xrange(1, 10000000)
cProfile.run('print sum(vector_op_lazy(v1,v2, op_add))')
99999990000000
20000045 function calls (20000039 primitive calls) in 89.603 seconds
ncalls tottime percall cumtime percall filename:lineno(function)
10000000 46.009 0.000 71.799 0.000 PythonApplication1.py:27()
9999999 25.790 0.000 25.790 0.000 PythonApplication1.py:29(op_add)
1 17.804 17.804 89.603 89.603 {sum}
Ok... last hope?! Let's inline the addition instead of calling operator function.
import cProfile
from itertools import izip
def vector_op_lazy(v1,v2):
return (x + y for x,y in izip(v1,v2))
v1 = xrange(1, 10000000)
v2 = xrange(1, 10000000)
cProfile.run('print sum(vector_op_lazy(v1,v2))')
99999990000000
10000046 function calls (10000040 primitive calls) in 45.078 seconds
ncalls tottime percall cumtime percall filename:lineno(function)
10000000 27.925 0.000 27.925 0.000 PythonApplication1.py:27()
1 17.153 17.153 45.077 45.077 {sum}
That seemed to help a lot, so beware of using too much abstraction when performance really matters. Not making the extra number of function calls reduces the execution time a lot. So now just for fun let us not return a generator expression but an array again.
import cProfile
from itertools import izip
def vector_op(v1,v2):
return [x + y for x,y in zip(v1,v2)]
v1 = xrange(1, 10000000)
v2 = xrange(1, 10000000)
cProfile.run('print sum(vector_op(v1,v2))')
What the f@ck... The generator expression in combination with the sum function seems like total killing performance. Storing the result from the zip call in an array and next calling sum boosted performance.
99999990000000
47 function calls (41 primitive calls) in 14.603 seconds
ncalls tottime percall cumtime percall filename:lineno(function)
1 12.678 12.678 13.733 13.733 PythonApplication1.py:25(vector_op)
1 0.810 0.810 0.810 0.810 {sum}
1 1.056 1.056 1.056 1.056 {zip}
This answer from stackoverflow is very useful to understand the outcome of this test. I conclude this article that it always makes good sense to profile your app and decide for yourself what is the best approach.
Monday, May 28, 2012
final exam CS212: Portmanteau
This is the sixth and last question from the Udacity CS212 final exam. I will update this article with my solution in 1 week. Please do not comment anything during the final exam week !!
My solution:
# Unit 6: Fun with Words
"""
A portmanteau word is a blend of two or more words, like 'mathelete',
which comes from 'math' and 'athelete'. You will write a function to
find the 'best' portmanteau word from a list of dictionary words.
Because 'portmanteau' is so easy to misspell, we will call our
function 'natalie' instead:
natalie(['word', ...]) == 'portmanteauword'
In this exercise the rules are: a portmanteau must be composed of
three non-empty pieces, start+mid+end, where both start+mid and
mid+end are among the list of words passed in. For example,
'adolescented' comes from 'adolescent' and 'scented', with
start+mid+end='adole'+'scent'+'ed'. A portmanteau must be composed
of two different words (not the same word twice).
That defines an allowable combination, but which is best? Intuitively,
a longer word is better, and a word is well-balanced if the mid is
about half the total length while start and end are about 1/4 each.
To make that specific, the score for a word w is the number of letters
in w minus the difference between the actual and ideal lengths of
start, mid, and end. (For the example word w='adole'+'scent'+'ed', the
start,mid,end lengths are 5,5,2 and the total length is 12. The ideal
start,mid,end lengths are 12/4,12/2,12/4 = 3,6,3. So the final score
is
12 - abs(5-3) - abs(5-6) - abs(2-3) = 8.
yielding a score of 12 - abs(5-(12/4)) - abs(5-(12/2)) -
abs(2-(12/4)) = 8.
The output of natalie(words) should be the best portmanteau, or None
if there is none.
Note (1): I got the idea for this question from
Darius Bacon. Note (2): In real life, many portmanteaux omit letters,
for example 'smoke' + 'fog' = 'smog'; we aren't considering those.
Note (3): The word 'portmanteau' is itself a portmanteau; it comes
from the French "porter" (to carry) + "manteau" (cloak), and in
English meant "suitcase" in 1871 when Lewis Carroll used it in
'Through the Looking Glass' to mean two words packed into one. Note
(4): the rules for 'best' are certainly subjective, and certainly
should depend on more things than just letter length. In addition to
programming the solution described here, you are welcome to explore
your own definition of best, and use your own word lists to come up
with interesting new results. Post your best ones in the discussion
forum. Note (5) The test examples will involve no more than a dozen or so
input words. But you could implement a method that is efficient with a
larger list of words.
"""
def natalie(words):
"Find the best Portmanteau word formed from any two of the list of words."
def test_natalie():
"Some test cases for natalie"
assert natalie(['adolescent', 'scented', 'centennial', 'always', 'ado']) == 'adolescented'
assert natalie(['eskimo', 'escort', 'kimchee', 'kimono', 'cheese']) == 'eskimono'
assert natalie(['kimono', 'kimchee', 'cheese', 'serious', 'us', 'usage']) == 'kimcheese'
assert natalie(['circus', 'elephant', 'lion', 'opera', 'phantom']) == 'elephantom'
assert natalie(['programmer', 'coder', 'partying', 'merrymaking']) == 'programmerrymaking'
assert natalie(['int', 'intimate', 'hinter', 'hint', 'winter']) == 'hintimate'
assert natalie(['morass', 'moral', 'assassination']) == 'morassassination'
assert natalie(['entrepreneur', 'academic', 'doctor', 'neuropsychologist', 'neurotoxin', 'scientist', 'gist']) == 'entrepreneuropsychologist'
assert natalie(['perspicacity', 'cityslicker', 'capability', 'capable']) == 'perspicacityslicker'
assert natalie(['backfire', 'fireproof', 'backflow', 'flowchart', 'background', 'groundhog']) == 'backgroundhog'
assert natalie(['streaker', 'nudist', 'hippie', 'protestor', 'disturbance', 'cops']) == 'nudisturbance'
assert natalie(['night', 'day']) == None
assert natalie(['dog', 'dogs']) == None
assert natalie(['test']) == None
assert natalie(['']) == None
assert natalie(['ABC', '123']) == None
assert natalie([]) == None
return 'tests pass'
print test_natalie()
My solution:
# Unit 6: Fun with Words
"""
A portmanteau word is a blend of two or more words, like 'mathelete',
which comes from 'math' and 'athelete'. You will write a function to
find the 'best' portmanteau word from a list of dictionary words.
Because 'portmanteau' is so easy to misspell, we will call our
function 'natalie' instead:
natalie(['word', ...]) == 'portmanteauword'
In this exercise the rules are: a portmanteau must be composed of
three non-empty pieces, start+mid+end, where both start+mid and
mid+end are among the list of words passed in. For example,
'adolescented' comes from 'adolescent' and 'scented', with
start+mid+end='adole'+'scent'+'ed'. A portmanteau must be composed
of two different words (not the same word twice).
That defines an allowable combination, but which is best? Intuitively,
a longer word is better, and a word is well-balanced if the mid is
about half the total length while start and end are about 1/4 each.
To make that specific, the score for a word w is the number of letters
in w minus the difference between the actual and ideal lengths of
start, mid, and end. (For the example word w='adole'+'scent'+'ed', the
start,mid,end lengths are 5,5,2 and the total length is 12. The ideal
start,mid,end lengths are 12/4,12/2,12/4 = 3,6,3. So the final score
is
12 - abs(5-3) - abs(5-6) - abs(2-3) = 8.
yielding a score of 12 - abs(5-(12/4)) - abs(5-(12/2)) -
abs(2-(12/4)) = 8.
The output of natalie(words) should be the best portmanteau, or None
if there is none.
Note (1): I got the idea for this question from
Darius Bacon. Note (2): In real life, many portmanteaux omit letters,
for example 'smoke' + 'fog' = 'smog'; we aren't considering those.
Note (3): The word 'portmanteau' is itself a portmanteau; it comes
from the French "porter" (to carry) + "manteau" (cloak), and in
English meant "suitcase" in 1871 when Lewis Carroll used it in
'Through the Looking Glass' to mean two words packed into one. Note
(4): the rules for 'best' are certainly subjective, and certainly
should depend on more things than just letter length. In addition to
programming the solution described here, you are welcome to explore
your own definition of best, and use your own word lists to come up
with interesting new results. Post your best ones in the discussion
forum. Note (5) The test examples will involve no more than a dozen or so
input words. But you could implement a method that is efficient with a
larger list of words.
"""
def natalie(words):
"Find the best Portmanteau word formed from any two of the list of words."
#we will represent a natalie with a tuple (word, score)
natalies = []
for (word1, word2) in getWordPairs(words):
mid = ''
endpos = 1
while word1.find(word2[0:endpos]) != -1 and endpos <= len(word2):
mid = word2[0:endpos]
endpos +=1
start = substringBefore(word1, mid)
end = substringAfter(word2, mid)
if mid == '' or len(start) == 0 or len(end) == 0:
#we didn't find a natalie
continue
word = start + mid + end
score = len(word) - abs(len(start)-(len(word)/4.0)) - abs(len(mid)-(len(word)/2.0)) - abs(len(end)-(len(word)/4.0))
natalies.append((word, score))
if len(natalies) == 0:
return None
max_score = max([natalie[1] for natalie in natalies])
#remark: I only return the first hit among several best matches
return [ele[0] for ele in natalies if ele[1] == max_score][0]
def substringAfter(text, substring):
start = text.find(substring)
if start == -1:
return ''
return text[start + len(substring):]
def substringBefore(text, substring):
start = text.find(substring)
if start == -1:
return ''
return text[:start]
def getWordPairs(words):
from itertools import permutations
perms = list(permutations(words, 2))
return perms
def test_natalie():
"Some test cases for natalie"
assert natalie(['adolescent', 'scented', 'centennial', 'always', 'ado']) == 'adolescented'
assert natalie(['eskimo', 'escort', 'kimchee', 'kimono', 'cheese']) == 'eskimono'
assert natalie(['kimono', 'kimchee', 'cheese', 'serious', 'us', 'usage']) == 'kimcheese'
assert natalie(['circus', 'elephant', 'lion', 'opera', 'phantom']) == 'elephantom'
assert natalie(['programmer', 'coder', 'partying', 'merrymaking']) == 'programmerrymaking'
assert natalie(['int', 'intimate', 'hinter', 'hint', 'winter']) == 'hintimate'
assert natalie(['morass', 'moral', 'assassination']) == 'morassassination'
assert natalie(['entrepreneur', 'academic', 'doctor', 'neuropsychologist', 'neurotoxin', 'scientist', 'gist']) == 'entrepreneuropsychologist'
assert natalie(['perspicacity', 'cityslicker', 'capability', 'capable']) == 'perspicacityslicker'
assert natalie(['backfire', 'fireproof', 'backflow', 'flowchart', 'background', 'groundhog']) == 'backgroundhog'
assert natalie(['streaker', 'nudist', 'hippie', 'protestor', 'disturbance', 'cops']) == 'nudisturbance'
assert natalie(['night', 'day']) == None
assert natalie(['dog', 'dogs']) == None
assert natalie(['test']) == None
assert natalie(['']) == None
assert natalie(['ABC', '123']) == None
assert natalie([]) == None
return 'tests pass'
final exam CS212: Darts probability
This is the fifth question from the Udacity CS212 final exam. I will update this article with my solution in 1 week. Please do not comment anything during the final exam week !!
My solution:
# Unit 5: Probability in the game of Darts
"""
In the game of darts, players throw darts at a board to score points.
The circular board has a 'bulls-eye' in the center and 20 slices
called sections, numbered 1 to 20, radiating out from the bulls-eye.
The board is also divided into concentric rings. The bulls-eye has
two rings: an outer 'single' ring and an inner 'double' ring. Each
section is divided into 4 rings: starting at the center we have a
thick single ring, a thin triple ring, another thick single ring, and
a thin double ring. A ring/section combination is called a 'target';
they have names like 'S20', 'D20' and 'T20' for single, double, and
triple 20, respectively; these score 20, 40, and 60 points. The
bulls-eyes are named 'SB' and 'DB', worth 25 and 50 points
respectively. Illustration (png image): http://goo.gl/i7XJ9
There are several variants of darts play; in the game called '501',
each player throws three darts per turn, adding up points until they
total exactly 501. However, the final dart must be in a double ring.
Your first task is to write the function double_out(total), which will
output a list of 1 to 3 darts that add up to total, with the
restriction that the final dart is a double. See test_darts() for
examples. Return None if there is no list that achieves the total.
Often there are several ways to achieve a total. You must return a
shortest possible list, but you have your choice of which one. For
example, for total=100, you can choose ['T20', 'D20'] or ['DB', 'DB']
but you cannot choose ['T20', 'D10', 'D10'].
"""
def test_darts():
"Test the double_out function."
assert double_out(170) == ['T20', 'T20', 'DB']
assert double_out(171) == None
assert double_out(100) in (['T20', 'D20'], ['DB', 'DB'])
"""
My strategy: I decided to choose the result that has the highest valued
target(s) first, e.g. always take T20 on the first dart if we can achieve
a solution that way. If not, try T19 first, and so on. At first I thought
I would need three passes: first try to solve with one dart, then with two,
then with three. But I realized that if we include 0 as a possible dart
value, and always try the 0 first, then we get the effect of having three
passes, but we only have to code one pass. So I creted ordered_points as
a list of all possible scores that a single dart can achieve, with 0 first,
and then descending: [0, 60, 57, ..., 1]. I iterate dart1 and dart2 over
that; then dart3 must be whatever is left over to add up to total. If
dart3 is a valid element of points, then we have a solution. But the
solution, is a list of numbers, like [0, 60, 40]; we need to transform that
into a list of target names, like ['T20', 'D20'], we do that by defining name(d)
to get the name of a target that scores d. When there are several choices,
we must choose a double for the last dart, but for the others I prefer the
easiest targets first: 'S' is easiest, then 'T', then 'D'.
"""
def double_out(total):
"""Return a shortest possible list of targets that add to total,
where the length <= 3 and the final element is a double.
If there is no solution, return None."""
# your code here
"""
It is easy enough to say "170 points? Easy! Just hit T20, T20, DB."
But, at least for me, it is much harder to actually execute the plan
and hit each target. In this second half of the question, we
investigate what happens if the dart-thrower is not 100% accurate.
We will use a wrong (but still useful) model of inaccuracy. A player
has a single number from 0 to 1 that characterizes his/her miss rate.
If miss=0.0, that means the player hits the target every time.
But if miss is, say, 0.1, then the player misses the section s/he
is aiming at 10% of the time, and also (independently) misses the thin
double or triple ring 10% of the time. Where do the misses go?
Here's the model:
First, for ring accuracy. If you aim for the triple ring, all the
misses go to a single ring (some to the inner one, some to the outer
one, but the model doesn't distinguish between these). If you aim for
the double ring (at the edge of the board), half the misses (e.g. 0.05
if miss=0.1) go to the single ring, and half off the board. (We will
agree to call the off-the-board 'target' by the name 'OFF'.) If you
aim for a thick single ring, it is about 5 times thicker than the thin
rings, so your miss ratio is reduced to 1/5th, and of these, half go to
the double ring and half to the triple. So with miss=0.1, 0.01 will go
to each of the double and triple ring. Finally, for the bulls-eyes. If
you aim for the single bull, 1/4 of your misses go to the double bull and
3/4 to the single ring. If you aim for the double bull, it is tiny, so
your miss rate is tripled; of that, 2/3 goes to the single ring and 1/3
to the single bull ring.
Now, for section accuracy. Half your miss rate goes one section clockwise
and half one section counter-clockwise from your target. The clockwise
order of sections is:
20 1 18 4 13 6 10 15 2 17 3 19 7 16 8 11 14 9 12 5
If you aim for the bull (single or double) and miss on rings, then the
section you end up on is equally possible among all 20 sections. But
independent of that you can also miss on sections; again such a miss
is equally likely to go to any section and should be recorded as being
in the single ring.
You will need to build a model for these probabilities, and define the
function outcome(target, miss), which takes a target (like 'T20') and
a miss ration (like 0.1) and returns a dict of {target: probability}
pairs indicating the possible outcomes. You will also define
best_target(miss) which, for a given miss ratio, returns the target
with the highest expected score.
If you are very ambitious, you can try to find the optimal strategy for
accuracy-limited darts: given a state defined by your total score
needed and the number of darts remaining in your 3-dart turn, return
the target that minimizes the expected number of total 3-dart turns
(not the number of darts) required to reach the total. This is harder
than Pig for several reasons: there are many outcomes, so the search space
is large; also, it is always possible to miss a double, and thus there is
no guarantee that the game will end in a finite number of moves.
"""
def outcome(target, miss):
"Return a probability distribution of [(target, probability)] pairs."
#your code here
def best_target(miss):
"Return the target that maximizes the expected score."
#your code here
def same_outcome(dict1, dict2):
"Two states are the same if all corresponding sets of locs are the same."
return all(abs(dict1.get(key, 0) - dict2.get(key, 0)) <= 0.0001
for key in set(dict1) | set(dict2))
def test_darts2():
assert best_target(0.0) == 'T20'
assert best_target(0.1) == 'T20'
assert best_target(0.4) == 'T19'
assert same_outcome(outcome('T20', 0.0), {'T20': 1.0})
assert same_outcome(outcome('T20', 0.1),
{'T20': 0.81, 'S1': 0.005, 'T5': 0.045,
'S5': 0.005, 'T1': 0.045, 'S20': 0.09})
assert same_outcome(
outcome('SB', 0.2),
{'S9': 0.01, 'S8': 0.01, 'S3': 0.01, 'S2': 0.01, 'S1': 0.01, 'DB': 0.04,
'S6': 0.01, 'S5': 0.01, 'S4': 0.01, 'S19': 0.01, 'S18': 0.01, 'S13': 0.01,
'S12': 0.01, 'S11': 0.01, 'S10': 0.01, 'S17': 0.01, 'S16': 0.01,
'S15': 0.01, 'S14': 0.01, 'S7': 0.01, 'S20': 0.01, 'SB': 0.76})
My solution:
# Unit 5: Probability in the game of Darts
"""
In the game of darts, players throw darts at a board to score points.
The circular board has a 'bulls-eye' in the center and 20 slices
called sections, numbered 1 to 20, radiating out from the bulls-eye.
The board is also divided into concentric rings. The bulls-eye has
two rings: an outer 'single' ring and an inner 'double' ring. Each
section is divided into 4 rings: starting at the center we have a
thick single ring, a thin triple ring, another thick single ring, and
a thin double ring. A ring/section combination is called a 'target';
they have names like 'S20', 'D20' and 'T20' for single, double, and
triple 20, respectively; these score 20, 40, and 60 points. The
bulls-eyes are named 'SB' and 'DB', worth 25 and 50 points
respectively. Illustration (png image): http://goo.gl/i7XJ9
There are several variants of darts play; in the game called '501',
each player throws three darts per turn, adding up points until they
total exactly 501. However, the final dart must be in a double ring.
Your first task is to write the function double_out(total), which will
output a list of 1 to 3 darts that add up to total, with the
restriction that the final dart is a double. See test_darts() for
examples. Return None if there is no list that achieves the total.
Often there are several ways to achieve a total. You must return a
shortest possible list, but you have your choice of which one. For
example, for total=100, you can choose ['T20', 'D20'] or ['DB', 'DB']
but you cannot choose ['T20', 'D10', 'D10'].
"""
def test_darts():
"Test the double_out function."
assert double_out(170) == ['T20', 'T20', 'DB']
assert double_out(171) == None
assert double_out(100) in (['T20', 'D20'], ['DB', 'DB'])
"""
My strategy: I decided to choose the result that has the highest valued
target(s) first, e.g. always take T20 on the first dart if we can achieve
a solution that way. If not, try T19 first, and so on. At first I thought
I would need three passes: first try to solve with one dart, then with two,
then with three. But I realized that if we include 0 as a possible dart
value, and always try the 0 first, then we get the effect of having three
passes, but we only have to code one pass. So I creted ordered_points as
a list of all possible scores that a single dart can achieve, with 0 first,
and then descending: [0, 60, 57, ..., 1]. I iterate dart1 and dart2 over
that; then dart3 must be whatever is left over to add up to total. If
dart3 is a valid element of points, then we have a solution. But the
solution, is a list of numbers, like [0, 60, 40]; we need to transform that
into a list of target names, like ['T20', 'D20'], we do that by defining name(d)
to get the name of a target that scores d. When there are several choices,
we must choose a double for the last dart, but for the others I prefer the
easiest targets first: 'S' is easiest, then 'T', then 'D'.
"""
def double_out(total):
"""Return a shortest possible list of targets that add to total,
where the length <= 3 and the final element is a double.
If there is no solution, return None."""
from itertools import ifilter
targets = getTargets()
#first we try to finish with a single shot (double out)
for target in targets:
if target[0] == total and target[2] == 'D':
return [target[1]]
#now we try to finish with two shots of which the last must be a double out
for target1 in targets:
for target2 in ifilter(lambda target: target[2] == 'D', targets):
if target1[0] + target2[0] == total:
return [target1[1], target2[1]]
#so it seems we are bad at darts and are going to need 3 throws
for target1 in targets:
if target1[0] < total:
for target2 in targets:
if target1[0] + target2[0] < total:
for target3 in ifilter(lambda target: target[2] == 'D', targets):
if target1[0] + target2[0] + target3[0] == total:
return [target1[1], target2[1], target3[1]]
return None
def getTargets():
#returns a sorted (high to low) array of tuples of all possible targets in form
#[(60, 'T20', 'T'), (57, 'T19', 'T'), ...,(50, 'DB', 'D'), ...,(19, 'S19', 'S'),...]
targets = [(3*i, 'T' + str(i), 'T') for i in range(1,21)] + [(2*i, 'D' + str(i), 'D') for i in range(1,21)] + [(i, 'S' + str(i), 'S') for i in range(1,21)] + [(25, 'SB', 'S'), (50, 'DB', 'D')]
targets.sort()
targets.reverse()
return targets
"""
def getTarget(targets, restvalue):
#return best target for remaining points to throw or None if we can't double out
for target in targets:
if target[0] < restvalue:
return target
if target[0] == restvalue and target[2] == 'D':
return target
return None
"""
"""
It is easy enough to say "170 points? Easy! Just hit T20, T20, DB."
But, at least for me, it is much harder to actually execute the plan
and hit each target. In this second half of the question, we
investigate what happens if the dart-thrower is not 100% accurate.
We will use a wrong (but still useful) model of inaccuracy. A player
has a single number from 0 to 1 that characterizes his/her miss rate.
If miss=0.0, that means the player hits the target every time.
But if miss is, say, 0.1, then the player misses the section s/he
is aiming at 10% of the time, and also (independently) misses the thin
double or triple ring 10% of the time. Where do the misses go?
Here's the model:
First, for ring accuracy. If you aim for the triple ring, all the
misses go to a single ring (some to the inner one, some to the outer
one, but the model doesn't distinguish between these). If you aim for
the double ring (at the edge of the board), half the misses (e.g. 0.05
if miss=0.1) go to the single ring, and half off the board. (We will
agree to call the off-the-board 'target' by the name 'OFF'.) If you
aim for a thick single ring, it is about 5 times thicker than the thin
rings, so your miss ratio is reduced to 1/5th, and of these, half go to
the double ring and half to the triple. So with miss=0.1, 0.01 will go
to each of the double and triple ring. Finally, for the bulls-eyes. If
you aim for the single bull, 1/4 of your misses go to the double bull and
3/4 to the single ring. If you aim for the double bull, it is tiny, so
your miss rate is tripled; of that, 2/3 goes to the single ring and 1/3
to the single bull ring.
Now, for section accuracy. Half your miss rate goes one section clockwise
and half one section counter-clockwise from your target. The clockwise
order of sections is:
20 1 18 4 13 6 10 15 2 17 3 19 7 16 8 11 14 9 12 5
If you aim for the bull (single or double) and miss on rings, then the
section you end up on is equally possible among all 20 sections. But
independent of that you can also miss on sections; again such a miss
is equally likely to go to any section and should be recorded as being
in the single ring.
You will need to build a model for these probabilities, and define the
function outcome(target, miss), which takes a target (like 'T20') and
a miss ration (like 0.1) and returns a dict of {target: probability}
pairs indicating the possible outcomes. You will also define
best_target(miss) which, for a given miss ratio, returns the target
with the highest expected score.
If you are very ambitious, you can try to find the optimal strategy for
accuracy-limited darts: given a state defined by your total score
needed and the number of darts remaining in your 3-dart turn, return
the target that minimizes the expected number of total 3-dart turns
(not the number of darts) required to reach the total. This is harder
than Pig for several reasons: there are many outcomes, so the search space
is large; also, it is always possible to miss a double, and thus there is
no guarantee that the game will end in a finite number of moves.
"""
def outcome(target, miss):
"Return a probability distribution of [(target, probability)] pairs."
from itertools import product
ring = ring_outcome(target,miss)
section = section_outcome(target, miss)
if target in ['SB', 'DB']:
result = {'SB': ring['SB'] * section['B'] + ring['S'] * section['B'], 'DB': ring['DB'] * section['B']}
for i in range(1,21):
result['S' + str(i)] = section[str(i)]
return result
else:
return dict((combi[0][0] + combi[1][0], combi[0][1] * combi[1][1]) for combi in product(ring.items(), section.items()))
def ring_outcome(target, miss):
#target = 'T20' | 'SB' | 'D5' | ...
#returns dictionary of ring outcomes
if target[0:2] == 'SB': #single bull
return {'SB': 1 - miss, 'DB': miss * 0.25, 'S': miss * 0.75}
elif target[0:2] == 'DB': #double bull
return {'DB': 1 - miss * 3, 'SB': miss * 3 * 1./3, 'S': miss * 3 * 2./3}
elif target[0] == 'T': #triple ring
return {'T': 1 - miss, 'S': miss}
elif target[0] == 'D': #double ring
return {'D': 1- miss, 'S': miss * 0.5} #OFF = miss * 0.5
elif target[0] == 'S': #single ring
return {'S': 1 - (miss / 5), 'D': (miss / 5) * 0.5, 'T': (miss / 5) * 0.5}
else: #should not occur
return {}
def section_outcome(target, miss):
#target = 'T20' | 'SB' | 'D5' | ...
#returns dictionary of section outcomes
clockwise = [20, 1, 18, 4, 13, 6, 10, 15, 2, 17, 3, 19, 7, 16, 8, 11, 14, 9, 12, 5]
misstargets = dict((str(clockwise[i]), (str(5),str(1))) if i == 1 else (str(clockwise[i]), (str(12),str(20))) if i == 19 else (str(clockwise[i]), (str(clockwise[i - 1]),str(clockwise[i + 1]))) for i in range(20))
if target[0:2] in ['SB', 'DB']:
d = dict((str(i), miss / 20) for i in range(1,21))
d['B'] = 1 - miss
return d
else:
section = target[1:]
leftmiss = misstargets[section][0]
rightmiss = misstargets[section][1]
return {section: 1 - miss, leftmiss: miss * 0.5, rightmiss: miss * 0.5}
def best_target(miss):
"Return the target that maximizes the expected score."
#[(60, 'T20', 'T'), (57, 'T19', 'T'), ...,(50, 'DB', 'D'), ...,(19, 'S19', 'S'),...]
targets = getTargets()
#first we create a dictionary (target: value) e.g. ('T20': 60)
valueDict = dict((target[1], target[0]) for target in targets)
return max(getAverageScore(target[1], miss, valueDict) for target in targets)[1]
def getAverageScore(target, miss, valueDict):
#returns a tuple of format (score, target) e.g. (17.5,'T20') so we can use max function easily
target_outcome = outcome(target, miss)
score = 0
for entry in target_outcome.items():
miss_target = entry[0]
probability = entry[1]
score += probability * valueDict[miss_target]
return (score, target)
def same_outcome(dict1, dict2):
"Two states are the same if all corresponding sets of locs are the same."
return all(abs(dict1.get(key, 0) - dict2.get(key, 0)) <= 0.0001
for key in set(dict1) | set(dict2))
def test_darts2():
assert best_target(0.0) == 'T20'
assert best_target(0.1) == 'T20'
assert best_target(0.4) == 'T19'
assert same_outcome(outcome('T20', 0.0), {'T20': 1.0})
assert same_outcome(outcome('T20', 0.1),
{'T20': 0.81, 'S1': 0.005, 'T5': 0.045,
'S5': 0.005, 'T1': 0.045, 'S20': 0.09})
final exam CS212: parking lot search
This is the fourth question from the Udacity CS212 final exam. I will update this article with my solution in 1 week. Please do not comment anything during the final exam week !!
My solution:
"""
UNIT 4: Search
Your task is to maneuver a car in a crowded parking lot. This is a kind of
puzzle, which can be represented with a diagram like this:
| | | | | | | |
| G G . . . Y |
| P . . B . Y |
| P * * B . Y @
| P . . B . . |
| O . . . A A |
| O . S S S . |
| | | | | | | |
A '|' represents a wall around the parking lot, a '.' represents an empty square,
and a letter or asterisk represents a car. '@' marks a goal square.
Note that there are long (3 spot) and short (2 spot) cars.
Your task is to get the car that is represented by '**' out of the parking lot
(on to a goal square). Cars can move only in the direction they are pointing.
In this diagram, the cars GG, AA, SSS, and ** are pointed right-left,
so they can move any number of squares right or left, as long as they don't
bump into another car or wall. In this diagram, GG could move 1, 2, or 3 spots
to the right; AA could move 1, 2, or 3 spots to the left, and ** cannot move
at all. In the up-down direction, BBB can move one up or down, YYY can move
one down, and PPP and OO cannot move.
You should solve this puzzle (and ones like it) using search. You will be
given an initial state like this diagram and a goal location for the ** car;
in this puzzle the goal is the '.' empty spot in the wall on the right side.
You should return a path -- an alternation of states and actions -- that leads
to a state where the car overlaps the goal.
An action is a move by one car in one direction (by any number of spaces).
For example, here is a successor state where the AA car moves 3 to the left:
| | | | | | | |
| G G . . . Y |
| P . . B . Y |
| P * * B . Y @
| P . . B . . |
| O A A . . . |
| O . . . . . |
| | | | | | | |
And then after BBB moves 2 down and YYY moves 3 down, we can solve the puzzle
by moving ** 4 spaces to the right:
| | | | | | | |
| G G . . . . |
| P . . . . . |
| P . . . . * *
| P . . B . Y |
| O A A B . Y |
| O . . B . Y |
| | | | | | | |
You will write the function
solve_parking_puzzle(start, N=N)
where 'start' is the initial state of the puzzle and 'N' is the length of a side
of the square that encloses the pieces (including the walls, so N=8 here).
We will represent the grid with integer indexes. Here we see the
non-wall index numbers (with the goal at index 31):
| | | | | | | |
| 9 10 11 12 13 14 |
| 17 18 19 20 21 22 |
| 25 26 27 28 29 30 31
| 33 34 35 36 37 38 |
| 41 42 43 44 45 46 |
| 49 50 51 52 53 54 |
| | | | | | | |
The wall in the upper left has index 0 and the one in the lower right has 63.
We represent a state of the problem with one big tuple of (object, locations)
pairs, where each pair is a tuple and the locations are a tuple. Here is the
initial state for the problem above in this format:
"""
puzzle1 = (
('@', (31,)),
('*', (26, 27)),
('G', (9, 10)),
('Y', (14, 22, 30)),
('P', (17, 25, 33)),
('O', (41, 49)),
('B', (20, 28, 36)),
('A', (45, 46)),
('|', (0, 1, 2, 3, 4, 5, 6, 7, 8, 15, 16, 23, 24, 32, 39,
40, 47, 48, 55, 56, 57, 58, 59, 60, 61, 62, 63)))
# A solution to this puzzle is as follows:
# path = solve_parking_puzzle(puzzle1, N=8)
# path_actions(path) == [('A', -3), ('B', 16), ('Y', 24), ('*', 4)]
# That is, move car 'A' 3 spaces left, then 'B' 2 down, then 'Y' 3 down,
# and finally '*' moves 4 spaces right to the goal.
# Your task is to define solve_parking_puzzle:
N = 8
def solve_parking_puzzle(start, N=N):
"""Solve the puzzle described by the starting position (a tuple
of (object, locations) pairs). Return a path of [state, action, ...]
alternating items; an action is a pair (object, distance_moved),
such as ('B', 16) to move 'B' two squares down on the N=8 grid."""
# But it would also be nice to have a simpler format to describe puzzles,
# and a way to visualize states.
# You will do that by defining the following two functions:
def locs(start, n, incr=1):
"Return a tuple of n locations, starting at start and incrementing by incr."
def grid(cars, N=N):
"""Return a tuple of (object, locations) pairs -- the format expected for
this puzzle. This function includes a wall pair, ('|', (0, ...)) to
indicate there are walls all around the NxN grid, except at the goal
location, which is the middle of the right-hand wall; there is a goal
pair, like ('@', (31,)), to indicate this. The variable 'cars' is a
tuple of pairs like ('*', (26, 27)). The return result is a big tuple
of the 'cars' pairs along with the walls and goal pairs."""
def show(state, N=N):
"Print a representation of a state as an NxN grid."
# Initialize and fill in the board.
board = ['.'] * N**2
for (c, squares) in state:
for s in squares:
board[s] = c
# Now print it out
for i,s in enumerate(board):
print s,
if i % N == N - 1: print
# Here we see the grid and locs functions in use:
puzzle1 = grid((
('*', locs(26, 2)),
('G', locs(9, 2)),
('Y', locs(14, 3, N)),
('P', locs(17, 3, N)),
('O', locs(41, 2, N)),
('B', locs(20, 3, N)),
('A', locs(45, 2))))
puzzle2 = grid((
('*', locs(26, 2)),
('B', locs(20, 3, N)),
('P', locs(33, 3)),
('O', locs(41, 2, N)),
('Y', locs(51, 3))))
puzzle3 = grid((
('*', locs(25, 2)),
('B', locs(19, 3, N)),
('P', locs(36, 3)),
('O', locs(45, 2, N)),
('Y', locs(49, 3))))
# Here are the shortest_path_search and path_actions functions from the unit.
# You may use these if you want, but you don't have to.
def shortest_path_search(start, successors, is_goal):
"""Find the shortest path from start state to a state
such that is_goal(state) is true."""
if is_goal(start):
return [start]
explored = set() # set of states we have visited
frontier = [ [start] ] # ordered list of paths we have blazed
while frontier:
path = frontier.pop(0)
s = path[-1]
for (state, action) in successors(s).items():
if state not in explored:
explored.add(state)
path2 = path + [action, state]
if is_goal(state):
return path2
else:
frontier.append(path2)
return []
def path_actions(path):
"Return a list of actions in this path."
return path[1::2]
My solution:
"""
UNIT 4: Search
Your task is to maneuver a car in a crowded parking lot. This is a kind of
puzzle, which can be represented with a diagram like this:
| | | | | | | |
| G G . . . Y |
| P . . B . Y |
| P * * B . Y @
| P . . B . . |
| O . . . A A |
| O . S S S . |
| | | | | | | |
A '|' represents a wall around the parking lot, a '.' represents an empty square,
and a letter or asterisk represents a car. '@' marks a goal square.
Note that there are long (3 spot) and short (2 spot) cars.
Your task is to get the car that is represented by '**' out of the parking lot
(on to a goal square). Cars can move only in the direction they are pointing.
In this diagram, the cars GG, AA, SSS, and ** are pointed right-left,
so they can move any number of squares right or left, as long as they don't
bump into another car or wall. In this diagram, GG could move 1, 2, or 3 spots
to the right; AA could move 1, 2, or 3 spots to the left, and ** cannot move
at all. In the up-down direction, BBB can move one up or down, YYY can move
one down, and PPP and OO cannot move.
You should solve this puzzle (and ones like it) using search. You will be
given an initial state like this diagram and a goal location for the ** car;
in this puzzle the goal is the '.' empty spot in the wall on the right side.
You should return a path -- an alternation of states and actions -- that leads
to a state where the car overlaps the goal.
An action is a move by one car in one direction (by any number of spaces).
For example, here is a successor state where the AA car moves 3 to the left:
| | | | | | | |
| G G . . . Y |
| P . . B . Y |
| P * * B . Y @
| P . . B . . |
| O A A . . . |
| O . . . . . |
| | | | | | | |
And then after BBB moves 2 down and YYY moves 3 down, we can solve the puzzle
by moving ** 4 spaces to the right:
| | | | | | | |
| G G . . . . |
| P . . . . . |
| P . . . . * *
| P . . B . Y |
| O A A B . Y |
| O . . B . Y |
| | | | | | | |
You will write the function
solve_parking_puzzle(start, N=N)
where 'start' is the initial state of the puzzle and 'N' is the length of a side
of the square that encloses the pieces (including the walls, so N=8 here).
We will represent the grid with integer indexes. Here we see the
non-wall index numbers (with the goal at index 31):
| | | | | | | |
| 9 10 11 12 13 14 |
| 17 18 19 20 21 22 |
| 25 26 27 28 29 30 31
| 33 34 35 36 37 38 |
| 41 42 43 44 45 46 |
| 49 50 51 52 53 54 |
| | | | | | | |
The wall in the upper left has index 0 and the one in the lower right has 63.
We represent a state of the problem with one big tuple of (object, locations)
pairs, where each pair is a tuple and the locations are a tuple. Here is the
initial state for the problem above in this format:
"""
puzzle0 = (
('@', (31,)),
('*', (26, 27)),
('G', (9, 10)),
('Y', (14, 22, 30)),
('P', (17, 25, 33)),
('O', (41, 49)),
('B', (20, 28, 36)),
('A', (45, 46)),
('|', (0, 1, 2, 3, 4, 5, 6, 7, 8, 15, 16, 23, 24, 32, 39,
40, 47, 48, 55, 56, 57, 58, 59, 60, 61, 62, 63)))
# A solution to this puzzle is as follows:
# path = solve_parking_puzzle(puzzle1, N=8)
# path_actions(path) == [('A', -3), ('B', 16), ('Y', 24), ('*', 4)]
# That is, move car 'A' 3 spaces left, then 'B' 2 down, then 'Y' 3 down,
# and finally '*' moves 4 spaces right to the goal.
# Your task is to define solve_parking_puzzle:
N = 8
def solve_parking_puzzle(start, N=N):
"""Solve the puzzle described by the starting position (a tuple
of (object, locations) pairs). Return a path of [state, action, ...]
alternating items; an action is a pair (object, distance_moved),
such as ('B', 16) to move 'B' two squares down on the N=8 grid."""
return shortest_path_search(start, csuccessors, isGoal, N)
def csuccessors(state, N, goal):
if isGoal(state, goal):
return {}
cars = [car for car in 'ABCDEFGHIJKLMNOPQRSTUVWXYZ*']
successors = []
currentStateDict = getStateDict(state, N)
successors += [successor for symbol, coords in state if symbol in cars for successor in getSuccessors(symbol, coords, state, currentStateDict, N)]
#print successors
return dict(successors)
def getStateDict(state, N):
dict = {}
for i in range(1, N*N + 1):
dict[i] = '.' #mark as freespot
for symbol, coords in state:
for coord in coords:
dict[coord] = symbol #override used spots
return dict
def drivesHorizontally(coords):
#we check if first two coordinates are adjacent
return True if coords[0] == coords[1] - 1 else False
def getSuccessors(car, coords, state, dict, N):
#[('A', -3), ('B', 16), ('Y', 24), ('*', 4)]
# That is, move car 'A' 3 spaces left, then 'B' 2 down, then 'Y' 3 down,
# and finally '*' moves 4 spaces right to the goal.
result = []
if drivesHorizontally(coords):
#move left
i = 1
while (coords[0] - i) in dict and dict[coords[0] - i] in ['.', '@']:
#print car + ' moves left'
result.append((getNewState(state, car, -i), (car, -i)))
i += 1
#move right
i = 1
while (coords[-1] + i) in dict and dict[coords[-1] + i] in ['.', '@']:
#print car + ' moves right'
result.append((getNewState(state, car, i), (car, i)))
i += 1
else:
#move up
i = 1
while (coords[0] -i * N) in dict and dict[coords[0] -i * N] in ['.', '@']:
#print car + ' moves up'
result.append((getNewState(state, car, -i * N), (car, -i * N)))
i += 1
#move down
i = 1
while (coords[-1] + i * N) in dict and dict[coords[-1] + i * N] in ['.', '@']:
#print car + ' moves down'
result.append((getNewState(state, car, i * N), (car, i * N)))
i += 1
return result
def getNewState(oldState, car, movedBy):
newState = []
for symbol, coords in oldState:
if symbol == car:
newState.append((symbol, addToVector(coords, movedBy)))
else:
newState.append((symbol, coords))
return tuple(newState)
def isGoal(state, goal):
#The solution state should contain both the car and the goal ('*', (30, 31)), ('@', (31,))
# ('@', (31,)),
# ('*', (26, 27)),
for symbol, coords in state:
if symbol == '*':
if goal in coords:
return True
return False
def getGoal(state):
#return the goal coordinate
for symbol, coords in state:
if symbol == '@':
return coords[0]
# But it would also be nice to have a simpler format to describe puzzles,
# and a way to visualize states.
# You will do that by defining the following two functions:
def locs(start, n, incr=1):
"Return a tuple of n locations, starting at start and incrementing by incr."
return tuple([start] + [start + incr*i for i in range(1,n)])
def getGoalCoordinate(N):
if N%2 == 0:
return N-1 + (N/2 -1)*N
else:
return N-1 + (N-1)/2 *N
def getWallCoords(N):
goal = getGoalCoordinate(N)
for i in range(1,N +1):
if i == 1:
for x in range(N):
yield x
elif i == N:
start = (N-1) * N
for x in range(start, start +N):
yield x
else:
start = (i-1) * N
end = start + N - 1
yield start
if end != goal:
yield end
def grid(cars, N=N):
"""Return a tuple of (object, locations) pairs -- the format expected for
this puzzle. This function includes a wall pair, ('|', (0, ...)) to
indicate there are walls all around the NxN grid, except at the goal
location, which is the middle of the right-hand wall; there is a goal
pair, like ('@', (31,)), to indicate this. The variable 'cars' is a
tuple of pairs like ('*', (26, 27)). The return result is a big tuple
of the 'cars' pairs along with the walls and goal pairs."""
return tuple([('@', (getGoalCoordinate(N),)), ('|', tuple(getWallCoords(N)))] + [car for car in cars])
def show(state, N=N):
"Print a representation of a state as an NxN grid."
# Initialize and fill in the board.
board = ['.'] * N**2
for (c, squares) in state:
for s in squares:
board[s] = c
# Now print it out
for i,s in enumerate(board):
print s,
if i % N == N - 1: print
# Here we see the grid and locs functions in use:
puzzle1 = grid((
('*', locs(26, 2)),
('G', locs(9, 2)),
('Y', locs(14, 3, N)),
('P', locs(17, 3, N)),
('O', locs(41, 2, N)),
('B', locs(20, 3, N)),
('A', locs(45, 2))))
puzzle2 = grid((
('*', locs(26, 2)),
('B', locs(20, 3, N)),
('P', locs(33, 3)),
('O', locs(41, 2, N)),
('Y', locs(51, 3))))
puzzle3 = grid((
('*', locs(25, 2)),
('B', locs(19, 3, N)),
('P', locs(36, 3)),
('O', locs(45, 2, N)),
('Y', locs(49, 3))))
# Here are the shortest_path_search and path_actions functions from the unit.
# You may use these if you want, but you don't have to.
def shortest_path_search(start, successors, is_goal, N):
"""Find the shortest path from start state to a state
such that is_goal(state) is true."""
goal = getGoal(start)
if is_goal(start, goal):
return [start]
explored = set() # set of states we have visited
frontier = [ [start] ] # ordered list of paths we have blazed
while frontier:
path = frontier.pop(0)
s = path[-1]
for (state, action) in successors(s,N, goal).items():
if state not in explored:
explored.add(state)
path2 = path + [action, state]
#print action
#print show(state,N)
if is_goal(state, goal):
return path2
else:
frontier.append(path2)
return []
def path_actions(path):
"Return a list of actions in this path."
return path[1::2]
def addToVector(v1, n):
v2 = tuple((n for i in range(len(v1))))
return add_vectors(v1,v2)
def add_vectors(v1, v2):
return tuple(x + y for x,y in zip(v1,v2))
def showAndSolve(puzzle):
show(puzzle, N)
solution = solve_parking_puzzle(puzzle, N)
print path_actions(solution)
final exam CS212: Polynomials
This is the third question from the Udacity CS212 final exam. I will update this article with my solution in 1 week. Please do not comment anything during the final exam week !!
My solution:
"""
UNIT 3: Functions and APIs: Polynomials
A polynomial is a mathematical formula like:
30 * x**2 + 20 * x + 10
More formally, it involves a single variable (here 'x'), and the sum of one
or more terms, where each term is a real number multiplied by the variable
raised to a non-negative integer power. (Remember that x**0 is 1 and x**1 is x,
so 'x' is short for '1 * x**1' and '10' is short for '10 * x**0'.)
We will represent a polynomial as a Python function which computes the formula
when applied to a numeric value x. The function will be created with the call:
p1 = poly((10, 20, 30))
where the nth element of the input tuple is the coefficient of the nth power of x.
(Note the order of coefficients has the x**n coefficient neatly in position n of
the list, but this is the reversed order from how we usually write polynomials.)
poly returns a function, so we can now apply p1 to some value of x:
p1(0) == 10
Our representation of a polynomial is as a callable function, but in addition,
we will store the coefficients in the .coef attribute of the function, so we have:
p1.coef == (10, 20, 30)
And finally, the name of the function will be the formula given above, so you should
have something like this:
>>> p1
>>> p1.__name__
'30 * x**2 + 20 * x + 10'
Make sure the formula used for function names is simplified properly.
No '0 * x**n' terms; just drop these. Simplify '1 * x**n' to 'x**n'.
Simplify '5 * x**0' to '5'. Similarly, simplify 'x**1' to 'x'.
For negative coefficients, like -5, you can use '... + -5 * ...' or
'... - 5 * ...'; your choice. I'd recommend no spaces around '**'
and spaces around '+' and '*', but you are free to use your preferences.
Your task is to write the function poly and the following additional functions:
is_poly, add, sub, mul, power, deriv, integral
They are described below; see the test_poly function for examples.
"""
def poly(coefs):
"""Return a function that represents the polynomial with these coefficients.
For example, if coefs=(10, 20, 30), return the function of x that computes
'30 * x**2 + 20 * x + 10'. Also store the coefs on the .coefs attribute of
the function, and the str of the formula on the .__name__ attribute.'"""
# your code here (I won't repeat "your code here"; there's one for each function)
def test_poly():
global p1, p2, p3, p4, p5, p9 # global to ease debugging in an interactive session
p1 = poly((10, 20, 30))
assert p1(0) == 10
for x in (1, 2, 3, 4, 5, 1234.5):
assert p1(x) == 30 * x**2 + 20 * x + 10
assert same_name(p1.__name__, '30 * x**2 + 20 * x + 10')
assert is_poly(p1)
assert not is_poly(abs) and not is_poly(42) and not is_poly('cracker')
p3 = poly((0, 0, 0, 1))
assert p3.__name__ == 'x**3'
p9 = mul(p3, mul(p3, p3))
assert p9 == poly([0,0,0,0,0,0,0,0,0,1])
assert p9(2) == 512
p4 = add(p1, p3)
assert same_name(p4.__name__, 'x**3 + 30 * x**2 + 20 * x + 10')
assert same_name(poly((1, 1)).__name__, 'x + 1')
assert (power(poly((1, 1)), 10).__name__ ==
'x**10 + 10 * x**9 + 45 * x**8 + 120 * x**7 + 210 * x**6 + 252 * x**5 + 210' +
' * x**4 + 120 * x**3 + 45 * x**2 + 10 * x + 1')
assert add(poly((10, 20, 30)), poly((1, 2, 3))) == poly((11, 22, 33))
assert sub(poly((10, 20, 30)), poly((1, 2, 3))) == poly((9, 18, 27))
assert mul(poly((10, 20, 30)), poly((1, 2, 3))) == poly((10, 40, 100, 120, 90))
assert power(poly((1, 1)), 2) == poly((1, 2, 1))
assert power(poly((1, 1)), 10) == poly((1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1))
assert deriv(p1) == poly((20, 60))
assert integral(poly((20, 60))) == poly((0, 20, 30))
p5 = poly((0, 1, 2, 3, 4, 5))
assert same_name(p5.__name__, '5 * x**5 + 4 * x**4 + 3 * x**3 + 2 * x**2 + x')
assert p5(1) == 15
assert p5(2) == 258
assert same_name(deriv(p5).__name__, '25 * x**4 + 16 * x**3 + 9 * x**2 + 4 * x + 1')
assert deriv(p5)(1) == 55
assert deriv(p5)(2) == 573
def same_name(name1, name2):
"""I define this function rather than doing name1 == name2 to allow for some
variation in naming conventions."""
def canonical_name(name): return name.replace(' ', '').replace('+-', '-')
return canonical_name(name1) == canonical_name(name2)
def is_poly(x):
"Return true if x is a poly (polynomial)."
## For examples, see the test_poly function
def add(p1, p2):
"Return a new polynomial which is the sum of polynomials p1 and p2."
def sub(p1, p2):
"Return a new polynomial which is the difference of polynomials p1 and p2."
def mul(p1, p2):
"Return a new polynomial which is the product of polynomials p1 and p2."
def power(p, n):
"Return a new polynomial which is p to the nth power (n a non-negative integer)."
"""
If your calculus is rusty (or non-existant), here is a refresher:
The deriviative of a polynomial term (c * x**n) is (c*n * x**(n-1)).
The derivative of a sum is the sum of the derivatives.
So the derivative of (30 * x**2 + 20 * x + 10) is (60 * x + 20).
The integral is the anti-derivative:
The integral of 60 * x + 20 is 30 * x**2 + 20 * x + C, for any constant C.
Any value of C is an equally good anti-derivative. We allow C as an argument
to the function integral (withh default C=0).
"""
def deriv(p):
"Return the derivative of a function p (with respect to its argument)."
def integral(p, C=0):
"Return the integral of a function p (with respect to its argument)."
"""
Now for an extra credit challenge: arrange to describe polynomials with an
expression like '3 * x**2 + 5 * x + 9' rather than (9, 5, 3). You can do this
in one (or both) of two ways:
(1) By defining poly as a class rather than a function, and overloading the
__add__, __sub__, __mul__, and __pow__ operators, etc. If you choose this,
call the function test_poly1(). Make sure that poly objects can still be called.
(2) Using the grammar parsing techniques we learned in Unit 5. For this
approach, define a new function, Poly, which takes one argument, a string,
as in Poly('30 * x**2 + 20 * x + 10'). Call test_poly2().
"""
def test_poly1():
# I define x as the polynomial 1*x + 0.
x = poly((0, 1))
# From here on I can create polynomials by + and * operations on x.
newp1 = 30 * x**2 + 20 * x + 10 # This is a poly object, not a number!
assert p1(100) == newp1(100) # The new poly objects are still callable.
assert p1.__name__ == newp1.__name__
assert (x + 1) * (x - 1) == x**2 - 1 == poly((-1, 0, 1))
def test_poly2():
newp1 = Poly('30 * x**2 + 20 * x + 10')
assert p1(100) == newp1(100)
assert p1.__name__ == newp1.__name__
assert Poly('x + 1') * Poly('x - 1') == Poly('x**2 - 1')
My solution:
"""
UNIT 3: Functions and APIs: Polynomials
A polynomial is a mathematical formula like:
30 * x**2 + 20 * x + 10
More formally, it involves a single variable (here 'x'), and the sum of one
or more terms, where each term is a real number multiplied by the variable
raised to a non-negative integer power. (Remember that x**0 is 1 and x**1 is x,
so 'x' is short for '1 * x**1' and '10' is short for '10 * x**0'.)
We will represent a polynomial as a Python function which computes the formula
when applied to a numeric value x. The function will be created with the call:
p1 = poly((10, 20, 30))
where the nth element of the input tuple is the coefficient of the nth power of x.
(Note the order of coefficients has the x**n coefficient neatly in position n of
the list, but this is the reversed order from how we usually write polynomials.)
poly returns a function, so we can now apply p1 to some value of x:
p1(0) == 10
Our representation of a polynomial is as a callable function, but in addition,
we will store the coefficients in the .coef attribute of the function, so we have:
p1.coef == (10, 20, 30)
And finally, the name of the function will be the formula given above, so you should
have something like this:
>>> p1
>>> p1.__name__
'30 * x**2 + 20 * x + 10'
Make sure the formula used for function names is simplified properly.
No '0 * x**n' terms; just drop these. Simplify '1 * x**n' to 'x**n'.
Simplify '5 * x**0' to '5'. Similarly, simplify 'x**1' to 'x'.
For negative coefficients, like -5, you can use '... + -5 * ...' or
'... - 5 * ...'; your choice. I'd recommend no spaces around '**'
and spaces around '+' and '*', but you are free to use your preferences.
Your task is to write the function poly and the following additional functions:
is_poly, add, sub, mul, power, deriv, integral
They are described below; see the test_poly function for examples.
"""
def poly(coefs):
"""Return a function that represents the polynomial with these coefficients.
For example, if coefs=(10, 20, 30), return the function of x that computes
'30 * x**2 + 20 * x + 10'. Also store the coefs on the .coefs attribute of
the function, and the str of the formula on the .__name__ attribute.'"""
# your code here (I won't repeat "your code here"; there's one for each function)
formula = getFormula(coefs)
def p(x):
return eval(formula.replace('x', str(x)))
p.coefs = coefs
p.__name__ = formula
return p
def getFormula(coefs):
coefs_list = list(coefs)
coefs_list.reverse()
#(10, 20, 30) results in '30 * x**2 + 20 * x + 10'
return ' + '.join(map_coef(coef, len(coefs_list) - index - 1) for index, coef in enumerate(coefs_list) if map_coef(coef, len(coefs_list) - index - 1) != '')
def map_coef(coef, index):
"""
map_coef(7, 3) = '7 * x**3'
map_coef(10, 2) = '10 * x**2'
map_coef(5, 1) = '5 * x'
map_coef(0, 2) = '' <-- 0 * x**2
map_coef(1, 2) = 'x**2' <-- 1 * x**2
map_coef(3, 0) = '3'
"""
if index == 0:
return str(coef) if coef != 0 else ''
elif index == 1:
if coef == 0:
return ''
elif coef == 1:
return 'x'
else:
return str(coef) + ' * x'
else:
if coef == 0:
return ''
if coef == 1:
return 'x**' + str(index)
else:
return str(coef) + ' * x**' + str(index)
def coefsDictionaryToTuple(coefs_dict):
dict_to_list = coefs_dict.items()
dict_to_list.sort()
return tuple((ele[1] for ele in dict_to_list))
def coefs_product(coefs1, coefs2):
#coef1=(10, 20, 30), coefs2=(1, 2, 3)
prod = []
for index1,coef1 in enumerate(coefs1):
for index2, coef2 in enumerate(coefs2):
prod.append((index1 + index2, coef1 * coef2))
#prod => [(0, 10), (1, 20), (2, 30), (1, 20), (2, 40), (3, 60), (2, 30), (3, 60), (4, 90)]
coefs_dict = {} #we will store tuples of same format in this dict but add up coefs with same index
for ele in prod:
if ele[0] not in coefs_dict:
#we create entry
coefs_dict[ele[0]] = ele[1]
else:
#we update entry by adding current coef
coefs_dict[ele[0]] = coefs_dict[ele[0]] + ele[1]
return coefsDictionaryToTuple(coefs_dict)
def coefs_add(coefs1, coefs2):
return coefs_op(coefs1, coefs2, lambda x,y: x + y)
def coefs_sub(coefs1, coefs2):
return coefs_op(coefs1, coefs2, lambda x,y: x - y)
def coefs_op(coefs1, coefs2, op):
coefs_dict = {}
for index, coef in enumerate(coefs1):
coefs_dict[index] = coef
for index, coef in enumerate(coefs2):
if index not in coefs_dict:
#we create entry
coefs_dict[index] = coef
else:
#we update entry by adding current coef
coefs_dict[index] = op(coefs_dict[index], coef)
return coefsDictionaryToTuple(coefs_dict)
def test_poly():
global p1, p2, p3, p4, p5, p9 # global to ease debugging in an interactive session
p1 = poly((10, 20, 30))
assert p1(0) == 10
for x in (1, 2, 3, 4, 5, 1234.5):
assert p1(x) == 30 * x**2 + 20 * x + 10
assert same_name(p1.__name__, '30 * x**2 + 20 * x + 10')
assert is_poly(p1)
assert not is_poly(abs) and not is_poly(42) and not is_poly('cracker')
p3 = poly((0, 0, 0, 1))
assert p3.__name__ == 'x**3'
p9 = mul(p3, mul(p3, p3))
assert p9(2) == 512
p4 = add(p1, p3)
assert same_name(p4.__name__, 'x**3 + 30 * x**2 + 20 * x + 10')
assert same_name(poly((1, 1)).__name__, 'x + 1')
assert (power(poly((1, 1)), 10).__name__ ==
'x**10 + 10 * x**9 + 45 * x**8 + 120 * x**7 + 210 * x**6 + 252 * x**5 + 210' +
' * x**4 + 120 * x**3 + 45 * x**2 + 10 * x + 1')
assert add(poly((10, 20, 30)), poly((1, 2, 3))).coefs == (11,22,33)
assert sub(poly((10, 20, 30)), poly((1, 2, 3))).coefs == (9,18,27)
assert mul(poly((10, 20, 30)), poly((1, 2, 3))).coefs == (10, 40, 100, 120, 90)
assert power(poly((1, 1)), 2).coefs == (1, 2, 1)
assert power(poly((1, 1)), 10).coefs == (1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1)
assert deriv(p1).coefs == (20, 60)
assert integral(poly((20, 60))).coefs == (0, 20, 30)
p5 = poly((0, 1, 2, 3, 4, 5))
assert same_name(p5.__name__, '5 * x**5 + 4 * x**4 + 3 * x**3 + 2 * x**2 + x')
assert p5(1) == 15
assert p5(2) == 258
assert same_name(deriv(p5).__name__, '25 * x**4 + 16 * x**3 + 9 * x**2 + 4 * x + 1')
assert deriv(p5)(1) == 55
assert deriv(p5)(2) == 573
def same_name(name1, name2):
"""I define this function rather than doing name1 == name2 to allow for some
variation in naming conventions."""
def canonical_name(name): return name.replace(' ', '').replace('+-', '-')
return canonical_name(name1) == canonical_name(name2)
def is_poly(x):
"Return true if x is a poly (polynomial)."
## For examples, see the test_poly function
return hasattr(x, '__call__') and isinstance(x(1), int) and x.__name__.find('x') != -1
def add(p1, p2):
"Return a new polynomial which is the sum of polynomials p1 and p2."
return poly(coefs_add(p1.coefs, p2.coefs))
def sub(p1, p2):
"Return a new polynomial which is the difference of polynomials p1 and p2."
return poly(coefs_sub(p1.coefs, p2.coefs))
def mul(p1, p2):
"Return a new polynomial which is the product of polynomials p1 and p2."
return poly(coefs_product(p1.coefs, p2.coefs))
def power(p, n):
"Return a new polynomial which is p to the nth power (n a non-negative integer)."
if n in [0,1]:
return p
else:
return mul(p, power(p,n-1))
"""
If your calculus is rusty (or non-existant), here is a refresher:
The deriviative of a polynomial term (c * x**n) is (c*n * x**(n-1)).
The derivative of a sum is the sum of the derivatives.
So the derivative of (30 * x**2 + 20 * x + 10) is (60 * x + 20).
The integral is the anti-derivative:
The integral of 60 * x + 20 is 30 * x**2 + 20 * x + C, for any constant C.
Any value of C is an equally good anti-derivative. We allow C as an argument
to the function integral (withh default C=0).
"""
def deriv(p):
"Return the derivative of a function p (with respect to its argument)."
coefs_new = tuple((index * coef for index, coef in enumerate(p.coefs) if index > 0))
return poly(coefs_new)
def integral(p, C=0):
"Return the integral of a function p (with respect to its argument)."
#The integral of 60 * x + 20 == 30 * x**2 + 20 * x + C, for any constant C.
#assert integral(poly((20, 60))).coefs == (0, 20, 30)
coefs_new = tuple([C] + [coef / (index + 1) for index, coef in enumerate(p.coefs)])
return poly(coefs_new)
"""
Now for an extra credit challenge: arrange to describe polynomials with an
expression like '3 * x**2 + 5 * x + 9' rather than (9, 5, 3). You can do this
in one (or both) of two ways:
(1) By defining poly as a class rather than a function, and overloading the
__add__, __sub__, __mul__, and __pow__ operators, etc. If you choose this,
call the function test_poly1(). Make sure that poly objects can still be called.
(2) Using the grammar parsing techniques we learned in Unit 5. For this
approach, define a new function, Poly, which takes one argument, a string,
as in Poly('30 * x**2 + 20 * x + 10'). Call test_poly2().
"""
def test_poly1():
# I define x as the polynomial 1*x + 0.
x = poly((0, 1))
# From here on I can create polynomials by + and * operations on x.
newp1 = 30 * x**2 + 20 * x + 10 # This is a poly object, not a number!
assert p1(100) == newp1(100) # The new poly objects are still callable.
assert p1.__name__ == newp1.__name__
assert (x + 1) * (x - 1) == x**2 - 1 == poly((-1, 0, 1))
def test_poly2():
newp1 = Poly('30 * x**2 + 20 * x + 10')
assert p1(100) == newp1(100)
assert p1.__name__ == newp1.__name__
assert Poly('x + 1') * Poly('x - 1') == Poly('x**2 - 1')
final exam CS212: Logic Puzzle
This is the second question from the Udacity CS212 final exam. I will update this article with my solution in 1 week. Please do not comment anything during the final exam week !!
My solution:
"""
UNIT 2: Logic Puzzle
You will write code to solve the following logic puzzle:
1. The person who arrived on Wednesday bought the laptop.
2. The programmer is not Wilkes.
3. Of the programmer and the person who bought the droid,
one is Wilkes and the other is Hamming.
4. The writer is not Minsky.
5. Neither Knuth nor the person who bought the tablet is the manager.
6. Knuth arrived the day after Simon.
7. The person who arrived on Thursday is not the designer.
8. The person who arrived on Friday didn't buy the tablet.
9. The designer didn't buy the droid.
10. Knuth arrived the day after the manager.
11. Of the person who bought the laptop and Wilkes,
one arrived on Monday and the other is the writer.
12. Either the person who bought the iphone or the person who bought the tablet
arrived on Tuesday.
You will write the function logic_puzzle(), which should return a list of the
names of the people in the order in which they arrive. For example, if they
happen to arrive in alphabetical order, Hamming on Monday, Knuth on Tuesday, etc.,
then you would return:
['Hamming', 'Knuth', 'Minsky', 'Simon', 'Wilkes']
(You can assume that the days mentioned are all in the same week.)
"""
def logic_puzzle():
"Return a list of the names of the people, in the order they arrive."
## your code here; you are free to define additional functions if needed
My solution:
"""
UNIT 2: Logic Puzzle
You will write code to solve the following logic puzzle:
1. The person who arrived on Wednesday bought the laptop.
2. The programmer is not Wilkes.
3. Of the programmer and the person who bought the droid,
one is Wilkes and the other is Hamming.
4. The writer is not Minsky.
5. Neither Knuth nor the person who bought the tablet is the manager.
6. Knuth arrived the day after Simon.
7. The person who arrived on Thursday is not the designer.
8. The person who arrived on Friday didn't buy the tablet.
9. The designer didn't buy the droid.
10. Knuth arrived the day after the manager.
11. Of the person who bought the laptop and Wilkes,
one arrived on Monday and the other is the writer.
12. Either the person who bought the iphone or the person who bought the tablet
arrived on Tuesday.
You will write the function logic_puzzle(), which should return a list of the
names of the people in the order in which they arrive. For example, if they
happen to arrive in alphabetical order, Hamming on Monday, Knuth on Tuesday, etc.,
then you would return:
['Hamming', 'Knuth', 'Minsky', 'Simon', 'Wilkes']
(You can assume that the days mentioned are all in the same week.)
"""
def logic_puzzle():
import itertools
"Return a list of the names of the people, in the order they arrive."
days = monday, tuesday, wednesday, thursday, friday = [1,2,3,4,5]
orderings = list(itertools.permutations(days))
(Wilkes, Hamming, Minsky, Knuth, Simon) = next((Wilkes, Hamming, Minsky, Knuth, Simon)
for (Wilkes, Hamming, Minsky, Knuth, Simon) in orderings
if Knuth == Simon + 1 # from [6]
for (programmer, writer, manager, designer, unemployed) in orderings
if programmer == Hamming # from [2] and [3]
and writer != Minsky # from [4]
and Knuth != manager # from [5]
and designer != thursday # from [7]
and Knuth == manager + 1 # from [10]
and monday != writer # from [11]
for (laptop, droid, tablet, iphone, nothing) in orderings
if laptop == wednesday # from [1]
and programmer != droid # from [2] and [3]
and droid == Wilkes # from [2] and [3]
and tablet != manager # from [5]
and tablet != friday # from [8]
and designer != droid # from [9]
and monday in [laptop, Wilkes] # from [11]
and writer in [laptop, Wilkes] # from [11]
and tuesday in [iphone, tablet] # from [12]
)
solution = [
(Wilkes, 'Wilkes'),
(Hamming, 'Hamming'),
(Minsky, 'Minsky'),
(Knuth, 'Knuth'),
(Simon, 'Simon')
]
solution.sort()
return [element[1] for element in solution]
print logic_puzzle()
Final exam CS212: bowling puzzler
This is the first question from the Udacity CS212 final exam. I will update this article with my solution in 1 week. Please do not comment anything during the final exam week !!
"""
UNIT 1: Bowling:
You will write the function bowling(balls), which returns an integer indicating
the score of a ten-pin bowling game. balls is a list of integers indicating
how many pins are knocked down with each ball. For example, a perfect game of
bowling would be described with:
>>> bowling([10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10])
300
The rules of bowling are as follows:
(1) A game consists of 10 frames. In each frame you roll one or two balls,
except for the tenth frame, where you roll one, two, or three. Your total
score is the sum of your scores for the ten frames.
(2) If you knock down fewer than ten pins with your two balls in the frame,
you score the total knocked down. For example, bowling([8, 1, 7, ...]) means
that you knocked down a total of 9 pins in the first frame. You score 9 point
for the frame, and you used up two balls in the frame. The second frame will
start with the 7.
(3) If you knock down all ten pins on your second ball it is called a 'spare'
and you score 10 points plus a bonus: whatever you roll with your next ball.
The next ball will also count in the next frame, so the next ball counts twice
(except in the tenth frame, in which case the bonus ball counts only once).
For example, bowling([8, 2, 7, ...]) means you get a spare in the first frame.
You score 10 + 7 for the frame; the second frame starts with the 7.
(4) If you knock down all ten pins on your first ball it is called a 'strike'
and you score 10 points plus a bonus of your score on the next two balls.
(The next two balls also count in the next frame, except in the tenth frame.)
For example, bowling([10, 7, 3, ...]) means that you get a strike, you score
10 + 7 + 3 = 20 in the first frame; the second frame starts with the 7.
"""
def bowling(balls):
"Compute the total score for a player's game of bowling."
#TODO: write your code here
def test_bowling():
assert 0 == bowling([0] * 20)
assert 20 == bowling([1] * 20)
assert 80 == bowling([4] * 20)
assert 190 == bowling([9,1] * 10 + [9])
assert 300 == bowling([10] * 12)
assert 200 == bowling([10, 5,5] * 5 + [10])
assert 11 == bowling([0,0] * 9 + [10,1,0])
assert 12 == bowling([0,0] * 8 + [10, 1,0])
test_bowling()
My solution:
"""
UNIT 1: Bowling:
You will write the function bowling(balls), which returns an integer indicating
the score of a ten-pin bowling game. balls is a list of integers indicating
how many pins are knocked down with each ball. For example, a perfect game of
bowling would be described with:
>>> bowling([10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10])
300
The rules of bowling are as follows:
(1) A game consists of 10 frames. In each frame you roll one or two balls,
except for the tenth frame, where you roll one, two, or three. Your total
score is the sum of your scores for the ten frames.
(2) If you knock down fewer than ten pins with your two balls in the frame,
you score the total knocked down. For example, bowling([8, 1, 7, ...]) means
that you knocked down a total of 9 pins in the first frame. You score 9 point
for the frame, and you used up two balls in the frame. The second frame will
start with the 7.
(3) If you knock down all ten pins on your second ball it is called a 'spare'
and you score 10 points plus a bonus: whatever you roll with your next ball.
The next ball will also count in the next frame, so the next ball counts twice
(except in the tenth frame, in which case the bonus ball counts only once).
For example, bowling([8, 2, 7, ...]) means you get a spare in the first frame.
You score 10 + 7 for the frame; the second frame starts with the 7.
(4) If you knock down all ten pins on your first ball it is called a 'strike'
and you score 10 points plus a bonus of your score on the next two balls.
(The next two balls also count in the next frame, except in the tenth frame.)
For example, bowling([10, 7, 3, ...]) means that you get a strike, you score
10 + 7 + 3 = 20 in the first frame; the second frame starts with the 7.
"""
def bowling(balls):
"Compute the total score for a player's game of bowling."
balls.reverse()
score = countScore(balls, 0, 0)
return score
def countScore(balls, frame, score):
frame += 1
if not balls:
return score
ball1 = balls.pop()
if ball1 == 10:
#we got a strike so we need to add two next balls
newscore = score + ball1 + (balls.pop() if frame == 10 else balls[-1]) + (balls.pop() if frame == 10 else balls[len(balls) - 2])
return countScore(balls, frame, newscore)
if ball1 + balls[-1] == 10:
#we got a spare so we pop the second one as well
ball2 = balls.pop()
newscore = score + ball1 + ball2 + (balls.pop() if frame == 10 else balls[-1])
return countScore(balls, frame, newscore)
#else we just add up the scores of the first two balls for this frame
ball2 = balls.pop()
newscore = score + ball1 + ball2
return countScore(balls, frame, newscore)
def test_bowling():
assert 0 == bowling([0] * 20)
assert 20 == bowling([1] * 20)
assert 80 == bowling([4] * 20)
assert 190 == bowling([9,1] * 10 + [9])
assert 300 == bowling([10] * 12)
assert 200 == bowling([10, 5,5] * 5 + [10])
#here we throw in frame 10 a strike and we add 1 and 0 to the final score
assert 11 == bowling([0,0] * 9 + [10,1,0])
#here we throw a strike in frame 9 (10 + 1) and in frame 10 we throw 1 and 0 so this adds up to 11 + 1
assert 12 == bowling([0,0] * 8 + [10, 1,0])
test_bowling()
Saturday, May 26, 2012
Using named and default arguments
This time I inlined some comments in REPL output just in case you wondered how that got there.
scala> def orderBigMacMenu(drink: String = "Coke", withMayo: Boolean = true): String = {
| /** the meal gets prepared **/
| "You ordered BigMac menu with " + drink + (if (withMayo) " with " else " without ") + "mayonaise"
| }
orderBigMacMenu: (drink: String, withMayo: Boolean)String
scala>
scala>
scala> orderBigMacMenu("drinkA", false)
res4: String = You ordered BigMac menu with drinkA without mayonaise
scala> orderBigMacMenu("drinkB", true)
res5: String = You ordered BigMac menu with drinkB with mayonaise
/** code below works because compiler tries to use arguments provided from left to right
and "drinkc" is of type String **/
scala> orderBigMacMenu("drinkC")
res6: String = You ordered BigMac menu with drinkC with mayonaise
/** code below fails because compiler tries to use arguments provided from left to right
and "true" is of type Boolean **/
scala> orderBigMacMenu(true)
:9: error: type mismatch;
found : Boolean(true)
required: String
Error occurred in an application involving default arguments.
orderBigMacMenu(true)
/** When using named arguments, there is no issue. You can even swap them around if you want to. **/
scala> orderBigMacMenu(withMayo=false)
res8: String = You ordered BigMac menu with Coke without mayonaise
scala> orderBigMacMenu(withMayo=false, drink="Beer")
res9: String = You ordered BigMac menu with Beer without mayonaise
Friday, May 25, 2012
Importance of immutability for concurrency
Let's take a look at how much performance we may gain when applying immutability to our code. I will let the code snippets speak for themselves. Below a generic trait with 3 methods of which two are important: the find and save method.
Now we create a mutable and immutable store (both insert key-value pairs (index mapped to fibonacci number) in respectively a mutable and immutable hashmap.
Below a very simple profiler which will give some idea of how long both stores will take to do a lot of calls to find and save.
package robbypelssers.concurrency
trait Store[K,V] {
def find(k: K): Option[V]
def save(v: V): Unit
def getName(): String
}
Now we create a mutable and immutable store (both insert key-value pairs (index mapped to fibonacci number) in respectively a mutable and immutable hashmap.
package robbypelssers.concurrency
import collection.mutable.HashMap
class MutableFibonacciStore extends Store[Int, Int] with Fibonacci {
val store = new HashMap[Int, Int]
def find(n: Int): Option[Int] = synchronized(store.get(n))
def save(n: Int): Unit = synchronized(store.put(n, fibonacci(n)))
def getName() = "MutableFibonacciStore"
}
package robbypelssers.concurrency
import collection.immutable.HashMap
class ImmutableFibonacciStore extends Store[Int, Int] with Fibonacci {
var store = new HashMap[Int,Int]
def find(n: Int): Option[Int] = store.get(n)
def save(n: Int): Unit = synchronized(store = store + ((n, fibonacci(n))))
def getName() = "ImmutableFibonacciStore"
}
package robbypelssers.concurrency
trait Fibonacci {
def fibonacci(n: Int): Int = n match {
case 0 => 0
case 1 => 1
case _ => fibonacci(n-2) + fibonacci(n-1)
}
}
Below a very simple profiler which will give some idea of how long both stores will take to do a lot of calls to find and save.
package robbypelssers.concurrency
import java.util.Date
object Profiler {
/**
* profile method works like an Around Advice in AspectOriented programming
*/
def profile[T](body: => T) = {
val start = new Date()
val result = body
val end = new Date()
println("Execution took " + (end.getTime() - start.getTime()) / 1000 + " seconds")
result
}
}
package robbypelssers.concurrency
import java.util.Date
import Profiler._
object StoreConcurrencyTest extends App {
val immutableStore = new ImmutableFibonacciStore()
val mutableStore = new MutableFibonacciStore()
val numberOfOperations = 40
def useStore(store: Store[Int,Int], n: Int) {
println("using " + store.getName())
for (x:Int <- 1 to n) {
store.save(x)
}
for (x:Int <- 1 to n) {
store.find(x)
}
}
profile(useStore(immutableStore, numberOfOperations))
profile(useStore(mutableStore, numberOfOperations))
}
using ImmutableFibonacciStore Execution took 1 seconds using MutableFibonacciStore Execution took 4 seconds
3 golden rules for implementing equality
- If two objects are equal, they should have the same hashCode.
- A hashCode computed for an object won’t change for the life of the object.
- When sending an object to another JVM, equality should be determined using attributes available in both JVMs.
Overloading primary constructor in Scala
scala> class Employee(employeeId: Int, isFulltimeEmployed: Boolean) {
| /** we can overload the primary constructor **/
| def this(employeeId: Int) = this(employeeId, true)
| override def toString() = employeeId + " works " + (if (isFulltimeEmployed) "fulltime" else "parttime")
| }
defined class Employee
scala>
scala> val john = new Employee(12545, true)
john: Employee = 12545 works fulltime
scala> val belinda = new Employee(32567, false)
belinda: Employee = 32567 works parttime
scala> val herman = new Employee(56784)
herman: Employee = 56784 works fulltime
Reference Immutatibility vs object immutability
It's important to understand the difference between immutability of reference and immutability of objects.
scala> class Car(brand:String, year:Int) {
| var fuelPercentage = 20
| def getBrand() = brand
| def getYear() = year
| def fillTank(): Unit = fuelPercentage = 100
| def getFuelPercentage() = fuelPercentage
| override def toString = getBrand() + " from " + year
| }
defined class Car
scala>
scala> val mycar = new Car("Citroen", 2011)
mycar: Car = Citroen from 2011
scala> mycar.getFuelPercentage()
res2: Int = 20
scala> mycar.fillTank()
scala> mycar.getFuelPercentage()
res4: Int = 100
scala> mycar = new Car("Audi", 2012)
:9: error: reassignment to val
mycar = new Car("Audi", 2012)
^
scala> var hiscar = new Car("Audi", 2012)
hiscar: Car = Audi from 2012
scala> hiscar = new Car("Porsche", 2012)
hiscar: Car = Porsche from 2012
scala>
As you can see mycar is an immutable reference because we used the 'val' keyword. hiscar on the other hand is a mutable reference due to using the keyword 'var'. Both car objects are mutable. We can fill the tank up and the fuelpercentage will change.
Scala promotes expression oriented programming
Scala promotes immutability and expression-oriented programming. Most scala constructs return a value as we will come to see in following examples.
First let's take a look at type matching:
Next let's take a look regular pattern matching using Fibonacci as an example
Also the if-else construct returns a value
As a rule of thumb, just remember that you can omit the return statement. Scala will return the last expression in a block statement.
First let's take a look at type matching:
scala> def replyMessage(message:Any): Any = message match {
| case s:String => "Your message is the string " + s
| case i:Int => "Your message is the number " + i
| case _ => "Your message is of unknown type"
| }
replyMessage: (message: Any)Any
scala> replyMessage("Hello world")
res1: Any = Your message is the string Hello world
scala> replyMessage(5)
res2: Any = Your message is the number 5
scala> replyMessage(true)
res3: Any = Your message is of unknown type
Next let's take a look regular pattern matching using Fibonacci as an example
scala> def fibonacci(n: Int): Int = n match {
| case 0 => 0
| case 1 => 1
| case _ => fibonacci(n-2) + fibonacci(n-1)
| }
fibonacci: (n: Int)Int
scala>
scala> (for (n <- 0 to 10) yield fibonacci(n)).toList
res15: List[Int] = List(0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55)
Also the if-else construct returns a value
scala> val fname = "Robby"
fname: java.lang.String = Robby
scala> if (fname.startsWith("B")) "Your name does start with B" else "Hey Mr R"
res19: java.lang.String = Hey Mr R
As a rule of thumb, just remember that you can omit the return statement. Scala will return the last expression in a block statement.
scala> def someComplexMethod(n: Int): Int = {
| val x = "blabla"
| n * 2
| }
someComplexMethod: (n: Int)Int
scala>
scala> someComplexMethod(3)
res22: Int = 6
Using Scala interpreter from Eclipse
It is possible to use the Scala interpreter from most IDE's. Here is a short description how to use it in Eclipse. You will first need to create a new Scala project however.
- Window => Open perspective => Scala
- Create new Scala project
- Window => Show view => Scala interpreter
- Select the newly created project and press ok
- Start typing any expression in the Evaluate textbox and press enter
Scala project management and continuous compilation
There are a few ways to manage your Scala projects among to name a few:
Expect more articles about usage some time in the future.
Expect more articles about usage some time in the future.
How to workaround Class and Companion object in REPL
As you can see below the companion object can't access the private variable from the Class Person. This is due to a different scope. To fix it you will need to define a container (can be called something else by the way).
Important side note!! As of Scala 2.9.x you can use the :paste command and everything will work normally.
scala> class Person {
| private var firstname = "Robby"
| private var lastname = "Pelssers"
| }
defined class Person
scala> object Person {
| def getFirstName(p: Person) = p.firstname
| }
:9: error: variable firstname in class Person cannot be accessed in Person
def getFirstName(p: Person) = p.firstname
scala> object container {
| class Person {
| private var firstname = "Robby"
| private var lastname = "Pelssers"
| }
|
| object Person {
| def getFirstName(p: Person) = p.firstname
| }
| }
defined module container
scala> import container.Person
import container.Person
scala> val person = new Person()
person: container.Person = container$Person@1a71d29a
scala> Person.getFirstName(person)
res0: java.lang.String = Robby
Important side note!! As of Scala 2.9.x you can use the :paste command and everything will work normally.
scala> :paste
// Entering paste mode (ctrl-D to finish)
class Person {
private var firstname = "Robby"
private var lastname = "Pelssers"
}
object Person {
def getFirstName(p: Person) = p.firstname
}
// Exiting paste mode, now interpreting.
defined class Person
defined module Person
scala> Person.getFirstName(new Person())
res0: java.lang.String = Robby
Scala REPL is smart
Apparently the REPL is smart enough to figure out when you declare a function as it waits for the closing bracket to return. It auto-injects the | for you as you enter new lines. I read that there are a few use cases it can't handle very well yet like defining a class and companion object. Once I've tested this with current version of the REPL I will blog about these cases.
scala> def addTwo(x: Int) = {
| x + 2
| }
addTwo: (x: Int)Int
scala> addTwo(5)
res5: Int = 7
scala>
Playing with the Scala REPL
The first thing you want to verify is that Scala is installed correctly. During installation you get asked if Scala should be added to your path, make sure to do so. If you're working on windows like me, you can now open the command line. Press Start button and type cmd into the 'Search programs and files'. This will open a DOS shell. Next type scala and the REPL (Read Eval Print Loop) will start.
C:\Users\nxp10009>scala Welcome to Scala version 2.9.2 (Java HotSpot(TM) 64-Bit Server VM, Java 1.6.0_30). Type in expressions to have them evaluated. Type :help for more information. scala> "Hello" res0: java.lang.String = Hello scala> "Hello".filter(_ != 'l') res1: String = Heo scala> 5 res2: Int = 5
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