During Python Brasil [12] in FLorianópolis - Brazil I took a cab and the taxi driver, called Luiz, listened me and my friends talking about computer science stuff. Then he look at us with a smiling face and said: "Are you from google generation?".
We obviosly answered: "hell yeah, we are! rsrs" Then he said: "Let's see if you can solve this problem...", and he described the Dominó problem. Luiz is a former economist. Nowadays he is a taxi driver that drops puzzles in his passengers!
There is a table game called Dominó. This game pieces are divided in two parts. Both parts has a number between 0 and 6. Each pair of numbers never repeat and there is no inverse pieace with this numbers.
Based on that, how many pieces has an odd number as a result of the sum of its two parts?
First of all, we freaked out and try to run a brute force algorithm together. One of my friends started saying loud the combinations of numbers starting from the lowest to the highest pairs. Another friend calculates if the sum of numbers were odd. And finally, I was counting the number of odd resulted numbers.
Basically, it did not work!
At this point, the taxi driver was in a very funny work day! I picked up a paper and a pen inside my backpack and started drawing the possibilities matrix as follows:
As can be seen in the above figure, we have a matrix of dominó pieces combinations. Each position represents a piece. Each coordinate pair (x, y), are the top and bottom parts of a piece.
The values inside matrix cells represents booleans values of the sum of the two parts (x + y). If the sum is odd, then the value is 1. Otherwise the value is 0. Through that, is possible to count how many sums resulted in an odd number.
The first observation about the problem was that if what we were looking for was odd resulted numbers, then pieces that have the same number as the top and bottom values must be excluded from counting. The sum of two even numbers are always an even number. Also, the sum of two odd numbers are always an even number. Based on that, looking at the possibilities matrix, we can exclude the matrix determinant values from counting:
Another property of the problem is that each pair of number only exist once. So, we do not have to iterate over the bottom part of the matrix. For that, we remove the mirrored values from the possibility matrix:
Knowing all of those constraints, to solve the problem we have to iterate over the top part of the matrix excluding the determinant. On each iteration step, we check if the sum is odd and count it. Otherwise we continue to the next iteration step until there is no steps left. The output result is the total number of odd resulted numbers.
At this point, Luiz was having great times!
After drawing the matrix and observing the problem we could count the odd sums and answer the problem correctly. Also, we find a pattern that makes the counting proccess unnecessary and by applying a calculation the result was found independent of the matrix size. The following paragraphs will discuss these solutions.
Using a brute force approach we implemented the following algorithm to calculate the number of odd results:
total = 0 # Total odd numbers
rows = cols = len(matrix)
increment = 1 # Starts from 1 because determinant values are always even
for i in range(rows):
for j in range(increment, cols):
if (i + j) % 2: # If sum is odd we increment total
total += 1
increment += 1
return totalAs a metric of the implementation, we tried to increment the matrix size assuming that could exist a dominó with more pieces. As the input matrix grows, we noted a pattern of results. For any matrix size, to solve that problem we can just apply a simple formula to get the total number of odd resulted values.
f(x) = odd_numbers * even_numbers
Based on that, we create a function to solve the problem using this formula and compare it with the brute force approach result to check correctness.
n_even = n_odd = int(n / 2)
if n % 2: # Matrix size is odd?
n_even = int(n / 2) + 1
# Domino problem is soved by simple multiplication of even and odd number
# inside a matrix of size n
return int(n_even * n_odd)To verify both implementation we create tests scripts that runs the same matrix for each implementation. There is also a script that computes 300 matrices with sizes starting from 1 to 300. For each computation a time execution log is written to future analysis.
The brute force algorithm takes n^2/2 - n iteration to complete. By the Big O Notation we have:
Time complexity: O(n^2)
Storage complexity: n * m = O(n^2) , because n = m. It is a square matrix
The formula based algorithm takes a constant time to execute. As it do not iterate over the matrix, it just takes the time that the running machine needs to do the calculation. So, for the constant algorithm we have:
Time complexity: O(1)
Storage complexity: - Irrelevant
The following image shows the time execution complexity of each implemented algorithm. As can be seen, the math function based one runs in a constant way and the iterative goes exponentially as the input matrix grows.
As the input matrix grows, the brute force solution tends to be slower. In the other hand the constant approach does not have impact in its time execution as the matrix grows.
This project aims to demonstrate that we can look at most simple problems and see and apply computer science on it. On the everyday tasks like taking a cab, a bus or at lunch we can come across nice problems. Toy problems are always a good place to start a study!
By the way, the result is 12.
The project has a Makefile to rule computation, plot and cleaning. The following target rules exists:
- compute - Computes execution times for implemented algorithms
- plot - Run GnuPlot to generate the computation analysis
- show - Open resulted image using eog
- test - Run tests using nose
- clean - Removes log and image files
The project uses Python 3.5 for computation, GNUPlot to generate graphs and Python Nose for tests.
Run the following command to compute and plot execution analysis:
$> make clean compute plot showWe use Python nose to run tests:
$> make testWe are not stringent with contributions. Just do a fork, make your modifications, write tests for it and send me a Pull Request. It will be very welcome!
Written by Gustavo Pantuza [email protected]
The following solution is for those who like One Line of Code programming. Kudos for Christian Clauss for contributing with the following code:
# one liner...
sum((a + b) % 2 for a in range(7) for b in range(a, 7))
# show your work...
>>> dominos = [[(a, b) for a in range(b+1)] for b in range(7)]
>>> print('\n'.join(str(row) for row in dominos))
[(0, 0)]
[(0, 1), (1, 1)]
[(0, 2), (1, 2), (2, 2)]
[(0, 3), (1, 3), (2, 3), (3, 3)]
[(0, 4), (1, 4), (2, 4), (3, 4), (4, 4)]
[(0, 5), (1, 5), (2, 5), (3, 5), (4, 5), (5, 5)]
[(0, 6), (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6)]
>>> for row in dominos:
... for domino in row:
... print(sum(domino) % 2, end=' ')
... print()
...
0
1 0
0 1 0
1 0 1 0
0 1 0 1 0
1 0 1 0 1 0
0 1 0 1 0 1 0
>>> for row in dominos:
... print(sum(sum(domino) % 2 for domino in row))
...
0
1
1
2
2
3
3
>>> sum(sum(sum(domino) % 2 for domino in row) for row in dominos)
12