ML homework 3

count_squares.m

%###################################################################
% Homework #3. Number of features in a 24x24 pixel image
% Student: Pablo J. Estrada. ID: 2014-25245
%###################################################################

% Here are our patterns. We need only specify their dimensions.
patt_x = [2,1,2,3,1]
patt_y = [1,2,2,1,3]

function res = count(x, y, patt_len,patt_x,patt_y)
    res = 0;
    if x == 0 || y == 0
        res = 0;
    else
        for i=1:x
            for j = 1:y
                for k = 1:patt_len
                    if mod(i,patt_x(k)) == 0 && mod(j,patt_y(k)) == 0
                        res = res + 1;
                    end
                end
            end
        end
    end
end

function r = total_count(x,y,patt_len,patt_x,patt_y)
    r = 0;
    if x == 0 || y == 0
        r = 0;
    else
        for i=0:x-1
            r = r + count(x-i,y,patt_len,patt_x,patt_y);
        end
        r = r + total_count(x,y-1,patt_len,patt_x,patt_y);
    end
end

res = total_count(24,24,4,patt_x,patt_y);
disp("Total features with abcd patterns: ")
res
res = total_count(24,24,5,patt_x,patt_y);
disp("Total features with abcde patterns: ")
res

%#############
% RUN RESULTS
%#############
% >>> 
% Total of features with 'abcd' patterns: 134,736
% Total of features with 'abcde' patterns: 162,336
%#############

count_squares.py

####################################################################
# Homework #3. Number of features in a 24x24 pixel image
# Student: Pablo J. Estrada. ID: 2014-25245
####################################################################

# Here are our patterns. We need only specify their dimensions.
allpatterns = {'a':[2,1], 'b':[1,2], 'c':[2,2],'d':[3,1],'e':[1,3]}

# We get only our 'abcd' patterns. Exclude pattern 'e'.
abcdpatterns = {key : allpatterns[key] 
                for key in allpatterns 
                if key in {'a','b','c','d'}}

# This function returns the count of possible patterns with their
# upper left corner grounded on the origin.
def count(x,y,patterns):
    if x == 0 or y == 0:
        return 0
    elems = ( 1
              if (i+1) % patterns[k][0] == 0 and (j+1) % patterns[k][1] == 0 
              else 0
              for i in range(x) for j in range(y) for k in patterns)
    return sum(elems)

# This function uses the function 'count' to scan for 
# and count all the possible patterns through the whole image
def total_count(x,y,patterns):
    if x == 0 or y == 0:
        return 0
    sum = 0

    # We count the whole upper row; and add the patterns found
    for i in range(x):
        sum += count(x-i,y,patterns)

    # And then we go down to the next row, start over; and add
    # the patterns found
    sum += total_count(x,y-1,patterns);

    return sum

print("Total of features with 'abcd' patterns: " + 
      "{:,}".format(total_count(24,24,abcdpatterns)))
print("Total of features with 'abcde' patterns: " + 
      "{:,}".format(total_count(24,24,allpatterns)))

##############
# RUN RESULTS
##############
# >>> 
# Total of features with 'abcd' patterns: 134,736
# Total of features with 'abcde' patterns: 162,336
##############