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Pythagora (A)

p104 The goal of incremental development is to avoid long debugging. By Pythagorean Theorem, the distance between two points is: \[d = \sqrt{(x2-x1)^2 + (y2-y1)^2} \]

First step

The first step will be to define the function, with wrong answer but syntactically correct.
 
# Incremental development - Pythagorean Theorem
#
#    d = square( (x2-x1)^2 + (y2-y1)^2 ) 

# First: Define the function, with wrong answer but syntactically correct

def distance(x1, y1, x2, y2):
    return 0.0

assert distance(1, 2, 4, 6) == 0.0

Second

The next step is to find the differences x2-x1, y2-y1
 
# Incremental development - Pythagorean Theorem
#
#    d = square( (x2-x1)^2 + (y2-y1)^2 ) 
#
# Second: Find the differences x2-x1, y2-y1

def distance(x1, y1, x2, y2):
    dx = x2 - x1
    dy = y2 - y1
    return 0.0

assert distance(1, 2, 4, 6) == 0.0

Third

Then we compute the sum of squares.
 
# Incremental development - Pythagorean Theorem
#
#    d = square( (x2-x1)^2 + (y2-y1)^2 ) 
#
# Third: Compute the sum of squares

def distance(x1, y1, x2, y2):
    dx = x2 - x1
    dy = y2 - y1
    sum = dx**2 + dy**2
    return 0.0

assert distance(1, 2, 4, 6) == 0.0

Forth

Finally, you can use math.sqrt() to compute the result.
 
# Incremental development - Pythagorean Theorem 
#
#    d = square( (x2-x1)^2 + (y2-y1)^2 )
#
# Third: Use math.sqrt() to compute the result

import math

def distance(x1, y1, x2, y2):
    dx = x2 - x1
    dy = y2 - y1
    sum = dx**2 + dy**2
    return math.sqrt(sum)

d = distance(1, 2, 4, 6)

assert distance(1, 2, 4, 6) == 5.0

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Last update: 13 days ago
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