Php
Pythagora (A)
p104 The goal of incremental development is to avoid long debugging. By Pythagorean Theorem, the distance betwee 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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