# Revision history [back]

### Infinity point how it defined - Secp256K Curve

I have problem with point infinity. How to solve it? Curve Secp256k

# A Python3 program to check if a given point
# lies inside a given polygon
#
# for explanation of functions onSegment(),
# orientation() and doIntersect()

# Define Infinite (Using INT_MAX
# caused overflow problems)
INT_MAX = 9000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

# Given three colinear points p, q, r,
# the function checks if point q lies
# on line segment 'pr'
def onSegment(p:tuple, q:tuple, r:tuple) -> bool:

if ((q <= max(p, r)) &
(q >= min(p, r)) &
(q <= max(p, r)) &
(q >= min(p, r))):
return True

return False

# To find orientation of ordered triplet (p, q, r).
# The function returns following values
# 0 --> p, q and r are colinear
# 1 --> Clockwise
# 2 --> Counterclockwise
def orientation(p:tuple, q:tuple, r:tuple) -> int:

val = (((q - p) *
(r - q)) -
((q - p) *
(r - q)))

if val == 0:
return 0
if val > 0:
return 1 # Collinear
else:
return 2 # Clock or counterclock

def doIntersect(p1, q1, p2, q2):

# Find the four orientations needed for
# general and special cases
o1 = orientation(p1, q1, p2)
o2 = orientation(p1, q1, q2)
o3 = orientation(p2, q2, p1)
o4 = orientation(p2, q2, q1)

# General case
if (o1 != o2) and (o3 != o4):
return True

# Special Cases
# p1, q1 and p2 are colinear and
# p2 lies on segment p1q1
if (o1 == 0) and (onSegment(p1, p2, q1)):
return True

# p1, q1 and p2 are colinear and
# q2 lies on segment p1q1
if (o2 == 0) and (onSegment(p1, q2, q1)):
return True

# p2, q2 and p1 are colinear and
# p1 lies on segment p2q2
if (o3 == 0) and (onSegment(p2, p1, q2)):
return True

# p2, q2 and q1 are colinear and
# q1 lies on segment p2q2
if (o4 == 0) and (onSegment(p2, q1, q2)):
return True

return False

# Returns true if the point p lies
# inside the polygon[] with n vertices
def is_inside_polygon(points:list, p:tuple) -> bool:

n = len(points)

# There must be at least 3 vertices
# in polygon
if n < 3:
return False

# Create a point for line segment
# from p to infinite
extreme = (INT_MAX, p)    # here point infinity - how to solve it? i
count = i = 0

while True:
next = (i + 1) % n

# Check if the line segment from 'p' to
# 'extreme' intersects with the line
# segment from 'polygon[i]' to 'polygon[next]'
if (doIntersect(points[i],
points[next],
p, extreme)):

# If the point 'p' is colinear with line
# segment 'i-next', then check if it lies
# on segment. If it lies, return true, otherwise false
if orientation(points[i], p,
points[next]) == 0:
return onSegment(points[i], p,
points[next])

count += 1

i = next

if (i == 0):
break

# Return true if count is odd, false otherwise
return (count % 2 == 1)

# Driver code
if __name__ == '__main__':
#i: 1
x= 0x79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798
#i:  2
x2= 0xc6047f9441ed7d6d3045406e95c07cd85c778e4b8cef3ca7abac09b95c709ee5
y2= 0x1ae168fea63dc339a3c58419466ceaeef7f632653266d0e1236431a950cfe52a
#i:  3
x3= 0xf9308a019258c31049344f85f89d5229b531c845836f99b08601f113bce036f9
y3= 0x388f7b0f632de8140fe337e62a37f3566500a99934c2231b6cb9fd7584b8e672
#i:  4
x4= 0xe493dbf1c10d80f3581e4904930b1404cc6c13900ee0758474fa94abe8c4cd13
y4= 0x51ed993ea0d455b75642e2098ea51448d967ae33bfbdfe40cfe97bdc47739922
#i:  5
x5= 0x2f8bde4d1a07209355b4a7250a5c5128e88b84bddc619ab7cba8d569b240efe4
y5= 0xd8ac222636e5e3d6d4dba9dda6c9c426f788271bab0d6840dca87d3aa6ac62d6
#i:  6
x6= 0xfff97bd5755eeea420453a14355235d382f6472f8568a18b2f057a1460297556
y6= 0xae12777aacfbb620f3be96017f45c560de80f0f6518fe4a03c870c36b075f297

polygon1 = [ (x,y), (x2,y2), (x4, y4), (x5, x5),(x6,y6) ]

p = (x3, y3)
if (is_inside_polygon(points = polygon1, p = p)):
print ('Yes')
else:
print ('No')


### Infinity point how it defined - Secp256K Curve

I have problem with point infinity. How to solve it? Curve Secp256k

# A Python3 program to check if a given point
# lies inside a given polygon
#
# for explanation of functions onSegment(),
# orientation() and doIntersect()

# Define Infinite (Using INT_MAX
# caused overflow problems)
INT_MAX = 9000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

# Given three colinear points p, q, r,
# the function checks if point q lies
# on line segment 'pr'
def onSegment(p:tuple, q:tuple, r:tuple) -> bool:

if ((q <= max(p, r)) &
(q >= min(p, r)) &
(q <= max(p, r)) &
(q >= min(p, r))):
return True

return False

# To find orientation of ordered triplet (p, q, r).
# The function returns following values
# 0 --> p, q and r are colinear
# 1 --> Clockwise
# 2 --> Counterclockwise
def orientation(p:tuple, q:tuple, r:tuple) -> int:

val = (((q - p) *
(r - q)) -
((q - p) *
(r - q)))

if val == 0:
return 0
if val > 0:
return 1 # Collinear
else:
return 2 # Clock or counterclock

def doIntersect(p1, q1, p2, q2):

# Find the four orientations needed for
# general and special cases
o1 = orientation(p1, q1, p2)
o2 = orientation(p1, q1, q2)
o3 = orientation(p2, q2, p1)
o4 = orientation(p2, q2, q1)

# General case
if (o1 != o2) and (o3 != o4):
return True

# Special Cases
# p1, q1 and p2 are colinear and
# p2 lies on segment p1q1
if (o1 == 0) and (onSegment(p1, p2, q1)):
return True

# p1, q1 and p2 are colinear and
# q2 lies on segment p1q1
if (o2 == 0) and (onSegment(p1, q2, q1)):
return True

# p2, q2 and p1 are colinear and
# p1 lies on segment p2q2
if (o3 == 0) and (onSegment(p2, p1, q2)):
return True

# p2, q2 and q1 are colinear and
# q1 lies on segment p2q2
if (o4 == 0) and (onSegment(p2, q1, q2)):
return True

return False

# Returns true if the point p lies
# inside the polygon[] with n vertices
def is_inside_polygon(points:list, p:tuple) -> bool:

n = len(points)

# There must be at least 3 vertices
# in polygon
if n < 3:
return False

# Create a point for line segment
# from p to infinite
extreme = (INT_MAX, p)    # **


# here point infinity - how to solve it? it?

** i count = i = 0 0

    while True:
next = (i + 1) % n

# Check if the line segment from 'p' to
# 'extreme' intersects with the line
# segment from 'polygon[i]' to 'polygon[next]'
if (doIntersect(points[i],
points[next],
p, extreme)):

# If the point 'p' is colinear with line
# segment 'i-next', then check if it lies
# on segment. If it lies, return true, otherwise false
if orientation(points[i], p,
points[next]) == 0:
return onSegment(points[i], p,
points[next])

count += 1

i = next

if (i == 0):
break

# Return true if count is odd, false otherwise
return (count % 2 == 1)

# Driver code
if __name__ == '__main__':
#i: 1
x= 0x79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798
#i:  2
x2= 0xc6047f9441ed7d6d3045406e95c07cd85c778e4b8cef3ca7abac09b95c709ee5
y2= 0x1ae168fea63dc339a3c58419466ceaeef7f632653266d0e1236431a950cfe52a
#i:  3
x3= 0xf9308a019258c31049344f85f89d5229b531c845836f99b08601f113bce036f9
y3= 0x388f7b0f632de8140fe337e62a37f3566500a99934c2231b6cb9fd7584b8e672
#i:  4
x4= 0xe493dbf1c10d80f3581e4904930b1404cc6c13900ee0758474fa94abe8c4cd13
y4= 0x51ed993ea0d455b75642e2098ea51448d967ae33bfbdfe40cfe97bdc47739922
#i:  5
x5= 0x2f8bde4d1a07209355b4a7250a5c5128e88b84bddc619ab7cba8d569b240efe4
y5= 0xd8ac222636e5e3d6d4dba9dda6c9c426f788271bab0d6840dca87d3aa6ac62d6
#i:  6
x6= 0xfff97bd5755eeea420453a14355235d382f6472f8568a18b2f057a1460297556
y6= 0xae12777aacfbb620f3be96017f45c560de80f0f6518fe4a03c870c36b075f297

polygon1 = [ (x,y), (x2,y2), (x4, y4), (x5, x5),(x6,y6) ]

p = (x3, y3)
if (is_inside_polygon(points = polygon1, p = p)):
print ('Yes')
else:
print ('No')


### Infinity point how it defined - Secp256K Curve

I have problem with point infinity. How to solve it? Curve Secp256k

# A Python3 program to check if a given point
# lies inside a given polygon
#
# for explanation of functions onSegment(),
# orientation() and doIntersect()

# Define Infinite (Using INT_MAX
# caused overflow problems)
INT_MAX = 9000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

# Given three colinear points p, q, r,
# the function checks if point q lies
# on line segment 'pr'
def onSegment(p:tuple, q:tuple, r:tuple) -> bool:

if ((q <= max(p, r)) &
(q >= min(p, r)) &
(q <= max(p, r)) &
(q >= min(p, r))):
return True

return False

# To find orientation of ordered triplet (p, q, r).
# The function returns following values
# 0 --> p, q and r are colinear
# 1 --> Clockwise
# 2 --> Counterclockwise
def orientation(p:tuple, q:tuple, r:tuple) -> int:

val = (((q - p) *
(r - q)) -
((q - p) *
(r - q)))

if val == 0:
return 0
if val > 0:
return 1 # Collinear
else:
return 2 # Clock or counterclock

def doIntersect(p1, q1, p2, q2):

# Find the four orientations needed for
# general and special cases
o1 = orientation(p1, q1, p2)
o2 = orientation(p1, q1, q2)
o3 = orientation(p2, q2, p1)
o4 = orientation(p2, q2, q1)

# General case
if (o1 != o2) and (o3 != o4):
return True

# Special Cases
# p1, q1 and p2 are colinear and
# p2 lies on segment p1q1
if (o1 == 0) and (onSegment(p1, p2, q1)):
return True

# p1, q1 and p2 are colinear and
# q2 lies on segment p1q1
if (o2 == 0) and (onSegment(p1, q2, q1)):
return True

# p2, q2 and p1 are colinear and
# p1 lies on segment p2q2
if (o3 == 0) and (onSegment(p2, p1, q2)):
return True

# p2, q2 and q1 are colinear and
# q1 lies on segment p2q2
if (o4 == 0) and (onSegment(p2, q1, q2)):
return True

return False

# Returns true if the point p lies
# inside the polygon[] with n vertices
def is_inside_polygon(points:list, p:tuple) -> bool:

n = len(points)

# There must be at least 3 vertices
# in polygon
if n < 3:
return False

# **


# Create a point for line segment segment

    # from p to infinite
infinite - hot defined this point?


** extreme = (INT_MAX, p) **

# here point infinity - how to solve it?

** i
count = i = 0

    while True:
next = (i + 1) % n

# Check if the line segment from 'p' to
# 'extreme' intersects with the line
# segment from 'polygon[i]' to 'polygon[next]'
if (doIntersect(points[i],
points[next],
p, extreme)):

# If the point 'p' is colinear with line
# segment 'i-next', then check if it lies
# on segment. If it lies, return true, otherwise false
if orientation(points[i], p,
points[next]) == 0:
return onSegment(points[i], p,
points[next])

count += 1

i = next

if (i == 0):
break

# Return true if count is odd, false otherwise
return (count % 2 == 1)

# Driver code
if __name__ == '__main__':
#i: 1
x= 0x79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798
#i:  2
x2= 0xc6047f9441ed7d6d3045406e95c07cd85c778e4b8cef3ca7abac09b95c709ee5
y2= 0x1ae168fea63dc339a3c58419466ceaeef7f632653266d0e1236431a950cfe52a
#i:  3
x3= 0xf9308a019258c31049344f85f89d5229b531c845836f99b08601f113bce036f9
y3= 0x388f7b0f632de8140fe337e62a37f3566500a99934c2231b6cb9fd7584b8e672
#i:  4
x4= 0xe493dbf1c10d80f3581e4904930b1404cc6c13900ee0758474fa94abe8c4cd13
y4= 0x51ed993ea0d455b75642e2098ea51448d967ae33bfbdfe40cfe97bdc47739922
#i:  5
x5= 0x2f8bde4d1a07209355b4a7250a5c5128e88b84bddc619ab7cba8d569b240efe4
y5= 0xd8ac222636e5e3d6d4dba9dda6c9c426f788271bab0d6840dca87d3aa6ac62d6
#i:  6
x6= 0xfff97bd5755eeea420453a14355235d382f6472f8568a18b2f057a1460297556
y6= 0xae12777aacfbb620f3be96017f45c560de80f0f6518fe4a03c870c36b075f297

polygon1 = [ (x,y), (x2,y2), (x4, y4), (x5, x5),(x6,y6) ]

p = (x3, y3)
if (is_inside_polygon(points = polygon1, p = p)):
print ('Yes')
else:
print ('No') 4 None

### Infinity point how it defined - Secp256K Curve

I have problem with point infinity. How to solve it? it?

Curve Secp256k Secp256k

# A Python3 program to check if a given point
point
# lies inside a given polygon
polygon
#
# for explanation of functions onSegment(),
onSegment(),
# orientation() and doIntersect()
doIntersect()

# Define Infinite Infinity (Using INT_MAX
INT_MAX
# caused overflow problems)
INT_MAX = 9000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

# Given three colinear points p, q, r,
# the function checks if def onSegment(p: tuple, q: tuple, r: tuple) -> bool:
r"""
Check whether point q lies
# on line segment 'pr'
def onSegment(p:tuple, q:tuple, r:tuple) -> bool:

if ((q 'pr'
"""
if (q <= max(p, r)) &
(q r) and
q >= min(p, r)) &
(q r) and
q <= max(p, r)) &
(q r) and
q >= min(p, r))):
r)):
return True
return False

# To find def orientation(p: tuple, q: tuple, r: tuple) -> int:
r"""
Return the orientation of the ordered triplet triple (p, q, r).
# r).

The function returns following values
# orientation is encoded as:

- 0 --> if p, q and r are colinear
# collinear
- 1 --> Clockwise
# if they turn clockwise
- 2 --> Counterclockwise
def orientation(p:tuple, q:tuple, r:tuple) -> int:

if they turn counterclockwise
"""
val = (((q ((q - p) *
* (r - q)) -
((q q)
- (q - p) *
* (r - q)))

q))
if val == 0:
return 0
if val > 0:
return 1 # Collinear
else:
return 2 # Clock or counterclock
1
return 2

def doIntersect(p1, q1, p2, q2):
# Find the four orientations needed for
for
# general and special cases
cases
o1 = orientation(p1, q1, p2)
o2 = orientation(p1, q1, q2)
o3 = orientation(p2, q2, p1)
o4 = orientation(p2, q2, q1)

# General case
if (o1 o1 != o2) o2 and (o3 o3 != o4):
o4:
return True

# Special Cases
cases
# p1, q1 and p2 are colinear and
collinear and
# p2 lies on segment p1q1
if (o1 p1q1
if o1 == 0) 0 and (onSegment(p1, onSegment(p1, p2, q1)):
q1):
return True

# p1, q1 and p2 are colinear and
collinear and
# q2 lies on segment p1q1
if (o2 p1q1
if o2 == 0) 0 and (onSegment(p1, onSegment(p1, q2, q1)):
q1):
return True

# p2, q2 and p1 are colinear and
collinear and
# p1 lies on segment p2q2
if (o3 p2q2
if o3 == 0) 0 and (onSegment(p2, onSegment(p2, p1, q2)):
q2):
return True

# p2, q2 and q1 are colinear and
collinear and
# q1 lies on segment p2q2
if (o4 p2q2
if o4 == 0) 0 and (onSegment(p2, onSegment(p2, q1, q2)):
q2):
return True

return False

# Returns true if def is_inside_polygon(points: list, p: tuple) -> bool:
r"""
Check whether the point p p lies
# inside inside
the polygon[] with n vertices
def is_inside_polygon(points:list, p:tuple) -> bool:

polygon defined by points.
"""
n = len(points)

# There must be at least 3 vertices
# in polygon
if n < 3:
return False

**


# # Create a point for line segment

segment
# from p to infinite infinity - hot defined how to define this point?


** extreme = (INT_MAX, p)
p) count = i = 0

0

while True:
next = (i + 1) % n

# Check if the line segment from 'p' to
to
# 'extreme' intersects with the line
line
# segment from 'polygon[i]' to 'polygon[next]'
if (doIntersect(points[i],
points[next],
'polygon[next]'
if doIntersect(points[i],
points[next],
p, extreme)):
extreme):

# If the point 'p' is colinear collinear with line
line
# segment 'i-next', then check if it lies
lies
# on segment. If it lies, return true, otherwise false
false
if orientation(points[i], p,
p,
points[next]) == 0:
return onSegment(points[i], p,
p,
points[next])

count += 1

i = next

if (i i == 0):
0:
break

# Return true True if count is odd, false otherwise
return (count False otherwise
return count % 2 == 1)
1

# Driver code
if __name__ == '__main__':
#i: 1
x=     # i: 1
x = 0x79be667ef9dcbbac55a06295ce870b07029bfcdb2dce28d959f2815b16f81798
y= y = 0x483ada7726a3c4655da4fbfc0e1108a8fd17b448a68554199c47d08ffb10d4b8
#i:      # i: 2
x2= x2 = 0xc6047f9441ed7d6d3045406e95c07cd85c778e4b8cef3ca7abac09b95c709ee5
y2= y2 = 0x1ae168fea63dc339a3c58419466ceaeef7f632653266d0e1236431a950cfe52a
#i:      # i: 3
x3= x3 = 0xf9308a019258c31049344f85f89d5229b531c845836f99b08601f113bce036f9
y3= y3 = 0x388f7b0f632de8140fe337e62a37f3566500a99934c2231b6cb9fd7584b8e672
#i:      # i: 4
x4= x4 = 0xe493dbf1c10d80f3581e4904930b1404cc6c13900ee0758474fa94abe8c4cd13
y4= y4 = 0x51ed993ea0d455b75642e2098ea51448d967ae33bfbdfe40cfe97bdc47739922
#i:      # i: 5
x5= x5 = 0x2f8bde4d1a07209355b4a7250a5c5128e88b84bddc619ab7cba8d569b240efe4
y5= y5 = 0xd8ac222636e5e3d6d4dba9dda6c9c426f788271bab0d6840dca87d3aa6ac62d6
#i:      # i: 6
x6= x6 = 0xfff97bd5755eeea420453a14355235d382f6472f8568a18b2f057a1460297556
y6= y6 = 0xae12777aacfbb620f3be96017f45c560de80f0f6518fe4a03c870c36b075f297

polygon1 = [ (x,y), (x2,y2), [(x, y), (x2, y2), (x4, y4), (x5, x5),(x6,y6) ]
x5), (x6, y6)]

p = (x3, y3)
if (is_inside_polygon(points = polygon1, p = p)):
is_inside_polygon(points=polygon1, p=p):
print ('Yes')
else:
print ('No')