Closed. This question needs to be more focused. It is not currently accepting answers.
Want to improve this question? Update the question so it focuses on one problem only by editing this post.
Closed 6 years ago.
Improve this question
I have a problem simplified into the following:
Xn+1 = Xn + Yn
Yn+1= Yn + Zn
Zn+1= Zn+ Xn
I know the values of X0,Y0,Z0 to be equal to 1.
I want to tell python to find the values of X1,Y1,Z1 and then X2,Y2,Z2,...etc. Can anyone help me with that? I think I have to use a nested loop but I am not sure exactly how to go about it. Thanks!
Are we talking about something as simple as:
x, y, z = 1, 1, 1
for i in range(10):
print("X{i} = {x}, Y{i} = {y}, Z{i} = {z}".format(**locals()))
x, y, z = x + y, y + z, z + x
which doesn't seem right as the output is so uninteresting:
X0 = 1, Y0 = 1, Z0 = 1
X1 = 2, Y1 = 2, Z1 = 2
X2 = 4, Y2 = 4, Z2 = 4
X3 = 8, Y3 = 8, Z3 = 8
X4 = 16, Y4 = 16, Z4 = 16
X5 = 32, Y5 = 32, Z5 = 32
X6 = 64, Y6 = 64, Z6 = 64
X7 = 128, Y7 = 128, Z7 = 128
X8 = 256, Y8 = 256, Z8 = 256
X9 = 512, Y9 = 512, Z9 = 512
Below is the sample function to achieve it:
def solve_equation(n):
X = {0: 1}
Y = {0: 1}
Z = {0: 1}
for i in range(n):
print 'For n: ', i+1
X[i+1] = X[i] + Y[i]
Y[i+1] = Y[i] + Z[i]
Z[i+1] = Z[i] + X[i]
print 'X = ', X[i+1], ' Y = ', Y[i+1], ' Z = ', Z[i+1]
Sample run:
>>> solve_equation(3)
For n: 1
X = 2 Y = 2 Z = 2
For n: 2
X = 4 Y = 4 Z = 4
For n: 3
X = 8 Y = 8 Z = 8
You can use recursive functions. For a faster and better code you can consider using yield
def Cvalue(c,xyz , n):
if n == 0 :
res = c
else :
if xyz == 1: res = Cvalue(c,1, n-1) + Cvalue(c,2, n-1)
elif xyz == 2: res = Cvalue(c,2, n-1) + Cvalue(c,3, n-1)
elif xyz == 3: res = Cvalue(c,3, n-1) + Cvalue(c,1, n-1)
else: print("Error\n")
return res
def XYZvalues(x0, y0, z0, n):
x = Cvalue(x0,1, n)
y = Cvalue(y0,2, n)
z = Cvalue(z0,3, n)
return (x, y , z)
print(XYZvalues(1,1,1, 3))
Related
I'm trying to implement a simple elliptic curve encryption program but I can't get the expected output of doubling and adding a Point P till 12P .The curve equation isy^2 = x^3 +ax + b mod p. According to this site 3P = [10, 6] when P = [5, 1] while I get 3p = [10, 5]. The equations I use can be found on Wikipedia.
P = [5, 1]
prime = 17
a = 2
b = 2
def gcdExtended(a, b):
if a == 0:
return b, 0, 1
gcd, x1, y1 = gcdExtended(b % a, a)
x = y1 - (b // a) * x1
y = x1
return gcd, x, y
def double_point(point: list):
x = point[0]
y = point[1]
s = ((3*(x**2)+a) * (gcdExtended(2*y, prime)[1])) % prime
newx = (s**2 - x - x) % prime
newy = (s * (x - newx) - y) % prime
return [newx, newy]
def add_points(P: list, Q: list):
x1 = P[0]
y1 = P[1]
x2 = Q[0]
y2 = Q[1]
s = ((y2 - y1) * ((gcdExtended(x2-x1, prime))[1] % prime)) % prime
newx = (s**2 - x1 - x2) % prime
newy = (s * (x1 - newx) - y1) % prime
return [newx, newy]
Q = P
index = 2
while True:
if Q[0] == P[0] and Q[1] == P[1]:
print("doubling")
Q = double_point(P)
else:
print("adding")
Q = add_points(Q, P)
if index == 12 :
break
print(f"{index}P = {Q}")
index += 1
If the point [5,1] is added successively, the following sequence is obtained:
1P = [ 5, 1]
2P = [ 6, 3]
3P = [10, 6]
4P = [ 3, 1]
5P = [ 9, 16]
6P = [16, 13]
7P = [ 0, 6]
8P = [13, 7]
9P = [ 7, 6]
10P = [ 7, 11]
11P = [13, 10]
12P = [ 0, 11]
13P = [16, 4]
14P = [ 9, 1]
15P = [ 3, 16]
16P = [10, 11]
17P = [ 6, 14]
18P = [ 5, 16]
19P = point at infinity
This can be verified e.g. here.
The problem in the posted code is that the method to determine the modular inverse, gcdExtended(a, b), is only valid for positive a and b. While in double_point and add_points b has the value prime (= 17 > 0), a can take negative values.
gcdExtended generally returns wrong values for negative a:
The modular inverse of 5 or -12 is 7: 5 x 7 mod17 = 35 mod17 = 1 and 7 x (-12) mod17 = -84 mod17 = 85 mod17 = 1.
The gcdExtended returns for these values: gcdExtended(5, 17)[1] = 7 (which is true) and gcdExtended(-12, 17)[1] = -7 (which is false).
To allow negative values for a, e.g. the following methods can be defined, see here:
def sign(x):
return 1 if x >= 0 else -1
def gcdExtendedGeneralized(a, b):
gcd, x1, y1 = gcdExtended(abs(a), b)
return gcd, (sign(a) * x1) % b, y1 % b
Replacing gcdExtended with gcdExtendedGeneralized in double_point and add_points provides the correct values (note that the current implementation does not consider the point at infinity).
You interchanged P and Q in add_points. Also a small simplification in your calculation of s:
def add_points(P: list, Q: list):
x1 = P[0]
y1 = P[1]
x2 = Q[0]
y2 = Q[1]
#s = ((y2 - y1) * ((gcdExtended(x2-x1, prime))[1] % prime)) % prime
s = (y2-y1) * (gcdExtended(x2-x1, prime)[1] % prime)
newx = (s**2 - x1 - x2) % prime
newy = (s * (x1 - newx) - y1) % prime
return [newx, newy]
Q = P
index = 2
while True:
if Q[0] == P[0] and Q[1] == P[1]:
print("doubling")
Q = double_point(P)
else:
print("adding")
Q = add_points(P, Q)
if index == 12 :
break
print(f"{index}P = {Q}")
index += 1
which results in
doubling
2P = [6, 3]
adding
3P = [10, 6]
adding
4P = [3, 1]
adding
5P = [9, 16]
adding
6P = [16, 13]
adding
7P = [0, 6]
adding
8P = [13, 8]
adding
9P = [8, 7]
adding
10P = [8, 10]
adding
11P = [13, 9]
adding
Here is my dataframe,
df = pd.DataFrame({'Id': [102,103,104,303,305],'ExpG_Home':[1.8,1.5,1.6,1.8,2.9],
'ExpG_Away':[2.2,1.3,1.2,2.8,0.8],
'HomeG_Time':[[93, 109, 187],[169], [31, 159],[176],[16, 48, 66, 128]],
'AwayG_Time':[[90, 177],[],[],[123,136],[40]]})
First, I need to create an array y, for a given Id number, it takes values from same row (ExpG_Home & ExpG_Away).
y = [1 - (ExpG_Home + ExpG_Away), ExpG_Home, ExpG_Away]
Second, I found this much harder, for the Id used in creating y, the function below takes the corresponding lists from HomeG_Time & AwayG_Time and creates an array. Unfortunately, my function takes one row at a time. I need to do this for a large dataset.
x1 = [1,0,0]
x2 = [0,1,0]
x3 = [0,0,1]
total_timeslot = 200 # number of timeslot per game.
k = 1 # constant
#For Id=102 with ExpG_Home=2.2 and ExpG_Away=1.8
HomeG_Time = [93, 109, 187]
AwayG_Time = [90, 177]
y = np.array([1-(2.2 + 1.8)/k, 2.2/k, 1.8/k])
# output of y = [0.98 , 0.011, 0.009]
def squared_diff(x1, x2, x3, y):
ssd = []
for k in range(total_timeslot):
if k in HomeG_Time:
ssd.append(sum((x2 - y) ** 2))
elif k in AwayG_Time:
ssd.append(sum((x3 - y) ** 2))
else:
ssd.append(sum((x1 - y) ** 2))
return ssd
sum(squared_diff(x1, x2, x3, y))
Out[37]: 7.880400000000012
This output is for the first row only.
Here is the complete snippet given,
>>> import numpy as np
>>> x1 = np.array( [1,0,0] )
>>> x2 = np.array( [0,1,0] )
>>> x3 = np.array( [0,0,1] )
>>> total_timeslot = 200
>>> HomeG_Time = [93, 109, 187]
>>> AwayG_Time = [90, 177]
>>> ExpG_Home=2.2
>>> ExpG_Away=1.8
>>> y = np.array( [1 - (ExpG_Home + ExpG_Away), ExpG_Home, ExpG_Away] )
>>> def squared_diff(x1, x2, x3, y):
... ssd = []
... for k in range(total_timeslot):
... if k in HomeG_Time:
... ssd.append(sum((x2 - y) ** 2))
... elif k in AwayG_Time:
... ssd.append(sum((x3 - y) ** 2))
... else:
... ssd.append(sum((x1 - y) ** 2))
... return ssd
...
>>> sum(squared_diff(x1, x2, x3, y))
4765.599999999989
Assuming this. Calculate y as (N,3) using pandas.DataFrame.apply
>>> y = np.array( df.apply(lambda row: [1 - (row.ExpG_Home + row.ExpG_Away),
... row.ExpG_Home, row.ExpG_Away ],
... axis=1).tolist() )
>>> y.shape
(5, 3)
Now calcualte squared error for a given x
>>> def squared_diff(x, y):
... return np.sum( np.square(x - y), axis=1)
In your case, if error2 is squared_diff(x2,y) you are adding this the number of occuerences of HomeG_Time
>>> n3 = df.AwayG_Time.apply(len)
>>> n2 = df.HomeG_Time.apply(len)
>>> n1 = 200 - (n2 + n3)
The final sum of squared error is (as per your calculation)
>>> squared_diff(x1, y) * n1 + squared_diff(x2, y) * n2 + squared_diff(x3, y) * n3
0 4766.4
1 2349.4
2 2354.4
3 6411.6
4 4496.2
dtype: float64
>>>
try this,
import pandas as pd
import numpy as np
df = pd.DataFrame({'Id': [102,103,104,303,305],'ExpG_Home':[1.8,1.5,1.6,1.8,2.9],
'ExpG_Away':[2.2,1.3,1.2,2.8,0.8],
'HomeG_Time':[[93, 109, 187],[169], [31, 159],[176],[16, 48, 66, 128]],
'AwayG_Time':[[90, 177],[],[],[123,136],[40]]})
x1 = [1,0,0]
x2 = [0,1,0]
x3 = [0,0,1]
k=1
total_timeslot = 200 # number of timeslot per game.
def squared_diff(x1, x2, x3,AwayG_Time,HomeG_Time, y):
ssd = []
for k in range(total_timeslot):
if k in HomeG_Time:
ssd.append(sum((x2 - y) ** 2))
elif k in AwayG_Time:
ssd.append(sum((x3 - y) ** 2))
else:
ssd.append(sum((x1 - y) ** 2))
return ssd
s=pd.DataFrame( pd.concat([df,1-(df['ExpG_Home']+df['ExpG_Away'])/k,df['ExpG_Home']/k,df['ExpG_Away']/k],axis=1).values)
df['res']=s.apply(lambda x: sum(squared_diff(x1,x2,x3,x[0],x[3],np.array([x[5],x[6],x[7]]))),axis=1)
del s
print df
Output:
AwayG_Time ExpG_Away ExpG_Home HomeG_Time Id res
0 [90, 177] 2.2 1.8 [93, 109, 187] 102 4766.4
1 [] 1.3 1.5 [169] 103 2349.4
2 [] 1.2 1.6 [31, 159] 104 2354.4
3 [123, 136] 2.8 1.8 [176] 303 6411.6
4 [40] 0.8 2.9 [16, 48, 66, 128] 305 4496.2
def squared_diff(row):
y = np.array([1 - (row.ExpG_Home + row.ExpG_Away), row.ExpG_Home, row.ExpG_Away])
HomeG_Time = row.HomeG_Time
AwayG_Time = row.AwayG_Time
x1 = np.array([1, 0, 0])
x2 = np.array([0, 1, 0])
x3 = np.array([0, 0, 1])
total_timeslot = 200
ssd = []
for k in range(total_timeslot):
if k in HomeG_Time:
ssd.append(sum((x2 - y) ** 2))
elif k in AwayG_Time:
ssd.append(sum((x3 - y) ** 2))
else:
ssd.append(sum((x1 - y) ** 2))
return sum(ssd)
df.apply(squared_diff, axis=1)
Out[]:
0 4766.4
1 2349.4
2 2354.4
3 6411.6
4 4496.2
I am a newbie to machine learning. I am trying a simple prediction using linear regression with "made up" data that follows a specific pattern. For some reason, the prediction is not matching the training data. Can you let me know what I need to fix? The sample code is below
from sklearn import linear_model
import numpy as np
X = np.random.randint(3, size=(3, 1000))
Y = np.random.randint(10, size=(1, 1000))
# f1, f2, f3 - min = 0, max = 2
# f1 = 0 and f2 = 1 then 7 <= Y < 10, irrespective of f3
# f1 = 1 and f2 = 2 Y is 0, irrespective of f3
# f1 = 0 and f2 = 2 if f3 = 2 then 3 <= Y < 7 else Y = 0
for i in range(1000):
if ((X[0][i] == 0 and X[1][i] == 1) or (X[0][i] == 1 and X[1][i] == 0)):
Y[0][i] = np.random.randint(7, 10)
elif ((X[0][i] == 1 and X[1][i] == 2) or (X[0][i] == 2 and X[1][i] == 1)):
Y[0][i] = 0
elif ((X[0][i] == 0 and X[1][i] == 2 and X[2][i] == 2) or
(X[0][i] == 2 and X[1][i] == 0 and X[2][i] == 2)):
Y[0][i] = np.random.randint(3, 7)
else:
Y[0][i] = 0
X1 = X.transpose()
Y1 = Y.reshape(-1, 1)
print zip(X1, Y1)
# create and fit the model
clf = linear_model.LinearRegression()
clf.fit(X1, Y1)
Z = np.array([[0, 0, 0, 0, 1, 1],
[1, 1, 2, 2, 2, 2],
[1, 2, 1, 2, 1, 2]])
Z1 = Z.transpose()
print Z1
y_predict = clf.predict(Z1)
print y_predict
And why would it match the training data? Your X->Y relation is clearly non-linear, and only for perfect linear relation, meaning that Y = AX + b, you can expect linear regression to fit training data perfectly. Otherwise, you can get arbitrary far away from the solution - see for example an Anscombe's quartet (image belowo from wiki).
I'm looking for a pythonic (1-line) way to extract a range of values from an array
Here's some sample code that will extract the array elements that are >2 and <8 from x,y data, and put them into a new array. Is there a way to accomplish this on a single line? The code below works but seems kludgier than it needs to be. (Note I'm actually working with floats in my application)
import numpy as np
x0 = np.array([0,3,9,8,3,4,5])
y0 = np.array([2,3,5,7,8,1,0])
x1 = x0[x0>2]
y1 = y0[x0>2]
x2 = x1[x1<8]
y2 = y1[x1<8]
print x2, y2
This prints
[3 3 4 5] [3 8 1 0]
Part (b) of the problem would be to extract values say 1 < x < 3 and 7 < x < 9 as well as their corresponding y values.
You can chain together boolean arrays using & for element-wise logical and and | for element-wise logical or, so that the condition 2 < x0 and x0 < 8 becomes
mask = (2 < x0) & (x0 < 8)
For example,
import numpy as np
x0 = np.array([0,3,9,8,3,4,5])
y0 = np.array([2,3,5,7,8,1,0])
mask = (2 < x0) & (x0 < 8)
x2 = x0[mask]
y2 = y0[mask]
print(x2, y2)
# (array([3, 3, 4, 5]), array([3, 8, 1, 0]))
mask2 = ((1 < x0) & (x0 < 3)) | ((7 < x0) & (x0 < 9))
x3 = x0[mask2]
y3 = y0[mask2]
print(x3, y3)
# (array([8]), array([7]))
import numpy as np
x0 = np.array([0,3,9,8,3,4,5])
y0 = np.array([2,3,5,7,8,1,0])
list( zip( *[(x,y) for x, y in zip(x0, y0) if 1<=x<=3 or 7<=x<=9] ) )
# [(3, 9, 8, 3), (3, 5, 7, 8)]
Is it possible to do magic squares with the Siamese/De La Loubere method without using modulo?
I would like to make odd n x n magic squares using it.
Yes, it's possible. Written on Python 3.5:
def siamese_method(n):
assert(n % 2 != 0), 'Square side size should be odd!'
square = [[0 for x in range(n)] for x in range(n)]
x = 0
y = int((n + 1) / 2 - 1)
square[x][y] = 1
for i in range(2, n * n + 1):
x_old = x
y_old = y
if x == 0:
x = n - 1
else:
x -= 1
if y == n - 1:
y = 0
else:
y += 1
while square[x][y] != 0:
if x == n - 1:
x = 0
else:
x = x_old + 1
y = y_old
square[x][y] = i
for j in square:
print(j)
siamese_method(3)
I've got following on output:
[8, 1, 6]
[3, 5, 7]
[4, 9, 2]