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labyrinth solver on python language

just a small labyrinth solver. try it to on python but dont even understand why it doesnt print anything. Im just started with python and trying to actually translate my teachers code from cpp just for interest.
need to reach to the destination after visiting each cell
m = [
[2,0,0,0,0,0],
[0,0,0,0,0,0],
[0,0,0,0,0,0],
[0,0,0,0,0,0],
[0,0,0,2,0,0],
[0,0,0,0,0,0]
]
STOP = 34
r = 0
c = 0
def init( r_: int = 0, c_: int = 0) -> None:
r = 0
c = 0
def step(r,c,level):
if level==STOP-1:
if r == 3 and c == 5:
for i in range(0,level):
print("("+init(int(r),int(c))+"," + init(int(r),int(c)) + ") - ")
print("(3,5)\n")
return True
else:
return False
m[r][c]=1
if c > 0 and m[r][c - 1] == 0:
if step(r, c - 1, level + 1):
return True
if c < 5 and m[r][c + 1] == 0:
if (step(r, c + 1, level + 1)):
return True
if r > 0 and m[r - 1][c] == 0:
if (step(r - 1, c, level + 1)):
return True
if r < 5 and m[r + 1][c] == 0:
if (step(r + 1, c, level + 1)):
return True
m[r][c] = 0
return False
def main(argc, argv):
step(0, 1, 0)

Python doesn't automatically call main like c does, nor does it require a main function. Your code can just live in the top level (e.g. print('hello world') is a valid, complete program)
Change this line:
def main(argc, argv):
to this:
if __name__ == '__main__':

got it
import matplotlib.pyplot as plt
S, F, W, X = 1, 2, 0, None
FIELD = [[X,S,W,W,W,W],
[W,W,W,W,W,W],
[W,W,W,W,W,W],
[W,W,W,W,W,F],
[W,W,W,X,W,W],
[W,W,W,W,W,W]]
ROWS = len(FIELD)
COLS = len(FIELD[0])
START_POS = [(i, line.index(S)) for i, line in enumerate(FIELD) if S in line][0]
FINISH_POS = [(i, line.index(F)) for i, line in enumerate(FIELD) if F in line][0]
STEPS = sum(line.count(W) for line in FIELD) + 1
print("Start:", START_POS, "\nFinish:", FINISH_POS, "\nTotal steps:", STEPS)
def do_step(level, path, points):
global FIELD
r, c = points[-1]
if r<0 or c<0 or r>=ROWS or c>=COLS or FIELD[r][c] == X:
return
if level == STEPS and (r, c) == FINISH_POS:
print("Found path:", path)
yy, xx = zip(*points)
plt.plot(xx, yy, "-ok")
plt.gca().invert_yaxis()
plt.show()
exit()
current = FIELD[r][c]
FIELD[r][c] = X
for dr, dc, dir in ((-1,0,'^'), (0,1,'>'), (1,0,'v'), (0,-1,'<')):
do_step(level+1, path+dir, points+[(r+dr,c+dc)])
FIELD[r][c] = current
do_step(0, "", [START_POS])
print("No path's found.")

How to change my code to solve modular equations with division?

I wrote the Extended Euclidean Algorithm, but i'm having trouble with using it to solve the following equation:
which the Modular Equation Solver reduces to
and
Here is my code:
def fastlinearcongruenceSO(powx, divmodx, N, withstats=False):
x, y, z = egcditerx2SO(powx, N, withstats)
answer = (y*divmodx)%N
if withstats == True:
print(f"answer = {answer}")
if x > 1:
powx//=x
divmodx//=x
N//=x
if withstats == True:
print(f"powx = {powx}, divmodx = {divmodx}, N = {N}")
x, y, z = egcditerx2SO(powx, N, withstats)
if withstats == True:
print(f"x = {x}, y = {y}, z = {z}")
answer = (y*divmodx)%N
if withstats == True:
print(f"answer = {answer}")
return answer
def egcditerx2SO(a, b, withstats=False):
s = 0
r = b
old_s = 1
old_r = a
quotient = 0
if withstats == True:
print(f"quotient = {quotient}, old_r = {old_r}, r = {r}, old_s = {old_s}, s = {s}")
while r!= 0:
quotient = old_r // r
old_r, r = r, old_r - quotient * r
old_s, s = s, old_s - quotient * s
if withstats == True:
print(f"quotient = {quotient}, old_r = {old_r}, r = {r}, old_s = {old_s}, s = {s}")
if b != 0:
bezout_t = quotient = (old_r - old_s * a) // b
if withstats == True:
print(f"bezout_t = {bezout_t}")
else:
bezout_t = 0
if withstats == True:
print("Bézout coefficients:", (old_s, bezout_t))
print("greatest common divisor:", old_r)
return old_r, old_s, bezout_t
To solve for inverses, i used this form:
IN: fastlinearcongruenceSO(327, 1, 1009)
OUT: 108
I'm not sure how or what changes i need to make to modify it to solve with the form of division, does anyone have any ideas of what changes i need to make or if i can solve it using my existing code? I really want to modify my code to handle it solving this equation: 743360/1008x = 272 (mod 1009) and x == 116 or for someone to let me know how that equation was solved, and i can then with those steps modify my code to solve for division based modulus equations. Thanks for anyone who knows how to solve these division based equations.

im trying use bisection method to find the root of the function x^4 - 5x^3 + (22/3)x^2 - (116/27)x + (8/9) in [0,1] with e = 0.01

here's the code i made
def bisection(f, a, b, e):
step = 1
condition = True
while condition:
c = (a+b)/2
if f(a) * f(c) < 0:
b = c
else:
a = c
if f(c) == 0:
break
step = step + 1
condition = abs(f(a)-f(b)) > e
print(c)
return(c)
if f(a)* f(b) > 0:
print("no roots")
else:
bisection(f, a, b, e)
a, b, e = input().split(' ')
a = float(a)
b = float(b)
print(round(root, 4))
but I'm getting a root of c = 0.171875
instead of 0.6875

Python pow() function returns wrong answer

I am creating a basic RSA encryption program without using an RSA library that takes a secret message, converts each character in the string to its ASCII value, encrypts with a public key and concatenates the values, and then decrypts it using a private key and returns it to a string.
All with the principle of cipher = pow(plain,e,n) and plain = pow(cipher,d,n). My issue is that when the numbers get very large as I need d and n to be 16 digits minimum, the pow() function seems to result in an error in calculation that yields an ASCII value that is out of range to convert to a character. I've been struggling to figure out where I'm going wrong for days now. Any help is appreciated. Code below:
from random import randrange, getrandbits
def is_prime(n, k=128):
# Test if n is not even.
# But care, 2 is prime !
if n == 2 or n == 3:
return True
if n <= 1 or n % 2 == 0:
return False
# find r and s
s = 0
r = n - 1
while r & 1 == 0:
s += 1
r //= 2
# do k tests
for q in range(k):
a = randrange(2, n - 1)
x = pow(a, r, n)
if x != 1 and x != n - 1:
j = 1
while j < s and x != n - 1:
x = pow(x, 2, n)
if x == 1:
return False
j += 1
if x != n - 1:
return False
return True
def generate_prime_candidate(length):
# generate random bits
p = getrandbits(length)
#p = randrange(10**7,9*(10**7))
# apply a mask to set MSB and LSB to 1
p |= (1 << length - 1) | 1
return p
def generate_prime_number(length=64):
p = 4
# keep generating while the primality test fail
while not is_prime(p, 128):
p = generate_prime_candidate(length)
return p
def gcd(a, b):
while b != 0:
a, b = b, a % b
return a
def generate_keypair(p, q):
n = p * q
#Phi is the totient of n
phi = (p-1) * (q-1)
#Choose an integer e such that e and phi(n) are coprime
e = randrange(1,65537)
g = gcd(e, phi)
while g != 1:
e = randrange(1,65537)
g = gcd(e, phi)
d = multiplicative_inverse(e, phi)
return ((e, n), (d, n))
def multiplicative_inverse(e, phi):
d = 0
k = 1
while True:
d = (1+(k*phi))/e
if((round(d,5)%1) == 0):
return int(d)
else:
k+=1
def encrypt(m,public):
key, n = public
encrypted = ''
print("Your original message is: ", m)
result = [(ord(m[i])) for i in range(0,len(m))]
encryption = [pow(result[i],key,n) for i in range(0,len(result))]
for i in range(0,len(encryption)):
encrypted = encrypted + str(encryption[i])
#encrypted = pow(int(encrypted),key,n)
print("Your encrypted message is: ", encrypted)
#return result,encrypted
return encrypted, encryption
def decrypt(e,c,private):
key, n = private
print("Your encrypted message is: ", c)
print(e)
decryption = [pow(e[i],key,n) for i in range(0,len(e))]
print(decryption)
result = [chr(decryption[i])for i in range(0,len(decryption)) ]
decrypted = ''.join(result)
print("Your decrypted message is: ",decrypted)
return result,decrypted
def fastpow(x,y,p):
res = 1
x = x%p
while(y>0):
if((y&1) == 1):
res = (res*x)%p
y = y>>1
x = (x*x)%p
return res
message = input("Enter your secret message: ")
p1 = generate_prime_number()
p2 = generate_prime_number()
public, private = generate_keypair(p1,p2)
print("Your public key is ", public)
print("Your private key is ", private)
encrypted,cipher = encrypt(message,public)
decrypt(cipher,encrypted,private)
Traceback:
File "<ipython-input-281-bce7c44b930c>", line 1, in <module>
runfile('C:/Users/Mervin/Downloads/group2.py', wdir='C:/Users/Mervin/Downloads')
File "C:\Users\Mervin\Anaconda3\lib\site-packages\spyder\util\site\sitecustomize.py", line 705, in runfile
execfile(filename, namespace)
File "C:\Users\Mervin\Anaconda3\lib\site-packages\spyder\util\site\sitecustomize.py", line 102, in execfile
exec(compile(f.read(), filename, 'exec'), namespace)
File "C:/Users/Mervin/Downloads/group2.py", line 125, in <module>
decrypt(cipher,encrypted,private)
File "C:/Users/Mervin/Downloads/group2.py", line 100, in decrypt
result = [chr(decryption[i])for i in range(0,len(decryption)) ]
File "C:/Users/Mervin/Downloads/group2.py", line 100, in <listcomp>
result = [chr(decryption[i])for i in range(0,len(decryption)) ]
OverflowError: Python int too large to convert to C long

Your method multiplicative_inverse is not correct. I'm not certain what you are doing in that method with the round method and floating point division, but even if you got that part correct it would be too slow anyway, requiring O(φ) steps. The usual method for computing modular multiplicative inverses is an adaptation of the extended Euclidean algorithm, which runs in O(log(φ)2) steps. Here is a straightforward mapping from the psuedocode in the Wikipedia article to python 3 code:
def multiplicative_inverse(a, n):
t, newt = 0, 1
r, newr = n, a
while newr:
quotient = r // newr
t, newt = newt, t - quotient * newt
r, newr = newr, r - quotient * newr
if r > 1:
raise ZeroDivisionError("{} is not invertible".format(a))
if t < 0:
t = t + n
return t

Russian Peasant Multiplication Python using recursive-tail

I tried making this code and run it, but the result showed "None".
I've edited by adding another else in the loop function, then it actually produces a int result of '57' which is not exactly the answer.
def double(n):
return n*2
def halve(n):
return n//2
def mult(m,n):
def loop(m,n):
if n>1:
if n%2 != 0:
return m + loop(double(m),halve(n))
else:
return m #added another else
else:
return m
if n>0:
return loop(m,n)
else:
return 0
print(mult(57,86))

by simple recursion
def double(n):
return n * 2
def halve(n):
return n // 2
def mult(m, n, a = 0):
if n % 2 != 0:
a = a + m
m = double(m)
n = halve(n)
if n % 2 == 0:
m = double(m)
n = halve(n)
if n != 0:
return mult(m, n, a)
return a
print(mult(57, 86))
by nested function
def double(n):
return n * 2
def halve(n):
return n // 2
def mult(m, n):
def loop(m, n, a = 0):
if n % 2 != 0:
a = a + m
m = double(m)
n = halve(n)
if n % 2 == 0:
m = double(m)
n = halve(n)
if n != 0:
return loop(m, n, a)
return a
return loop(m, n)
print(mult(57, 86))