def main():
start = time()
A = 1777
B = 1855
phiLimit = phi(LIMIT)
'''
we use efficient modular exponentiation to do this.
essentially, there are several highly efficient ways
of computing (a ^ b) mod n. Python just so happens to use
binary exponentiation in its built-in pow() function.
http://en.wikipedia.org/wiki/Modular_exponentiation
we also use Euler's theorem:
http://en.wikipedia.org/wiki/Euler's_theorem
'''
residual = A
for i in xrange(B - 2): residual = pow(A, residual, phiLimit)
residual = pow(A, residual, LIMIT)
print "Last ", DIGITS_TO_SEE, " digits from the tetration (", A, ", ", B, "): ", residual
end = time()
print "Runtime: ", end - start, " seconds."
def main():
start = time()
A = 1777
B = 1855
phiLimit = phi(LIMIT)
'''
we use efficient modular exponentiation to do this.
essentially, there are several highly efficient ways
of computing (a ^ b) mod n. Python just so happens to use
binary exponentiation in its built-in pow() function.
http://en.wikipedia.org/wiki/Modular_exponentiation
we also use Euler's theorem:
http://en.wikipedia.org/wiki/Euler's_theorem
'''
residual = A
for i in xrange(B - 2):
residual = pow(A, residual, phiLimit)
residual = pow(A, residual, LIMIT)
print "Last ", DIGITS_TO_SEE, " digits from the tetration (", A, ", ", B, "): ", residual
end = time()
print "Runtime: ", end - start, " seconds."
def main():
start = time()
solution = 0
LOOP_LIMIT = 6
MOD_LIMIT = 14**8
PHI_LIMIT = phi(MOD_LIMIT)
powDict = {}
for i in xrange(LOOP_LIMIT + 1):
newSolution = ackermann(i, MOD_LIMIT, PHI_LIMIT, powDict)
print i, " : ", newSolution
solution += newSolution
print "Solution: ", solution % MOD_LIMIT
end = time()
print "Runtime: ", end - start, " seconds."
def main():
start = time()
solution = 0
LOOP_LIMIT = 6
MOD_LIMIT = 14 ** 8
PHI_LIMIT = phi(MOD_LIMIT)
powDict = {}
for i in xrange(LOOP_LIMIT + 1):
newSolution = ackermann(i, MOD_LIMIT, PHI_LIMIT, powDict)
print i, " : ", newSolution
solution += newSolution
print "Solution: ", solution % MOD_LIMIT
end = time()
print "Runtime: ", end - start, " seconds."
def main():
# generate primes.
p = ProjectEulerPrime()
primes = {2}
for i in xrange(3, 100000, 2):
if p.isPrime(i):
primes.add(i)
'''
We use the following facts:
R(n) mod(n)
=> ((10 ** n - 1) / 9) mod n
=> (10 ** n) mod (9 * n) = 1
x ** y (mod n) == x ** (y mod phi(n)) (mod n)
We stop once we detect a repeated residual
(which means that we've entered a period).
By sheer luck, I used a similar technique in problem 282
to collapse much larger power towers.
'''
results = set()
for prime in primes:
phi_mod = phi(9 * prime)
mod = 9 * prime
seen_residuals = set()
j = 1
while True:
residual = pow(10, pow(10, j, phi_mod), mod)
if residual in seen_residuals:
break
if residual == 1:
results.add(prime)
break
seen_residuals.add(residual)
j += 1
print "Sum: {}".format(sum(primes - results))
def main():
# generate primes.
p = ProjectEulerPrime()
primes = {2}
for i in xrange(3, 100000, 2):
if p.isPrime(i):
primes.add(i)
'''
We use the following facts:
R(n) mod(n)
=> ((10 ** n - 1) / 9) mod n
=> (10 ** n) mod (9 * n) = 1
x ** y (mod n) == x ** (y mod phi(n)) (mod n)
We stop once we detect a repeated residual
(which means that we've entered a period).
By sheer luck, I used a similar technique in problem 282
to collapse much larger power towers.
'''
results = set()
for prime in primes:
phi_mod = phi(9 * prime)
mod = 9 * prime
seen_residuals = set()
j = 1
while True:
residual = pow(10, pow(10, j, phi_mod), mod)
if residual in seen_residuals:
break
if residual == 1:
results.add(prime)
break
seen_residuals.add(residual)
j += 1
print "Sum: {}".format(sum(primes - results))
def main():
RATIO = (15499, 94744)
cache = dict()
default = float('inf')
bestSoFar = default
'''
So, here are the facts:
* phi(n) gives the number of positive integers < n that are relatively prime to n.
* if m and n are relatively prime to each other, phi(m * n) = phi(m) * phi(n).
* phi(n) is always nearly n.
So we want to focus on doing as much work as possible with small n. So, the algo is as follows:
1). iterate from 2 to infinity. Call this i.
2). calculate phi(i). store this in a cache.
3). find phi(m * i) for each m in the cache.
4). if we find an element that works, stop the iteration (since the keys are unordered, go through
all keys first and pick the smallest m * i).
'''
i = 2
while True:
value = phi(i)
cache[i] = value
for num in cache.keys():
if gcd(num, i) == 1:
cache[num * i] = cache[num] * value
if num * i < bestSoFar and (RATIO[1] * cache[num * i]) < (
(num * i - 1) * RATIO[0]):
bestSoFar = num * i
if bestSoFar != default:
break
else:
i += 1
print "Solution: {}".format(bestSoFar)
def main():
RATIO = (15499, 94744)
cache = dict()
default = float('inf')
bestSoFar = default
'''
So, here are the facts:
* phi(n) gives the number of positive integers < n that are relatively prime to n.
* if m and n are relatively prime to each other, phi(m * n) = phi(m) * phi(n).
* phi(n) is always nearly n.
So we want to focus on doing as much work as possible with small n. So, the algo is as follows:
1). iterate from 2 to infinity. Call this i.
2). calculate phi(i). store this in a cache.
3). find phi(m * i) for each m in the cache.
4). if we find an element that works, stop the iteration (since the keys are unordered, go through
all keys first and pick the smallest m * i).
'''
i = 2
while True:
value = phi(i)
cache[i] = value
for num in cache.keys():
if gcd(num, i) == 1:
cache[num * i] = cache[num] * value
if num * i < bestSoFar and (RATIO[1] * cache[num * i]) < ((num * i - 1) * RATIO[0]):
bestSoFar = num * i
if bestSoFar != default:
break
else:
i += 1
print "Solution: {}".format(bestSoFar)
for n in range(2, 1000000):
#phi_val = mpz(-1)
#factors = factorize(n, True)
#for i in factors:
# for j in factors:
# if i == j:
# continue
# if i*j == n and gcd(i,j) == 1:
# phi_val = phi_cache[i]*phi_cache[j]
# print('+ '),
# break
# if int(phi_val) != -1:
# break
#if int(phi_val) == -1:
# print('- '),
phi_val = phi(n)
#phi_cache[n] = phi_val
val = n/mpq(phi_val)
print('%d ' % n),
print('%d ' % phi_val),
print('%f' % val)
if val > max:
max = val
maxN = n
print(maxN),
print(max)
