def problem113():
DIGITS = 100
increasing = eulerlib.binomial(DIGITS + 9, 9) - 1
decreasing = eulerlib.binomial(DIGITS + 10, 10) - (DIGITS + 1)
flat = DIGITS * 9
ans = increasing + decreasing - flat
return ans
def problem106():
SET_SIZE = 12
def catalan(n):
return eulerlib.binomial(n * 2, n) // (n + 1)
ans = sum(
eulerlib.binomial(SET_SIZE, i * 2) *
(eulerlib.binomial(i * 2, i) // 2 - catalan(i))
for i in range(2, SET_SIZE // 2 + 1))
return ans
def explore(remain, limit, history):
if remain == 0:
hist = list(history)
while len(hist) < num_colors:
hist.append(0)
histogram = [0] * (balls_per_color + 1)
for _ in hist:
histogram[_] += 1
count = math.factorial(num_colors)
for _ in histogram:
count = divide_exactly(count, math.factorial(_))
for _ in hist:
count *= eulerlib.binomial(balls_per_color, _)
distinctcolors = len(history)
numerator[0] += count * distinctcolors
elif len(history) < num_colors:
for i in range(min(limit, remain), 0, -1):
history.append(i)
explore(remain - i, i, history)
history.pop()
def problem053():
"""
There are exactly ten ways of selecting three from five, 12345:
123, 124, 125, 134, 135, 145, 234, 235, 245, and 345
In combinatorics, we use the notation, ^5C[3] = 10.
In general,
^nC[r] = n! ,where r ≤ n, n! = n×(n−1)×...×3×2×1, and 0! = 1.
r!(n−r)!
It is not until n = 23, that a value exceeds one-million: ^23C[10] =
1144066.
How many, not necessarily distinct, values of ^nC[r], for 1 ≤ n ≤ 100,
are greater than one-million?
"""
ans = sum(1 for n in range(1, 101) for k in range(0, n + 1)
if eulerlib.binomial(n, k) > 1000000)
return ans
def problem323():
SIZE = 32
DECIMALS = 10
assert SIZE >= 0
assert DECIMALS >= 0
# Calculate the answer
expect = [fractions.Fraction(0)]
for n in range(1, SIZE + 1):
temp = sum(eulerlib.binomial(n, k) * expect[k] for k in range(n))
expect.append((2**n + temp) / (2**n - 1))
ans = expect[-1]
# Round the fraction properly. This is the pedantically
# correct version of doing "{:.10f}".format(float(ans))
assert ans >= 0
scaled = ans * 10**DECIMALS
whole = scaled.numerator // scaled.denominator
frac = scaled - whole
assert 0 <= frac < 1
HALF = fractions.Fraction(1, 2)
if frac > HALF or (frac == HALF and whole % 2 == 1):
whole += 1
temp = str(whole)
if DECIMALS == 0:
return temp
temp = temp.zfill(DECIMALS + 1)
return "{}.{}".format(temp[:-DECIMALS], temp[-DECIMALS:])
def problem015():
"""
Starting in the top left corner of a 2×2 grid, and only being able to move
to the right and down, there are exactly 6 routes to the bottom right
corner.
How many such routes are there through a 20×20 grid?
p_015.gif
This is a classic combinatorics problem. To get from the top left corner to the bottom right corner of an N*N grid,
it involves making exactly N moves right and N moves down in some order. Because each individual down or right move
is indistinguishable, there are exactly 2N choose N (binomial coefficient) ways of arranging these moves.
"""
return eulerlib.binomial(40, 20)
def problem203():
# Collect unique numbers in Pascal's triangle
numbers = set(eulerlib.binomial(n, k) for n in range(51) for k in range(n + 1))
maximum = max(numbers)
# Prepare list of squared primes
primes = eulerlib.list_primes(eulerlib.sqrt(maximum))
primessquared = [p * p for p in primes]
def is_squarefree(n):
for p2 in primessquared:
if p2 > n:
break
if n % p2 == 0:
return False
return True
# Sum up the squarefree numbers
ans = sum(n for n in numbers if is_squarefree(n))
return ans
def problem267():
# Heuristic sampling algorithm.
# At level 1 we test {1/2}. At level 2 we test {1/4, 3/4}.
# At level 3 we test {1/8, 3/8, 5/8, 7/8}. Et cetera.
TRIALS = 1000
maxindex = -1
prevchangelevel = 1
level = 1
while level - prevchangelevel <= 8:
scaler = 0.5**level
for i in range(1, 1 << level, 2):
index = calc_billionaire_probability(i * scaler, TRIALS)
if index > maxindex:
maxindex = index
prevchangelevel = level
level += 1
# Calculate the cumulative probability: binomialSum = sum (n choose k) for 0 <= k < maxIndex
binomialsum = sum(eulerlib.binomial(TRIALS, i) for i in range(maxindex))
return round_to_decimal(fractions.Fraction(binomialsum, 1 << TRIALS), 12)
def problem493():
"""
:return:
"""
num_colors = 7
balls_per_color = 10
num_picked = 20
decimals = 9
numerator = [0]
def explore(remain, limit, history):
if remain == 0:
hist = list(history)
while len(hist) < num_colors:
hist.append(0)
histogram = [0] * (balls_per_color + 1)
for _ in hist:
histogram[_] += 1
count = math.factorial(num_colors)
for _ in histogram:
count = divide_exactly(count, math.factorial(_))
for _ in hist:
count *= eulerlib.binomial(balls_per_color, _)
distinctcolors = len(history)
numerator[0] += count * distinctcolors
elif len(history) < num_colors:
for i in range(min(limit, remain), 0, -1):
history.append(i)
explore(remain - i, i, history)
history.pop()
explore(num_picked, balls_per_color, [])
denominator = eulerlib.binomial(num_colors * balls_per_color, num_picked)
ans = fractions.Fraction(numerator[0], denominator)
return float(format_fraction(ans, decimals))
def catalan(n):
return eulerlib.binomial(n * 2, n) // (n + 1)
def test_binomial(input_n, input_k, expected_output):
output = eulerlib.binomial(input_n, input_k)
print(output)
assert output == expected_output