#Let S(A) represent the sum of elements in set A of size n. We shall call it a special sum set if for any two non-empty disjoint subsets, B and C, the following properties are true:
#1. S(B) ≠ S(C); that is, sums of subsets cannot be equal.
#2. If B contains more elements than C then S(B) > S(C).
#For this problem we shall assume that a given set contains n strictly increasing elements and it already satisfies the second rule.
#Surprisingly, out of the 25 possible subset pairs that can be obtained from a set for which n = 4, only 1 of these pairs need to be tested for equality (first rule). Similarly, when n = 7, only 70 out of the 966 subset pairs need to be tested.
#For n = 12, how many of the 261625 subset pairs that can be obtained need to be tested for equality?
#NOTE: This problem is related to problems 103 and 105.
#Answer:
#21384
from time import time; t=time()
from mathplus import combinations
M = 12
s = 0
n = M
for k in range(2, n//2+1):
for p in combinations(range(n), k):
qq = [i for i in range(n) if i not in p and i > p[0]]
for q in combinations(qq, k):
if not (all(q[i] > p[i] for i in range(k)) or all(q[i] < p[i] for i in range(k))): s += 1
print(s)#, time()-t
from time import time; t=time()
from mathplus import Cnr, combinations
NN = N = 7
l, h = 17, 50
#l, h = 11, 26
min_s = h * N
min_k = []
cache = {}
for i in range(l, h):
cache[(i,)] = True
for N in range(2, NN+1):
for k in combinations(range(l, h - NN + N), N):
kk = sorted(k)
if tuple(kk) in cache: continue
if tuple(kk[:N-1]) not in cache: continue
if N == NN:
s = sum(kk)
if s >= min_s: continue
for i in range(1, (N+1)//2):
if sum(kk[-i:]) >= sum(kk[:i+1]): break
else:
for r in range(2, N-1):
ss = set()
for j in combinations(kk, r):
ss.add(sum(j))
if len(ss) != Cnr(N, r): break
else:
#But because we are allowing 6 and 9 to be reversed, the two distinct sets in the last example both represent the extended set {1, 2, 3, 4, 5, 6, 9} for the purpose of forming 2-digit numbers.
#How many distinct arrangements of the two cubes allow for all of the square numbers to be displayed?
#Answer:
#1217
from time import time
t = time()
from mathplus import combinations
def test(p, a, b):
if p[0] in a and p[1] in b: return True
if p[0] in b and p[1] in a: return True
return False
s = 0
for b in combinations(range(0, 10), 6):
if 2 not in b and 5 not in b: continue
if 6 in b or 9 in b: b += (6, 9)
for a in combinations(range(0, 10), 6):
if 6 in a or 9 in a: a += (6, 9)
for p in [(0, 1), (0, 4), (0, 9), (1, 6), (2, 5), (3, 6), (4, 9),
(6, 4), (8, 1)]:
if not test(p, a, b): break
else: s += 1
print(s // 2) #, time()-t
def get_ret2(a, b):
r = [a+b, a-b, a*b, b-a]
if a != 0: r.append(b/a)
if b != 0: r.append(a/b)
return r
@memorize
def get_ret3(abc):
ss = set()
for i in permutations(abc):
a, b, c = i[0], i[1], i[2]
for r in get_ret2(a, b):
ss.add(tuple(sorted((r, c))))
return set(r for ab in ss for r in get_ret2(*ab))
for k in combinations(range(1, 10), 4):
ss = set()
for i in permutations(k):
a, b, c, d = i[0], i[1], i[2], i[3]
for r in get_ret2(a, b):
ss.add(tuple(sorted((r, c, d))))
ret = set(r for abc in ss for r in get_ret3(abc))
i = 1
while True:
if i not in ret: break
i += 1
i -= 1
if i > s:
s = i
abcd = sorted(k)
if len_active_k > 10: continue
full_f = [0] * 26
def test(s):
if full_f[s[0]] == 0: return 0
c = full_f[s[-1]]
if c in (0, 2, 3, 7, 8): return 0
d = full_f[s[-2]]
if c == 5 and d != 2: return 0
if c == 6 and d % 2 == 0: return 0
if d % 2 == 1: return 0
n = reduce(lambda x, y: x * 10 + y, (full_f[c] for c in s))
return n if isqrt(n)**2 == n else 0
def test_xy(x, y):
global max_value
for f in permutations(range(10), len_active_k):
for j in range(len_active_k):
full_f[active_k[j]] = f[j]
if test(x) > 0 and test(y) > 0:
max_value = max(max_value, test(x), test(y))
#print(v, max_value)
if len(v) == 2:
test_xy(v[0], v[1])
else:
for x, y in combinations(v, 2):
test_xy(x, y)
print(max_value) #, time()-t)
#{1, 2, 3, 4, 5, 6} is equivalent to {3, 6, 4, 1, 2, 5}
#{1, 2, 3, 4, 5, 6} is distinct from {1, 2, 3, 4, 5, 9}
#But because we are allowing 6 and 9 to be reversed, the two distinct sets in the last example both represent the extended set {1, 2, 3, 4, 5, 6, 9} for the purpose of forming 2-digit numbers.
#How many distinct arrangements of the two cubes allow for all of the square numbers to be displayed?
#Answer:
#1217
from time import time; t=time()
from mathplus import combinations
def test(p, a, b):
if p[0] in a and p[1] in b: return True
if p[0] in b and p[1] in a: return True
return False
s = 0
for b in combinations(range(0, 10), 6):
if 2 not in b and 5 not in b: continue
if 6 in b or 9 in b: b += (6, 9)
for a in combinations(range(0, 10), 6):
if 6 in a or 9 in a: a += (6, 9)
for p in [(0,1),(0,4),(0,9),(1,6),(2,5),(3,6),(4,9),(6,4),(8,1)]:
if not test(p, a, b): break
else: s += 1
print(s//2)#, time()-t
#For this problem we shall assume that a given set contains n strictly increasing elements and it already satisfies the second rule.
#Surprisingly, out of the 25 possible subset pairs that can be obtained from a set for which n = 4, only 1 of these pairs need to be tested for equality (first rule). Similarly, when n = 7, only 70 out of the 966 subset pairs need to be tested.
#For n = 12, how many of the 261625 subset pairs that can be obtained need to be tested for equality?
#NOTE: This problem is related to problems 103 and 105.
#Answer:
#21384
from time import time
t = time()
from mathplus import combinations
M = 12
s = 0
n = M
for k in range(2, n // 2 + 1):
for p in combinations(range(n), k):
qq = [i for i in range(n) if i not in p and i > p[0]]
for q in combinations(qq, k):
if not (all(q[i] > p[i]
for i in range(k)) or all(q[i] < p[i]
for i in range(k))):
s += 1
print(s) #, time()-t
#print(cnt)
active_k = [i for i in range(26) if k[i] > 0]
len_active_k = len(active_k)
if len_active_k > 10: continue
full_f = [0]*26
def test(s):
if full_f[s[0]] == 0: return 0
c = full_f[s[-1]]
if c in (0, 2, 3, 7, 8): return 0
d = full_f[s[-2]]
if c == 5 and d != 2: return 0
if c == 6 and d % 2 == 0: return 0
if d % 2 == 1: return 0
n = reduce(lambda x,y: x*10+y, (full_f[c] for c in s))
return n if isqrt(n)**2 == n else 0
def test_xy(x, y):
global max_value
for f in permutations(range(10), len_active_k):
for j in range(len_active_k):
full_f[active_k[j]] = f[j]
if test(x) > 0 and test(y) > 0:
max_value = max(max_value, test(x), test(y))
#print(v, max_value)
if len(v) == 2:
test_xy(v[0], v[1])
else:
for x, y in combinations(v, 2):
test_xy(x, y)
print(max_value)#, time()-t)
