numbers = [[], [], [], [], [], [], [], [], []]

n=1
while True:
tri = n*(n+1)/2
squ = n*n
penta = n*(3*n-1)/2
hexa = n*(2*n-1)
hepta = n*(5*n-3)/2
octa = n*(3*n-2)
if tri > 9999:
break
if tri > 999 and tri 999 and squ 999 and penta 999 and hexa 999 and hepta 999 and octa < 10000:
numbers[8].append([n, octa, 8])
n=n+1

def isCyclic(n1, n2):
return str(n1)[2:4] == str(n2)[0:2]

def solve(solution):
if len(solution) == 6:
if isCyclic(solution[len(solution)-1][1], solution[0][1]) and solution[len(solution)-1][0] != solution[0][0]:
print "Found solution:"
print solution
s = 0
for sol in solution:
s = s + sol[1]
print "Sum is: " + str(s)
return
else:
for n in range(3,9):
Found = False
for sol in solution:
if sol[2] == n:
Found = True
break
if Found:
continue
for i in range(0,len(numbers[n])):
if len(solution) == 0 or numbers[n][i][0] != solution[len(solution)-1][0]:
if len(solution) == 0 or isCyclic(solution[len(solution)-1][1], numbers[n][i][1]):
solution.append(numbers[n][i])
solve(solution)
solution.pop()

solve([])

I noticed you have solutions for some of the project Euler problems published.

They look nice, but I would kindly like you to reconsider if it is appropriate / in the spirit of Project Euler to publish solutions in public?

There are many cheaters spamming the high score lists in PE due to the easy accessibility of solutions on the net. I think it is a pity that solutions are available in public. It takes away a lot of the fun for those of us who work trhough it ourselves to see users climb the high score lists in a pace which implies that they have to be cheating.

I do realize that the javaist blog is not the only one having solutions published, I will attempt to contact those of them I manage to find as well.

Please help the potential cheaters but not making it possible to cheat

Best wishes,

Kim aka user Slaunger on PE

(define (3-or-5 limit)
(3-or-5-helper limit 0 0))

(define (3-or-5-helper limit start sum)
(cond ((= limit start) sum)
((or (= (remainder start 3) 0)
(= (remainder start 5) 0))
(3-or-5-helper limit (+ 1 start) (+ start sum)))
(else (3-or-5-helper limit (+ 1 start) sum))))

(3-or-5 1000)

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?- findall(N, (N mod 3 #= 0 #\/ N mod 5 #= 0, N in 0..999, indomain(N)), Ns), sum(Ns, #=, Sum).
Ns = [0, 3, 5, 6, 9, 10, 12, 15, 18|...],
Sum = 233168.