Programming

[_madness_]

Looking at it, I realize that I called factorial() twice for each iteration, which is too inefficient. Here, I fixed it:

def nth_permutation(n,objects):
    perm = []
    n -= 1
    while len(objects) > 0:
        k = factorial(len(objects)-1)
        m = int(n/k)
        n -= m * k
        perm.append(objects[m])
        objects.pop(m)
    return perm

@ByteInBits, you said

Normally I would smile and let it pass but I have to take issue with you
to come to some understanding.

In post #98171 you state:
@ByteInBits, your MD5 sum is for the NEXT permutation. By my example, you should start with 1 and not 0.

This make you incorrect with the entry number!!
(standard practice doese not start at 1, and you never stated the deviation)

But I did. When I said:

Notice that I listed them in a sort of numerical order: 123, 132, 213, 231, 312, 321. There are 3! = 6 of them. If I were to ask you what is the fourth permutation of 1, 2, 3, you should answer “2, 3, 1”. OK, got it?

I was very careful about making sure the list started at 1, so I said “fourth” instead of “third”. I even stopped to ask if everyone understood before I went on.

Never wager against a Sicilian when death is on the line.

[robb]

@ByteInBits PROGRAMMING QUEST #4 #98187 and #98369

Sorting the list of random numbers was an interesting challenge in this task; for example, the correct ordering of 6531, 65 in set 4. Here’s how I tackled that part:

def cnvrtsort(s):
#convert set s to a list with string values sorted in descending order
sstr = [str(elem) for elem in s]
sl = list(sorted(sstr, reverse=True))
sc =compare(sl)
#Combine sc to get a single number and print
sf=”.join(sc)
print(sf)

def compare(s):
#compare consecutive elements and swap if needed
for i in range(0,len(s)-1):
if s[i][:len(s[i+1])] == s[i+1]:
a = s[i]
b = s[i+1]
rmv=a[len(str(b)):]

if s[i+1]>rmv:
#swap order
s[i] = b
s[i+1]=a
return s

[ByteInBits]

@_madness_ Thank you, and sorry for my misunderstanding of your narrative which now make sense. @rob also helped get me there.

[_madness_]

@ByteInBits, It’s all cool.

[_madness_]

If anyone is interested in something almost as much fun as a seriated Polybius cipher, this programming challenge may or may not be hard. I don’t know. The challenge is to find a Pythagorean triple (x^2 + y^2 = z^2) where each number has more than 100 digits. To show you that it might be easy, here is one:

5448316437483445534218637633951372103721472133084302800181196711700804618365736182804131733102336602000582925889704780559079553695392126450638360766858239249449202384020719285919431179599065584204861405445366919281541626876

590954136213084844483425568023560779816003827176678161335509531818673831091741858041079741290341646336920339282126397662570865581372141104223408019010508444113688638386186080310786949311722657164655061901989505154837335335001963211536629992088337093907

590954136213084844483425568023560779816003827176678161335534647263497664985816974232810815094373455008058948195003268779709416780329753566378275824425812622524697460051934566226668094855294530020882890470426060467302586456555110110607191094703028457845

But then again, maybe it’s hard.

What’s harder? Constructing examples or checking they are right?! Harry

[Author]

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