-nal
sub p{@_?map{my$e=$_;map[$e,@$_],p(grep{$e ne$_}@_)}@_:0}map{print"@$_"}p@F
def f(n,v=[]):
if[f([j for j in n if j!=i],v+[i])for i in n]==n:print v
Usage: >>> f(list)
4 bytes shorter including line break and indentation ;)
I don't know Python (don't like the concept of indenting as block syntax), so how do you actually use this from command line? Note how the perl version also does the parsing of arguments and intial invocation of the recursion.
You have to invoke it yourself, that is just the function definition. (Invoke it like f([1, 2, 3])) Parsing the arguments would take a bit more of work. At least, importing sys and invoking sys.argv, like:
import sys
def f(n, v=[]):
if[f([j for j in n if j!=i],v+[i])for i in n]==n:print v
f(map(int, sys.argv))
No, it's built into the language. It doesn't require the loading of any packages like <<DiscreteMath`Combinatorica`.
I interpreted Gorash's challenge to mean that you were supposed to write your own permutation function/procedure/etc.
That's how I believe he meant for it to be interpreted, as well, but I think it is our responsibility to punish him for insufficiently detailed instructions and an insufficiently limited context. In the Mathematica Programming Language, Permutation[] is not a package. <<DiscreteMath`Combinatorica` is, but luckily you don't need it!
No, it's built into the language. It doesn't require the loading of any packages like <<DiscreteMath`Combinatorica`.
I interpreted Gorash's challenge to mean that you were supposed to write your own permutation function/procedure/etc.
That's how I believe he meant for it to be interpreted, as well, but I think it is our responsibility to punish him for insufficiently detailed instructions and an insufficiently limited context. In the Mathematica Programming Language, Permutation[] is not a package. <<DiscreteMath`Combinatorica` is, but luckily you don't need it!
In the context of this challenge, the Mathematica Programming Language is a package.
How fleeting are all human passions compared with the massive continuity of ducks.
In general, this has no elementary solution. Using the Lambert W function, the solution can be expressed as W((1/50)*y*log(1.1))/log(1.1). But the best idea is to solve it numerically.
I got one (It's not really a problem but more of a question that I want an answer to):
Why is it that x^2-1=(x+1)*(x-1) ?
Because (x+1)*(x-1) = X^2 + 1*x - 1*x + (-1*1) = X^2-1
You could also explain it visually:
(x-a)(x+a) = x^2-a^2. Basically, take a square that is length x by x, and subtract an arbitrary length, and add the same arbitrary length to each side (x-a) and (x+a) repectively. Compare the two areas (x^2 and (x-a)(x+a), and you'll see that x^2 is larger than (x-a)(x+a) by an amount of a^2.
Edit: I made an image
Oh, I got a question! Found this in my math text and couldn't quite solve it even until now:
The positive integers a and b, where a<b, satisfy the equation a^b=b^a
Show that there is only 1 possible value for a and one for b, and find these values.
Couldn't quite do the first part, though the second one came almost intuitively.
PS. for the answer of the second part: a=2, b=4
edit: the answer was still quite visible on tiny, so I coloured it out :)
Joined: 4/11/2006
Posts: 487
Location: North of Russia :[
a^b = b^a (a < b, a and b are integers)
b = a^k (k>1)
a^(a^k) = (a^k)^a (=a^(ak))
ak = a^k
a = k^(1/k-1)
so k^(1/k-1) should be integer with k>1...
but (k^(1/k-1)) when increasing k goes to 1 from above, and the ony integer value it takes is 2 (you can build graph (y(x) = x^(1/x-1)) and see yourself)
a = 2 ^ (1/(2-1)) = 2
b = 2^2 = 4
so, a ^ b = b ^ a only when a = 2 and b = 4 or b = 2 and a = 4
P.S. Thanks, solving this question was quite interesting )
That's an interesting solution... yay i finally can submit my assignment haha.
but (k^(1/k-1)) when increasing k goes to 1 from above, and the ony integer value it takes is 2 (you can build graph (y(x) = x^(1/x-1)) and see yourself)
Though I personally hope that there is another way of working around this. It doesn't really look very convincing, unless of course i use graphamatica. Any other solutions?
Thanks zefiris anyway, you showed me how to proceed past a=k^(1/k-1) :)
Joined: 4/11/2006
Posts: 487
Location: North of Russia :[
Okay
let's prove that k^(1/(k-1)) takes only one integer value when k>1
f(k) = k^((k-1)^-1)
lim f(k) = e
k -> 1
lim f(k) = 1
k -> infinity
f'(k) = -k^(k-2) * (k-1)^-3
k>1 -> k^(k-2) > 0
k>1 -> (k-1)^-3 > 0
f'(k) < 0 when k > 1
therefore it takes values from e downto 1 always decreasing. We can conclude it takes values from range (1;e). 2 is the only integer there.
Joined: 4/11/2006
Posts: 487
Location: North of Russia :[
DrJones wrote:
Prove that no number on the sequence:
5
15
115
1115
... (and so on)
can be divided by seven.
It's simple...
N1 = 5
N1 mod 7 = 5
Nn = (Nn-1 - 4) * 10 + 5
Nn = 10*Nn-1 - 35
35 mod 7 = 0
10*Nn-1 can not be divided by 7 as Nn-1 can not and 10 can not
10*Nn-1 mod 7 = (10*Nn-1 - 35) mod 7
therefore for any n Nn can not be divided by 7
zefiris: your method is sort of an induction right? not sure about your notation(function?) mod; when you use it, the value it presents is the remainder of the quotient, am I right? Nice solution though. I like how you make easy work of our questions haha.
Alright, let's now take a break from all our brain wrecking math, and chill with some math humor. Enjoy :)
Haha... Wonder where MahaTma is. Busy with his own life I guess?
Anyway, I've got another question, which probably most people here would probably be more familiar than I am. Can the binary number system denote all integers? And is there some form of reasoning behind it?
eh... anyone care to try? :)
