I want to produce the following string of numbers:

OP never said it was a programming assignment. He just said he wanted to "produce" those numbers.

So OP, you could carve them into a piece of wood, write them down, count them out loud, take n steps and then n steps backwards, etc.

Now this is elegant and simple, thanks

Top kek

He's right tho

If your starting i is a negative number

hiding a check in a conditional is just deceiving yourself that you've achieved simplicity

I mean hiding a check in an absolute value

No? Let's say n is 10, and my starting i is -12.
10 - |(10 - (-12))| = 10 - (22) = -12

Let's say i is -2 then:
10 - |(10 - (-2))| = 10 - 12 = -2

It literally just produces i. That's the joke.
Maybe I misunderstood what you meant though.

Your solution is correct, I just means it's not simpler than actually writing an IF statement

Getting the absolute value is as simple as adding a 0 in the righ-most position of a string of bytes

There's no if involved

Do you even machine code?

i is just an index so it can't be negative. But yeah you're right, I assumed it always started at 1.

n-|n-i|+j-1
where j is the starting number.
in your case it would be -12

10-|10-i|-13 for the i-th number
so the first number in the sequence would be
10-|10-1|-13=-12

...

Ah, right

thank god for haskell
[code]
make n = map show $ [1..n] ++ [n-1,n-2..1]
[/code]

not elegant or simple

print str(range(n)+reversed(range(n-1)))

>not simple
. brainlet

>probably uses a hundred of unnecessary clock cycles
>simple

the only other one here is more complicated and the result is longer

That's because your programming language is garbage

Try it in assembly

>1,2,3 ... n ... 3,2,1

Oh, that's just [math]\{(A_0, A_1, A_2, ... , A_n) \, | \, A_0 = 1, \, A_1 = 2, \, A_2 = 3, \, A_{n-2} = 3, \, A_{n-1} = 2, \, A_n = 1\}.[/math]

pls do this in assembly
[code]
rpn = head . foldl f [] . words
where f (x:y:stk) "-" = y - x : stk
f (x:y:stk) "+" = y + x : stk
f (x:y:stk) "*" = y * x : stk
f (x:y:stk) "/" = y / x : stk
f (x:stk) "log" = log x : stk
f stk n = read n : stk
[/code]

>[code]

not an argument

>wanting to "produce" a mathematical abstraction that cannot exist in the corporeal realm

lol

k = 1;
b = 0;
while(k>0)
{
print "k ";
b = b+1;
k = k + (-1)^(b/100);
}

[1:n,n-1:1] or is it [1:n;n-1:1] in MATLAB

#include "stdafx.h"
#include

using namespace std;

int main()
{
cout > number;

cout

How'd I do senpai?

int[] counter = new int[2*num - 1];
for(int i = 1; i

ls = []
t = []
ls.extend(range(0, 100))
t.extend(range(0, 99))
t.reverse()
ls.extend(t)
print (ls)

[code]
int n;
printf("Type n: ");
scanf("%d", &n);
int a = 0;
int add = 1;
do{
a += add;
if(a != 0){
printf("%d", a);
}
if(a == n){
add = -1;
}
} while(a > 0);
[/code]

I tried. Haven't used C in a while.

Python and two loops:
[code]
n = input("Type n: ")
for a in range(n):
print a + 1
for a in range(n-1):
print n-a
[/code]

Ah fuck, no code tags on Veeky Forums? Lame.

Take your pedophile cartoons back to and your programming homework back to .

Fucking degenerate.