looping in two directions - php

Question

hey I'm looking for are clean solution to this problem:
i start the loop with i = 0 in the second loop step the i = 1, then i = -1 and then i = 2 ect.
how to programm this with a for loop in a clean way?

Answer 1

f(0); //do stuff with 0
for(var i = 1; i<len; i++) //where len = positive boundary
{
f(i); //do stuff with i
f(-i); //do stuff with -i
}
Should do what you want

Answer 2

If you don't mind having the inner loop appear 3 times:
f(0);
for (var i = 1; i <= 3; ++ i) {
f(i);
f(-i);
}
2 times with an if:
for (var i = 0; i <= 3; ++ i) {
f(i);
if (i > 0)
f(-i);
}
single time but with an ugly expression:
for (var j = 1; j <= 7; ++ j) {
var i = j / 2;
if (j % 2) i = -i;
f(i);
}

Answer 3

Each loop, you appear to be adding n*(-1)^(n+1), where n is the step you are currently taking, starting at 1, and starting at i=0.
initialize i = 0
n=0, i+=0*(-1)^1 # 0
n=1, i+=1*(-1)^2 # 1
n=2, i+=2*(-1)^3 # -1
n=3, i+=3*(-1)^4 # 2
etc.
From here, it depends on what language you would wish to write in. Iterate from n = 0 to wherever you are stopping.
edit this is a bad answer. but fun =D
(I added that last bit because as soon as I made that edit, someone downvoted me =( )

Answer 4

Here is implementation in javascript
for ( var i = 0; Math.abs(i)<10; i=(i<=0?Math.abs(i)+1:-i)) {
console.debug(i)
}
Hope it helps.

Answer 5

Just one addition one subtraction and a negation:
for(int i=0, d=1, f=-1; i<10; i+=d, d=f-d, f=-f)
{
printf("%d\n", i);
}
generates an inner loop of:
push esi
push offset string "%d\n" (0D20F4h)
call dword ptr [__imp__printf (0D20A4h)]
mov eax,ebx
add esi,edi
sub eax,edi
add esp,8
neg ebx
mov edi,eax
cmp esi,0Ah
jl wmain+10h (0D1010h)

Answer 6

I used the sine function:
for ($i = 0; $i < 10; $i++)
{
echo round(0.5 * $i * sin((0.5 + $i) * M_PI))."\n";
}

Answer 7

for (int i = 0; i < 10; i++)
{
int div = i / 2;
int mod = i % 2;
int offset = mod == 0 ? div : -div;
}

Answer 8

There is a pattern to this loop. Looking at it on the number line - it goes like:
0 steps backward
1 step forward
2 steps backward
3 steps forward
4 steps backward
Here's one solution - keep incrementing the step size in each iteration of the loop, and flip direction (forward/backward) every time. Keep adding to the current value.
// n is the number of elements to generate
for(var i = 0, value = 0, dir = -1; i < n; i++) {
value = value + (dir * i);
console.log(value);
dir = dir * -1; // reverse direction
}
Another solution using generators in JavaScript 1.7 which is identical to #FallingBullet's solution but more aesthetically pleasing to my eye :)
function sequence() {
var i = 0;
yield i;
while(true) {
i++;
yield i;
yield -i;
}
}
var seq = sequence();
seq.next(); // 0
seq.next(); // 1
seq.next(); // -1
seq.next(); // 2
...

Answer 9

For what it's worth, here is my own interpretation of the problem.
for (var i = 0; i > -8; i = (i<=0) - i) // arbitrary condition stops loop at -8

Answer 10

A modification of falling bullet's solution, that will handle the 0 index case without a special condition.
//do stuff with 0
for(int i = 0; i< (arrayLength/2); i++)
{
//do stuff with i
if(-i != i)
{
//do stuff with negIndex
}
}

Answer 11

In C. The value of N is the total number of values in the sequence you wish to yield.
int i, n = 0, m = 1;
for (i = 1; i < N; i++, m = -m) {
/* n is the next in the sequence */
n += m * i;
}

Answer 12

I'd probably go with:
for (var i = 0; i <= max; i = ( i <= 0 ) ? -i + 1 : -i)
{
f( i );
}