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# difficult series ?

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Hi there, I wanna describe a formula for a series that happened to appear in my program but was unable to find a good one until now. The series looks like the following: Input (''x'') Output (''y'') x = 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 ... y = 0 1 2 3 4 1 2 3 4 5 6 7 8 1 2 3 4 ... and so on. Any hint would be very appreciated. In the meanwhile I''m looking for my math book. Lorenz

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You''ll have to give us more than that. I''d say at LEAST three ''cycles''.

i.e.
0 1 2 3 4 -> 1st cycle
1 2 3 4 5 6 7 8 -> 2nd cycle

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Ok,Ok,

0 0
1 1
2 2
3 3
4 4
5 1
6 2
7 3
8 4
9 5
10 6
11 7
12 8
13 1
14 2
15 3
16 4
17 5
18 6
19 7
20 8
21 9
22 10
23 11
24 12
25 13
26 14
27 15
28 16
29 1
30 2
31 3
32 4
33 5
.. ..
60 32
61 1
................. and so on ...

There you go.

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I think the function to produce your series would look something like:

  DWORD GetNum( DWORD x ){        DWORD d = 4;        while( x / d )        {                x -= d;                d <<= 1;        }        return x;}

Check it out and post back any questions u have

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The first thing you want to do when looking for a relationship is identify a pattern in the output, since the input is just sequencial from 0. From what I can see, excluding the first element, it starts at 1 then goes to 4. It then repeats from 1 and goes to 8, and next repeats from 1 to 16, 1 to 32. I'd say that the number it counts to is increasing by a power of two each iteration.

The hardest part now it coming up with a formula that maps your sequencial input to this pattern. The first step is to break down each iteration and try to look for relationships in the input, now that we have identified the output pattern:
1 1     5 1     13 1    29 172 2     6 2     14 2    30 183 3     7 3     15 3      .4 4     8 4     16 4     etc.        9 5     17 5        10 6    18 6        11 7    19 7        12 8    20 8                21 9                22 10                23 11                24 12                25 13                26 14                27 15                28 16

I looked at it for about 15 minutes, and I couldn't come up with anything off the top of my head. It's hard to go backwards

[edited by - Zipster on January 21, 2003 5:18:17 AM]

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int yourValue(int x) { int nextPowerOf2 = findNextPowerOf2(x); return x - nextPowerOf2;}

code for next power of two can be found on flipcode..

"take a look around" - limp bizkit

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int fn(int x)
{
if (x == 0) return 0;
int y = x + 4;
return y - highestpowerof2LT(y);
}

''highestpowerof2LT'' is simply the highest power of 2 that is less than y. Some processors have a function to do this in a cycle or two. Otherwise you''ll need a lookup or loop, e.g.

int highestpowerof2LT(int y)
{
z = 1;
while (z < y) z <<= 1;
return z >> 1;
}

If you are generating a sequence you can avoid this loop by incrementing z whenever it is needed as the series advances.

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Just my usual caution against posting homework questions here. Not sure if this is one, since I know sprinter_trueno has been doing game related stuff, but please be clear what your purpose is and avoid asking homework questions.

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