How can I generate this pattern of numbers? - algorithm

Question

Given inputs 1-32 how can I generate the below output?
in. out
1
1
1
1
2
2
2
2
1
1
1
1
2
2
2
2
...
Edit Not Homework.. just lack of sleep.
I am working in C#, but I was looking for a language agnostic algorithm.
Edit 2 To provide a bit more background... I have an array of 32 items that represents a two dimensional checkerboard. I needed the last part of this algorithm to convert between the vector and the graph, where the index aligns on the black squares on the checkerboard.
Final Code:
--Index;
int row = Index >> 2;
int col = 2 * Index - (((Index & 0x04) >> 2 == 1) ? 2 : 1);

Answer 1

Assuming that you can use bitwise operators you can check what the numbers with same output have in common, in this case I preferred using input 0-31 because it's simpler (you can just subtract 1 to actual values)
What you have?
0x0000 -> 1
0x0001 -> 1
0x0010 -> 1
0x0011 -> 1
0x0100 -> 2
0x0101 -> 2
0x0110 -> 2
0x0111 -> 2
0x1000 -> 1
0x1001 -> 1
0x1010 -> 1
0x1011 -> 1
0x1100 -> 2
...
It's quite easy if you notice that third bit is always 0 when output should be 1 and viceversa it's always 1 when output should be 2
so:
char codify(char input)
{
return ((((input-1)&0x04)>>2 == 1)?(2):(1));
}
EDIT
As suggested by comment it should work also with
char codify(char input)
{
return ((input-1 & 0x04)?(2):(1));
}
because in some languages (like C) 0 will evaluate to false and any other value to true. I'm not sure if it works in C# too because I've never programmed in that language. Of course this is not a language-agnostic answer but it's more C-elegant!

Answer 2

in C:
char output = "11112222"[input-1 & 7];
or
char output = (input-1 >> 2 & 1) + '1';
or after an idea of FogleBird:
char output = input - 1 & 4 ? '2' : '1';
or after an idea of Steve Jessop:
char output = '2' - (0x1e1e1e1e >> input & 1);
or
char output = "12"[input-1>>2&1];
C operator precedence is evil. Do use my code as bad examples :-)

Answer 3

You could use a combination of integer division and modulo 2 (even-odd): There are blocks of four, and the 1st, 3rd, 5th block and so on should result in 1, the 2nd, 4th, 6th and so on in 2.
s := ((n-1) div 4) mod 2;
return s + 1;
div is supposed to be integer division.
EDIT: Turned first mod into a div, of course

Answer 4

Just for laughs, here's a technique that maps inputs 1..32 to two possible outputs, in any arbitrary way known at compile time:
// binary 1111 0000 1111 0000 1111 0000 1111 0000
const uint32_t lu_table = 0xF0F0F0F0;
// select 1 bit out of the table
if (((1 << (input-1)) & lu_table) == 0) {
return 1;
} else {
return 2;
}
By changing the constant, you can handle whatever pattern of outputs you want. Obviously in your case there's a pattern which means it can probably be done faster (since no shift is needed), but everyone else already did that. Also, it's more common for a lookup table to be an array, but that's not necessary here.

Answer 5

The accepted answer return ((((input-1)&0x04)>>2 == 1)?(2):(1)); uses a branch while I would have just written:
return 1 + ((input-1) & 0x04 ) >> 2;

Answer 6

Python
def f(x):
return int((x - 1) % 8 > 3) + 1
Or:
def f(x):
return 2 if (x - 1) & 4 else 1
Or:
def f(x):
return (((x - 1) & 4) >> 2) + 1

Answer 7

In Perl:
#!/usr/bin/perl
use strict; use warnings;
sub it {
return sub {
my ($n) = #_;
return 1 if 4 > ($n - 1) % 8;
return 2;
}
}
my $it = it();
for my $x (1 .. 32) {
printf "%2d:%d\n", $x, $it->($x);
}
Or:
sub it {
return sub {
my ($n) = #_;
use integer;
return 1 + ( (($n - 1) / 4) % 2 );
}
}

Answer 8

In Haskell:
vec2graph :: Int -> Char
vec2graph n = (cycle "11112222") !! (n-1)

Answer 9

Thats pretty straightforward:
if (input == "1") {Console.WriteLine(1)};
if (input == "2") {Console.WriteLine(1)};
if (input == "3") {Console.WriteLine(1)};
if (input == "4") {Console.WriteLine(1)};
if (input == "5") {Console.WriteLine(2)};
if (input == "6") {Console.WriteLine(2)};
if (input == "7") {Console.WriteLine(2)};
if (input == "8") {Console.WriteLine(2)};
etc...
HTH

Answer 10

It depends of the language you are using.
In VB.NET, you could do something like this :
for i as integer = 1 to 32
dim intAnswer as integer = 1 + (Math.Floor((i-1) / 4) mod 2)
' Do whatever you need to do with it
next
It might sound complicated, but it's only because I put it into a sigle line.

Answer 11

In Groovy:
def codify = { i ->
return (((((i-1)/4).intValue()) %2 ) + 1)
}
Then:
def list = 1..16
list.each {
println "${it}: ${codify(it)}"
}

Answer 12

char codify(char input)
{
return (((input-1) & 0x04)>>2) + 1;
}

Answer 13

Using Python:
output = 1
for i in range(1, 32+1):
print "%d. %d" % (i, output)
if i % 4 == 0:
output = output == 1 and 2 or 1

Answer 14

JavaScript
My first thought was
output = ((input - 1 & 4) >> 2) + 1;
but drhirsch's code works fine in JavaScript:
output = input - 1 & 4 ? 2 : 1;
and the ridiculous (related to FogleBird's answer):
output = -~((input - 1) % 8 > 3);

Answer 15

Java, using modulo operation ('%') to give the cyclic behaviour (0,1,2...7) and then a ternary if to 'round' to 1(?) or 2(:) depending on returned value.
...
public static void main(String[] args) {
for (int i=1;i<=32;i++) {
System.out.println(i+"="+ (i%8<4?1:2) );
}
Produces:
1=1 2=1 3=1 4=2 5=2 6=2 7=2 8=1 9=1
10=1 11=1 12=2 13=2 14=2 15=2 16=1
17=1 18=1 19=1 20=2 21=2 22=2 23=2
24=1 25=1 26=1 27=1 28=2 29=2 30=2
31=2 32=1