BCS wrote:
> 
> char a=0,b=0,c=0,d=0;
> while(a<5 || b<9 || c<5 || d<9)
> {
>   writef("%d%d:%d%d\n", a,b,c,d);
> 
>   // add a single /expression/ here
> }
> 
> make that count from "00:00" to "59:59" (minutes and seconds)
> 
> I seem to remember I figured this out once, but I don't remember how I did it.
> 
> 




import std.stdio;

void main()
{


char a=0,b=0,c=0,d=0;

do{
  writef("%d%d:%d%d\n", a,b,c,d);

   ( d == 9 ? d = 0 || (c == 5 ? c = 0 || ( b == 9 ? b = 0 || ++a : ++b )  : ++c) : ++d);

}while(a<6 || b<0 || c<0 || d<0)

}

Reply to wyverex,

> BCS wrote:
> 
>> char a=0,b=0,c=0,d=0;
>> while(a<5 || b<9 || c<5 || d<9)
>> {
>> writef("%d%d:%d%d\n", a,b,c,d);
>> // add a single /expression/ here
>> }
>> make that count from "00:00" to "59:59" (minutes and seconds)
>> 
>> I seem to remember I figured this out once, but I don't remember how
>> I did it.
>> 
> import std.stdio;
> 
> void main()
> {
> char a=0,b=0,c=0,d=0;
> 
> do{
> writef("%d%d:%d%d\n", a,b,c,d);
> ( d == 9 ? d = 0 || (c == 5 ? c = 0 || ( b == 9 ? b = 0 || ++a :
> ++b )  : ++c) : ++d);
> 
> }while(a<6 || b<0 || c<0 || d<0)
> 
> }
> 

yah, that looks right. I think I used a comma expression or some sutch.

Reply to Koroskin,

> On Thu, 07 Aug 2008 01:50:56 +0400, Wyverex <wyverex.cypher@gmail.com>
> wrote:
> 
>> just some fun little programming puzzles I found around online...
>> 
>> Write a "Hello World" program in 'C' without using a semicolon.
>> (Note: #include in C doesn't need a semicolon but import does)
>> 
>> Problem #1 Write a "Hello World" program in D with only a semicolon
>> on  import statement.
>> 
>> Problem #2 Test if an int is even or odd without looping or if
>> statement  (Cant use: do, while, for, foreach, if).
>> 
>> Problem #3 Write a program without using any loop (if, for, while
>> etc)  to print numbers from 1 to 100 and 100 to 1;
>> 
>> Problem #4 Find if the given number is a power of 2.
>> 
> // 0 is considered a power of two here
> bool isPowerOfTwo(int i) {
> return (i & (i-1) == 0);
> }

isPowerOfTwo(int.min);


0b10000000  == byte.min
0b01111111  == byte.min -1

0b00000000

BCS wrote:
> Reply to Koroskin,
> 
>> On Thu, 07 Aug 2008 01:50:56 +0400, Wyverex <wyverex.cypher@gmail.com>
>> wrote:
>>
>>> just some fun little programming puzzles I found around online...
>>>
>>> Write a "Hello World" program in 'C' without using a semicolon.
>>> (Note: #include in C doesn't need a semicolon but import does)
>>>
>>> Problem #1 Write a "Hello World" program in D with only a semicolon
>>> on  import statement.
>>>
>>> Problem #2 Test if an int is even or odd without looping or if
>>> statement  (Cant use: do, while, for, foreach, if).
>>>
>>> Problem #3 Write a program without using any loop (if, for, while
>>> etc)  to print numbers from 1 to 100 and 100 to 1;
>>>
>>> Problem #4 Find if the given number is a power of 2.
>>>
>> // 0 is considered a power of two here
>> bool isPowerOfTwo(int i) {
>> return (i & (i-1) == 0);
>> }
> 
> isPowerOfTwo(int.min);
> 
> 
> 0b10000000  == byte.min
> 0b01111111  == byte.min -1
> 
> 0b00000000
> 
> 



What are you trying to say?  I imagine your validating Koroskin solution.

10000000  == byte.min
01111111  == byte.min - 1
=
00000000  == is power of 2

-Joel

Wyverex wrote:
> just some fun little programming puzzles I found around online...
> 
> 
> Problem #2 Test if an int is even or odd without looping or if statement (Cant use: do, while, for, foreach, if).
> 
> Problem #4 Find if the given number is a power of 2.
> 
> Post Solutions to this root, comments to someones solution in that thread.


Some of these are pretty standard interview questions. Although I don't personally like to ask these sort of questions because they are often about knowing a "trick" which you an easily lookup.  The can be fun to figure out though.

Here's another common one:

| Write a bitcount for a 32-bit number.

And a little more challenging:

| Write a bitcount for a 32-bit number that is less then 15 operations without using a lookup table.

| Can you do that in 12 or less?

-Joel

Wyverex wrote:

> Sweet! Didn't know printf was accessible without an import! 10pts!

Actually this is due to a bad decision early on, where printf was put into object.d. It is not part of Tango's object.

-- 
Lars Ivar Igesund
blog at http://larsivi.net
DSource, #d.tango & #D: larsivi
Dancing the Tango

On Thu, 07 Aug 2008 10:50:25 +0400, JAnderson <ask@me.com> wrote:

> Wyverex wrote:
>> just some fun little programming puzzles I found around online...
>>    Write a "Hello World" program in 'C' without using a semicolon.
>> (Note: #include in C doesn't need a semicolon but import does)
>>    Problem #1 Write a "Hello World" program in D with only a semicolon on import statement.
>>  Problem #2 Test if an int is even or odd without looping or if statement (Cant use: do, while, for, foreach, if).
>>  Problem #3 Write a program without using any loop (if, for, while etc) to print numbers from 1 to 100 and 100 to 1;
>>  Problem #4 Find if the given number is a power of 2.
>>    Post Solutions to this root, comments to someones solution in that thread.
>
> These are pretty standard interview questions. Although I don't personally like to ask these sort of questions because they are often about knowing a "trick" which you an easily lookup.  The can be fun to figure out though.
>
> Here's another common one:
>
> | Write a bitcount for a 32-bit number.
>
> And a little more challenging:
>
> | Write a bitcount for a 32-bit number that is less then 15 operations without using a lookup table.
>
> | Can you do that in 12 or less?
>
> -Joel

I know the solution for 15 operations, but it is impossible to come up with solution quickly. Besides, it requires 64 bit arithmetic support.

Koroskin Denis wrote:
> On Thu, 07 Aug 2008 10:50:25 +0400, JAnderson <ask@me.com> wrote:
> 
>> Wyverex wrote:
>>> just some fun little programming puzzles I found around online...
>>>    Write a "Hello World" program in 'C' without using a semicolon.
>>> (Note: #include in C doesn't need a semicolon but import does)
>>>    Problem #1 Write a "Hello World" program in D with only a semicolon on import statement.
>>>  Problem #2 Test if an int is even or odd without looping or if statement (Cant use: do, while, for, foreach, if).
>>>  Problem #3 Write a program without using any loop (if, for, while etc) to print numbers from 1 to 100 and 100 to 1;
>>>  Problem #4 Find if the given number is a power of 2.
>>>    Post Solutions to this root, comments to someones solution in that thread.
>>
>> These are pretty standard interview questions. Although I don't personally like to ask these sort of questions because they are often about knowing a "trick" which you an easily lookup.  The can be fun to figure out though.
>>
>> Here's another common one:
>>
>> | Write a bitcount for a 32-bit number.
>>
>> And a little more challenging:
>>
>> | Write a bitcount for a 32-bit number that is less then 15 operations without using a lookup table.
>>
>> | Can you do that in 12 or less?
>>
>> -Joel
> 
> I know the solution for 15 operations, but it is impossible to come up with solution quickly. Besides, it requires 64 bit arithmetic support.

The 12 op solution doesn't require 64-bit.  Your right though to come up with something quickly for a question like this is extremely difficult when its been worked on for years, unless you "know" it.

-Joel

JAnderson wrote:
> Wyverex wrote:
>> just some fun little programming puzzles I found around online...
>>
>>
>> Problem #2 Test if an int is even or odd without looping or if statement (Cant use: do, while, for, foreach, if).
>>
>> Problem #4 Find if the given number is a power of 2.
>>
>> Post Solutions to this root, comments to someones solution in that thread.
> 
> 
> Some of these are pretty standard interview questions. Although I don't personally like to ask these sort of questions because they are often about knowing a "trick" which you an easily lookup.  The can be fun to figure out though.
> 
> Here's another common one:
> 
> | Write a bitcount for a 32-bit number.
> 
> And a little more challenging:
> 
> | Write a bitcount for a 32-bit number that is less then 15 operations without using a lookup table.
> 
> | Can you do that in 12 or less?
> 
> -Joel

int count( in int i )
{
   int c = (i & 1) + (i & 2 ? 1 : 0) + (i & 4 ? 1 : 0) + (i & 8 ? 1 : 0);
   i >>= 4;
   c += (i & 1) + (i & 2 ? 1 : 0) + (i & 4 ? 1 : 0) + (i & 8 ? 1 : 0);
   i >>= 4;
   c += (i & 1) + (i & 2 ? 1 : 0) + (i & 4 ? 1 : 0) + (i & 8 ? 1 : 0);
   i >>= 4;
   c += (i & 1) + (i & 2 ? 1 : 0) + (i & 4 ? 1 : 0) + (i & 8 ? 1 : 0);

   return c;
}

int count2( in int i )
{
   int c = (i & 1) + ((i >> 1) & 1) + ((i >> 2) & 1) + ((i >> 3) & 1);
   i >>= 4;
   c += (i & 1) + ((i >> 1) & 1) + ((i >> 2) & 1) + ((i >> 3) & 1);
   i >>= 4;
   c += (i & 1) + ((i >> 1) & 1) + ((i >> 2) & 1) + ((i >> 3) & 1);
   i >>= 4;
   c += (i & 1) + ((i >> 1) & 1) + ((i >> 2) & 1) + ((i >> 3) & 1);

   return c;
}

On Thu, 07 Aug 2008 19:37:27 +0400, Wyverex <wyverex.cypher@gmail.com> wrote:

> JAnderson wrote:
>> Wyverex wrote:
>>> just some fun little programming puzzles I found around online...
>>>
>>>
>>> Problem #2 Test if an int is even or odd without looping or if statement (Cant use: do, while, for, foreach, if).
>>>
>>> Problem #4 Find if the given number is a power of 2.
>>>
>>> Post Solutions to this root, comments to someones solution in that thread.
>>   Some of these are pretty standard interview questions. Although I don't personally like to ask these sort of questions because they are often about knowing a "trick" which you an easily lookup.  The can be fun to figure out though.
>>  Here's another common one:
>>  | Write a bitcount for a 32-bit number.
>>  And a little more challenging:
>>  | Write a bitcount for a 32-bit number that is less then 15 operations without using a lookup table.
>>  | Can you do that in 12 or less?
>>  -Joel
>
> int count( in int i )
> {
>     int c = (i & 1) + (i & 2 ? 1 : 0) + (i & 4 ? 1 : 0) + (i & 8 ? 1 : 0);
>     i >>= 4;
>     c += (i & 1) + (i & 2 ? 1 : 0) + (i & 4 ? 1 : 0) + (i & 8 ? 1 : 0);
>     i >>= 4;
>     c += (i & 1) + (i & 2 ? 1 : 0) + (i & 4 ? 1 : 0) + (i & 8 ? 1 : 0);
>     i >>= 4;
>     c += (i & 1) + (i & 2 ? 1 : 0) + (i & 4 ? 1 : 0) + (i & 8 ? 1 : 0);
>
>     return c;
> }
>
> int count2( in int i )
> {
>     int c = (i & 1) + ((i >> 1) & 1) + ((i >> 2) & 1) + ((i >> 3) & 1);
>     i >>= 4;
>     c += (i & 1) + ((i >> 1) & 1) + ((i >> 2) & 1) + ((i >> 3) & 1);
>     i >>= 4;
>     c += (i & 1) + ((i >> 1) & 1) + ((i >> 2) & 1) + ((i >> 3) & 1);
>     i >>= 4;
>     c += (i & 1) + ((i >> 1) & 1) + ((i >> 2) & 1) + ((i >> 3) & 1);
>
>     return c;
> }

Much simpler:

int getBitCount32(int i) {
    return getBitCount16(i) + getBitCount16(i >> 16);
}

int getBitCount16(int i) {
    return getBitCount8(i) + getBitCount8(i >> 8);
}

int getBitCount8(int i) {
    return getBitCount4(i) + getBitCount4(i >> 4);
}

int getBitCount4(int i) {
    return getBitCount2(i) + getBitCount2(i >> 2);
}

int getBitCount2(int i) {
    return (i & 2) + (i & 1);
}