flat assembler
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revolution
windwakr: ∞^∞^∞^∞^∞


31 Dec 2009, 03:06 

shoorick
what is "^"  "power" or "xor" ?


31 Dec 2009, 09:22 

revolution
shoorick wrote: what is "^"  "power" or "xor" ? 

31 Dec 2009, 09:30 

r22
"largest #" is still my favorite solution. Although the stuff with Grahm's number and a hyper cube of the Grahm's number dimension were also entertaining.


31 Dec 2009, 14:20 

Borsuc
revolution wrote:
because otherwise it would be 0 _________________ Previously known as The_Grey_Beast 

31 Dec 2009, 15:30 

windwakr
I think revolution means it doesn't matter whether it's pow or xor because the result from either would be undefined.


31 Dec 2009, 16:38 

Borsuc
I thought infinity xor infinity is 0... after all, that's what XORing something with itself yields no matter what it is.


31 Dec 2009, 17:53 

windwakr
Well, you would think infinity minus infinity would be 0....but guess what? It's not.
Infinity is too cool to follow the rules. EDIT: Wolfram Alpha(the only computing thing online I could find that understands the infinity symbol) says "∞ xor ∞" is 0......I guess you're right. Infinity is a very confusing concept. 

31 Dec 2009, 18:28 

revolution
No, I just extended windwakr's example where ∞=7F800000h.
So ∞^∞^∞^∞^∞ = ∞; where ^ = xor, or ^ = power. Same answer. 

31 Dec 2009, 21:16 

edemko
example 1:
system: hexadecimal radix: 0x10 used chars: '0'..'9' + 'a'..'f' rank count(for a byte): 0x02 num. variations: 0x10^0x02 = 0xff +1 example 2: system: binary radix: 10b used chars '0'..'1' rank count(for a byte): 1000b num. variations: 10b^1000b = 11111111b +1 example 3: system: 0x20_0x7e_revolution radix: 0x7e  0x20 +1 = 0x5f used chars: 0x20..0x7e rank count(as U desired): 0x09 num. variations: radix^rank count +1 = 0x5f^0x09 = 0x8BF187FBA88F35F +1 As you have mentioned we get a qword value. BTW, are you making a su perior keygen? Last edited by edemko on 17 May 2010, 21:42; edited 4 times in total 

01 Jan 2010, 02:07 

edemko
Happy New You Know 2010!


01 Jan 2010, 02:12 

revolution
Yes, of course, there are 95^9 (630249409724609375) possible different answers. Most of which will be nonsense. Therefore there is a highest value and thus a definitive answer to this puzzle. The trick is in finding which of those 630249409724609375 possible answers is the correct one.


01 Jan 2010, 02:56 

tom tobias
borsuc wrote: I thought infinity xor infinity is 0... after all, that's what XORing something with itself yields no matter what it is. windwakr wrote: Well, you would think infinity minus infinity would be 0....but guess what? It's not. Revolution wrote: No, I just extended windwakr's example where ∞=7F800000h Happy New Year 

01 Jan 2010, 13:37 

revolution
tom tobias: The IEEE754 spec defines infinity with specific encodings.
But you knew that right? 

01 Jan 2010, 13:41 

tom tobias
Thanks for asking revolution, and Happy New Year.
Nope, didn't know of that error. Infinity cannot be bounded. There are no limits for an infinite number, and accordingly, there are no specific values, which may be assigned to a number defined as infinitely large or small. 

01 Jan 2010, 21:27 

revolution
tom tobias wrote: Sorry, that is an illegal assignment. Infinity can not be assigned a numeric value. It can not be limited, hence, no Boolean operator can act upon it. Code: include 'win32ax.inc' .data align 16 infinities: dd 7F800000h,7F800000h,7F800000h,7F800000h .code begin: movaps xmm0,[infinities] movaps xmm1,xmm0 pxor xmm0,xmm1 pxor xmm0,xmm1 pxor xmm0,xmm1 pxor xmm0,xmm1 movaps [infinities],xmm0 sub esp,1024 mov eax,esp invoke wvsprintf,eax,<'Result is:%lx,%lx,%lx,%lx'>,infinities mov eax,esp invoke MessageBox,0,eax,eax,0 invoke ExitProcess,0 .end begin 

01 Jan 2010, 21:40 

LocoDelAssembly
Code: format pe console include 'win32ax.inc' movapd xmm0,[infinities] movapd xmm1,xmm0 pxor xmm0,xmm1 pxor xmm0,xmm1 pxor xmm0,xmm1 pxor xmm0,xmm1 movapd [infinities],xmm0 cinvoke printf, <'Result is: %f, %f', 10>, double [infinities], double [infinities+8] cinvoke system, "pause" invoke ExitProcess, 0 align 16 infinities: dq $7ff0'0000'0000'0000,\ ;+INF $fff0'0000'0000'0000 ;INF (This is only to see what printf prints, the property doesn't hold for INF) data import library kernel32, 'kernel32.dll',\ msvcrt,'msvcrt.dll' import kernel32,\ ExitProcess, 'ExitProcess' import msvcrt,\ printf, 'printf',\ system, 'system' end data Quote: Result is: 1.#INF00, 1.#INF00 [edit] Changed code to use 64bit infinite from start and also added minus infinity to the test[/edit] 

01 Jan 2010, 22:24 

edfed
∞^∞^∞^∞.1


02 Jan 2010, 15:00 

revolution


02 Jan 2010, 16:38 

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