challenging task

[aal_fareed]

Dear All ,
Its been while since i posted here ..
Am trying to achieve a task using alpha using xbasic or java-script it doesn't matter .
I have dim 26 variables twice assuming it's key-bored letters ...
26 Small Letter ( English )
26 Capital Letters ( English )
Beside the above i have 10 digits ( 0-9 ) the numbers in keybored
13 symbols (assuming Others key-bored strike like , . ? #$% etc )
Finally i have the password length ( assuming it's 10 )
What i need to achieve are combination of all the above key bored strike so i can record all key bored strike possibilities ...
I thought about this as loop for each set ..
the equation should be followed by each number factorial ...26 ! x 26 ! x 10! x 13! x 10! = ( Huge Result Numer )
Factorial example :
5! = 5x4x3x2x1 to have an idea about it ...

I know this is almost infinity but i still want to do it and for each loop i want to store the generated into my sql table ...
Can you help me getting this done ?

thanks in advance

Ahmed Shahin

[madtowng]

Re: challenging task

Not infinity. you are looking for 839 299 365 868 340 224 different combinations.
With almost 839.3 quadrillion combinations, you'll have to choose where to store them, but the code
below should produce each possible password.

Good luck with this, and have fun.
Gregg

debug(1)
dim keystroke[0] as p
for pressed = 48 to 57
	keystroke[].key = chr(pressed)
	keystroke[..].asciivalue = pressed
next
for pressed = 65 to 90
	keystroke[].key = chr(pressed)
	keystroke[..].asciivalue = pressed
next
for pressed = 97 to 122
	keystroke[].key = chr(pressed)
	keystroke[..].asciivalue = pressed
next
debug(1)

for each firstkey in keystroke
	pw = ""
	pw = firstkey.key
	
	for each secondkey in keystroke
		pw = left(pw,1) + secondkey.key
			
		for each thirdkey in keystroke
			pw = left(pw,2) + thirdkey.key
			
			for each fourthkey in keystroke
				pw = left(pw,3) + fourthkey.key
				
				for each fifthkey in keystroke
					pw = left(pw,4) + fifthkey.key
					
					for each sixthkey in keystroke
						pw = left(pw,5) + sixthkey.key	
					
						for each seventhkey in keystroke
							pw = left(pw,6) + seventhkey.key
							
							for each eighthkey in keystroke
								pw = left(pw,7) + eighthkey.key
									
								for each ninethkey in keystroke
									pw = left(pw,8) + ninethkey.key
										
									for each tenthkey in keystroke
										pw = left(pw,9) + tenthkey.key
										' pw should now have a complete 10 character password
										' choose where to store the password
										debug(1)	
									next
								next
							next
						next
					next
				next
			next
		next
	next
next

[madtowng]

Re: challenging task

After reviewing my answer, I see I missed the punctuation keys.
Those should be relatively easy to add in.

[aal_fareed]

Re: challenging task

wow ... i will definitely go with this today ,

thanks a lot Gregg...

[madtowng]

Re: challenging task

Glad to help.

[bob9145]

Re: challenging task

Sounds like you'll dim the lights at the power station. Is this a job for a Quantum machine? What is the purpose of this challenge? Hmmmm
Bob

[CraigSchumacker]

Re: challenging task

Are we hacking with Alpha? If so, even with an i9 9th gen, you won't be starting any time soon...

Nicely done code, Gregg. I'm impressed you got that together in 3 hours. I haven't tried it, but it looks well thought out despite my feeling it seems demonic in nature!

[aal_fareed]

Re: challenging task

Are we hacking with Alpha? If so, even with an i9 9th gen, you won't be starting any time soon...

Nicely done code, Gregg. I'm impressed you got that together in 3 hours. I haven't tried it, but it looks well thought out despite my feeling it seems demonic in nature!

i always have an arguments when it comes to hacking tools ,

to be honest !
hacking exist with alpha or without it ... you or i or any one cannot stop it .. in my opinion .. bringing awareness to people is the best method to fight it , beside that i think its better to do ethical hacking in public
so everyone can consider security .. the above code is old technic .. used as brutal force .. it exist probably 30 years ago ...

there is no harsh doing such a thing ... and to avoid the above method , you may use pictures while entering wrong password 3 times .. hence , that's why any email providers would give that picture to confirm you are not a robot using brutal force to hack emails ..

i consider the above code as achievement with alpha although the method is old .. but Gregg was kind engaph to write it .. it shows the different between normal people and brilliant ones that he done it in few hours ...
thanks Gregg again ...

so please chill and lets enjoy coding :)

[aal_fareed]

Re: challenging task

Sounds like you'll dim the lights at the power station. Is this a job for a Quantum machine? What is the purpose of this challenge? Hmmmm
Bob

i think ... quantum computing exist ..
after all , these kind of knowledge do not go to public , but let me share this with you ...

there is achieved project alredy that would share all computers cpu proccess, ram and even storage ...

i already saw that and i know it exist ...
there is even more than this ,

the problem with such subjects is , most people would bring personal issues , hate etc ..

if you are interested add me and i will tell you how you would control the entire world wide web ... and i mean this ..
it takes only team work , understanding and thinking out of the box ...

thanks for your comment ..

[aal_fareed]

Re: challenging task

Not infinity. you are looking for 839 299 365 868 340 224 different combinations.
With almost 839.3 quadrillion combinations, you'll have to choose where to store them, but the code
below should produce each possible password.

Good luck with this, and have fun.
Gregg

debug(1)
dim keystroke[0] as p
for pressed = 48 to 57
	keystroke[].key = chr(pressed)
	keystroke[..].asciivalue = pressed
next
for pressed = 65 to 90
	keystroke[].key = chr(pressed)
	keystroke[..].asciivalue = pressed
next
for pressed = 97 to 122
	keystroke[].key = chr(pressed)
	keystroke[..].asciivalue = pressed
next
debug(1)

for each firstkey in keystroke
	pw = ""
	pw = firstkey.key
	
	for each secondkey in keystroke
		pw = left(pw,1) + secondkey.key
			
		for each thirdkey in keystroke
			pw = left(pw,2) + thirdkey.key
			
			for each fourthkey in keystroke
				pw = left(pw,3) + fourthkey.key
				
				for each fifthkey in keystroke
					pw = left(pw,4) + fifthkey.key
					
					for each sixthkey in keystroke
						pw = left(pw,5) + sixthkey.key	
					
						for each seventhkey in keystroke
							pw = left(pw,6) + seventhkey.key
							
							for each eighthkey in keystroke
								pw = left(pw,7) + eighthkey.key
									
								for each ninethkey in keystroke
									pw = left(pw,8) + ninethkey.key
										
									for each tenthkey in keystroke
										pw = left(pw,9) + tenthkey.key
										' pw should now have a complete 10 character password
										' choose where to store the password
										debug(1)	
									next
								next
							next
						next
					next
				next
			next
		next
	next
next

Gregg, do you mind if you can include nonasci strikes as well ?

also i am thinking to loop the 10 length as well so it can cumulatively store possibility from 1 to 10 length pass .. so i get the possibilities of everything from 1 to 10 ..

thanks again

[aal_fareed]

Re: challenging task

Not infinity. you are looking for 839 299 365 868 340 224 different combinations.
With almost 839.3 quadrillion combinations, you'll have to choose where to store them, but the code
below should produce each possible password.

Good luck with this, and have fun.
Gregg

debug(1)
dim keystroke[0] as p
for pressed = 48 to 57
	keystroke[].key = chr(pressed)
	keystroke[..].asciivalue = pressed
next
for pressed = 65 to 90
	keystroke[].key = chr(pressed)
	keystroke[..].asciivalue = pressed
next
for pressed = 97 to 122
	keystroke[].key = chr(pressed)
	keystroke[..].asciivalue = pressed
next
debug(1)

for each firstkey in keystroke
	pw = ""
	pw = firstkey.key
	
	for each secondkey in keystroke
		pw = left(pw,1) + secondkey.key
			
		for each thirdkey in keystroke
			pw = left(pw,2) + thirdkey.key
			
			for each fourthkey in keystroke
				pw = left(pw,3) + fourthkey.key
				
				for each fifthkey in keystroke
					pw = left(pw,4) + fifthkey.key
					
					for each sixthkey in keystroke
						pw = left(pw,5) + sixthkey.key	
					
						for each seventhkey in keystroke
							pw = left(pw,6) + seventhkey.key
							
							for each eighthkey in keystroke
								pw = left(pw,7) + eighthkey.key
									
								for each ninethkey in keystroke
									pw = left(pw,8) + ninethkey.key
										
									for each tenthkey in keystroke
										pw = left(pw,9) + tenthkey.key
										' pw should now have a complete 10 character password
										' choose where to store the password
										debug(1)	
									next
								next
							next
						next
					next
				next
			next
		next
	next
next

Worked like a charm ... its currently inserting values to mysql table .. not sure when it will end but .. wow ..
this is going to be really exciting when its done ...

thanks a billion again Gregg .

[madtowng]

Re: challenging task

Greg , do you mind if you can include nonasci strikes as well ?

also i am thinking to loop the 10 length as well so it can cumulatively store possibility from 1 to 10 length pass .. so i get the possibilities of everything from 1 to 10 ..

thanks again

I don't mind at all, but you'll have to code the non-ascii characters yourself.
I'm pretty sure I designed this to create pw codes with a length of 10, but I will admit I did my actual testing setting the length to 3, and then running the test just long
enough to make sure it did what I expected.

As a last item, there is another g at the end of my name. It's a minor deal, but kind of like me changing your name to al_fareed.

[aal_fareed]

Re: challenging task

I don't mind at all, but you'll have to code the non-ascii characters yourself.
I'm pretty sure I designed this to create pw codes with a length of 10, but I will admit I did my actual testing setting the length to 3, and then running the test just long
enough to make sure it did what I expected.

As a last item, there is another g at the end of my name. It's a minor deal, but kind of like me changing your name to al_fareed.

Please accept my warm apology for misspelling your name .. it’s adjusted now .

thanks a lot Gregg ,

My name is ahmed btw aal_fareed is just a nick name

Please call me Ahmed :)

Thanks a billion for your kind help sir .

[aal_fareed]

Re: challenging task

I don't mind at all, but you'll have to code the non-ascii characters yourself.
I'm pretty sure I designed this to create pw codes with a length of 10, but I will admit I did my actual testing setting the length to 3, and then running the test just long
enough to make sure it did what I expected.

As a last item, there is another g at the end of my name. It's a minor deal, but kind of like me changing your name to al_fareed.

i needed example on NoneAsci Code is what i meant ...
you can define one to give the idea ..
when i run the code first time alpha crashed ...

am gonna play and tweak the code and keep you posted ...
actually your code have no problem at all , i think its the alpha version which i am having what caused this issue ..