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Sane · Jul 7th, 2005, 4:39 PM

I'm creating an encryption program for some time off, and I haven't learned this stuff so I'm having a little bit of trouble.

It's supposed to use abc of the quadratic equation to alter the ascii, and then convert back. But I can't figure out how to reverse a quadratic equation.

I know to change an x to a y in a quadratic equation is
y = x**a + x*b + c

and to get the intercepts of x is

x = -b +- ([b**2 - 4ac]**0.5 / 2a)

but how can I convert y back to x?

For example.

Say the character is 's', when converted to ascii that is 115 (alt + 115).

Now the quadratic equation is:
y = x**a + x*b + c
and we'll say a =1, b = 2, c = 3

so now it is 233.

converted back into ascii is Θ .

But now when it's time to decrypt, how do I get 233 back to 115?

I think it's something like:

x = -b +- ([b**2 - 4ac]**0.5 / 2a)

original X = x * 233

Can someone help?

mackenga · Jul 7th, 2005, 5:10 PM

(Toggle Plain Text)

y = x**a + x * b + c

I assume you're using the ** operator to denote exponentiation. First up, in your example, if x = 115, y != 233; y == 115 ** 1 + 2 * 115 + 3 == 115 + 230 + 3 == 348. Chars are just one byte, but 348 > 2 ** 8, so you will need more than one byte per character for your ciphertext to do it this way (unless there's some funky trick I'm not thinking of off the top of my head, which is possible since the ciphertext will still only contain 256 distinct values).

As for reversing the encryption; first, you can subtract 3 (the 3 added at the end of the encryption process) to get x ** a + x * b == 345. a and b are 1 and 2 respectively, so that's x ** 1 + x * 2 = 345. Any number n to the first power is just n, so you have 3 * x = 345, => x = 345/3 = 115 = your plaintext character s.

This encryption system has a few problems, the most serious of which is that each character of plaintext will still always map to a particular value in the ciphertext, which means this is trivial to crack using frequency analysis and all the mathematical stuff is beside the point.

Sane · Jul 7th, 2005, 7:44 PM

First Paragraph: Ya, I was rushing an example, sorry I missed the initial 115. Once it gets passed 256 it loops to 0, and vice-versa.

Second Paragraph: Okay, thanks. But that still doesn't give me a formula because it won't work if the encryptor chooses > 1 for a.

Third Paragraph: Incorrect, I didn't mention the entire scheme, of falses bytes scattered around in a partially hidden but organised manner. So frequency analysis will lead false. And even if they open the source code, they would need to know the initial value of the text to find out how the false characters were organized.

Btw, this is going to be my completely unecessary method of keeping my programs with no source, everything is going to be an encrypted in a data file. The CD comes with a serial id that happens to be the A B C. When they run the program they enter the number, and if it's correct it translates the text file. Save them all as .pyc, runs them, then deletes them. And the decryptor will be fully obfuscated and compiled.

I'm so good at being unecessary

Sane · Jul 7th, 2005, 8:17 PM

It would take a while, but I could create a loop that adds 0.0000001 to itself, and keeps testing the equation until it reaches approximately Y. *shrug* I might just do that ...

Edit:

That was considerably easier:

(Toggle Plain Text)

y = ?
a = ?
b = ?
c = ?


x = 0.000
TestY = x**a + x*b + c
while int(round(TestY*10)*10) != int(round(y*100)):
    x += 0.01
    TestY = x**a + x*b + c

print x

skuinders · Jul 7th, 2005, 9:51 PM

in general, if you want to push a value through a function "backwards" you will need to have an inverse of the original function. an obvious illustration:

(Toggle Plain Text)

f(x) = e^x
g(x) = ln x

so you need to research methods for finding the inverse of polynomials. the quadratic formula is not relevant to what you are trying to do.

Sane · Jul 7th, 2005, 10:10 PM

HAH THIS WORKS GREAT!!! I bet you no one here can crack this!!!!! I challenge everyone here to actually try to get the correct CD Key.

Run this decoder program, and try to get the message to pop up by entering the correct CD key. Use any methods of Brute Force as you wish but you won't get it!

(Toggle Plain Text)

Encrypted_Code = """`���'��-��-6�08���4V�*'�+��38�'�+��38���+?�5��3��48���$8���3'�+~�.�1��-��'8�~�.��'��3��3��$��)��'8���(��26�0~�.8�~�.��'��8����$Y�/��'r�8�����'��4$���38���3��'��'8�'�+��(8���$Y�/E�96�0Y�/��'8���$`�&��4?�5��$��-��-E�98���)6�0��48�'�+��4x�"""

import math,sys
class EG(object):
    def __init__(self,e,g,d):self.e=e;self.g=g;self.d=d     
    def D(self, TrF):
        print'Decrypting:';yUT=map(ord,TrF)
        for d in range(len(yUT)):
            if str(yUT[d])=='%s%s%s'%('2',['5','4','3','2','1'][int('0')],str(4+1)):yUT[d-1]=int(yUT[d-1])+((128+127)*(int(yUT[d+1])));yUT[d]='';yUT[d+1]=''
        for d in range(len(yUT)):
            for e in range(len(str(yUT[d]))):
                if int(str(yUT[d])[e])>4:
                    try:yUT[d+1]=''
                    except:pass
        yuI=' '
        while yuI:
            try:yUT.remove('')
            except:yuI=False
        KJjh=[]
        for item in yUT:
            try:
                y=int(str(item));a=self.e;b=self.g;c=self.d;x=0.000;Gh=x**a+x*b+c
                while int(round(Gh*10)*10)!=int(round(y*100)):x+=0.01;Gh=x**a+x*b+c
                KJjh.append(int(round(x)))
            except:pass
        yUT=map(chr,KJjh);return''.join(yUT)
hYh=raw_input('\n\nCD Key? (e.g. 1-1-259)\n');HhH=[]
for a in range(len(hYh)):
    try:
        if hYh[a-1]!='-'and hYh[a]=='-':HhH.append(a)
    except:
        if hYh[a]=='-':HhH.append(a)
try:a=int(hYh[:HhH[0]]);b=int(hYh[HhH[0]+1:HhH[1]]);c=int(hYh[HhH[1]+1:]);EG=EG(a,b,c)
except:raw_input('Invalid CD Key Format!');sys.exit()
try:d=EG.D(Encrypted_Code)
except:raw_input('Invalid CD Key!');sys.exit()
raw_input(d)

BWAHAHAHAHA

HUmarMatt · Jul 8th, 2005, 12:50 AM

Man, I have such a generic name ... *sigh*
BTW, Sane, I'm a complete noob at programming, and I'm a new member here but damn that was easy, here's the first 8 words of it: 'Hello this is just a simple message from'

Sane · Jul 8th, 2005, 1:01 AM

OWNED!!

>.>
<.<

Errr good job, mind explaining how you did it so easily exactly? It would be better if it was closed-source.

Did I miss out on some simple opening?

Jul 7th, 2005, 4:39 PM	#1
Sane Programming Guru Join Date: Apr 2005 Location: Waterloo, Ontario Posts: 1,868 Rep Power: 5	Grrrr I need Math Programming help. I'm creating an encryption program for some time off, and I haven't learned this stuff so I'm having a little bit of trouble. It's supposed to use abc of the quadratic equation to alter the ascii, and then convert back. But I can't figure out how to reverse a quadratic equation. I know to change an x to a y in a quadratic equation is y = x*a + xb + c and to get the intercepts of x is x = -b +- ([b2 - 4ac]0.5 / 2a) but how can I convert y back to x? For example. Say the character is 's', when converted to ascii that is 115 (alt + 115). Now the quadratic equation is: y = x*a + xb + c and we'll say a =1, b = 2, c = 3 so now it is 233. converted back into ascii is Θ . But now when it's time to decrypt, how do I get 233 back to 115? I think it's something like: x = -b +- ([b2 - 4ac]0.5 / 2a) original X = x * 233 Can someone help? __________________ Waterloo's Canadian Computing Competition (CCC) - Stage 2 Problems, Solutions, and Test Data

Jul 7th, 2005, 5:10 PM	#2
mackenga Professional Programmer Join Date: Mar 2005 Location: Glasgow, Scotland Posts: 317 Rep Power: 4	(Toggle Plain Text) y = x*a + x b + c y = x*a + x b + c I assume you're using the operator to denote exponentiation. First up, in your example, if x = 115, y != 233; y == 115 1 + 2 * 115 + 3 == 115 + 230 + 3 == 348. Chars are just one byte, but 348 > 2 8, so you will need more than one byte per character for your ciphertext to do it this way (unless there's some funky trick I'm not thinking of off the top of my head, which is possible since the ciphertext will still only contain 256 distinct values). As for reversing the encryption; first, you can subtract 3 (the 3 added at the end of the encryption process) to get x a + x * b == 345. a and b are 1 and 2 respectively, so that's x ** 1 + x * 2 = 345. Any number n to the first power is just n, so you have 3 * x = 345, => x = 345/3 = 115 = your plaintext character s. This encryption system has a few problems, the most serious of which is that each character of plaintext will still always map to a particular value in the ciphertext, which means this is trivial to crack using frequency analysis and all the mathematical stuff is beside the point.

Jul 7th, 2005, 7:44 PM	#3
Sane Programming Guru Join Date: Apr 2005 Location: Waterloo, Ontario Posts: 1,868 Rep Power: 5	First Paragraph: Ya, I was rushing an example, sorry I missed the initial 115. Once it gets passed 256 it loops to 0, and vice-versa. Second Paragraph: Okay, thanks. But that still doesn't give me a formula because it won't work if the encryptor chooses > 1 for a. Third Paragraph: Incorrect, I didn't mention the entire scheme, of falses bytes scattered around in a partially hidden but organised manner. So frequency analysis will lead false. And even if they open the source code, they would need to know the initial value of the text to find out how the false characters were organized. Btw, this is going to be my completely unecessary method of keeping my programs with no source, everything is going to be an encrypted in a data file. The CD comes with a serial id that happens to be the A B C. When they run the program they enter the number, and if it's correct it translates the text file. Save them all as .pyc, runs them, then deletes them. And the decryptor will be fully obfuscated and compiled. I'm so good at being unecessary __________________ Waterloo's Canadian Computing Competition (CCC) - Stage 2 Problems, Solutions, and Test Data Last edited by Sane; Jul 7th, 2005 at 7:54 PM.

Jul 7th, 2005, 8:17 PM	#4
Sane Programming Guru Join Date: Apr 2005 Location: Waterloo, Ontario Posts: 1,868 Rep Power: 5	It would take a while, but I could create a loop that adds 0.0000001 to itself, and keeps testing the equation until it reaches approximately Y. shrug I might just do that ... Edit: That was considerably easier: (Toggle Plain Text) y = ? a = ? b = ? c = ? x = 0.000 TestY = x*a + xb + c while int(round(TestY10)10) != int(round(y100)): x += 0.01 TestY = xa + xb + c print x y = ? a = ? b = ? c = ? x = 0.000 TestY = x*a + xb + c while int(round(TestY10)10) != int(round(y100)): x += 0.01 TestY = xa + xb + c print x __________________ Waterloo's Canadian Computing Competition (CCC) - Stage 2 Problems, Solutions, and Test Data Last edited by Sane; Jul 7th, 2005 at 8:29 PM.

Jul 7th, 2005, 9:51 PM	#5
skuinders Hobbyist Programmer Join Date: Jun 2005 Location: MA, US Posts: 204 Rep Power: 4	in general, if you want to push a value through a function "backwards" you will need to have an inverse of the original function. an obvious illustration: (Toggle Plain Text) f(x) = e^x g(x) = ln x f(x) = e^x g(x) = ln x so you need to research methods for finding the inverse of polynomials. the quadratic formula is not relevant to what you are trying to do. __________________ "A stupid man's report of what a clever man says can never be accurate, because he unconciously translates what he hears into something he can understand." - B. Russell http://web.bryant.edu/~srk2

Jul 7th, 2005, 10:10 PM	#6
Sane Programming Guru Join Date: Apr 2005 Location: Waterloo, Ontario Posts: 1,868 Rep Power: 5	HAH THIS WORKS GREAT!!! I bet you no one here can crack this!!!!! I challenge everyone here to actually try to get the correct CD Key. Run this decoder program, and try to get the message to pop up by entering the correct CD key. Use any methods of Brute Force as you wish but you won't get it! (Toggle Plain Text) Encrypted_Code = """`��'��-��-6�08��4V�'�+��38�'�+��38��+?�5��3��48��$8��3'�+~�.�1��-��'8�~�.��'��3��3��$��)��'8��(��26�0~�.8�~�.��'��8��$Y�/��'r�8��'��4$��38��3��'��'8�'�+��(8��$Y�/E�96�0Y�/��'8��$`�&��4?�5��$��-��-E�98��)6�0��48�'�+��4x�""" import math,sys class EG(object): def __init__(self,e,g,d):self.e=e;self.g=g;self.d=d def D(self, TrF): print'Decrypting:';yUT=map(ord,TrF) for d in range(len(yUT)): if str(yUT[d])=='%s%s%s'%('2',['5','4','3','2','1'][int('0')],str(4+1)):yUT[d-1]=int(yUT[d-1])+((128+127)(int(yUT[d+1])));yUT[d]='';yUT[d+1]='' for d in range(len(yUT)): for e in range(len(str(yUT[d]))): if int(str(yUT[d])[e])>4: try:yUT[d+1]='' except:pass yuI=' ' while yuI: try:yUT.remove('') except:yuI=False KJjh=[] for item in yUT: try: y=int(str(item));a=self.e;b=self.g;c=self.d;x=0.000;Gh=x*a+xb+c while int(round(Gh10)10)!=int(round(y100)):x+=0.01;Gh=xa+xb+c KJjh.append(int(round(x))) except:pass yUT=map(chr,KJjh);return''.join(yUT) hYh=raw_input('\n\nCD Key? (e.g. 1-1-259)\n');HhH=[] for a in range(len(hYh)): try: if hYh[a-1]!='-'and hYh[a]=='-':HhH.append(a) except: if hYh[a]=='-':HhH.append(a) try:a=int(hYh[:HhH[0]]);b=int(hYh[HhH[0]+1:HhH[1]]);c=int(hYh[HhH[1]+1:]);EG=EG(a,b,c) except:raw_input('Invalid CD Key Format!');sys.exit() try:d=EG.D(Encrypted_Code) except:raw_input('Invalid CD Key!');sys.exit() raw_input(d) Encrypted_Code = """`��'��-��-6�08��4V�'�+��38�'�+��38��+?�5��3��48��$8��3'�+~�.�1��-��'8�~�.��'��3��3��$��)��'8��(��26�0~�.8�~�.��'��8��$Y�/��'r�8��'��4$��38��3��'��'8�'�+��(8��$Y�/E�96�0Y�/��'8��$`�&��4?�5��$��-��-E�98��)6�0��48�'�+��4x�""" import math,sys class EG(object): def __init__(self,e,g,d):self.e=e;self.g=g;self.d=d def D(self, TrF): print'Decrypting:';yUT=map(ord,TrF) for d in range(len(yUT)): if str(yUT[d])=='%s%s%s'%('2',['5','4','3','2','1'][int('0')],str(4+1)):yUT[d-1]=int(yUT[d-1])+((128+127)(int(yUT[d+1])));yUT[d]='';yUT[d+1]='' for d in range(len(yUT)): for e in range(len(str(yUT[d]))): if int(str(yUT[d])[e])>4: try:yUT[d+1]='' except:pass yuI=' ' while yuI: try:yUT.remove('') except:yuI=False KJjh=[] for item in yUT: try: y=int(str(item));a=self.e;b=self.g;c=self.d;x=0.000;Gh=x*a+xb+c while int(round(Gh10)10)!=int(round(y100)):x+=0.01;Gh=xa+xb+c KJjh.append(int(round(x))) except:pass yUT=map(chr,KJjh);return''.join(yUT) hYh=raw_input('\n\nCD Key? (e.g. 1-1-259)\n');HhH=[] for a in range(len(hYh)): try: if hYh[a-1]!='-'and hYh[a]=='-':HhH.append(a) except: if hYh[a]=='-':HhH.append(a) try:a=int(hYh[:HhH[0]]);b=int(hYh[HhH[0]+1:HhH[1]]);c=int(hYh[HhH[1]+1:]);EG=EG(a,b,c) except:raw_input('Invalid CD Key Format!');sys.exit() try:d=EG.D(Encrypted_Code) except:raw_input('Invalid CD Key!');sys.exit() raw_input(d) BWAHAHAHAHA __________________ Waterloo's Canadian Computing Competition (CCC) - Stage 2 Problems, Solutions, and Test Data

Jul 8th, 2005, 12:50 AM	#7
HUmarMatt Newbie Join Date: Jul 2005 Posts: 0 Rep Power: 0	Man, I have such a generic name ... sigh BTW, Sane, I'm a complete noob at programming, and I'm a new member here but damn that was easy, here's the first 8 words of it: 'Hello this is just a simple message from'

Jul 8th, 2005, 1:01 AM	#8
Sane Programming Guru Join Date: Apr 2005 Location: Waterloo, Ontario Posts: 1,868 Rep Power: 5	OWNED!! >.> <.< Errr good job, mind explaining how you did it so easily exactly? It would be better if it was closed-source. Did I miss out on some simple opening? __________________ Waterloo's Canadian Computing Competition (CCC) - Stage 2 Problems, Solutions, and Test Data

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Programming Forums > Community Forums > Coder's Corner Lounge

Grrrr I need Math Programming help.