Comments for Oxaric's Blog

Comment on Fun With BASH Piping! by Nick Warrington

Nick Warrington — Tue, 12 May 2009 20:27:36 +0000

This is my kind of tutorial. Thanks!

Comment on Hidden Image Watermarking by Linda

Linda — Tue, 10 Mar 2009 08:28:34 +0000

Hai louis….. your code is very helpfull for me..
i’m linda from Indonesia…
I’m in a project of watermarking… and combine two image, exactly the same what u’ve done on this…

u really a smart guy… but i wanna ask u…
what method that u use in this?? i mean what’s the method’s name??

i hear about some methods like “spread spectrum”, “Least significant bit (LSB)”, “Singular Value decomposition”, “DCT”, “DWT”

Please answer me… and may i use your code for my exam project??
i am a student college… and i need to know the name of your methods :)

Comment on Euler Project Problem #104 Solution by oxaric

oxaric — Sun, 28 Dec 2008 05:54:18 +0000

Sorry it has taken such a long time for a response!

You never directly define k but from your words it appears you mean k to be the final answer. I’ve already put up the code! I’m not going to give out answers too. :)

If you are using dynamic programming I would suggest you create a dynamic program that can dynamically handle any array length needed to find the answer to the solution. If a dynamic program is hard coded then it’s not really dynamic right?

Comment on Euler Project Problem #59 Solution by oxaric

oxaric — Sun, 28 Dec 2008 05:48:43 +0000

That’s a good method that is probably much more functional for secret messages without so many known constraints on the final message. I was just hacking for speed and answers. :)

Comment on Euler Project Problem #59 Solution by Ori

Ori — Fri, 26 Dec 2008 04:27:52 +0000

Clever solution! What I did was test for the frequency of space, ‘e’, and ‘t’ (the three most common letters in English) with each code letter, and picked the codes that deviated least from their standard frequency (as given here). Here it is in Python:

message = [ 79....73 ] # snipped for brevity

likely_spaces = 0.1874 * len(message)
likely_es = 0.0960 * len(message)
likely_ts = 0.0702 * len(message)

code = [0,0,0]

for pos in range(3): # first, second, third code letter
mindiv = 1000
for possible_codeletter in range(97, 123): # possiblities
spaces = 0
es = 0
ts = 0

for letter in message[pos::3]:
decrypted_letter = letter ^ possible_codeletter
if decrypted_letter == 32: spaces += 1
elif decrypted_letter == 69 or decrypted_letter == 101: es += 1
elif decrypted_letter == 84 or decrypted_letter == 116: ts += 1
spaces *= 3
es *= 3
ts *= 3
deviation = abs( likely_spaces – spaces ) + abs ( likely_ts – ts ) + abs( likely_es + es )
if deviation < mindiv:
code[pos] = possible_codeletter
mindiv = deviation

codeword = chr(code[0]) + chr(code[1]) + chr(code[2])
print “The code is: “, codeword, ‘(‘, code, ‘)’

sum = 0
for i in range(len(message)):
if (i – 1) % 3 == 0:
message[i] = message[i] ^ code[1]
elif (i – 2) % 3 == 0:
message[i] = message[i] ^ code[2]
else:
message[i] = message[i] ^ code[0]
sum += message[i]

for i in range(len(message)):
message[i] = chr(message[i])

decrypted = ”.join(message)

print decrypted
print sum

Comment on Euler Project Problem #104 Solution by YuYevon

YuYevon — Tue, 16 Dec 2008 12:14:07 +0000

Hi Louis “oxaric” Casillas,

(forgive me if my English isn’t very good…)

I was trying to write the code to solve this problem, and I found a solution in C++, using the NTL library, but now I’m “in trouble”:

even if my algorithm (which doesn’t take too much time) is correct, I fail in finding ‘k’ because the algorithm implements dynamic programming and I don’t know how big the array must be in order to get the correct output.

In sum (to avoid bothering you), could you tell me the value of ‘k’ you find out with your algorithm?
I would try it by myself with your code but I haven’t the GMP library installed and it would be too boring to download and install it only for this.

Thank you for your eventual answer!

YuYevon