1. Inverse Wavelet Transform
A New Method of Cryptography:
Jacqueline Christy  Dr. James Walker  Mathematics University of Wisconsin-Eau Claire
This project was funded by UWEC Differential Tuition and by the UWEC Foundation.
Decoding:
Introduction:
Encoded Message
De-altered
Inverse Wavelet Transform
Plain Text
Using Visual Basic 6.0, a program implementing a new cryptographic scheme was developed. The new cryptographic scheme combines the wavelet transform based scheme with a series of random alteration values to encode finite sequences of integers.
The encoded message of integers is converted into bytes.
These bytes are de-altered, first by subtracting 100 and the seven sets of alteration values obtained from δ , and then by using the inverse wavelet transform.
These de-altered bytes are converted back into text using the key set up earlier.
The decoded message is displayed.
The Program:
Coding:
Plain Text (bytes)
Wavelet Transform
Randomly Altered
Encoded Message
Wavelet Transform:
In this program, a wavelet based transform often used with video compression, the Daub 5/3 integer-to-integer transform, was used to compress the messages into an encoded set of integers. This was done by applying the following lifting equations:
A key is set up in which an integer is assigned to every symbol on the common 104-key US English QWERTY keyboard (for each letter, the uppercase and lowercase versions are considered).
Using this key, each character from the text is converted into an integer, and these integers are then converted into bytes.
The bytes are then altered using the wavelet transform.*
Seven random integers are chosen between the integers 1 and These integers are used to generate seven sets of integers (of length n / 7, where n is the number of characters in the original text) from the irrational number δ . These sets are generated into random sequences by multiplying each digit by (-1) if it is odd, and by (+1) if it is even. These integers are then added to the wavelet transform values.
dk = f2k – ½(f2k-1 + f2k+1) + ½
ak = f2k ¼(dk-1 + dk) + ½
Because this transform combines five adjacent values in the ak values, and these values are iterated on four times, it removes any short-term relations between characters that are present in the English language.
Results:
δ = …
Original Message
Encoded Message
Each number is bumped up by 100 to avoid overflow.
The encoded message is displayed.
*See explanation in later section