authenticated communication through insecure channel using visual channel

AUTHENTICATED COMMUNICATION THROUGH INSECURE CHANNEL USING VISUAL CHANNEL

Cheh Carmen

OVERVIEW Introduction

Protocol Project

Design Issues Existing barcodes and algorithms Proposed barcode and analysis Future development

INTRODUCTION Computers located in many public

places Use of public computers is plagued with

many security problems

PROTOCOL Visual channel

Visual Channel

visual inspection

Server

Mobile Device equipped with camera

A-101-30-003

K-354-90-981

C-000-05-011

PROJECT Implement 2D barcode Two criteria

u ( vw , kb ) = c d ( vw , vo ) < δ

Proof-of-concept

DESIGN ISSUES(1) Choosing an image Color, gray-scale or binary?

Binary

TYPE OF IMAGE Picture or barcode?

Barcode One barcode or multiple?

Multiple Visual cue

DESIGN ISSUES(3) Digital watermark Fragile or robust?

Fragile Spatial domain or frequency domain?

Spatial domain Edge pixel hiding or block data hiding?

Block data hiding

ANALYSIS OF BARCODES L-shaped barcode

ANALYSIS OF L-BARCODE High data embedding capacity =2/3 Not very secure

Key space = 4!x4! Quite poor image quality

PROPERTIES OF BARCODE High data capacity Moderate image quality High security

PROPOSED ALGORITHM 1 F = original image (3x3 block) Watermarking key (K,M)

K is 9-tuple : permuted coordinates of 3x3 block

M is 3x3 binary matrix : mask B = 3-bit message : b0b1b2

Maximum 3 bits flipped in the block

ALGORITHM 1 (CONT.) F = 1 1 0 0 1 1 0 0 0 K = [(3,3),(3,2),(3,1),(2,3),(2,2),(2,1),(1,3),

(1,2),(1,1)] M = 0 1 0 1 1 0 1 0 0 B = 0 0 0

ALGORITHM 1(CONT.) F XNOR M (masking F – confusion) F = 1 1 0 M = 0 1 0 0 1 1 1 1 0 0 0 0 1 0 0

Result F’: 0 1 1 0 1 0 0 1 1

ALGORITHM 1(CONT.) Shuffle F’ using K(diffusion) K = [(3,3),(3,2),(3,1),(2,3),(2,2),(2,1),

(1,3),(1,2),(1,1)] F’ = 0 1 1

0 1 0 0 1 1

Result F’’ : 1 1 0 0 1 0 1 1 0

ALGORITHM 1(CONT.) Invariant: MSB of F’’ = b0 B = 0 0 0 F’’ = 1 1 0 0 1 0 1 1 0

Result F’’ = 0 1 0 0 1 0 1 1 0

ALGORITHM 1(CONT.) right-shift continuous: A binary string is

right-shift continuous when MSB=LSB. right-shift continuous length: Right-

shift the binary string. Right-shift continuous length is the number of digits of the same value starting from MSB without interruption by the opposite digit.

0 0 1 0 1 0 1 0 0

ALGORITHM 1(CONT.)

F’’ = 0 1 0 0 1 0 1 1 0 Current RCL = 2 b1b2 Ξ 0 mod 4

Result : 0 0 0 1 1 0 1 1 0

INVARIANT: RCL F’’ Ξ b1b2 mod 4

ALGORITHM 1(CONT.) Modify original F Flipped bits : 1 1 0 0 1 0 1 1 0 F = 1 1 0

0 1 1 0 0 0

K = [(3,3),(3,2),(3,1),(2,3),(2,2),(2,1),(1,3),(1,2),(1,1)]

Result : 1 1 0 0 1 0 0 1 1

ALGORITHM 1(DECODING) Fw XNOR M (mask it again) Fw = 1 1 0 M = 0 1 0

0 1 0 1 1 0 0 1 1 1 0 0

Result Fw’ : 0 1 0 0 1 1

0 0 0

ALGORITHM 1(DECODING) Shuffle Fw’ using K Fw’ = 0 1 0

0 1 1 0 0 0 K = [(3,3),(3,2),(3,1),(2,3),(2,2),(2,1),

(1,3),(1,2),(1,1)]

Result Fw’’: 0 0 0 1 1 0 1 0

ALGORITHM 1(DECODING) Fw’’ = 0 0 0 1 1 0 1 0

b0 = 0 b1b2 = 4 mod 4 = 0 B = b0b1b2 = 000

ANALYSIS OF ALGORITHM 1 Embedding capacity: 3 bits/block = 3/9

= 1/3 Average number of bit flipped/block =

2 Security: Key space = 28 x 9!

Main disadvantage: Embedding capacity too low

ALGORITHM 2 Example: F=3 bits B=2 bits

Divide F into 4 cosets

Entries in table represent all possible F Header represents all possible B

0 1 2 3000 001 010 011111 110 101 100

ALGORITHM 2(CONT.)

F = 000 B = 11

F belongs to coset 0. Move to coset 112=3 100 Modify F to 100

0 1 2 3000 001 010 011111 110 101 100

ALGORITHM 2(CONT.) In general, F = n bits, B = n-1 bits Partition F into 2n-1 cosets Each coset has 2 elements Choose codeword in coset B s.t. d(F,c)

is minimum among all other codewords in coset B

ANALYSIS OF ALGORITHM 2 Embedding capacity: 2 bits/block = 2/3 In general = n-1/n Average number of bits flipped = ¾ Security: Can apply M and K in the

same manner as algo 1

Embedding capacity is higher than algo 1

FUTURE DEVELOPMENT Compromising a bit of embedding

capacity for visual effect Experimenting with different kind of

distance formula Qgram Edit distance

Simulation

END