on a generalization of the gcd for intervals in r +

On a Generalization of the GCD for Intervals in R+

Stan Baggen June 4, 2014

orhow can a camera see at least 1 tone for unkown Texp

Philips Research Stan Baggen

Contents

• Introduction

• Cameras, exposure times and problem definition

• Introduction to Solution using GCD for Integer Frequencies

• Extension of GCD to intervals over R+

• Application to the Original Problem

• Discussion

• Yet another generalization

2

Philips Research Stan Baggen 3

Introduction

• Transmit digital information from a luminaire to a smartphone or tablet using Visible Light Communication (VLC)

– Bits are encoded in small intensity variations of the emitted light

– Detect bits using the camera of a smartphone

• We consider an FSK-based system

– Symbols correspond to frequencies (tones)

– Emitted light variations are sinusoidal

• Problem: camera may be “blind” for certain frequencies


Camera Image divided into lines and pixels

original image sequence

lines covering source

lines per frame

hidden lines

active lines

• Each line consists of a row of pixels


• A camera can set its exposure time Texp • typically, Texp ranges from 1/30 to 1/2500 [s]

• Each pixel “sees” the average light during Texp seconds before read-out– smearing of intensity variations of received light

• If an integer number of periods of a sinusoid fit into Texp, the camera cannot detect such a sinusoid

Exposure Time

time

Texp

ISI filter (moving average)

0 1 2 3 4 5 6 7 8 9 100

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

f

|sin

c(f)|

Transfer Function of Exposure Time

|sinc(f)||sinc(0.7f)|

f1 f2


• Due to the exposure time Texp of a camera, certain frequencies cannot be detected by it (multiples of fexp = 1/Texp)

• Can we have sets of 2 frequencies each, such that not both can be blocked for any fexp ≥ 30 Hz

• Each set then forms an fexp-independentdetection set for a light source that emitsboth frequencies

0 1 2 3 4 5 6 7 8 9 100

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

f

|sin

c(f)|



Exposure Time

6

f1 f2

time

Texp

ISI filter


Discrete Solution

• If the involved frequencies can only take on integer values, we can find solutions using the GCD (Greatest Common Divisor) from number theory

• We would like to have 2 frequencies f1 and f2, such that not both can be integer multiples of any fexp ≥ 30

• Suppose that both f1 and f2 are integer multiples of fexp

• If GCD(f1,f2) < 30 no solution possible for fexp ≥ 30

pair (f1,f2) is a good choice7

),GCD(||

|21exp

2exp

1exp fffff

ff


Discrete Solution: Example

• f1 = 290; f2 = 319

• Largest integer that divides both f1 and f2 equals GCD(f1,f2) = 29

• No integer fexp ≥ 30 exists for which multiples are simultaneously equal to f1 and f2

8


Problem with Discrete Solution

• GCD(300,301) = 1; GCD(300,300) = 300

• Physically: due to the nature of the Texp-filter and detection algorithms, if a pair of frequencies (f1,f2) is bad for detection, then a real interval (f1±ε,f2 ±ε) is also bad

• We need a method that allows us to eliminate bad intervals over R+

90 1 2 3 4 5 6 7 8 9 10

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

f

|sin

c(f)|



f1 f2


GCD for intervals in R+

• Consider 2 half-open intervals I1 and I2 in R+

• Definition:

• Note that the concept I1,I2: GCD(I1,I2) < 30 solves our original problem:

• There can be no real fexp≥30 such that integer multiples are simultaneously close to F1 and F2 10

21,21 |max:),GCD( ImaInaRaII Nmn

0( ] ( ]

I1 I2

0( ] ( ]

1F 2F

30


GCD for intervals in R+

• How to find GCD(I1,I2)?

• Define divisor sets D1,D2 in R:

• Theorem 1:

• Proof: □ 11

21,21 |max:),GCD( ImaInaRaII Nmn

0( ] ( ]

I1 I2

2121 max),GCD( DDII

22

11

||

ImdRdDIndRdD

Nm

Nn

21

2121

|||ImdIndRd

ImdRdIndRdDD

NmNn

NmNn


Example

12

0 50 100 150 200 250 300-1

0

1

2

3

4

5

frequency

divisor sets, their intersection and the GCD, f1 = 240, f2 = 256, interval = 16

f1 = 240f2 = 256GCD = 28.4444

]240,16240(1 I

]256,16256(2 I

21 DD

),GCD( 21 II

11 | IndRdD Nn

22 | ImdRdD Nm


10 15 20 25 30

0

0.5

1

1.5

2

2.5

frequency

divisor sets, their intersection and the GCD, f1 = 240, f2 = 256, interval = 16

f1 = 240f2 = 256GCD = 28.4444

Enlargement of Example

1D

21 DD ),GCD( 21 II

2D


Overlap of Intervals in Divisor Sets• Consider divisor set

• Let where

• Theorem 2: for w>0, D consists of a finite number n0 of disjunct intervals, where

• Proof: overlap of consecutive intervals happens if

• Corollary: 14

0( ]

I

],(,| fwfIIndRdD Nn

2n3nfwf

NnnIIndRdDn ,/|

1D2/12 DD

n

nDD

wfn0

wfn

wfn

nwf

nf

0.1

00

,0,0,,0nfwD

nf

3/13 DD


Another Theorem

• Suppose that we have 2 intervals I1=(f1-w1,f1] and I2=(f2-w2,f2]

• Theorem 3: For w1,w2> 0, GCD(f1,f2 ; w1,w2) equals an integer sub-multiple of either f1, f2 or both

• Proof:

equals a right limit point of for some i and j.

Each is the intersection of 2 half-open intervals (...], where the right limit point of each half-open interval is an integer sub-multiple of either f1 or f2 or both. □

• Note: f1 and f2 are real numbers 15

ji

ji

j

j

i

i DDDDDD,

212121

21max DD ji DD 21

ji DD 21


Some Interesting Examples

• Numbers in N+

– For w sufficiently small, we find the classical solutions for f1, f2 in N+

– GCD(15,21; w≤1) = 3

– GCD(15,21; w=1.1) = 7 • w too large for finding the classical solution

• Numbers in Q+

– GCD(0.9,1.2; w=0.1) = 0.3

• Numbers in R+ (computed with finite precision)– GCD(7π,8π; w=0.1) = 3.1416– GCD(6π,8π; w=0.1) = 6.2832

16


Application to the Original Problem• Suppose that we find that for a certain (f1, f2; w1,w2) :

GCD (f1, f2; w1,w2) < 30

• Then there exists no real number fexp≥30 such that integer multiples of fexp fall simultaneously in (f1-w1, f1] and (f2-w2, f2]

• By picking F1= f1-w1/2 and F2= f2-w2/2, we can insure that if one multiple of fexp≥30 falls within a range of wi/2 of Fi for some i, then the other interval is free from any multiple of fexp

17

2f22 wf

0( ] ( ]

30f

11 wf 1f1F 2F


Numerical Examples (1)

100 200 300 400 500 600 7000

200

400

600

800

1000

1200

frequency

acce

ptab

le fr

eque

ncy

pairs

Acceptable integer frequency pairs for w = 15, GCD < 30, 100 < f < 700

acceptable_frequencies_2012_10_20_1


100 120 140 160 180 200 220

100

150

200

250

300

frequency

acce

ptab

le fr

eque

ncy

pairs



typical solutions: (f1,f2) = (f1, f1+15)

Numerical Examples (1) detail


100 200 300 400 500 600 7000

500

1000

1500

2000

2500

frequency

acce

ptab

le fr

eque

ncy

pairs





90 100 110 120 130 140 150 160 170 180 190

0

50

100

150

200

250

300

350

frequency

acce

ptab

le fr

eque

ncy

pairs





100 200 300 400 500 600 7000

500

1000

1500

2000

2500

3000

3500

4000

4500

5000

frequency

acce

ptab

le fr

eque

ncy

pairs





100 200 300 400 500 600 7000

1000

2000

3000

4000

5000

6000

7000

8000

9000

frequency

acce

ptab

le fr

eque

ncy

pairs





100 120 140 160 180 200 220 240 260 280 3000

100

200

300

400

500

600

700

800

900

frequency

acce

ptab

le fr

eque

ncy

pairs




typical solutions: (f1,f2) = (f1, f1+15), (f1, 2f1-20), ), (f1, 2f1+15)


Discussion (1)

• It is convenient to use half open intervals (…] and have the right limit point as a characterizing number, since then

– We can reproduce the familiar results from number theory– The maximum in the definition of GCD exists– We do not obtain subsets in having measure 0

• The concept of GCD can be generalized to an arbitrary number of K intervals over R+

• Theorem 2 shows that the complexity of the computation of a GCD is reasonable

• Can we have an efficient algorithm like Euclid’s algorithm for computing the GCD of real intervals?

25

21 DD

K

kkK DII

11 max),...,GCD(


Discussion (2)

• It can be shown that GCD(f1, f2;w) is non-decreasing as w increases

• For rational numbers a/b and p/q, where a,b,p,q are in N+, we find for sufficiently small w:

where LCM(.) is the Least Common Multiple.

How small must w be as a function of a,b,p and q to find this solution?

• Conjecture: for incommensurable numbers a and b

• Effects of finite precision computations

26

,),LCM(

),LCM(,),LCM(;,

qbqpqb

baqbGCD

wqp

baGCD

0;,lim0

wbaGCDw


Yet Another Generalization

• GCD(f1,f2;w) on intervals still makes hard decisions on frequencies being in or out of intervals

• Can we make some sensible reasoning that leads to “smooth” decisions concerning acceptable frequency pairs

• We have to use a more friendly measure on the intervals

• We start by re-phrasing the previous approach in a different manner

27


• GCD(f1,f2;w) on intervals as discussed previously, effectively uses indicator functions as a measure of membership:

• Divisor Measure DM1,DM2 in R+

28

0( ] ( ]

I1 I2

1

1)()(max:),( 2121

fDMfDMIIGCDRf

2,1,),(max:)(

iRfnfMfDM

iINni

1IM

2IM

RffMfM IIII ,1)(;1)(2211

f


0 2 4 6 8 10 12 140

1

2

3

4

5

6

7

f

cost

s

Cost Functions

D1D2D1 D2Example

f1 = 9; f2 = 12

w = 0.5

GCD(f1,f1;w) = 3

0 0.5 1 1.5 2 2.5 3 3.50

1

2

3

4

5

6

f

cost

s

Cost Functions

D1D2D1 D2

9

12

3


Using a Different Measure

• Suppose that we change the definition of the measure of membership for the fundamental interval

• Divisor Measure:

• Common Divisor Measure:

30

),;(),;(:),,,;( 22112121 ffDMffDMfffCDM

2,1,),,;(max:),;(

iRffnfMffDM iiNnii

2,1,,

2exp:),;( 2

2

iRf

ffffM

i

iii

0 2 4 6 8 10 12 14

0

0.5

1

1.5

2

2.5

3

f

cost

s

Cost Functions of Fundamental Intervals I1 and I2

)25.0,12;( fM

)25.0,9;( fM

example


0 2 4 6 8 10 12 140

1

2

3

4

5

6

7

f

cost

s

Cost Functions

D1D2D1 D2

0 0.5 1 1.5 2 2.5 3 3.5

0

0.2

0.4

0.6

0.8

1

1.2

1.4

f

cost

s

Cost Functions

D1D2D1 D2

)25.0,25.0,12,9;( fCDM

)25.0,9;( fDM

)25.0,12;( fDM

Example

Multiples of frequencies in the neighborhood of 3(and 3/n) also end up both near 9 and 12

For frequencies f>3.2, no multiples end up both near 9 and 12 according to the measure

Multiples of 1.1, 1.3 and 1.7 come somewhat close to both 9 and 12 (c.f. other measure)

)25.0,25.0,12,9;( fCDM


0 2 4 6 8 10 12 140

1

2

3

4

5

6

7

f

cost

s

Cost Functions

D1D2D1 D2

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5

0

0.2

0.4

0.6

0.8

1

1.2

f

cost

s

Cost Functions

D1D2D1 D2

)5.0,5.0,12,9;( fCDM

)5.0,9;( fDM

)5.0,12;( fDM

Example

If we increase σ, it becomes more difficult to“avoid” the intervals around 9 and 12 for integer multiples of f

For σ=0.5, some multiples of 4.16 also come close to both 9 and 12 according to the measure


0 2 4 6 8 10 12 14

0

0.5

1

1.5

2

2.5

3

fco

sts

Cost Functions of Fundamental Intervals I1 and I2

f1 = 9; f2 = 12σ = 0.5fexp = 4.16

Example

samples taken at integer multiples of 4.16

CDM(4.16;.) equals product of largest “red” sample (n=3) and largest “blue” sample (n=2)

)5.0,12;( fM

)5.0,9;( fM

on a generalization of the gcd for intervals in r +

Documents

integer fexp

integer multiples of

pair of frequencies

certain frequencies

involved frequencies

pair f1

f2 equals gcdf1

r definition