aa r X i v : . [ c s . I T ] J un On Matched Metric and Channel Problem
Artur Poplawski
Abstract —The sufficient condition for partial function from thecartesian square of the finite set to the reals to be ”compatible”with some metric on this set is given. It is then shown, thatwhen afforementioned set and function are respectively a spaceof binary words of length n and probalities of receiving someword after sending the other word through Binary AssymetricChannel the condition is satisfied so the required metrics exist.This proves under the slightly weaker definition of matchedmetric and channel conjecture stated in [1] Keywords — metric, channel. I. I
NTRODUCTION
In [1] authors consider following problem originating fromInformation Theory: for which channel models there is metric d on the space of the codewords such that following matchingcondition holds for each codewords x, y, z: P r ( x | y ) >P r ( x | z ) if and only if d ( x, y ) < d ( x, z ) Here P ( x | y ) meansprobablity of the receiving codeword x assuming that code-word y was sent. [1] gives extensive overview of the history,current literature and state of knowledge on the subject. Italso proves existence of such a metric in case of so called Z-channels and for the codewords of length 2 and 3 in the caseof Binary Assymetric Channel (BAC). Finally authors of [1]state the conjecture that such a compatible metric exists forthe space of the codewords of arbitrary length n for BAC.In section II existence of metrics compatible to certainfunction will be considered (in slightly more general casecomparing to [1]). In this case necessary and sufficient condi-tion for existence of such a metrics will be given. In sectionIII we will discuss relation between the theorem from II andinformation theoretic case and also with some of the notionsintroduced in [1]. Section IV applies result from II to the caseof BAC and proves (with some corrections discussed in III)conjecture from [1].II. C OMPATIBLE METRICS
Let X be a finite set and let S ⊂ X − ∆ where ∆ = { ( x, x ) : x ∈ X } Let f be a function: f : S → R We have follwiong theorem:
Theorem 1:
If for each n and sequence x , . . . x n − suchthat: f (( x , x n − )) > f (( x , x )) f (( x , x )) > f (( x , x )) . . .f (( x n − , x n − )) > f (( x n − , x n − )) and ( x n − , x n − ) ∈ S and ( x n − , x ) ∈ S we have: f (( x n − , x n − )) ≤ f (( x n − , x )) then there exists metric d on X such that for all x, y, zf ( x, y ) > f ( x, z ) ⇒ d ( x, y ) < d ( x, z ) Proof:
Let U be the set of unordered pairs of elementsof X so: U = { s ⊂ X : | s | = 2 } Let’s define relation R ⊂ U in the following manner: for a, b ∈ S aRb if one of the two conditions below holds: a = b or a = { x, y } , b = { x, z } , y = z, f ( x, y ) > f ( x, z ) R is reflexive (what is obvious) and antisymmetric. Let R be the transitive closure of the R . It is, again, reflexive andby definition transitive. We will show that it is antisymmetric.Let’s assume that this is not true, so we have a and b , suchthat a = b and aRb and bRa but a = b . Since R is transitiveclosure of R we would have, that there exists n and m andtwo sequences (after appropriate allignement of idices: a = z , z , . . . , z n = b and b = z n , z n +1 , . . . , z n + m − , z n + m = a = z such that z Rz , z Rz , . . . z n + m − Rz From the definition of R and because z i ∈ U we havefor ≤ i < n + m z i = { x i , x ( i −
1) mod n + m } where x i = z i ∩ z ( i +1) mod ( n + m ) It means, that we have respectively: f (( x , x n + m − )) > f (( x , x )) f (( x , x )) > f (( x , x )) . . .f (( x n + m − , x n + m − )) > f (( x n + m − , x n + m − )) so, by the assumption on f we must have: f (( x n + m − , x n + m − ) ≤ f (( x n + m − , x )) at the other hand, since z n + m − Rz n + m − we have: f (( x n + m − , x n + m − ) > f (( x n + m − , x )) what is contradiction.This proves, that R is a antysymmetric so a partial order in U . As a partial order it can be extended to total (linear) order R , and, since the X so also U is finite there is a function g : U → R + such that xRy iff g ( x ) < g ( y ) Now we will usethe trick from [1] Lemma 7 (for the selfcontainedness of thework, Appendix Lemma 1 of this note repeats statement andproof of the Lemma 7 from [1]). Let’s define the e : X → R + as e ( x, x ) = 0 and for x = y e ( x, y ) = g ( { x, y } ) . e issymmetric and e ( x, y ) = 0 iff x = y , so e is semimetric. From[1] Lemma 7 there is a metric d such that d ( x, y ) < d ( x, z ) if and only if e ( x, y ) < e ( x, z ) This finishes the proof.
Remark 1:
The extension of the relation R to partial orderis a special case of the Suzumura’s Extension Theorem seee.g. [2] III. I NFORMATION - THEORETIC CONTEXT
To bring results of II into the information-theoretic context,for certain channel model we will consider the partial function f defined on subspace the space X where X is a space ofcodewords of the length n by f ( x, y ) = P r ( x | y ) There is one delicate point related to Theorem 1 and [1]that needs to be discussed. Elements of a, b ∈ U (as definedin the proof) are incomparable by R in three cases:1) if a ∩ b = ∅ a = { x, y } , b = { x, z } , y = z but either ( x, y ) / ∈ S or ( x, z ) / ∈ S where S is domain of f a = { x, y } , b = { x, z } , y = z but f ( x, y ) = f ( x, z ) In information-theoretic interpretation of f second casenever happens. For the third case, construction of the R relation introduces order between such a { x, y } and { x, z } so,when we move to the construction of d there is implication P r ( x | y ) > P r ( x | z ) ⇒ d ( x, y ) < d ( x, z ) but no implication d ( x, y ) < d ( x, z ) ⇒ P r ( x | y ) > P r ( x | z ) claimed in [1].Metric is still matched to the channel according to slightlymodified Definition 1 from [1]: Definition 1:
Let W : X → X be a channel with inputand output alphabets X and let d be a metric on X Wesay that W and d are matched if maximum for every code C ⊂ X and every x ∈ X arg min y ∈ X −{ x } d ( x, y ) ⊂ arg max y ∈ X −{ x } P r ( x | y ) where we interpret, that arg max returns list of size at least 1, not the single element.One (not essential) modification is that we require that rangeof arg max and arg min exclude x : this is because [1] makesassumption that channel is reasonable, so P r ( x | x ) > P r ( x | y ) whenever x = y so without this exclusion operators triviallyreturn { x } More subtle is weakening the requirement arg min y ∈ X −{ x } d ( x, y ) = arg max y ∈ X −{ x } P r ( x | y ) expressed in the same context in corresponding Definition 1in [1]. This one is essential. In fact in case of the metricsbuilt as in the Theorem 1, expression arg min y ∈ X −{ x } d ( x, y ) will always evaluate to a single element list. Some of the channels that do not have matched metrics inthe sense of Defintion 1 from [1] do have in the sense ofDefintion 1 as stated above. It is such in the following case:Let X = { a, b, c } and let P r ( a | a ) = P r ( b | b ) = P r ( c | c ) = 12 P r ( a | b ) = P r ( a | c ) = 14 P r ( b | a ) = P r ( c | b ) = 16 P r ( b | c ) = P r ( c | a ) = 13 Let’s also observe, that in the case whe we assume, that
P r ( x | z ) = P r ( x | t ) whenever x = z = t = x both definitionscoincide. It would be interesting to further explore this relationbetween definitions in context of some perturbation argument,where we modify slighly the channel to assure condition abovein consistent manner and go to the limit.IV. B INARY A SSYMETRIC C HANNEL HAS MATCHEDMETERICS
Now, let’s apply theorem from previous section and provefollowing:
Theorem 2:
Binary Assymetric Channel has matched metric(in sense of definition 1).
Proof:
Let X be a space of the codewords of length m .According to theorem from section I, if for each sequence ofcodewords x , x , . . . , x n − we will have: P r ( x | x n − ) > P r ( x : x ) P r ( x | x ) > P r ( x | x ) . . .P r ( x n − | x n − ) > P r ( x n − | x n − ) implies: P r ( x n − | x n − ) < P ( x n − | x ) there is a metric with required propery. Le’s assume that thisis not true, so there is a sequence x i which satisfies all theinequalities from the premise but for which P r ( x n − | x n − ) ≥ P ( x n − | x ) Let x i ( j ) for i ∈ { , . . . , n − } , j ∈ { , . . . , m − } be the j − th symbol in i − th codeword. Probability of reception of thesymbol codeword x i when x ( i + k ) mod n , where k ∈ { , − } ,was sent is then: P r ( x i | x ( i + k ) mod n ) = m − Y j =0 P r ( x i ( j ) | x ( i + k ) mod m ( j )) So inequalities from the premise can be written as m − Y j =0 P r ( x i ( j ) | x ( i −
1) mod m ( j )) × m − Y j =0 P r ( x i ( j ) | x ( i +1) mod m ( j )) − > for i ∈ { , . . . , n − } For convenience, let’s move tologarithms, so we have for the same i : m − X j =0 ( log ( P r ( x i ( j ) | x ( i −
1) mod m ( j ))) − log ( P r ( x i ( j ) | x ( i +1) mod m ( j )))) > and also m − X j =0 ( log ( P r ( x n − ( j ) | x n − ( j ))) − log ( P r ( x n − ( j ) | x ( j )))) > summing over i ∈ { , . . . , n − } we have: n − X i =0 m − X j =0 ( log ( P r ( x i ( j ) | x ( i −
1) mod m ( j ))) − log ( P r ( x i ( j ) | x ( i +1) mod m ( j )))) > Let’s change the order of summation, and we will have m − X j =0 n − X i =0 ( log ( P r ( x i ( j ) | x ( i −
1) mod m ( j ))) − log ( P r ( x i ( j ) | x ( i +1) mod m ( j )))) > We claim however, that for each j the inner sum is : s j = n − X i =0 ( log ( P r ( x i ( j ) | x ( i −
1) mod m ( j ))) − log ( P r ( x i ( j ) | x ( i +1) mod m ( j )))) = 0 so the total sum is , what leads to a contradiction.To show that the claim is true this, let’s represent { x i ( j ) } i ofsmbols s and s as concatenation of sequences consisting ofidentical symbols in such a way, that sequences we concatenatecontain different syblos, e.g. sequence (0 , , , , , will berepresented as conctatenation of sequences (0 , , (1 , , , and let’s call this sequences respectively X , X , . . . , X k − and it’s elements by the the duble indexed x l,i wich represents. i -th element of l -th sequence. Because sum we consider iscyclic, we can also assume without loss of generality, that firstand the last sequence consists of different symbols (shiftingsequence cyclically if needed). Let’s denote by | X l | length ofthe sequence X l We will also fix j so we will not write it. Wenow have: s j = n − X i =0 [ log ( P r ( x i | x (( i −
1) mod m ) )) − log ( P r ( x i | x (( i +1) mod m ) ))] = n − X i =0 log ( P r ( x i | x (( i −
1) mod m ) )) − n − X i =0 log ( P r ( x i | x (( i +1) mod m ) )) = k − X l =0 [ log ( P r ( x l, | x (( l −
1) mod k ) , | X (( l −
1) mod k ) |− )+ | X l |− X i =1 log ( P r ( x l,i | x l, ( i − )))] − k − X l =0 [ | X l |− X i =0 log ( P r ( x l,i | x l, ( i +1) ))+ log ( P r ( x l, | X l |− | x (( l +1) mod k ) , ] Let’s observe, that for each l : log ( P r ( x l, | x (( l −
1) mod k ) , | X (( l −
1) mod k ) |− )+ | X l |− X i =1 log ( P r ( x l,i | x l, ( i − ))) = | X l |− X i =0 log ( P r ( x l,i | x l, ( i +1) ))+ log ( P r ( x l, | X l |− | x (( l +1) mod k ) , )) bcause number of terms under the sign P is the same and theseterms are equal (since concatenated subsequences consisted ofthe same symbol. The terms outside the sum are equal, sinceneighbour groups consists of different simbols, so on both sidestheir will be either P r (0 | or P r (1 | This, together withcontradiction pointed out earlier completes the proof.A
PPENDIX AL EMMA FROM [1]Following repeats the Lemma 7 from [1] and it’s proof
Lemma 1:
Let X be a finite set and e : X → R be asemimetric, i.e. satisfy following conditions: e ( x, y ) ≥ forall x, y ∈ X e ( x, y ) = 0 if and only if x = y e ( x, y ) = e ( y, x ) for all x, y ∈ X then, there is a metric d on X such that d ( x, y ) < d ( x, z ) if and only if e ( x, y ) < e ( x, z ) Proof:
Let m = min { e ( x, y ) : x, y ∈ X, x = y } > Let M = max { e ( x, y ) : x, y ∈ X, x = y } Let δ satisfying < δ < be some number. Let f be a strictly increasingbijective function f : [ m, M ] → [1 − δ, δ ] . We define d : X → R + following manner: d x,y = (cid:26) if x = y is odd ,f ( e ( x, y )) otherwise . Symmetry, nonnegativity and accordance of inequalities be-tween e and d is immediate consequence of the definition of d . It also satisfies the triangle inequalirty since d ( x, y ) + d ( y, z ) ≥ − δ ) > −
13 ) =43 > δ ≥ d ( x, z ) so d is a metric what finishes the proof of the lemmaR EFERENCES[1] M. Firer and L. L. Walker,
Matched Metrics and Channels
IEEE Trans.Infor. Theory vol. 62 no. 3, pp. 1150-1156, Mar. 2016[2] S. Lahiri