I nternat. J. ath. & Math. Sci.
(1980)15-28
Vol. 3 No.
15
ON BLOCK IRREDUCIBLE FORMS
OVER EUCLIDEAN DOMAINS
W. EDWIN CLARK and J.J. LIANG
Department of Mathematics
University of South Florida
U.S.A.
33620
Tampa, Florida
(Received December ii, 1978)
ABSTRACT.
In this paper a general canonical form for elements in a ring Euclidean
It is also shown that this form
with respect to a real valuation is established.
is unique and minimal thus gives the arithmetical weight of an element with
respect to a radix.
KEY WORDS AND PHRASES.
Euclidean Domains, Canonical Forms, Arithmetical Coding.
1980 MATHEMATICS SUBJECT CLASSIFICATION CODES.
i.
94A10.
INTRODUCTION.
In this paper we shall establish a general canonical form for elements in
a ring Euclidean with respect to a real valuation.
We show this form is unique
and minimal and thus gives us the arithmetical weight of an element with respect
to a radix r.
Throughout
v satisfying:
R
will denote a commutative ring Euclidean for a real valuation
W.E. CLARK AND J.J. LIAN
16
v(R)
(i)
is well-rdered hy the usual ordering of the real numbers.
a, 5 # O
for
R, there exists q, r in R such that a
in
+ r
hq
v(r) < v(5).
and
For completeness we recall that an element
a base) for
R if every element
a
R
of
R
of
r
(or
is called a radix
can 5e represented as a finite sum of
the form
[alri
a
where
v(a i) < v(r)
(i.I)
and we call such a representation a weak radlx-r form (or representation) for
For convnlence we often write
(i.I).
a
(an_l,...,a0)
or
The form (i.I) is said to he a mnlmal weak
number of indices
a
with
i
#
i
0
is minimal.
the radlx-r form is the number of nonzero
ai’s
in lleu of
an_l,...,ala0
radix
form for
Te weight of
a
a.
a
if the
relative to
in a mnlmal weak radlx-r form.
Some canonical minimal forms were given by Reltwiesner [i] for integers with
radix
r
2, Clark and Lang [2], Boyarinov [3], Kabatyanskll [4] for integers
and Clark and Liang [5] for Gaussian integers with
wth general radls r
radix
r
+ 1
+-
i,
We shall etahlish here a more general canonical minimal form for radix r
of
R whlc we call a 51ock irreducible form.
LEMMA i.
(am,...,al,a0)
Let
he an element of
r
(5m,...,51,b 0)
R
such that
v(r) > 3.
if and only if there exists
such that
b
0
5j
h
and
m
a
0
c0r
aj + cj_ 1
=a -Cm_
1
m
cjr,
for
0 < j < m
Then
c0,...,cj,..,
in
R
BLOCK IRREDUCIBLE FORMS OVER EUCLIDEAN DOMAINS
PROOF.
50
hence
Assume
a
<
V(Co)
< 2
3
(am,....,a l,aO)
Now,
c0r.
0
v(c t)
i
(bin,.. ,,5l,b0).
a
c0r
for all
50
0
mples
(am,...,al,a0) (.am,...,aI + Co, 50)
(bm,...,b I) (am,...,aI + Co). We thus have 51 -=
i al + c0 clr or Clr a I 51 + co Hence,
Therefore,
v(_e.)
v(r)
since
Therefore,
a
c2r
2
aj
hj
a
This implies
b
0
v-(-r) + v(r)
v(c0) <
vfr)
(5m,...,5 l,b0) which
a
c od r.
I + o
0
rood r
2
implies
Again, let
vCr) + 2 < 2 + 2 < 3
< vO:) +V)
’v(Ir
(am,. ,a2,aI + co (am,...,a2 + Cl,bl) (bm,...,b2,bl).
(am,...,a 2 + Cl) (bm,..,,b2). As before a 2 + c I c2r h 2 or
2v(r) + 3 < 3.
h + c
Proceeding in this way we
2
I. We have vc2) <
v(r)
>_
Now,
3.
aj + cj_1
If
17
0
we have
cjr
0
cj_ I
v(cj)
bj,
since
cj_I
cj r
< 3
J.
for all
implies
v
(cj_I)
> v(r) > 3,
a contradiction.
For the converse, we must assume v(a i)
We call the
DEFINITION 0.
in Lemma i carries.
a
Note that if
thus all carries are digits.
i
and
v(h i)
are both less than
v(r).
in (I.i) and in Lemma i digits, and the c
v(r) > 3 then all carries
However if
c.
satisfy
v(cj)
DEFINITION i.
(bin,... ,b 0)
Henceforth all carries are assumed to be digits.
The form
such that
(an,
,a0)
< 3
v(r) < 3 then a carry may not be a digit.
To avoid this complication we make the following
ASSUMPTION.
i
is reducible if there exists a form
W.E. CLARK AND J.J. LIANG
18
(1)
b.1
(2)
(hm,...,bO)
0
for some
,n}
{0,1,
i B
ad
(an,... ,aO) is
form (an,...a0)
Otherwise the form
LEMMA 2.
The
#
(an,...,aO)
(i)
a
(2)
there exists no
i
0
called irreducible.
is irreducible if and only if
O,...,n
for all i
and
,b 0)
(bk_ I,
where
<__ n
(bk+l,0,bk_l,...,b 0)
(,...,al,aO)
such that
is irreducible.
(an,...,a 0)
Let
PROOF.
k
If (2) fails
be irreducible then clearly (!) holds.
then
(bk+l,0,Sk_l,...,b0)
(ak,...,a0)
k
If
n
we get a contradiction so we. may assume
a r
n
Therefore, (an
a r
n
+
n
+
for some
+
n
+ ak+irk+l +
+
,...,%+i,%,...,a01
la’+Ir
k+l
+
m
K’I r
+
k+l
(anrn +
+
.(bk+l,0,bk_ l,...,b 0)_
.(0,bk_ I, ...,b 0)_ emr +
(Cm,... ,ek+l,0,bk_l,... ,b 0)
+
c,_+ir
k+l
+
m
+
<_
n.
i < n.
/
+
We can write
k+l
Ck+l r
ak+Irk+l) + (ak
anr
n
+
"’ao)
+ ak+ Ir k+l +
bk+Ir
Conversely, let
a
(an,
Then
(an,...,al,a0)-- (bm,...,bj,...,b0)
where
b.
0
for some
j, 0 < j < n
and
k+l
(0,bk_l,...,b 0)
a contradiction.
satisfy (i) and (2) and being reducible.
c r
m
k
k
j
being smallest possible.
Now
a l,a
0)
BLOCK IRREDUCIBLE FORMS OVER EUCLIDEAN DOMAINS
5
0
a
0
must
e
cor
c
I + o
51
a
5j
aj + cj_l
0 +
cj 0
cj
We have
0
clr
c
jr
r
(aj,aj_l,...,a0)
irreducible
19
(cj,0,Sj_l,...,b0). By the choice of
otherwise we would have
(cj ,O,5j_l,... ,b 0)
j,
bj_l,...b 0
(bm,... ,b’.3_l,...,b’s
0’’’"b0) and we could use this to find a smaller "j".
(hm,o..,bs,0,5_2,o..,5 ), then we can write
If (aj,aj_l,...,aO)
0
By
"addition’:,
an,...,a O) 65,.,.,,0,Bs2,...,0).
(an,...,aj+l,O,O ,0)
+ (0,...,0,aj,a_l,...,a0) (an,.,.,aj+l,0,...,0) + (...,bs,0,bs_2,...,b 0)
(bt,. ,bs,0,Bs_2,
DEFINITION 2.
aj #
0
irreduciSle.
or
t < j < s
hut
as at
(an,,., ,al,a0)
In otherwords
is called block irreducible if whenever
0, we must have
(as_ I ,...,at+I)
is composed of irreduciSle sequences
blocks) separated by sequences (or blocks) of zeros.
LEMMA 3.
v(q)
J,
for all
(an,...,al,a 0)
The form
<
If
a
qr
+
c
where
and
v(a) >_ v(r) > 2, then
2
-),-v(.a,
Te following corollary is an immediate consequence of lemm 3.
COROLLARY.
If
v (r) > 2, then the sequence
a
ql r + a0,
ql
q2 r +
qi
qi r + ai’
al’
v(al)
< v(r),
W.E. CLARK AND J.J. LIANG
20
contains an element
REMARK.
DEFINITION.
Let
n,...,al,a0)
(a
a
0
if
a
a
0
# O.
0
# O,
qo
qi
qi+l r + ai’
qi+l
.
ql
I # 0,...,
a
Can,,., aO)
ai_ 1 # O,
If
a
air +
+
anr
a
Then
0.
+ aI
+ a2
n- (i+2)
+
+ ai+ 2
a
a
is the soonest possible zero after
and for no smaller
0
i
with
aj
O, J <
is irreducible if and only if
is the soonest possible zero after
REMARK.
0
i
possible to find a representation for
a
+
r+an
We shall say that
a
n
+
anrn-I
n-2
anr +
q0
qn=0
Suppose
a r
n
q0 r + a0,
ql r + al’
=-I
(-1,2,2,2)
(I,0,0,-i), etc.
(i,0,-i),(2,2,2)
a
a
(-1,2,2)
(-1,2)
r-- 3, we have
(i,-i), (2,2)
2
< v(r).
v(qk)
such that
The sequence given above need not be bounded since e.g. in the ring
of integers for base
since
qk
a
to the following
q0 r +
a
qt-2
qt-i r +
qt-i
qt r +
qt
is it
i.
a
0
#
0
and
an+I
0
0.
.,.,as+2,0,as,,,,at,0,at_l,...,
a
i
0
at-2
0
qt+l r +
at
then the sequence corresponds
21
BLOCK IRREDUCIBLE FORMS OVER EUCLIDEAN DOMAINS
qs-i
qs
Clearly,
(as,...,a t)
possible zero after
for
r
b
a
and
t
0
Since
b
I
THEOREM i.
a
qs+l r +
a
t
0.
0
as+1
n+ 2
is the soonest
We shall show in theorem 3 that this
where
(an,...,ak,0,0,...,0)
a
b
0
-=
(an,...,ak,0,...,0)
b
as
V(qn)
< v(r)).
(bm,...,bk,...,b 0)
then
b
i
0
0, I,..., k-1.
i
PROOF
__a
If
+
is irreducible if and only if
process must stop (at or before,
LEMMA 4.
qs r
v(b 0) < r, this implies b
0. Thus
0
0
By induetion,
(bm,..,bk,.. ,bl and
0 mod r
and
bl
0.
k
(Uniqueness of Block Irreducible Form)
(an,...,al,a0)
Let
v(r) > 3
and
be a block irreducible form with non zero blocks.
(ak4 ,ak3
etc.
Then these blocks are unique in the sense that if
are the
and
(bm,...,b t)
irreducible blocks in two different block irreducible representa-
i-th
tlons, then
(a,...,ak)
k
t,
Z
m
and
m
j=k
m
W.E. CLARK AND J. J
22
LIANG
(...,0,a,..0,ak: 0,...,0) and a
where (a,...,ak) and (bm,...,b t) are both irreducible.
< m, then (bm,...,b k)
iff b # 0, hence t
k and if
t
PROOF.
Let
not irreducible.
a
Therefore,
We may assume
m.
k
m
m
=- --o airj rood rm+2
b r
j=o
or
m
J=O
(bj
aj)r
j
0 mod r
m+2
Therefore, either
m
Z
(b
S
aj) rj
0
in which case we have
m
j =o
bj rj
m
Z
j =o
aj
rj
or
2
J
0
(bj- aj)r
j
>_v(r)
m+2
which implies
2[v(r) m+l +
or
+ v(r)] > v(r) m+2
0.
0,bm,..,,ht,0,..,0)
By lemma 4, ak # 0
(...,0,a,...,ak)
Then we have
23
BLOCK IRREDUCIBLE FORMS OVER EUCLIDEAN DOMAINS
v(r)m+l
2v(r)
i
v(r)
v(r)m+2
>
i
or
v(r)m+l
2(v(r) m+l
i
i) > 2
2
v(r)m+l
-v(r)I
ii
> v(r) m+l
Therefore,
a contradiction.
m
a.r
m
j
3
j=k
j=k
bj rj
By induction one may show that the next irreducible block is also unique and all
blocks are unique.
(Minimality of Block Irreducible Form)
THEOREM 2.
a block irreducible form, then it is minimal.
(bm,...,bi,...,b 0)
:(ai,. ,a0)
a
PROOF.
b
b.
b0)
(bi,
(an,
a
Furthermore for each
a
0, some
j e
i,
(b i,...,b
0)
has no more
i
{l,2,...,k}.
b
b
0
s
0
c.
(aj,...,a0)
By lemma i
a
a
c0r
0
+
c
s-i
Csr
aj + cj_ I
c.r3
0
s
+ c.
0
is
i, if
,a0). By lemma 4, we may assume a 0 # 0, b 0 # 0.
have a
(...,0,ak,...,a0) where (ak, a0) is irreducible. If
(...,bk,...,b 0) then b.3 # 0 for j 0,...,k, for suppose not, let
(a
O)
has weight at least the weight of
It suffices to show that for each
zero terms than
we
then
If
r
(cj,0,bj_l,...,b 0)
0 < s < j
i
Thus
W.E. CLARK AND J.J. LIANG
24
(ak,...,a0)
(bp, b 0)
which cannot happen since
1
i mapping of zeros of
where
and
p
a
P
is irreducible.
b
P
a
to
However, if
p
(ap,...,a0)
into zeros of
(an,...,a0).
is beyond the first irreducible block of
0, we map
a
P
Now, suppose we have a
# 0 and b
p
for some
0
bp
If
0, we then
P
have the following situation:
(ap,... ,aZ)
0=a
p
is irreducible
+ Cp-1
-cr
p
bp_ I ap_ 1 + Cp_2 Cp_ir
b
S
b
I
b_ 1
Suppose
b.3
0
for some
J
a
S
+ cj_ I
cjr
I + c/_ I c/r
0 + c_
c_ir
2
a
e {p-l,...,}, we have
aj
cjr
Hence
cj_ I.
(Cp,0,bp_l,...,bi) (ap,...,a). Since we can begin the carrying at aj [with
aj cjr] and this will allow us to get 0 at the p-th digit, we obtain a
contradiction to the fact that
(ap,...,a) is irreducible. Hence bp-i # 0
bp_ 2 # 0,...,b # 0. Now if b_ 1 0 we have cz_ 2 c_ir which implies
c_ 2 c_ 1 0 and so we have (0,bp_l,...,bZ) (ap,...,a) since we do not
need the carry from
mapped to
a_ 1
THEOREM 3.
(-l)st digit (it
is
zero).
Therefore
b
P
0
can be
0.
(Existence of Block Irreducible Form)
has a block irreducible form with respect to a radix
Every element
r
if
v(r)
>_
2.
a
in
R
25
BLOCK IRREDUCIBLE FORMS OVER EUCLIDEAN DOMAINS
that
PROOF.
Let
a
aj #
but
a
0
(az,...,a0)
(ak+l,%,...,aj) reducible. Then
where (a,...,
a) is irreduclble.
and induction yield for
Note that
(am,...,an,...,a0)
since
V(qn
qn- an
qn+l r
pick any
n
such that
< v(r)
as
a"
(am,... an,....,a 0)
where
leads to the sequence of
q0 r +
a
and at some point
a
Now we want to show the process will stop.
is block irreducible.
a
Assume
(a+2,0,,...,a)
the above to
as large as desired, a
n
a.
irreducible but
Now, we can rewrlte
,a’.3’0’’’’’0)" Applying
2,0,
(an,...,a0)
(ak,...,aj)
(ak+l,ak,...,aj
t < j, also
0,
t
be any weak radlx-r form for
a
0
q0
ql r + al
qn
qn+l r +
an
which implies that
2v(r)
V(qn+l) < v(r)V(qn) < v(r) and a
so
2
v(qj)
<_ v(r)
(...
< v(r)
and by induction.
,an,... ,a0)
an # 0.
for all
j
_>
n
Now
where
qn rqn+l + an’
qn+l r 0 + qn+l and 0 r 0 + 0. So a (0,qn+l,an,...,az,0,...) where
a # 0
a # 0 and (an, az) is irreducible. If (qn+l,an, az) is
n
and
(a’.,0,a’
If not (0, qn+l,an,..., a)
irreducible, we are done
n ’’’’’a
nz
(a,... ,a) is irreducible so a (an+ 2,0,a’n, ...,a,0, ,a l,a0) is block
(an,...,a0)
irreducible.
smallest
is block irreducible.
Now if
n < j
a
n
such that
0
Suppose
we claim
aj #
0
a.
3
we have
0
for
We then have
j > n.
Otherwise for
W.E. CLARK AND J.J. LIANG
26
qn+l r +
qn
qjr + 0
qj+l r + aj
qj-i
qj
but
implies
qjr
qj-I
0
qj
0
and
aj
-qj+l r
implies
0, a contra-
a.3
diction.
In what follows we shall give an algorithm for finding the block irreducible
v(r) > 3.
form for
Actually these are just some ideas on how to possibly
simplify the search for block irreducible forms.
LEMMA 5.
be the set of all representatives of the form
Let
(ak,ak_l,... ,a 0)
where all proper subsequences are irreducible but the sequence
itself is reducible.
{l,2,...,k},
j
Alj A
a
0)
I #
(a l,a 0)
If
(ak_l,...,a 0)
is
(ak, ak_l,... ,ak_ j
A.3
# 0.
k
is reducible.
irreducible so are all proper subsequences.
then ther is a smallest
j
such that
No proper subsequences will be reducible since it
would contradict to the choice of
ALGORITHM.
U
a
Since
(ak,... ,ak_ j)
2
is irreducible iff
(ak_l,...,a0) is
(ak,...,a0) were reducible
PROOF.
Thus, if
A
(ak,ak_ I,
irreducible then
for all
Let
j.
(For finding block irreducible form)
We may assume a 0 # 0,
(al,a0) AI iff (al,al) is irreducible. If
A, consider (a 2,al,a0). WOLG, assume ai # 0, i 0, i, 2. It is
irreducible iff (a2,a I)
A, and (a2,al,a0) e A 2. In general if we have chosen
(ak_ I, a I) irreducible then (ak,... ,aI) is also irreducible iff
Thus if we find
(ak,ak_ I) { AI, (ak,ak_l,ak_ 2) A2,...,(ak,...,a0)
a
0.
By definition
.
BLOCK IRREDUCIBLE FORMS OVER EUCLIDEAN DOMAINS
(ak,...,aj) A t, then we replace
(bk+l,0,bk_l,...,bj,aj_l,...,a0)
irreducible.
Reduce the rest of
(ak,...,a 0)
by
and we know
a
27
(bk_l,...,bj,aj_l,...,a0)
by carrlng
and then begin the same process with the new
bk+l
is
to the left as necessary
(k+l)st
term if it is non zero
(or the next non zero term).
LEMMA 6.
If the form
0,1,...,k+l
j
(ak+l’’’"a0)
ak+l
(2)
v(a 0
c0r)
(3)
and
v(r)
for
j e
but
v(aj
Let
0 <
cjr)
cjr + cj_ 1)
(ak+l""’a0)
bj aj + cj_ I cjr,
v(bj) < v(r) for j _< k
reducible, a contradiction.
j
+I
e
v(aj
v(r)
<
cj)
then
a
Let
(ak+l,...,a0)
then
k+l,...,l
otherwise
Also, v(a.
and
0
b.3
v
i
e
0
a
(bk+2,0,bk,
c0r.
0
would imply
3
because
is irreducible since
{0,+i}.
being
Since
would be reducible, again a
(ak+l,,...,a0)
a
4
Now
i < j < k.
for
b0)
(aj,...,0)
R be the ring of Gaussian integers and
(4 + 711, 50 + 50i)
ui,
b
c.r) > v(r)
(ak+l,ak,...,aj)
[-(I+i),4 + 71i,50 + 50i] e A 2
for any
{i,...,}
< v(r)
contradiction since no proper subsequence of
EXAMPLE.
k
>__ v(r)
3
if
c
Ck+ir
<
v(aj
with
cj,
such that
(1)
PROOF.
then there exist carries
+l’
(0,-95
+ 711 + u I +
r
is reducible.
i00.
281, -50
u21
The element
501)
# 100(vI +
and
v21)
W.E. CLARK AND J.J. LIANG
2.8
REFERENCES
I.
Relwlesner, G. H. Advances in Computers (F. L. Alt, ed.), Vol. i,
Academic Press, New York, 1960.
2.
Clark, W. E. and J. J. Liang.
Oft arithmetic weight for a general radix
presentation of integers, IEEE Trans. Information Theory (Nov. 1973)
823-826.
3.
Boyarlnov, I. M. Nonblnary arithmetic codes with large minimum distance,
March 1975)
Vol. ii, No. i, Problemy Peredachl Informatsli (Jan.
57-63.
(Russian)
4.
Kabatyanskll, G. A. Bounds on the number of code words in binary arithmetic
codes, Vol. 12, No. 4, Problemy Peredachl Informatsil (Oct. -Dec. 1976)
46-54. (Russian)
5.
Clark, W. E. and J. J. Liang. Weak radix representation and cyclic codes
over Euclidean domains, Communicat.lons in Algebra 4(11) (1976) 9991028.