(I) Notation:
1. N, n are two large integers. In our paper, we take N = n + j.
2. K is a subset of {1, 2, …, N}. Total number of set K is k.
3. A block bi of K is a subset in K with the consecutive number, and out
end point must not belong to the set K. For example, bi = {a. a+1. a+2} is
a block, which means a-1 and a+3 do not belong to K. and the number of
bi = 3..
4. A bar C h is an edge in the circle N. We call ‘the bar C h cover bi‘ that
means all numbers i between two end points belong to bi (in the small part,
i.e. its length is less than N/2).
Theorem 1: K={b1, b2,....bj}, b1, b2,....bj are partitions of K. C1 = 1, Cj =
2^(j+1)= 0<j <g with 2^g <= K < 2^(j+2). For any bi, there is at least
one bar Cj covers bi, and bi <= Cj/2.
Theorem 2: G[1, 2^1. …2^g: n+2k] is an K-ft of G[1; n]. Here 2^(g-1)
<=K < 2^G.
If a set with K broken nodes in the circle, It is a K set. So we could
suppose that broken node form a broken set {b1, b2,....bj} Try to find
cover bar in C. In Theorem 1, we could find C such that total length < 2K.
Therefore we could get an N nodes circle each node with a offset connect.
Proof of Theorem 1: A set of bj ’s we call it a good set, if we could find a
bar(s) C which covers all bj with its length less than doble??. Union of
two good sets is a good set. Denot the “maximum good set” by G which
means there is no good set is bigger than G.. We want to prove G = K.
1. Out of G, no bi = 1, or 2.: 1 and 2 are good sets.
Let a set N = {1,2,…,N},and K is a subset of N, the number in K use the
same notation K. a block in K, it is a subset{a, a+1,…a+b} with, a-1,
A+b+1 are not in K. We could write K = {b1,b2…bj}bi are blocks in K.,
We call a partition of K, So b1 + b2 +…+bi = K.
Let Ci be a set {a+1, …, a+h}., We call Ch cover bg it means bg contain in
Ch without a, a+h+1. We could use offset to get an edge to cross the
block.
Lemma 1. If b1 = 3, b1={a, a+a, a+2} is a block in K, we could find a Ci =
4,cover b1. Or a-2, and a+4 in K. So between a-2 to a+4, at least there are
5 numbers in K.
Proof: Lengths of [a-1,a+4}, [a-2, a+3] both are 4. So that there is no
4-bar covers b1, i.e., one of end points in above two intervals must belong
K
Lemma 2: G[1,2,4,8;N+8] is a 5-ft of G[1,N].
Proof: The broken node up to K, we could say that it is partition of K. If
the longest block of K is 4, or 5. It is easy to find a bridge which covers
those with length of bar less than 8. If the length of longest bar is 3, from
lemma 1, we could find bar(s) whos total length less then 8, which covers
it .
Lemma 3: G[1,2,4,8;N+8] is a 7-ft of G[1,N].
Proof: Use Lemma2, if bi = 1, 2, we could find a bridge 1, or 2 cover it
and no except any good points.
Lemma 4: G[1,2,4,8;N+8] is a 8-ft of G[1,N].
Let N = {1,2,…N}. K be a subset of N with K numbers. Bi (1<=j) be a
subset of K, such that
1, with the consecutive numbers for example {a, a+1,…a+h}
2. a-1, and a+h+1 is not in K.
We call bi by a block of K.
Denote K = {b1, b2,…bj} bh (1 < =h <+j) be block of K. we use bh
denote the length of block bh also. So K = b1 +b2 +…+bj. In fact {bh} be
a partition of K.We want to prove:
Theorem: For any K, we could find a circle {d1,d2,…dN} with:
1 d1 – dN and di – d(i-1) = 2^I, 0 <= I <= j+1, 2^j<= K <2^(j+1).
2. {d1,d2,…dN} and K no dommon point.
This is to prove G[1,2,2^2,…2^J;N+ 2k] is a K-ft of G[1;N].
We call di-d(i-1) by “bridge., between two number di, d(i-1) might be
belong to K,or not, we call the points belong to K by “bad points” If it is
not we call it “good points”
Theorem: G[1, 2,2^2,…, 2^j; N + 2K] is a k-ft of G[1;N], where
2^(j-1) < K < 2^j.
Denote:
N be a set {1, 2,…N}. K be a subset of N, total number in K be K. So if
broken nodes be <= K, it must be a set like K (total number <= K and
subset of N + 2K)
So we only need to prove, for any set K we could find a subgraph : a
circle with the nodes in {N + 2k} \ {K}.
Some consequtive number we call it by “bar”, for examble {b,+1,
b+2,…b+h} it is a bar, denote it by bh. From now we suppose a bar in K
mean the numbers b and b+b+h+1 are not in {K}. It means bar bh still in
the grath if we take K nodes in K off. We could use it to be a part in the
circle we want.
Now we want to finda punch of bars{bh1, b1h2,…bdhd}Such that:
1. Those bars no common poind each other.
2. h1 + h2 + …+hd <= 2K.
3. Bars cover all points in K.
Lemma 1: If we could find the bars s.t. above conditions, we could find a
circle with N nodes and is a subgraph of G[1, 2,2^2,…, 2^j; N + 2K].
So G[1, 2,2^2,…, 2^j; N + 2K] is a k-ft of G[1;N].
Proof: We choose the bars as a part of circle, add the nodes out of bars,
then the points in the circle must no one in the set K. Total nodes is N
+2K – Z. where z be sum of bars i. e. h1 + h2 +… + hd. Since z < = 2K,
so N + 2K – z >= N. If N is big enough, we could use lengh 2 bar instead
lengh 1 bar to cut some node and keep graph be a circle. So we could get
a N nodes circle with the bars as edges. So must be not include any point
in K.
So we want to prove our theorem it is only to show, we could get some
bars s. t. three conditions abouv.
Now we denote set K as a punch of chain: K = {Cd1, Cd2,… Cdj} Chain
Cdi, be di consecutive number C+1, C+2, … C+di, s.t.
1. all number in K,
2. 2.C and C(di+1) are not in K.
3. So we could choose a Bar C to C+4 to cover a chain C3, i.e. {C+1,
C+2, C+3}. We try to show, for any K set, we could find some bar s. t.
above conditions. As an example we proof:
Lemma 2. G[1,2,3, 5,9;N+10]is a 5-ft of G[1,N]
ProoF: K = 5, so write as chain, if only show the lengh of chain, it will be:
{5}, {4,1}, {3,2} {3,1,1} {2,2,1} ,{2,1,1,1} or {1,1,1,1,1}. Chain 1, 2 4
we could find bar 2,3, or 5, cover it s.t. above conditions. For Chain 5, we
could use bar9, 8 < 2 time 5. Only for chain3: a+1, a+2, a+3. We could
use bar (a, a+5),(a-1, a+4) cover it, if a+5, and a-1 not contain in K. If
they are in K, then we could use (a-2, a+7) to cover 5 points in K.
Lemma 2.1. G[1,2,4,8;N+10]is a 5-ft of G[1,N]
Proof. The proff is similar.
Lemma 2.2. G[1,2,4:N+3] is a 3-ft of G[1;N].
Proof:3-broken nodes might be one of the following:
I.
{1,2,3}
II.
{1,2,4}
III.
{1,2,b} b is not 3 and not 4.
IV.
{a,B,c}, ab-a >=2, c-b>=2.
We want to show for each of above, we could find a circle without the
broken nodes.
I; We choose, {0,4,5 …N}.
II We choose{1,5,3 7,6 8,…N}
III. We choose {1 5,4,3,7,6,8. >>> N}
IV we choose a-1, a+1, b-1, b+1, c-1, C+1, and all between that except a,
b,c.
So for any Kset, K=3, there is a N-node circle.