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 hauger 1:06am on Sunday, April 18th, 2010 Efficient, attractive looking, large fridge, fast freeze. Bargain Nothing really Good Styling, Cheap Price. Very Very noisy Jclijsen 5:31pm on Wednesday, March 31st, 2010 for my family of five the size of both fridge and freezerare great.its not an obtrusive size for 12.5 ltrs. easy instructions, holding capacity enough for a family of four, easy to clean, does what it says on the box, easy to move, looks good none

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### Documents

#### princeton u. sp02

cos 598B: algorithms and complexity
Lecture 10: Bourgains Theorem
Lecturer: Sanjeev Arora Scribe:Daniel J. Peng
The goal of this weeks lecture is to prove the Theorem 1 (Bourgain) Every n-point metric embeds into
version of Bourgains Theorem:

#### with distortion O(lg n).

The proof will depend on the partitioning idea from Fakcharoenphol, Rao and Talwar (2003). The idea is similar to the proof of Bourgains Theorem for 1 from Lecture 3. For each length scale s, we dene a partition Ps by the following algorithm: 1. Pick uniformly at random a number R [2s1 , 2s ). 2. Pick uniformly at random an order on the elements of X. 3. Partition the items of X into at most n = |X| blocks as follows. (a) Proceed with the elements of X according to the order. (b) For each element x X, pick all non-assigned elements within distance R from it, and form a new block. We call Ps the partition created above, and denote by Ps (x) the block in which x was placed. Just as in the
case, we can apply the Padded Decomposition Property.
Theorem 2 (Padded Decomposition Property) Let Ps be a partition of X, and let x X. If 2s3 , B(x, 2s+1 ) lg |B(x, 2s3 )| (1)
1 then Pr,R [B(x, ) Ps (x)] 2.
Proof: We proved this theorem in lecture 3. 2 We also have a corollary; since the growth ratio is less than the number of nodes, that is, B(x, 2s+1 ) n, |B(x, 2s3 )| the corollary follows: Corollary 3 (Corollary to the Padded Decomposition Property) Let Ps be a partition of X. For all x X and any constant c R, we have Pr,R B x, 2s |c lg n| Ps (x) . 2 (3) (2)
2 Before proving Bourgains Theorem, we rst establish a simpler result, embedding an n-point

#### metric into

with distortion O (lg n) 2.

#### Simpler Embedding into

with distortion O (lg n)(3/2).
First, we construct zero sets, so named because they will have coordinate 0 in the embedding. Definition 1 (Zero sets) To construct the zero set Zs , merge nodes whose distance is less than 2s 10n. Then, pick each block in Ps independently with probability 2 , and take the union. Now, we proceed to construct the embedding into 2. First, scale the distance function so that the minimum distance is 1, and let be the maximum distance. Then, since the radius R of a block for Zs is at least 2s1 , there are lg nontrivial zero sets: Z1 ,. , Zlg . Then, we construct the embedding f by Frechets technique: Definition 2 Let f be the function from X into Rlg as follows: f : x (d(x, Z1 ), d(x, Z2 ),. , d(x, Zlg )) We claim that this embedding f has O (lg n)(3/2) distortion, that is, d(x, y) E[|f (x) f (y)|] (lg n)(3/2) d(x, y). This is immediate from the following theorem: Theorem 4 If (X, d) is an n-point metric space and f is an embedding as described above, then for all x, y X, we have: d(x, y)2 (E[|f (x) f (y)|])2 (lg n)d(x, y)2. (lg n)2 Proof: The trivial upper bound follows from the triangle inequality:

s=1 lg

|d(x, y)|2

#### (8) (9)

d(x, y) lg
2 so (E[|f (x) f (y)|])2 d(x, y) lg d(x, y)2 (lg ). We improve this bound by observing that only lg n terms contribute to the summation in Equation 7. Nodes x and y are merged if 2s 10n(d(x, y)), so the contribution is zero. On the
2 That is, if d(x, y) < 10n , then merge them into one node z. The metric on the new set is is the shortest path metric d , where for any u,v, d (u, z) = min(d(u, x), d(u, y)) and d (u, v) = min(d (u, z) + d (z, v), d(u, v)). 1
3 other hand, when 2s d(x, y), the distance d(x, Zs ) and d(y, Zs ) will tend to be less than 2s , so this contribution falls o geometrically. Thus, the terms matter only for the lg n values of s satisfying d(x, y)) 2s 10n(d(x, y)). We conclude that (E[|f (x) f (y)|])2 d(x, y)2 (lg n). For the lower bound, we show that just one coordinates contribution makes the required contribution. Fix x X and y X. Consider the coordinate s where 2s d(x,y). 4 Since the diameter of each block in Ps is at most 2s+1 and the distance d(x, y) is 4(2s ) = 2s+2 , it must be the case that x and y are in dierent blocks. Thus, the zero set Zs contains x with probability 2 , and it contains y with independent probability 2. By Corollary 3 to the Padded 2s 1 Decomposition Property, we have that B(x, 10 lg n ) P (x) with probability at least 2. So, with s probability 1 , the zero set Zs contains x but not y, and d(y, Zs ) 102lg n. Since d(x, y) 2s+2 , we 8 have d(y, Zs )

#### d(x,y) 40 lg n

with probability 1. It follows that 8 (E[|f (x) f (y)|])2

#### d(x,y)2. 8(40 lg n)2

2 Remark 1 To show this property without the expectation, we simply repeat the process and concatenate the embeddings from each iteration. Cherno bounds show that this process will bring the embedding arbitrarily close (1 + ) to the expectation. Remark 2 The scale where 2s d(x, y) is important because it is the only scale where the distances are large enough to make a real contribution, but small enough that x and y are in dierent blocks.

#### Bourgains

Theorem (full)
We will now prove Bourgains theorem: any n-point metric can be embedded into distortion. This will be immediate from the following theorem:

#### with O(lg n)

Theorem 5 If (X, d) is an n-point metric space and f is an embedding as described below, then for all x, y X, we have: d(x, y)2 (E[|f (x) f (y)|])2 (lg n)d(x, y)2. (11) lg n To prove this theorem, we rst apply a technique due to KLMN 2004; we glue the lg scales into lg n coordinates. Definition 3 Let R(x, t) be the maximum radius R for which |B(x, R)| 2t. Definition 4 Let K(x, t) = lg R(x, t). We now dene a growth ratio that reects how quickly the density of vertices changes around x. Definition 5 For any small natural numbers c and c , we let GR = lg This denition is ambiguous, but we will x this later.

#### m+c B(x,2

|B(x,2mc )|
4 Remark 3 Observe that if lg |B(x, 2m )| lg B(x, 2m3 ) , then R(x, t) stays around 2m for many |B(x,2m )| values of t. More precisely, R(x, t) 2m (and K(x, t) m) for about lg |B(x,2m3 )| = lg(GR) values of t. We have Zs as before, except that we do not need to merge nodes before constructing Zs. The Zs will not be the zero sets for this theorem, however. Instead, we will now dene the zero sets Wt as follows, by gluing the Zs together. In order to decide whether or not to join Wt , each node x snis around its neighborhood to determine K(x, t) and then checks if it lies in ZK(x,t). If so, it joins Wt. Definition 6 Wt = x : x ZK(x,t). This is not the precise denition we will nally use, but it will convey the general idea of the proof. We will dene Wt more precisely later. Remark 4 Observe that when t = lg n, the maximum radius R for which |B(x, R)| 2t is the maximum distance. Thus, t goes from 1 to lg n, the function R(x, t) goes to , the function K(x, t) goes to lg , and ZK(x,t) goes to Zlg , as we would expect. Again, we dene the embedding function f in Frechets style: Definition 7 Let f be the function from X into Rlg as follows: f : x (d(x, W1 ), d(x, W2 ),. , d(x, Wlg n )) We now present the proof of Theorem 5. Proof: The upper bound follows trivially from the triangle inequality:

#### t=1 lg n

|d(x, Wt ) d(y, Wt )|2 |d(x, y)|2

#### (14) (15)

(lg n)d(x, y)2
For the lower bound, we consider the scale m where 2m d(x, y). This is the important scale that we noted in Remark 2; however, in contrast to the O((lg n)3/2 ) embedding, the gluing now gives us multiple coordinates that involve this scale. In the previous embedding, there was just one zero set, Zm , that involved this scale; now, since Remark 3 gives us lg(GR) values of t for which K(x, t) m, we have lg(GR) zero sets Wt = x : x ZK(x,t) that involve Zm = ZK(x,t). m We apply the Padded Decomposition Property to show that B(x, lg2GR ) Pm (x) with proba1 bility at least 2. Then, we apply the same logic from the previous proof to each coordinate: with d(x,y) 1 2s probability 8 , the zero set Wt contains x but not y, and d(y, Wt ) lg(GR) lg(GR). Thus, it follows that for each t for which K(x, t) m, we have: E[|d(x, Wt ) d(y, Wt )|] d(x, y). lg(GR) (16)
5 This allows us to conclude the proof of the upper bound: (E[|f (x) f (y)|])2 E[|f (x) f (y)|2 ]

#### (17) (18) (19)

E[|d(x, Wt ) d(y, Wt )|2 ] E[|d(x, Wt ) d(y, Wt )|2 ]

#### t:K(x,t)m

By the padding property from Equation (16):

#### d(x, y)2 (lg(GR))2

Since there are log(GR) such coordinates: (lg(GR)) Since the growth ratio is at most n: This proves the theorem. 2 There are two subtleties that we overlooked in the proof of the theorem. We consider them here. subtlety 1: The Padded Decomposition Property does not strictly apply to Wt. A point x is not in Wt because it is not in Zm , where m = K(x, t). The point x hopes that, with good probability, it is a distance at least 2m / log GR from Wt , so that it will be far away from a distant point y, d(x, y) > 2m+1 that lands in Wt. This hope is jeopardized by the possibility that a point z in B(x, ) might be in Wt because it is deciding looking at a dierent Zm , where m = K(z, t) and m is not necessarily the same as m. We deal with this by showing that the K(, ) has a certain smoothness property which implies that m is necessarily close to m. Lemma 6 (Smoothness Lemma) Let x X and m = K(x, t). Let z B(x, ), where

(21) (22)

#### d(x, y)2 lg(GR)

d(x, y)2 lg n
Finally, let m = K(z, t). Then
m {m 4, m 3, m 2, m 1, m, m + 1}. Proof: We know that B(x, 2m ) contains roughly 2t points. Since d(x, z) 2m , we know that B(x, 2m ) B(z, 2m + 2m ) = B(z, 2m+1 ), and there must be 2t points in B(z, 2m+1 ). Thus, m m + 1.
6 For the lower bound, we know that B(z, 2m ) contains roughly 2t points. Since d(x, z) m we know that B(z, 2m ) B(x, 2m + 2 ). 2 10

#### 2m 10 ,

So, the concern is warranted, but there are only 6 dierent scales to worry about. Thus, we simply assert that with probability 26 , we have d(x, Zm ) lg 2m B(x, 2m +1 ) |B(x, 2m 3 )| (24)
for all m between m 4 and m + 1. This simply reduces the constant factor in our padding for Wt in Equation 16. subtlety 2: The growth ratio GR stands for all of the dierent growth ratios |B(x,2mc )| with dierent constants c and c. Most notably, in the size of the paddings (Equation 24), we have various m , and in the number of coordinates (Remark 3), the growth ratios have dierent constants. We resolve this by picking a large constant c. Then, we expand the number of coordinates by a factor of 2c + 1 by dening the zero sets: Wi,t = x : x ZK(x,t)i , where i ranges from c to c and t ranges from 1 to lg n. Since K(x, t) i m when K(x, t) [m c, m + c], the summation in Equation 19 includes lg |B(x,2m+c )|
values of (i, t). By making c suciently large, we can add more and more coordinates into the sum, until we cancel a lg(GR) in the denominator in Equation 21: (E[|f (x) f (y)|])2

#### t:K(x,t)im

d(x, y)2 (lg(GR))2 d(x, y)2 (lg(GR))2

#### |B(x,2m+c )|

(26) (27) (28)
d(x, y)2 lg(GR) d(x, y)2 lg n

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