Optimal Proof - Nisoku in SRM 463

I'll prove a subtle part of the problem:

Given a sorted array of even length $\{a_1, a_2, \ldots, a_{2n}\}$, we need to pair these elements and multiply the sum of each pair. The largest value should be $(a_1 + a_{2n}) \times (a_2 + a_{2n-1}) \cdots$.

Let's consider an easier case. Suppose we have 4 numbers $a<b<c<d$. It's obvious that $(a+d)(b+c)>(a+b)(c+d)$ and $(a+d)(b+c)>(a+c)(b+d)$. How can we generalize this property?

Suppose $a_1$ is paired with some $a_i$, and $a_{2n}$ is paired with some $a_j$. By a single exchange, we could get a better solution. After that, the pair of the smallest & largest element remains the same, and we can use induction here for inner elements.

Sometimes it's just essential to prove pairwise inequalities for sequence optimality.

Iterate through the Subsets of Subsets

Bit-mask DP

The main loop:

for (int mask = 0; mask < (1 << n); ++mask)

go(mask);

In the memoization process:

for (int i = mask; i != 0; i = (i - 1) & mask)

// the status transformation

For a certain bit which is 1 in mask, all possible combinations of lower 1 bits are enumerated, ending in all 0s. Then this bit becomes 0 with other initial lower 1 bits recovered by the bit-wise and. Again combinations of those bits are enumerated, ending in all 0s, while this bit remains 0.

Complexity

The overall time complexity measured in the number of transformations is $latex \sum_{i=0}^{n}(\binom{n}{i} \sum_{j=0}^{i}\binom{i}{j}) - 2^n = 3^n - 2^n$.

SRM536 Div1 Level2 500 Problem

The solution is ready for download at https://sites.google.com/site/boyouepii/document/SRM-536-Div1-Medium.pdf?attredirects=0&d=1.
I'm practicing $latex \textrm{\LaTeX}$ these days and this solution happens to be quite mathematical.

Enable LaTeX in Blogger

The method is as follows:

1) Registry on Blogger: http://www.blogger.com
2) Create a site on Google sites for keeping your files: https://sites.google.com
(Take a look at mine: https://sites.google.com/site/boyouepii/)
3) Create a Picassa album for including photos inside your blog.(Optional)

You can do all of this steps with your same Google Account.

Now if you want to add $latex \textrm{\LaTeX} $ functionality then:

Download this file: https://sites.google.com/site/boyouepii/mathtex.js?attredirects=0&d=1

And, if you want, change its name to for example: mylatex.js.

Upload this file to your Google sites account and copy the link URL.
Then inside Blogger, on the configuration:

Design>Page Elements>Add a Gadget>HTML/JavaScript

Create a Gadget named, for example, LaTeX and include inside it this HTML Code:

<script src="here include the link to your mylatex.js file" type="text/javascript">
</script>
<script type="text/javascript">
replaceMath( document.body );
</script>

And that′s everything, now you can try to write a post including(without line breaks):

$latex \displaystyle \lim_{n \to \infty} \sum_{k=1}^{n} \frac{1}{k^2} = \frac{\pi^2}{6} $

One Possible Implementation of the Generic Push-Relabel Algorithm

The book said that we can select an applicable operation in O(1) time, and use O(V) and O(1) time for relabel and push operation respectively.

First note that all operations are performed on overflowing vertices other than s or t. We may store all these vertices in a data structure and when that structure becomes empty, the algorithm terminates. When nonempty, pick an arbitrary vertex u from it. If there is some admissible edge (u,v)(that is, (u,v) is in the residual network and h[u]=h[v]+1), a push operation is applicable. Otherwise, a relabel operation is applicable.

So we may wish to maintain a list of admissible edges for every vertex. Every time, either a push or relabel operation may be applicable depending on whether the corresponding list is empty or not. After a push, the only possible influence on these lists is that (u, v) has to be removed from the list corresponding to u after a saturating push. If we simply associate every edge a pointer to the nodes in these lists, the removal can be done in O(1) time. Each edge (u,v) may appear in these lists at most once, for h[u]=h[v]+1 and h[v]=h[u]+1 cannot hold at the same time.

But after a relabel operation on u, some edges (u,v) has to be inserted into the list corresponding to u, and all edges (v,u) appearing in other lists has to be removed. The insertion is simple. But we have to check the lists for every other vertex in order to remove edges ending at u, thus resulting in O(E) time complexity. The solution is to use orthogonal lists. Intuitively, all admissible edges leaving u are on the horizontal line, while those entering u are on the vertical line. Now, the insertion and removal operations take time proportional to the out degrees and in degrees of vertex u. A single operation of both types is now a little bit more complicated, but requires only O(1) time.

Finally, a simple list suffices for the data structure mentioned in the second paragraph. Every time, pick the vertex at the end of that list. Relabel doesn't change the flow of any vertex, so no updates are necessary. After a push from u to v, remove u if it is no longer overflowing and insert v if its flow is 0 previously.

Now, the generic push-relabel algorithm is implemented using a total time of O(V²E).

The Initialization of the Generic Push-Relabel Algorithm

When we apply the generic push-relabel algorithm, we first label s as |V| and t as 0, and all edges starting from s are pushed in as many units of flow as possible, i.e they are saturated. But why these operations are necessary and reasonable is not stated explicitly.

What if we label s and t with different values? The difference between h[s] and h[t], h[s]-h[t], must be no smaller than |V|, for otherwise the statement that there is no path from s to t in the residual network never holds. And when the preflow becomes a flow, it is not necessarily a maximum flow since there may be augmenting paths. If s and t are labeled with greater values other than |V| and 0, the bounds on the number of relabel, saturating push and thus non-saturating push will become higher. Intuitively, there must be redundant relabel operations before extra flows are pushed to t or back to s since h[s] and h[t] become greater, and during that process, extra push operations between vertices other than s and t occur.

Suppose for some neighbor u of s, we initially push less flow into edge (s,u) than c(s,u). Then (s,u) is in the residual network. But h[s]=|V|>=2>h[u]+1=0+1=1, i.e. h is no longer a valid height function.

SRM522 Div1 Level1 250 Problem

Given a board consisting of n(n>=2) cells, each marked with either an 'A' or 'B', Alice and Bob alternatively place coins on some continuous empty cells, with the restriction that at least one cell must be left empty. When there is only one cell left, Alice wins if that cell is marked with 'A' and Bob wins otherwise.

First of all, if at least one end is marked with an 'A', then Alice wins. Without loss of generality, suppose the board is of the form "A??...?"('?' means empty cells marked with 'A' or 'B' arbitrarily). Alice can obtain "ACC...C"('C' means covered by a coin) in one step and win.

Then what if the board is of the form "B??...?B"?
If Alice covered either end with a coin, again without loss of generality, the board becomes "CC...CB" or "CC...C??..?B". Bob wins immediately for the former, by covering all other empty cells except the rightmost 'B' for the latter. So Alice would have to choose "B??...?CC...C??...?B" if she wanted to win.
Now let's consider a simpler situation, "B??...?CC...CB". (I)If it is now Bob's turn, he only needs to cover all the empty cells on the left and wins. (II)If it is now Alice's turn, she has up to four choices. If she choose to cover the rightmost 'B' or either end of the empty cells on the left, Bob can obviously win. Then again she has to choose to make the board of the form "B??...?CC...C??...?CC...CB". If Bob covers all empty cells of the second interval, then we reach the same configuration except that the number of empty cells decreases. So we can prove by induction that no matter whose turn is it, the winner would be Bob if the board is of the form "B??...?CC...CB".
So let's return to "B??...?CC...C??...?B". If the second interval is empty, the it corresponds to (I). Otherwise, Bob could cover all empty cells on the right except the rightmost 'B', and now it corresponds to (II). Either way, Bob wins.

So all we have to do is to check the two cells on both ends.

I code a bit mask DP during the contest. Finding patterns and proving them is still a hard task for me. Need more exercise.