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## Double cosets and contingency tables

Like my previous post, this blog is also motivated by a comment by Professor Persi Diaconis in his recent Stanford probability seminar. The seminar was about a way of “collapsing” a random walk on a group to a random walk on the set of double cosets. In this post, I’ll first define double cosets and then go over the example Professor Diaconis used to make us probabilists and statisticians more comfortable with all the group theory he was discussing.

### Double cosets

Let $G$ be a group and let $H$ and $K$ be two subgroups of $G$. For each $g \in G$, the $(H,K)$-double coset containing $g$ is defined to be the set

$HgK = \{hgk : h \in H, k \in K\}$

To simplify notation, we will simply write double coset instead of $(H,K)$-double coset. The double coset of $g$ can also be defined as the equivalence class of $g$ under the relation

$g \sim g' \Longleftrightarrow g' = hgk$ for some $h \in H$ and $g \in G$

Like regular cosets, the above relation is indeed an equivalence relation. Thus, the group $G$ can be written as a disjoint union of double cosets. The set of all double cosets of $G$ is denoted by $H\backslash G / K$. That is,

$H\backslash G /K = \{HgK : g \in G\}$

Note that if we take $H = \{e\}$, the trivial subgroup, then the double cosets are simply the left cosets of $K$, $G / K$. Likewise if $K = \{e\}$, then the double cosets are the right cosets of $H$, $H \backslash G$. Thus, double cosets generalise both left and right cosets.

### Double cosets in $S_n$

Fix a natural number $n > 0$. A partition of $n$ is a finite sequence $a = (a_1,a_2,\ldots, a_I)$ such that $a_1,a_2,\ldots, a_I \in \mathbb{N}$, $a_1 \ge a_2 \ge \ldots a_I > 0$ and $\sum_{i=1}^I a_i = n$. For each partition of $n$, $a$, we can form a subgroup $S_a$ of the symmetric group $S_n$. The subgroup $S_a$ contains all permutations $\sigma \in S_n$ such that $\sigma$ fixes the sets $A_1 = \{1,\ldots, a_1\}, A_2 = \{a_1+1,\ldots, a_1 + a_2\},\ldots, A_I = \{n - a_I +1,\ldots, n\}$. Meaning that $\sigma(A_i) = A_i$ for all $1 \le i \le I$. Thus, a permutation $\sigma \in S_a$ must individually permute the elements of $A_1$, the elements of $A_2$ and so on. This means that, in a natural way,

$S_a \cong S_{a_1} \times S_{a_2} \times \ldots \times S_{a_I}$

If we have two partitions $a = (a_1,a_2,\ldots, a_I)$ and $b = (b_1,b_2,\ldots,b_J)$, then we can form two subgroups $H = S_a$ and $K = S_b$ and consider the double cosets $H \backslash G / K = S_a \backslash S_n / S_b$. The claim made in the seminar was that the double cosets are in one-to-one correspondence with $I \times J$ contingency tables with row sums equal to $a$ and column sums equal to $b$. Before we explain this correspondence and properly define contingency tables, let’s first consider the cases when either $H$ or $K$ is the trivial subgroup.

### Left cosets in $S_n$

Note that if $a = (1,1,\ldots,1)$, then $S_a$ is the trivial subgroup and, as noted above, $S_a \backslash S_n / S_b$ is simply equal to $S_n / S_b$. We will see that the cosets in $S_n/S_b$ can be described by forgetting something about the permutations in $S_n$.

We can think of the permutations in $S_n$ as all the ways of drawing without replacement $n$ balls labelled $1,2,\ldots, n$. We can think of the partition $b = (b_1,b_2,\ldots,b_J)$ as a colouring of the $n$ balls by $J$ colours. We colour balls $1,2,\ldots, b_1$ by the first colour $c_1$, then we colour $b_1+1,b_1+2,\ldots, b_1+b_2$ the second colour $c_2$ and so on until we colour $n-b_J+1, n-b_J+2,\ldots, n$ the final colour $c_J$. Below is an example when $n$ is equal to 6 and $b = (3,2,1)$.

Note that a permutation $\sigma \in S_n$ is in $S_b$ if and only if we draw the balls by colour groups, i.e. we first draw all the balls with colour $c_1$, then we draw all the balls with colour $c_2$ and so on. Thus, continuing with the previous example, the permutation $\sigma_1$ below is in $S_b$ but $\sigma_2$ is not in $S_b$.

It turns out that we can think of the cosets in $S_n \setminus S_b$ as what happens when we “forget” the labels $1,2,\ldots,n$ and only remember the colours of the balls. By “forgetting” the labels we mean only paying attention to the list of colours. That is for all $\sigma_1,\sigma_2 \in S_n$, $\sigma_1 S_b = \sigma_2 S_b$ if and only if the list of colours from the draw $\sigma_1$ is the same as the list of colours from the draw $\sigma_2$. Thus, the below two permutations define the same coset of $S_b$

To see why this is true, note that $\sigma_1 S_b = \sigma_2 S_b$ if and only if $\sigma_2 = \sigma_1 \circ \tau$ for some $\tau \in S_b$. Furthermore, $\tau \in S_b$ if and only if $\tau$ maps each colour group to itself. Recall that function composition is read right to left. Thus, the equation $\sigma_2 = \sigma_1 \circ \tau$ means that if we first relabel the balls according to $\tau$ and then draw the balls according to $\sigma_1$, then we get the result as just drawing by $\sigma_2$. That is, $\sigma_2 = \sigma_1 \circ \tau$ for some $\tau \in S_b$ if and only if drawing by $\sigma_2$ is the same as first relabelling the balls within each colour group and then drawing the balls according to $\sigma_1$. Thus, $\sigma_1 S_b = \sigma_2 S_b$, if and only if when we forget the labels of the balls and only look at the colours, $\sigma_1$ and $\sigma_2$ give the same list of colours. This is illustrated with our running example below.

### Right cosets of $S_n$

Typically, the subgroup $S_a$ is not a normal subgroup of $S_n$. This means that the right coset $S_a \sigma$ will not equal the left coset $\sigma S_a$. Thus, colouring the balls and forgetting the labelling won’t describe the right cosets $S_a \backslash S_n$. We’ll see that a different type of forgetting can be used to describe $S_a \backslash S_n = \{S_a\sigma : \sigma \in S_n\}$.

Fix a partition $a = (a_1,a_2,\ldots,a_I)$ and now, instead of considering $I$ colours, think of $I$ different people $p_1,p_2,\ldots,p_I$. As before, a permutation $\sigma \in S_n$ can be thought of drawing $n$ balls labelled $1,\ldots,n$ without replacement. We can imagine giving the first $a_1$ balls drawn to person $p_1$, then giving the next $a_2$ balls to the person $p_2$ and so on until we give the last $a_I$ balls to person $p_I$. An example with $n = 6$ and $a = (2,2,2)$ is drawn below.

Note that $\sigma \in S_a$ if and only if person $p_i$ receives the balls with labels $\sum_{k=0}^{i-1}a_k+1,\sum_{k=0}^{i-1}a_k+2,\ldots, \sum_{k=0}^i a_k$ in any order. Thus, in the below example $\sigma_1 \in S_a$ but $\sigma_2 \notin S_a$.

It turns out the cosets $S_a \backslash S_n$ are exactly determined by “forgetting” the order in which each person $p_i$ received their balls and only remembering which balls they received. Thus, the two permutation below belong to the same coset in $S_a \backslash S_n$.

To see why this is true in general, consider two permutations $\sigma_1,\sigma_2$. The permutations $\sigma_1,\sigma_2$ result in each person $p_i$ receiving the same balls if and only if after $\sigma_1$ we can apply a permutation $\tau$ that fixes each subset $A_i = \left\{\sum_{k=0}^{i-1}a_k+1,\ldots, \sum_{k=0}^i a_k\right\}$ and get $\sigma_2$. That is, $\sigma_1$ and $\sigma_2$ result in each person $p_i$ receiving the same balls if and only if $\sigma_2 = \tau \circ \sigma_1$ for some $\tau \in S_a$. Thus, $\sigma_1,\sigma_2$ are the same after forgetting the order in which each person received their balls if and only if $S_a \sigma_1 = S_a \sigma_2$. This is illustrated below,

We can thus see why $S_a \backslash S_n \neq S_n / S_a$. A left coset $\sigma S_a$ correspond to pre-composing $\sigma$ with elements of $S_a$ and a right cosets $S_a\sigma$ correspond to post-composing $\sigma$ with elements of $S_a$.

### Contingency tables

With the last two sections under our belts, describing the double cosets $S_a \backslash S_n / S_b$ is straight forward. We simply have to combine our two types of forgetting. That is, we first colour the $n$ balls with $J$ colours according to $b = (b_1,b_2,\ldots,b_J)$. We then draw the balls without replace and give the balls to $I$ different people according $a = (a_1,a_2,\ldots,a_I)$. We then forget both the original labels and the order in which each person received their balls. That is, we only remember the number of balls of each colour each person receives. Describing the double cosets by “double forgetting” is illustrated below with $a = (2,2,2)$ and $b = (3,2,1)$.

The proof that double forgetting does indeed describe the double cosets is simply a combination of the two arguments given above. After double forgetting, the number of balls given to each person can be recorded in an $I \times J$ table. The $N_{ij}$ entry of the table is simply the number of balls person $p_i$ receives of colour $c_j$. Two permutations are the same after double forgetting if and only if they produce the same table. For example, $\sigma_1$ and $\sigma_2$ above both produce the following table

By the definition of how the balls are coloured and distributed to each person we must have for all $1 \le i \le I$ and $1 \le j \le J$

$\sum_{j=1}^J N_{ij} = a_i$ and $\sum_{i=1}^I N_{ij} = b_j$

An $I \times J$ table with entries $N_{ij} \in \{0,1,2,\ldots\}$ satisfying the above conditions is called a contingency table. Given such a contingency table with entries $N_{ij}$ where the rows sum to $a$ and the columns sum to $b$, there always exists at least one permutation $\sigma$ such that $N_{ij}$ is the number of balls received by person $p_i$ of colour $c_i$. We have already seen that two permutations produce the same table if and only if they are in the same double coset. Thus, the double cosets $S_a \backslash S_n /S_b$ are in one-to-one correspondence with such contingency tables.

### The hypergeometric distribution

I would like to end this blog post with a little bit of probability and relate the contingency tables above to the hyper geometric distribution. If $a = (m, n-m)$ for some $m \in \{0,1,\ldots,n\}$, then the contingency tables described above have two rows and are determined by the values $N_{11}, N_{12},\ldots, N_{1J}$ in the first row. The numbers $N_{1j}$ are the number of balls of colour $c_j$ the first person receives. Since the balls are drawn without replacement, this means that if we put the uniform distribution on $S_n$, then the vector $Y=(N_{11}, N_{12},\ldots, N_{1J})$ follows the multivariate hypergeometric distribution. Thus, if we have a random walk on $S_n$ that quickly converges to the uniform distribution on $S_n$, then we could use the double cosets to get a random walk that converges to the multivariate hypergeometric distribution (although there are smarter ways to do such sampling).

## Combing groups

Two weeks ago I gave talk titled “Two Combings of $\mathbb{Z}^2$“. The talk was about some material I have been discussing lately with my honours supervisor. The talk went well and I thought it would be worth sharing a written version of what I said.

## Geometric Group Theory

Combings are a tool that gets used in a branch of mathematics called geometric group theory. Geometric group theory is a relatively new area of mathematics and is only around 30 years old. The main idea behind geometric group theory is to use tools and ideas from geometry and low dimensional topology to study and understand groups. It turns out that some of the simplest questions one can ask about groups have interesting geometric answers. For instance, the Dehn function of a group gives a natural geometric answer to the solvability of the word problem.

## Generators

Before we can define what a combing is we’ll need to set up some notation. If $A$ is a set then we will write $A^*$ for the set of words written using elements of $A$ and inverses of elements of $A$. For instance if $A = \{a,b\}$, then $A^* = \{\varepsilon, a,b,a^{-1},b^{-1},aa=a^2,abb^{-1}a^{-3},\ldots\}$ (here $\varepsilon$ refers to the empty word). If $w$ is a word in $A^*$, we will write $l(w)$ for the length of $w$. Thus $l(\varepsilon)=0, l(a)=1, l(abb^{-1}a^{-3})=6$ and so on.

If $G$ is a group and $A$ is a subset of $G$, then we have a natural map $\pi : A^* \rightarrow G$ given by:

$\pi(a_1^{\pm 1}a_2^{\pm 1}\ldots a_n^{\pm 1}) = a_1^{\pm 1}\cdot a_2^{\pm 1} \cdot \ldots \cdot a_n^{\pm 1}$.

We will say that $A$ generates $G$ if the above map is surjective. In this case we will write $\overline{w}$ for $\pi(w)$ when $w$ is a word in $A^*$.

## The Word Metric

The geometry in “geometric group theory” often arises when studying how a group acts on different geometric spaces. A group always acts on itself by left multiplication. The following definition adds a geometric aspect to this action. If $G$ is a group with generators $A$, then the word metric on $G$ with respect to $A$ is the function $d_G : G \times G \rightarrow \mathbb{R}$ given by

$d_G(g,h) = \min \{ l(w) \mid w \in A^*, \overline{w} = g^{-1}h \}.$

That, is the distance between two group elements $g,h \in G$ is the length of the shortest word in $A^*$ we can use to represent $g^{-1}h$. Equivalently the distance between $g$ and $h$ is the length of the shortest word we have to append to $g$ to produce $h$. This metric is invariant under left-multiplication by $G$ (ie $d_G(g\cdot h,g\cdot h') =d_G(h,h')$ for all $g,h,h' \in G$). Thus $G$ acts on $(G,d_G)$ by isometries.

## Words are Paths

Now that we are viewing the group $G$ as a geometric space, we can also change how we think of words $w \in A^*$. Such a word can be thought of as discrete path in $G$. That is we can think of $w$ as a function from $\mathbb{N}$ to $G$. This way of thinking of $w$ as a discrete path is best illuminated with an example. Suppose we have the word $w = ab^2a^{-1}b$, then

$w(0) = e,$
$w(1) = \overline{a}$
$w(2) = \overline{ab}$
$w(3) = \overline{ab^2}$
$w(4) = \overline{ab^2a^{-1}}$
$w(5) = \overline{ab^2a^{-1}b}$
$w(t) = \overline{ab^2a^{-1}b},$ $t \ge 5$.

Thus the path $w : \mathbb{N} \rightarrow G$ is given by taking the first $t$ letters of $w$ and mapping this word to the group element it represents. With this interpretation of word in $A^*$ in mind we can now define combings.

## Combings

Let $G$ be a group with a finite set of generators $A$. Then a combing of $G$ with respect to $A$ is a function $\sigma : G \rightarrow A^*$ such that

1. For all $g \in G$, $\overline{\sigma_g} = g$ (we will write $\sigma_g$ for $\sigma(g))$.
2. There exists $k >0$ such that for all $g, h \in G$ with $g \cdot \overline{a} = h$ for some $a \in A$, we have that $d_G(\sigma_g(t),\sigma_h(t)) \le k$ for all $t \in \mathbb{N}$.

The first condition says that we can think of $\sigma$ as a way of picking a normal form $\sigma_g \in A^*$ for each $g \in G$. The second condition is a bit more involved. It states that if the group elements $g, h \in G$ are distance 1 from each other in the word metric, then the paths $\sigma_g,\sigma_h : \mathbb{N} \rightarrow G$ are within distance $k$ of each other at any point in time.

## An Example

Not all groups can be given a combing. Indeed if we have a combing of $G$, then the word problem in $G$ is solvable and the Dehn function of $G$ is at most exponential. One group that does admit a combing is $\mathbb{Z}^2 = \{(m,n) \mid m,n \in \mathbb{Z}\}$. This group is generated by $A = \{(1,0),(0,1)\} = \{\beta,\gamma\}$ and one combing of $\mathbb{Z}^2$ with respect to this generating set is

$\sigma_{(m,n)} = \beta^m\gamma^n$.

The first condition of being a combing is clearly satisfied and the following picture shows that the second condition can be satisfied with $k = 2$.

## A Non-Example

The discrete Heisenberg group, $H_3$, can be given by the following presentation

$H_3 = \langle \alpha,\beta,\gamma \mid \alpha\gamma = \gamma\alpha, \beta\gamma = \gamma\beta, \beta\alpha = \alpha\beta\gamma \rangle$.

That is, the group $H_3$ has three generators $\alpha,\beta$ and $\gamma$. The generator $\gamma$ commutes with both $\alpha$ and $\beta$. The generators $\alpha$ and $\beta$ almost commute but don’t quite as seen in the relation $\beta\alpha = \alpha\beta\gamma$.

Any $h \in H_3$ can be represented uniquely as $\sigma_h = \alpha^p\beta^m\gamma^n$ for $p,m,n \in \mathbb{Z}$. To see why such a representation exists it’s best to consider an example. Suppose that $h = \overline{\gamma\beta\alpha\gamma\alpha\beta\gamma}$. Then we can use the fact that $\gamma$ commutes with $\alpha$ and $\beta$ to push all $\gamma$‘s to the right and we get that $h = \overline{\beta\alpha\alpha\beta\gamma^3}$. We can then apply the third relation to switch the order of $\alpha$ and $\beta$ on the right. This gives us that that $h = \overline{\alpha\beta\gamma\alpha\beta\gamma^3}=\overline{\alpha\beta\alpha\beta\gamma^4}$. If we apply this relation once more we get that $h = \overline{\alpha^2\beta^2\gamma^5}$ and thus $\sigma_h = \alpha^2\beta^2\gamma^5$. The procedure used to write $h$ in the form $\alpha^p\beta^m\gamma^n$ can be generalized to any word written using $\alpha^{\pm 1}, \beta^{\pm 1}, \gamma^{\pm 1}$.

The fact the such a representation is unique (that is if $\overline{\alpha^p\beta^m\gamma^n} = \overline{\alpha^{p'}\beta^{m'}\gamma^{n'}}$, then $(p,m,n) = (p',m',n')$) is harder to justify but can be proved by defining an action of $H_3$ on $\mathbb{Z}^3$. Thus we can define a function $\sigma : H_3 \rightarrow \{\alpha,\beta,\gamma\}^*$ by setting $\sigma_h$ to be the unique word of the form $\alpha^p \beta^m\gamma^n$ that represents $h$. This map satisfies the first condition of being a combing and has many nice properties. These include that it is easy to check whether or not a word in $\{\alpha,\beta,\gamma\}^*$ is equal to $\sigma_h$ for some $h \in H_3$ and there are fast algorithms for putting a word in $\{\alpha,\beta,\gamma\}^*$ into its normal form. Unfortunately this map fails to be a combing.

The reason why $\sigma : H_3 \rightarrow \{\alpha,\beta,\gamma\}^*$ fails to be a combing can be seen back when we turned $\overline{ \gamma\beta\alpha\gamma\alpha\beta\gamma }$ into $\overline{\alpha^2\beta^2\gamma^5}$. To move $\alpha$‘s on the right to the left we had to move past $\beta$‘s and produce $\gamma$‘s in the process. More concretely fix $m \in \mathbb{N}$ and let $h = \overline{\beta^m}$ and $g = \overline{\beta^m \alpha} = \overline{\alpha \beta^m\gamma^m}$. We have $\sigma_h = \beta^m$ and $\sigma_g = \alpha \beta^m \gamma^m$. The group elements $g$ and $h$ differ by a generator. Thus, if $\sigma$ was a combing we should be able to uniformly bound $d_{H_3}(\sigma_g(t),\sigma_h(t))$ for all $t \in \mathbb{N}$ and all $m \in \mathbb{N}$.

If we then let $t = m+1$, we can recall that

$d_{H_3}(\sigma_g(t),\sigma_h(t)) = \min\{l(w) \mid w \in \{\alpha,\beta,\gamma\}^*, \overline{w} = \sigma_g(t)^{-1}\sigma_h(t)\}.$

We have that $\sigma_h(t) = \overline{\beta^m}$ and $\sigma_g(t) = \overline{\alpha \beta^m}$ and thus

$\sigma_g(t)^{-1}\sigma_h(t) = (\overline{\alpha\beta^m})^{-1}\overline{\beta^m} = \overline{\beta^{-m}\alpha^{-1}\beta^m} = \overline{\alpha^{-1}\beta^{-m}\gamma^{m}\beta^m}=\overline{\alpha^{-1}\gamma^{m}}.$

The group element $\overline{\alpha^{-1}\gamma^{m}}$ cannot be represented as a shorter element of $\{\alpha,\beta,\gamma\}^*$ and thus $d_{H_3}(\sigma_g(t),\sigma_h(t)) = m+1$ and the map $\sigma$ is not a combing.

## Can we comb the Heisenberg group?

This leaves us with a question, can we comb the group $H_3$? It turns out that we can but the answer actually lies in finding a better combing of $\mathbb{Z}^2$. This is because $H_3$ contains the subgroup $\mathbb{Z}^2 \cong \langle \beta, \gamma \rangle \subseteq H_3$. Rather than using the normal form $\sigma_h = \alpha^p \beta^m \gamma^n$, we will use $\sigma'_h = \alpha^p \tau_{(m,n)}$ where $\tau : \mathbb{Z}^2 \rightarrow \{\beta,\gamma \}^*$ is a combing of $\mathbb{Z}^2$ that is more symmetric. The word $\tau_{(m,n)}$ is defined to be the sequence of $m$ $\beta$‘s and $n$ $\gamma$‘s that stays closest to the straight line in $\mathbb{R}^2$ that joins $(0,0)$ to $(n,p)$ (when we view $\beta$ and $\gamma$ as representing $(1,0)$ and $(0,1)$ respectively). Below is an illustration:

This new function isn’t quite a combing of $H_3$ but it is the next best thing! It is an asynchronous combing. An asynchronous combing is one where we again require that the paths $\sigma_h,\sigma_g$ stay close to each other whenever $h$ and $g$ are close to each other. However we allow the paths $\sigma_h$ and $\sigma_g$ to travel at different speeds. Many of the results that can be proved for combable groups extend to asynchronously combable groups.

## References

Hairdressing in Groups by Sarah Rees is a survey paper that includes lots examples of groups that do or do not admit combings. It also talks about the language complexity of a combing, something I didn’t have time to touch on in my talk.

Combings of Semidirect Products and 3-Manifold Groups by Martin Bridson contains a proof that $H_3$ is asynchornously combable. He actually proves the more general result that any group of the form $\mathbb{Z}^n \rtimes \mathbb{Z}^m$ is asynchronously combable.

Thank you to my supervisor, Tony Licata, for suggesting I give my talk on combing $H_3$ and for all the support he has given me so far.


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# One week a certain truck rental lot had a total of 20 trucks

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The Official Guide For GMAT® Quantitative Review, 2ND Edition

One week a certain truck rental lot had a total of 20 trucks, all of which were on the lot Monday morning. If 50 percent of the trucks that were rented out during the week were returned to the lot on or before Saturday morning of that week, and if there were at least 12 trucks on the lot that Saturday morning, what is the greatest number of different trucks that could have been rented out during the week?

(A) 18
(B) 16
(C) 12
(D) 8
(E) 4

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One week a certain truck rental lot had a total of 20 trucks, all of which were on the lot Monday morning. If 50 percent of the trucks that were rented out during the week were returned to the lot on or before Saturday morning of that week, and if there were at least 12 trucks on the lot that Saturday morning, what is the greatest number of different trucks that could have been rented out during the week?

(A) 18
(B) 16
(C) 12
(D) 8
(E) 4

First how to deal with "at least" and "greatest number" part of the question.

General rule for such kind of problems:
to maximize one quantity, minimize the others;
to minimize one quantity, maximize the others.

So to maximize the # of trucks rented we should minimize # of trucks at the lot on Saturday. We are told that # of trucks at the lot on Saturday was at least 12, so to minimize it, we should consider this number to be 12 (minimum possible).

Next, the # of trucks at the lot on Saturday, 12, equals to {the # of trucks not rented} plus {half of the # of trucks rented} --> $$(20-R)+\frac{1}{2}R=12$$ --> $$R=16$$.

Or: as "50% of the trucks that were rented out during the week were returned" then another 50% of the trucks that were rented were not returned --> not returned = 20-12=8 trucks, which is 50% of the trucks that were rented --> # of truck were rented = 2*8 = 16.

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Rented Trucks (R)

20-R+.5R>=12
R<=16

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Total trucks = 20
Say rented trucks = x
Balance trucks = (20-x)

Half of the rented trucks returned early $$= \frac{x}{2}$$ returned

So now the balance trucks are $$(20-x) + \frac{x}{2}$$

Availability = 12 trucks, so equating

$$20-x + \frac{x}{2} = 12$$

$$\frac{x}{2} = 8$$

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Re: One week a certain truck rental lot had a total of 20 trucks [#permalink]

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28 Feb 2014, 06:29
4
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The difficulty of this problem is the way it has been written. So lets diagonise each sentence and try to make an equation

One week a certain truck rental lot had a total of 20 trucks, all of which were on the lot Monday morning
Total trucks = 20 Day:Monday

If 50 percent of the trucks that were rented out during the week were returned to the lot on or before Saturday morning of that week, and

Lets say 'x' trucks were rented. We can conclude two things :
No. of trucks remaining in lot = 20 - x Day:unknown
No. of trucks returned to lot = 0.5x Day: Various days (at this point we get a hint that days might not be important)

if there were at least 12 trucks on the lot that Saturday morning
No. of truck in lot >= 12 Day:Saturday

what is the greatest number of different trucks that could have been rented out during the week?

No. of trucks rented out i.e. x

It should be fairly easy to analyze

20-x + 0.5x >= 12
20 -0.5x >= 12
0.5x =< 8
x =< 16

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Re: One week a certain truck rental lot had a total of 20 trucks [#permalink]

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20 May 2014, 13:06
I mean this is quite obvious, but has anyone else also noticed that we could solve this problem far more quickly if we just plug in answer choices
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Re: One week a certain truck rental lot had a total of 20 trucks [#permalink]

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20 May 2014, 22:32
MensaNumber wrote:
I mean this is quite obvious, but has anyone else also noticed that we could solve this problem far more quickly if we just plug in answer choices

I think "plugging choices" is based upon how comfortable you are in any topic and it's the best method if you, indeed, are comfortable.

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Re: One week a certain truck rental lot had a total of 20 trucks [#permalink]

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20 May 2014, 23:17
b2bt wrote:
MensaNumber wrote:
I mean this is quite obvious, but has anyone else also noticed that we could solve this problem far more quickly if we just plug in answer choices

I think "plugging choices" is based upon how comfortable you are in any topic and it's the best method if you, indeed, are comfortable.

Youre right. I am not comfortable with 'plugging choices' myself and I learning this important skill. I could save the day if you get stuck or running out of time
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Re: One week a certain truck rental lot had a total of 20 trucks [#permalink]

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12 May 2015, 05:30
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Expert's post
This problem seems difficult at the first glance because of the lengthy sentences in which the information has been conveyed. In questions like this, it's a good idea to try and represent the given information visually as you go through each sentence of the question. Here's one mode of visually representing the information given here:

The trucks on the lot on the Saturday morning = (Trucks that were not rented) + (Trucks that were rented and returned by Saturday)
= $$(20-r) + (\frac{r}{2})$$

As given:

$$(20-r) + (\frac{r}{2})$$ > = 12

Upon solving this inequality, we get: r < = 16

Thus, the maximum possible number of rented trucks = 16

Hope this was useful!

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Re: One week a certain truck rental lot had a total of 20 trucks [#permalink]

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02 Nov 2015, 03:04
Rented=x
Returned=0,5x
Not Returned=0,5x
20-0,5x(not returned)=12
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One week a certain truck rental lot had a total of 20 trucks [#permalink]

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12 Sep 2016, 09:11
Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

One week a certain truck rental lot had a total of 20 trucks, all of which were on the lot Monday morning. If 50 percent of the trucks that were rented out during the week were returned to the lot on or before Saturday morning of that week, and if there were at least 12 trucks on the lot that Saturday morning, what is the greatest number of different trucks that could have been rented out during the week?

(A) 18
(B) 16
(C) 12
(D) 8
(E) 4

Problem Solving
Question: 120
Category: Arithmetic; Algebra Percents; Applied problems
Page: 77
Difficulty: 600

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Thank you!

I tried the plug and play approach (it just came to my mind first instead of the conventional one)

If we assume all 20 truck were out, then only 10 truck would have returned Saturday morning.

But we know that at least 12 truck were there on a Saturday morning,

We also know that even number of trucks left else 50% of them could not return (you cannot return half truck can you )

so again, moving to next possible number:

assume 2 were not rented and 18 were rented. On Saturday morning = 2+9= 11

ahh still not there but close:

assume 4 were not rented this time and 16 were rented. On Saturday morning trucks = 4(not rented) + 8(16\2) = 12 -> this is our guy. Note that we need to find max number of truck rented, if we go ahead in the flow i.e 6 were not rented and 14 were (well doesn't matter because 14 is anyways less than 16).

So 16 is the answer, beware of the trap answer 8! (8 were returned but rented were 8*2 and not 8).

thanks.
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Re: One week a certain truck rental lot had a total of 20 trucks [#permalink]

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16 Oct 2017, 15:19
Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

One week a certain truck rental lot had a total of 20 trucks, all of which were on the lot Monday morning. If 50 percent of the trucks that were rented out during the week were returned to the lot on or before Saturday morning of that week, and if there were at least 12 trucks on the lot that Saturday morning, what is the greatest number of different trucks that could have been rented out during the week?

(A) 18
(B) 16
(C) 12
(D) 8
(E) 4

Problem Solving
Question: 120
Category: Arithmetic; Algebra Percents; Applied problems
Page: 77
Difficulty: 600

GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition - Quantitative Questions Project

Each week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution.

We'll be glad if you participate in development of this project:
2. Please vote for the best solutions by pressing Kudos button;
3. Please vote for the questions themselves by pressing Kudos button;
4. Please share your views on difficulty level of the questions, so that we have most precise evaluation.

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this problem can be elegantly paraphrased as:

total number of trucks - number of trucks rented out + number of trucks returned = 12

cheers

thanks

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Re: One week a certain truck rental lot had a total of 20 trucks   [#permalink] 16 Oct 2017, 15:19
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## Thursday, December 29, 2011

### The Theories and Math's that have led to BIGTOE

An overt statement was made to say: "The boson is a natural outcrop of the Higgs mechanism . Why is it so difficult to corner the cagey Higgs Boson?"  To which I replied:

How is the Boson, a natural outcrop of the Higgs mechanism? You may ask better,

"It appears that, the Boson, is a natural outcrop of the Higgs mechanism, how is it that it is so difficult to corner the cagey Higgs boson?"

Otherwise I am going to have to ask you for your math proof for your statement, or reference to whom has proved the boson to be, ..."a natural outcrop of the Higgs mechanism".

It is simple, Bosons, by definition, are slippery. They carry no effective electrically charged surface differential providers, which makes them slippery. Having an effective electrically charged surface differential provider, that's created dynamically, what we view as a spiny constant mono static field of charged electric potential, that is perpendicular to the surface, with the magnetic field counterpart, that is in concentric spheroidal shells, parallel to the electrically charged surface differential provider, is a fermion, like an electron, or Proton.

Neutral particles are Bosons, and are not easily affected by magnets, except for very, very strong magnets that will align the axis of the magnetic moment, that is bipolar in nature and results from the singular average spin axis, and the triangular spinning and rotating structure of the 3, point shaped quarks, that form the triangle, the spinning and rotation of said triangle structure, forms the larger neutral particle.

Photons are a strange case, because they lack a structure, such as that of a sphere that seems to hold for point particles, and larger, one that gives them a more 'solid feel', instead it is more of a 'solid rod'. Photons are more like a rod of a finite length, that spins and rotates through 4D hyperspace, that gives us up to 3 different ways, so far, to view the Photon, each dependant upon the measuring device used, and the relative position / direction of the rod, (photon), at impact / absorption, that gives us the concept of a particle, wave, or field.

Neutrinos are the strangest, in that they have next to zero interaction with anything the size of an atom and larger. They seem to be subnucleonic, and all matter is transparent to them, except to very slow moving ones, that will hit an electron, kicking out a photon out of the electron cloud, to be detected by very sensitive instruments. More at, IMHO, these neutrinos, with a dense flux streaming in from the Sun and elsewhere, actively work with the neutrons, to produce the electron and proton "pairs" that form the nuclei of atoms with the electron clouds around them, leaving an average of one neutron to one proton in all stable atoms, and a few extra neutrons in the not so stable isotopes.

From wikipedia:
In particle physics, the Higgs mechanism is the process in which gauge bosons in a gauge theory can acquire non-vanishing masses through absorption of Nambu-Goldstone bosons arising in spontaneous symmetry breaking.

The above leads to qualifying questions, "How is a gauge boson different from any other boson?"; "What, exactly, is a gauge boson?"; If mass can vanish, and then reappear, that would explain another 3 experiences I have had, if the mechanism is KNOWN in how the mass vanishes, then comes back, let alone be non-vanishing. Who are Nambu and Goldstone, how did they get their bosons, what makes them different from the standard boson, or the gauge boson, or the one I have discovered, an ⓒ"Intermediate Hypervector Boson"™ = IHB, and don't have it named after me as yet, that might be the affect that created an effect that I witnessed.

Hmm let's see ... Gauge Theory - In physics, a gauge theory is a type of field theory in which the Lagrangian is invariant under a **continuous group of local transformations**.

The term gauge refers to redundant degrees of freedom in the Lagrangian. The Lagrangian, L, of a dynamical system is a function that summarizes the dynamics of the system. It is named after Joseph Louis Lagrange. The concept of a Lagrangian was originally introduced in a reformulation of classical mechanics by Irish mathematician William Rowan Hamilton known as Lagrangian mechanics. In classical mechanics, the Lagrangian is defined as the kinetic energy, T, of the system minus its potential energy, V:: L = T - V;

Under conditions that are given in Lagrangian mechanics, if the Lagrangian of a system is known, then the equations of motion of the system may be obtained by a direct substitution of the expression for the Lagrangian into the Euler–Lagrange equation, a particular family of partial differential equations.

The transformations between possible gauges, called gauge transformations, form a Lie group which is referred to as the symmetry group or the gauge group of the theory. Associated with any Lie group is the Lie algebra of group generators. For each group generator there necessarily arises a corresponding vector field called the gauge field.

In mathematics, a Lie algebra ( /ˈliː/, not /ˈlaɪ/) is an algebraic structure whose main use is in studying geometric objects such as Lie groups and differentiable manifolds. Lie algebras were introduced to study the concept of infinitesimal transformations. The term "Lie algebra" (after Sophus Lie) was introduced by Hermann Weyl in the 1930s. In older texts, the name "infinitesimal group" is used.

A Lie algebra is a vector space over some field F together with a binary operation [·, ·]
[*,*] : g x g -> g, called the Lie bracket, which satisfies the following axioms:

Bi-linearity:
[ax + by , z] = a[x , z] + b[y , z] , [z , ax + by] = a[z , x] + b[z , y]
for all scalars a, b in F and all elements x, y, z in g.

Alternating on g :
[x , x] = 0 for all x in g.

The Jacobi identity:
[x , [y , z]] + [y , [z , x]] + [z , [x , y]] = 0
for all x, y, z in g.

Note that the bi-linearity and alternating properties imply anticommutativity, i.e., [x , y] = -[y , x] for all elements x, y in g, while anticommutativity only implies the alternating property if the field's characteristic is not 2.[1]

For any associative algebra A with multiplication * , one can construct a Lie algebra L(A). As a vector space, L(A) is the same as A. The Lie bracket of two elements of L(A) is defined to be their commutator in A: [a , b] = a * b - b * a;

Please see : xxx http://en.wikipedia.org/wiki/File:E8PetrieFull.svg xxx is a nice 2D spherical representation. Simple Lie groups include many classical Lie groups, which provide a group-theoretic underpinning for spherical geometry, projective geometry and related geometries in the sense of Felix Klein's Erlangen programme. It emerged in the course of classification of simple Lie groups that there exist also several exceptional possibilities not corresponding to any familiar geometry. These exceptional groups account for many special examples and configurations in other branches of mathematics, as well as contemporary theoretical physics. Secondly the Lie algebra only determines uniquely the simply connected (universal) cover G* of the component containing the identity of a Lie group G. It may well happen that G* isn't actually a simple group, for example having a non-trivial center. We have therefore to worry about the global topology, by computing the fundamental group of G (an abelian group: a Lie group is an H-space). This was done by Élie Cartan.

In mathematics, an H-space is a topological space X (generally assumed to be connected) together with a continuous map μ : X × X → X with an identity element e so that μ(e, x) = μ(x, e) = x for all x in X. Alternatively, the maps μ(e, x) and μ(x, e) are sometimes only required to be homo-topic to the identity (in this case e is called homotopy identity), sometimes through base point preserving maps. These three definitions are in fact equivalent for H-spaces that are CW complexes. Every topological group is an H-space; however, in the general case, as compared to a topological group, and from what I have seen, H-spaces do not lack associativity and inverses.

In topology, a CW complex is a type of topological space introduced by J. H. C. Whitehead to meet the needs of homotopy theory. This class of spaces is broader and has some better categorical properties than simplicial complexes, but still retains a combinatorial nature that allows for computation.

Roughly speaking, a CW-complex is made of basic building blocks called cells. The precise definition prescribes how the cells may be topologically glued together. The C stands for "closure-finite", and the W for "weak topology".

An n-dimensional closed cell is a topological space that is homeomorphic to an n-dimensional closed ball. For example, a simplex is a closed cell, and more generally, a convex polytope is a closed cell. An n-dimensional open cell is a topological space that is homeomorphic to the open ball. A 0-dimensional open (and closed) cell is a singleton space.
A CW complex is a Hausdorff space (H-space) X together with a partition of X into open cells (of perhaps varying dimension) that satisfies two additional properties:

• For each n-dimensional open cell C in the partition of X, there exists a continuous map f from the n-dimensional closed ball to X such that
• the restriction of f to the interior of the closed ball is a homeomorphism onto the cell C, and
• the image of the boundary of the closed ball is contained in the union of a finite number of elements of the partition, each having cell dimension less than n.
• A subset of X is closed if and only if it meets the closure of each cell in a closed set.
If the largest dimension of any of the cells is n, then the CW complex is said to have dimension n. If there is no bound to the cell dimensions then it is said to be infinite-dimensional. The n-skeleton of a CW complex is the union of the cells whose dimension is at most n. If the union of a set of cells is closed, then this union is itself a CW complex, called a sub-complex. Thus the n-skeleton is the largest sub-complex of dimension n or less.

A CW complex is often constructed by defining its skeleton inductively. Begin by taking the 0-skeleton to be a discrete space. Next, attach 1-cells to the 0-skeleton. Here, the 1-cells are attached to points of the 0-skeleton via some continuous map from unit 0-sphere, that is, S0. Define the 1-skeleton to be the identification space obtained from the union of the 0-skeleton, 1-cells, and the identification of points of boundary of 1-cells by assigning an identification mapping from the boundary of the 1-cells into the 1-cells. In general, given the n-1-skeleton and a collection of (abstract) closed n-cells, as above, the n-cells are attached to the n-1-skeleton by some continuous mapping from Sn − 1, and making an identification (equivalence relation) by specifying maps from the boundary of each n-cell into the n-1-skeleton. The n-skeleton is the identification space obtained from the union of the n-1-skeleton and the closed n-cells by identifying each point in the boundary of an n-cell with its image.
Up to isomorphism every n-dimensional complex can be obtained from its n-1 skeleton in this sense, and thus every finite-dimensional CW complex can be built up by the process above. This is true even for infinite-dimensional complexes, with the understanding that the result of the infinite process is the direct limit of the skeleta: a set is closed in X if and only if it meets each skeleton in a closed set.

Singular homology and cohomology of CW-complexes is readily computable via cellular homology. Moreover, in the category of CW-complexes and cellular maps, cellular homology can be interpreted as a homology theory. To compute an extraordinary (co)homology theory for a CW-complex, the Atiyah-Hirzebruch spectral sequence is the analogue of cellular homology.  There is a technique, developed by Whitehead, for replacing a CW-complex with a homotopy-equivalent CW-complex which has a simpler CW-decomposition.  Consider, for example, an arbitrary CW-complex. Its 1-skeleton can be fairly complicated, being an arbitrary graph. Now consider a maximal forest F in this graph. Since it is a collection of trees, and trees are contractible, consider the space X / ∼ where the equivalence relation is generated by xy if they are contained in a common tree in the maximal forest F.
The quotient map $X \to X/\sim$ is a homotopy equivalence. Moreover, X / ∼ naturally inherits a CW-structure, with cells corresponding to the cells of X which are not contained in F. In particular, the 1-skeleton of X / ∼ is a disjoint union of wedges of circles.  Another way of stating the above is that a connected CW-complex can be replaced by a homotopy-equivalent CW-complex whose 0-skeleton consists of a single Point.

In algebraic topology, a branch of mathematics, singular homology refers to the study of a certain set of algebraic invariants of a topological space X, the so-called homology groups Hn(X). Intuitively spoken, singular homology counts, for each dimension n, the n-dimensional holes of a space. Singular homology is a particular example of a homology theory, which has now grown to be a rather broad collection of theories. Of the various theories, it is perhaps one of the simpler ones to understand, being built on fairly concrete constructions.

In brief, singular homology is constructed by taking maps of the standard n-simplex to a topological space, and composing them into formal sums, called singular chains. The boundary operation on a simplex induces a singular chain complex. The singular homology is then the homology of the chain complex. The resulting homology groups are the same for all homo-topically equivalent spaces, which is the reason for their study. These constructions can be applied to all topological spaces, and so singular homology can be expressed in terms of category theory, where the homology group becomes a functor from the category of topological spaces to the category of graded abelian groups.  By dualizing the homology chain complex (i.e. applying the functor Hom(-, R), R being any ring) we obtain a cochain complex with co-boundary map δ. The cohomology groups of X are defined as the cohomology groups of this complex; in a quip, "cohomology is the homology of the co- (dual complex)".  The cohomology groups have a richer, or at least more familiar, algebraic structure than the homology groups. Firstly, they form a differential graded algebra as follows:
There are additional cohomology operations, and the cohomology algebra has addition structure mod p (as before, the mod p cohomology is the cohomology of the mod p cochain complex, not the mod p reduction of the cohomology), notably the Steenrod algebra structure.

The multiplicative structure of an H-space adds structure to its homology and cohomology groups. For example, the cohomology ring of a path-connected H-space with finitely generated and free cohomology groups is a Hopf algebra. Also, one can define the Pontryagin product on the homology groups of an H-space.

The fundamental group of an H-space is abelian. To see this, let X be an H-space with identity e and let f and g be loops at e. Define a map F: [0,1]×[0,1] → X by F(a,b) = f(a)g(b). Then F(a,0) = F(a,1) = f(a)e is homotopic to f, and F(0,b) = F(1,b) = eg(b) is homotopic to g. It is clear how to define a homotopy from [f][g] to [g][f].

Adams theorem: S0, S1, S3, S7 are the only spheres that are H-spaces (e.g., using multiplication restricted from the reals, complexes, quaternions, and octonions, respectively). In fact, S0, S1, and S3 are groups (Lie groups) with these multiplications. But S7 is not a group in this way because octonion multiplication is not associative, nor can it be given any other continuous multiplication for which it is a group. However S7 is associative and has other continuous multiplication for which it is a group, when an actual spherical coordinate system, in conjunction with my One equation, is used in place of the octonions with their linearity of cubism restrictions.

Then with the continuous group of local transformations in mind, I came up with: xxx http://youtu.be/59157zE-u6s  xxx

So, I would say that: 'The Higgs mechanism is the natural outcrop of the Gauge Field Boson Theory, so why is the Higgs Boson so cagey?' if I thought the Higgs field Boson was even real or remotely possible, let alone findable, because it is like trying to create a particle of space. Space is an empty concept, while particle is a fully solid concept. In my theory, and the One Equation, they are related, but different structures. The primary vectors form dynamic relationship hypervectors that spin on 4 axis', which form into a surface element triangle that rotates with 4 degrees of freedom, that depending on the scalar values of the dynamic relationship hypervectors, with each hypervector perpendicular to the others, and the ratios of any 2 hypervectors that form dynamically a third hypervector, form S7, as there are 7 geometric functions in the group, a group that exhibits continuous local transformations between the 7 geometries within the group.

Please note that in the above concepts: gauge theory, Lagrangian, Lie group and Lie algebra, H-space, CW complexes, were composited from information found on wikipedia, (don't forget to donate, no ads) to which I added what I know, and hopefully made a coherent contribution.  And on YouTube : http://youtu.be/1_HrQVhgbeo Higgs


text:
yi-core-0.19.2: Yi editor core library

Yi.Misc

Description

Various high-level functions to further classify.

Synopsis

# Documentation

Given a possible starting path (which if not given defaults to the current directory) and a fragment of a path we find all files within the given (or current) directory which can complete the given path fragment. We return a pair of both directory plus the filenames on their own that is without their directories. The reason for this is that if we return all of the filenames then we get a hint which is way too long to be particularly useful.

Given a path, trim the file name bit if it exists. If no path given, return current directory.

cd :: YiM () Source #

Like M-x cd, it changes the current working directory. Mighty useful when we don't start Yi from the project directory or want to switch projects, as many tools only use the current working directory.

pwd :: YiM () Source #

Shows current working directory. Also see cd.

Given a possible path and a prefix, return matching file names.

Place mark at current point. If there's an existing mark at point already, deactivate mark.

Select the contents of the whole buffer

A simple wrapper to adjust the current indentation using the mode specific indentation function but according to the given indent behaviour.

promptFile :: Text -> (Text -> YiM ()) -> YiM () Source #

Generic emacs style prompt file action. Takes a prompt and a continuation act and prompts the user with file hints.

Arguments

 :: Text Prompt -> (Text -> [Text] -> YiM [Text]) Hint transformer: current path, generated hints -> (Text -> YiM ()) Action over choice -> YiM ()

As promptFile but additionally allows the caller to transform the list of hints arbitrarily, such as only showing directories.

Prints out the rope of the current buffer as-is to stdout.

The only way to stop it is to close the buffer in question which should free up the BufferRef.


text:
# Evaluate the series 1+1/9+1/(25)+1/(49)+cdots

Evaluate the series
$1+\frac{1}{9}+\frac{1}{25}+\frac{1}{49}+\cdots$
You can still ask an expert for help

## Want to know more about Laplace transform?

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beshrewd6g
One approach is as follows:
Let $f\left(t\right)$ denote a triangle wave with the precise form described here (we'll take L=1). As you can see in the link, f can be expanded into its Fourier series as
$f\left(t\right)=\frac{8}{{\pi }^{2}}\sum _{n=1}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n-1}}{\left(2n-1{\right)}^{2}}\mathrm{sin}\left(\left(2n-1\right)\pi t\right)$
Now, just plug $t=\frac{1}{2}$ into both sides of the equation above.
###### Did you like this example?
Denisse Fitzpatrick
Hint: On way to evaluate your series is using Parseval theorem. At first you need to find the fourier series of a function like $f\left(x\right)=x$ in $\left[-\pi ,\pi \right]$ and then applying Parseval theorem.


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# American Institute of Mathematical Sciences

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## Immediate schedule adjustment and semidefinite relaxation

 1 School of Mathematics and Physics, University of Science and Technology Beijing, 30 Xueyuan Road, Haidian District, Beijing 100083, China 2 Business School, Nankai University, 94 Weijin Road, Nankai District, Tianjin 300071, China

* Corresponding author: Su Zhang

Received  November 2017 Revised  January 2018 Published  April 2018

Fund Project: The first author is supported by National Natural Science Foundation of China No. 11101028,11271206, and the Fundamental Research Funds for the Central Universities. The third author is supported by National Natural Science Foundation of China No. 11401322

This paper considers the problem of temporary shortage of some resources within a project execution period. Mathematical models for two different cases of this problem are established. Semidefinite relaxation technique is applied to get immediate solvent of these models. Relationship between the models and their semidefinite relaxations is studied, and some numerical experiments are implemented, which show that these mathematical models are reasonable and feasible for practice, and semidefinite relaxation can efficiently solve the problem.

Citation: Jinling Zhao, Wei Chen, Su Zhang. Immediate schedule adjustment and semidefinite relaxation. Journal of Industrial & Management Optimization, doi: 10.3934/jimo.2018062
##### References:
 [1] M. Bartusch, R. H. Mohring and F. J. Randermacher, Scheduling project networks with resource constraints and time windows, Annals of Operations Research, 16 (1988), 201-240. [2] A. Ben-Tal and A. Nemirovski, Lectures on Modern Convex Optimization, MOS-SIAM Series on Optimization, 2001. [3] J. Blazewicz, J. K. Lenstra and A. H. G. Kan Rinnooy, Scheduling subject to resource constraints: Classification and complexity, Discrete Applied Mathematics, 5 (1983), 11-24. doi: 10.1016/0166-218X(83)90012-4. [4] S. Boyd and L. Vandenberghe, Semidefinite programming relaxations of non-convex problems in control and combinatorial optimization, Communications, Computation, Control, and Signal Processing, Springer, 1997,279-287 doi: 10.1007/978-1-4615-6281-8_15. [5] S. Boyd and L. Vandenberghe, Convex Optimization, Cambridge University Press, first edition, 2004. [6] P. Brucker, Scheduling and constraint propagation, Discrete Applied Mathematics, 123 (2002), 227-256. doi: 10.1016/S0166-218X(01)00342-0. [7] M. Goemans and D. Williamson, Imporved approximation algorihtms for maximum cut and satisfiablity problems using semidefinite programming, J. Assoc. Comput. Mach., 42 (1995), 1115-1145. doi: 10.1145/227683.227684. [8] S. Hartmann and D. Briskorn, A survey of variants and extensions of the resource-constrained project scheduling problem, European Journal of Operational Research, 207 (2010), 1-14. doi: 10.1016/j.ejor.2009.11.005. [9] D. Henrion, J. Lasserre and J. Loefberg, GloptiPoly 3: moments, optimization and semidefinite programming, Optim. Methods Softw., 24 (2009), 761-779, http://homepages.laas.fr/henrion/software/gloptipoly3. doi: 10.1080/10556780802699201. [10] H. Li and N. K. Womer, Solving stochastic resource-constrained project scheduling problems by closed-loop approximate dynamic programming, European Journal of Operational Research, 246 (2015), 20-33. doi: 10.1016/j.ejor.2015.04.015. [11] U. Malik, I. M. Jaimoukha, G. D. Halikias and S. K. Gungah, On the gap between the quadratic integer programming problem and its semidefinite relaxation, Math. Program., 107 (2006), 505-515. doi: 10.1007/s10107-005-0692-2. [12] I. Pólik, Addendum to the SeDuMi user guide version 1.1, http://sedumi.ie.lehigh.edu/?page_id=58, 2005. [13] A. A. B. Pritsker, L. J. Watters and P. M. Wolfe, Multiproject scheduling with limited resources: A zero-one programming approach, Management Science, 16 (1969), 93-108. doi: 10.1287/mnsc.16.1.93. [14] F. Rendl, Semidefinite relaxations for integer programming, 50 Years of Integer Programming 1958-2008, (2009), 687-726. doi: 10.1007/978-3-540-68279-0_18. [15] N. Z. Shor, Quadratic optimization problems, Soviet. J. Comput. Systems Sci., 25 (1987), 1-11. [16] J. F. Sturm, Using SeDuMi 1.02, a matlab toolbox for optimizativer over smmetric cones, Optim. Methods Softw., 11/12 (1999), 625-653. [17] X. Sun and R. Li, New progress in integer programming, Operations Research Transactions, 18 (2014), 39-67. [18] L. Vandenerghe and S. Boyd, Semidefinite programming, SIAM Review, 38 (1996), 49-95. doi: 10.1137/1038003. [19] H. Waki, S. Kim, M. Kojima and M. Muramatsu, Sums of squares and semidefinite program relaxations for polynomial optimization problems with structured sparsity, SIAM Journal on Optimization, 17 (2006), 218-242. doi: 10.1137/050623802. [20] H. Wolkowicz, R. Saigak and L. Vandenerghe, Handbook of Semidefinite Programming, Kluwer's Publisher, 2000. [21] F. Zhang, The Schur Complement and Its Applications: Numerical Methods and Algorithms, Springer Science Business Media, 2005. [22] X. Zheng, X. Sun, D. Li and Y. Xia, Duality gap estimation of linear equality constraintd binary quadratic programming, Mathematics of Operations Research, 35 (2010), 864-880. doi: 10.1287/moor.1100.0472.

show all references

##### References:
 [1] M. Bartusch, R. H. Mohring and F. J. Randermacher, Scheduling project networks with resource constraints and time windows, Annals of Operations Research, 16 (1988), 201-240. [2] A. Ben-Tal and A. Nemirovski, Lectures on Modern Convex Optimization, MOS-SIAM Series on Optimization, 2001. [3] J. Blazewicz, J. K. Lenstra and A. H. G. Kan Rinnooy, Scheduling subject to resource constraints: Classification and complexity, Discrete Applied Mathematics, 5 (1983), 11-24. doi: 10.1016/0166-218X(83)90012-4. [4] S. Boyd and L. Vandenberghe, Semidefinite programming relaxations of non-convex problems in control and combinatorial optimization, Communications, Computation, Control, and Signal Processing, Springer, 1997,279-287 doi: 10.1007/978-1-4615-6281-8_15. [5] S. Boyd and L. Vandenberghe, Convex Optimization, Cambridge University Press, first edition, 2004. [6] P. Brucker, Scheduling and constraint propagation, Discrete Applied Mathematics, 123 (2002), 227-256. doi: 10.1016/S0166-218X(01)00342-0. [7] M. Goemans and D. Williamson, Imporved approximation algorihtms for maximum cut and satisfiablity problems using semidefinite programming, J. Assoc. Comput. Mach., 42 (1995), 1115-1145. doi: 10.1145/227683.227684. [8] S. Hartmann and D. Briskorn, A survey of variants and extensions of the resource-constrained project scheduling problem, European Journal of Operational Research, 207 (2010), 1-14. doi: 10.1016/j.ejor.2009.11.005. [9] D. Henrion, J. Lasserre and J. Loefberg, GloptiPoly 3: moments, optimization and semidefinite programming, Optim. Methods Softw., 24 (2009), 761-779, http://homepages.laas.fr/henrion/software/gloptipoly3. doi: 10.1080/10556780802699201. [10] H. Li and N. K. Womer, Solving stochastic resource-constrained project scheduling problems by closed-loop approximate dynamic programming, European Journal of Operational Research, 246 (2015), 20-33. doi: 10.1016/j.ejor.2015.04.015. [11] U. Malik, I. M. Jaimoukha, G. D. Halikias and S. K. Gungah, On the gap between the quadratic integer programming problem and its semidefinite relaxation, Math. Program., 107 (2006), 505-515. doi: 10.1007/s10107-005-0692-2. [12] I. Pólik, Addendum to the SeDuMi user guide version 1.1, http://sedumi.ie.lehigh.edu/?page_id=58, 2005. [13] A. A. B. Pritsker, L. J. Watters and P. M. Wolfe, Multiproject scheduling with limited resources: A zero-one programming approach, Management Science, 16 (1969), 93-108. doi: 10.1287/mnsc.16.1.93. [14] F. Rendl, Semidefinite relaxations for integer programming, 50 Years of Integer Programming 1958-2008, (2009), 687-726. doi: 10.1007/978-3-540-68279-0_18. [15] N. Z. Shor, Quadratic optimization problems, Soviet. J. Comput. Systems Sci., 25 (1987), 1-11. [16] J. F. Sturm, Using SeDuMi 1.02, a matlab toolbox for optimizativer over smmetric cones, Optim. Methods Softw., 11/12 (1999), 625-653. [17] X. Sun and R. Li, New progress in integer programming, Operations Research Transactions, 18 (2014), 39-67. [18] L. Vandenerghe and S. Boyd, Semidefinite programming, SIAM Review, 38 (1996), 49-95. doi: 10.1137/1038003. [19] H. Waki, S. Kim, M. Kojima and M. Muramatsu, Sums of squares and semidefinite program relaxations for polynomial optimization problems with structured sparsity, SIAM Journal on Optimization, 17 (2006), 218-242. doi: 10.1137/050623802. [20] H. Wolkowicz, R. Saigak and L. Vandenerghe, Handbook of Semidefinite Programming, Kluwer's Publisher, 2000. [21] F. Zhang, The Schur Complement and Its Applications: Numerical Methods and Algorithms, Springer Science Business Media, 2005. [22] X. Zheng, X. Sun, D. Li and Y. Xia, Duality gap estimation of linear equality constraintd binary quadratic programming, Mathematics of Operations Research, 35 (2010), 864-880. doi: 10.1287/moor.1100.0472.
Project Network Diagram in Example 1
Project Network Diagram in Example 2
Basic Notations
 Symbol Definition $Pred(i)$ set of direct predecessors of Activity $i$ $Succ(i)$ set of direct successors of Activity $i$ $d_i$ processing time (or duration) of Activity $i$ $s_i$ start time of Activity $i$ according to the existing schedule $f_i$ completion time of Activity $i$ according to the existing schedule $R_k$ amount of originally available units of renewable resource $k$ in unit time $r_{ik}$ usage of Activity $i$ of renewable resource $k$ in unit time $t_0$ start time of the temporary shortage of resources $T$ lasting time of the resources shortage $\Delta R_k$ amount of decrement of resource $k$ in unit time
 Symbol Definition $Pred(i)$ set of direct predecessors of Activity $i$ $Succ(i)$ set of direct successors of Activity $i$ $d_i$ processing time (or duration) of Activity $i$ $s_i$ start time of Activity $i$ according to the existing schedule $f_i$ completion time of Activity $i$ according to the existing schedule $R_k$ amount of originally available units of renewable resource $k$ in unit time $r_{ik}$ usage of Activity $i$ of renewable resource $k$ in unit time $t_0$ start time of the temporary shortage of resources $T$ lasting time of the resources shortage $\Delta R_k$ amount of decrement of resource $k$ in unit time
Comparing the 1/3-Method with the Rounding Method
 $\Delta R_k$ 1/3-Method Rounding Method SDR Opt.Val $t^*$ Opt.Val $t^*$ Delay Time $2$ 15 infeasible 15.0000 15 1 $5$ 15 15 15.0000 15 1 $8$ 18 infeasible 15.4583 18 4 $11$ 19 19 16.7083 19 5
 $\Delta R_k$ 1/3-Method Rounding Method SDR Opt.Val $t^*$ Opt.Val $t^*$ Delay Time $2$ 15 infeasible 15.0000 15 1 $5$ 15 15 15.0000 15 1 $8$ 18 infeasible 15.4583 18 4 $11$ 19 19 16.7083 19 5
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2017 Impact Factor: 0.994


text:
# Open PDE & Analysis Seminar

Cyril Letrouit (Ecole Normale Supérieure)

### Subelliptic wave equations are never observable

Friday 24th April 2020, 3pm (Paris time)

Abstract : In this talk, we explain a result we obtained recently, concerning the wave equation with a sub-Riemannian (i.e. subelliptic) Laplacian. Given a manifold $M$, a measurable subset $\omega\subset M$, a time $T_0$ and a subelliptic Laplacian $\Delta$ on $M$, we say that the wave equation with Laplacian $\Delta$ is observable on $\omega$ in time $T_0$ if any solution $u$ of $\partial_{tt}^2u-\Delta u=0$ with fixed initial energy satisfies $\int_0^{T_0}\int_\omega |u|^2dxdt\geq C$ for some constant $C>0$ independent on $u$. It is known since the work of Bardos-Lebeau-Rauch that the observability of the elliptic wave equation, i.e. with a Riemannian Laplacian, in time $T_0$ is almost equivalent to the geometric control condition (GCC), which stipulates that any geodesic ray meets $\omega$ within time $T_0$. We show that in the subelliptic case, as soon as $M\backslash \omega$ has non-empty interior and $\Delta$ is subelliptic but not elliptic", GCC is never verified, which implies that subelliptic wave equations are never observable. The proof is based on the construction of sequences of solutions of the wave equation concentrating on geodesics (for the distance on $M$ associated to $\Delta$) spending a long time in $M\setminus \omega$.


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Article

# Detection of a weak surface magnetic field on Sirius A: are all tepid stars magnetic ?

[more]
(Impact Factor: 4.48). 06/2011; 532. DOI: 10.1051/0004-6361/201117573
Source: arXiv

ABSTRACT We aim at a highly sensitive search for weak magnetic fields in main sequence
stars of intermediate mass, by scanning classes of stars with no previously
reported magnetic members. After detecting a weak magnetic field on the normal,
rapidly rotating A-type star Vega, we concentrate here on the bright star
Sirius A, taken as a prototypical, chemically peculiar, moderately rotating Am
star. We employed the NARVAL and ESPaDOnS high-resolution spectropolarimeters
to collect 442 circularly polarized spectra, complemented by 60 linearly
polarized spectra. Using a list of about 1,100 photospheric spectral lines, we
computed a cross correlation line profile from every spectrum, leading to a
signal-to-noise ratio of up to 30,000 in the polarized profile. We report the
repeated detection of circularly polarized, highly asymmetric signatures in the
line profiles, interpreted as Zeeman signatures of a large-scale photospheric
magnetic field, with a line-of-sight component equal to $0.2 \pm 0.1$ G. This
is the first polarimetric detection of a surface magnetic field on an Am star.
Using rough estimates of the physical properties of the upper layers of Sirius
A, we suggest that a dynamo operating in the shallow convective envelope cannot
account for the field strength reported here. Together with the magnetic field
of Vega, this result confirms that a new class of magnetic objects exists among
non Ap/Bp intermediate-mass stars, and it may indicate that a significant
fraction of tepid stars are magnetic.

### Full-text

Available from: Pascal Petit, May 21, 2015
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• ##### Article: Do the close binaries HD 22128 and HD 56495 contain Ap or Am stars?
[Hide abstract]
ABSTRACT: HD 22128 and HD 56495 are both double-lined spectroscopic binary systems with short orbital periods, which have been proposed to host magnetic Ap stars. Ap stars in short period binary systems are very rare, and may provide insight into the origin of magnetism in A-type stars. We study these two systems using high-resolution MuSiCoS spectropolarimetric data, in order to asses the presence of magnetic fields and study the atmospheric chemistry of the components. This represents the first modern magnetic measurements and careful spectroscopic analyses of these stars. We find no evidence of a magnetic field in any of the stars, with precise uncertainties on the longitudinal magnetic field of 50 and 80 G in the components of HD 22128, and 80 and 100 G in the components of HD 56495. We performed detailed abundance analyses of both stars in both systems, finding clear evidence of Am chemical peculiarities in both components of HD 22128, and in the brighter component of HD 56495, with overabundant iron peak elements and underabundant Sc and Ca. The less luminous component of HD 56495 is chemically normal. The atmospheric chemistry is consistent with the absence of magnetic fields, and consistent with the theory of Am star formation proposing that tidal interactions slow the rotation rate of the star, allowing atomic diffusion to proceed efficiently.
Monthly Notices of the Royal Astronomical Society 06/2013; 433(4). DOI:10.1093/mnras/stt1003 · 5.23 Impact Factor
• Source
##### Article: gamma Peg: testing Vega-like magnetic fields in B stars
[Hide abstract]
ABSTRACT: gam Peg is a bright B pulsator showing both p and g modes of beta Cep and SPB types. It has also been claimed to be a magnetic star by some authors while others do not detect a magnetic field. We aimed at checking for the presence of a field, characterise it if it exists or provide a firm upper limit of its strength if it is not detected. If gam Peg is magnetic, it would make an ideal asteroseismic target to test various theoretical scenarios. If it is very weakly magnetic, it would be the first observation of an extension of Vega-like fields to early B stars. Finally, if it is not magnetic and we can provide a very low upper limit on its non-detected field, it would make an important result for stellar evolution models. We acquired high resolution, high signal-to-noise spectropolarimetric Narval data at TBL. We also gathered existing dimaPol@DAO and Musicos@TBL spectropolarimetric data. We analysed the Narval and Musicos observations using the LSD technique to derive the longitudinal magnetic field and Zeeman signatures in lines. The longitudinal field strength was also extracted from the Hbeta line observed with the DAO. With a Monte Carlo simulation we derived the maximum strength of the field possibly hosted by gam Peg. We find that no magnetic signatures are visible in the very high quality spectropolarimetric data. The average longitudinal field measured in the Narval data is Bl=-0.1+/-0.4 G. We derive a very strict upper limit of the dipolar field strength of Bpol~40 G. We conclude that gamma Peg is not magnetic: it does not host a strong stable fossil field as observed in a fraction of massive stars, nor a very weak Vega-like field. There is therefore no evidence that Vega-like fields exist in B stars contrary to the predictions by fossil field dichotomy scenarios. These scenarios should thus be revised. Our results also provide strong constraints for stellar evolution models.
Astronomy and Astrophysics 12/2013; 562. DOI:10.1051/0004-6361/201323093 · 4.48 Impact Factor
• Source
##### Article: B fields in OB stars (BOB): on the detection of weak magnetic fields in the two early B-type stars β CMa and ϵ CMa
[Hide abstract]
ABSTRACT: Within the context of the "B fields in OB stars (BOB)" collaboration, we used the HARPSpol spectropolarimeter to observe the early B-type stars beta CMa (HD44743; B1 II/III) and epsilon CMa (HD52089; B1.5 II). For both stars, we consistently detected the signature of a weak (<30 G in absolute value) longitudinal magnetic field. We determined the physical parameters of both stars and characterise their X-ray spectrum. For beta CMa, our mode identification analysis led to determining a rotation period of 13.6+/-1.2 days and of an inclination angle of the rotation axis of 57.6+/-1.7 degrees, with respect to the line of sight. On the basis of these measurements and assuming a dipolar field geometry, we derived a best fitting obliquity of ~22 degrees and a dipolar magnetic field strength (Bd) of ~100 G (60<Bd<230 G within 1 sigma), below what is typically found for other magnetic massive stars. For epsilon CMa we could only determine a lower limit on the dipolar magnetic field strength of 13 G. For this star, we determine that the rotation period ranges between 1.3 and 24 days. Both stars are expected to have a dynamical magnetosphere. We also conclude that both stars are most likely core hydrogen burning and that they have spent more than 2/3 of their main sequence lifetime. A histogram of the distribution of the dipolar magnetic field strength for the magnetic massive stars known to date does not show the magnetic field "desert" observed instead for intermediate-mass stars. The biases involved in the detection of (weak) magnetic fields in massive stars with the currently available instrumentation and techniques imply that weak fields might be more common than currently observed. Our results show that, if present, even relatively weak magnetic fields are detectable in massive stars and that more observational effort is probably still needed to properly access the magnetic field incidence.
Astronomy and Astrophysics 11/2014; 574. DOI:10.1051/0004-6361/201424986 · 4.48 Impact Factor


text:
## Seminars and Colloquia by Series

### Topology, algebra, and combinatorics walk into a bar

Series
Time
Friday, November 4, 2022 - 15:30 for 1 hour (actually 50 minutes)
Location
Skiles 005
Speaker
Roberta ShapiroGeorgia Tech

One of the most beautiful aspects of math is the interplay between its different fields. We will discuss one such interaction by studying topology using tools from combinatorics and group theory. In particular, given a surface (two-dimensional manifold) S, we construct the curve complex of S, which is a graph that encodes topological data about the surface. We will then state a seminal result of Ivanov: the symmetries of a surface S are in a natural bijection with the symmetries of its curve complex. In the direction of the proof of Ivanov's result, we will touch on some tools we have when working with infinite graphs.

### A brief introduction to the circle method and sparse domination

Series
Time
Friday, October 28, 2022 - 12:00 for 1 hour (actually 50 minutes)
Location
Skiles 005
Speaker
Christina GiannitsiGeorgia Tech

In this talk we will go over the Hardly-Littlewood circle method, and the major and minor arc decomposition. We shall then see a toy-example of the High-Low decomposition, and proceed with defining sparse families and sparse domination. We will conclude by explaining why sparse domination is of interest to us  when studying $L^p$ bounds. This talk aims to be accessible to people without a strong background in the area. Some basic concepts of real and harmonic analysis will be useful (e.g. $L^p$ spaces, Fourier transform,  Holder inequality, the Hardy-Littlewood Maximal function, etc)

### The Entropy Compression Method

Series
Time
Friday, October 7, 2022 - 15:30 for 1 hour (actually 50 minutes)
Location
Skiles 005
Speaker
Abhishek DhawanGeorgia Tech Math

The Lovasz Local Lemma is a powerful tool to prove existence of combinatorial structures satisfying certain properties. In a constructive proof of the LLL, Moser and Tardos introduced a proof technique that is now referred to as the entropy compression method. In this talk I will describe the main idea of the method and apply it to a problem easily solved using the LLL. I will also describe recent applications of the idea to various graph coloring problems.

Series


text:
Map
WordNet

### noun

(1)   A diagrammatic representation of the earth's surface (or part of it)
(2)   A function such that for every element of one set there is a unique element of another set

### verb

(3)   To establish a mapping (of mathematical elements or sets)
(4)   Explore or survey for the purpose of making a map
"We haven't even begun to map the many galaxies that we know exist"
(5)   Make a map of; show or establish the features of details of
"Map the surface of Venus"
(6)   Depict as if on a map
"Sorrow was mapped on the mother's face"
(7)   Plan, delineate, or arrange in detail
"Map one's future"
(8)   Locate within a specific region of a chromosome in relation to known DNA or gene sequences
"Map the genes"
WiktionaryText

## English

### Etymology

Latin mappa “napkin, signal cloth”, via Old French mapemonde “map of the world”.

### Noun

1. A visual representation of an area, whether real or imaginary.
2. A function.
The discrete topology is always continuous, therefore functions with discrete domains are always maps.
3. A continuous function.
4. A diagram of components of an item.
5. The butterfly Araschnia levana.

### Verb

1. To create a visual representation of a territory, etc. via cartography.
2. To inform someone of a particular idea.
3. To act as a function on.
The discrete topology is always continuous, therefore functions with discrete domains are always mappings.
4. To act as a continuous function on.
f maps A to B, mapping $a\in A$ to $b\in B$.


text:
In honour to 2004

222444 is the smallest number that is divisible by 2004 and only contains the digits 2 and 4. What’s the next number that has the same properties?

Hint:The number is 11 digits long.

• Besides, is this possible to solve without computers in a reasonable amount of time? I did so with a computer, and the answer doesn't seem to have a lot of patterns in it (but I may be wrong). – my pronoun is monicareinstate Jun 6 at 4:07
• Its solvable by hands. – Display maths Jun 6 at 13:21
• Could be 44444222244444, if we require a palindrome. – Daniel Mathias Jun 6 at 15:10
• Not exactly.... – Display maths Jun 6 at 15:19
• 222444 = 2004 * 111, where 111 is a repunit. But the 11-digit one won't be so simple... – smci Jun 6 at 20:18

3 Answers

42244442244 = 21080061 times 2004

To answer, consider the number

222...222 modulo 2004.
As the length increases, the remainder follows the pattern 2, 22, 222, 218, 178, -222, -214, -134, 666, 650, 490...
The change to the remainder that we get from changing each of those 2's into a 4 likewise follows 2, 20, 200, -4, -40, -400, 8, 80, 800, -16, -160...
The problem of finding an 11-digit multiple of 2004 therefore reduces to finding a subset of the above remainders that sums to -490 (modulo 2004).
From the equation 2+20-40-400+8+80-160 = -490, we take the 2's with the remainders corresponding to the terms of the sum and turn them into 4's, yielding the answer.

Probable answer:

$$42244442244=21080061\times2004$$

Firstly

There are no such numbers below $$10^9$$. On the other hand, $$222444222444=222444\times1000001$$ works.

Explanations:

Let us consider a "divisibility by 1002" rule: a number $$N=1000000x+1000y+z$$ (so, $$x$$, $$y$$ and $$z$$ are 3-digit groups of $$N$$) has the same remainder modulo 1002 as $$4x-2y+z$$ (Proof: $$(1000000x+1000y+z)-(4x-2y+z)=999996x+1002y$$, and $$999996=1000000-4=(1000+2)(1000-2)=1002\times998$$. So, $$999996x+1002y$$ is divisible by 1002.) Since $$0\leqslant x\leqslant444$$, $$222\leqslant y\leqslant444$$ and $$222\leqslant z\leqslant444$$, we have that $$-222\leqslant4x-2y+z\leqslant1776$$. We know that $$z$$ is divisible by 4 (because we actually want our number to be divisible by 2004, rather that 1002), so the whole thing $$4x-2y+z$$ must be divisible by 4 (because $$y$$ is even, containing only 2s and 4s).
So, $$4x-2y+z=0$$ ($$0$$ is the only number between $$-222$$ and $$1776$$ which is divisible by both $$4$$ and $$1002$$). We know that $$z$$ ends with $$4$$ (to be composed only of 2s and 4s and be divisible by 4).
- If $$x$$ ends with 4, then $$4x$$ ends with 6, and $$4x+z$$ ends with 0, so $$y$$ must end with either 0 or 5, which is impossible.
- If $$x$$ ends with 2, then $$4x$$ ends with 8, and $$4x+z$$ ends with 2, so $$y$$ must end with either 1 or 6, which is again impossible.
So, we came to a contradiction. So, no such number exists which is 7 to 9 digits long.

Note

For the numbers up to 12 digits, we should consider and expression of the form $$8x-4y+2z-t$$, but there is a lot more possibilities. Trial and error gives the answer written above.

Partial Answer on how to do it manually:

We know for a fact that the next highest number will be $$222444 \times \text{Natural Number}$$.
Let that natural number be $$...N_3N_2N_1$$ (a number of arbitrary length as we don't know how long it will be)

Let's do some elementary school multiplication!
We multiply $$222444$$ by $$...N_3N_2N_1$$

       2  2  2  4  4  4
X  ..N6 N5 N4 N3 N2 N1
----------------------

Multiplying $$N_1$$ with $$222444$$ gives us the first line of the solution. Notice here that the unit's place of the final answer will be the unit's place of $$N_1 \times 4$$. The rest of the multiplication won't affect this place.
Since we want every number to be a $$2$$ or $$4$$, the possible values of $$N_1$$ are {$$1,3,6,8$$}.

Let's continue the multiplication with the possible values.

       2  2  2  4  4  4        2  2  2  4  4  4       2  2  2  4  4  4       2  2  2  4  4  4
X  ..N6 N5 N4 N3 N2 1    X ..N6 N5 N4 N3 N2 3   X ..N6 N5 N4 N3 N2 6   X ..N6 N5 N4 N3 N2 8
----------------------  ---------------------  ---------------------   --------------------
2  2  2  4  4  4        6  6  7  3  3  2    1  3  3  4  6  6  4    1  7  7  9  5  5  2
0                       0                      0                      0

Notice that the second place digit in $$N_1 = 3$$ and $$N_1 = 8$$ is odd. This means that in the next step. (when we multiply with $$N_2$$) the overall ten's digit will be $$\text{Odd} + N_2 \times 4$$.
$$N_2 \times 4$$ is always even hence, $$\text{Odd+Even = Odd}$$. In this case, the ten's digit cannot be either $$4$$ or $$2$$. Hence we eliminate the cases with $$N_1=3$$ and $$N_1=8$$
Matter of fact with every progression, (as we go up to higher $$N$$'s) every number with an odd digit at the $$L_{th}$$ place for the $$N_L$$ step can be discarded.
The next step then has to be done with both the remaining cases. We can say our question has branched. The same process is recursively used, making a number of branches, as we have to do it for every subbranch that is a possible solution.

Note: I admit this method is hard work. Even for the third step, it becomes very sophisticated to progress. And though we can develop some minor techniques to refine the process, I myself didn't have the courage to find further than $$N_4$$. The overall procedure remains what I have described above.

• It’s a very good explanation, however it’s wrong to say that 222444 divides the next number. – Display maths Jun 6 at 13:25
• This is going to find the next lowest number that is divisible by 222444, not just plain old 2004 ! Can you please correct it? The approach is great, though. – smci Jun 6 at 20:52


text:
2020 NA Regionals Practice Contest 13

#### Start

2021-02-13 09:00 AKST

## 2020 NA Regionals Practice Contest 13

#### End

2021-02-13 14:00 AKST
The end is near!
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Contest is starting in -347 days 2:05:20

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0:00:00

# Problem DDunglish

A group of confused Dutchmen. Public domain.
A confused Dutchman trying to speak English could say “I am in the war”, even though there is no hostile activity going on. The confusion1 here is that the English sentence “I am confused” is translated in Dutch as “Ik ben in de war”, which is phonetically (“sounding”) quite close to the first sentence. Such confusion leads to much enjoyment, but can complicate matters a bit.

Given a sentence in Dutch and a dictionary containing both correct translations as well as phonetic (incorrect) translations of individual words, find the translation of the sentence and indicate whether it is correct, or in case there is more than one find the total number of correct and incorrect translations. A sentence is correctly translated when each word of the sentence is correctly translated.

## Input

The input consists of:

• One line with an integer $n$ ($1 \leq n \leq 20$), the number of words in the Dutch sentence.

• One line with $n$ words, the Dutch sentence $s$.

• One line with an integer $m$ ($1 \leq m \leq 10^5$), the number of words in the dictionary.

• $m$ lines, each with three strings $d$, $e$ and $c$, a Dutch word, the English translation, and “correct” if this is the correct translation or “incorrect” otherwise.

A word consists of between $1$ and $20$ lowercase letters. Each word in $s$ appears at least once as a Dutch word in the dictionary, no word appears more than $8$ times as a Dutch word in the dictionary, and each combination of a Dutch and English word appears at most once.

## Output

In case there is only a single translation of $s$, output one line with the translation followed by one line with “correct” or “incorrect”. In case there is more than one translation, output one line with the number of possible correct translations followed by “correct”, and one line with the number of possible incorrect translations followed by “incorrect”.

Sample Input 1 Sample Output 1
7
als mollen mollen mollen mollen mollen mollen
4
als when correct
mollen moles correct
mollen destroy correct
mollen mills incorrect

64 correct
665 incorrect

Sample Input 2 Sample Output 2
5
de zuigers zijn buiten werking
6
zijn are correct
banaan banana correct
de the correct
zuigers suckers incorrect
buiten out correct
werking working incorrect

the suckers are out working
incorrect

Footnotes

1. Pun intended.


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Madee is one of the most sensitive bands in spanish indie scene. They have…

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