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# Graphs with Small Bandwidth and Cutwidth
## Graphs with Small Bandwidth and Cutwidth
Discrete Mathematics 75 (1989) 113-119 North-Holland 113 GRAPHS WITH SMALL BANDWIDTH AND CUTWIDTH F.R.K. CHUNG, P.D. SEYMOUR Bell Communications Res...
Discrete Mathematics 75 (1989) 113-119 North-Holland
113
GRAPHS WITH SMALL BANDWIDTH AND CUTWIDTH F.R.K. CHUNG, P.D. SEYMOUR Bell Communications Research, Morristown, NJ 07960,U.S.A.
We give counter-examples to the following conjecture which arose in the study of small bandwidth graphs. “For a graph G, suppose that IV(G‘)l< 1+ c1 . diameter (G‘) for any connected subgraph G’ of G, and that G does not contain any refinement of the complete binary tree of c, levels. Is it true that the bandwidth of G can be bounded above by a constant c depending only on c , and c,?” On the other hand, we show that if the maximum degree of G is bounded and G does not contain any refinement of a complete binary tree of a specified sue, then the cutwidth and the topological bandwidth of G are also bounded.
1. Introduction For a graph G with vertex set V ( G ) and edge set E(G), a numbering of G is a one-to-one mapping JG from V ( G )to the integers. The bandwidth of a numbering n is max{ln(u) - n(v)l: {u,v} E E ( G ) } . The bandwidth b ( G ) of G is the minimum bandwidth of all numberings. The cutwidth of a numbering n is max I{ {u, v } E E ( G ):n ( u ) i < x(v)}I. i
The cutwidth c ( G ) of G is the minimum cutwidth of all numberings. The bandwidth problem and the cutwidth problem are associated with many optimization problems in circuit layout. In a circuit design or a network system, the maximum length of the wire is often proportional to the delay for transmitting messages, and so bandwidth is a graph-invariant of importance in circuit design. On the other hand, the cutwidth problem is of particular interest in designing microchip circuits and is often associated with the area for the layout (see [7]). One of the interesting problems about bandwidth is to understand what substructures force up the bandwidth of a graph. There are two known factors which may make bandwidth large. The first is the density lower bound (see [I, 21):
where D ( G ) is the diameter of G, that is, the maximum distance among all pairs 0012-365X/89/\$3.50 01989, Elsevier Science Publishers B.V. (North-Holland)
114
F.R.K . Chung, P.D.Seymour
of vertices in G. A somewhat stronger lower bound, the so-called “local density” bound, can also be easily obtained:
where G’ ranges over all connected subgraphs of G with 3 2 vertices. One natural problem arises: “If local density is small, is it true that the bandwidth is small?’’ This question was answered in the negative by ChvAtalovA [4] by examining refinements of the complete binary tree Bk of k levels. A graph G’ is said to be a refinement of G if G’ can be formed by replacing some edges in C by paths. For each integer k, every refinement of Bzk has bandwidth ak, and there is a refinement of BU, with local density at most 3. Now if a graph contains a refinement of BZk,its bandwidth is at least k. Again, containing a large complete binary tree is sufficient but not necessary for the graph to have large bandwidth (as we see from the star K l , n . )That suggests the following question. Suppose that the local density of a graph G is no more than c , , and that G does not contain any refinement of Bc2. Is it true that the bandwidth of G is bounded above by a constant depending only on c1 and c,? We mention that ChvAtalovA and Opatrinq [ 5 ] proved a somewhat similar result for infinite trees. They showed that if an infinite countable tree T satisfies that (i) the maximum degree is at most c , , (ii) the number of edge-disjoint semi-infinite paths is at most c Z rand (iii) T does not contain a refinement of B,, as a subgraph, then some refinement of T has finite bandwidth bounded above by a function depending only on c,, c2 and c3. in this paper, we will prove two results. One of them answers the above question in the negative and identifies a third structure which drives up the bandwidth. The other result answers positively an analogous question for cutwidth (or “topological bandwidth”, defined below).
Theorem 1. For each integer k , there exists a tree with the following properties (i) its local density is at most 9 (ii) it does not contain any refinement of B4 (iii) its bandwidth is at least k.
Theorem 2. Suppose that G has maximum degree c l , and does not contain any refinement of Bc2. Then the cutwidth of G is bounded above by a constant depending only on c1 and c2. The topological bandwidth b * ( G ) of a graph G is the minimum bandwidth h(G’) over all refinements G‘ of G. The topological bandwidth problem can be viewed as the optimization problems of circuit layout when vertices of degree two (interpreted as “drivers” or “repeaters”) can be inserted to help minimize the length of the edges. Cutwidth and topological bandwidth are known to be closely
Graphs with small band- and cutwidths
115
related, and it has been shown [3, 61 that b * ( G ) S c ( G )for any graph G. In particular, for trees [3]
b*(T ) S C( T ) S b*(T ) + log2 b*(T ) + 2.
But it is not hard to see that for some graphs, such as G = K,,, the cutwidth c ( G ) can be much larger than b*(G). Nevertheless, Theorem 2 implies the following relation between c(G) and b*(G).
Theorem 3. There is a function f such that for any graph G , c ( G )S f ( b * ( G ) ) . (One interpretation of Theorem 3 is that if the topological bandwidth is bounded above by a constant c,, then the cutwidth is bounded above by another constant c2 which depends only on ct.) The paper is organized as follows. In Section 2, we will construct some special trees, the so-called Cantor combs, which imply Theorem 1. In Section 3 we will give the proof of Theorems 2 and 3.
2. Cantor combs
In this section, we will show that the two conditions, local density S C , and containing no binary tree of c2 levels, do not imply small bandwidth. A comb is a tree T with two special vertices, called its roots, such that every vertex of T with degree 3 3 has degree 3 and lies on the path of T between the roots. For k 3 1, we define the Cantor comb c k as follows. C , is the 2-vertex tree, where both vertices are roots. Inductively, having defined C k - 1 , we define c k as follows. Take two disjoint copies T,, T2 of c k - 1 with roots s,, tl and s2, t2. Let P and Q be paths with 4 l v ( c k - 1 ) l and 6(k - 1) I v ( c k - 1 ) I edges respectively, such that P, Q,T, and T2 are mutually vertex-disjoint except that p has ends t1 and t2, and one end of Q is the middle vertex of P. We define c k to be & U U P U Q, with roots s,, s2. This completes the inductive definition of c k . We observe that there is an automorphism of c k exchanging the roots. Let I v ( c k ) l = Nk ( k 2 1). We shall show that c k satisfies Theorem 1, by means of the following assertions.
(2.1) For k 3 1, the bandwidth of
c k
is at least k.
Proof. If possible, choose k 3 1 minimum such that c k has bandwidth
k a 2 , ; let T,,
F. R.K. Chung, P.D.Seymour
116
between w, and w2 uses neither e l nor e2. We may assume that n ( w l ) < n ( w 2 ) . Since IE(P)( = 4Nk-1 it follows that IE(R)I <6Nk-, and so n ( w 2 )- n ( w l )< 6(k - 1)Nk-1. Since IV(Q)I *6(k - 1)Nk-l, some vertex w E V ( Q ) does not satisfy n ( w , ) < ~ ( w<)n(wz),and we may assume that R(W) < n(w,).Let S be the path of c k between w and w2. Since n ( w ) < n ( w l ) < n ( w 2 ) and w1 t\$ V ( S ) , there are consecutive vertices u, v of S with n ( u ) < n ( w , ) < n ( v ) . Since n ( v ) - n ( u )s k - 1 (because u, u are adjacent) it follows that n ( v ) - n(w,)< k - 1 and n(wl) - n ( u )< k - 1; but then one of n ( u ) , n ( v ) lies between n ( u l ) and n ( v , ) ,a contradiction. This completes the proof. 0 For k 3 1, we define Lk to be the number of edges in the path of C, between its roots. Let u be a root; for r 3 0 we define &(r) to be the number of vertices of C, different from v and within distance r of v. (From the symmetry of c k ) this does not depend on the choice of v.)
Proof. We proceed by induction on k. The result holds for k = 1, and we assume k > 1. Let T I , T2, P, Q etc. be as in the definition of ck. (1) f’r6Lk-l t h e n X k ( r ) s 3 r . For every vertex of c k within distance r of si belongs to T I , and the result follows from the inductive hypothesis. (2) If Lk-l< r < 4 L k then Xk(r) 6 3r. For the number of vertices of TI within distance r of s 1 is at most 2r, from our inductive hypothesis; and there are at most r further vertices of c k within distance r of sI,all from P. (3) If 4Lk s r s Lk - Lk-1 then Xk(r) s 3r. For within distance r of s1 there are at most Nk-l vertices of T,, at most r further vertices of P, at most r further vertices of Q , and none from T2. Thus Xk(r)s Nk-i+ 2r < 3r since r P iLk 2 Nk-l. (4) If r > Lk - Lk-1 then Xk(r) c 2r. For P U TI U T2 has S6Nk-l vertices, and there are r - +Lk S r - 2Nk-l further vertices of Q within distance r of sl. Thus Xk(r) c 6NkPl + r - 2Nk-l < 2r since r 2 Lk IE(P)I =4Nk-,. This completes the proof of (2.2). 0
(2.3) Let k 2 1 and r 3 0 be vertices of
c k
integers and let u E V ( C k ) .There are at most 9r
within distance r of v.
+1
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Proof. We proceed by induction on k. We may assume that k 3 2. Let K , T,, P, Q, etc. be as before. Now there are three paths of C ( v ) , each starting at v, which include P U Q in their union, and at most r vertices different from v of each of these paths is within distance r of v. Thus at most 3r vertices of P U Q are different from v and are within distance r of v. If v E V ( P U Q ) then, by (2.2), for i = 1, 2 at most 3r vertices of are different from v and are within distance r of v, as required. Thus we may assume that ~ E V ( G )V- ( P ) . Let the number of edges in the path of from v to tl be L. If L a r the result follows from our inductive hypothesis applied to 7i.Thus we assume that L < r. Every vertex of TI within distance r of v is within distance r + L of tl; and every vertex of T2 within distance r of v is within distance r - L of tZ. Thus by (2.2), there are at most 3(r+ L) + 3 ( r - L ) vertices of T,U & different from t l , t,, v which are within distance r of v. Hence in total there are at most 9r + 1 vertices of c k within distance r of v as required. 0 From assertions (2.1) and (2.3) we deduce the following result, which implies Theorem 1.
Theorem 1’. For k 5 1, the comb c k has local density S9, it does not contain any refinement of B4,and its bandwidth i~ at least k.
Proof. Let G’ be a connected subgraph of C, with IV(G’)Ia2. Choose v E V ( G ’ ) .By (2.3), IV(G‘)l s 9 D ( G ‘ )+ 1 and so
Thus C, has local density S9. Moreover, it contains no refinement of B4since it is a comb, and its bandwidth is at least k by (2.1).
3. Bounded cutwidth or topological bandwidth Before we proceed to prove that having bounded degree and containing no refinement of some bounded complete binary tree imply bounded cutwidth and hence bounded topological bandwidth, we will first discuss the “path-width” of a graph, which was introduced in [8] for studying graph minors. The path-width of a graph G is the minimum k 3 0 such that its vertex set V ( G ) is a union of subsets V,, V,, . . . , V, with the following properties; (i) I V J s k + l for 1 G i G t . (ii) vnl\$:.Vmfor l S i S m G j S t (iii) for each edge {u, v}, there exists some & containing both u and v. Path-width and bandwidth can differ significantly; for example, a star K l , n has path-width G1 and bandwidth S i n .
F.R. K. Chung, P.D.Seymour
1 I8
In [S] it was shown that if a graph contains no refinement of B,, then its path-width is at most c z , where c2 depends only on c I .This will be used to prove Theorem 2.
Proof of Theorem 2. Since G contains no refinement of B,,, its path-width is at most cj where c3 depends only on c2. Let V,, V,, . . . , V, denote subsets of G with s c3 1 (1 =s i =s t ) , as in the definition of path-width. For each vertex v, we define a(v) and b ( v ) to be respectively the least and largest numbers i such that v is in 6 . Choose a numbering JT from V ( G ) to integers {I, 2 , . . . , IV(G)l} such that z ( u ) s n ( v ) if and only if a ( u ) s a ( v ) . (Ties in a ( v ) are broken in any arbitrary way.) We shall show that JT (and hence G) has cutwidth ScI(c3 1). Let i be any number between 1 and n = IV(G)l. Choose x E V ( C )with J C ( X ) = i. We for every edge { u , v } with n ( u ) S i < n(v). For a ( u ) s a ( x ) claim that u E VQcx, since J T ( U ) s n(x), and a ( x ) s a(v) since n(x)6 n(v). Moreover, u ( u ) =sb ( u ) since { u , u } is an edge. Hence a ( u ) S a ( x ) G h ( u ) and consequently u E Vat,,, as claimed. But there are at most c3 + 1 vertices in VQ(x) each of which is adjacent to at most c , vertices. So there are at most cI(c3 + 1) edges “crossing” i , that is,
1x1 +
+
for every i. This completes the proof of Theorem 2.
0
Armed with Theorem 2 it is easy to deduce Theorem 3.
Proof of Theorem 3. Let G be a graph with b * ( G ) = k . Then G contains no refinement of B2k+2(since every such refinement has bandwidth a k + 1) and G has maximum degree G2k 1. From Theorem 2, c(G) is at most some f (k),as required. 0
+
We conclude by proposing the following problem.
For u graph G , suppose all subtrees of G have bandwidth Sc. Is it true that the bandwidth of G is no more than c’ where c’ is a function of c?
Acknowledgement The authors would like to thank W.T. Trotter for some very helpful discussions. The authors also wish to thank R.L. Graham for naming the combs which are constructed in Section 2.
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References [l] P.Z. Chinn, J. ChavBtalovB, A.K. Dewdney and N.E.Gibbs, The bandwidth problem for graphs and matrices - a survey, J. Graph Theory 6 (1982) 223-254. [2] F.R.K. Chung, Labelings of graphs, a chapter in Selected Topics in Graph Theory, 111 (eds. L. Beineke and R. Wilson). [3] F.R.K. Chung, On the cutwidth and the toplogical bandwidth of a tree, SIAM J. Alg. Discrete Methods 6 (1985) 268-277. [4] J. ChvBtalovB, On the bandwidth problem for graphs, Ph.D. dissertation, University of Waterloo (1980). [5] J. ChvBtalovB and J. Opatrinf, Two results on the bandwidth of graphs, Proc. Tenth Southeastern Conf. on Combinatorics, Graph Theory and Computing 1, Utilitas Math. Winnipeg (1979) 263-274. [6] F. Makedon, C.H. Papadimitriou and I.H. Sudborough, Topological bandwidth, SIAM J. Alg. Discrete Mathods 6 (1985). [7] T. Lengauer, Upper and lower bounds on the complexity of the min cut linear arrangement problem on trees, SIAM J. Alg. Discrete Methods 3 (1982) 99-113. [8] N. Robertson and P.D. Seymour, Graph minors. I. Excluding a forest, J. Combinatorial Theory Ser. B, 35 (1983) 39-61.
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# Quick Answer: How To Use A Random Check Digit To Scan Barcode As400?
## How does a barcode check digit work?
The last digit of a barcode number is a calculated check digit. The check digit is calculated from all the other numbers in the barcode and helps to confirm the integrity of your barcode number. Simply enter the ID Number below and the Check Digit Calculator will calculate the last digit for you.
## How do you do modulo 11 check digit?
Multiply each digit in turn by its weighting factor: 0 21 6 35 32 12 14. Add these numbers together: 0+21+6+35+32+12+14 = 120. Divide this sum by the modulus 11: 120: 11 =10 remainder 10. Substract the remainder from 11: 11 -10 = 1.
## How do you validate a check digit?
Add the even number digits: 1+1+1+1+1=5. Add the two results together: 0 + 5 = 5. To calculate the check digit, take the remainder of (5 / 10), which is also known as (5 modulo 10), and if not 0, subtract from 10: i.e. (5 / 10) = 0 remainder 5; (10 – 5) = 5. Therefore, the check digit x value is 5.
You might be interested: Question: How To Purchase A Barcode?
## How do you validate a barcode?
In order to validate barcodes, fixed-mount barcode scanners are positioned to read the codes on a production or shipping line. A software solution lets the scanner know what barcodes to expect. The system generates an alert if the wrong barcode is present or if the barcode label is missing.
## What is the check digit for EAN 13 Barcode?
The check digit for an EAN – 13 code is calculated as follows:
1. Count digit positions from the left to the right, starting at 1.
2. Sum all the digits in odd positions.
3. Sum all the digits in even positions and multiply the result by 3.
## Is check digit validation or verification?
Types of validation
Validation type How it works
Check digit The last one or two digits in a code are used to check the other digits are correct
Format check Checks the data is in the right format
Length check Checks the data isn’t too short or too long
Lookup table Looks up acceptable values in a table
## How do you solve ISSN?
Take the first seven digits of the ISSN (the check digit is the eighth and last digit): 0 3 1 7 8 4 7. Take the weighting factors associated with each digit: 8 7 6 5 4 3 2. Multiply each digit in turn by its weighting factor: 0 21 6 35 32 12 14. Add these numbers together: 0+21+6+35+32+12+14 = 120.
## What is a modulus 11?
MOD 11 Check Digit. MOD 11 Check Digit. A check digit is a number that is used to validate a series of numbers whose accuracy you want to insure. Frequently the last digit of a number string such as identification number is a check digit. Lets say the identification number starts out at 6 digits.
You might be interested: Readers ask: Why Does The Starbucks Barcode Always Pop Up.On.Yhe App?
## What is the value of MOD 11?
To find 1 mod 11 using the Modulo Method, we first divide the Dividend (1) by the Divisor ( 11 ). Second, we multiply the Whole part of the Quotient in the previous step by the Divisor ( 11 ). Thus, the answer to “What is 1 mod 11?” is 1.
## Why is check digit validation?
The check digit is a particularly important method of validation. It is used to ensure that code numbers that are originally produced by a computer are re-entered into another computer correctly. The check digit is calculated from the other digits in the number.
## Can a check digit be zero?
Divide the sum by 10 and check if the remainder is zero. If the remainder is zero then that is the check digit. If the number is not zero, then subtract the remainder from 10. The resulting number will be the check digit.
## What is range check in data validation?
A range check ensures that data is between an upper and lower acceptable value, within a certain range. Range checks are useful for dates, not just numbers. E.g. we could check that a date of birth puts a customer within an acceptable age range.
## Is my barcode scannable?
If you don’t modify the barcode after generating it (for example, manually editing its dimensions in Illustrator) and it’s printed in a high quality way, it will be scannable. Buy a USB barcode scanner like IntelliScanner Pro, then scan your barcodes to make sure they scan and verify the value.
## Can barcodes be faked?
People counterfeiting a product can simply paste the QR Code used by the company on their product. Consumers will be directed to the information linked to the original product’s QR Code. The best way to make sure you don’t buy fakes is by purchasing items from verified retailers.
You might be interested: FAQ: How To Sign Barcode?
## How do you read a 13 digit barcode?
An EAN – 13 number includes a 3- digit GS1 prefix (indicating country of registration or special type of product). A prefix with a first digit of “0” indicates a 12- digit UPC-A code follows. A prefix with first two digits of “45” or “49” indicates a Japanese Article Number (JAN) follows.
## FAQ: What Information Is Stored In Oklahoma Drivers License Barcode?FAQ: What Information Is Stored In Oklahoma Drivers License Barcode?
Contents1 What shows up when an ID is scanned?2 What is the barcode on back of ID?3 What do the things on the back of a driving Licence mean?4 What
Barcode
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Pressure = force per unit area (Definition!)
# Force
Pressure = Area
Gas samples exert a force pushing against any surface they contact, due to the impact of molecules striking the surface. Per unit area, it’s the same for all surfaces in or walls containing the same gas sample, unless the gas is flowing somewhere collectively.
That pressure increases with molecular velocity and mass, and with the number of molecules per unit volume (number density).
A sheet suspended within a gas sample gets same force on both sides, so it does not move.
Average molecular velocity increases with temperature but decreases with molecular mass, in such a way that the mass effect cancels.
This leaves pressure nearly proportional to temperature and to number density, but independent of the molecular mass:
# P = R(T)(n/V), where R is the gas constant, n = # moles = N/NA
Document views 245 Page views 245 Page last viewed Mon Jan 16 15:39:48 UTC 2017 Pages 77 Paragraphs 871 Words 5446
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# A horizontal force of 10 N is necessary to just hold a block stationary against a wall. The coefficient of friction between the block and the wall is 0.2.
more_vert
A horizontal force of 10 N is necessary to just hold a block stationary against a wall. The coefficient of friction between the block and the wall is 0.2. The weight of the block is-
(a) 20 N
(b) 50 N
(c) 100 N
(d) 2 N
more_vert
verified
The correct option is d) 2N
Frictional force balances the weight of the body
Frictional force f = μN = mg
f= 0.2 X 10 = 2N
Therefore, weight of the block mg= 2N
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# Bijection for Rook placecement and Stirling number of 2nd kind
Say we have an nxn chessboard from which the squares below the diagonal are removed to obtain a new board $C_n$. The board $C_3$ is shown below.
Let the number of ways to place k non-attacking rooks on board $C_n$ is given by $r_k(C_n)$. What I want to prove is that:
$$\sum_{k=0}^{n} r_k(C_n)x^k = \sum_{k=0}^{n} S(n+1, n+1-k)x^k$$ where $S(n,k)$ is a stirling number of second kind.
So far I have the bijective proof for recurrence relation for the rook placement:
$$r_k(C_n) = r_k(C_{n-1}) + (n-(k-1))r_{k-1}(C_{n-1})$$
Proof I have goes as follow:
Either: Place k rooks on the board $C_{n-1}$, there are $r_k(C_{n-1})$ ways to do so where $C_{n-1}$ is obtained by deleting top row of $C_n$
Or: Place $k-1$ rooks on the board $C_{n-1}$ and place the last rook on the top row of $C_n$. Placing $k-1$ rooks on the board $C_{n-1}$ implies that we cannot place the last rook on $k-1$ columns. Hence we can place the last rook on $n-(k-1)$ squares in the top row of $C_n$. there are $(n-(k-1))r_{k-1}(C_{n-1})$ ways to do this. Using BCP2 we add either and or case to give:
$$r_k(C_n) = r_k(C_{n-1}) + (n-(k-1))r_{k-1}(C_{n-1})$$
The recursion for Stirling number is:
$S(n+1, n+1-k) = S(n, n - k) + (n+1-k)S(n, n+1-k)$
They both satisfy the same initial conditions:
When $k = 0$ we have $r_0(C_n) = 1$ and $S(n+1, n+1) = 1$
and when $k=n$ we have $r_n(C_n) = 1$ and $S(n+1, 1) = 1$
I also know that e.g. rook placement for 2 rooks on board $C_3$:
corresponds to the set partition {{4,3,2}, {1}}
I don't understand what to do after this to complete the proof. Am I even on the right track? Ultimately, what I want is a bijective proof for:
$$\sum_{k=0}^{n} r_k(C_n)x^k = \sum_{k=0}^{n} S(n+1, n+1-k)x^k$$
Let $C_n$ be the triangular board of side length $n$ with squares in positions $(i,j)$ for $i \le j$, so $C_3$ is as shown in the question.
Given a non-attacking placement of $k$-rooks on $C_n$, there is a corresponding function $f : \{1,2,\ldots,n,n+1\} \rightarrow \{2,\ldots,n,n+1\} \cup \{ \star\}$ defined by
$$f(i) = \begin{cases} j+1 &\text{if there is a rook in square (i,j)} \\ \star & \text{if there is no rook on row i.} \end{cases}$$
(In particular, $f(n+1) = \star$ for any placement, since row $n+1$ is empty.)
Bijection. Let $Y = f(\{1,2,\ldots,n,n+1\}) \backslash \{\star\}$. Let $Y' = \{1,2,\ldots,n,n+1\} \backslash Y$. Since $|Y| = k$, we have $|Y'| = n+1-k$. For each $i \in Y'$ let
$$A_{i} = \{i,f(i),f(f(i)), \ldots \} \backslash \{ \star \}.$$
If $i \in Y'$ then $i$ is not in the range of $f$, so the sets $A_i$ for $i \in Y'$ form a set partition of $\{1,2,\ldots,n,n+1\}$ into $n+1-k$ parts.
Inverse. Conversely given such a set partition, take $Y'$ to be the $(n+1-k)$-set of minimum elements of the parts. For each $i \in Y'$, define $f$ so that the displayed equation above holds for elements in the part containing $i$, setting $f(i^\star) = \star$ if $i^\star$ is the biggest element in this part. Then take the corresponding rook placement.
For example, if $n=5$ and $k=3$ then the set partition $\bigl\{ \{1,4,6\}, \{2\}, \{3,5\} \bigr\}$ corresponds to the function $f$ defined by $f(1)=4$, $f(4) = 6$, $f(6) = \star$, $f(2) = \star$, $f(3) = 5$, $f(5) = \star$, and so to the rook placement with rooks in positions $(1,4-1)= (1,3)$, $(4,6-1) = (4,5)$, $(3,5-1) = (3,4)$.
This is essentially the same as the bijection outlined on pages 77 and 78 of Loehr's textbook Bijective combinatorics.
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CC-MAIN-2020-05
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https://code.dennyzhang.com/rotate-string
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# Leetcode: Rotate String
Rotate String
Similar Problems:
We are given two strings, A and B.
A shift on A consists of taking string A and moving the leftmost character to the rightmost position. For example, if A = ‘abcde’, then it will be ‘bcdea’ after one shift on A. Return True if and only if A can become B after some number of shifts on A.
Example 1:
```Input: A = 'abcde', B = 'cdeab'
Output: true
```
Example 2:
```Input: A = 'abcde', B = 'abced'
Output: false
```
Note:
• A and B will have length at most 100.
Github: code.dennyzhang.com
Credits To: leetcode.com
Leave me comments, if you have better ways to solve.
```## Blog link: https://code.dennyzhang.com/rotate-string
class Solution:
def rotateString(self, A, B):
"""
:type A: str
:type B: str
:rtype: bool
"""
return len(A) == len(B) and B in A + A
```
Share It, If You Like It.
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https://docusaurus.io/docs/2.x/markdown-features/math-equations
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Version: 2.x
# Math Equations
Mathematical equations can be rendered using KaTeX.
## Usage
Write inline math equations by wrapping LaTeX equations between $: Let$f\colon[a,b]\to\R$be Riemann integrable. Let$F\colon[a,b]\to\R$be$F(x)=\int_{a}^{x} f(t)\,dt$. Then$F$is continuous, and at all$x$such that$f$is continuous at$x$,$F$is differentiable at$x$with$F'(x)=f(x)$. http://localhost:3000 Let $f\colon[a,b] \to \R$ be Riemann integrable. Let $F\colon[a,b]\to\R$ be $F(x)= \int_{a}^{x} f(t)\,dt$. Then $F$ is continuous, and at all $x$ such that $f$ is continuous at $x$, $F$ is differentiable at $x$ with $F'(x)=f(x)$. ### Blocks For equation block or display mode, use line breaks and $$:$$ I = \int_0^{2\pi} \sin(x)\,dx$\$
http://localhost:3000
$I = \int_0^{2\pi} \sin(x)\,dx$
## Configuration
To enable KaTeX, you need to install remark-math and rehype-katex plugins.
npm install --save remark-math@3 rehype-katex@5 [email protected]
warning
Use the exact same versions. The latest versions are incompatible with Docusaurus 2.
Import the plugins in docusaurus.config.js:
const math = require('remark-math');
const katex = require('rehype-katex');
Add them to your content plugin or preset options (usually @docusaurus/preset-classic docs options):
remarkPlugins: [math],
rehypePlugins: [katex],
Include the KaTeX CSS in your config under stylesheets:
stylesheets: [
{
href: 'https://cdn.jsdelivr.net/npm/[email protected]/dist/katex.min.css',
type: 'text/css',
integrity:
'sha384-odtC+0UGzzFL/6PNoE8rX/SPcQDXBJ+uRepguP4QkPCm2LBxH3FA3y+fKSiJ+AmM',
crossorigin: 'anonymous',
},
],
Overall the changes look like:
docusaurus.config.js
const math = require('remark-math');
const katex = require('rehype-katex');
module.exports = {
title: 'Docusaurus',
tagline: 'Build optimized websites quickly, focus on your content',
presets: [
[
'@docusaurus/preset-classic',
{
docs: {
path: 'docs',
remarkPlugins: [math],
rehypePlugins: [katex],
},
},
],
],
stylesheets: [
{
href: 'https://cdn.jsdelivr.net/npm/[email protected]/dist/katex.min.css',
type: 'text/css',
integrity:
'sha384-odtC+0UGzzFL/6PNoE8rX/SPcQDXBJ+uRepguP4QkPCm2LBxH3FA3y+fKSiJ+AmM',
crossorigin: 'anonymous',
},
],
};
## Self-hosting KaTeX assets
Loading stylesheets, fonts, and JavaScript libraries from CDN sources is a good practice for popular libraries and assets, since it reduces the amount of assets you have to host. In case you prefer to self-host the katex.min.css (along with required KaTeX fonts), you can download the latest version from KaTeX GitHub releases, extract and copy katex.min.css and fonts directory (only .woff2 font types should be enough) to your site's static directory, and in docusaurus.config.js, replace the stylesheet's href from the CDN URL to your local path (say, /katex/katex.min.css).
docusaurus.config.js
module.exports = {
stylesheets: [
{
href: '/katex/katex.min.css',
type: 'text/css',
},
],
};
tip
Only use the latest version if you actually need the bleeding-edge features of $\KaTeX$. Most users should find the older versions work just as well.
The latest versions of rehype-katex (starting from v6.0.0) has moved to ES Modules, a new module system of JavaScript, which Docusaurus doesn't officially support yet. However, it is possible to import rehype-katex dynamically, using an async config creator. Docusaurus will call this creator function and wait for it to return the config object.
docusaurus.config.js
async function createConfig() {
// ES Modules are imported with import() instead of require(), and are imported asynchronously
const katex = (await import('rehype-katex')).default;
return {
// ...
};
}
In this case, the overall configuration changes will look like:
docusaurus.config.js
const math = require('remark-math');
async function createConfig() {
const katex = (await import('rehype-katex')).default;
return {
title: 'Docusaurus',
tagline: 'Build optimized websites quickly, focus on your content',
presets: [
[
'@docusaurus/preset-classic',
{
docs: {
path: 'docs',
remarkPlugins: [math],
rehypePlugins: [katex],
},
},
],
],
stylesheets: [
{
href: 'https://cdn.jsdelivr.net/npm/[email protected]/dist/katex.min.css',
type: 'text/css',
integrity:
'sha384-MlJdn/WNKDGXveldHDdyRP1R4CTHr3FeuDNfhsLPYrq2t0UBkUdK2jyTnXPEK1NQ',
crossorigin: 'anonymous',
},
],
};
}
module.exports = createConfig;
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|
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http://forums.wolfram.com/mathgroup/archive/2005/Jul/msg00353.html
| 1,585,481,970,000,000,000
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text/html
|
crawl-data/CC-MAIN-2020-16/segments/1585370494331.42/warc/CC-MAIN-20200329105248-20200329135248-00271.warc.gz
| 68,218,288
| 9,217
|
Re: Modeling and Array Problem
• To: mathgroup at smc.vnet.net
• Subject: [mg58742] Re: Modeling and Array Problem
• From: Ken Levasseur <kenneth_levasseur at uml.edu>
• Date: Sun, 17 Jul 2005 03:03:56 -0400 (EDT)
• References: <200507160503.BAA14933@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com
```Peter:
The left side of Set can't be a part of a part of a symbol. It can
be a part of a symbol, so you can do this:
In[16]:=
array = {{1, 1}, {2, 2}, {3, 3}, {4, 4}, {5, 5}};
In[17]:=
array[[1,1]] = 42
Out[17]=
42
In[18]:=
array
Out[18]=
{{42,1},{2,2},{3,3},{4,4},{5,5}}
You were correct in referring to the first row , first col item
this way. It's just a restriction of Set.
In[19]:=
array[[1]][[1]]
Out[19]=
42
Ken Levasseur
Mathematical Sciences
UML
http://faculty.uml.edu/klevasseur
Please avoid sending me Word or PowerPoint attachments.
See http://www.gnu.org/philosophy/no-word-attachments.html
On Jul 16, 2005, at 1:03 AM, Sycamor at gmail.com wrote:
> Hello,
>
> I am a high school student and am interning/volunteering at a
> university over the summer. My ultimate goal is to model the movement
> of charge in a Nickel Metal Hydride Sphere using Mathematica. This
> goal is beyond my ability as it requires calculus and differential
> equations and so forth. But I am to progress as best I can, using
> iterative processes to replace calculus where possible. The professor
> I am working with has started me with the simpler task of finding the
> curvature of a set of data points (in this first easy case, the 'data'
> is the values of 101 points of the x^2 function).
>
> While programming, I have found it necessary to change the value of
> certain elements of an array of ordered pairs, but have been unable to
> do so.
>
> In[1]:= array = {{1,1},{2,2},{3,3},{4,4},{5,5}};
> In[2]:= array[[1]][[1]]
> Out[2]= 1
> In[2]:= array[[1]][[1]] = 42
> Out[2]= Set::setps: test in assignment of part is not a symbol.
>
> I suppose the error arises when I attempt to set the numerical
> contents
> of the array equal to a certain value. Is there anyway to simply
> overwrite the element at a certain address? Why does the syntax used
> above seem to work for lists?
>
> In[3]:= list = {1,2,3,4,5};
> In[4}:= list[[1]]
> Out[4]= 1
> In[5]:= list[[1]] = 42;
> In[6]:= list
> Out[6]= {42,2,3,4,5}
>
> I have included what I have written so far at the end of this message
> to put my problem in context. As I am new to Mathematica, I do not
> know hoe to write elegant, concise programs using the array of
> powerful
> built in functions, so I appologize for my clunky code. There is
> probably a much easier way to accomplish my task.
>
> I would appreciate any help on this matter.
>
>
> Thank you,
>
> Peter Hedman
>
>
>
>
>
>
> Bellow is what I have written so far. The comments are mostly meant
> for myself, and probably do not make much sense.
>
> g = 1;
> (*sets value of liberated proton constant*)
>
> d=0.1;
> (*sets value of diffusion constant*)
>
> v= 0.1;
> (*sets value of boundry velocity*)
>
> a = 0.1;
> (*sets the value of constant \[Alpha]*)
>
> j = 0;
>
> (*sets the initial value of j. This step shouldn't be necessary*)
>
> rprevb=rprev = Table[0z, {z,0,100}];
>
> r = 0; s = 0;
>
> (Do[rprev[([q])] = {r, s^2}; (r++); (s++), {q, 1, 101}];)
>
> (*These three lines create the initial list of ordered pairs*)
>
> Do[rprevb[[q]]={0,0}, {q,1,101}]
>
> jmax = 100;
> (*initially sets the maximum horizontal range*)
>
> iter[n_]:= If[
> n==jmax , (rprev[[n+1]][[2]] + v*(g+rprev[[n+1]][[2]]) -
> d*(rprev[[n+2]][[2]]-rprev[[n+1]][[2]])), (rprev[[n+1]]
> [[2]]+
> a*(rprev[[n]][[2]]-2rprev[[n+1]][[2]]+rprev[[n+2]][[2]]))] ;
>
> (*defines transformation function*)
>
> dim =Dimensions[rprev];
>
> dimb = Dimensions[rprevb];
>
> Print["Dimensions of initial list = ",dim]
>
> Print["Dimensions of initial list = ",dimb]
>
> initialplot = ListPlot[rprev]
>
> Do[Do[rprevb[[(j+1)]][[2]]=iter[j]; Print[rprevb]; {j,0,jmax,1}];
> rprev=rprevb; Print["time = ",t]; Print["jmax = ",jmax ];
> ListPlot[rprev];
> jmax--,{t,1,10}]
>
> (*With every iteration of the outside loop,
> jmax is decremented by one. As iterations progress to flatten the
> curve,
> the boundry moves. Should these two processes take place in this
> way?
> It appears this is not the ideal model, but it will work for
> now.*)
>
>
***********************************
Ken Levasseur
Mathematical Sciences
UML
http://faculty.uml.edu/klevasseur
Please avoid sending me Word or PowerPoint attachments.
See http://www.gnu.org/philosophy/no-word-attachments.html
```
• Prev by Date: Re: D[...] change in 5.1
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| 1,573
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https://www.thedrippingroot.com/blog/how-many-cups-are-in-18-oz
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# How Many Cups are in 18 Oz? | Get Your Answer Here
Are you ever confused by the many measurements for common ingredients? Do converting ounces to cups leave your head spinning? Don’t worry, we are here to solve this conundrum and make cooking with multiple measurements easy. In this post, we will answer one simple question: How many cups are in 18 oz? Read on for a step-by-step guide on how to correctly convert units so you can create delicious recipes with confidence.
## What Are Cups?
Cups are a common measurement used in cooking. They can be filled with any type of ingredient, including liquids and solids, depending on the recipe you’re following. The standard cup size is eight fluid ounces and is used to measure dry ingredients like flour or sugar, but it can also be used to measure liquid ingredients, such as water or oil. When using a measuring cup, it is important to remember to level off any excess ingredient that may have been added.
## What Are Ounces?
Ounces (oz) are another common unit used in cooking. Ounces measure the weight or mass of an ingredient and can be used when measuring both liquids and solids. One ounce equals 28.35 grams, so it is important to pay attention to this measurement when following a recipe that requires ounces to ensure accuracy.
## How Many Cups Are In 18 Oz?
The answer to “how many cups are in 18 oz?” question is 2.25 cups. To calculate the number of cups in 18 oz, divide the ounces by 8 (the amount of ounces in one cup). This calculation yields a result of 2.25 cups when rounded off to the nearest hundredth. For example, if you need 18 oz of flour for a recipe, use 2 1/4 cups of flour.
## Why Should You Know About How Many Cups Are In 18 Oz?
Knowing how many cups are in 18 oz can help you accurately measure ingredients when cooking or baking. Even the slightest miscalculation can change the texture and taste of a recipe, so it is important to be precise with your measurements. Additionally, using the correct amount of ingredients helps ensure that you don’t waste any food.
## How To Convert Cups To 18 Oz?
If you need to convert cups to 18 oz, it is just as easy. To calculate the equivalent number of ounces in 1 cup, multiply the number of cups by 8 (the amount of ounces in one cup). This calculation yields a result of 14.4 oz when rounded off to the nearest hundredth. For example, if you need 6 cups of flour for a recipe, use 14.4 oz of flour.
## How To Read A Measuring Cup?
Reading a measuring cup is easy. On the outside of the cup, you will see markings for different amounts of fluid ounces. To use the cup correctly, fill it with your ingredient and then level it off by running a flat edge (such as a metal ruler) along the top of the measuring cup. The ingredient should be even with the marking that corresponds to the amount of ounces you need.
## How To Measure Ounces?
Measuring ounces is just as simple! To measure ounces, you can use a kitchen scale. Place the ingredient on the scale and reset it to zero if necessary. Then, adjust the weight until it reads the amount of ounces needed for your recipe. Now that you know how many cups are in 18 oz, measuring ingredients accurately has never been easier.
## What Are Benefits Of Using Ounces And Cups Conversion?
Using ounces and cups conversion is an essential part of cooking and baking. By being able to accurately measure ingredients, you can easily create delicious recipes with confidence. Additionally, using proper measurements helps ensure that you don’t waste any food. Now that you know how many cups are in 18 oz, you can use this knowledge to your advantage when creating your favorite dishes.
## What Are Tools To Help With Ounces And Cups Conversion?
There are many tools to help with ounces and cups conversion. An online converter is one option that will save you time when figuring out how many cups are in 18 oz. Alternatively, there are also measuring cups and kitchen scales available for purchase if you prefer to measure your ingredients manually.
## What Are Tips For Remembering Ounce And Cup Conversion Formulas?
It can be difficult to remember the conversion formulas for ounces and cups. A great tip is to write them down somewhere you can easily refer back to, such as on a sticky note in your kitchen. You can also try using mnemonic devices like acronyms or rhymes to help you remember the formulas more easily.
## Conclusion: How Many Cups Are In 18 Oz?
In conclusion, 18 ounces is equivalent to 2.25 cups when rounded off to the nearest hundredth. Knowing how many cups are in 18 oz can help you accurately measure ingredients when cooking or baking, saving you time and helping you create delicious recipes with confidence. Now that you know the answer to this common question, measuring ingredients correctly has never been easier.
## FAQs: 18 Oz
### How many cups is 18 oz of food?
The conversion from 18 oz to cups yields 2.25 cups. It’s important to note that this conversion is based on US customary units, and the volume represented by a cup may vary in different countries.
### How many cups is 18 fluid ounces?
With the knowledge of the conversion factor, determining the equivalent of 18 fluid ounces in cups becomes effortless. By multiplying 0.10408422724619 with the given quantity of 18 fluid ounces, we arrive at the answer to the query, “What is 18 fluid ounces in cups?”—1.8735160904315 cup.
### Is 18 oz water bottle too small?
For optimal hydration throughout the day, the ideal water bottle size ranges from 16 to 24 ounces or 473 to 708 milliliters. This convenient size ensures easy portability while keeping you hydrated.
### Does 18 oz water bottle fit in cup holder?
Will it fit in my cupholders? Our Rambler® 18 oz Water Bottle is designed to fit most cupholder sizes. However, please note that some of our larger bottles may not fit in standard cupholders.
### How many cm is 18 oz aquaflask?
Volume: The capacity of this bottle is 532 ml or 18 oz, providing ample space for your hydration needs. Weight: With a weight of 310 g or 10.88 oz, this bottle is lightweight and easy to carry on your adventures. Mouth Dameter: The bottle features a 4.9 cm wide mouth, allowing for easy pouring and sipping. Bottle Diameter: The bottle has a diameter of 7.3 cm, providing a comfortable grip and stability. Whether you’re on-the-go or enjoying outdoor activities, this bottle is designed to meet your hydration requirements with its convenient size and weight.
### Does 18 oz Hydro Flask fit in cup holder?
Introducing our 18 oz Standard Mouth insulated water bottle: sleek stainless steel design, cup holder compatible, and fits effortlessly anywhere. This compact, reusable, and refillable bottle keeps your cold drinks refreshing for 24 hours, while your hot beverages stay piping hot for up to 12 hours. With ample hydration capacity, it’s perfect for your spontaneous adventures.
### How many cups of coffee is 18 oz?
As mentioned earlier, 1 cup is equivalent to 8 oz. Consequently, if you have a measurement of 18 oz, it would be equal to 2.25 cups. This conversion allows you to easily transform any item measured in 18 ounces into 2.25 cups.
### How big is a 18 oz of water?
An exact equivalence of 18 ounces is 2 and a 1⁄4 cups. It’s important to note that a U.S. fluid ounce, which is 1/128th of a U.S. gallon, differs from an ounce of weight or an Imperial fluid ounce. Similarly, a U.S. cup represents a volume equal to 1/16th of a U.S. gallon, approximately 236 milliliters.
### Is 18 oz water bottle enough?
The optimal water bottle size ranges from 16 to 24 ounces or 473 to 708 milliliters. This capacity is generally sufficient to keep you hydrated all day long while remaining conveniently portable.
### How do you measure 18 ounces?
The correct measurement is 2.25 cups. This indicates that if a recipe requires 18 ounces of an ingredient, you can accurately measure out 2.25 cups to achieve the desired quantity.
### How many 18 oz bottles of water should I drink a day?
According to the National Academies of Sciences, it is recommended that women consume approximately 2.7 liters of water per day, while men should aim for around 3.7 liters. This roughly translates to 5-6 bottles for women and 7-8 bottles for men. Ensuring an adequate intake of water is crucial for maintaining optimal hydration levels.
### How tall is a 18 oz tumbler?
The exact height of a 18 oz tumbler depends on the specific product. Generally, the tumblers are approximately 7 inches tall with an opening diameter of 3.25 inches and a base diameter of 3.5 inches. The total capacity of each tumbler is 18 ounces or 530 ml and they feature convenient double wall design for superior insulation.
### How tall is a 18 oz plastic cup?
The exact height of a 18 oz plastic cup depends on the specific product. Generally, these cups are approximately 5-6 inches tall with an opening diameter of 3.25 inches and a base diameter of 1.75 inches. The total capacity is 18 ounces or 530 ml making them ideal for cold drinks or hot beverages depending on your preferences. They also feature high quality construction and a textured non-slip grip for added convenience.
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# What answer is divided by 99 divided by 33?
99 divided by 33 = 3
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# Questions tagged [3d]
3D refers to three dimensional space where coordinates are represented with X, Y and Z values.
196 questions with no upvoted or accepted answers
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995 views
### How do i find the circumsphere of a tetrahedron?
I'm looking for the most minimized equation to find the center coordinates and the radius of a tetrahedron circumsphere given four 3D points. What I found on the internet mainly deal with the circum ...
506 views
### How to create skeletal animation data for a skirt?
Please think of this in an engine-agnostic way first, and then we can talk about specific engines. In my game, I have a 3D character which is wearing a hakama (think of it as a long pleated skirt). ...
3k views
### Matrix rotation wrong orientation LibGDX
I'm having a problem with matrix rotation in libgdx. I rotate it using the method matrix.rotate(Vector3 axis, float angle) but the rotation happens in the model orientation and I need it happens in ...
5k views
### Handling movement on sloped surfaces - clamping character to sloped surface
I've noticed that a lot of people seem to have this issue but I've yet to find an actual working solution - when a rigidbody-based character controller (I'm not using Unity's character controller) ...
484 views
### Implementing proportional navigation in 3D
Good afternoon guys, a = N * λ' * V is the formula for the commanded acceleration required to hit the target, where N is the proportionality constant, λ' is the change in line of sight and V is ...
1k views
### Displaying text on a 3D plane
I am busy with a pet project where I am trying to create a Collectible Card Game in 3D. These cards have Hp and Damage etc. but I cannot find a way to neatly display these values. I tried to use a ...
246 views
### Rigid body falls through static mesh even though collision is detected
I'm using Jitter Physics to create a game. For testing, I have ball (a spherical rigid body) falling onto a small box-like mesh. The problem is that the ball falls straight through the floor, even ...
261 views
### How to generate a multiplier map for radiosity
I am following this tutorial: here I am at the part where you are creating a hemicube. I have got the code to render the scene into a texture and therfore an array. Now how can I generate a so-called ...
208 views
### Perspective correct texture mapping in iOS Metal
I tried various ways of getting the interpolation work and generate a perspective correct image, but none of the suggested ways works. My current code is: ...
396 views
### Collision detection problem in XNA4/Monogame
I have been trying to get 3D collision detection working for a while and I think (hope) you people can help me. For my project I have created a 'sprite3d' kinda class which is called ...
912 views
### 3d Wall sliding collision
I'm working on collision for walls in my game and the way I currently have it I get stuck walking into a wall. I'm trying to make my character slide on the walls but still collide. My character moves ...
93 views
### most compact way to represent dice face translations rolling in cardinal directions
I'm making a 2-D dice rolling game in BASIC (think mechanics similar to Devil Dice on the Playstation, but top-down 2D graphics). I'm trying to accurately represent the transitions of a die when ...
147 views
### How do you find the collision between 3D boxes and 3D QUAD face meshes?
I have an axis aligned bounding boxes for 3D object collision, but I have a complex 3d mesh that's divided into 3d faces so I want to check for bounding boxes and collision with the faces of that ...
50 views
### Weird texture applied
So I applied the baked texture (on the right side of screen shot) but it shows in this manner. What is the problem? I made the model in Blender and also the baked image. The house consists of ...
388 views
### JavaFX import 3D models
I want to import some 3d models to my javaFX application. In javaFX I can create rectangles, traingles, circles, pentagons. Those figures can be filled with color or filled with image. They are ...
116 views
### Creating “Run over by car” effect in unity using spotlights only
I'm doing a project where I need to simulate a car running the character over. It's happening in a completely dark space, so I want to create it using spotlights moving towards the camera until the ...
332 views
### 3D Physics Engine Collision Response: Solving Simultaneous Collisions via equation
I have been implementing a basic physics engine for a small game project I'm working on. It has very specific requirements, so I decided to try and write my own physics engine to meet them. I found ...
118 views
### Retrieve 3d models from a server in unity
I have 10 3d models which are about 8mb each..Now, should I store these 3d models in my apk or will it be wise to store it in a server and then retrieve it?.If so, how can I do that?
175 views
### Get mesh height at position
Me and my friends are creating a simple 3D game in LibGDX and are currently trying to implement simple collision detection with the terrain (ground). For this I thought the best way would be to simply ...
99 views
### How to disable and enable a randomly generated prefab in an endless runner 3d
Hi all I have been experimenting the endless runner game for some time and needed to optimize it. I read some articles about optimizing my game and it says that once you instantiate the prefabs in the ...
172 views
### Recognize pattern in 3D building environment
I'm currently developing 3rd person building game (as bacheleor thesis). I need to recognize constructed patterns co I can mark corresponding structure as some building (so player can start using that ...
1k views
### Dual Contouring - what is it and how to use it?
I'm having trouble understanding Dual Contouring yet and I need some explanations from someone who has first hand experience. I searched all I could before writing this question. I've already read: ...
328 views
### 3D Zoom To Mouse Cursor accounting for Change in Camera Pitch
Overview: Simply put, I'm trying to make a zoom to cursor algorithm. This would normally be very simple, but the implementation of the camera is making it difficult. As the camera zooms in and out, ...
136 views
### Custom 3D Renderer: Frustum clipping
I have recently been toying with the idea of creating a 3D-renderer from scratch and, after looking into it, doesn't seem like such a behemoth of a task as it sounds. Last night, I was tossing and ...
208 views
### Texturing Objects in OpenGL
Using OpenGL tutorials found here, I constructed an .obj file loader as well as a texture file loader using SOIL. I have an issue with drawing a textured object. (Shown here) Assuming I created the ...
120 views
### Zoom to Mouse on a Globe using a Perspective Projection
I am trying to create a strategy game that takes place on a globe (i.e. a textured sphere). I would like to be able to zoom to mouse like on google maps. My camera always points towards the center of ...
113 views
### How would I warp a mesh with code?
I recently asked a question about putting a camera at the head of a 3D human model here and it worked out rather well and whatnot, but there is still a problem that bothers me: How would I contort/...
127 views
### UDK Modular construction (customisable vehicles etc)
I'd like to build an interface whereby the user can attach various components and blocks onto a vehicle, this is then saved (as object/coordinate, rotation, scale values - in database) and can be used ...
364 views
### How to create 3d picking ray with the cursor position
I'm new here and I'm not sure how to go about doing this. I am creating an application that is supposed to use the cursor's position in the window to perform 3d picking. The tutorials I have found ...
403 views
### How can I export a model using the CAT bone system into XNA?
Has anyone successfully used the CAT bone system in 3ds Max and exported the file into XNA? If so, what was your method of doing so? There are a number of methods of doing this, apparently, but the ...
326 views
### Combining skydome and fog
I am developing a 3D game in DarkBASIC Professional, and I have a theoretical problem with the combination of skydomes and fog in general. My question is not specific to a programming language, so I ...
359 views
### How to fix my camera collision, C# XNA
I am having some weird issues with my camera. I am trying to make it so it is a first person camera, that when the user presses the W or Up key, it moves towards where the camera is looking. My issue ...
563 views
### Computing the UV coordinates of a 3D point that lies within a 3D triangle
From a triangular mesh I have a 3D triangle T represented by 3D vertexes v0, v1 and v2. Each vertex has an associated UV coordinate into a common texture image, represented as uv0, uv1, uv2. ...
51 views
### How should a 3D map be stored?
I have an array of 3d vertices, Vertex3D[][] ground; so for each element in the array, I have an array of three 3D vertices, a triangle. But this means I cannot ...
39 views
### How can I get my player's character to slide along a rail?
I am creating a snowboarding game and would like my player to be able to jump on a rail and move along it if it lands on it. I have no idea how to approach this. I have already tried using ray casts ...
107 views
### Multi-level 3D grid based pathfinding in Unity
I'm trying to come up with a way to pathfind on multiple levels. See image: The orange walls would be considered a ladder, connecting you to the upper levels. At first I thought of looking at it as ...
24 views
### Scene for Computer Screen
We are developing a scene that will involve heavily interacting with a "touch screen monitor" in the scene.. Because the workflow of the touch screen will need to be dynamic we are creating a REST API ...
27 views
### Discrete line inside a 3D matrix
First of all a little premise: I'm following this tutorial on YouTube for procedural 2d cave generation in unity using cellular automata and I'm extending it in order to make caves in all three ...
63 views
### OBB vs Frustrum collision detection
Some definitions: An oriented bounding box (OBB) is a rectangular block which can be represented in many ways, as a collection of eight vertices, a collection of six planes or others, for this ...
50 views
### Convert bones by bone matrix
I have a 3d-models from old game. They have the following structure: ...
34 views
### Does it make sense for a view matrix to have a scale?
When building a view matrix from an SRT matrix multiply, examples (here and here) only ever compose it with a R and T. Does it make sense to have the S component be anything other than the identity? ...
122 views
### How to implement weighted blended order independent transparency in OpenGL?
I want to implement weighted blended OIT in my C++ OpenGL 3D game, but I didn't find any C++ examples about weighted blended order independent transparency. Paper How to clear accumTexture(...
29 views
### Switch Vector 3 x,y,z Components Based on Rotation
I have a cube that moves by rotating on its edge. I want my code to also work for cuboids. The following code is what I'm using to rotate my cube: ...
249 views
### Is there a way to convert a shader code to shader graph?
I would like to ask if there is a way to convert a Coded shader into the new Shader graph, beacause i want to edit my shaders in more efficient way. Any help is appreciated.
44 views
### How can I move the camera to create a parallax effect?
Right now I am making an application which can be described as model viewer for iPad Pro. The client wants to have an orbiting camera around the object and in addition he wants to have a parallax ...
349 views
Is there such a thing as screen-space shadows? I know that most shadow projection comes from rendering the world from the viewpoint of the light source, projecting the shadows onto a large square ...
32 views
### HTC Vive Grabbing Object Trigger
I'm doing a HTC Vive VR game with Unity. I have 3 objects, and I want to trigger different events when the player pick up different objects. Since I set the rigidbody to the controller and don't ...
53 views
### Unity3D: Simple animations ending with a bit of offset
I'm trying to make a simple game with some cubes rotating 90º when the user clicks on them. I've setup an empty game object, a simple cube nested, an animator, and two animations: RotateClockWise, ...
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# Happy number
A number is a happy number when the sum of the squares of its digits eventually becomes 1. Numbers that are not happy are called unhappy numbers.
Consider a number t, then define a sequence t0,t1,t2,.. where t0 = t, and ti+1 is the sum of the squares of the digits of ti. t is said to be happy if and only if this sequence is eventually 1, that is for some i, ti = 1.
Thus 7 is happy, as the associated sequence is:
(72) = 49
(42 + 92) = 97
(92 + 72) = 130
(12 + 32) = 10
(12) = 1
A number t is (un)happy iff all members of the above sequence are (un)happy.
The first 20 happy numbers are 1, 7, 10, 13, 19, 23, 28, 31, 32, 44, 49, 68, 70, 79, 82, 86, 91, 94, 97, 100.
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# Relative Velocity Part 3
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##### Author:
Parmanand Jagnandan (386)
##### Objective:
This is the 3rd packet in a series of 4 packets aimed at describing relative velocity. In this packet I review how to determine the relative velocities of two objects moving in opposite directions.
(more)
## Prerequisites
The student should go over Relative Velocity Part 1 and Relative Velocity Part 2 before beginning this packet.
## Relative Velocity Part 2
In this slideshow I provide some background information for the video, "Objects Moving In Opposite Directions."
## Objects Moving In Opposite Directions
In this video I go over an in-depth solution on how to calculate the relative velocities of two objects moving in opposite directions.
SOPHIA has reviewed the tutorial and found it academically sound.
"Well Done."
"Again, great worked-out example."
Angie Eilers (450) - about over 2 years ago
"Thorough!"
## Sophia Summer Rewards for College Readiness
### Earn \$250 in grants
for Sophia online college credit courses.
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# Tableau Area Chart Multiple Measures
In this video we walk through how to create a dual axis chart with two measures on one axis and one on the other. 22How to create a combination chart that shows multiple measures as one mark type and another measure as a different mark type.
Tableau Tip Displaying Multiple Disparate Measures On Multiple Rows Data Visualization Multiple Tips
### In this tutorial well see how to combine multiple measure in single chart in Tableau.
Tableau area chart multiple measures. Otherwise check out my first Tableau lesson. Use a separate bar for each dimension. In this article we will show you how to Create Area Chart with an example.
Here we will start by learning some basics about an area chart and then go into a stepwise process of learning how to create an area chart in Tableau. Following are some of the advanced charts in Tableau. As you can see in the previous image when the mark type was changed from line to area the values for each of the dimension members for Sub-Category.
The Total Order Dollars shipped from individual offices effectively showing the market share. For this Tableau Area chart demonstration we are going to use the Sample. How to create a stacked bar chart with multiple measures.
Area Charts with multiple measures – overlay instead of stacked. In the Marks card select Pie from the drop down menu. And while many people are sleeping on the release a few people have already started to realize.
September 1 2017 Rahul Tableau Tips 2. In Tableau area chart tutorial we are going to learn all about an area chart and its use in Tableau. But if youre just getting started with Tableau you might not know about a few more advanced line chart types.
Right click Measure Values or Measure Names on the Marks card and select Edit Filter. In this tableau tutorial video I have. The Show Me feature in Tableau automatically creates an appropriate chart based on the type of measures you drop in the row column and marks section.
Drag Measure Names to Color on the Marks card. In todays tip I will show you an effective way of creating an area chart with two colours. If you are already familiar with Tableau feel free to continue on.
Unselect show header for both the measure in the row axis and your dual area chart is ready. 30Drag Sales and Profit to the Rows shelf. On the Marks card labeled SUM Sales Click Size and then adjust the slider to change the width.
25Issue How to create a pie chart using multiple measures. Drag Measure Values to Size. How to Use Map Layers in Tableau 20204 to Build Next-Level Visualizations Tableaus 20204 will be its boldest release yet.
26Creating 100 Stacked Bar Graph with Multiple Measures. 26 Feb 2020 Question. 12I have gone through numerous resources but none seem to apply to only using 2 measures basically putting the actual data measure turned into a percent within the goal measure.
Drag Measure Names to Color. Combined measures in single chart Tableau. Drag a dimension to Columns.
Creating basic line charts in Tableau is very easy as youll see below. So open a blank worksheet on your Tableau software and create your very first area chart. 28Tableau Area Chart is a Line Chart where the Area between the Axis and the Lines fill with colors.
4This is the third part of a three-part series on Tableau Playbook – Area ChartIn the first part we delved into theoretical knowledgeIn the second part we practice with two advanced area chartsCheck it out in case you missed it. Repeat step 4 on the Marks card labeled SUM Profit. This lesson is a continuation of an earlier lesson.
Tableau will generate a raw discrete. Create individual axes for each measure. Recently I came across a situation where monthly sales where in different columns and the trend has to be displayed in single line chart in Tableau.
Environment Tableau Desktop Answer Option 1. Moreover you can add more features to these simple charts and make them more advanced. Drag SUMSales to Rows.
26 Feb 2020 Last Modified Date. Hold down the Control key Command key on Mac while clicking to multiple select Date Promo and Sales then choose area charts discrete in Show Me. These charts allow you to display two or mo.
Generally in some good tableau data visualization we have seen that people are using a thick line on top of area chart. Blend two measures to share an axis. There are few tips and tricks to create awesome Tableau line charts and this guide goes through everything you need.
For area charts discrete try 1 date 0 or more Dimensions 1 or more Measures. I have the chart working by using a calculated field of Missed Goal Goal-Actual but I am wanting into to look like a 100 stacked chart. As stacked area charts on one of the foundational Show Me options and are extremely easy to create in Tableau this post will focus more on the best applications of this chart type and a trick for using them most effectively.
Just dragging and dropping a few times. 22Click on Show Me and see the request for the discrete area chart. 3 or More Measures.
Drag Region to Columns. How to create a 100 stacked bar chart with measure values on row or column shelf. On Color right-click Measure Names select Filter select the check boxes for the measures to display and then click OK.
Line and Bar Charts If you want to add 3 or more measures to a line chart you need to take a different approach than in regular charts. Area Chart With Two Colours. On the Marks card labeled All set the mark type to Bar in the dropdown menu.
Hello Im creating some area charts and Id like to display the Total Market Order Dollars vs. Add dual axes where there are two independent axes layered in the same pane. Environment Tableau Desktop Resolution.
29Tableau provides a wide range of visualization tools and chart styles. 10There are several different ways to compare multiple measures in a single view. Environment Tableau Desktop Answer The following instructions use the Sample – Superstore data source.
Right-click the second measure on the Rows shelf and select Dual Axis. July 26 2018 Niket Kedia Leave a comment. Hi Everybody out there.
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# Properties
Label 64400bg1 Conductor $64400$ Discriminant $-7043750000$ j-invariant $$-\frac{15185664}{28175}$$ CM no Rank $0$ Torsion structure trivial
# Related objects
Show commands: Magma / Pari/GP / SageMath
## Minimal Weierstrass equation
sage: E = EllipticCurve([0, 0, 0, -325, -4625])
gp: E = ellinit([0, 0, 0, -325, -4625])
magma: E := EllipticCurve([0, 0, 0, -325, -4625]);
$$y^2=x^3-325x-4625$$
trivial
## Integral points
sage: E.integral_points()
magma: IntegralPoints(E);
None
## Invariants
sage: E.conductor().factor() gp: ellglobalred(E)[1] magma: Conductor(E); Conductor: $$64400$$ = $2^{4} \cdot 5^{2} \cdot 7 \cdot 23$ sage: E.discriminant().factor() gp: E.disc magma: Discriminant(E); Discriminant: $-7043750000$ = $-1 \cdot 2^{4} \cdot 5^{8} \cdot 7^{2} \cdot 23$ sage: E.j_invariant().factor() gp: E.j magma: jInvariant(E); j-invariant: $$-\frac{15185664}{28175}$$ = $-1 \cdot 2^{8} \cdot 3^{3} \cdot 5^{-2} \cdot 7^{-2} \cdot 13^{3} \cdot 23^{-1}$ Endomorphism ring: $\Z$ Geometric endomorphism ring: $$\Z$$ (no potential complex multiplication) Sato-Tate group: $\mathrm{SU}(2)$ Faltings height: $0.57957974406777500374134489761\dots$ Stable Faltings height: $-0.45618827233592362003144547616\dots$
## BSD invariants
sage: E.rank() magma: Rank(E); Analytic rank: $0$ sage: E.regulator() magma: Regulator(E); Regulator: $1$ sage: E.period_lattice().omega() gp: E.omega[1] magma: RealPeriod(E); Real period: $0.52973918567390980098839586923\dots$ sage: E.tamagawa_numbers() gp: gr=ellglobalred(E); [[gr[4][i,1],gr[5][i][4]] | i<-[1..#gr[4][,1]]] magma: TamagawaNumbers(E); Tamagawa product: $4$ = $1\cdot2\cdot2\cdot1$ sage: E.torsion_order() gp: elltors(E)[1] magma: Order(TorsionSubgroup(E)); Torsion order: $1$ sage: E.sha().an_numerical() magma: MordellWeilShaInformation(E); Analytic order of Ш: $1$ (exact) sage: r = E.rank(); sage: E.lseries().dokchitser().derivative(1,r)/r.factorial() gp: ar = ellanalyticrank(E); gp: ar[2]/factorial(ar[1]) magma: Lr1 where r,Lr1 := AnalyticRank(E: Precision:=12); Special value: $L(E,1)$ ≈ $2.1189567426956392039535834769164681053$
## Modular invariants
Modular form 64400.2.a.c
sage: E.q_eigenform(20)
gp: xy = elltaniyama(E);
gp: x*deriv(xy[1])/(2*xy[2]+E.a1*xy[1]+E.a3)
magma: ModularForm(E);
$$q - 3q^{3} - q^{7} + 6q^{9} + 6q^{11} + 5q^{13} + 2q^{17} + 2q^{19} + O(q^{20})$$
sage: E.modular_degree() magma: ModularDegree(E); Modular degree: 82944 $\Gamma_0(N)$-optimal: yes Manin constant: 1
## Local data
This elliptic curve is not semistable. There are 4 primes of bad reduction:
sage: E.local_data()
gp: ellglobalred(E)[5]
magma: [LocalInformation(E,p) : p in BadPrimes(E)];
prime Tamagawa number Kodaira symbol Reduction type Root number ord($N$) ord($\Delta$) ord$(j)_{-}$
$2$ $1$ $II$ Additive -1 4 4 0
$5$ $2$ $I_2^{*}$ Additive 1 2 8 2
$7$ $2$ $I_{2}$ Non-split multiplicative 1 1 2 2
$23$ $1$ $I_{1}$ Non-split multiplicative 1 1 1 1
## Galois representations
sage: rho = E.galois_representation();
sage: [rho.image_type(p) for p in rho.non_surjective()]
magma: [GaloisRepresentation(E,p): p in PrimesUpTo(20)];
The $\ell$-adic Galois representation has maximal image $\GL(2,\Z_\ell)$ for all primes $\ell$.
## $p$-adic regulators
sage: [E.padic_regulator(p) for p in primes(5,20) if E.conductor().valuation(p)<2]
All $p$-adic regulators are identically $1$ since the rank is $0$.
## Iwasawa invariants
$p$ Reduction type $\lambda$-invariant(s) $\mu$-invariant(s) 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 add ss add nonsplit ordinary ordinary ordinary ordinary nonsplit ordinary ordinary ordinary ordinary ordinary ordinary - 0,0 - 0 0 0 0 0 0 2 0 0 2 0 0 - 0,0 - 0 0 0 0 0 0 0 0 0 0 0 0
An entry - indicates that the invariants are not computed because the reduction is additive.
## Isogenies
This curve has no rational isogenies. Its isogeny class 64400bg consists of this curve only.
## Growth of torsion in number fields
The number fields $K$ of degree less than 24 such that $E(K)_{\rm tors}$ is strictly larger than $E(\Q)_{\rm tors}$ (which is trivial) are as follows:
$[K:\Q]$ $E(K)_{\rm tors}$ Base change curve $K$ $3$ 3.1.23.1 $$\Z/2\Z$$ Not in database $6$ 6.0.12167.1 $$\Z/2\Z \times \Z/2\Z$$ Not in database $8$ Deg 8 $$\Z/3\Z$$ Not in database $12$ Deg 12 $$\Z/4\Z$$ Not in database
We only show fields where the torsion growth is primitive. For fields not in the database, click on the degree shown to reveal the defining polynomial.
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Cookie Consent by FreePrivacyPolicy.com
Search a number
55555 = 541271
BaseRepresentation
bin1101100100000011
32211012121
431210003
53234210
61105111
7320653
oct154403
984177
1055555
1138815
1228197
131c396
1416363
15116da
hexd903
55555 has 8 divisors (see below), whose sum is σ = 68544. Its totient is φ = 43200.
The previous prime is 55547. The next prime is 55579.
55555 is nontrivially palindromic in base 10.
It is a sphenic number, since it is the product of 3 distinct primes.
It is not a de Polignac number, because 55555 - 23 = 55547 is a prime.
It is a nude number because it is divisible by every one of its digits.
It is an Ulam number.
It is a Duffinian number.
55555 is a modest number, since divided by 555 gives 55 as remainder.
55555 is a lucky number.
55555 is a nontrivial repdigit in base 10.
It is a plaindrome in base 10.
It is a nialpdrome in base 10.
It is a zygodrome in base 10.
It is a congruent number.
It is an unprimeable number.
It is a pernicious number, because its binary representation contains a prime number (7) of ones.
It is an upside-down number.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 70 + ... + 340.
It is an arithmetic number, because the mean of its divisors is an integer number (8568).
255555 is an apocalyptic number.
55555 is a deficient number, since it is larger than the sum of its proper divisors (12989).
55555 is a wasteful number, since it uses less digits than its factorization.
55555 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 317.
The product of its digits is 3125, while the sum is 25.
The square root of 55555 is about 235.7010818813. The cubic root of 55555 is about 38.1570142275.
The spelling of 55555 in words is "fifty-five thousand, five hundred fifty-five".
Divisors: 1 5 41 205 271 1355 11111 55555
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+0
# help algebra
0
122
2
My cupcake recipe makes 12 cupcakes and requires 2 sticks of butter. I can only buy whole sticks of butter. How many whole sticks of butter must I buy to make 200 cupcakes?
Sep 15, 2021
#1
0
So if 2 sticks of butter make 12 cupcakes... and we set the number of cupcakes as c and the number of sticks of butter as b, we get
12c = 2b
dividing each side by 2 we get
6c = 1b
Since we need 200 cupcakes, we divide 200 by 6 and get 33.3, which we round up because we need 200 cupcakes or more
so we multiply the equation 6c = 1b by 34, and get 34 sticks of butter
:)
Sep 15, 2021
#2
0
2 b / 12 * 200 c = 33.33 b ~~ 34 sticks of b utter ( notice how the 'c' cancels out and you are left with b utter)
Sep 15, 2021
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http://www.proz.com/kudoz/spanish_to_portuguese/tech_engineering/566502-cos_cosine.html
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# cos (cosine)
## Portuguese translation: cos
#### Login or register (free and only takes a few minutes) to participate in this question.
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GLOSSARY ENTRY (DERIVED FROM QUESTION BELOW)
Spanish term or phrase: cos (cosine) Portuguese translation: cos Entered by:
17:54 Nov 9, 2003
Spanish to Portuguese translations [Non-PRO]
Tech/Engineering / Technical User Manual
Spanish term or phrase: cos (cosine) I would like to verify the correct translation of this mathematical function (Cos) in Portuguese. It appears in the following sentence: Carga resistiva: 2.300W, 10 A máx., cos()=1
Jonathon
cos Explanation:One "cos" in context: == # Equações trigonométricas sen(kx) = sen a , cos(kx+a) = cos a , tg(kx) = tg a ; ==
Selected response from:
Ana Hermida
Spain
Local time: 11:10
Graded automatically based on peer agreement.4 KudoZ points were awarded for this answer
5 +3co-seno OR cossenorhandler
4 +3cos
Ana Hermida
9 mins confidence: peer agreement (net): +3
cos
Explanation:
One "cos" in context:
==
# Equações trigonométricas sen(kx) = sen a , cos(kx+a) = cos a , tg(kx) = tg a ;
==
Reference: http://www.prof2000.pt/users/folhalcino/estudar/objectiv/obj...
Ana HermidaSpainLocal time: 11:10Native speaker of: Spanish, GalicianPRO pts in pair: 65
Graded automatically based on peer agreement.
agree Marcelo Fogaccia: Dentro da equação: cos. Nome da função: co-seno ou cosseno.
19 hrs
-> Pois, mas no contexto da pergunta é dentro da equação. Obrigada!
agree
21 hrs
agree
5 days
28 mins confidence: peer agreement (net): +3
co-seno OR cosseno
Explanation:
Both forms are acceptable: co-seno and cosseno. The acronym is the same as in Spanish or English. Here's how it's defined in the Aurelio dictionary:
[De co-2 + seno.]
S. m.
1. Trig. Função de um ângulo orientado, definida como o quociente da abscissa da extremidade dum arco de circunferência subtendido por esse ângulo pelo raio da circunferência.
2. Mat. Função periódica de uma variável, igual a um, quando a variável é zero, e cuja derivada segunda lhe é simétrica [símb.: cos ] .
u Co-seno hiperbólico. Mat.
1. Função transcendente igual à média aritmética entre duas exponências com os expoentes simétricos.
u Co-seno hiperbólico inverso. Mat.
1. Arco co-seno hiperbólico.
u Co-seno inverso. Mat.
1. Arco co-seno.
rhandlerLocal time: 06:10Native speaker of: PortuguesePRO pts in pair: 1906
agree
2 hrs
agree Marcelo Fogaccia: Dentro da equação: cos. Nome da função: co-seno ou cosseno.
19 hrs
-> Obrigado, Marcelo. Na equação, é cosseno de fi (letra grega).
agree
1 day18 hrs
#### KudoZ™ translation helpThe KudoZ network provides a framework for translators and others to assist each other with translations or explanations of terms and short phrases.
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Ukraine
+380 57 7281624
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## Th Card Faro Check
This is an idea of many years that I have used to be sure of cutting at exactly the 26 th card before proceeding into any miracles dependent on it. The idea of peeking a 26th card for a 26th card location, was a Bert Allerton subtlety that won him a prize many years ago for the best card effect at a convention.
The use of the Faro Check is my idea to insure that the 26th card is actually being used thus insuring definite success with the 26th card principle.
1. The 26th Card Faro Check consists merely in cutting the pack at 26, then starting a Faro Shuffle as in Figure 29.
1. The 26th Card Faro Check consists merely in cutting the pack at 26, then starting a Faro Shuffle as in Figure 29.
2. Remember, you only start the Faro Shuffle. If the cut and weave has been perfect, every card will be weaved with no cards left over, thus you will be sure that the cards have been cut at 26.
3. Spot the bottom card of the right hand packet and remember it.
4. As if changing you mind about a shuffle, merely pull the halves away from each other thus undoing the weave.
5. From here you can replace the upper half back on top and thus you now know the 26th card. You can also give each half to the two spectators for the Automatic Placement, later in this chapter, where both cards will end up 13th in each packet of 26.
6. Use of the 26th Card Faro Check will be evident in such effects as Faro Foolers in Chapter Seven, Faro Notes.
7. Psychologically, the 26th Card Faro Check is quite sound as it has all the elements of an attempted Faro Shuffle that did not quite come off. Your whole manner, upon completion of the 26th Card Faro Check, should be one of, "Oh, well, let's try something else."
### 4th Finger Table
I consider this an Faro aid. Not only does it eliminate any possible flare out of the bottom cards during the starting of the weave, but also steadies the packets as well as being a freat aid in such Faros as the Partial aro Check, Above the Crimp Faro, Off Center Faro, etc.. It is one that I constantly use in connection with the First Technique for Faro described above.
1. The pack is held in the same position as for the First Technique except that the left fourth finger moves in under the cards until it comes to rest directly beneath the joined corners of the pack.
2. The joined corners actually rest on the back or nail of the left fourth finger. The picture in Figure 30 shows how it looks from the operator's view.
Figure 30
3. Slightly moving the left fourth finger either upwards or downwards allows a better control of the start of In or Out Faros.
4. Once the Faro is started, the left fourth finger moves out of the way to join the others alongside the pack.
5. For the present, this will suffice but its usefulness will become more appa-rant further on in this work. For the present, let us now delve into the technique of the -
### Faro Riffle Shuffle
Here again the techniques are strictly our own. These Faro Riffle Shuffles can be done on any type of surface which | in itself is a great step forward.
1. Hold the pack between both hands in a manner exactly to that of the First Technique.
2. Place the deck's lower end against the table thus also squaring them.
3. The right thumb breaks the side of the pack in a manner similar to Figure 2 except here the pack is upright on the table with the operator looking down at the top end.
4. Having split the side of the deck that is towards the body, you separate the halves like a book; i.e. at the back side only. The front side stays together.
5. At this stage the cards are held momentarily for comparison of one half against the other. This comparison is
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##### Finding the initial number
Algebra Tutor: None Selected Time limit: 1 Day
Carli spent two-thirds of her money and then spent \$4 more. Then she spent half of her remaining money. It cost her \$1 for the bus ride home. She then had \$5 left over. How much money did she start with?
May 14th, 2015
Go from back to front:
She had \$5 left, add her \$1 from the bus journey: \$6.
If she spent half of her remaining money, that means what she had before, the value x, satisfies:
(1/2) * x = \$6
Multiplying each side by 2 gives: x = \$12.
Then add the \$4 more she spent: \$16.
Then the money she started with, the value y, satisfies:
(1/3) * y = \$16: (she spent two thirds of her money so she only has a third of the original value left)
Multiplying each side by 3 gives: y = 16 * 3 = \$48.
Therefore, Carli started with \$48.
May 14th, 2015
...
May 14th, 2015
...
May 14th, 2015
Dec 4th, 2016
check_circle
Mark as Final Answer
check_circle
Unmark as Final Answer
check_circle
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+0
# geometry question
0
34
2
A rectangle is inscribed in a circle of radius 5 cm. If the area of the rectangle is equal to 1/3 the area of the circle, what is the perimeter of the rectangle? Express your answer as a decimal to the nearest tenth.
Jul 15, 2022
#1
+124525
+1
Call the sides of the rectangle a, b
The diagonal of the rectangle = 2r = 10
And
a^2 + b^2 = 100
The area of the circle = 25pi
So.....the area of the rectangle = (25/3) pi = ab → b = (25pi) / (3a) ...b^2 = 625 pi^2 / (9a^2)
So
a^2 + 625pi^2 / (9a^2) = 100 multiply through by a^2 and rearrange
a^4 - 100a^2 + (625/9)pi^2 = 0
a^4 -100a^2 = - (625pi^2)/9 complete the square on a
a^4 -100a^2 + 2500 = -(625/9) pi^2 + 2500
(a^2 - 50)^2 = - (625/9)pi^2 + 2500 take the positive root
a^2 - 50 = sqrt [- (625/9)pi^2 + 2500 ] ≈ 42.6 cm
a^2 = 42.6 + 50
a^2 ≈ 92.6
a = sqrt (92.6) ≈ 9.6 cm
b = sqrt ( 100 - 92.6) ≈ 2.7 cm
Perimeter of rectangle = 2 ( a + b) = 2 ( 9.6 + 2.7) ≈ 24.6 cm
Jul 15, 2022
#2
+1105
+7
Here's my go at this!
Area of circle: $$\pi r^2$$= 6.25pi
Rectangle area = 1/3 of 6.25pi, right? So, here's an equation. Let's call the height of the rectangle x.
$$5x=6.25\pi$$
Just solve for x!
Jul 15, 2022
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https://socratic.org/questions/how-do-i-find-the-vertical-component-of-a-vector
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# How do I find the vertical component of a vector?
Aug 19, 2014
The formula for the vertical component of a vector ai + bj is as follows:
v_y = ||A|| sin(θ)
First, calculate the magnitude of the vector A which is $| | A | |$:
||A|| = $\sqrt{{a}^{2} + {b}^{2}}$
Next, determine $\theta$
If you draw a triangle where a is the x axis and b is the y axis, you get a right triangle. The angle $\theta$ has the following measurement below:
$\tan \left(\theta\right) = \frac{b}{a}$
$\theta = a r t c a n \left(\frac{b}{a}\right)$
Finally, we have the vertical component formula:
v_y = ||A|| sin(θ)
For calculator assistance, use the component button here:
http://www.mathcelebrity.com/vector.php
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# Bodies of constant width?
In two-dimensional case one can generalize figures of constant width as figures which can rotate in a covex polygon. Here is one example which can be used to drill triangular holes:
I would like to know what happens with this generalization in dimension $3$ and maybe higher. Obviously body of constant width $1$ can rotate arbitrary in a unit cube. More formally, given a body $B$ of constant width $1$ and $A\in SO(3)$ there is $v\in \mathbb R^3$ such that $$A(B)+v\subset\square,$$ where $\square$ is unit cube. On the other hand, except for the cube, I do not see any other examples of convex polyhedron which have nontrivial rotating bodies (i.e. distinct from the inscribed ball).
I hope that the answer is known. (= I hope I should wait for the answer and I do not have to think.)
The question is inpired by this one: "Local minimum from directional derivatives in the space of convex bodies."
-
These bodies are called "rotors" of the corresponding cavity. A full classification of non-trivial (i.e. other than bodies of constant width) rotors is known. A good reference on these is "Geometric Applications of Fourier Series and Spherical Harmonics" by Helmut Groemer. As I recall, only the regular simplex and cross-polytope admit non-trivial rotors in dimensions four and above. In three-dimensions, there are more cases. – Yoav Kallus Jan 26 '12 at 18:10
@Yoav, why don't you write it as an answer? – Anton Petrunin Jan 26 '12 at 18:57
Sorry, I'm still getting used to how things work here on MO. Thanks for the MOtiquette pointer. – Yoav Kallus Jan 26 '12 at 19:30
I found the reference I was looking for. The full list of cases under which $K$ is a rotor in a cavity shaped like the polytope $P$ is available on page 27 of the notes title "The use of spherical harmonics in convex geometry" by Rolf Schneider. They are available under "Course Materials" on his website. As I recall, there is one more non-trivial case in $d=3$ if the cavity is allowed to be open (e.g. a cone), and this case appears in the more complete list in Groemer's book.
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Home > Standard Deviation > Error Ratio Formula
# Error Ratio Formula
## Contents
more... Later Messance (~1765) and Moheau (1778) published very carefully prepared estimates for France based on enumeration of population in certain districts and on the count of births, deaths and marriages as Sorry for the confusion, but changing D.Hinkley formulas would be a bit too much. –Severin Pappadeux Feb 10 at 17:33 | show 1 more comment up vote 3 down vote With Springer, §7.3.1 (iii) ^ Pearson K (1897) On a form of spurious correlation that may arise when indices are used for the measurement of organs. http://stylescoop.net/standard-deviation/rsd-calculation-formula.html
Is it unethical of me and can I get in trouble if a professor passes me based on an oral exam without attending class? A correction of the bias accurate to the first order is[3] r c o r r = r − s [ y / x ] x m x {\displaystyle r_{\mathrm {corr} error t1* 1.520313 0.04465751 0.2137274 PS: I thing this issue has come up earlier in this forum [FONT=Fixedsys] Visit : [/FONT][URL="http://www.statsravingmad.com/"]Site[/URL]|[URL="http://www.statsravingmad.com/blog"]Stats Blog[/URL]|[URL="http://parzakonis.info/"]Personal mini-site[/URL] [FONT=Fixedsys]Contact : [/FONT][URL="http://www.statsravingmad.com/contact/"]Contact[/URL] Reply With Quote + Reply The same procedure for the same desired sample size is carried out with the y variate.
## Error Ratio Formula
share|improve this answer edited Feb 6 '13 at 23:15 answered Feb 6 '13 at 16:14 Glen_b♦ 151k20250519 Sven Hohenstein: I know how X and Y are distributed. –Lucy Feb In other words Var(Y/X)/(m(Y/X))^2 = Var(X)/m(X)^2 + Var(Y)/m(Y)^2 or rearranging sd(Y/X) = sqrt[ Var(X)*m(Y/X)^2/m(X)^2 + Var(Y)*m(Y/X)^2/m(Y)^2 ] For random variables with the mean well away from zero, this is a reasonable I don't think it gets confused, it honestly integrates symmetric over 0 term 1/(1+x^2) multiplied by antisymmetric value of x. Related 3How to compute standard deviation of difference between two data sets?3Sum standard deviation vs standard error0Standard Deviation of percentages0Calculating standard deviation associated with percentage change0Calculating Standard deviation of percentages?0Averaging standard
asked 3 years ago viewed 21309 times active 3 years ago Get the weekly newsletter! Can you help me to find out the book source of this? How do I Turbo Boost in Macbook Pro Why does Fleur say "zey, ze" instead of "they, the" in Harry Potter? Ratio Of Two Standard Deviations Very helpful here. Oct 14, 2015 Diogenes de Souza Bido · Universidade Presbiteriana Mackenzie from Pearson (1897): http://www.ssc.wisc.edu/cde/cdewp/72-18.pdf Apr 17, 2016 Can you help by adding an answer?
The variance of the ratio using these methods differs from the estimates given previously. Variance Of A Ratio I found the relative variance of R in Cochran(1977), i.e., V(R)/R^2, at the bottom of page 183. He references Hansen, Hurwitz, and Madow. Cochran, W.G(1977), Sampling Techniques, 3rd ed., John Wiley & Sons. That section goes on to talk about If I am told a hard percentage and don't get it, should I look elsewhere? More questions If I have two mean values with their standard errors, how do i calculate the standard error of their quotient?
Uses Although the ratio estimator may be of use in a number of settings it is of particular use in two cases: when the variates x and y are highly correlated Normal Ratio Cauchy Random noise based on seed Is the ability to finish a wizard early a good idea? NCBISkip to main contentSkip to navigationResourcesHow ToAbout NCBI AccesskeysMy NCBISign in to NCBISign Out PMC US National Library of Medicine National Institutes of Health Search databasePMCAll DatabasesAssemblyBioProjectBioSampleBioSystemsBooksClinVarCloneConserved DomainsdbGaPdbVarESTGeneGenomeGEO DataSetsGEO ProfilesGSSGTRHomoloGeneMedGenMeSHNCBI Web The precise relationship between the variances depends on the linearity of the relationship between the x and y variates: when the relationship is other than linear the ratio estimate may have
## Variance Of A Ratio
On a method of bias reduction in ratio estimation. Hot Network Questions Secret of the universe Stainless Steel Fasteners Pythagorean Triple Sequence Should non-native speakers get extra time to compose exam answers? Error Ratio Formula I am aware that there is limits to what I am doing, but it's exactly those limits I wish to investigate. Standard Deviation Of A Ratio Of Independent Variables Using the standard noninformative prior for the Gaussian linear model, we do not need MCMC techniques and we probably obtain a good frequentist-matching property: a $95\%$-posterior credibility interval is approximately a
I have a question, in line 50 and 51 of your code you write print(mean(x/y)) and print(sd(x/y)), I can't really tell if x/y is assumed elsewhere in your code, but what http://stylescoop.net/standard-deviation/pert-standard-deviation-formula.html These versions differ only in the factor in the denominator ( N - 1 ). What is your larger goal here? Expand» Details Details Existing questions More Tell us some more Upload in Progress Upload failed. Standard Deviation Of A Ratio Excel
I hope this would be helpful. A second-order correction is[4] r c o r r = r [ 1 + 1 n ( 1 m x − s x y m x m y ) + 1 I strongly suspect that your statement ("doesn't have a SD") is only true sometimes (e.g. Source The Midzuno-Sen technique described below is recommended instead.
DDoS: Why not block originating IP addresses? Ratio Of Standard Deviation Calculator Ordinary least squares regression If a linear relationship between the x and y variates exists and the regression equation passes through the origin then the estimated variance of the regression equation Join them; it only takes a minute: Sign up Estimating the Standard Deviation of a ratio using Taylor expansion up vote 4 down vote favorite 1 I am interested to build
## I'll see if I can dig up a reference; there's sure to be something in a standard old reference like Cox and Hinkley or Kendall and Stuart or Feller or something
If I knew it, what would be the way to calculate the standard error? You can only upload videos smaller than 600MB. Why don't C++ compilers optimize this conditional boolean assignment as an unconditional assignment? Ratio Of Standard Deviation To Mean If the sample size is decreased, what is the effect on the standard error?
Nature 174: 270–271 ^ a b c Ogliore RC, Huss GR, Nagashima K (2011) Ratio estimation in SIMS analysis. share|improve this answer edited Jan 16 '12 at 16:39 answered Jan 16 '12 at 15:47 jbowman 13.9k12859 add a comment| up vote 2 down vote With the Bayesian approach it is e.g. http://stylescoop.net/standard-deviation/standard-error-anova-formula.html If I have two means, each with a standard error (same units), is there a simple way to approximate the standard error of the ratio?
asked 8 months ago viewed 578 times active 8 months ago Visit Chat Related 0Estimating Parameters in R from a Weighted Sum of Lagged Variables0R - functions of an estimated vector I think the mean and SD do exist under some cases. –Ben Bolker Feb 9 at 15:57 1 @BenBolker it gets confused and reports a mean of 0. Sign up today to join our community of over 11+ million scientific professionals. Can this formula be applied for my case in which (x=100*(y-z)/z)? –Lucy Feb 6 '13 at 16:47 see the edit to my answer –Glen_b♦ Feb 6 '13 at 23:09
You could sample more and more values, but from sampling std.dev is growing as well, as noted by @G.Grothendieck library(ggplot2) m.x <- 5; s.x <- 4 m.y <- 4; s.y <- La base de un prisma es un rombo cuyas diagonales miden 70cm y 40cm. You can only upload files of type 3GP, 3GPP, MP4, MOV, AVI, MPG, MPEG, or RM. Choose i at random from a uniform distribution on [1,N].
Nuclear Instruments and Methods in Physics Research Section B: Beam Interactions with Materials and Atoms 269 (17) 1910–1918 ^ Pascual JN (1961) Unbiased ratio estimators in stratified sampling. Browse other questions tagged r estimation taylor-series or ask your own question. How on earth do i find the zeroes of (x^3-2x^2+1)/(x^2+3)??? So I have $((\bar Y-\bar X)/\bar X)*100$.
Midzuno-Sen's method In 1952 Midzuno and Sen independently described a sampling scheme that provides an unbiased estimator of the ratio.[16][17] The first sample is chosen with probability proportional to the size This won't give a meaningful answer in the situation posed in the question but might be OK in certain situations. –G. m2 <- integrate(function(x) x*x*PDF(x), -Inf, Inf) Thus, it is impossible to test any Taylor expansion for std.dev. To find an error, you need to see that is the maximum deviation from the base value what both x and y change: d(x/y) = dx*1/y - dy*x/y^2 = (dx/x)*(x/y) -
If I have two independent variables (say X and Y) with their respective variances (or standard deviation), how could I calculate the mean and variance of the resultant division of variable Is it good to call someone "Nerd"? more stack exchange communities company blog Stack Exchange Inbox Reputation and Badges sign up log in tour help Tour Start here for a quick overview of the site Help Center Detailed Forum Normal Table StatsBlogs How To Post LaTex TS Papers FAQ Forum Actions Mark Forums Read Quick Links View Forum Leaders Experience What's New?
In it, you'll get: The week's top questions and answers Important community announcements Questions that need answers see an example newsletter By subscribing, you agree to the privacy policy and terms Thanks! –Eric Fail Feb 8 at 22:38 I don't think a Taylor series approximation is going to be useful here. (1) The s.d.
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```----------------------------------------------------------------------------
-- |
-- Module : Data.Group
-- Copyright : 2007-2009 Edward Kmett
-- License : BSD
--
-- Maintainer : Edward Kmett <ekmett@gmail.com>
-- Stability : experimental
-- Portability : portable
--
-- Extends 'Monoid' to support 'Group' operations
-----------------------------------------------------------------------------
module Data.Group
( Group
, gnegate
, gsubtract
, minus
, MultiplicativeGroup
, over
, under
, grecip
) where
import Data.Monoid (Monoid, Sum(..), Product(..), Dual(..))
import Data.Monoid.Additive (plus, zero)
import Data.Monoid.Multiplicative (Multiplicative, one, times, Log(..), Exp(..))
import Data.Monoid.Self (Self(Self,getSelf))
infixl 6 `minus`
-- | Minimal complete definition: 'gnegate' or 'minus'
class Monoid a => Group a where
-- additive inverse
gnegate :: a -> a
minus :: a -> a -> a
gsubtract :: a -> a -> a
gnegate = minus zero
a `minus` b = a `plus` gnegate b
a `gsubtract` b = gnegate a `plus` b
instance Num a => Group (Sum a) where
gnegate = Sum . negate . getSum
Sum a `minus` Sum b = Sum (a - b)
instance Fractional a => Group (Product a) where
gnegate = Product . negate . getProduct
Product a `minus` Product b = Product (a / b)
instance Group a => Group (Dual a) where
gnegate = Dual . gnegate . getDual
instance Group a => Group (Self a) where
gnegate = Self . gnegate . getSelf
Self a `minus` Self b = Self (a `minus` b)
-- | Minimal definition over or grecip
class Multiplicative g => MultiplicativeGroup g where
-- | @x / y@
over :: g -> g -> g
-- | @x \ y@
under :: g -> g -> g
grecip :: g -> g
x `under` y = grecip x `times` y
x `over` y = x `times` grecip y
grecip x = one `over` x
instance MultiplicativeGroup g => Group (Log g) where
Log x `minus` Log y = Log (x `over` y)
Log x `gsubtract` Log y = Log (x `under` y)
gnegate (Log x) = Log (grecip x)
instance Group g => MultiplicativeGroup (Exp g) where
Exp x `over` Exp y = Exp (x `minus` y)
Exp x `under` Exp y = Exp (x `gsubtract` y)
grecip (Exp x) = Exp (gnegate x)
instance MultiplicativeGroup g => MultiplicativeGroup (Self g) where
Self x `over` Self y = Self (x `over` y)
Self x `under` Self y = Self (x `under` y)
grecip (Self x) = Self (grecip x)
instance MultiplicativeGroup a => MultiplicativeGroup (Dual a) where
grecip = Dual . grecip . getDual
```
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# Daily Archives: January 1, 2015
## Find min sublist
This problem is given by Junmin Liu. I dicussed with him, and thought about this for 2 days, and finally came up with this the O(n) time, O(1) soluiton.
Problem:
Give a list of unsorted number, find the min window or min sublist of the list, such as if sublist is sorted, the whole list is sorted too. e.g. input 123 5433 789, output 5433, basically when 5433 is sorted, then whole list is sorted.
Idea 1: use max/min stack
use 2 stacks, which can indicate min/max value. Push all elements from right-to-left min stack. And then pop from the top. When it encounters an element, where the element is larger than current min value in stack. Stop. This is the left boundary. The right boundary is similar to it, which use a max stack. Both time and space take O(n).
Idea 2: Solve by finding a certain boundary.
For arr[0,…,n-1] try to find the first element which is unsorted from left side, and first element which is unsorted from right side. Besides, we need to make sure the right side element is always larger than the left side.
So we make it into: arr[0,…,i,i+1,…,j-1,j,…,n-1].
arr[0,…,i] is ascend, and arr[i]>arr[i+1].
arr[j,…,n-1] is ascend, arr[j-1]>a[j].
And arr[i] < arr[j].
In my code, I call the i left_bound, and j right_bound. The arr[i+1,…,j-1] is the unsorted array that we met. But it is not the final result. Because inside of arr[i+1,…j-1], there could be value falling in arr[0,…,i], and value falling in arr[j,…,n-1]. We find the min and max in arr[i+1,…,j-1].
Then we use binary search find the min position in arr[0,…,min,…,i] and max position in arr[j,…,max,…,n-1]. The arr[min,…,max] list is the result. Time is O(n),space is O(1).
Below is my code. Because the boundary condition for this binary search is kinda complex, so used normal comparison. But it still gives O(n) time.
1. public class MinSublist {
2. public static void main(String[] args) {
3. int[] arr1 = {1,2,3,4,5,4,2,3,7,8,6,9};
4. int[] arr2 = {1,2,3,4,5,8,2,3,7,9,8};
5. int[] arr3 = {1,5,2,6,5,4,3,8,6,7,8};
6. int[] arr4 = {1,5,2,6,5,4,3,8,4,6,5,8};
7. int[] arr5 = {1,2,3,5,8,2,2,3,2,7,9};
8. int[] arr6 = {1,2,3,5,4,3,3,7,8,9};
9. int[] arr7 = {1,2,3,5,4,3,3,7,8,9,2};
10. int[] arr = arr7;
11. findMinSublist(arr);
12. }
13. public static void findMinSublist(int[] arr){
14. Bound bound = findBound(arr);
15. printArray(arr, 0, arr.length-1);
16. MaxMinPos maxMin = findMaxMinPos(arr, bound.left_bound+1, bound.right_bound-1);
17. int left = getLeftPos(arr, arr[maxMin.min], bound.left_bound);
18. int right = getRightPos(arr, arr[maxMin.max], bound.right_bound);
19. System.out.println(“(“+left+”,”+right+”)”);
20. }
21. /*
22. * This function return a left_bound, right_bound of the arr[], where arr[0,…,left_bound], arr[right_bound,…,len-1] are ascend.
23. * And arr[left_bound] is less than arr[right_bound].
24. * The idea is to find the right value in left and right side to compare. If left_compare is larger than right_compare, we found the boundes, and stop.
25. * If left_bound is found, then we should always use arr[left_bound] to compare.
26. * If right_bound is found, then we should always use arr[right_bound] to compare.
27. * This function should pass the below cases:
28. * int[] arr1 = {1,2,3,4,5,4,2,3,7,8,6,9};
29. * int[] arr2 = {1,2,3,4,5,8,2,3,7,8,6,9};
30. * int[] arr3 = {1,5,2,6,5,4,3,8,6,7,8,9};
31. * int[] arr4 = {1,5,2,6,5,4,3,8,4,6,7,8,9};
32. * int[] arr5 = {1,2,3,5,8,2,7,6,5,2,3,4,7,9};
33. */
34. public static Bound findBound(int[] arr){
35. if(arr==null||arr.length<2){
36. return null;
37. }
38. int left_bound = 0, right_bound = arr.length-1;
39. boolean found_left = false, found_right = false;
40. while(left_bound
41. int left_compare, right_compare;
42. if(!found_left){
43. if(arr[left_bound+1]
44. found_left = true;
45. left_compare = arr[left_bound];
46. }
47. else{
48. left_compare = arr[left_bound+1]; //if not, current comparison is arr[left_bound+1]
49. }
50. }
51. else{
52. left_compare = arr[left_bound]; //already found, left_compare is the left_bound value.
53. }
54. if(!found_right){
55. if(arr[right_bound-1]>arr[right_bound]){ //if right_bound is found
56. found_right = true;
57. right_compare = arr[right_bound];
58. }
59. else{
60. right_compare = arr[right_bound-1]; //if not, current comparison is arr[right_bound-1]
61. }
62. }
63. else{
64. right_compare = arr[right_bound]; //already found, right_compare is the right_bound value.
65. }
66. if(found_left&&found_right||left_compare>right_compare){ //check if left, right bound are both found.
67. break;
68. }
69. if(!found_left){
71. }
72. if(!found_right){
74. }
75. }
76. return new Bound(left_bound, right_bound);
77. }
78. /*
79. * Find the max, min pos in arr[]
80. */
81. public static MaxMinPos findMaxMinPos(int[] arr, int start, int end){
82. int max_pos = arr[start], min_pos = arr[start];
83. for(int i=start;i<=end;i++){
84. if(arr[i]
85. min_pos = i;
86. }
87. if(arr[i]>arr[max_pos]){
88. max_pos = i;
89. }
90. }
91. return new MaxMinPos(max_pos, min_pos);
92. }
93. /*
94. * Get the left position for final result.
95. */
96. public static int getLeftPos(int[] arr, int value, int end){
97. int i;
98. for(i=0;i<=end;i++){
99. if(arr[i]>value){
100. break;
101. }
102. }
103. return i;
104. }
105. /*
106. * Get the right position for final result.
107. */
108. public static int getRightPos(int[] arr, int value, int start){
109. int i;
110. for(i=arr.length-1;i>=start;i–){
111. if(arr[i]
112. break;
113. }
114. }
115. return i;
116. }
117. public static void printArray(int[] arr, int start, int end){
118. for(int i=start;i<=end;i++){
119. System.out.print(arr[i]);
120. }
121. System.out.println();
122. }
123. public static void printArray2(int[] arr, int left, int right){
124. for(int i=0;i<=left;i++){
125. System.out.print(arr[i]);
126. }
127. System.out.print(” “);
128. for(int i=left+1;i<=right-1;i++){
129. System.out.print(arr[i]);
130. }
131. System.out.print(” “);
132. for(int i=right;i<=arr.length-1;i++){
133. System.out.print(arr[i]);
134. }
135. System.out.println();
136. }
137. }
138. class Bound{
139. int left_bound;
140. int right_bound;
141. public Bound(int left_bound, int right_bound){
142. this.left_bound = left_bound;
143. this.right_bound = right_bound;
144. }
145. }
146. class MaxMinPos{
147. int max;
148. int min;
149. public MaxMinPos(int max, int min){
150. this.max = max;
151. this.min = min;
152. }
153. public String toString(){
154. return “(“+max+”,”+min+”)”;
155. }
156. }
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https://www.mathvids.com/browse/college/calculus-ii/sequences-and-series/sequences-and-series/440-sequences-ii
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# Sequences II
Taught by Houston
• Currently 4.0/5 Stars.
3302 views | 2 ratings
Part of video series
Meets NCTM Standards:
Lesson Description:
Precise definition of the limit of a sequence. Monotonicity and boundedness; convergence of bounded, monotonic sequences. Recursively defined sequences, fixed points, and web plots.
Copyright 2005, Department of Mathematics, University of Houston. Created by Selwyn Hollis. Find more information on videos, resources, and lessons at http://online.math.uh.edu/HoustonACT/videocalculus/index.html.
Questions answered by this video:
• What is the epsilon definition for the limit of a sequence?
• What does it mean for a sequence to be bounded?
• What is monotonicity?
• What does it mean for a sequence to be decreasing, increasing, nondecreasing, nonincreasing, strictly decreasing, and strictly increasing?
• What does it mean for a sequence to be monotonic?
• What does it mean for a sequence to be eventually monotonic?
• What is a recursively generated sequence?
• What is a fixed point of a sequence?
• What is a web plot of a sequence?
• How do you find the intersection between y = x and a sequence using a web plot?
• How can you tell if a sequence converges or diverges using a web plot?
• What is alternating convergence?
• How do you find the limit of a sequence using a web plot?
• #### Staff Review
• Currently 4.0/5 Stars.
This video explains limits of sequences and shows that a limit is equal to a number in a way that is similar to what is done in an undergraduate Analysis course. Boundedness and monotonicity are also explained and shown with examples. Some really great examples are shown and explained. Web plots are shown with examples also.
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# 10++ Graphing Quadratic Equations Worksheet
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### Now this is Interesting Browsers Puzzle – Brainteaser Math Puzzles Image
Now this is Interesting Browsers Puzzle - Brainteaser Math Puzzles Image - http://picsdownloadz.com/puzzles/now-this-is-interesting-browsers-puzzle-brainteaser-math-puzzles-image/
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Solve this Simple Math Quiz- Cool Math Puzzles that can trick your mind.
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## Nickadamsinamerica Just gotta have Worksheet.
By . Worksheet. At Saturday, October 16th 2021, 03:01:00 AM.
These addition worksheets will produce 12 vertical or horizontal addition problems using dot figures to represent the numbers. You may select the numbers for the addition worksheets to be used from 0 to 10.
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### All inequalities on this worksheet have addition or subtraction on the side of the inequality as the variable.
Solving one step inequalities worksheet answer key pdf. Solving And Graphing Inequalities Worksheet Answer Key Pdf Algebra 2. On these problems students need to isolate the variable using only a single step. 66 6 x 4 3.
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Solving one step inequalities worksheet answer key pdf. R o 3AwlvlR 1r Si Ogdh WtasW Crge Ns1evruvce nd w6 X 5M ma cdhe M 6w Ei Jt jh a eIWn7f3iQn ZiGt ne7 tP Fr1eZ-GATlmgIedbtr 0ai. Two step inequalities date period solve each inequality and graph its solution.
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27 solving and graphing inequalities worksheet answer key pdf rather than inserting the exact same text modifying font styles or adjusting margins every time you begin a new document opening a personalized template can let you get straight to work on the material. Solve inequality and match to the graph. Answers to One-Step Inequalities ID.
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Solving and graphing inequalities we know 4 is greater than 3 so is 5 so is 6 so is 7 and 31 works also. 8 Worksheet by Kuta Software LLC Kuta Software - Infinite Pre-Algebra Name_____ Solving One-Step Inequalities by AddingSubtracting Date_____ Period____. Use inverse operations or mental math to solve for x.
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Dividing by a negative means switch the sign. 6 5A HlQlZ yr BiEg 1hZtqsz vrneTs2e0rxvre fd LX y MMard be l fw si rt Phh DIwnUflicndiDtAej YAmligKeybUr9a G J1 vn Worksheet by Kuta Software LLC 13 b 6 4 32 30 28 26 24 22. 1 A cMOaMdse e pw pi etHhY 7I xnEf ticn1iNtTeV yP dr 9eO- LANl Jg8e FbKr OaHh Worksheet by Kuta Software LLC.
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The three-in-a-tree problem
The three-in-a-tree problem Maria Chudnovsky1 and Paul Seymour2 Princeton University, Princeton NJ 08544 May 10, 2006; revised July 21, 2009 1 2 Thi...
Author: Bruce Payne
The three-in-a-tree problem Maria Chudnovsky1 and Paul Seymour2 Princeton University, Princeton NJ 08544 May 10, 2006; revised July 21, 2009
1 2
This research was conducted while the author served as a Clay Mathematics Institute Research Fellow. Supported by ONR grant N00014-01-1-0608, and NSF grant DMS-0070912.
Abstract We show that there is a polynomial time algorithm that, given three vertices of a graph, tests whether there is an induced subgraph that is a tree, containing the three vertices. (Indeed, there is an explicit construction of the cases when there is no such tree.) As a consequence, we show that there is a polynomial time algorithm to test whether a graph contains a “theta” as an induced subgraph (this was an open question of interest) and an alternative way to test whether a graph contains a “pyramid” (a fundamental step in checking whether a graph is perfect).
1
Introduction
All graphs in this paper are finite and simple. If G is a graph, its vertex- and edge-sets are denoted V (G), E(G). If X ⊆ V (G), the subgraph with vertex set X and edge set all edges of G with both ends in X is denoted G|X, and called the subgraph induced on X. There are many algorithmic questions of interest concerning the existence of an induced subgraph of some specific type containing some specified vertices, but almost all of them seem to be NPcomplete, by virtue of the following result of Bienstock [1]: 1.1 The following problem is NP-complete: Input: A graph G and two edges e, f of G. Question: Is there a subset X ⊆ V (G) such that G|X is a cycle containing e, f ? Bienstock’s result leaves very little room between the trivial problems and the NP-complete problems, but in this paper we report on a problem that falls into the gap. We call the following the “three-in-a-tree” problem: Input: A graph G, and three vertices v1 , v2 , v3 of G. Question: Does there exist X ⊆ V (G) with v1 , v2 , v3 ∈ X such that G|X is a tree? For most graphs one would expect a “yes” answer, but there are interesting graphs for which the answer is “no”; for instance, if e1 , e2 , e3 are edges of a graph H each incident with a vertex of degree one, and G is the line graph of H, then e1 , e2 , e3 are vertices of G and there is no induced tree in G containing e1 , e2 , e3 . Nevertheless, we will show that the three-in-a-tree problem can be solved in time O(|V (G)|4 ). We shall give an explicit construction of all instances (G, v1 , v2 , v3 ) such that the desired tree does not exist, and the proof that all such instances must fall under this construction can be converted to an algorithm to check whether the desired tree exists or not.
2
Thetas, pyramids and prisms
We were led to the three-in-a-tree problem while working on the question of deciding if a graph contains a theta, so let us describe that. First we need some definitions. If G, H are graphs, and H is isomorphic to G|X for some X ⊆ V (G), we say that G contains H as an induced subgraph. A path is a graph P whose vertex set and edge set can be labeled as V (P ) = {v1 , . . . , vk } and E(P ) = {e1 , . . . , ek−1 } for some k ≥ 1, such that ei is incident with vi , vi+1 for 1 ≤ i ≤ k − 1. A cycle is a graph C with V (C) = {v1 , . . . , vk } and E(C) = {e1 , . . . , ek } for some k ≥ 3, such that ei is incident with vi , vi+1 for 1 ≤ i ≤ k − 1, and ek is incident with v1 , vk . The length of a path or cycle is the number of edges in it, and a path or cycle if odd or even if its length is odd or even respectively. A path or cycle of G means a subgraph (not necessarily induced) of G that is a path or cycle. A hole of G means a cycle in G that is an induced subgraph and has length at least four. A triangle is a set of three pairwise adjacent vertices. Here are three types of graph that will be important to us: • A pyramid is a graph consisting of a vertex a and a triangle {b1 , b2 , b3 }, and three paths P1 , P2 , P3 , such that: Pi is between a and bi for i = 1, 2, 3; for 1 ≤ i < j ≤ 3 Pi , Pj are vertex-disjoint except for a and the subgraph induced on V (Pi ) ∪ V (Pj ) is a cycle; and at most one of P1 , P2 , P3 has only one edge. 1
• A theta is a graph consisting of two nonadjacent vertices a, b and three paths P1 , P2 , P3 , each joining a, b and otherwise vertex-disjoint, such that for 1 ≤ i < j ≤ 3 the subgraph induced on V (Pi ) ∪ V (Pj ) is a cycle. • A prism is a graph consisting of two vertex-disjoint triangles {a1 , a2 , a3 } and {b1 , b2 , b3 }, and three paths P1 , P2 , P3 , pairwise vertex-disjoint, such that for 1 ≤ i < j ≤ 3 the subgraph induced on V (Pi ) ∪ V (Pj ) is a cycle. Perhaps the main reason for interest in pyramids, thetas and prisms is that every graph containing a pyramid as an induced subgraph has an odd hole, and every graph containing a theta or prism has an even hole; and there seems to be some parallel between pyramids and “thetas or prisms”. Yet although pyramids, thetas and prisms are superficially similar, there is a real difference in the difficulty of detecting their presence. We showed in [3] that 2.1 There is an algorithm with running time O(|V (G)|9 ), that, with input a graph G, tests whether G contains a pyramid as an induced subgraph. Motivated by the parallel between pyramids and thetas-or-prisms, Chudnovsky and Kapadia [2, 4] proved the following: 2.2 There is a polynomial-time algorithm to test whether a graph G contains either a theta or a prism as an induced subgraph. In contrast, Maffray and Trotignon[5] showed that 2.3 It is NP-complete to test whether a graph contains a prism as an induced subgraph. There are two useful applications of the three-in-a-tree algorithm here. First, until now the complexity of testing whether G contains a theta has been open; but it can be solved in polynomial time as follows. 2.4 There is an algorithm to test if a graph G contains a theta as an induced subgraph, with running time O(|V (G)|11 ). Proof. Enumerate all four-tuples (a, b1 , b2 , b3 ) of distinct vertices such that a is adjacent to b1 , b2 , b3 and b1 , b2 , b3 are pairwise nonadjacent. For each such four-tuple (a, b1 , b2 , b3 ), enumerate all subsets X ⊆ V (G) such that a has no neighbour in X, and b1 , b2 , b3 each have exactly one neighbour in X, and each member of X is adjacent to at least one of b1 , b2 , b3 (it follows that |X| ≤ 3). For each such choice of X, let G′ be obtained from G by deleting a and all vertices adjacent to one of a, b1 , b2 , b3 except for the members of {b1 , b2 , b3 } ∪ X; and test whether there is an induced tree in G′ containing all of b1 , b2 , b3 . Thus we have to run the three-in-a-tree algorithm at most |V (G)|7 times, and each one takes time O(|V (G)|4 ). It is easy to see that there is some choice of a, b1 , b2 , b3 and X such that the tree exists, if and only if G contains a theta. Second, the algorithm of 2.1 depended heavily on some fortuitous properties of the smallest pyramid in a graph, and this was a little disturbing because testing for a pyramid was a crucial step in our algorithm to test whether a graph is perfect. The three-in-a-tree algorithm can be used to give another, less miraculous, way to test for pyramids, as follows. 2
2.5 There is an algorithm to test if a graph G contains a pyramid as an induced subgraph, with running time O(|V (G)|10 ). Proof. Enumerate all six-tuples (a1 , a2 , a3 , b1 , b2 , b3 ) of distinct vertices such that {b1 , b2 , b3 } is a triangle, and ai is adjacent to bj if and only if i = j (for 1 ≤ i, j ≤ 3). For each such six-tuple, let G′ be obtained from G by deleting all vertices with a neighbour in {b1 , b2 , b3 } except for a1 , a2 , a3 , and test whether there is an induced tree in G′ that contains all of a1 , a2 , a3 . It is easy to see that G contains a pyramid if and only if for some six-tuple there is an induced tree as described.
3
Strip structures
Let G be a connected graph. We say that Z ⊆ V (G) is constricted if |Z ∩ V (T )| ≤ 2 for every induced tree T of G. We wish to study which three-vertex subsets of V (G) are constricted; but we might as well study which sets are constricted in general, because that question is no more difficult, and does seem to be strictly more general. We will prove that if |Z| ≥ 2, then Z is constricted if and only if G admits a certain decomposition with respect to Z, that we call an “extended strip decomposition”. Our next goal is to define this; but before we do so, let us motivate the definition a little. First, let us say a leaf edge of a graph is an edge incident with a vertex of degree one. We observe that for any graph H, with line graph G say, if Z ⊆ E(H) = V (G) and every member of Z is a leaf edge of H, then Z is constricted in G. Our main result is an attempt at a converse of this; that for any graph G, if Z ⊆ V (G) is constricted then G is a sort of modified line graph of a graph H in which the members of Z are the leaf edges. Second, to see that there is some hope of a converse, let us mention the following “toy” version of our main theorem. (This is essentially a reformulation of 5.1, and we postpone the proof.) 3.1 Let G be a connected graph and let Z ⊆ V (G) be constricted with |Z| ≥ 2. Suppose that there is no proper subset X ⊂ V (G) with Z ⊆ X such that G|X is connected. Then there is a tree H with E(H) = X, such that G is the line graph of H, and Z is the set of leaf edges of H. Thus, we need to modify 3.1, eliminating the hypothesis about the nonexistence of X, and modifying the conclusion appropriately. It is natural to try changing the conclusion from “there is a tree H...” to “there is a connected graph H...”, but this is not sufficiently general; the graph G might admit a 2-join with all the members of Z on one side, and if so then we do not have much information or control about the other side of the 2-join. Such far sides of 2-joins must be incorporated into the theorem as general pieces of graph about which we know nothing, except we know that they attach to the controlled part of the graph via a 2-join structure. (We call these pieces “strips”). Now let us define an “extended strip decomposition”. Let G be a graph and Z ⊆ V (G). Let H also be a graph, let W be the set of vertices of H that have degree one in H, and let η be a map with domain the union of E(H) and the set of all pairs (e, v) where e ∈ E(H), v ∈ V (H) and e incident with v, satisfying the following conditions: • for each edge e ∈ E(H), η(e) ⊆ V (G), and for each v ∈ V (H) incident with e, η(e, v) ⊆ η(e) • η(e) ∩ η(f ) = ∅ for all distinct e, f ∈ E(H)
3
• for all distinct e, f ∈ E(H), let x ∈ η(e) and y ∈ η(f ); then x, y are adjacent in G if and only if e, f share an end-vertex v in H, and x ∈ η(e, v) and y ∈ η(f, v) • |Z| = |W |, and for each z ∈ Z there is a vertex v ∈ W such that η(e, v) = {z}, where e is the (unique) edge of H incident with v. Let us call such a map η an H-strip structure in (G, Z). (Thus, we have now incorporated the “far sides of 2-joins”, as discussed above. Unfortunately, this still is not enough, and we need a little more, as follows.) Let η be an H-strip structure, and let us extend the domain of η by adding to it the union of V (H) and the set of all triangles of H, as follows. For each vertex v ∈ V (H), let η(v) ⊆ V (G), and for each triangle D of H let η(D) ⊆ V (G), satisfying the following: • all the sets η(e) (e ∈ E(H)), η(v) (v ∈ V (H)) and η(D) (for all triangles D of H) are pairwise disjoint, and their union is V (G) • for each v ∈ V (H), if x ∈ η(v) and y ∈ V (G) \ η(v) are adjacent in G then y ∈ η(e, v) for some e ∈ E(H) incident in H with v • for each triangle D of H, if x ∈ η(D) and y ∈ V (G) \ η(D) are adjacent in G then y ∈ η(e, u) ∩ η(e, v) for some distinct u, v ∈ D, where e is the edge uv of H. In this case we say that η is an extended H-strip decomposition of (G, Z). Our main theorem asserts the following: 3.2 Let G be a connected graph and let Z ⊆ V (G) with |Z| ≥ 2. Then Z is constricted if and only if for some graph H, (G, Z) admits an extended H-strip decomposition. Proof of the “if ” half of 3.2. Suppose that Z is not constricted, and yet (G, Z) admits an extended H-strip-decomposition η. Choose an induced tree T in G with |V (T ) ∩ Z| ≥ 3, with V (T ) minimal. It follows that every vertex of T that has degree one in T belongs to Z (for otherwise it could be deleted from T ); and since there is such a vertex, and it cannot be deleted from T , it follows that |V (T ) ∩ Z| = 3. Now either T is an induced path with both end-vertices in Z (in this case there is a unique vertex of Z ∩ V (T ) that is an internal vertex of P , say y) or the three members of Z ∩ V (T ) all have degree one in T , and T has a unique vertex y of degree three (and possibly some vertices of degree two). If z ∈ Z ∩ V (T ) \ {y}, a path P of T is said to be a z-limb if y ∈ / V (P ) and z ∈ V (P ) (and consequently z is an end-vertex of P , and no other vertex of P belongs to Z). For each v ∈ V (H), let N (v) denote the union of all the sets η(e, v), as e ranges over all edges of H incident with v. (1) For each v ∈ V (H), there do not exist distinct z1 , z2 ∈ Z ∩ V (T ) \ {y} such that some z1 limb contains a vertex in N (v) and some z2 -limb contains a vertex in N (v). For suppose this is false; then for i = 1, 2 we may choose a zi -limb Qi , with end-vertices zi , yi say, and v ∈ V (H), such that y1 , y2 ∈ N (v). Choose Q1 , Q2 , v so that the sum of the lengths of Q1 , Q2 is minimum. It follows that for i = 1, 2, no vertex of Qi belongs to N (v) except yi . Since y1 is not adjacent to y2 (from the definition of z1 -, z2 -limb) it follows that for some e ∈ E(H), 4
y1 , y2 ∈ η(e, v). Let e be incident with v, u ∈ V (H) say. Since |η(e, v)| > 1, it follows that not both z1 , z2 ∈ η(e), so we may assume that z1 ∈ / η(e). Let R1 be a maximal z1 -limb such that R1 is a subpath of Q1 and no vertex of R1 belongs to η(e). Let R1 have ends r1 , z1 say; and let s1 be the neighbour of r1 in Q1 that does not belong to R1 . Thus s1 ∈ η(e). If z2 ∈ / η(e) define R2 , r2 , s2 similarly. Since r1 ∈ / η(e), it follows that either r1 ∈ η(D) for some triangle D of H, or r1 ∈ η(w) for some w ∈ V (H), or r1 ∈ η(f ) for some f ∈ E(H) \ {e}. Suppose that the first holds, that is, r1 ∈ η(D) for some triangle D of H. Since s1 ∈ η(e) and r1 , s1 are adjacent, it follows that u, v ∈ D, and s1 ∈ η(e, u) ∩ η(e, v). Let D = {u, v, w} say. Since z1 ∈ / η(D), there is an edge ab of R1 such that a ∈ η(D) and b ∈ / η(D). Consequently b belongs to one of η(uv, u) ∩ η(uv, v), η(vw, v) ∩ η(vw, w), η(uw, u) ∩ η(uw, w). But from the choice of R1 , b ∈ / η(e), and from the choice of Q1 , b ∈ / N (v). Hence b ∈ η(uw, u) ∩ η(uw, w); but then s1 ∈ η(uv, u) and b ∈ η(uw, u), and so s1 , b are adjacent, contradicting that T is an induced tree. Thus the first case cannot occur. Suppose that the second holds, and r1 ∈ η(w) for some w ∈ V (H). Since r1 , s1 are adjacent, it follows that s1 ∈ N (w), and so s1 ∈ η(f, w) for some edge f incident with w; and since s1 ∈ η(e), it follows that f = e, so w is one of u, v. Moreover, since z1 ∈ / η(w), there is an edge ab of R1 such that a ∈ η(w) and b ∈ / η(w), and therefore b ∈ N (w). Since b ∈ / η(v) it follows that w 6= v, and so w = u, and hence s1 ∈ η(e, u). Moreover, since T is an induced tree it follows that b and s1 are not adjacent. Since s1 , b ∈ N (u), it follows that b ∈ η(e, u), contrary to the choice of R1 . Thus the second case cannot hold. We deduce that the third holds, and so r1 ∈ η(f ) for some f ∈ E(H) \ {e}. Since r1 , s1 are adjacent, it follows that there is a vertex w of H incident with e, f such that s1 ∈ η(e, w) and r1 ∈ η(f, w). Since r1 ∈ / N (v), it follows that w = u, and so s1 ∈ η(e, u). In particular, η(e, u) 6= {z2 }, and since η(e, v) 6= {z2 }, it follows that z2 ∈ / η(e), and so R2 , r2 , s2 are defined. By the same argument with z1 , z2 exchanged, it follows that s2 ∈ η(e, u). Since r1 ∈ η(f, u) for some f ∈ E(H) \ {e}, and s2 ∈ η(e, u), it follows that r1 , s2 are adjacent, a contradiction. This proves (1). Let Z ∩ V (T ) = {z1 , z2 , z3 }. (2) y ∈ η(e) for some e ∈ E(H). For either y ∈ η(D) for some triangle D of H, or y ∈ η(v) for some v ∈ V (H), or y ∈ η(e) for some e ∈ E(H). Suppose that the first holds. Now for i = 1, 2, 3, zi ∈ / η(D); let Ri be a maximal zi -limb containing no vertex of η(D), with ends zi , ri say, and ri has a neighbour in η(D). Let D = {u, v, w}; then for i = 1, 2, 3, ri belongs to at least two of N (u), N (v), N (w). Hence one of N (u), N (v), N (w) meets both of R1 , R2 , contrary to (1). Next suppose the second holds, and y ∈ η(v) for some v ∈ V (H). For i = 1, 2, 3, since zi ∈ / η(v), there is a maximal zi -limb Ri with no vertex in η(v), with ends zi , ri say, and ri has a neighbour in η(v). Hence r1 , r2 , r3 all belong to N (v), contrary to (1). Thus the third holds, that is, y ∈ η(e) for some e ∈ E(H). This proves (2). Let e be as in (2). For i = 1, 2, 3, if zi ∈ / η(e), let Ri be a maximal zi -limb containing no vertex of η(e), with ends zi , ri say, and let ri be adjacent in T to si ∈ η(e). Let e be incident in H with 5
u, v ∈ V (H). (3) For i = 1, 2, 3, if zi ∈ / η(e), then there exists w ∈ {u, v} and an edge f ∈ E(H) \ {e} incident with w such that ri ∈ η(f, w) and si ∈ η(e, w). For either ri ∈ η(D) for some triangle D of H, or ri ∈ η(w) for some w ∈ V (H), or ri ∈ η(f ) for some f ∈ E(H) \ {e}. Suppose that the first holds, and ri ∈ η(D). Since si ∈ / η(D), and ri , si are adjacent, and si ∈ η(e), it follows that u, v ∈ D, and si ∈ η(e, u) ∩ η(e, v). Let D = {u, v, w} say. Since zi ∈ / η(D), there is an edge ab of Ri with a ∈ η(D) and b ∈ / η(D). Consequently b belongs to one of η(uv, u) ∩ η(uv, v), η(vw, v) ∩ η(vw, w), η(uw, u) ∩ η(uw, w). Since b, si are nonadjacent, it follows that b ∈ / η(uw, u) ∪ η(vw, v), and since b ∈ / η(e), this is impossible. Thus the first case cannot occur. Suppose that the second holds, and ri ∈ η(w) for some w ∈ V (H). Since ri , si are adjacent, and si ∈ η(e), it follows that w ∈ {u, v} and si ∈ η(e, w); and we may assume that w = v. Since zi ∈ / η(v), there is an edge ab of Ri with a ∈ η(v) and b ∈ / η(v). Hence b ∈ N (v), and so b ∈ η(f, v) for some edge f ∈ E(H) incident with v. Since si , b are nonadjacent, it follows that f = e, contrary to the definition of Ri . Thus the second case cannot hold. We deduce that the third holds, and ri ∈ η(f ) for some f ∈ E(H) \ {e}. Since ri , si are adjacent, and si ∈ η(e), there exists w ∈ V (H) incident with both e, f , such that ri ∈ η(f, w) and si ∈ η(e, w). Hence w ∈ {u, v}. This proves (3). Now if none of z1 , z2 , z3 belong to η(e), then by (3), we may assume that ri ∈ η(f, u) and si ∈ η(e, u) for i = 1, 2, contrary to (1). Thus we may assume that z3 ∈ η(e), and therefore we may assume that N (u) = η(e, u) = {z3 }. If z1 , z2 ∈ / η(e), then for i = 1, 2, ri ∈ / N (u), and so ri ∈ N (v) by (3), contrary to (1). Consequently one of z1 , z2 ∈ η(e), and so we may assume that N (v) = η(e, v) = {z2 }. But then z1 ∈ / η(e), and yet r1 ∈ / N (u) ∪ N (v), contrary to (3). This completes the proof.
4
A lemma
In this section we prove a lemma needed for the “only if” half of 3.2. If G′ is a subgraph of a graph G, and C is a subgraph of G \ V (G′ ), and v ∈ V (G′ ) has a neighbour in V (C), we say that v is an attachment of C (in G′ ). A separation of a graph K is a pair (A, B) of subsets of V (K) with union V (K), such that no edge joins a vertex in A \ B and a vertex in B \ A. We call |A ∩ B| the order of the separation. Let W ⊆ V (K). We say that (K, W ) is a frame if • every vertex in W has degree one in K • |W | ≥ 3 • K is connected • for every separation (A, B) of K of order at most two with W ⊆ B 6= V (K), we have that |A ∩ B| = 2 and K|A is a path between the two members of A ∩ B.
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If (K, W ) is a frame, we see that W is the set of all vertices of K that have degree one. A branch of (K, W ) is a path of K with distinct ends, such that both its ends have degree in K different from two, and all its internal vertices have degree two in K. Since (K, W ) is a frame, it follows that every branch is an induced subgraph of K, and every edge of K belongs to a unique branch. If v ∈ V (K), δK (v) or δ(v) denotes the set of all edges of K incident with v. Let (K, W ) be a frame and let F ⊆ E(K). An F -line is a path P in K such that one end of P belongs to W and some edge of P belongs to F . A double F -line is a path P such that both ends of P belong to W and exactly one edge of P belongs to F . We say that F ⊆ E(K) is focused if • there do not exist three F -lines that are pairwise vertex-disjoint, and • there do not exist an F -line and a double F -line that are vertex-disjoint. We need to study which subsets F are focused. We shall prove the following: 4.1 Let (K, W ) be a frame and let F ⊆ E(K) be focused. Then either: 1. there exists x ∈ V (K) with F ⊆ δ(x), or 2. |F | = 3 and the three edges of F form a triangle, or 3. there exist x, y ∈ V (K), not in the same branch of K, such that F = δ(x) ∪ δ(y), or 4. there is a branch B of (K, W ) with F ⊆ E(B), or 5. there is a branch B of (K, W ) with ends x, y such that x ∈ / W and F \ E(B) = δ(x) \ E(B) and F ∩ E(B) 6⊆ δ(x), or 6. there is a branch B of (K, W ) with ends x, y such that x, y ∈ / W and F \ E(B) = (δ(x) ∪ δ(y)) \ E(B). Proof. We proceed by induction on |E(K)|. (1) We may assume that for every X ⊆ V (K) with |X| ≤ 2, there is an F -line with no vertex in X. For suppose that X ⊆ V (K) with |X| ≤ 2, and every F -line has a vertex in X. Choose such a set X with |X| minimum. Since all vertices in W have degree one and their neighbours are not in W , we may also choose X with X ∩ W = ∅. If |X| = 2 and the two members of X belong to the same branch, let B0 be the path in this branch between the two members of X, and otherwise let B0 be the subgraph with vertex set X and no edges. Suppose that there exists f ∈ F \ E(B0 ) not incident with any member of X. There is no path between f and W in K \ X, from the property of X, and so there is a separation (A, B) of K with A ∩ B = X and W ⊆ B such that both ends of f belong to A. Since f has no end in X, it follows that B 6= V (K), and so |X| = 2 and K|A is a path between the two members of X; but then this path is B0 and contains f , a contradiction. This proves that every member of F either belongs to E(B0 ) or is incident with a member of X. If |X| ≤ 1 then the first outcome holds, so we may assume that X = {x1 , x2 } say. For i = 1, 2, let Vi be the set of all neighbours of xi that are not in V (B0 ), let Yi be the set of all y ∈ Vi such that the 7
edge yxi belongs to F , and let Zi = Vi \ Yi . If one of Y1 , Z1 is empty, and one of Y2 , Z2 is empty, then one of the outcomes of the theorem holds; so we may assume that Y1 , Z1 are both nonempty. Let J = K \ V (B0 ). Since X ∩ W = ∅, it follows that W ⊆ V (J). If G is a graph and X, Y ⊆ V (G), we denote by κ(G, X, Y ) the minimum order of a separation (A, B) of G with X ⊆ A and Y ⊆ B. Since (K, W ) is a frame, it follows that • κ(J, Y1 , W ) ≥ 1, for otherwise there would be a separation (A, B) of K with A ∩ B = X and V (B0 ) ⊆ A and Y1 ⊆ A, which is impossible since (K, W ) is a frame and no branch of (K, W ) includes V (B0 ) ∪ Y1 , since Z1 6= ∅ • κ(J, Z1 , W ) ≥ 1, similarly • κ(J, V2 , W ) ≥ 1; indeed, κ(J, V2 , W ) = κ(K \ {x1 }, V2 , W ), and therefore is at least two unless |V2 | = 1 • κ(J, Y1 ∪ Z1 , W ) ≥ 2, similarly • κ(J, Y1 ∪ V2 , W ), κ(J, Z1 ∪ V2 , W ) ≥ 2, since no branch of (K, W ) includes both V (B0 ) and one of Y1 , Z1 • κ(J, Y1 ∪ Z1 ∪ V2 , W ) ≥ 3, since κ(J, Y1 ∪ Z1 ∪ V2 , W ) = κ(K, V1 ∪ V2 , W ), and the latter is at least three since no branch of K includes x1 , x2 and all their neighbours. From this and Menger’s theorem (applied to the graph obtained from J by adding three new vertices with neighbour sets Y1 , Z1 , V2 respectively, and asking for three vertex-disjoint paths between the three new vertices and W ), we deduce that there are three vertex-disjoint paths P1 , P2 , P3 of J, from some y1 ∈ Y1 to w1 ∈ W , from z1 ∈ Z1 to w2 ∈ W , and from v2 ∈ V2 to w3 ∈ W respectively. We may assume that v2 is the only vertex of P3 in V2 . The path w1 -P1 -y1 -x1 -z1 -P2 -w2 (with the obvious notation) is a double F -line Q say. Hence the path x2 -v2 -P3 -w3 is not an F -line, since F is focused, and so v2 ∈ Z2 . For y2 ∈ Y2 , the path y2 -x2 -v2 -P3 -w3 is an F -line, and therefore is not disjoint from Q; and so Y2 ⊆ V (Q); and by a similar argument, every edge of B0 in F is incident with x1 . Thus if Y2 = ∅ then the first outcome of the theorem holds, so we assume that Y2 6= ∅. Let y2 ∈ Y2 ; then we have seen that y2 ∈ V (P1 ) ∪ V (P2 ). If y2 ∈ V (P2 ), then the path w3 -P3 -v2 -x2 -y2 -P2 -w2 is a double F -line (here the notation y2 -P2 -w2 at the end of this sequence of concatenations means that we take the subpath of P2 between y2 and w2 ; we will use this and similar notation repeatedly without further explanation); and this double F -line is vertex-disjoint from the F -line x1 -y1 -P1 -w1 , a contradiction. Thus y2 ∈ V (P1 ). If y2 6= y1 , then w3 -P3 -v2 -x2 -y2 -P1 -w1 is a double F -line vertex-disjoint from the F -line y1 -x1 -z1 -P2 -w2 , a contradiction; and so y2 = y1 . This proves that Y2 = {y1 }. Since Y2 , Z2 6= ∅, this restores the symmetry between x1 , x2 , and so it follows that Y1 = {y1 }, and every edge of B0 in F is incident with x2 (as well as with x1 ). If no edge of B0 belongs to F , then every edge in F is incident with y1 , contrary to the minimality of X; so F ∩ E(B0 ) 6= ∅, and therefore B0 is a path of length one, and the second outcome holds. This proves (1). Henceforth, therefore, we make the assumption of (1), and will obtain a contradiction. (2) K is not a tree.
8
For suppose K is a tree. It follows that for any set C of trees of K, either there are k members of C pairwise vertex-disjoint, or there is a set X ⊆ V (K) with |X| < k meeting every member of C. Since there do not exist three pairwise disjoint F -lines, we deduce (by taking C to be the set of all F -lines, and k = 3) that there exists X ⊆ V (K) with |X| ≤ 2, such that every F -line contains a member of X. But this contradicts (1), and so proves (2). (3) There is a branch B of (K, W ) such that, if K ′ denotes the graph obtained from K by deleting the edges and internal vertices of B, then (K ′ , W ) is a frame. For let T be a minimal connected subgraph of K with W ⊆ V (T ); then T is a tree, and it is easy to see that (T, W ) is a frame. Since K is not a tree, it follows that T 6= K, and so there is a frame (K ′ , W ) with K ′ a proper subgraph of K. Choose such a frame with as many branches as possible. Suppose that there exist u, v ∈ V (K ′ ) that are joined by a path P of K such that no edges or internal vertices of P belong to K ′ , such that u, v are not in the same branch of (K ′ , W ). It follows that (K ′ ∪ P, W ) is a frame, and from our choice of (K ′ , W ), we deduce that K ′ ∪ P = K. But then P is a branch of (K, W ), and (2) holds taking B = P . We may therefore assume that for every two vertices u, v that are joined by a path with no edges or internal vertices in K ′ , some branch of (K ′ , W ) contains u, v. In particular every edge of K that does not belong to E(K ′ ) but has both ends in V (K ′ ) is between two vertices in the same branch. For any component C of K \ V (K ′ ), we have seen that every two attachments of C belong to the same branch. If for every such C there is a branch of (K ′ , W ) that contains every attachment of C, this contradicts that (K, W ) is a frame. Thus there is a component C of K \ V (K ′ ) such that no branch of (K ′ , W ) contains all attachments of C. In particular C has at least two attachments, and every two of them belong to a branch; let B1 be a branch, with ends x2 , x3 , containing at least two attachments v2 , v3 of C. Let v1 be an attachment of C that is not in B1 . Since some branch B2 contains v1 , v3 , it follows that v3 is one of x2 , x3 , say v3 = x3 , and B2 has ends x1 , x3 say. Similarly there is a branch B3 containing v1 , v2 ; so v2 = x2 , and v1 is a common end of B2 , B3 . Since v1 , v2 , v3 are attachments of C, we may choose a vertex c of C and three paths P1 , P2 , P3 from c to v1 , v2 , v3 respectively, pairwise vertex-disjoint except for c, such that V (Pi ) ⊆ V (C) ∪ {vi }. But then if K ′′ denotes the graph obtained from K ′ by deleting the edges and internal vertices of B1 , and adding P1 ∪ P2 ∪ P3 , then (K ′′ , W ) is a frame, and K ′′ 6= K, since the edges of B1 do not belong to K ′′ , contrary to the choice of K ′ . This proves (3). Henceforth, let B, K ′ be as in (3). Let the ends of B be x1 , x2 . Let F ′ = F ∩ E(K ′ ). (4) There does not exist v ∈ V (K ′ ) such that F ′ ⊆ δK ′ (v). For suppose that v has this property. Let Y be the set of neighbours y of v in K such that the edge vy belongs to F . By (1) there is an F -line disjoint from {x1 , x2 }, and in particular, v ∈ / V (B), and Y 6⊆ V (B). Since (K, W ) is a frame, there are three paths P1 , P2 , P3 from x1 , x2 , v respectively to W , pairwise vertex-disjoint. Consequently P3 , B are vertex-disjoint. For i = 1, 2, 3 let wi be the end of Pi in W . By (1), no vertex of B meets every edge in E(B) ∩ F (since otherwise this vertex together with v would meet every edge in F ), and so there are two disjoint edges in E(B) ∩ F , and therefore there are two vertex-disjoint F -lines Q1 , Q2 , both subpaths of P1 ∪ B ∪ P2 . Hence P3 is not an F -line, and for each y ∈ Y \ V (B), y-v-P3 -w3 is not disjoint from both Q1 , Q2 , since y-v-P3 -w3 9
includes an F -line. In particular, Y ⊆ V (P1 ∪ B ∪ P2 ). We have already seen that there exists y ∈ Y \ V (B), and we may therefore assume that y ∈ V (P1 ) \ {x1 }. The path w3 -P3 -v-y-P1 -w1 is a double F -line, and it is disjoint from the F -line B ∪ P2 , a contradiction. This proves (4). (5) It is not the case that |F ′ | = 3, and the three edges in F ′ form a triangle. For suppose there is a triangle {v1 , v2 , v3 } of K ′ such that F ′ consists of the three edges v1 v2 , v2 v3 , v3 v1 . (Possibly v1 , v2 , v3 , x1 , x2 are not all distinct.) By (1) (applied to {v1 , v2 }) there is an edge of B in F . For any M ⊆ {v1 , v2 , v3 , x1 , x2 } with |M | = 3, there is no branch of (K, W ) including all members of M , and so, since (K, W ) is a frame, there exist three vertex-disjoint paths in K between M and W . Consequently there are three vertex-disjoint paths P1 , P2 , P3 from {v1 , v2 , v3 , x1 , x2 } to W , such that at least one of them has first vertex in {x1 , x2 }, and at least two of them have first vertex in {v1 , v2 , v3 }. Choose three such paths P1 , P2 , P3 with minimal union, and let Pi be between ui and wi ∈ W say. We may assume that u1 = x1 , u2 = v2 and u3 = v3 . Moreover, from the minimality of P1 ∪ P2 ∪ P3 , v1 is not a vertex of P2 ∪ P3 , and x2 is not a vertex of P1 . Hence B is not a path of any of P1 , P2 , P3 . The path w2 -P2 -v2 -v3 -P3 -w3 is a double F -line, disjoint from P1 ∪ B \{x2 }; so the latter is not an F -line. But some edge of B is in F , and so the edge of B incident with x2 is the unique edge of B in F . Since the F -line B ∪ P1 is not disjoint from the double F -line w2 -P2 -v2 -v3 -P3 -w3 , it follows that x2 belongs to one of P2 , P3 , say P2 . The double F -line w1 -P1 -x1 -B-x2 -P2 -w2 is not vertex-disjoint from the F -line v2 -v3 -P3 -w3 , so v2 = x2 . But then every edge in F is incident with one of v2 , v3 , contrary to (1). This proves (5). (6) There do not exist two vertices x3 , x4 ∈ V (K ′ ) such that F ′ = δK ′ (x3 ) ∪ δK ′ (x4 ). For suppose such x3 , x4 exist. By (1) (applied to {x3 , x4 }) there is an edge f of B in F and incident with neither of x3 , x4 . Also by (1) {x1 , x2 } = 6 {x3 , x4 }, so there are three pairwise vertex-disjoint paths P2 , P3 , P4 from {x1 , x2 , x3 , x4 } to W , where Pi is from ui ∈ {x1 , x2 , x3 , x4 } to wi ∈ W , and u3 = x3 , u4 = x4 , and u2 ∈ {x1 , x2 }. Choose such paths with P2 minimal; then we may assume that u2 = x2 , and x1 ∈ / V (P2 ). Hence B is not a path of any of P2 , P3 , P4 . Moreover, P3 , P4 are disjoint F -lines, and P2 ∪ B \ {x1 } is disjoint from both of them; so the latter is not an F -line. Hence f is incident with x1 , and therefore x1 6= x3 , x4 , and so x1 , . . . , x4 are all distinct. Since P2 ∪ B is an F -line, it meets one of P3 , P4 , and so x1 ∈ V (P3 ) ∪ V (P4 ), and we may assume that x1 ∈ V (P3 ) say. But then the double F -line w2 -P2 -x2 -B-x1 -P3 -w3 is disjoint from the F -line P4 , a contradiction. This proves (6). (7) There is no branch B ′ of (K ′ , W ) such that F ′ ⊆ E(B ′ ). For suppose that B ′ is such a branch, with ends x3 , x4 . First suppose that one of x1 , x2 , say x1 , belongs to V (B ′ ). Since (K, W ) is a frame, it follows that x2 ∈ / V (B ′ ), and there are three paths Pi of K from xi to wi ∈ W for i = 2, 3, 4, pairwise vertex-disjoint. Consequently P 2, P 3, P 4 contain no edges of B or of B ′ . For i = 3, 4, let Qi be the subpath of B ′ between x1 and xi , and let Q2 = B. Now F ⊆ E(Q2 ) ∪ E(Q3 ) ∪ E(Q4 ). By (1), not every edge in F is incident with x1 , so we may assume that Pi ∪ Qi \ {x1 } is an F -line for some i ∈ {2, 3, 4}. Let {i, j, k} = {2, 3, 4}. Consequently there do not exist two vertex-disjoint F -lines in Pj ∪ Qj ∪ Qk ∪ Pk , and so there are at most two edges in
10
F ∩ E(Qj ∪ Qk ), and if there are two then they have a common end. By (1) (applied to {x1 , xi }) at least one edge of Qj ∪ Qk is in F , and if there is only one then Pj ∪ Qj ∪ Qk ∪ Pk is a double F -line, disjoint from the F -line Pi ∪ Qi \ {x1 }, a contradiction. Hence exactly two edges of Qj ∪ Qk belong to F , and they have a common end yj ∈ V (Qj ) say. By (1) applied to {x1 , xi } it follows that yj 6= x1 , so yj is an internal vertex of Qj , and in particular F ∩ E(Qk ) = ∅, and F ∩ E(Qj ) = δ(yj ). Then Pj ∪ Qj \ {x1 } is an F -line, and so by the same argument there is an internal vertex yi of Qi such that F ∩ E(Qi ) = δ(yi ). But this contradicts (1) (applied to {yi , yj }). This completes the proof of (7) in the case when one of x1 , x2 belongs to B ′ . Thus we may assume that B, B ′ are vertex-disjoint. There is symmetry between B and B ′ (for we will not use any more that (K ′ , W ) is a frame). By two applications of (1), both of B, B ′ contain an edge in F . There are three vertex-disjoint paths between {x1 , . . . , x4 } and W , and we may assume that none of them has an internal vertex in {x1 , . . . , x4 }; and from the symmetry we may assume that these paths are P1 , P2 , P3 , where Pi is between xi and wi ∈ W . Now P3 ∪ B ′ is an F -line, and so P1 ∪B ∪P2 is not a double F -line and does not include two disjoint F -lines; so there is an internal vertex y of B such that F ∩ E(B) = δ(y). There are three disjoint paths Q2 , Q3 , Q4 from {x1 , . . . , x4 } to W , such that for i = 3, 4, Qi has first vertex xi ; choose them with Q2 minimal, then we may assume that Q2 has first vertex x2 and x1 ∈ / V (Q2 ) (possibly x1 ∈ V (Q3 ∪ Q4 )). The path y-B-x2 -Q2 -w2′ (where Q2 is from x2 to w2′ ∈ W ) is an F -line, disjoint from the path Q3 ∪ B ′ ∪ Q4 ; so the latter is not a double F -line, and does not include two disjoint F -lines. Hence there is an internal vertex y ′ of B ′ such that F ∩ E(B ′ ) = δ(y ′ ); but this contradicts (1) (applied to {y, y ′ }). This proves (7). (8) There is no branch B ′ of (K ′ , W ) with ends x3 , x4 , such that F ′ \ E(B ′ ) = δK ′ (x3 ). For suppose B ′ is such a branch. Again there are two cases depending whether B, B ′ are vertexdisjoint or not. First suppose that x1 ∈ V (B ′ ) say, and as in (7) we may choose three paths Pi of K from xi to wi ∈ W for i = 2, 3, 4, pairwise vertex-disjoint. For i = 3, 4, let Qi be the subpath of B ′ between x1 and xi , and let Q2 = B. Suppose that x1 = x3 . By (1) (applied to {x2 , x3 }), there is an edge of B ′ \{x3 } in F , and similarly an edge of B\{x3 } in F . But then P2 ∪Q2 \{x3 }, P3 , P4 ∪Q4 \{x3 } are three disjoint F -lines, a contradiction. Thus x1 6= x3 . By (1) (applied to {x1 , x3 }), at least one edge of Q2 ∪ Q4 is in F and not incident with x1 . Since P3 is an F -line, the path P2 ∪ Q2 ∪ Q4 ∪ P4 is not a double F -line, and does not include two disjoint F -lines; so exactly two edges of Q2 ∪ Q4 belong to F and they have a common end y 6= x1 . From the symmetry we may assume that y belongs to the interior of Q2 say, and so F ∩ E(Q4 ) = ∅. By (1) (applied to {x3 , y}) there is an edge of Q3 \ {x3 } in F ; but then P2 ∪ Q2 \ {x1 }, P3 , P4 ∪ Q4 ∪ Q3 \ {x3 } are three vertex-disjoint F -lines, a contradiction. This proves (8) in the case that B, B ′ are not disjoint. We may therefore assume that B, B ′ are disjoint. By (1) (applied to {x3 , x4 }) at least one edge of B is in F . By (4) at least one edge of B ′ \ {x3 } belongs to F . There are three vertex-disjoint paths P2 , P3 , P4 from {x1 , . . . , x4 } to W , such that for i = 3, 4, Pi is from xi to wi ∈ W say. We may assume that P2 is from x2 to w2 , and x1 ∈ / V (P2 ) (possibly x1 ∈ V (P3 ∪ P4 )). Thus P3 , P4 ∪ B ′ \ {x3 } are disjoint F -lines, and so P2 ∪ B \ {x1 } is not an F -line; and therefore the edge of B incident with x1 is the unique edge of B in F . Since B ∪ P2 is an F -line, it follows that x1 belongs to one of P3 , P4 . If x1 ∈ V (P3 ), then w2 -P2 -x2 -B-x1 -P3 -w3 is a double F -line, and B ′ ∪ P4 is an F -line, and they are disjoint, a contradiction. If x1 ∈ V (P4 ), then w2 -P2 -x2 -B-x1 -P4 -w4 is a double F -line, and P3 is an X-line, and they are disjoint, a contradiction. This proves (8).
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(9) There is no branch B ′ of (K ′ , W ) with ends x3 , x4 , such that F ′ \ E(B ′ ) = δK ′ (x3 ) ∪ δK ′ (x4 ). For suppose that B ′ is such a branch. Again there are two cases depending whether B, B ′ are vertex-disjoint or not. First suppose that x1 ∈ V (B ′ ) say, and as in (7) we may choose three paths Pi of K from xi to wi ∈ W for i = 2, 3, 4, pairwise vertex-disjoint. For i = 3, 4, let Qi be the subpath of B ′ between x1 and xi , and let Q2 = B. Thus P3 , P4 are F -lines. Suppose that x1 = x3 . Then from (1) (applied to {x3 , x4 }) there is an edge of Q2 \ {x3 } in F , and so P2 ∪ Q2 \ {x3 } is an F -line disjoint from P3 , P4 , a contradiction. Thus x1 6= x3 , and similarly x1 6= x4 . By (1) (applied to {x3 , x4 }) there is an edge of F in Q2 ∪ Q3 ∪ Q4 not incident with either of x3 , x4 ; but hence there is an F -line in P2 ∪ Q2 ∪ Q3 ∪ Q4 \ {x3 , x4 }, and it is disjoint from P3 , P4 , a contradiction. This proves (9) in the case that B, B ′ are not disjoint. Thus we may assume that B, B ′ are disjoint. By (1) (applied to {x3 , x4 }) at least one edge of B is in F . There are three vertex-disjoint paths P2 , P3 , P4 from {x1 , . . . , x4 } to W , such that for i = 3, 4, Pi is from xi to wi ∈ W say. We may assume that P2 is from x2 to w2 , and x1 ∈ / V (P2 ) (possibly x1 ∈ V (P3 ∪ P4 )). Thus P3 , P4 are disjoint F -lines, and so P2 ∪ B \ {x1 } is not an F -line; and therefore the edge of B incident with x1 is the unique edge of B in F . Since B ∪ P2 is an F -line, it follows that x1 belongs to one of P3 , P4 , and we may assume it belongs to P3 from the symmetry. Then w2 -P2 -x2 -B-x1 -P3 -w3 is a double F -line, and P4 is an F -line, and they are disjoint, a contradiction. This proves (9). But (4)–(9) are contrary to the inductive hypothesis applied to the frame (K ′ , W ). This proves that our assumption of (1) was false, and so proves 4.1.
5
The main proof
In this section we prove the “only if” half of 3.2. We need to show that if G is a connected graph and Z ⊆ V (G) with |Z| ≥ 2 is constricted, then (G, Z) admits an extended H-strip decomposition for some graph H. The result is trivial if |Z| = 2, so we may assume that |Z| ≥ 3. Therefore, throughout this section we assume that G is a connected graph, Z ⊆ V (G) with |Z| ≥ 3, and Z is constricted in G. We shall prove a series of lemmas about the pair (G, Z). Let (K, W ) be a frame. We say it is a frame for (G, Z) if E(K) ⊆ V (G), and • for all distinct e, f ∈ E(K), e, f have a common end in K if and only if e, f ∈ V (G) are adjacent in G • Z is the set of edges of K incident with a vertex in W . We begin with: 5.1 There is a frame for (G, Z). Proof. Since G is connected, we may choose X ⊆ V (G) with Z ⊆ X, minimal such that G|X is connected.
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(1) For each v ∈ X \ Z, G|(X \ {v}) has exactly two components, and they both contain at least one vertex of Z. For G|(X \ {v}) is not connected, from the minimality of X. Let its components be C1 , . . . , Ck say where k ≥ 2. If Ci ∩ Z = ∅, let X ′ = X \ V (Ci ); then G|X ′ is connected and Z ⊆ X ′ , contrary to the minimality of X. Thus each Ci contains at least one vertex of Z. Suppose that k ≥ 3, and choose zi ∈ V (Ci ) ∩ Z for i = 1, 2, 3. Since G|X is connected, there are paths P1 , P2 , P3 of G|X between v and z1 , z2 , z3 respectively, with V (Pi ) ⊆ Ci ∪ {v}, and if we choose P1 , P2 , P3 with minimal union then their union is an induced tree of G, containing three members of Z, contradicting that Z is constricted in G. This proves (1). For each v ∈ X \ Z, let Av , Bv be the vertex sets of the two components of G|(X \ {v}). (2) For each v ∈ X \ Z, the set of neighbours of v in Av is a clique, and so is the the set of neighbours of v in Bv . For suppose that u1 , u2 ∈ Av are nonadjacent, and are both adjacent to v. Choose z3 ∈ Bv ∩ Z, and let P3 be an induced path between v and z3 with vertex set in Bv ∪ {v}. From the minimality of X, for i = 1, 2 there exists zi ∈ Z such that every path of G|X between v and zi contains ui ; let Pi be some such path, induced. Consequently u1 ∈ / V (P2 ), since P2 is induced and u2 ∈ V (P2 ), and similarly u2 ∈ / V (P1 ). Hence z1 6= z2 , and V (P1 ), V (P2 ) ⊆ Av ∪ {v}. Since every path of G|X between v and z1 contains u1 , it follows that V (P1 ) \ {v, u1 } is disjoint from V (P2 ) \ {v}, and there is no edge between these two sets. Similarly there is no edge between V (P1 ) \ {v} and V (P2 ) \ {v, u2 }; and therefore there is no edge between V (P1 ) \ {v} and V (P2 ) \ {v}, since u1 , u2 are nonadjacent. Hence P1 ∪ P2 ∪ P3 is an induced tree in G, contradicting that Z is constricted. This proves (2). (3) For each z ∈ Z, the set of neighbours of z in X \ {z} is a clique. The proof is similar to that of (2). Suppose that u1 , u2 ∈ X \ {z} are nonadjacent, and both adjacent to z. From the minimality of X, there exist zi ∈ Z such that every path of G|X between z and zi contains ui ; let Pi be such a path, induced, for i = 1, 2. Then z1 , z2 6= z, and as in (2), there are no edges between V (P1 ) \ {z} and V (P2 ) \ {z}. But then P1 ∪ P2 is an induced tree containing z, z1 , z2 , a contradiction. This proves (3). From (2) and (3) it follows that G|X is the line graph of a tree K; thus E(K) = X, and for x, y ∈ X, x, y are adjacent in G if and only if some vertex of K is incident with them both. Let W be the set of vertices of K that have degree one in K. By (3), every z ∈ Z is incident in K with a member of W . Moreover, if x ∈ E(K) is incident with a member of W , and x ∈ / Z, then one of Ax , Bx is empty, which is impossible; so Z is equal to the set of edges of K incident with members of W . But then (K, W ) is a frame for (G, Z). This proves 5.1. Let η be an H-strip structure in (G, Z). If e ∈ E(H) with ends u, v, an e-rung of η means an induced path G|η(e) with vertices p1 , . . . , pk in order, where for 1 ≤ i ≤ k, pi ∈ η(e, u) if and only if i = 1, and pi ∈ η(e, v) if and only if i = k. (Possibly k = 1.) An H-strip structure η in (G, Z) is said to be connected if for every e ∈ E(H), η(e) is nonempty, and η(e) is the union of the vertex sets of the e-rungs of η. 13
5.2 There is a graph H with the following properties, where W denotes the set of vertices of H of degree one: • (H, W ) is a frame • no vertex of H has degree two • there is a connected H-strip structure in (G, Z) • subject to these three conditions, |E(H)| is maximum. Proof. By 5.1, there is a frame (K, W ) for (G, Z). Let W2 be the set of vertices of K that have degree two, and let W3 be the set that have degree at least three; thus W, W2 , W3 are pairwise disjoint and have union V (K). Let H be the graph with vertex set W ∪ W3 , in which vertices u, v are adjacent if there is a branch of K with ends u, v. Hence for each edge e ∈ E(H) there is a branch Be of K with the same ends as e. Thus (H, W ) is a frame (though no longer a frame for (G, Z), in general), and no vertex of H has degree two. Define η as follows: • for each e ∈ E(H), η(e) = E(Be ) ⊆ V (G) • for each e ∈ E(H) incident with v ∈ V (H), η(e, v) = {f } where f ∈ E(Be ) ⊆ V (G) is the edge of Be incident with v. It follows that η is a connected H-strip structure in (G, Z). Thus the first three conditions of the theorem are satisfied. Since the sets η(e) (e ∈ E(H)) are nonempty (since η is connected) and pairwise disjoint, it follows that |E(H)| ≤ |V (G)| for every choice of H satisfying the first three conditions above, and therefore the fourth can also be satisfied. This proves 5.2. Henceforth in the section, H, W will be as in 5.2. Moreover, η will be a connected H-strip structure in (G, Z), chosen with ∪η maximal, where ∪η denotes the union of all the sets η(e) (e ∈ E(H)). For each e = uv ∈ E(H), define M (e) = η(e, u) ∩ η(e, v). For each v ∈ V (H), define [ N (v) = η(e, v), e∈δH (v)
and for every triangle D = {v1 , v2 , v3 } of H, define N (D) = M (v1 v2 ) ∪ M (v2 v3 ) ∪ M (v3 v1 ). 5.3 Let p ∈ V (G) \ ∪η, and let Y denote the set of all neighbours of p in ∪η. Then either • there is an edge e of H such that Y ⊆ η(e), or • there is a vertex v of H such that Y ⊆ N (v), or • there is a triangle D of H such that Y ⊆ N (D). Proof. (1) For each e ∈ E(H), let Re be an e-rung. Let R be the union of all the sets V (Re ) (e ∈ E(H)). Then one of the following holds: 14
• there exists e ∈ E(H) with Y ∩ R ⊆ V (Re ), or • there exists v ∈ V (H) such that Y ∩ R ⊆ N (v), or • there is a triangle {u, v, w} of H such that Ruv , Rvw , Rwu all have length zero, and Y ∩ R = V (Ruv ) ∪ V (Rvw ) ∪ V (Rwu ), or • there exists e = uv ∈ E(H) such that u ∈ / W and Y ∩ (R \ V (Re )) = N (u) ∩ (R \ V (Re )) and Y ∩ V (Re ) 6⊆ N (u), or • there exists e = uv ∈ E(H) such that u, v ∈ / W and Y ∩ (R \ V (Re )) = (N (u) ∪ N (v)) ∩ (R \ V (Re )). For let K be obtained from H by replacing each edge e ∈ E(H) by a path with edges the vertices of Re in order, in the natural way, so that G|R is the line graph of K. Thus (K, W ) is a frame for (G, Z). Let F = R ∩ Y ; then F ⊆ E(K). Moreover, F is focused, since Z is constricted in G. Hence one of the six outcomes of 4.1 holds. If 4.1.3 holds and x, y are as in 4.1.3, then there is a frame (K ′ , W ) for G, where K ′ is obtained from K by adding p to K as a new edge incident with x, y, contrary to the maximality of |E(H)|. If 4.1.1 holds, then the second outcome of (1) holds. Similarly if one of 4.1.2, 4.1.4, 4.1.5, 4.1.6 holds then respectively the third, first, fourth and fifth outcome of (1) holds. This proves (1). (2) If there is an edge e = v1 v2 of H such that Y ⊆ η(e) ∪ N (v1 ) ∪ N (v2 ) then the theorem holds. Suppose there is such an edge e = v1 v2 . For each f ∈ E(H) choose an f -rung Rf . For i = 1, 2, let Ei be the set of all edges of H that are incident with vi and different from e; thus |Ei | = 6 1. Let Ai be the set of all f ∈ Ei such that Y contains the end of Rf in N (vi ), and let Bi = Ei \ Ai . Suppose first that Y ∩ N (v2 ) ⊆ η(e). If also Y ∩ N (v1 ) ⊆ η(e) then Y ⊆ η(e) and the theorem holds, so we may assume that Y ∩N (v1 ) 6⊆ η(e). Moreover, we may assume that Y ∩η(e) 6⊆ N (v1 ), for otherwise Y ⊆ N (v1 ) and the theorem holds. Hence we may choose the rungs Rf (f ∈ E(H)) such that there exists a1 ∈ A1 and Y ∩ V (Re ) 6⊆ N (v1 ). By (1), it follows routinely that B1 = ∅. Since this holds for all choices of Rf (f 6= a1 , e), we deduce that Y ∩ η(f ) = η(f, v1 ) for all f ∈ E(H) \ {a1 , e} incident with v1 . Since |E1 | = 6 1, it follows by exchanging the roles of a1 and some other member of E1 that Y ∩ η(a1 ) = η(a1 , v1 ), and so N (v1 ) \ η(e) = Y \ η(e). But then a can be added to η(e) and to η(e, v1 ), contrary to the maximality of ∪η. Thus we may assume that Y ∩ N (v2 ) 6⊆ η(e), and similarly Y ∩ N (v1 ) 6⊆ η(e). Hence we may choose the Rf (f ∈ E(H)) such that A1 , A2 are both nonempty. Suppose that there is a choice of the Rf (f ∈ E(H)) such that for some a1 ∈ A1 and a2 ∈ A2 , either a1 , a2 are disjoint edges of H, or not both Ra1 , Ra2 have length zero. By (1), B1 , B2 are both empty. Since this holds for all choices of Rf (f 6= a1 , a2 ), we deduce that Y ∩ η(f ) = η(f, vi ) for i = 1, 2 and for all f ∈ E(G) \ {e, a1 , a2 } incident with vi . Now there exist a′1 ∈ A1 and a′2 ∈ A2 such that a′1 6= a1 and a′1 , a′2 are disjoint edges of H, since |E1 |, |E2 | ≥ 2 (possibly a′2 = a2 ). We have seen that we can choose Ra′1 , Ra′2 such that they both meet Y , and so by the same argument with a1 , a2 replaced by a′1 , a′2 , it follows that Y ∩ η(a1 ) = η(a1 , v1 ), and so N (vi ) \ η(e) ⊆ Y for i = 1, and also for i = 2 by the symmetry. But then a can be added to η(e), η(e, v1 ) and η(e, v2 ), contrary to the maximality of ∪η. 15
Hence for every choice of the Rf (f ∈ E(H)) with A1 , A2 both nonempty, we have that |A1 | = |A2 | = 1, say Ai = {ai } for i = 1, 2, and Ra1 , Ra2 both have length zero, and a1 , a2 have a common end w in H. Take some such choice of the Rf (f ∈ E(H)). For f ∈ B1 , we deduce that we cannot replace Rf with some other f -rung that meets Y , and therefore Y ∩ (N (v1 ) \ η(e)) ⊆ η(a1 , v1 ). Moreover, there is no a1 -rung that meets Y with length greater than zero, and so Y ∩η(a1 ) ⊆ η(a1 , v1 )∩η(a1 , w). Similarly Y ∩ (N (v2 ) \ η(e)) ⊆ η(a2 , v2 ), and Y ∩ η(a2 ) ⊆ η(a2 , v2 ) ∩ η(a2 , w). We may assume that Y 6⊆ N (w), and so Y ∩ η(e) 6= ∅. Choose Re with Y ∩ V (Re ) nonempty. By (1), every such choice of Re has length zero. But then Y ⊆ N (D) where D is the triangle {v1 , v2 , w}, and the theorem holds. This proves (2). (3) For each e ∈ E(H), let Re be an e-rung. If either the fourth or fifth outcome of (1) holds, then the theorem holds. For let R be the union of the sets V (Re ) (e ∈ E(H)). Suppose first that there exists e = uv ∈ E(H) such that u ∈ / W and Y ∩ (R \ V (Re )) = N (u) ∩ (R \ V (Re )) and Y ∩ V (Re ) 6⊆ N (u). Then (1) implies that η(f ) ∩ Y = ∅ for all f ∈ E(H) not incident with u (because otherwise we could make another choice of Rf so that (1) was violated); and η(f ) ∩ Y ⊆ N (u) for each f 6= e incident with u, for the same reason. But then Y ⊆ η(e) ∪ N (u) ∪ N (v), and so the theorem holds by (2). Next suppose that there exists e = uv ∈ E(H) such that u, v ∈ / W and Y ∩ (R \ V (Re )) = (N (u) ∪ N (v)) ∩ (R \ V (Re )). Again by (1), Y ∩ η(f ) = ∅ for each edge f of H not incident with u, v, and Y ∩ η(f ) ⊆ η(f, u) for every f 6= e incident with u, and a similar result holds with u, v exchanged. But then Y ⊆ η(e) ∪ N (u) ∪ N (v) and the theorem holds by (2). This proves (3). Suppose that there exists e = v1 v2 ∈ E(H) such that Y ∩ η(e) 6⊆ η(e, v1 ) ∪ η(e, v2 ). Choose an e-rung Re such that some internal vertex of Re belongs to Y . By (1) and (3), Y ∩ η(f ) = ∅ for all f ∈ E(H) \ {e}, and so the theorem holds. Hence every vertex in Y belongs to at least one of the sets N (v) (v ∈ V (H)). Suppose that some y ∈ Y belongs to exactly one such set; say y ∈ η(e, v) \ M (e), where e = uv. By (1), Y \ η(e) ⊆ N (v), and so the theorem holds by (2). Thus we may assume that every vertex in Y belongs to M (e) for some e ∈ E(H). Let F be the set of all f ∈ E(H) with Y ∩ M (f ) 6= ∅. If there exist e, f ∈ F with no common end in H, then the theorem holds by (1) and (3); if there is some vertex v of H incident with every edge in F , then Y ⊆ N (v) and the theorem holds; and if neither of these hold, then |F | = 3, and the three members of F are the edges of a triangle D of H, and Y ⊆ N (D) and the theorem holds. This proves 5.3. 5.4 Let X ⊆ V (G) \ ∪η such that G|X is connected, and let Y be the set of all attachments of G|X in ∪η. Then either • there is an edge e of H such that Y ⊆ η(e), or • there is a vertex v of H such that Y ⊆ N (v), or • there is a triangle D of H such that Y ⊆ N (D). Proof. Suppose that this is false for some X, and choose X minimal such that 5.4 is false for X. (1) There exist y1 , y2 in Y such that {y1 , y2 } 6⊆ η(e) for each e ∈ E(H), and {y1 , y2 } 6⊆ N (v) 16
for each v ∈ V (H), and {y1 , y2 } 6⊆ N (D) for each triangle D of H. For suppose first that for some e = v1 v2 ∈ E(H), there exists y1 ∈ Y ∩η(e) with y1 ∈ / η(e, v1 )∪η(e, v2 ). Now Y 6⊆ η(e); choose y2 ∈ Y \ η(e), and then y1 , y2 satisfy (1). We may therefore assume that Y ⊆ ∪v∈V (H) N (v). Next suppose that some y1 ∈ Y belongs to exactly one of the sets N (v) (v ∈ V (H)); say y1 ∈ η(e, v1 ) \ M (e), where e = v1 v2 ∈ E(H). Since Y 6⊆ N (v1 ), there exists y2 ∈ Y \ N (v2 ). If also y2 ∈ / η(e), then the pair y1 , y2 satisfies (1), so we may assume that y2 ∈ η(e) \ N (v1 ). We already assumed that every member of Y belongs to one of the sets N (v) (v ∈ V (H)), and so y2 ∈ η(e, v2 ) \ M (e). This restores the symmetry between v1 and v2 . Since Y 6⊆ η(e), there exists y3 ∈ Y with y3 ∈ / η(e). Since we may assume that the pair y1 , y3 does not satisfy (1), it follows that y3 ∈ N (v1 ), and similarly y3 ∈ N (v2 ). Let f ∈ E(H) with y3 ∈ η(f ); then f is incident with v1 since y3 ∈ N (v1 ), and similarly f is incident with v2 , and so f = e, a contradiction since y3 ∈ / η(e). We may therefore assume that every y ∈ Y belongs to M (e) for some e ∈ E(H). Let F be the set of all edges f ∈ E(H) such that Y ∩ M (f ) 6= ∅. Suppose that there exist e, f ∈ F with no common end in H. Choose y1 ∈ Y ∩ M (e) and y2 ∈ Y ∩ M (f ); then y1 , y2 satisfy (1). Thus we may assume that every two edges in F share an end. Consequently either there is a vertex v ∈ V (H) incident with every member of F , or |F | = 3 and the three edges in F form a triangle D of H. In the first case Y ⊆ N (v), and in the second Y ⊆ N (D), in either case a contradiction. This proves (1). For each p ∈ X, let Y (p) denote the set of all v ∈ ∪η adjacent to p; and for P ⊆ X, let Y (P ) = ∪p∈P Y (p). Thus Y = Y (X). Let y1 , y2 be as in (1). Then y1 , y2 are nonadjacent. Since G|X is connected, there is an induced path of G with vertices y1 -p1 -p2 - · · · -pk -y2 in order. By 5.3 it follows that k > 1. From the minimality of X, X = {p1 , . . . , pk }. Let P1 = X \ {pk }, and P2 = X \ {p1 }. Then for i = 1, 2, the minimality of X implies that either • there is an edge e of H such that Y (Pi ) ⊆ η(e), or • there is a vertex v of H such that Y (Pi ) ⊆ N (v), or • there is a triangle D of H such that Y (Pi ) ⊆ N (D). Thus there are three possibilities for Y (P1 ) and three for Y (P2 ), and we need to check these nine possibilities individually. For each e ∈ E(H) choose an e-rung Re (in some cases we shall need to choose the e-rungs subject to some further conditions), and let K be the graph obtained from H by replacing every edge e of H by a path whose edges are the vertices of Re in the corresponding order. Then (K, W ) is a frame for (G, Z). (Thus K depends on the choice of the rungs Re . In what follows it is sometimes useful to change one or more of the paths Re , and K is assumed to change correspondingly, although we may not say so explicitly.) We remark that the subgraph of G induced on the union of the sets V (Re ) (e ∈ E(H)) is the line graph of K. (2) There do not exist v1 , v2 ∈ V (H) such that Y (P1 ) ⊆ N (v1 ) and Y (P2 ) ⊆ N (v2 ). For suppose that such v1 , v2 exist. Since Y = Y (P1 ) ∪ Y (P2 ), it follows that v1 6= v2 . Now v1 , v2
17
may or may not be adjacent in H. If they are adjacent, let f = v1 v2 ∈ E(H), and otherwise f is undefined. For 1 < i < k, Y (pi ) ⊆ Y (P1 ) ∩ Y (P2 ) ⊆ N (v1 ) ∩ N (v2 ) = M (f ), (where M (f ) = ∅ if f is not defined) and so Y ⊆ Y (p1 )∪Y (pk )∪M (f ). Suppose first that for i = 1, 2, either N (vi ) \ η(f, vi ) ⊆ Y (pi ) or (N (vi ) \ η(f, vi )) ∩ Y (pi ) = ∅ (where η(f, vi ) = ∅ if f is undefined). If (N (vi ) \ η(f, vi )) ∩ Y (pi ) = ∅ for i = 1, 2, then Y ⊆ η(f ) (where η(f ) = ∅ if f is undefined), a contradiction, so we may assume that N (v1 ) \ η(f, v1 ) ⊆ Y (p1 ) say. If N (v2 ) \ η(f, v2 ) ∩ Y (pk ) = ∅, then f is defined since Y (pk ) 6= ∅, and we can add p1 , . . . , pk to η(f ), and add p1 to η(f, v1 ), contrary to the maximality of ∪η; if N (v2 ) \ η(f, v2 ) ⊆ Y (pk ) and f is defined, we can add p1 , . . . , pk to η(f ), add p1 to η(f, v1 ), and add pk to η(f, v2 ), again contrary to the maximality of ∪η; so we may assume that N (v2 ) \ η(f, v2 ) ⊆ Y (pk ) and f is undefined. Thus Y (p1 ) = N (v1 ) and Y (pk ) = N (v2 ). Let K ′ be obtained from K by adding a path between v1 , v2 with edges p1 , . . . , pk in order; then (K ′ , W ) is a frame for (G, Z), contrary to the maximality of |E(H)|. Thus we may assume that for some i ∈ {1, 2}, N (vi ) \ η(f, vi ) 6⊆ Y (pi ) and (N (vi ) \ η(f, vi )) ∩ Y (pi ) 6= ∅. For i = 1, 2, let Ei be the set of edges of K incident with vi if f is undefined, and let Ei is the set of edges of K incident with vi not in the branch of K between v1 , v2 if f is defined. Therefore we may choose the paths Re (e ∈ E(H)) such that at least three of the sets A1 , B1 , A2 , B2 are nonempty, where for i = 1, 2, Ai = Ei ∩ Y (Pi ), and Bi = Ei \ Ai . We may also assume that for i = 1, 2, either yi ∈ E(K), or |Bi | = 1 and one of Aj , Bj is empty, where {i, j} = {1, 2}. (To see this, observe that if say y1 ∈ / E(K), let y1 ∈ η(e) say, and choose an e-rung Re′ containing y1 ; then if either |B1 | = 6 1, or A2 , B2 are both nonempty, we may replace Re by Re′ , and still satisfy all the other requirements.) From the symmetry we may assume that A1 , B1 6= ∅, and hence y2 ∈ E(K). Since (K, W ) is a frame, there are three paths Q1 , Q2 , Q3 of K such that • Q1 is between v1 and w1 ∈ W , • Q2 is between v1 and w2 ∈ W , • Q3 is between v2 and w3 ∈ W , • V (Q1 ∩ Q2 ) = {v1 }, • Q3 is vertex-disjoint from both Q1 , Q2 , • the edge a1 = u1 v1 of Q1 incident with v1 belongs to A1 , • the edge of Q2 incident with v1 belongs to B1 , and • Q3 is an induced subgraph of K. If the edge of Q3 incident with v2 belongs to A2 , then G|(X ∪ E(Q1 ) ∪ E(Q2 ) ∪ E(Q3 )) is an induced tree containing three vertices of Z, a contradiction. Thus the first edge of Q3 is in B2 . If y2 ∈ / A2 , then f is defined and y2 belongs to the branch of K between v1 , v2 , and this branch has length at least two (since y2 ∈ / N (v1 )); but then G|(X ∪ E(Q1 ) ∪ E(Q2 ) ∪ E(Q3 ) ∪ {y2 }) 18
Let e be incident with v2 , v3 in H. Since (K, W ) is a frame, there are three paths Q1 , Q2 , Q3 of K, such that Q1 is from v1 to some w1 ∈ W , Q2 is from v1 to some w2 ∈ W , Q3 is from one of v2 , v3 to some w3 ∈ W , V (Q1 ) ∩ V (Q2 ) = {v1 }, Q3 is vertex-disjoint from both Q1 and Q2 , the edge of Q1 incident with v1 belongs to A1 , and the edge of Q2 incident with v1 belongs to B1 . Moreover, we may assume that only one of v2 , v3 belongs to V (Q3 ), say v3 . It follows that v1 6= v3 . Since none of Q1 , Q2 , Q3 contain both v2 , v3 , we may alter our choice of Re without affecting Q1 , Q2 , Q3 ; and since Y (P2 ) 6⊆ N (v2 ) by (2), we may choose Re such that some vertex of Re belongs to Y (P2 ) \ N (v2 ). Let S be a minimal subpath of Re that meets both Y (P2 ) and N (v3 ). Then G|(E(Q1 ) ∪ E(Q2 ) ∪ E(Q3 ) ∪ V (S) ∪ {p1 , . . . , pk }) is an induced tree of G containing three vertices of Z, a contradiction. This proves (3). (4) There do not exist edges e1 , e2 of H such that Y (Pi ) ⊆ η(ei ) for i = 1, 2. For suppose that such edges exist. Then e1 6= e2 ; let ei have ends ui , vi for i = 1, 2. For i = 1, 2, let Ti be the branch of K with edge set V (Rei ). Since η(e1 ) ∩ η(e2 ) = ∅, it follows that Y = Y (p1 ) ∪ Y (p2 ). We may assume that v1 6= u2 , v2 and v2 6= u1 , v1 ; that is, u1 , u2 , v1 , v2 are all distinct except that possibly u1 = u2 . For i = 1, 2, choose Rei such that some vertex of Rei belongs to Y (Pi ), and in addition, choose Rei such that some vertex of Rei belongs to Y (Pi ) and not to N (ui ) (this is possible since Y (Pi ) 6⊆ N (ui ) by (3)). Suppose first that for i = 1, 2, exactly two vertices xi , yi of Rei belong to Y (Pi ) and they are adjacent. Thus xi , yi are edges of Ti with a common end ti say. Let K ′ be obtained from K by adding a new path between t1 , t2 with edges p1 , . . . , pk in order; then (K ′ , W ) is a frame for (G, Z), contrary to the maximality of |E(H)|. We may therefore assume that either exactly one vertex of Re1 belongs to Y (p1 ), or two nonadjacent vertices of Re1 belong to Y (p1 ). Let Q1 , Q2 , Q3 be vertex-disjoint paths of K between {u1 , v1 , v2 } and W , where Q1 is between u1 and some w1 ∈ W , and Q2 is between v1 and some w2 in W , and Q3 is between v2 and some w3 ∈ W . (Possibly u2 belongs to one of these paths.) Then some edge of T2 belongs to Y (p2 ) and is not incident with u2 , from the choice of Re2 . Hence there is a path S3 of T2 ∪ Q3 , with first vertex in V (T2 ) and last vertex w3 , such that the first edge and no other edge of S3 belongs to Y (p2 ), and Q1 , Q2 , S3 are pairwise vertex-disjoint. If only one vertex of Re1 is in Y (p1 ), then G|(E(Q1 ) ∪ E(Q2 ) ∪ E(S3 ) ∪ {p1 , . . . , pk }) is an induced tree of G containing three members of Z, a contradiction. Thus there are two nonadjacent vertices in V (Re1 ) ∩ Y (p1 ), and so there are vertex-disjoint subpaths S1 , S2 of Q1 ∪ T1 ∪ Q2 , such that for i = 1, 2, Si has first vertex in V (T1 ), first edge and no other edge in Y (p1 ), and last vertex wi . But then G|(E(S1 ) ∪ E(S2 ) ∪ E(S3 ) ∪ {p1 , . . . , pk }) is an induced tree of G containing three members of Z, a contradiction. This proves (4). From (2),(3),(4), we may assume that Y (P2 ) ⊆ N (D) for some triangle D = {u1 , u2 , u3 } of H, and that M (u1 u2 ), M (u2 u3 ), M (u1 u3 ) all contain at least one member of Y (P2 ). Let e1 , e2 , e3 be the edges u2 u3 , u3 u1 , u1 u2 of H respectively. Choose Re1 of length zero such that its vertex (r1 say) is in Y (P2 ), and choose Re2 , Re3 similarly. Thus r1 is the edge of K joining u2 , u3 . For i = 1, 2, 3, 20
let Qi be a path of K between ui and some wi ∈ W , such that Q1 , Q2 , Q3 are pairwise vertex-disjoint. (5) Y (P1 ) ∩ E(K) ⊆ {r1 , r2 , r3 }. For let e ∈ Y (P1 ) ∩ E(K), and suppose first that at most one of Q1 , Q2 , Q3 contains an end of e. Since K is connected, we may choose a path S of K with first edge in Y (p1 ) such that S meets one of Q1 , Q2 , Q3 ; and by choosing S minimal, we may assume that S meets Q3 and not Q1 , Q2 , and only its first edge is in Y (p1 ). In particular, u1 , u2 ∈ / V (S), and so no edge of S is in N (D); and therefore no edge of S except the first is adjacent in G to any of p1 , . . . , pk . Let S ′ be a path of S ∪ Q3 between the first vertex of S and w3 . Then G|(E(Q1 ) ∪ E(Q2 ) ∪ E(S ′ ) ∪ {p1 , . . . , pk }) is an induced tree in G containing three members of Z, a contradiction. This proves that two of Q1 , Q2 , Q3 contain ends of e. Let e = v1 v2 where v1 ∈ V (Q1 ) and v2 ∈ V (Q2 ) say. Suppose that v2 6= u2 . For i = 1, 2, let Si be a subpath of Qi between vi and wi . Then G|(E(S1 ) ∪ E(S2 ) ∪ E(Q3 ) ∪ {e, r1 } ∪ {p1 , . . . , pk }) is an induced tree in G containing three members of Z, a contradiction. Thus v2 = u2 , and similarly v1 = u1 and so e = r3 . This proves (5). From (5), we deduce that Y (P1 ) ⊆ η(e1 ) ∪ η(e2 ) ∪ η(e3 ) (for otherwise we could make a choice of the Rf for f 6= u1 u2 , u1 u3 , u2 u3 that would violate (5)). Since Y (P1 ) 6⊆ N (D), we may assume that there is some e3 -rung Re′ 3 such that some vertex of Re′ 3 belongs to Y (P1 ) and not to η(e3 , u2 ) (and so Re′ 3 has length at least one). Let S be a minimal subpath of Re′ 3 that meets both Y (P1 ) and η(e3 , u1 ). Then G|(E(Q1 ) ∪ E(Q2 ) ∪ E(Q3 ) ∪ V (S) ∪ {r1 } ∪ {p1 , . . . , pk }) is an induced tree in G containing three members of Z, a contradiction. This proves 5.4. Proof of 3.2. We have already seen the proof of the “if” half. To prove the “only if” half, we may assume that |Z| ≥ 3. We choose H, W as in 5.2. Choose η as before; that is, η is a connected H-strip structure in (G, Z), chosen with ∪η maximal. Let C be the set of all vertex sets of components of G \ ∪η. For C ∈ C we define its home as follows. Let Y be the set of attachments of G|C in G| ∪ η. We say that e ∈ E(H) is the home of C if Y ⊆ η(e); v ∈ V (H) is the home of C if Y ⊆ N (v) and no edge of H is the home of C; and a triangle D of H is the home of C if Y ⊆ N (D) and there is no vertex or edge of H that is the home of C. By 5.4, each C ∈ C has a (unique) home. For each e ∈ E(H), let η ′ (e) be the union of η(e) and all C ∈ C with home e. Define η ′ (e, v) = η(e, v) if v ∈ V (H) is incident with e. For each v ∈ V (H), define η ′ (v) to be the union of all C ∈ C with home v, and for each triangle D of H, let η ′ (D) be the union of all C ∈ C with home D. It follows that η ′ is an extended H-strip decomposition of (G, Z). This proves 3.2.
6
The algorithm
So far, we have proved our main result, the description of the structure of the constricted pairs; now we present a polynomial-time algorithm for the following question: 21
CONSTRICTED. With input a graph G and Z ⊆ V (G), decide whether Z is constricted. Our main theorem 3.2 asserts that for a given pair (G, Z), either there is an induced subtree containing three members of Z, or (G, Z) admits an extended H-strip decomposition for some H, and not both. Therefore we could ask for a polynomial algorithm for the following more demanding question: TREE-OR-DECOMPOSITION. With input a connected graph G and Z ⊆ V (G) with |Z| ≥ 2, output either an induced subtree of G containing at least three members of Z, or a graph H and an extended H-strip decomposition η of (G, Z). We will first give an algorithm for CONSTRICTED, and then discuss how to modify it to solve TREE-OR-DECOMPOSITION. It would be nice to use 3.2 to show that some simple algorithm works, but so far we have not been able to do this. The best method we can see is just to convert the proof of the theorem to an algorithm. Thus, we have input a graph G and a subset Z ⊆ V (G), and we wish to decide whether Z is constricted in G. It is easy to reduce this question to the special case when G is connected and |Z| ≥ 3, so from now on we assume that. Let |V (G)| = n. Let η be a connected H-strip structure in (G, Z), let X ⊆ V (G) \ ∪η, and let Y be the set of all members of ∪η that have at least one neighbour in X. We say that X is local (with respect to (H, η)) if either Y ⊆ η(e) for some e ∈ E(H), or Y ⊆ N (v) for some v ∈ V (H), or Y ⊆ N (D) for some triangle D of H (using the notation of 5.3 and 5.4). It is easy to modify the first part of the proof of 5.4, and the proof of the “only if” half of 3.2, to yield the following subroutine. 6.1 Algorithm. • Input: A pair (G, Z) as above, a graph H, and a connected H-strip structure η in (G, Z). • Output: One of the following: – An extended H-strip decomposition of (G, Z), or – a subset X ⊆ V (G) \ ∪η, such that G|X is connected, X is not local, and X is minimal with these properties. • Running time: O(n2 ). Let η be some connected H-strip structure in (G, Z). We define its worth to be |∪η|+(n+1)|E(H)|. Thus every H-strip structure has worth at most (n + 1)2 , since | ∪ η| ≤ n and |E(H)| ≤ n. We need another subroutine, as follows. 6.2 Algorithm. • Input: A pair (G, Z) as above, a graph H, a connected H-strip structure η in (G, Z), and a subset X ⊆ V (G) \ ∪η, such that G|X is connected, X is not local, and X is minimal with these properties. • Output: One of the following: 22
– A graph H ′ and a connected H ′ -strip structure η ′ in (G, Z) with worth greater than that of η, or – a (true) statement that Z is not constricted in G|((∪η) ∪ X), and therefore not in G. • Running time: O(n2 ). This can be obtained by modifying the proof of 5.3 and the latter part of the proof of 5.4 appropriately. More exactly, if |X| = 1 we modify the proof of 5.3, and otherwise we modify the proof of 5.4. Suppose first that |X| = 1. We test whether one of the five statements of step (1) of the proof of 5.3 holds. The first three are impossible since X is not local. If either the fourth or fifth holds, we can add the vertex in X to one of the strips η(e) and produce a connected H-strip structure in (G, Z) with worth greater than that of η as required. If none of the outcomes of step (1) of the proof of 5.3 holds, it follows that either Z is not constricted in G|((∪η) ∪ X) is not constricted and we are done, or H, η does not satisfy the hypotheses of 5.3; and in this case it is easy to modify the proof of 5.3 to yield a connected H ′ -strip structure η ′ in (G, Z) with worth greater than that of η. When |X| > 1 we modify the proof of 5.4 in an analogous way. Combining these two subroutines yields: 6.3 Algorithm. • Input: A pair (G, Z) as above, a graph H, and a connected H-strip structure η in (G, Z). • Output: One of the following: – An extended H-strip decomposition of (G, Z), or – a graph H ′ and a connected H ′ -strip structure η ′ in (G, Z) with worth greater than that of η, or – a statement that Z is not constricted in G. • Running time: O(n2 ). To make use of 6.3 to solve CONSTRICTED, we first choose some frame (K, W ) for (G, Z), by using the method of 5.1. (Or we find some induced tree containing at least three members of Z, and then we output this and stop.) This takes time O(n2 ), where n = |V (G)|. We convert this to a connected H1 -strip structure η1 say. Inductively, having produced a graph Hi and a connected Hi -strip structure ηi of worth at least i, we apply 6.3 to Hi , ηi . If we obtain either the first or third output of 6.3 we are done, and if we obtain the second output this defines Hi+1 , ηi+1 , of worth at least i + 1. Since no ηi has worth more than (n + 1)2 , this process iterates at most (n + 1)2 times before terminating, and since each iteration takes time O(n2 ), the total running time is O(n4 ). To solve TREE-OR-DECOMPOSITION instead of just CONSTRICTED, note first that the algorithm just described outputs an extended H-decomposition for some H when Z is constricted, so we have half of what we need for free. We just need to modify the last output of 6.2. At that stage, instead of just the statement that Z is not constricted in G, we need to find an induced subtree of G containing three members of Z; and to do so we need to look more carefully at the proofs of 4.1, 5.3 and 5.4 at the appropriate points. Since we will exit the main recursion at this stage, we can afford to spend time O(n4 ) instead of just O(n2 ) to exhibit the desired tree, and this is easily done (we
23
omit the details). Thus the running time of the algorithm to solve TREE-OR-DECOMPOSITION is also O(n4 ). There is also a simpler method to solve TREE-OR-DECOMPOSITION (but with running time O(n5 )). We repeatedly use the algorithm of CONSTRICTED as a subroutine. We may assume that initially Z is not constricted in G (for otherwise the algorithm above for CONSTRICTED outputs an extended H-strip decomposition and we are done). For each vertex v in turn, test whether Z \ {v} is constricted in G \ {v}, and if not then remove v from G, Z. What remains at the end is the desired tree.
References [1] D. Bienstock, “On the complexity of testing for even holes and induced odd paths”, Discrete. Math. 90 (1991), 85-92. Corrigendum in Discrete Math. 102 (1992), 109. [2] M. Chudnovsky and R. Kapadia, “Detecting a theta or a prism”, SIAM J. Discrete Math. 22 (2008), 1–18. [3] M. Chudnovsky, G. Cornu´ejols, X. Liu, P. Seymour and K. Vuˇskovi´c, “Recognizing Berge graphs”, Combinatorica 25 (2005), 143–186. [4] R. Kapadia, Detecting a Theta or a Prism, Senior Thesis, Princeton, May 2006. [5] F. Maffray and N. Trotignon, “Algorithms for perfectly contractile graphs”, SIAM J. Discrete Math. 19 (2005), 553–574.
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Monty Hall Problem
The Monty Hall Problem is a problem in probabilistic reasoning. It is often stated as follows.
Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice? [Whitaker/vos Savant 1990]
You can read more about the Monty Hall Problem in a great Wikipedia article.
Play
Door No. 1 Door No. 2 Door No. 3
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final_practice_07
# final_practice_07 - Physics 9B Practice Exam...
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Physics 9B Practice Exam There are six problems. Work on the test for an hour and we will go over it. [1] A three-dimensional model of a molecule treats the atoms as point particles located at the corners of a cube and connected by massless rigid rods. a) (6 pts) List the degrees of freedom this molecule has. b) (14 pts) If 3 moles of a gas of these molecules undergoes an adiabatic expansion from a pressure of 5.0 atm and a volume of 1.0 L to a volume of 3.0 L, find the final pressure. [2] A transverse wave on a string is described by y = 3 mm ( ) cos π m 1 ( ) x 9 π s 1 ( ) t ( ) . At t = 0, find the position at which the transverse velocity of the string is a maximum and positive. [3] A spherical source of water emits water at a constant rate of 5.0 m 3 /s. Derive an expression for the velocity of the water flow at a distance r from the center of the source. Assume no mass loses and use conservation of mass. For water ® = 1000.0 kg/m 3 [4] In a double slit experiment the separation distance between the centers of the slits is five times the width of each slit. Find the ratio of the intensity of the third interference maximum relative to intensity of the central maximum. You are to assume the double slit interference pattern is modulated by the single
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# Problem 59636. Swap between first and last column
The idea is to swap between first and last column
Ex = [1 2 3 4 5;
1 2 3 4 5;
1 2 3 4 5;
1 2 3 4 5;
1 2 3 4 5];
will be:
Ex = [5 2 3 4 1;
5 2 3 4 1;
5 2 3 4 1;
5 2 3 4 1;
5 2 3 4 1];
If you like the problem, please like it :)
### Solution Stats
69.86% Correct | 30.14% Incorrect
Last Solution submitted on Apr 13, 2024
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Find the treasures in MATLAB Central and discover how the community can help you!
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NurbsSolve Class Reference
Solving the linear system of equations using Eigen or UmfPack. (can be defined in on_nurbs.cmake) More...
`#include <opennurbs_fit.h>`
## Public Member Functions
NurbsSolve ()
Empty constructor. More...
void assign (unsigned rows, unsigned cols, unsigned dims)
Assign size and dimension (2D, 3D) of system of equations. More...
void K (unsigned i, unsigned j, double v)
Set value for system matrix K (stiffness matrix, basis functions) More...
void x (unsigned i, unsigned j, double v)
Set value for state vector x (control points) More...
void f (unsigned i, unsigned j, double v)
Set value for target vector f (force vector) More...
double K (unsigned i, unsigned j)
Get value for system matrix K (stiffness matrix, basis functions) More...
double x (unsigned i, unsigned j)
Get value for state vector x (control points) More...
double f (unsigned i, unsigned j)
Get value for target vector f (force vector) More...
void resizeF (unsigned rows)
Resize target vector f (force vector) More...
void printK ()
Print system matrix K (stiffness matrix, basis functions) More...
void printX ()
Print state vector x (control points) More...
void printF ()
Print target vector f (force vector) More...
bool solve ()
Solves the system of equations with respect to x. More...
ON_Matrix * diff ()
Compute the difference between solution K*x and target f. More...
void setQuiet (bool val)
Enable/Disable debug outputs in console. More...
## Detailed Description
Solving the linear system of equations using Eigen or UmfPack. (can be defined in on_nurbs.cmake)
Definition at line 235 of file opennurbs_fit.h.
## Constructor & Destructor Documentation
NurbsSolve ( )
inline
Empty constructor.
Definition at line 239 of file opennurbs_fit.h.
## Member Function Documentation
void assign ( unsigned rows, unsigned cols, unsigned dims )
Assign size and dimension (2D, 3D) of system of equations.
Definition at line 252 of file opennurbs_fit.cpp.
Referenced by FittingSurface::assemble().
void K ( unsigned i, unsigned j, double v )
Set value for system matrix K (stiffness matrix, basis functions)
Definition at line 267 of file opennurbs_fit.cpp.
References OSL::Strings::v.
void x ( unsigned i, unsigned j, double v )
Set value for state vector x (control points)
Definition at line 274 of file opennurbs_fit.cpp.
References OSL::Strings::v.
void f ( unsigned i, unsigned j, double v )
Set value for target vector f (force vector)
Definition at line 279 of file opennurbs_fit.cpp.
References OSL::Strings::v.
double K ( unsigned i, unsigned j )
Get value for system matrix K (stiffness matrix, basis functions)
Definition at line 286 of file opennurbs_fit.cpp.
double x ( unsigned i, unsigned j )
Get value for state vector x (control points)
Definition at line 291 of file opennurbs_fit.cpp.
double f ( unsigned i, unsigned j )
Get value for target vector f (force vector)
Definition at line 296 of file opennurbs_fit.cpp.
void resizeF ( unsigned rows )
Resize target vector f (force vector)
Definition at line 303 of file opennurbs_fit.cpp.
void printK ( )
Print system matrix K (stiffness matrix, basis functions)
void printX ( )
Print state vector x (control points)
void printF ( )
Print target vector f (force vector)
bool solve ( )
Solves the system of equations with respect to x.
• Using UmfPack incredibly speeds up this function.
Definition at line 345 of file opennurbs_fit.cpp.
References solveSparseLinearSystemLQ().
Referenced by FittingSurface::solve().
Here is the call graph for this function:
ON_Matrix* diff ( )
Compute the difference between solution K*x and target f.
void setQuiet ( bool val )
inline
Enable/Disable debug outputs in console.
Definition at line 292 of file opennurbs_fit.h.
Referenced by FittingSurface::setQuiet().
The documentation for this class was generated from the following files:
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# Types of Flows in a Pipe /Fluid Mechanics -Basic Of Fluid Mechanics
Types of Flows in a Pipe – The type of flow of a liquid depends upon the manner in which the particles unite and move. Though there are many types of flows, yet the following are important :
1. Uniform flow: A flow, in which the liquid particles at all sections of a pipe or channel have the same velocities, is called a uniform flow.
2. Non-uniform flow: A flow, in which the liquid particles at different sections of a pipe or channel have different velocities, is called a non-uniform flow.
3. Streamline flow (Laminar Flow) : A flow, in which each liquid particle has a definite path and the paths of individual particles do not cross each other, is called a streamline flow.
4. Turbulent flow: A flow, in which each liquid particle does not have a definite path and the paths of individual particles also cross each other, is called a turbulent flow.
5. Steady flow: A flow, in which the quantity of liquid flowing per second is constant, is called a steady flow. A steady flow may be uniform or non – uniform.
6. Unsteady flow: A flow, in which the quantity of liquid flowing per second is not constant, is called an unsteady flow.
7. Compressible flow: A flow, in which the volume of a fluid and its density changes during the flow, is called a compressible flow. All the gases are considered to have compressible flow.
8. In-compressible flow: A flow, in which the volume of a fluid and its density does not change during the flow, is called an incompressible flow. All the liquids are considered to have incompressible flow.
9. Rotational flow: A flow, in which the fluid particles also rotate (i.e. have some angular velocity) about their own axes while flowing, is called a rotational flow.
10. Irrotational flow: A flow, in which the fluid particles do not rotate about their own axes and retain their original orientations, is called an irrotational flow.
11. One-dimensional flow: A flow, in which the streamlines of its moving particles are represented by straight line, is called an one-dimensional flow.
12. Two-dimensional flow: A flow, whose streamlines of its moving particles are represented by a curve, is called a two-dimensional flow.
13. Three – dimensional flow: A flow, whose streamlines are represented in space i.e. along the three mutually perpendicular directions, is called a three – dimensional flow.
Sachin Thorat
Sachin is a B-TECH graduate in Mechanical Engineering from a reputed Engineering college. Currently, he is working in the sheet metal industry as a designer. Additionally, he has interested in Product Design, Animation, and Project design. He also likes to write articles related to the mechanical engineering field and tries to motivate other mechanical engineering students by his innovative project ideas, design, models and videos.
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## Elementary Statistics (12th Edition)
$0.00484$
We know that $probability=\frac{number \ of \ good \ outcomes}{number\ of\ all\ outcomes}$, therefore $P(not \ guilty)=\frac{58+14}{392+58+564+14}=\frac{72}{1028}\approx0.07003$. By using the multiplication rule: $\frac{72}{1028}\times\frac{71}{1027}=\frac{1278}{263939}\approx0.00484$.
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# SIMPLIFYING EXPRESSIONS
"Simplifying Expressions" is the basic topic required for the students who would like to study Algebra in math.
## What is expression?
Expression is an algebraic or mathematical statement involves numbers,one or more variables and the operation symbols like +,-, x ,÷.
Simplifying expressions means combining all x terms as one x term and combining all constant terms as one constant term.
## Distributing and simplifying expressions
Distribution means sharing something among other recipients. In other words multiply the number or a variable which is outside the parenthesis with the inner terms.
Procedure of multiplying two polynomials with exponents
We will multiply two or more polynomials in the following order.
(1) Symbol
(2) Number
(3) Variable
Let us see how it works
Multiply ( 5 x² ) and (-2 x³)
= ( 5 x² ) x (-2 x³)
10 x
if we have only one x term then we have to put that x term as it is.
Problem 1:
Simplify the expression 7 (x - 3) + 2 (2x - 5) - 3(x - 5)
Solution:
7 (x - 3) + 2 (2x - 5) - 3(x - 5)
let us see how to simplify the given expression step by step
Step 1 :
Distributing the number which is outside the parenthesis with inner terms
= 7x - 21 + 4x - 10 - 3x + 15
Step 2 :
Combine the like terms
= 7x + 4x - 3x - 21 - 10 + 15
= 11x - 3x - 31 + 15
= 8x - 16
Problem 2:
Simplify the expression 4x - (2 + 4x) - 2 (x - 1) - 8 (x -3)
Solution:
= 4x - (2 + 4x) - 2 (x - 1) - 8 (x -3)
= 4x - 2 - 4x - 2x + 2 - 8x + 24
= 4x - 4x - 2x - 8x - 2 + 2 + 24
= -10x + 24
## Simplifying expressions with fractions
Problem 3 :
Simplify the expression (2x - 7)/5 + (x + 9)/15 - (2x -2)/5
Solution :
= (2 x - 7)/5 + (x + 9 )/15 - (2 x - 2 )/5
Now we are going to combine the like terms
Problem 4 :
Simplify the expression (3x+41)/2 + (x-3 )/5 -(9-2x)/6
Solution :
Now we are going to distribute the numbers which is out side the parenthesis to the inner terms.Then combine the like terms
## Related topics
Apart from the stuff given above, if you want to know more about "Simplifying expressions", please click here
Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.
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# Logs
The trunk diameter is 52 cm. Is it possible to inscribe a square prism with side 36 cm?
Result
#### Solution:
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## Next similar math problems:
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What is the smallest diameter of a tree trunk that we can cut a square-section square with a side length of 20 cm?
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It is given a rhombus with side a = 6 cm and the radius of the inscribed circle r = 2 cm. Calculate the length of its two diagonals.
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Calculate the perimeter and area of a rhombus whose diagonals are 39 cm and 51 cm long.
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The prism has a square base with a side length of 3 cm. The diagonal of the sidewall of the prism/BG/is 5 cm. Calculate the surface of this prism in cm square and the volume in liters
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The log has diameter 30 cm. What's largest beam with a rectangular cross-section can carve from it?
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A diagonal of a rhombus is 20 cm long. If it's one side is 26 cm find the length of the other diagonal.
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• Rhombus IV
Calculate the length of the diagonals of the rhombus, whose lengths are in the ratio 1: 2 and a rhombus side is 35 cm.
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In the rectangular triangle ABC has known the length of the legs a = 15cm and b = 36cm. Calculate the length of the median to side c (to hypotenuse).
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## 559603
559,603 (five hundred fifty-nine thousand six hundred three) is an odd six-digits composite number following 559602 and preceding 559604. In scientific notation, it is written as 5.59603 × 105. The sum of its digits is 28. It has a total of 2 prime factors and 4 positive divisors. There are 508,720 positive integers (up to 559603) that are relatively prime to 559603.
## Basic properties
• Is Prime? No
• Number parity Odd
• Number length 6
• Sum of Digits 28
• Digital Root 1
## Name
Short name 559 thousand 603 five hundred fifty-nine thousand six hundred three
## Notation
Scientific notation 5.59603 × 105 559.603 × 103
## Prime Factorization of 559603
Prime Factorization 11 × 50873
Composite number
Distinct Factors Total Factors Radical ω(n) 2 Total number of distinct prime factors Ω(n) 2 Total number of prime factors rad(n) 559603 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 559,603 is 11 × 50873. Since it has a total of 2 prime factors, 559,603 is a composite number.
## Divisors of 559603
4 divisors
Even divisors 0 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 4 Total number of the positive divisors of n σ(n) 610488 Sum of all the positive divisors of n s(n) 50885 Sum of the proper positive divisors of n A(n) 152622 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 748.066 Returns the nth root of the product of n divisors H(n) 3.66659 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 559,603 can be divided by 4 positive divisors (out of which 0 are even, and 4 are odd). The sum of these divisors (counting 559,603) is 610,488, the average is 152,622.
## Other Arithmetic Functions (n = 559603)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 508720 Total number of positive integers not greater than n that are coprime to n λ(n) 254360 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 45936 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 508,720 positive integers (less than 559,603) that are coprime with 559,603. And there are approximately 45,936 prime numbers less than or equal to 559,603.
## Divisibility of 559603
m n mod m 2 3 4 5 6 7 8 9 1 1 3 3 1 2 3 1
559,603 is not divisible by any number less than or equal to 9.
## Classification of 559603
• Arithmetic
• Semiprime
• Deficient
• Polite
• Square Free
### Other numbers
• LucasCarmichael
## Base conversion (559603)
Base System Value
2 Binary 10001000100111110011
3 Ternary 1001102122001
4 Quaternary 2020213303
5 Quinary 120401403
6 Senary 15554431
8 Octal 2104763
10 Decimal 559603
12 Duodecimal 22ba17
20 Vigesimal 39j03
36 Base36 bzsj
## Basic calculations (n = 559603)
### Multiplication
n×i
n×2 1119206 1678809 2238412 2798015
### Division
ni
n⁄2 279802 186534 139901 111921
### Exponentiation
ni
n2 313155517609 175242767120549227 98066378208960709076881 54878239444869039681549838243
### Nth Root
i√n
2√n 748.066 82.4062 27.3508 14.1116
## 559603 as geometric shapes
### Circle
Diameter 1.11921e+06 3.51609e+06 9.83807e+11
### Sphere
Volume 7.34055e+17 3.93523e+12 3.51609e+06
### Square
Length = n
Perimeter 2.23841e+06 3.13156e+11 791398
### Cube
Length = n
Surface area 1.87893e+12 1.75243e+17 969261
### Equilateral Triangle
Length = n
Perimeter 1.67881e+06 1.356e+11 484630
### Triangular Pyramid
Length = n
Surface area 5.42401e+11 2.06526e+16 456914
## Cryptographic Hash Functions
md5 a7819f182dcb08b353f18eb828b6da9b 921127481cb9835fca46b36e931dc2997e1d0897 1a6893135be237789ff5e48f3847d32d0992c715d9e5ca250234244c94e0d87e efa95cb5311d55faba0de0e931fa04b80825fa24808f4300ab6a1b3227d310582684c1717fc02477278880d25e671fd3d5513b4a4656386cd150f0b9484a3a54 afa779d176f112d20a7e5421da1e6ad4690930bd
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# Thread: Linear Equation using Gaussian Elimination or Gauss-Jordan Elimination
1. ## Linear Equation using Gaussian Elimination or Gauss-Jordan Elimination
Problem 1:
x + 2y - z = -2
x + z = 0
2x - y - z = -3
Problem 2:
2x - 3y - z = 13
-x + 2y - 5z = 6
5x - y - z = 49
The answer to problem 3 (that I got) is: x = -1, y = 0, z = 1
Is this correct?
For problem 4, do I use Gaussian elimination or Gauss-Jordan elimination?
2. Originally Posted by play60
Problem 1:
x + 2y - z = -2
x + z = 0
2x - y - z = -3
Problem 2:
2x - 3y - z = 13
-x + 2y - 5z = 6
5x - y - z = 49
The answer to problem 3 (that I got) is: x = -1, y = 0, z = 1
Is this correct?
For problem 4, do I use Gaussian elimination or Gauss-Jordan elimination?
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CFD Online Discussion Forums (http://www.cfd-online.com/Forums/)
- OpenFOAM Running, Solving & CFD (http://www.cfd-online.com/Forums/openfoam-solving/)
- - InterFoam mass conservation question (http://www.cfd-online.com/Forums/openfoam-solving/59449-interfoam-mass-conservation-question.html)
msrinath80 September 11, 2007 20:13
I'm trying to simulate the mot
I'm trying to simulate the motion of a single air bubble through a column of water.
When the simulation starts, the log file reads:
MULES: Solving for gamma
Liquid phase volume fraction = 0.0013888 Min(gamma) = 0 Max(gamma) = 1
MULES: Solving for gamma
After about 0.8 seconds (with time-step size of approximately 0.0001, Max Co = 0.15):
MULES: Solving for gamma
Liquid phase volume fraction = 0.0013888 Min(gamma) = -2.7143954e-24 Max(gamma) = 0.9551684
MULES: Solving for gamma
Why does the maximum gamma in the domain decrease to 0.95? Surely there are significant regions in the domain where only pure water is present. If I'm mistaken can someone explain what Max(gamma) represents. Thanks!
hsieh September 11, 2007 21:24
Hi, Did you plot the result
Hi,
Did you plot the results to check if the results are reasonable? Do you have surface tension turned on?
pei
msrinath80 September 11, 2007 21:28
Yup, results look reasonable.
Yup, results look reasonable. Surface tension is defined in transportProperties. How is it turned on?
hsieh September 13, 2007 11:37
Hi, Any luck? If not, you
Hi,
Any luck? If not, you can send me your case. I will take a look at it when I have the spare time.
pei
msrinath80 September 13, 2007 13:33
Hi Pei, I am running a slig
Hi Pei,
I am running a slightly different case with a very well refined mesh. Now Max(gamma) is at 0.99999993 which I believe is not a problem. However should it decrease to something like 0.9 or 0.95 I will package the case and email it to you. Thanks for your help.
Nazanin December 9, 2013 02:19
Quote:
Originally Posted by msrinath80 (Post 186096) Hi Pei, I am running a slightly different case with a very well refined mesh. Now Max(gamma) is at 0.99999993 which I believe is not a problem. However should it decrease to something like 0.9 or 0.95 I will package the case and email it to you. Thanks for your help.
Hi Srinath
Do you found your problem?I have a problem like you....
All times are GMT -4. The time now is 14:18.
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Connect with us
# FEDERAL COOPERATIVE COLLEGE, ENUGU STATE (FEDCOOP-OJI) – List Of Courses
Published
on
## FEDERAL COOPERATIVE COLLEGE, ENUGU STATE (FEDCOOP-OJI) – List Of Courses
FEDCOOP-OJI Courses are listed below. The JAMB subject combination, subject requirements and combinations for each course, and other institutions that offer them can be found in the “COURSES” section. You can as well click on each course to view its info.
### “FEDCOOP-OJI” – Programmes(courses):
• BANKING AND FINANCE
• CO-OPERATIVE ECONOMICS & MANAGEMENT
• COMPUTER SCIENCE
# PHY 203 Quiz For OAU
Published
on
0%
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268
Created by Oluwaferanmi Akinyele
PHY 203 Quiz
These are likely questions for our test based on our class notes, few questions our lecturers asked in class, examples in the notes and some online questions.
This quiz consists of 95 questions covering all the topics we have been taught in PHY 203 till present. I hope we all find it helpful.
You have 50 mins for this quiz.
Contributors: Zainab, Verah & Lammylex (CHM department)
and Shukroh, Great Blakky and Ebun (PHY department)
Goodluck!❤️
1 / 95
1. A node is a collection point for two or more banches
2 / 95
2. The core in a transformer provides ?
3 / 95
3. The principle behind the working of a transformer is _________
4 / 95
4. Which of this is true
5 / 95
5. No electrons can stay in the forbidden gap because
6 / 95
6. Which of the statements about PN junction is true
I. A PN junction presents a high resistance if forward biased.
II. A PN junction presents a high resistance if reversed biased.
III A PN junction presents a low resistance if forward biased.
IV. A PN junction presents a low resistance if reversed biased.
7 / 95
7. A semiconductor in an extremely pure form is known as ………….
8 / 95
8. Circuit elements are connected in parallel when a common ……………. is applied to each elements
9 / 95
9. A capacitor C and an Inductor L store…………….. and ……………….. respectively
10 / 95
10. ________ converts alternating currents into direct currents
11 / 95
11. A transformer only increases the voltage with corresponding increase in current
12 / 95
12. A variable resistor in an electric circuit is for ……………
13 / 95
13. To convert a multiplier to an ammeter, a ………. resistance is connected in ……………… with a galvanometer
14 / 95
14. The separation between conduction band and valence band on the energy level diagram is called ________
15 / 95
15. The voltage signal at the collector is out of phase with the signal at the base by:
16 / 95
16. The arrangement of atoms in a repetitive manner is known as
17 / 95
17. Which of the following is true about the effect of temperature on semiconductors
I. At absolute zero, there are no free electrons.
II. Above absolute zero, bonds break and electrons move under the influence of the applied electric field.
III. The semiconductor has its highest conductivity at absolute zero
IV. The semiconductor behaves as a perfect insulator at absolute zero
18 / 95
18. Which of this must be connected to multimeter to be able to measure AC voltages?
19 / 95
19. Loading effect of a meter happens because……………
20 / 95
20. Any transistor designed must operate in the ________ region of the characteristic curve
21 / 95
21. Which of these will result in forward biasing
I. when certain dc voltage is applied to the junction in such a direction that it cancels the potential barrier thus permitting the flow of current
II. Connecting the negative terminal of the battery to the p-type and positive terminal to the n-type
III. Connecting the positive terminal of the battery to the p-type and negative terminal to the n-type
IV. when certain dc voltage is applied to the junction in such a direction that the potential barrier is increased
22 / 95
22. Which of the following is true about transformers?
I. Open circuit test are done with a load on the transformer.
II. Copper loss is due to the ohmic resistance of the transformer windings.
III. Open circuit test is done to determine the ‘no load loss’ of the transformer.
IV. A transformer is electrically linked.
23 / 95
23. A meter is protected in the case of accidental workload by connecting …………… in ………….. with a voltmeter
24 / 95
24. In a transformer, most flux which get linked to the secondary winding is called?
25 / 95
25. Circuit elements connected in series have common …………….. passing through each element
26 / 95
26. In electronics, _____ refers to the use of DC voltage to establish certain operating conditions for an electronic device.
27 / 95
27. In a multimeter, when the variable resistance and the fixed resistance are connected in series with the battery and galvanometer, the multimeter functions as
28 / 95
28. Which of the following transformers has a high rating?
29 / 95
29. when the sensitivity of a meter is high, it means the meter has ………….
30 / 95
30. A resistor is regarded as a circuit element which puts energy into a circuit
31 / 95
31. A p-n junction is also known as
32 / 95
32. The Norton’s equivalent circuit contains a Norton equivalent source “I(N)” in parallel with an equivalent resistor R(N)
33 / 95
33. Depletion layer serves as barrier to the movement of free electrons.
34 / 95
34. Which of the following statements is/are true? I. A multimeter can be used as an ammeter when low resistance is connected in parallel with the galvanometer II. A low sensitive meter will measure Voltage perfectly III. A multimeter cannot be used to determine circuit continuity IV. A multimeter can be used as an voltmeter when high resistance is connected in series with the galvanometer
35 / 95
35. A transformer consists of two inductive coils namely;
36 / 95
36. The e–h+ pair can also be called ____
37 / 95
37. The majority carriers in an n-type semiconductor is ?
38 / 95
38. Which of the following is correct
I. Small changes in V(BE) will produce large current changes in I(C) & I(E)
II. Small changes in V(CE) have little effect on I(B)
III. Small changes in V(CE) will produce large current changes in I(C) & I(E)
I, II, III & IVIV. Small changes in V(BE) will produce large current changes in I(B)
39 / 95
39. The loss due to the reversal of magnetization in the transformer core is known as
40 / 95
40.
How many branches, nodes, loop and mesh do we have here?
41 / 95
41. p-type and n-type semiconductor join together to form a …………….. known as …………….
42 / 95
42. The transistor is a three-terminal device that we can use to form a _____________ circuit
43 / 95
43. An ammeter also detects the direction of the flow of current
44 / 95
44. A emitter-follower circuit has Vcc =12V, Rm = 1KΩ and a 2mA meter. If the amplification factor of the resistor is 80 and Vbe = 0.7. Calculate the suitable r for Rs to give fsd when E= 5V.
45 / 95
45. Which of the following is/are true
I. Reverse biasing prevents the flow of charge carriers across the junction.
II. Forward biasing is created when the positive terminal is connected to the n-type part of the PN junction.
III. In practice, when we apply reverse biasing to a PN junction, all the charge carriers stops to flow.
46 / 95
46. What are the examples of passive elements?
47 / 95
47. …………………. are responsible for the conduction in a conductor
48 / 95
48. Inductors smoothen sharp sudden changes in ……….
49 / 95
49. AC Current and Voltage may vary with time. True or False?
50 / 95
50. Select the correct options only
I. Pentavalent impurities provide large number of free electrons to the semiconductor.
II. Trivalent impurities provide large number of free electrons to the semiconductor.
III. Trivalent impurities provide large number of holes to the semiconductor.
IV. Pentavalent impurities provide large number of holes to the semiconductor.
51 / 95
51. What type of bonds exist in semiconductors
52 / 95
52. In a RC circuit, voltage across the resistor ………………. exponentially with time
53 / 95
53. If a circuit has the following parameters Vcc =20V, Rs +Rm = 9.3KΩ, Vbe = 0.7, Im =1mA, and Β = 100. Calculate the meter current when E =10V and the voltmeter input r with or without the transistor.
54 / 95
54. The collector current is determined by the _________
55 / 95
55. A capacitor and an Inductor in an AC circuit is referred to as
56 / 95
56. When a small amount of trivalent impurity is added to a pure semi conductor, the semi conductor becomes
57 / 95
57. Calculate Rth, Vth, and I5
58 / 95
58. In checking for continuity of circuit which on these is used
59 / 95
59. What is the number of branches, nodes, loops, and meshes?
60 / 95
60. The process of setting up a current to operate a transistor at a desired point on its characteristic curve is known as __________
61 / 95
61. A multiplier measures ………………, ………………….. and …………………
62 / 95
62. Consider the following statements and choose only correct answers
I. The resistivity of a semiconductor is less than that of an insulator and more than a conductor
II. As temperature increases, the conductivity of semiconductors decreases
III. We can consider a semiconductor as a resistance material
IV. Semiconductors have a negative temperature coefficient of resistance
63 / 95
63. A perfect amplifier has all of the following properties except:
64 / 95
64. In a semiconductor, as temperature increases, conductivity ………………
65 / 95
65. Calculate the equivalent resistance
66 / 95
66. Coils are …………… separated, but ……………. linked together
67 / 95
67. Calculate the current flowing in the circuit
68 / 95
68. Copper loss is usually larger than other losses
69 / 95
69. Voltmeter draws …………………… current
70 / 95
70. Inductors are very common in high-frequency circuit
71 / 95
71. A Thevenin’s equivalent circuit contains an equivalent source Vth in parallel with an equivalent resistor Rth
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72. In an RL circuit voltage across the resistor increases exponentially
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73. A galvanometer measures and detects extremely small currents both in ………………… and ………………………
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74. A capacitor stores charge and hence, electric energy
75 / 95
75. Which of the following is/are correct?
I. Bipolar junction transistors control the current by varying the number of charge carriers
II. Voltage between 2 terminals are indicated by repeated letters
III. The emitter is lightly doped to reduce junction capacitance
IV. The emitter-base junction is forward biased
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76. Extrinsic semiconductors have more current conducting capacity than intrinsic semiconductors
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77. Why is a laminated steel core used in a transformer instead of a solid core?
78 / 95
78. A 6600/550V, 25KVA transformer has iron losses of 350W and its 1º and 2° winding resistances are 14.5 and 0.1Ω respectively. Determine the full load efficiency at a power factor of unity.
79 / 95
79. A hole can also be described as a _________
80 / 95
80. What is known as eddy current loss in a transformer?
81 / 95
81. An Inductor tend to smooth sudden change in ……………..
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82. A mesh is any closed path in a circuit which does not enclose any other closed path inside it
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83. The most versatile circuit configuration for transistors is the
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84. On the basis of their use, transformers are classified as
85 / 95
85. The transistor is a current __________ device.
86 / 95
86. A material that releases an electron from the valence band into the conduction band when a photon of light is incident on it is a ___________
87 / 95
87. The resistivity of a semiconductor is ………. that of an insulator, and …………that of conductors
88 / 95
88. The potential barrier of a P-N junction depends on which of the following except
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89. The emf that gets induced in the secondary winding according to Faraday’s law of electromagnetic induction is called?
90 / 95
90. A laminated steel core that has high silicon content provides
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91. A component that allows the flow of current in only one direction is called ……….
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92. In pure inductors, current ……… the voltage by ……
93 / 95
93. The loss due to ohmic resistance of the transformer windings
94 / 95
94. A inductor stores energy in its _________
95 / 95
95. Calculate Ia, Ib and Ic
The average score is 53%
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# Quiz for CSC 201
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You have 30 mins for this quiz
Time up!
Created by Oluwaferanmi Akinyele
CSC 201
Preparatory Quiz for CSC 201 Test 2022/2023. This quiz contains 40 multiple choice questions.
The questions are compiled by Physics department gurus;
Ebun (Class rep)
Levi
Oluwaferanmi
Great
and Blakky
1 / 40
What will be the output of the following Python expression?
“`round(4.576)“`
2 / 40
The ______ is the brain of the computer that performs simple arithmetic & logical operations
3 / 40
Which type of Programming does Python support?
4 / 40
The expression
“`2**2**3“` is equal to “`(2**2)**3“`
5 / 40
The following are high-level languages except
6 / 40
Machines developed in transistor era include the following except
7 / 40
A computer software comprises of ___________ software & ___________ software
8 / 40
The following are program elements except
9 / 40
Application software includes the following except
10 / 40
Python was created in what year
11 / 40
Which of the following character is used to give single-line comments in Python?
12 / 40
Who developed Python Programming Language?
13 / 40
Which among the following list of operators has the highest precedence?
“` +, -, **, %, /, <<, >>, |“`
14 / 40
Which of the following is true for variable names in Python?
15 / 40
Which of the following is an invalid statement?
16 / 40
In flow charts, a terminal is represented with what shape
17 / 40
Is Python case sensitive when dealing with identifiers?
18 / 40
‘Bit’ stands for
19 / 40
Which of the following is used to define a block of code in Python language?
20 / 40
Is Python code compiled or interpreted?
21 / 40
The correct sequence of the history of modern era is
22 / 40
What are the values of the following Python expressions?
“` 2**(3**2)“`
“` (2**3)**2“`
“` 2**3**2“`
23 / 40
Which of the following functions is a built-in function in python?
24 / 40
What is the value of the following expression?
“`float(22//3+3/3)“`
25 / 40
The outcome of a programming activity is a _________
26 / 40
Which of the following is the correct extension of the Python file?
27 / 40
Which of the following variable naming is correct
28 / 40
The following are python reserved words excerpt
29 / 40
What is the maximum possible length of an identifier in Python?
30 / 40
Operating Systems includes the following except
31 / 40
A ________ is a sequence of instructions telling the computer what to do.
32 / 40
What will be the value of X in the following Python expression?
“`X = 2+9*((3*12)-8)/10“`
33 / 40
What is the order of precedence in python?
34 / 40
What are the two main types of functions in Python?
35 / 40
Which of the following is an invalid variable?
36 / 40
Which one of the following is the use of function in python?
37 / 40
Which keyword is used to call a function in Python language?
38 / 40
What is the average value of the following Python code snippet?
“`average = (grade1 + grade2) / 2“`
39 / 40
The following is true about python except
40 / 40
Machines developed in mechanical era include the following except
The average score is 66%
0%
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# PHY 205 Quiz
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48
OAU PHY 205 Quiz
PHY 205
1 / 34
1. What are electromagnetic waves?
2 / 34
2. In quantum mechanics, what role does a photon play in the electromagnetic field?
3 / 34
3. What is the primary reason for the scattering of alpha particles in Rutherford's model?
4 / 34
4. Arrange the following types of electromagnetic radiation in order of increasing wavelength
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5. If the dimensions of the nucleus are small enough, what can happen to alpha particles passing very near the nucleus in Rutherford's model?
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6. According to quantum mechanics, what is the alternate view of electromagnetic radiation (EMR)?
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7. How do the oscillations of the electrical and magnetic components in electromagnetic waves relate to the direction of propagation?
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8. What is the significance of alpha particle scattering experiments in Rutherford's model?
9 / 34
9. In Thomson's model, where are the negatively charged electrons located?
10 / 34
10. What is a photon?
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11. How are the electrons distributed in Thomson's model?
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12. In Rutherford's model, where is the positive charge concentrated?
13 / 34
13. Arrange the following types of electromagnetic radiation in order of increasing frequency:
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14. In electromagnetic waves, how do the electrical and magnetic components oscillate?
15 / 34
1. What happens when a metal wire is heated in a vacuum during thermionic emission?
16 / 34
2. Why is it necessary for the tube to be in a vacuum during thermionic emission?
17 / 34
3. What type of radiation can be produced using cathode ray tubes?
18 / 34
4. What is thermionic emission?
19 / 34
5. High beam of cathode ray in vacuum tube can be steered and manipulated by
20 / 34
6. In what devices are cathode ray tubes commonly used?
21 / 34
7. In Thompson’s experiment, a a beam of electron travelling at 6.8 x 10^7m/s is bent into a circular path of radius 4cm in the magnetic of induction 10^-2 W/m2. Find the specific charge to mass (e/m)
22 / 34
1. Which of the following statements about electrons within an atom is true?
23 / 34
2. What type of process is represented by this equation e† + e¯ → ϒ + ϒ
24 / 34
3. Which field properties does an electron exhibit due to its charge and motion?
25 / 34
4. Electrons, being fermions, adhere to which principle that prohibits two electrons from occupying the same quantum state?
26 / 34
5. What is the intrinsic angular momentum (spin) value of an electron?
27 / 34
6. Wavelength is inversely proportional to an object’s mass
28 / 34
7. What term is used to describe the lowest energy state that an electron can occupy within an atom?
29 / 34
8. Which of the following is not quantized?
30 / 34
9. What is the electric charge of a positron?
31 / 34
10. Which branch of physics is concerned with the quantization of energy and its effects on the interaction of matter and energy?
32 / 34
11. Electrons absorbs energy in form of photons as they are accelerated
33 / 34
12. Which of the following statements is TRUE?
34 / 34
13. What is the relative mass of an electron compared to a proton?
The average score is 66%
0%
# OAU PHY 203 Class Notes
Published
on
This PDF is the class notes for OAU PHY 203 during the 2022/2023 session.
It encompasses topics ranging from circuit theory and network theorems to Thevenin’s and Norton’s theorems, progressing further into AC circuits, transformers, physics of active devices, semiconductors, and more.
I will continually update this file as we proceed through the classes. Thank you.
### RECOMMENDED:
[mailpoet_form id="1"]
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# 7564 (number)
7,564 (seven thousand five hundred sixty-four) is an even four-digits composite number following 7563 and preceding 7565. In scientific notation, it is written as 7.564 × 103. The sum of its digits is 22. It has a total of 4 prime factors and 12 positive divisors. There are 3,600 positive integers (up to 7564) that are relatively prime to 7564.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 4
• Sum of Digits 22
• Digital Root 4
## Name
Short name 7 thousand 564 seven thousand five hundred sixty-four
## Notation
Scientific notation 7.564 × 103 7.564 × 103
## Prime Factorization of 7564
Prime Factorization 22 × 31 × 61
Composite number
Distinct Factors Total Factors Radical ω(n) 3 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 3782 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 7,564 is 22 × 31 × 61. Since it has a total of 4 prime factors, 7,564 is a composite number.
## Divisors of 7564
1, 2, 4, 31, 61, 62, 122, 124, 244, 1891, 3782, 7564
12 divisors
Even divisors 8 4 2 2
Total Divisors Sum of Divisors Aliquot Sum τ(n) 12 Total number of the positive divisors of n σ(n) 13888 Sum of all the positive divisors of n s(n) 6324 Sum of the proper positive divisors of n A(n) 1157.33 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 86.9713 Returns the nth root of the product of n divisors H(n) 6.53571 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 7,564 can be divided by 12 positive divisors (out of which 8 are even, and 4 are odd). The sum of these divisors (counting 7,564) is 13,888, the average is 11,57.,333.
## Other Arithmetic Functions (n = 7564)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 3600 Total number of positive integers not greater than n that are coprime to n λ(n) 60 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 964 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 3,600 positive integers (less than 7,564) that are coprime with 7,564. And there are approximately 964 prime numbers less than or equal to 7,564.
## Divisibility of 7564
m n mod m 2 3 4 5 6 7 8 9 0 1 0 4 4 4 4 4
The number 7,564 is divisible by 2 and 4.
• Deficient
• Polite
## Base conversion (7564)
Base System Value
2 Binary 1110110001100
3 Ternary 101101011
4 Quaternary 1312030
5 Quinary 220224
6 Senary 55004
8 Octal 16614
10 Decimal 7564
12 Duodecimal 4464
20 Vigesimal ii4
36 Base36 5u4
## Basic calculations (n = 7564)
### Multiplication
n×y
n×2 15128 22692 30256 37820
### Division
n÷y
n÷2 3782 2521.33 1891 1512.8
### Exponentiation
ny
n2 57214096 432767422144 3273452781097216 24760396836219341824
### Nth Root
y√n
2√n 86.9713 19.6299 9.32584 5.96692
## 7564 as geometric shapes
### Circle
Diameter 15128 47526 1.79743e+08
### Sphere
Volume 1.81277e+12 7.18974e+08 47526
### Square
Length = n
Perimeter 30256 5.72141e+07 10697.1
### Cube
Length = n
Surface area 3.43285e+08 4.32767e+11 13101.2
### Equilateral Triangle
Length = n
Perimeter 22692 2.47744e+07 6550.62
### Triangular Pyramid
Length = n
Surface area 9.90977e+07 5.10021e+10 6175.98
## Cryptographic Hash Functions
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Flow on Revolved Surface 1-20 21-38
From: Mike K4ICY (MAJIKMIKE) 25 Sep 2011 (1 of 38)
I'm now starting to play around with V3 Beta's Flow command, and I tested a theory today: - Flowing objects on a revolved surface. Yes, usually when you think of flowing objects onto a revolved surface like a cylinder or sphere, you take the objects you want to flow and place them on a simple rectangle as the reference surface. The result is as expected - it starts the objects at the seam line and sweeps them around like a record (axially) until it reaches the line again. But the flow command works by matching points in the reference surface's "grid structure" to that, by percentage of the target's. With that being said, why not consider matching a reference surface THAT IS ALREADY IN THE FORM OF THE TARGET? Consider a simple revolved profile: You simply start off with a revolve profile made from a free-form spline. Not just considering the control points, but how much distance is in between each control point or relative control point on the surface or curve. - Considering the resultant U/W layout if everything was "unwrapped" Now draw another line for the reference revolve profile, but straight and making each control point spaced out to represent the true distance along the target's profile curve. Distance horizontally is considered, but remember if the curve moves downward or at an angle or in an arc - there is a distance there too. Represent these distances in the spacing between control points along the reference profile. Note: Ensure that the directions and orientations of the revolves and control profiles are in the same directions so that the Flow command will match them as you desire. Also: Your flat profile curve for the reference revolve has to have a few of it's control points slightly offset from the straight line, or MoI will generate a planar surface with no swept control points. Now Revolve each profile, using the same orientation. You have a revolve with a nice shape of your liking, and one that is relatively flat. Take the flat-like reference revolve in a side view and make it "Flat" using the handle bars. It will now be perfectly flat, but retain the knots and surface information. Now work the Flow command with the original objects on the surface or relatively spaced to the reference surface. Since each corner of the swept surface's control cages will line up, the original objects will be translated to the characteristic shape of the target surface. Notice how these objects are NOT sweeping in a rotational manner on the target surface! Issues: In this beta edition there is a problem with the flow result curves and surfaces "rippling" as the distance increases relative to some point of origin. This is an issue that I think Michael said he was working on for the next beta release. I tried the telephone chord spiral along a path thing and got the same strange ripple that increase in severity more to one end. You must also consider that the Flow command does not have supernatural powers and will flow well over seams, tight bends and the degenerated center point on a revolve as shown above. But it will get you in the ball park. Also - Flow is sloooooooooow.... eesh. My goal was to make a speaker grill with well placed holes and other detail, as such would be done if a metal sheet of grill metal was stamped into a shape. I gave up after it sat for two hours and nothing happened... I'll take it to work and try on the faster PC. BTW, Michael, I thought it was cute when my brother was looking at the Task Manager's Performance graphs on the work PC - while I was working with MoI, and asked me why only one of the cores was the only one working.... Well, someday. ;-) From what we can learn from my experiment is that there may be some power in considering the original reference surface with the Flow command as well as the target.
From: Michael Gibson 25 Sep 2011 (2 of 38)
4549.2 In reply to 4549.1 Hi Mike, yup you can actually use any surface as the base surface - it doesn't have to be a plane although mapping from a plane as the base tends to be the most simple one. But the way Flow works is every point that is evaluated on the base object is dropped to the closest point on the surface. That then gives a UV location and an offset distance - then that same UV and offset distance is evaluated on the target surface. You may have some difficulty with using revolves in this particular way though because of the compression in UV space that happens as you near the pole area. Possibly the new "Projective" option in the next v3 beta would be easier to do that kind of thing that you're showing. Also as a side note, when the new "Projective" option is enabled there will also be a "Straight" option that will do the kind of shaping that you had asked about a while ago where the shaping is constrained in one direction. It turned out to be a good fit with that option since in that mode there is a particular direction set up for projection already which could be used without any extra picking required. When that option is set the result of the projection will expand like an extrusion rather than like an offset. > In this beta edition there is a problem with the flow result curves and > surfaces "rippling" as the distance increases relative to some point of origin. > This is an issue that I think Michael said he was working on for the next beta release. > I tried the telephone chord spiral along a path thing and got the same strange > ripple that increase in severity more to one end. Are you talking about using base curve to target curve flow here? There is a bug right now where curve to curve flow (meaning using curve backbones rather than surface backbones) doesn't work properly unless both curves are flat in the world X/Y plane. If the curves are in some other orientation like standing up vertically you'll get that weird rippling effect. I've been working on fixing that just recently here and almost have it tracked down. But for now lay both your base curve and target curve down in the world x/y plane and it should behave better. > Also - Flow is sloooooooooow.... eesh. My goal was to make a speaker grill > with well placed holes and other detail, as such would be done if a metal > sheet of grill metal was stamped into a shape. Yeah basically every single surface and edge in the model is reconstructed with a new refitted version, and it goes in and refines each of those rebuilt surfaces until it is within a good tolerance. If you have a complex model with a whole lot of little surface pieces in it it may take a while to crunch through it all. You can reduce the amount of time taken by reducing the number of surfaces in the model that you are deforming. > BTW, Michael, I thought it was cute when my brother was looking at the > Task Manager's Performance graphs on the work PC - while I was working > with MoI, and asked me why only one of the cores was the only one > working.... Well, someday. ;-) Unfortunately things that are coded in the most straightforward manner only work with one core - it takes a considerable amount of special work and special attention to make an algorithm work well with multiple cores, it doesn't just happen on its own. Flow could be a good candidate for using multiple cores in the future at some point though. In the current v3 time frame I'm much more focused on just getting it to generate a good result. - Michael
From: Mike K4ICY (MAJIKMIKE) 25 Sep 2011 (3 of 38)
Thanks Michael! The Flow tool system has the potential to become a major design staple with MoI. Especially in streamlining workflows to produce more complicated structures. I've got a few things in mind for future projects already. Mike
From: Mauro (M-DYNAMICS) 25 Sep 2011 (4 of 38)
=Michael= You may have some difficulty with using revolves in this particular way though because of the compression in UV space that happens as you near the pole area. Possibly the new "Projective" option in the next v3 beta would be easier to do that kind of thing that you're showing. I can't wait next beta..so i'm working on a model by a complex pattern flowed on a revolved solid (i'll post soon..) i got compression on the pole areas and i hope next beta can solve this distorsion Magic..it's true..FLOW is slow when you use complex surfaces to flow (in my case i wait 4 minutes each pattern row but i got time to wait if the results are these ..!! ) hope will benefit multiple cores in the future my hardware: INTEL QUAD CORE 9450 overclocked 3.203 Mhz 8 Giga RAM NVIDIA GTX 460 1Giga video mem HD WD RAPTOR 10000 RPM S.O. WINDOWS 7 PROFESSIONAL 64 bit you see this is strong workstation
From: Michael Gibson 25 Sep 2011 (5 of 38)
4549.5 In reply to 4549.4 Hi Mauro, > so i'm working on a model by a complex pattern flowed > on a revolved solid (i'll post soon..) > i got compression on the pole areas and i hope next beta > can solve this distorsion The surface-to-surface mapping Flow will just follow the structure of the target surface. So if there is compression in the target surface's structure like there is with a default sphere then that causes compression of the pattern. If you want to avoid that, you would need to work with a surface that does not compress its top or bottom areas to a single point which is what a default sphere surface will do. Maybe you need to use a rounded surface dome that is constructed in a different way than a sphere, something like a sweep of a profile and a rail like this: - Michael Attachments:
From: Michael Gibson 25 Sep 2011 (6 of 38)
4549.6 In reply to 4549.4 Or if you have a pattern that repeats at certain intervals maybe you can trim out a small section the middle area of a sphere where it is not so highly compressed as at the poles and then repeat that around, something like the attached sphere. Although that may not quite work right because each sector has some trimmed away area in the corners, and Flow will be working on the underlying full surface. But it depends on what kind of pattern you are trying to create. - Michael Attachments:
From: Mauro (M-DYNAMICS) 26 Sep 2011 (7 of 38)
Thanks Michael for suggestions almost tried many solutions,i think you cannot flow everything you want..at the end i'll post it check out this screenshot i made during flow operation CPU's behavior : you see 4 cores but only one at 100% computation Image Attachments:
From: Michael Gibson 26 Sep 2011 (8 of 38)
4549.8 In reply to 4549.7 Hi Mauro, > Thanks Michael for suggestions > almost tried many solutions,i think you cannot flow everything > you want..at the end Yeah not every design has a 2D element to it - if your pattern does not map well to a 2D pattern then Flow won't really be a help for that particular case. > CPU's behavior : you see 4 cores but only one at 100% computation Yes, that's normal behavior. It takes special and careful work to make an algorithm make use of multiple CPU cores, it is not something that happens automatically. Implementing a parallel version of Flow could be possible in the future at some point but for v3 the main focus is just on making Flow give a good result. - Michael
From: Mike K4ICY (MAJIKMIKE) 26 Sep 2011 (9 of 38)
Behold! (This is a rough screen capture shown with a little Photoshop) Here, I was able to wrap multiple surfaces trimmed with a pattern, around the revolved target surface. The center was crudy as was expected, so I did a little work on it. The total surface took about 25 minutes in calculation time with Flow. However, when I tried a solid version with some depth, I started last night and when I came home tonight, there was still no result. Which leads me to believe that for every single surface or curve, Flow has to make a separate calculation, noting that the time seems exponential. I can't fathom the workings of the logarithms behind Flow, but I imagine this to be so. Since the resultant Flow were single layer surfaces I performed a simple extrusion on them to give them a little thickness. I could have simply trimmed the hexagons from the target shape in one direction, but the point here was to be able to accomplish this result on surfaces oriented in many directions. Now I'd like to build upon this object.
From: mjs (MSHIDELER) 26 Sep 2011 (10 of 38)
4549.10 In reply to 4549.9 Not to ask a dumb question, but for geometry like the one shown, have you tried doing a small section and then making the rest of the model with a cirular array?
From: Michael Gibson 26 Sep 2011 (11 of 38)
4549.11 In reply to 4549.9 Hi Mike, > Which leads me to believe that for every single surface or > curve, Flow has to make a separate calculation, Yup, that's correct - each surface or curve goes through a fitting process that reconstructs it. On something like you're showing there where each little hole is not really deformed so much you could try something like use the "Rigid" option for Flow to just replicate a planar hexagon over all those areas and then use boolean difference and use all those planar hexagons as cutting objects. The "rigid" option only takes a point at the bounding box center of the object and deforms only that point to determine where to move and rotate the object, so it positions the object along the target surface but does not actually deform it when that option is enabled. - Michael
From: Mike K4ICY (MAJIKMIKE) 27 Sep 2011 (12 of 38)
4549.12 In reply to 4549.10 Yes MJS, Rotational array can produce beautiful results. In this case I wanted to establish a method to use Flow to "wrap" a pattern of objects along a rotated of non-conventional surface. Here's a good example: ... And I finally achieved a design effect on my wish list. :-) This woven metal basket was made by flowing each strand to the bowl shape by referencing them off of a flat rotate shape with the same control point spacing: The grid pattern maintains it's relative shape as it is translated across U and V. The wire object used in each run was made from one single sweep of a circle. A wavy curve was made for the sweep rail, repeated multiple times, joined and Rebuilt to form a "seamless" object - so that Flow would work quicker. As two surfaces of half length take twice as long as one of a single length. The wire mesh here is ad-hoc and nothing really lines up like a mesh should, but the process worked well to show. Michael, I noticed that after selecting more than a dozen or so surfaces to Flow, the flow time was not incremental but seemed exponential. ...Or even as if it was stuck. May I suggest placing a simple curves counter on the script to tell you how far along it is? Such as: "Object: 23 of 400" It could also be divided into the time spent to estimate a "remaining" time. Also, you did talk about the intricacies of using multiple processor cores - for future thought, could the total objects to flow be simply divided up and batched into separately spawned running instances of the Flow command script? But I'm not a programming expert, just a humble fan of MoI. ;-)
From: Mauro (M-DYNAMICS) 27 Sep 2011 (13 of 38)
4549.13 In reply to 4549.12 Great Magic !!
From: Frenchy Pilou (PILOU) 27 Sep 2011 (14 of 38)
4549.14 In reply to 4549.12 it's touching or pentrating between elements of the grid? --- Pilou Is beautiful that please without concept! My Gallery
From: Mike K4ICY (MAJIKMIKE) 27 Sep 2011 (15 of 38)
4549.15 In reply to 4549.14 The wires of the grid in this model are most likely colliding everywhere. This was a rough layout. I'd like to make a cleaner one later. Each wire or sets of wire were Flowed on separate instances.
From: Michael Gibson 27 Sep 2011 (16 of 38)
From: Mike K4ICY (MAJIKMIKE) 27 Sep 2011 (17 of 38)
4549.17 In reply to 4549.16 Thanks Michael, Yes, I understand... and I hope you have increased success with V3's developments of course - we're there to test them out for ya. ;-) At work I'm using Win7pro, 16 gig of DDR3 on an ASUS P8Z68 64bit system with an i7 - 4 core x2 with hyper-threading (8), and a GeFore GTX 560 Ti with 4 gig ram. The HD is a WD Black, 7200rpm ??? A modest system where MoI kicks serious butt, even using one core. With Flow running, It barely registered. ;-) My home system however... sigh... the newest part on it was made before 2004. There's only 1 gig of ram on it... and three empty ram slots... hmm. :-/ ...But I've made some killer MoI models on it, none the less.
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# Function NotationInstructor: Dr.Jo Steig
DEFINITION: A function can be defined in a variety of ways. Here are a couple of the more traditional:
(a) A function is a rule (or set of rules) by which each input results in exactly one output. The set of all acceptable inputs in called the domain. The set of all resulting outputs is called the range.
(b) A function is a set of ordered pairs in which each first component is matched up with exactly one second component.
For specific information about finding the domain of a function, see the Supplementary Reading entitled Domain
What both of these definitions has in common is that each input to the function can result in EXACTLY one output. The usefulness of this idea is that the function has predictable, deterministic outcomes.
EXAMPLES OF FUNCTIONS:
(a) The amount of your take home pay.
(b) The amount of your electric bill
(c) Mixing paint to match a particular color
(d) Making cookies from a recipe
(e) The amount of weight an elevator can lift
Because the concept of a function is used so often, a shorthand has been developed...we call it functional notation.
(1) the name of the function is g
(2) the letter, x, stands for the input.
(3) the output (based on an input of x) is referred to as g(x)
(4) what the function does (its process) is to take the input, multiply it by 3, add 4 , then divide the result by the original input
To indicate that we want to put the number 5 into the function g and calculate the result we would write the following:
In other words, when we put the number 5 into the function called g, the output is 19/5.
In the function g, the letter x is actually being used as a place holder. To illustrate this point, we can remove the x. This leads to the following:
We indicate that something is going into the function by putting it into the parentheses.
So, g(x) represents the output of the function when x represents the input.
And g(5) represents the output of the function when 5 is the input,
and g(a + b) represents the output of the function when a + b is the input,
Here are some examples of functions. In each case notice the function has a name, a letter representing the input, and a specific process by which the output is determined.
The name of the function is f, the input is represented by the letter x, and the output (when x is the input) is denoted by f(x). f( -3) represents the output of the function when - 3 is the input.
The process that is described by the function is described by the algebraic expression (x3 - 2x + 1)/(x + 1).
The name of the function is g, the input is represented by the letter t, and the output (when t is the input) is denoted by g(t). g(a+b) represents the output when a + b is the input to the function.
See if you can describe the process that this function represents?
The name of the function is h, the input is represented by the letter x, and the output (when x is the input) is denoted by h(x). h(t + 1) represents the output to the function when t+1 has been the input.
In the next example, the input to the function is the expression x + h
PARENTHESES PLACEMENT: In Example 4 the input was the expression, x + h. That means that x + h goes into the function. We say that the function operates on the input x + h. This is VERY different from the expression f(x) + h, which means just add h to the function.
Again, look closely at the two different statements:
When working with functions it is important to notice where the parentheses are placed as they change the meaning of the expression.
In the next example we will subtract two functions, simplifying where possible.
In the last example we make use of a particular combination of functions that includes subtraction and division. It is called the difference quotient. The problem from EXAMPLE 5 forms the numerator of the quotient.
You should try to repeat the steps that I have outlined here. In a effort to save space, I have not shown all of the steps that you might require to work this problem yourself. See if you can put in the intermediate steps so that you understand how to move from one step to the next.
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# Mix Review Worksheet-18
Mix Review Worksheet-18
1. 30% of Rs.12 is:
A. Rs.36.0 B. Rs.3.60 C. Rs.0.36 D. Rs.360
1. 8/125 is equal to ___ %.
A. 8 B. 64 C. 125 D. 6.4
1. 138% is equal to:
A. 68/50 B. 69/50 C. 13.8 D. 1.83
1. 68/100 = ____ %
A. 0.68 B. 6.8 C. 68 D. 68.01
1. 89% is written as:
A. 89/100 B. 98/100 C. 8.9 D. 8(9/100)
1. Decimal for 79% is:
A. 7.9 B. 0.79 C. 79.00 D. 1.79
1. 0.97 is equal to _____ %
A. 9.7 B. 9.71 C. 97 D. 0.97
1. 7/5 is equal to ____ %.
A. 14 B. 104 C. 140 D. 20
1. 6/20 is equal to ____ %.
A. 5 B. 30 C. 20 D. 6
1. 75% of 35 kg is:
A. 2625 kg B. 26.25 kg C. 262.5 kg D. 2.625 kg
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# 3 Phase Wiring Installation In Multi Story Building
In house wiring, we install two types of wiring of which one 3 phase wiring installation for the house and the other one single-phase installation. In this post, you will learn how to install three-phase electrical wiring using a simple diagram in which I have shown how to install a three-phase 4-wire system in a multi-story building. One thing more that I have shown 3 distribution boards in which every story has its own electrical main distribution board.
## 3 Phase Wiring Installation In Multi-Story Building / House
The three-phase 4 wire system installation at home is very simple but to understand you must study each part of this class and the devices that I use in the below diagram. I start this from step one.
#### Utility Pole 3 Phase 4 Wire System
The diagram starts from the utility pole and I get the service line supply from the utility pole to 3 phase kWh energy meter. This utility pole connects to a step-down transformer (supply from the substation). This means that the pole supply comes from the substation or step-down transformer.
#### 3 Phase kWh Energy meter
I also have shown the wiring diagram for 3 phase 4 wire system kWh (kilowatt-hours) meter, I also published a post about the wiring and installation of three phase four wire system kWh energy meter. So kindly read the below article if you did not understand the method of 3 phase kWh energy meter connection.
#### 4 Pole MCCB Breaker
After the connection of the energy meter, I connect the 3 Phase 4 wire power supply to the MCCB (module case circuit breaker). However, if you can do this connection then read the below post.
How to wire 3 Pole or 4 Pole MCCB Breakers?
Now come to the connection for the distribution board for one story. The method of installation 3 phase wiring is simple, you divide your load into three parts and each part has there own Line (phase). My means that divide your load into 3 parts and provide one phase to each part.
Now here I have a multi-story building in which each story has there own single-phase, distribution board.
So we will provide one phase or line to each story. And our neutral will be common but this will be common in circuit breaker output points and each story has its own neutral wire. So provide my Line 1 or phase 1 to the first distribution board as you can see in the diagram. So come to the Distribution board.
### Distribution Board Wiring Connection
In the DB board, I use the following electrical devices.
#### Double Pole MCB Circuit Breaker 63 Amps
I get phase and neutral wire from the main circuit breaker( 4 pole MCCB) and connect to double pole MCB (miniature circuit breaker). Note that double pole MCB will be considered the main circuit breaker for the single phase distribution board below the diagram.
RCD Breaker.
I also install and wire double pole RCD (residual current device) in each distribution board. If you did not understand the wiring and installation of RCD then read the following post.
#### Earthing or ground connection
I also did the earthing connection in 3 phase wiring installation diagram in multi-story house. I have shown this connection by using an earth symbol and a green wire. In the below diagram the earth, symbol consider as the earth electrode.
#### AC Voltmeter Wiring connection for single phase
I also do the voltmeter wiring connection is each distribution board and you can see it in the diagram. However, here is some post which helps you more than this. (Note that the VM denotes the ACV voltmeter below diagram.
#### Ampere meter / Ammeter connection with CT (current transformer)
I also have shown the ammeter connection with the current transformer in 3 phase installation diagram for a multi-story house. The connection of CT is simple with ammeter but this will be better for you if you read the following articles.
### Complete Explanation of Three – 3 phase wiring installation in a multi story building
The installation of three phase 4 wire system in multi-story building is simple if you want to use single phase supply for each story. Note that by doing this type of connection you must wire 4 distribution boards. where one for the main MCCB circuit breaker and the other for each story. And you can also do these connections in a single main DB but you must use a big size distribution board box where you can easily install all circuit breakers. But if use their owns distribution board in every story then this will be a better option.
So let’s do it step by step.
In the 3-phase wiring diagram, I have shown a utility pole from where I get a service line connection for 3 phase kWh meter (energy meter). Now I connect the supply wire Red, Yellow, Blue, and Black to the energy meter input connection terminals and get the supply from the energy meter outgoing connection to load.
One thing more is that we use a colour code of red, yellow, and blue when doing the three-phase wiring, but when it comes to the 3 phase four-wire system, we use black for neutral. And when we are gonna do connection of single phase ( current between phase and neutral) then we use red for phase (hot wire) and black for neutral.
After the connections of three phase kWh meter, we provide the 3 phase 4 wire supply to 4 pole MCCB circuit breaker which is on the- 100 amp. Now the next step is outgoing supply from the 4-pole MC CB circuit breaker. The N neutral wire will be common for each story or load and you see it in the below diagram, that I connect the neutral black color wire from the breaker and wire it with a connection point connector.
So the neutral will be common for each story or distribution board and one-one phase will be distributed to each story. So I connect the neutral wire to a neutral connection point or connector from where we can get a neutral connection for the load.
I connect a neutral wire to this neutral connection point and connect to the story 3 distribution board double pole MCB breaker. And then I get the Phase wire from the MCCB and connect it to this MCB breaker. Then MCB supply goes to the double pole RCD and you can see it in the below diagram. Note these two breakers are 63 amps.
Now from the RCD breaker, the neutral wire goes to the neutral connection point as I have shown in the above 3 phase wiring diagram, and the phase (hot wire) red colour goes to all single pole MCB breakers which I have shown in the above diagram.
In the above3 phase wiring diagram the 2 single pole breakers are 16 amp, 2 are 10 amps and the other sex are 6ampere breakers. Note that this is just an example and you must install or wire breaker regarding your total load. Note that use high-quality wire for your wiring system. And always chose or use the wire regarding load (amps).
I also wire the voltmeter in the above three / 3 phase wiring third story building wiring, for which I get hot wire connection from input points of single pole MCB breakers and neutral from neutral connection connector point. And same I wire the light indicator.
In the above diagram, I show the ammeter connection with the current transformer (CT coil) and also provide the complete method of wiring and installation of the ammeter with the current transformer in the above links. However, use always the current transformer regarding the ammeter requirements. If the ammeter is 60/5 then use a current transformer with the same value 60/5.
You can also use the AC series type of ammeter but the CT type ammeter is better than the ammeter which we use between supply in series connection.
As I have shown in the above 3-phase wiring diagram I do the earthing connection, and provide the earth wire to each point. I have shown at bottom of the diagram the earth symbol with the name of the earth electrode and from here I provide the ground connection to all story loads/rooms etc. And in each story, I use the earth connection point, from where a person can provide the earth wire to any load / electrical machine, outlets etc.
In the above 3 phase wiring diagram 1,2,3,4….. in the circle means that these wire goes to load 1 , load 2 and load 3. It means that if the 1 hot wire (phase – red) wire goes to room A then the neutral 1 (black wire) is also for room A and the earth connection 1 number wire is also for the same point (load- room A).
Follow the same method for story 1 and story 2, but provide the L2 (phase 2 – Hotwire 2 ) to story 2 and Line 3 to story 1 which is the bottom of the 3 phase wiring diagram. However, let’s get another review of the diagram step by step.
3 Phase Wiring Step-By-Step Guide Multi-Story Building
• First of all, provide the service line wire to the 3 phase kWh meter from the utility pole.
• Then provide the 3-phase 4 wire supply to the 4 pole MC CB Circuit breaker.
• Then connect the neutral point from MC CB to the neutral connection point.
• Then get connections from the main neutral point and connect to story 3 or another story double pole MCB circuit breaker.
• The connect the L1 to the story three double pole MCB between the CT (current transformer) and from MCB connect neutral and phase to the RCD circuit breaker. (Note some people know RCD by name of ELCB (earth leakage circuit breaker).
• Then from RCD connect the neutral wire to the neutral connection point and phase to circuit single pole breakers.
• Then provide the supply to the voltmeter and indicator light.
• Then Do connections between the CT coil and the ammeter.
• Then provide the earth connection (green wire) to story three earthing connection point.
• Then provide the supply to each load from the circuit breaker, neutral connection point and earth/ground connection connectors.
That’s all, same follow this method for each story. but provide their own different phase wire and neutral wire.
#### Message:
This is an example of 3 phase wiring installation in multi-story building, and I use different ampere circuit breakers, just for your understanding. However, always use breakers regarding your load. Now I hope you understood the 3 phase wiring in multi-story houses or buildings, now if want to give us suggestions or have any questions then use the below comments section.
### 7 thoughts on “3 Phase Wiring Installation In Multi Story Building”
• October 13, 2016 at 9:38 am
Kindly share with 3 phase selector
• October 19, 2016 at 7:13 am
Hi, I'm using three phase line. I would like connect ROTARY SWITCH,. Is that possible to connect (4P ISOLATOR- ROTARY SWITCH- 4p MCB). OR By (4P ISOLATOR-4P MCB- ROTARY SWITCH). COULD YOU explain 4p ISOLATOR- ROTARY SWITCH – 4mcb CONNECTION with diagram.
• August 10, 2021 at 5:33 am
Great post, very educative and interesting information.
Please can you help me with this diagram.
Am connecting five story buildings to five 3phase meters each and connecting a solar system to only one of the five story buildings with an isolator(change-over switch). You can kindly send it to my email (bojoelawson1@gmail.com)
Thank you.
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Question Solved1 AnswerProblem begins here: A tank with a capacity of 74 liters, initially contains 20 kilograms of salt dissolved in 50 liters of water. Another solution, containing 3 kilograms of dissolved salt per liter, runs into the tank at 4 liters every minute and that the mixture (kept uniform by stirring runs out of the tank at the rate of 1 liter per minute. What is the concentration of the solution in the tank at point of overflow? Intermediate values should be rounded only to two decimal places. Round off the final answer to two decimal places. Choose the correct answer from among the given choices.
GHPDCG The Asker · Calculus
Transcribed Image Text: Problem begins here: A tank with a capacity of 74 liters, initially contains 20 kilograms of salt dissolved in 50 liters of water. Another solution, containing 3 kilograms of dissolved salt per liter, runs into the tank at 4 liters every minute and that the mixture (kept uniform by stirring runs out of the tank at the rate of 1 liter per minute. What is the concentration of the solution in the tank at point of overflow? Intermediate values should be rounded only to two decimal places. Round off the final answer to two decimal places. Choose the correct answer from among the given choices.
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Transcribed Image Text: Problem begins here: A tank with a capacity of 74 liters, initially contains 20 kilograms of salt dissolved in 50 liters of water. Another solution, containing 3 kilograms of dissolved salt per liter, runs into the tank at 4 liters every minute and that the mixture (kept uniform by stirring runs out of the tank at the rate of 1 liter per minute. What is the concentration of the solution in the tank at point of overflow? Intermediate values should be rounded only to two decimal places. Round off the final answer to two decimal places. Choose the correct answer from among the given choices.
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let y(t) be the amount of Salt af ary tinie t:.(dy)/(dt)= (rete of Sat flowing into tenk) -(vete of Salt out of tank)(dy^('))/(dt)=(3(kg)//urxx4inim//wi^('))-((y)/(v(t))*1" wirf/in ")Hene V(t)= ehaying velune of Solution{:[v(t)=50+(4-1)t],[v(t)=" So "+3t quad A+0" verflow "quad" So "+3t=74],[3t=74-50],[" overflow at "t=8" min "]:}:.(dy)/(dt)=12-(y)/((50+3t))ov (dy)/(dt)+(y)/(50+3t)=12 rarr1 let mu(t)=e^(int(1)/((50+3t))dt)Mutiply eqn (c) on b/s by mu(t){:[root(3)(50+3t)(dy)/(dt)+(y(t))/(sqrt(3t+50)(50+3t))=12root(3)(50+3t)],[root(3)(50+3t)(dy)/(dt)+(y(t))/((50+3t)^(2//3))=12^(3)sqrt(50+3t)]:}Apply reverse product nuee f(dg)/(dt)+g(df)/(dt)=(d)/(dt)fga ... See the full answer
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Home > Error Bars > Plotting Mean And Standard Error In Excel
# Plotting Mean And Standard Error In Excel
## Contents
Close Yeah, keep it Undo Close This video is unavailable. Press DELETE. Watch Queue Queue __count__/__total__ Find out whyClose Adding Standard Deviation Error Bars to Line Plot Excel 2012 BIO204 sbbiovideos SubscribeSubscribedUnsubscribe139139 Loading... Amol Patel 6,625 views 11:40 How to create bar chart with error bars (multiple variables) - Duration: 12:41. my review here
sbbiovideos 41,384 views 7:11 Simple Custom Error Bars Excel 2013 - Duration: 5:17. Reply Excel Tips and Tricks from Pryor.com says: January 21, 2016 at 8:57 pm A standard deviation is stated this way, in a cell =STDEV(C5:F43) This will return the standard deviation Reply Rob Douglass says: September 14, 2016 at 8:47 pm HI, I have calculated the means and my std dev of my data points. Once you have calculated the mean for the -195 values, then copy this formula into the cells C87, etc.
## Add Standard Deviation To Excel Graph
How can we improve our confidence? Without going into detail, the mean is a way of summarizing a group of data and stating a best guess at what the true value of the dependent variable value is Sign in Share More Report Need to report the video? What Are Error Bars By dividing the standard deviation by the square root of N, the standard error grows smaller as the number of measurements (N) grows larger.
It was created for BIO204, an introductory biology class for biology majors at MiraCosta College, Oceanside, CA Category Education License Standard YouTube License Show more Show less Loading... Standard Deviation Graph Excel Bozeman Science 176,049 views 7:05 Plotting Graph in Excel with Error Bars - Duration: 14:20. One is with the standard deviation of a single measurement (often just called the standard deviation) and the other is with the standard deviation of the mean, often called the standard
These cells contain a formula that calculates the error value based on a margin of error that is unique to each type of value.
Autoplay When autoplay is enabled, a suggested video will automatically play next. How To Add Error Bars In Excel 2010 All data points in the series display the error amount in the same height for Y error bars and the same width for X error bars.Custom uses values in a worksheet Peter Stanley 15,735 views 6:38 Standard Deviation Error Bars Excel 2010 - Duration: 1:31. While we were able to use a function to directly calculate the mean, the standard error calculation is a little more round about.
## Standard Deviation Graph Excel
To follow using our example below, download Standard Deviation Excel Graphs Template1 and use Sheet 1. These steps will apply to Excel 2013. http://www.uvm.edu/~jleonard/AGRI85/spring2004/Standard_Error_Bars_in_Excel.html James Lim 20,003 views 5:30 Adding X and Y Error Bars in a Scatter Plot in Excel 2008 - Duration: 3:36. Add Standard Deviation To Excel Graph Compare these error bars to the distribution of data points in the original scatter plot above.Tight distribution of points around 100 degrees - small error bars; loose distribution of points around How To Add Error Bars In Excel 2013 The +/- value is the standard error and expresses how confident you are that the mean value (1.4) represents the true value of the impact energy.
MrNystrom 588,368 views 17:26 Loading more suggestions... this page Add to Want to watch this again later? This displays the Chart Tools, adding the Design, Layout, and Format tabs. Sir Prime 144,187 views 1:31 J Reimer shows students how make column graph with error bars in Excel 2013 - Duration: 7:15. Custom Error Bars Excel
The standard deviation shows the dispersion of the values of a data set from their average. Math Meeting 347,798 views 8:26 Standard Deviation and Z-scores - Duration: 20:00. If your column represents 100,000,000 and your error is only 10, then the error bar would be very small in comparison and could look like it's either missing or the same http://setiweb.org/error-bars/plotting-standard-error-in-excel.php Sign in Share More Report Need to report the video?
Standard Error and Standard Deviation use the following equations to calculate the error amounts that are shown on the chart. How To Add Individual Error Bars In Excel Loading... Standard Error Bars in Excel Enter the data into the spreadsheet.
## Stephanie Castle 4,212 views 3:29 What is a "Standard Deviation?" and where does that formula come from - Duration: 17:26.
You can remove either of these error bars by selecting them, and then pressing Delete.The ease with which our tutorials provide students with all the necessary information enables our students to References Office.com: Average FunctionOffice.com: STDEVOffice.com: Change Data Markers in a Line, Scatter, or Radar Chart About the Author Laura Gittins has been writing since 2008 and is an expert in document Any other feedback? How To Add Error Bars In Excel Mac These cells contain a formula that calculates the error value based on a margin of error that is unique to each type of bird species.
Click the "Layout" tab. Add to Want to watch this again later? Choose the style of the error bar. useful reference Watch Queue Queue __count__/__total__ Find out whyClose Adding standard error bars to a column graph in Microsoft Excel Stephanie Castle SubscribeSubscribedUnsubscribe9,4029K Loading...
The dialog box should look like: Click OK and the graph should be complete. Sign in Search Microsoft Search Products Templates Support Products Templates Support Support Apps Access Excel OneDrive OneNote Outlook PowerPoint SharePoint Skype for Business Word Install Subscription Training Admin Add, change, or One such feature includes adding error bars to any excel generated chart, a skill guaranteed to boost any Excel user’s confidence in the data provided for that particular chart. Error bars Loading...
jasondenys 188,823 views 2:07 How to make a line graph in Excel (Scientific data) - Duration: 6:42.
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# International Association for Hydro-Environment Engineering and Research
IAHR, founded in 1935, is a worldwide independent member-based organisation of engineers and water specialists working in fields related to the hydro-environmental sciences and their practical application. Activities range from river and maritime hydraulics to water resources development and eco-hydraulics, through to ice engineering, hydroinformatics, and hydraulic machinery.
You are here : eLibrary : IAHR World Congress Proceedings : 36th Congress - The Hague (2015) FULL PAPERS : THEME 1- MANAGING DELTAS : WAVE OVERTOPPING NUMERICAL SIMULATION USING OPENFOAM
WAVE OVERTOPPING NUMERICAL SIMULATION USING OPENFOAM Author : KARAGIANNIS N. , KARAMBAS TH. & KOUTITAS CHR. A numerical model has been developed and implemented with the open source toolbox OpenFoam and the additional toolbox waves2Foam (Jacobsen et al. 2012), in order to investigate the wave overtopping physical process on inclined impermeable structures. RANS (Reynolds averaged Navier Stokes) equations are solved in conjunction with the k-? SST turbulence model ones, while VOF (Volume of fluid) method is used to track the free surface. Surface elevation and depth averaged velocities are obtained from the model and the average overtopping rate is calculated and compared to semi- empirical formulae from literature. Iribarren number as a criterion of breaking waves, the slope angle, freeboard in relation to wave height, and the wave steepness are the most important factors which determine the average overtopping rate. o According to Kofoed (2002), a slope of 1:1.73, namely 30 angle, is considered optimal for the wave overtopping and that angle is implemented in the model. A wave steepness of around 0.033 is regarded as optimal for the maximum overtopping rate as well. Extended investigation on wave characteristics and freeboard was made and the results, which are in good agreement with empirical formulae, are presented herein. File Size : 914,580 bytes File Type : Adobe Acrobat Document Chapter : IAHR World Congress Proceedings Category : 36th Congress - The Hague (2015) FULL PAPERS Article : THEME 1- MANAGING DELTAS Date Published : 18/04/2016 Download Now
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https://pl.tradingview.com/script/wSMZKu7B-Indicator-Trend-Trigger-Factor/
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# Indicator: Trend Trigger Factor
22204 obejrzeń
Introduced by M.H.Pee, Trend Trigger Factor is designed to keep the trader trading with the trend.
System rules according to the developer:
* If the 15-day TTF is above 100 (indicating an uptrend), you will want to be in long positions.
* If the 15-day TTF is below -100, you will want to be short.
* If it is between -100 and 100, you should remain with the current position.
```//
// @author LazyBear
//
study("Trend Trigger Factor [LazyBear]", shorttitle="TTF_LB")
length=input(15)
bt = input( 100, title="Buy Trigger")
st = input( -100, title="Sell Trigger")
markCrossovers=input(false, type=bool)
prev(s,i) =>
y=abs(round(i))
s[y]
calc_ttf( periods ) =>
bp = highest( high, periods ) - prev( lowest( low, periods ), - periods )
sp = prev( highest( high, periods ), - periods ) - lowest( low, periods )
100 * (bp - sp) / ( 0.5*( bp + sp) )
ttf = calc_ttf( length )
plot(0, color=gray)
btl=plot(bt, color=gray, style=3)
stl=plot(st, color=gray, style=3)
long_f = cross( ttf, st ) and rising(ttf, 1)
short_f = cross(ttf, bt ) and falling(ttf, 1)
bs = (ttf > bt) ? bt : ttf
us = (ttf < st) ? st : ttf
bl=plot(bs, color=white)
ul=plot(us, color=white)
tl=plot(ttf, title="TTF", color=markCrossovers ? (long_f ? green : short_f ? red : blue) : maroon, linewidth=2)
fill(bl, tl, color=green, transp=75)
fill(ul, tl, color=red, transp=75)
```
List of my free indicators: http://bit.ly/1LQaPK8
List of my indicators at Appstore: http://blog.tradingview.com/?p=970
## Komentarze
Interesting in understanding the following formula, for which values of s and i dont understand. prev(s,i) =>
y=abs(round(i))
s Thanks.
Odpowiedz
telepatico
@telepatico, prev(s,i)=s
Odpowiedz
You should be advanced into a Mod position:) You would be the only one that earned it in a hard way. Great and appreciate it work.
Odpowiedz
sublimares2
thanks sublimares2. :)
Odpowiedz
sublimares2
Totally agree about LazyBear, he is doing a marvellous job with the indicators.
Odpowiedz
LudmilaHanania
Thanks
Odpowiedz
what kind of statement the following codes are? A switch statement: #prev(s,i) =>
# y=abs(round(i))
# s
#calc_ttf( periods ) =>
# bp = highest( high, periods ) - prev( lowest( low, periods ), - periods )
# sp = prev( highest( high, periods ), - periods ) - lowest( low, periods )
# 100 * (bp - sp) / ( 0.5*( bp + sp) )
Thanks.
Odpowiedz
Yet another useful indicator... Thanks LB
Odpowiedz
primaindustria
yw :)
Odpowiedz
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5 Replies Latest reply on Apr 3, 2018 7:45 AM by Michael Hesser
# Filtering \$ Amounts from Largest to Smallest within 2 Countries
I am trying to figure out how I need to go about creating a filter that can apply to sorting \$ amount from largest to smallest in each individual country and not only to the overall \$ amount. Below in the picture Supplier A has the largest \$ spend amount in the US but not in Canada. In Canada, Supplier B has the largest spend amount but does not in the US. Is there a way that I can apply this to only sort each individual country and not overall \$ amount?
• ###### 1. Re: Filtering \$ Amounts from Largest to Smallest within 2 Countries
hello,
Create a measure : RANK(SUM(Total Delivered))
Right click go to compute using --> Pane down.
This will give your desired result.
Below is the sample from superstore data
• ###### 2. Re: Filtering \$ Amounts from Largest to Smallest within 2 Countries
Hi Chase;
There are many different ways to get the results you're after: it's a little more difficult because I don't know what your data looks like.
One way is to just sort by negative sales:
• Make a calc
SortSales
-[Sales]
• Drop onto rows
• Change it into Discrete
• Move the now blue pill BETWEEN Plant and Supplier Parent
• Hide it by de-selecting Show in Header
• Your sheet should now be sorted by Plant Country, by Sales Rank for Supplier Parent for the Plant Country, by Plant name
Easy!
Other option is to rank them:
• Create a rank calculation. You can do this as quick calc, or make your own formula:
RANKER
RANK(SUM([Sales]))
• Change the Default Table Calculation so that results are computed along Supplier. In my example, they're done via Category.
• Drop RANKER into rows
• Transform it into Discrete
• Move the now blue pill BETWEEN Plant and Supplier Parent
• Hide it by de-selecting Show in Header
• Your sheet should now be sorted by Plant Country, by Sales Rank for Supplier Parent for the Plant Country, by Plant name
My results, using Superstore, look like this:
Good luck!
• ###### 3. Re: Filtering \$ Amounts from Largest to Smallest within 2 Countries
Solution verified! Thanks!
• ###### 4. Re: Filtering \$ Amounts from Largest to Smallest within 2 Countries
Thanks for the quick response!
• ###### 5. Re: Filtering \$ Amounts from Largest to Smallest within 2 Countries
We're here to help!
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# Statistics
posted by .
The call letter of a radio station must have 4 letters.The first letter must be a K or a W. How many different station call setters can be made if repetitions are not allowed? If repetitions are allowed?
• Statistics -
With no repetition, there are 2 choices for the first letter
25 choices for the second
24 choices for the third
23 choices for the last.
Use the multiplication rule to find the total number of possible names.
## Similar Questions
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2. ### MATH Prob.
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6. ### Math/ Probability
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7. ### math
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More Similar Questions
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Unit of study_
# MATH4313: Functional Analysis
## Overview
Functional analysis is one of the major areas of modern mathematics. It can be thought of as an infinite-dimensional generalisation of linear algebra and involves the study of various properties of linear continuous transformations on normed infinite-dimensional spaces. Functional analysis plays a fundamental role in the theory of differential equations, particularly partial differential equations, representation theory, and probability. In this unit you will cover topics that include normed vector spaces, completions and Banach spaces; linear operators and operator norms; Hilbert spaces and the Stone-Weierstrass theorem; uniform boundedness and the open mapping theorem; dual spaces and the Hahn-Banach theorem; and spectral theory of compact self-adjoint operators. A thorough mechanistic grounding in these topics will lead to the development of your compositional skills in the formulation of solutions to multifaceted problems. By completing this unit you will become proficient in using a set of standard tools that are foundational in modern mathematics and will be equipped to proceed to research projects in PDEs, applied dynamics, representation theory, probability, and ergodic theory.
### Details
Academic unit Mathematics and Statistics Academic Operations MATH4313 Functional Analysis Semester 1, 2023 Normal day Camperdown/Darlington, Sydney 6
### Enrolment rules
Prohibitions ? None None None Real Analysis and abstract linear algebra (e.g., MATH2X23 and MATH2X22 or equivalent), and, preferably, knowledge of Metric Spaces Yes
### Teaching staff and contact details
Coordinator James Parkinson, james.parkinson@sydney.edu.au Daniel Daners James Parkinson
## Assessment
Type Description Weight Due Length
Supervised exam
Final Exam
Written mathematical formulae and arguments.
55% Formal exam period 2 hours
Outcomes assessed:
Assignment Assignment 1
Written assignment
15% Week 04
Due date: 15 Mar 2023 at 23:59
Closing date: 22 Mar 2023
14 days
Outcomes assessed:
Assignment Assignment 2
Written assignment
15% Week 08
Due date: 19 Apr 2023 at 23:59
Closing date: 26 Apr 2023
14 days
Outcomes assessed:
Assignment Assignment 3
Written assignment
15% Week 12
Due date: 17 May 2023 at 23:59
Closing date: 24 May 2023
14 days
Outcomes assessed:
• Writing task: 3 assignments worth 10% each. These assignments will require you to synthesise information from lectures and tutorials to create concise written arguments.
• Final exam: The exam will cover all material in the unit from both lectures and tutorials.
• Final Exam: If a second replacement exam is required, this exam may be delivered via an alternative assessment method, such as a viva voce (oral exam). The alternative assessment will meet the same learning outcomes as the original exam. The format of the alternative assessment will be determined by the unit coordinator.
### Assessment criteria
The University awards common result grades, set out in the Coursework Policy 2014 (Schedule 1).
As a general guide, a high distinction indicates work of an exceptional standard, a distinction a very high standard, a credit a good standard, and a pass an acceptable standard.
Result name
Mark range
Description
High distinction
85 - 100
At HD level, a student demonstrates a flair for the subject as well as a detailed and comprehensive understanding of the unit material. A ‘High Distinction’ reflects exceptional achievement and is awarded to a student who demonstrates the ability to apply their subject knowledge and understanding to produce original solutions for novel or highly complex problems and/or comprehensive critical discussions of theoretical concepts.
Distinction
75 - 84
At DI level, a student demonstrates an aptitude for the subject and a well-developed understanding of the unit material. A ‘Distinction’ reflects excellent achievement and is awarded to a student who demonstrates an ability to apply their subject knowledge and understanding of the subject to produce good solutions for challenging problems and/or a reasonably well-developed critical analysis of theoretical concepts.
Credit
65 - 74
At CR level, a student demonstrates a good command and knowledge of the unit material. A ‘Credit’ reflects solid achievement and is awarded to a student who has a broad general understanding of the unit material and can solve routine problems and/or identify and superficially discuss theoretical concepts.
Pass
50 - 64
At PS level, a student demonstrates proficiency in the unit material. A ‘Pass’ reflects satisfactory achievement and is awarded to a student who has threshold knowledge.
Fail
0 - 49
When you don’t meet the learning outcomes of the unit to a satisfactory standard.
### Late submission
In accordance with University policy, these penalties apply when written work is submitted after 11:59pm on the due date:
• Deduction of 5% of the maximum mark for each calendar day after the due date.
• After ten calendar days late, a mark of zero will be awarded.
### Special consideration
If you experience short-term circumstances beyond your control, such as illness, injury or misadventure or if you have essential commitments which impact your preparation or performance in an assessment, you may be eligible for special consideration or special arrangements.
The Current Student website provides information on academic integrity and the resources available to all students. The University expects students and staff to act ethically and honestly and will treat all allegations of academic integrity breaches seriously.
We use similarity detection software to detect potential instances of plagiarism or other forms of academic integrity breach. If such matches indicate evidence of plagiarism or other forms of academic integrity breaches, your teacher is required to report your work for further investigation.
You may only use artificial intelligence and writing assistance tools in assessment tasks if you are permitted to by your unit coordinator, and if you do use them, you must also acknowledge this in your work, either in a footnote or an acknowledgement section.
Studiosity is permitted for postgraduate units unless otherwise indicated by the unit coordinator. The use of this service must be acknowledged in your submission.
## Weekly schedule
WK Topic Learning activity Learning outcomes
Week 01 Review of the topology of metric spaces, compactness and continuity Lecture (3 hr)
Week 02 Normed linear spaces and basic properties of linear operators Lecture and tutorial (4 hr)
Week 03 Finite and infinite dimensional Banach spaces Lecture and tutorial (4 hr)
Week 04 Banach algebras and the Stone-Weierstrass Theorem Lecture and tutorial (4 hr)
Week 05 Hilbert spaces, orthonormal systems, projections and abstract Fourier series Lecture and tutorial (4 hr)
Week 06 Baire's theorem and the open mapping theorem Lecture and tutorial (4 hr)
Week 07 The closed graph theorem, the uniform boundedness theorem Lecture and tutorial (4 hr)
Week 08 Closed operators Lecture and tutorial (4 hr)
Week 09 Duality and the Hahn-Banach theorem Lecture and tutorial (4 hr)
Week 10 Weak convergence Lecture and tutorial (4 hr)
Week 11 Duality in Hilbert spaces and the Lax-Milgram Theorem Lecture and tutorial (4 hr)
Week 12 Basics of spectral theory, ascent and descent of linear operators Lecture and tutorial (4 hr)
Week 13 The spectrum of compact operators Lecture and tutorial (4 hr)
### Study commitment
Typically, there is a minimum expectation of 1.5-2 hours of student effort per week per credit point for units of study offered over a full semester. For a 6 credit point unit, this equates to roughly 120-150 hours of student effort in total.
Daniel Daners, Introduction to Functional Analysis, University of Sydney, 2021 (available on Canvas)
## Learning outcomes
Learning outcomes are what students know, understand and are able to do on completion of a unit of study. They are aligned with the University’s graduate qualities and are assessed as part of the curriculum.
At the completion of this unit, you should be able to:
• LO1. demonstrate a coherent and advanced understanding of the key concepts of geometry of normed spaces, Hilbert Space Theory, Abstract Fourier Analysis, Hahn-Banach Theory and Spectral Theory, and how they provide a unified approach to infinite-dimensional linear problems in mathematics
• LO2. apply the fundamental ideas and results in functional analysis to solve given problems
• LO3. distinguish and compare the properties of different types of linear operators, analysing their spectra and deriving their main properties
• LO4. formulate analytic problems in functional-analytic terms and determine the appropriate framework to solve them
• LO5. communicate coherent mathematical arguments appropriately to student and expert audiences, both orally and through written work
• LO6. devise computational solutions to complex problems in functional analysis
• LO7. compose correct proofs of unfamiliar general results in functional analysis.
The graduate qualities are the qualities and skills that all University of Sydney graduates must demonstrate on successful completion of an award course. As a future Sydney graduate, the set of qualities have been designed to equip you for the contemporary world.
GQ1 Depth of disciplinary expertise Deep disciplinary expertise is the ability to integrate and rigorously apply knowledge, understanding and skills of a recognised discipline defined by scholarly activity, as well as familiarity with evolving practice of the discipline. GQ2 Critical thinking and problem solving Critical thinking and problem solving are the questioning of ideas, evidence and assumptions in order to propose and evaluate hypotheses or alternative arguments before formulating a conclusion or a solution to an identified problem. GQ3 Oral and written communication Effective communication, in both oral and written form, is the clear exchange of meaning in a manner that is appropriate to audience and context. GQ4 Information and digital literacy Information and digital literacy is the ability to locate, interpret, evaluate, manage, adapt, integrate, create and convey information using appropriate resources, tools and strategies. GQ5 Inventiveness Generating novel ideas and solutions. GQ6 Cultural competence Cultural Competence is the ability to actively, ethically, respectfully, and successfully engage across and between cultures. In the Australian context, this includes and celebrates Aboriginal and Torres Strait Islander cultures, knowledge systems, and a mature understanding of contemporary issues. GQ7 Interdisciplinary effectiveness Interdisciplinary effectiveness is the integration and synthesis of multiple viewpoints and practices, working effectively across disciplinary boundaries. GQ8 Integrated professional, ethical, and personal identity An integrated professional, ethical and personal identity is understanding the interaction between one’s personal and professional selves in an ethical context. GQ9 Influence Engaging others in a process, idea or vision.
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# NPDA for accepting the language L = {amb(2m+1) | m ≥ 1}
Last Updated : 01 Nov, 2022
Prerequisite – Pushdown automata, Pushdown automata acceptance by final state
Problem – Design a non deterministic PDA for accepting the language L = {| m ≥ 1}, or,L = {| m ≥ 1}, i.e.,
L = {abbb, aabbbbb, aaabbbbbbb, aaaabbbbbbbbb, ......}
In each of the string, the number of ‘b’ is one more than the twice of the number of ‘a’.
Explanation – Here, we need to maintain the order of a’s and b’s.That is, all the a’s are coming first and then all the b’s are coming. Thus, we need a stack along with the state diagram. The count of a’s and b’s is maintained by the stack. We will take 2 stack alphabets:
= { a, z }Where, = set of all the stack alphabets z = stack start symbol
Approach used in the construction of PDA – As we want to design a NPDA, thus every time ‘a’ comes before ‘b’. We will push one ‘a’ into the stack for every ‘a’s and again for the next ‘a’, we will push one ‘a’ into the stack. And then when ‘b’ comes, for the first ‘b’ we will do nothing only state will change. For the next two ‘b’ we will pop one ‘a’ and again for next two ‘b’, we pop one ‘a’. And similarly we perform this alternatively. i.e., For third ‘b’ we pop first ‘a’ For fifth ‘b’ we pop second ‘a’ For seventh ‘b’ we pop third ‘a’ So, at the end if the stack becomes empty then we can say that the string is accepted by the PDA.
Stack transition functions –
(q0, a, z) (q0, az) (q0, a, a) (q0, aa) (q0, b, a) (q1, a) [ Indicates no operation only state change ](q1, b, a) (q2, a) [ Indicates no operation only state change ](q2, b, a) (q3, ) [Indicates pop operation ](q3, b, a) (q2, a ) [ Indicates no operation only state change ](q3, , z) (qf, z )
Where, q0 = Initial state qf = Final state = indicates pop operation
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What is the probability that the product xy less than 9, Mathematics
Assignment Help:
A number x is selected from the numbers 1,2,3 and then a second number y is randomly selected from the numbers 1,4,9. What is the probability that the product xy of the two numbers will be less than 9? (Ans: 5/9 )
Ans : Number X can be selected in three ways and corresponding to each such way there are three ways of selecting number y . Therefore , two numbers can be selected in 9 ways as listed below:
(1,1), (1,4), (2,1), (2,4), (3,1)
∴ Favourable number of elementary events = 5
Hence, required probability = 5/9
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calculates the value of the following limit. Solution Now, notice that if we plug in θ =0 which we will get division by zero & so the function doesn't present at this
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From Sharp Baboon, 10 Months ago, written in Plain Text.
Embed
Hits: 107
1. c=randi(10,3);
2. b=[2; 5; 4];
3. %c=[2 1 3; 4 5 -7; -1 2 2]; SPRAWDZENIE
4. %b=[4 ;7; -3]; SPRAWDZENIE
5. AA=gaussian(c,b);
6. z=length(b)+1;
7.
8. bb=AA(:,z);
9. xxx=backward(AA,bb);
10. function Ag = gaussian(A,b)
11. Ag = [A b];
12. n = size(Ag,1);
13. for k=1:n-1
14. for i = k+1:n
15. l = Ag(i,k) / Ag(k,k);
16. for j = k:n+1
17. Ag(i,j) = Ag(i,j) - l * Ag(k,j);
18. end
19. end
20. end
21. end
22.
23. function x = backward(U,c)
24. n = size(U-1,1);
25. x = zeros(n,1);
26. for i = n:-1:1
27. s = 0;
28. for j = i+1:n
29. s = s + U(i,j)*x(j);
30. end
31. x(i) = (c(i) - s) / U(i,i);
32. end
33. end
34.
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### The Mathematical Link Between Gears in a Gear Train
Wednesday, May 14th, 2014
Last time we analyzed the angular relationship between the Force and Distance vectors in this simple gear train. Today we’ll discover a commonality between the two gears in this train which will later enable us to develop individual torque calculations for them. From the illustration it’s clear that the driving gear is mechanically linked to the driven gear by their teeth. Because they’re linked, force, and hence torque, is transmitted by way of the driving gear to the driven gear. Knowing this we can develop a mathematical equation to link the driving gear Force vector F1 to the driven gear Force vector F2, then use that linking equation to develop a separate torque formula for each of the gears in the train. We learned in the previous blog in this series that F1 and F2 travel in opposite directions to each other along the same line of action. As such, both of these Force vectors are situated in the same way so that they are each at an angle value ϴ with respect to their Distance vectors D1 and D2. This fact allows us to build an equation with like terms, and that in turn allows us to use trigonometry to link the two force vectors into a single equation: F = [F1 × sin(ϴ)] – [F2 × sin(ϴ)] where F is called a resultant Force vector, so named because it represents the force that results when the dead, or inert, weight that’s present in the resisting force F2 cancels out some of the positive force of F1. Next week we’ll simplify our gear train illustration and delve into more math in order to develop separate torque computations for each gear in the train. _______________________________________
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1
AIEEE 2012
+4
-1
Let $$a,b \in R$$ be such that the function $$f$$ given by $$f\left( x \right) = In\left| x \right| + b{x^2} + ax,\,x \ne 0$$ has extreme values at $$x=-1$$ and $$x=2$$
Statement-1 : $$f$$ has local maximum at $$x=-1$$ and at $$x=2$$.
Statement-2 : $$a = {1 \over 2}$$ and $$b = {-1 \over 4}$$
A
Statement - 1 is false, Statement - 2 is true.
B
Statement - 1 is true , Statement - 2 is true; Statement - 2 is a correct explanation for Statement - 1.
C
Statement - 1 is true, Statement - 2 is true; Statement - 2 is not a correct explanation for Statement - 1.
D
Statement - 1 is true, Statement - 2 is false.
2
AIEEE 2012
+4
-1
A line is drawn through the point $$(1, 2)$$ to meet the coordinate axes at $$P$$ and $$Q$$ such that it forms a triangle $$OPQ,$$ where $$O$$ is the origin. If the area of the triangle $$OPQ$$ is least, then the slope of the line $$PQ$$ is :
A
$$-{1 \over 4}$$
B
$$-4$$
C
$$-2$$
D
$$-{1 \over 2}$$
3
AIEEE 2011
+4
-1
For $$x \in \left( {0,{{5\pi } \over 2}} \right),$$ define $$f\left( x \right) = \int\limits_0^x {\sqrt t \sin t\,dt.}$$ Then $$f$$ has
A
local minimum at $$\pi$$ and $$2\pi$$
B
local minimum at $$\pi$$ and local maximum at $$2\pi$$
C
local maximum at $$\pi$$ and local minimum at $$2\pi$$
D
local maximum at $$\pi$$ and $$2\pi$$
4
AIEEE 2011
+4
-1
Out of Syllabus
The shortest distance between line $$y-x=1$$ and curve $$x = {y^2}$$ is
A
$${{3\sqrt 2 } \over 8}$$
B
$${8 \over {3\sqrt 2 }}$$
C
$${4 \over {\sqrt 3 }}$$
D
$${{\sqrt 3 } \over 4}$$
EXAM MAP
Medical
NEET
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# formula
Discussion in 'N / Z Scale Model Trains' started by willy4, Feb 4, 2008.
1. ### willy4Member
what is the formula for changing O scale to N scale. If I have a drawing in O
scale, and I measure it with a O scale ruler and it is 30 ft, can I take my N scale ruler and measure it to 30 ft.
2. ### MasonJarIt's not rocket surgery
That is the easiest way to change the scale (no matter what scale) to any other. If it is 30 real feet long, it is also 30 O scale feet, or 30 N scale feet or whatever.
If you need to change the scale of a drawing, you can calculate approximately how much by comparing the scales.
Reducing O scale (1:48 ) to N (1:160), use 30% (48/160 * 100)
Enlarging N to O, use 333% (160/48 * 100)
And so on.
Andrew
3. ### willy4Member
Thanks masonjar
I see you are from Ottawa Canada. I live 30 miles from Ottawa on the Que side (Luskville). I am in Ottawa alot, also I am a member of the Ottawa valley N-Trak club. Now I am in Florida for 3 months to get away from those cold Canadian winters. If you go to logging under willy4 you will see one of scrachbuilt models, a Lombard Tractor. I read about so many formulas that I just wanted to reasure myself.
Thank you
Bill
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# WBJEE 2014 Solved Mathematics Question Paper – Part 1
Find WBJEE Solved Mathematics Question Paper for the year 2014. This solved paper will help students in their final level of preparation for WBJEE Exam.
Created On: Mar 9, 2017 17:37 IST
Find WBJEE 2014 Solved Mathematics Question Paper – Part 1 in this article. This paper consists of 5 questions (#1 to #5) from WBJEE 2014 Mathematics paper. Detailed solution of these questions has been provided so that students can match their solutions.
Importance of Previous Years’ Paper:
Previous years’ question papers help aspirants in understanding exam pattern, question format, important topics and assessing preparation. It has also been seen that sometimes questions are repeated in WBJEE Exam. So, this paper will certainly boost your confidence.
WBJEE is a common entrance examinations held at state level for admission to the Undergraduate Level Engineering and Medical courses in the State of West Bengal. The Mathematics section of WBJEE 2014 engineering entrance exam consists of 40 questions.
2. If y = 4x + 3 is parallel to a tangent to the parabola y2 = 12x, then its distance from the normal parallel to the given line is
Ans: (B)
Sol:
We have equation of line,
y = 4x + 3
So, slope of line = 4
Also, a = 3
Equation of normal
y = mx – 2am – am3
y = 4x – 216
3. In a ΔABC, tanA and tanB are the roots of pq(x2 + 1) = r2x. Then ΔABC is
(A) a right angled triangle
(B) an acute angled triangle
(C) an obtuse angled triangle
(D) an equilateral triangle
Ans: (A)
Sol:
We have equation, pq(x2 + 1) = r2x.
So, pqx2 – r2x + pq = 0
tanA and tanB are the roots of the above equation
Product of roots = tanA. tanB = pq/pq = 1
tanA tanB = 1
So, tan(A+B) is undefined
∴∠C = π / 2
So, ABC is right angled triangle.
WBJEE 2016 Solved Physics and Chemistry Question Paper
4. Let the number of elements of the sets A and B be p and q respectively. Then the number of relations from the set A to the set B is
(A) 2p+q
(B) 2pq
(C) p + q
(D) pq
Ans : (B)
Sol:
Number of elements in set A = p
Number of elements in set B = q
Number of relations from the set A to the set B is pq.
Also Get:
WBJEE Sample Papers
WBJEE Previous Years' Question Papers
WBJEE Online Test
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# Check if a Binary Tree contains duplicate subtrees of size 2 or more
Given a Binary Tree, check whether the Binary tree contains a duplicate sub-tree of size 2 or more.
Note : Two same leaf nodes are not considered as subtree size of a leaf node is one.
```Input : Binary Tree
A
/
B C
/
D E B
/
D E
Output : Yes
```
Tree with duplicate Sub-Tree [ highlight by blue color ellipse ]
[ Method 1]
A simple solution is that, we pick every node of tree and try to find is any sub-tree of given tree is present in tree which is identical with that sub-tree. Here we can use below post to find if a subtree is present anywhere else in tree.
Check if a binary tree is subtree of another binary tree
[Method 2 ]( Efficient solution )
An Efficient solution based on tree serialization and hashing. The idea is to serialize subtrees as strings and store the strings in hash table. Once we find a serialized tree (which is not a leaf) already existing in hash-table, we return true.
Below The implementation of above idea.
## C++
`// C++ program to find if there is a duplicate ` `// sub-tree of size 2 or more. ` `#include ` `using` `namespace` `std; ` ` ` `// Separator node ` `const` `char` `MARKER = ``'\$'``; ` ` ` `// Structure for a binary tree node ` `struct` `Node ` `{ ` ` ``char` `key; ` ` ``Node *left, *right; ` `}; ` ` ` `// A utility function to create a new node ` `Node* newNode(``char` `key) ` `{ ` ` ``Node* node = ``new` `Node; ` ` ``node->key = key; ` ` ``node->left = node->right = NULL; ` ` ``return` `node; ` `} ` ` ` `unordered_set subtrees; ` ` ` `// This function returns empty string if tree ` `// contains a duplicate subtree of size 2 or more. ` `string dupSubUtil(Node *root) ` `{ ` ` ``string s = ``""``; ` ` ` ` ``// If current node is NULL, return marker ` ` ``if` `(root == NULL) ` ` ``return` `s + MARKER; ` ` ` ` ``// If left subtree has a duplicate subtree. ` ` ``string lStr = dupSubUtil(root->left); ` ` ``if` `(lStr.compare(s) == 0) ` ` ``return` `s; ` ` ` ` ``// Do same for right subtree ` ` ``string rStr = dupSubUtil(root->right); ` ` ``if` `(rStr.compare(s) == 0) ` ` ``return` `s; ` ` ` ` ``// Serialize current subtree ` ` ``s = s + root->key + lStr + rStr; ` ` ` ` ``// If current subtree already exists in hash ` ` ``// table. [Note that size of a serialized tree ` ` ``// with single node is 3 as it has two marker ` ` ``// nodes. ` ` ``if` `(s.length() > 3 && ` ` ``subtrees.find(s) != subtrees.end()) ` ` ``return` `""``; ` ` ` ` ``subtrees.insert(s); ` ` ` ` ``return` `s; ` `} ` ` ` `// Driver program to test above functions ` `int` `main() ` `{ ` ` ``Node *root = newNode(``'A'``); ` ` ``root->left = newNode(``'B'``); ` ` ``root->right = newNode(``'C'``); ` ` ``root->left->left = newNode(``'D'``); ` ` ``root->left->right = newNode(``'E'``); ` ` ``root->right->right = newNode(``'B'``); ` ` ``root->right->right->right = newNode(``'E'``); ` ` ``root->right->right->left= newNode(``'D'``); ` ` ` ` ``string str = dupSubUtil(root); ` ` ` ` ``(str.compare(``""``) == 0) ? cout << ``" Yes "``: ` ` ``cout << ``" No "` `; ` ` ``return` `0; ` `} `
## Java
`// Java program to find if there is a duplicate ` `// sub-tree of size 2 or more. ` `import` `java.util.HashSet; ` `public` `class` `Main { ` ` ` ` ``static` `char` `MARKER = ``'\$'``; ` ` ` ` ``// This function returns empty string if tree ` ` ``// contains a duplicate subtree of size 2 or more. ` ` ``public` `static` `String dupSubUtil(Node root, HashSet subtrees) ` ` ``{ ` ` ``String s = ``""``; ` ` ` ` ``// If current node is NULL, return marker ` ` ``if` `(root == ``null``) ` ` ``return` `s + MARKER; ` ` ` ` ``// If left subtree has a duplicate subtree. ` ` ``String lStr = dupSubUtil(root.left,subtrees); ` ` ``if` `(lStr.equals(s)) ` ` ``return` `s; ` ` ` ` ``// Do same for right subtree ` ` ``String rStr = dupSubUtil(root.right,subtrees); ` ` ``if` `(rStr.equals(s)) ` ` ``return` `s; ` ` ` ` ``// Serialize current subtree ` ` ``s = s + root.data + lStr + rStr; ` ` ` ` ``// If current subtree already exists in hash ` ` ``// table. [Note that size of a serialized tree ` ` ``// with single node is 3 as it has two marker ` ` ``// nodes. ` ` ``if` `(s.length() > ``3` `&& subtrees.contains(s)) ` ` ``return` `""``; ` ` ` ` ``subtrees.add(s); ` ` ``return` `s; ` ` ``} ` ` ` ` ``//Function to find if the Binary Tree contains duplicate ` ` ``//subtrees of size 2 or more ` ` ``public` `static` `String dupSub(Node root) ` ` ``{ ` ` ``HashSet subtrees=``new` `HashSet<>(); ` ` ``return` `dupSubUtil(root,subtrees); ` ` ``} ` ` ` ` ``public` `static` `void` `main(String args[]) ` ` ``{ ` ` ``Node root = ``new` `Node(``'A'``); ` ` ``root.left = ``new` `Node(``'B'``); ` ` ``root.right = ``new` `Node(``'C'``); ` ` ``root.left.left = ``new` `Node(``'D'``); ` ` ``root.left.right = ``new` `Node(``'E'``); ` ` ``root.right.right = ``new` `Node(``'B'``); ` ` ``root.right.right.right = ``new` `Node(``'E'``); ` ` ``root.right.right.left= ``new` `Node(``'D'``); ` ` ``String str = dupSub(root); ` ` ``if``(str.equals(``""``)) ` ` ``System.out.print(``" Yes "``); ` ` ``else` ` ``System.out.print(``" No "``); ` ` ``} ` `} ` ` ` `// A binary tree Node has data, ` `// pointer to left child ` `// and a pointer to right child ` `class` `Node { ` ` ``int` `data; ` ` ``Node left,right; ` ` ``Node(``int` `data) ` ` ``{ ` ` ``this``.data=data; ` ` ``} ` `}; ` `//This code is contributed by Gaurav Tiwari `
## C#
`// C# program to find if there is a duplicate ` `// sub-tree of size 2 or more. ` `using` `System; ` `using` `System.Collections.Generic; ` ` ` `class` `GFG ` `{ ` ` ` ` ``static` `char` `MARKER = ``'\$'``; ` ` ` ` ``// This function returns empty string if tree ` ` ``// contains a duplicate subtree of size 2 or more. ` ` ``public` `static` `String dupSubUtil(Node root, ` ` ``HashSet subtrees) ` ` ``{ ` ` ``String s = ``""``; ` ` ` ` ``// If current node is NULL, return marker ` ` ``if` `(root == ``null``) ` ` ``return` `s + MARKER; ` ` ` ` ``// If left subtree has a duplicate subtree. ` ` ``String lStr = dupSubUtil(root.left,subtrees); ` ` ``if` `(lStr.Equals(s)) ` ` ``return` `s; ` ` ` ` ``// Do same for right subtree ` ` ``String rStr = dupSubUtil(root.right,subtrees); ` ` ``if` `(rStr.Equals(s)) ` ` ``return` `s; ` ` ` ` ``// Serialize current subtree ` ` ``s = s + root.data + lStr + rStr; ` ` ` ` ``// If current subtree already exists in hash ` ` ``// table. [Note that size of a serialized tree ` ` ``// with single node is 3 as it has two marker ` ` ``// nodes. ` ` ``if` `(s.Length > 3 && subtrees.Contains(s)) ` ` ``return` `""``; ` ` ` ` ``subtrees.Add(s); ` ` ``return` `s; ` ` ``} ` ` ` ` ``// Function to find if the Binary Tree contains ` ` ``// duplicate subtrees of size 2 or more ` ` ``public` `static` `String dupSub(Node root) ` ` ``{ ` ` ``HashSet subtrees = ``new` `HashSet(); ` ` ``return` `dupSubUtil(root,subtrees); ` ` ``} ` ` ` ` ``// Driver code ` ` ``public` `static` `void` `Main(String []args) ` ` ``{ ` ` ``Node root = ``new` `Node(``'A'``); ` ` ``root.left = ``new` `Node(``'B'``); ` ` ``root.right = ``new` `Node(``'C'``); ` ` ``root.left.left = ``new` `Node(``'D'``); ` ` ``root.left.right = ``new` `Node(``'E'``); ` ` ``root.right.right = ``new` `Node(``'B'``); ` ` ``root.right.right.right = ``new` `Node(``'E'``); ` ` ``root.right.right.left= ``new` `Node(``'D'``); ` ` ``String str = dupSub(root); ` ` ``if``(str.Equals(``""``)) ` ` ``Console.Write(``" Yes "``); ` ` ``else` ` ``Console.Write(``" No "``); ` ` ``} ` `} ` ` ` `// A binary tree Node has data, ` `// pointer to left child ` `// and a pointer to right child ` `public` `class` `Node ` `{ ` ` ``public` `int` `data; ` ` ``public` `Node left,right; ` ` ``public` `Node(``int` `data) ` ` ``{ ` ` ``this``.data = data; ` ` ``} ` `}; ` ` ` `// This code is contributed by 29AjayKumar `
Output:
```Yes
```
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http://mytestbook.com/worksheet.aspx?test_id=1393&grade=8&subject=Math&topics=Data%20Interpretation%20Graphs%20and%20Charts
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go to: myTestBook.com You must use Internet Explorer if you want to print this worksheet in a proper format.
print help! Print: Use this to print without Ads and Toolbar (taks a few seconds). dotted fields in the header are editable. Report an error
The following text/image is for 1 through 5 Scientific observation of Force and Acceleration on an object.
Question 1 The range of force data in the graph is ________. A. 0 - 350 B. 50 - 300 C. 100 - 600 D. 0 - 700
Question 2 How much acceleration is observed at 200 Newtons force? A. 100 m/sq. sec B. 110 m/sq. sec C. 150 m/sq. sec D. 200 m/sq. sec
Question 3 Which data is an outlier? A. at force = 100 B. at force = 300 C. at force = 600 D. at force = 700
Question 4 What kind of trend do you see in the graph? A. Acceleration increases with increase in force B. Acceleration decreases with increase in force C. Acceleration increases with decrease in force D. There is no trend
Question 5 Based on the trend, how much acceleration is expected at a force = 800 Newtons? A. 200 m/sq. sec B. 300 m/sq. sec C. 350 m/sq. sec D. 400 m/sq. sec
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The following text/image is for 6 through 10
Question 6 What product was exported most in year 2004? A. Rice B. Wheat C. Corn D. Soy
Question 7 For how many years did the corn production dominated the export? A. 4 years B. 6 years C. 3 years D. 5 years
Question 8 Which product suffered the sharpest decline in export and when? A. Rice from 2003 to 2004 B. Rice from 2004 to 2005 C. Corn from 2007 to 2008 D. Wheat from 2007 to 2008
Question 9 Calculate the total export of three products in year 2007. (approximately) A. 2000 metric ton B. 2400 metric ton C. 2600 metric ton D. 2200 metric ton
Question 10 Calculate the percentage of rice export compare to the total export in year 2007. A. 40% B. 50% C. 60% D. 30%
Free Worksheets From myTestBook.com ------ © myTestBook.com, Inc.
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https://www.justintools.com/unit-conversion/area.php?k1=square-miles&k2=dunams
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# AREA Units Conversionsquare-miles to dunams
1 Square Miles
= 2589.98811 Dunams
Embed this to your website/blog
Category: area
Conversion: Square Miles to Dunams
The base unit for area is square meters (Non-SI/Derived Unit)
[Square Miles] symbol/abbrevation: (mi2, sq mi)
[Dunams] symbol/abbrevation: (dunam)
How to convert Square Miles to Dunams (mi2, sq mi to dunam)?
1 mi2, sq mi = 2589.98811 dunam.
1 x 2589.98811 dunam = 2589.98811 Dunams.
Always check the results; rounding errors may occur.
Definition:
In relation to the base unit of [area] => (square meters), 1 Square Miles (mi2, sq mi) is equal to 2589988.11 square-meters, while 1 Dunams (dunam) = 1000 square-meters.
1 Square Miles to common area units
1 mi2, sq mi = 2589988.11 square meters (m2, sq m)
1 mi2, sq mi = 25899881100 square centimeters (cm2, sq cm)
1 mi2, sq mi = 2.58998811 square kilometers (km2, sq km)
1 mi2, sq mi = 27878411.999612 square feet (ft2, sq ft)
1 mi2, sq mi = 4014489599.4792 square inches (in2, sq in)
1 mi2, sq mi = 3097599.9995981 square yards (yd2, sq yd)
1 mi2, sq mi = 1 square miles (mi2, sq mi)
1 mi2, sq mi = 4.0144895994792E+15 square mils (sq mil)
1 mi2, sq mi = 258.998811 hectares (ha)
1 mi2, sq mi = 639.99943412918 acres (ac)
Square Milesto Dunams (table conversion)
1 mi2, sq mi = 2589.98811 dunam
2 mi2, sq mi = 5179.97622 dunam
3 mi2, sq mi = 7769.96433 dunam
4 mi2, sq mi = 10359.95244 dunam
5 mi2, sq mi = 12949.94055 dunam
6 mi2, sq mi = 15539.92866 dunam
7 mi2, sq mi = 18129.91677 dunam
8 mi2, sq mi = 20719.90488 dunam
9 mi2, sq mi = 23309.89299 dunam
10 mi2, sq mi = 25899.8811 dunam
20 mi2, sq mi = 51799.7622 dunam
30 mi2, sq mi = 77699.6433 dunam
40 mi2, sq mi = 103599.5244 dunam
50 mi2, sq mi = 129499.4055 dunam
60 mi2, sq mi = 155399.2866 dunam
70 mi2, sq mi = 181299.1677 dunam
80 mi2, sq mi = 207199.0488 dunam
90 mi2, sq mi = 233098.9299 dunam
100 mi2, sq mi = 258998.811 dunam
200 mi2, sq mi = 517997.622 dunam
300 mi2, sq mi = 776996.433 dunam
400 mi2, sq mi = 1035995.244 dunam
500 mi2, sq mi = 1294994.055 dunam
600 mi2, sq mi = 1553992.866 dunam
700 mi2, sq mi = 1812991.677 dunam
800 mi2, sq mi = 2071990.488 dunam
900 mi2, sq mi = 2330989.299 dunam
1000 mi2, sq mi = 2589988.11 dunam
2000 mi2, sq mi = 5179976.22 dunam
4000 mi2, sq mi = 10359952.44 dunam
5000 mi2, sq mi = 12949940.55 dunam
7500 mi2, sq mi = 19424910.825 dunam
10000 mi2, sq mi = 25899881.1 dunam
25000 mi2, sq mi = 64749702.75 dunam
50000 mi2, sq mi = 129499405.5 dunam
100000 mi2, sq mi = 258998811 dunam
1000000 mi2, sq mi = 2589988110 dunam
1000000000 mi2, sq mi = 2589988110000 dunam
(Square Miles) to (Dunams) conversions
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# algebra
posted by on .
A cleaning company charges \$125 to visit a home and \$40 to clean each room. If the total bill was \$365, how many rooms were cleaned?
• algebra - ,
(365 - 125) / 40 = ?
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http://www.av8n.com/physics/rotating-frame.htm
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[Contents]
Motion in a Rotating Frame
John Denker
## 1 Overview
### 1.1 A First Example: The Centrifugal Field
At the park near my house, there is a small merry-go-round. A cartoon of the situation is shown in figure 1.
1) The merry-go-round is rotating. We imagine a reference frame attached to the merry-go-round, as indicated by the circular light-blue area.
2) We also imagine a non-rotating reference frame, as indicated by the rectangular green area. Loosely speaking, we imagine this reference frame to be attached to the ground. That is, we ignore the fact that the earth is slowly rotating. This is OK because the earth’s rotation is very slow compared to the rotation of the merry-go-round, in the situations of interest.
Figure 1: Rotating and Non-Rotating Coordinate Systems
The following is common knowledge among the children in the neighborhood: Suppose you are riding the merry-go-round, sitting somewhere not too near the pivot, analyzing things relative to the rotating frame. You set a baseball on the floor of the merry-go-round and hold it stationary relative to the rotating frame for a moment. When you let go, the ball does not remain stationary; it accelerates outward. The initial acceleration is directly outward, radially away from the pivot.
The technical name for this is centrifugal acceleration. Please don’t tell me there is no such thing as centrifugal acceleration. It is a readily-observable physical fact. It is characteristic of rotating reference frames.
The magnitude and direction of the centrifugal acceleration varies from place to place in the rotating frame. Therefore we call it the centrifugal field. It is zero at the pivot. The farther we go from the pivot, the stronger it gets. It is everywhere directed radially outwards from the pivot.
It must be emphasized that the centrifugal field is associated with the rotation of the reference frame. It exists in rotating frames and not otherwise. In particular when describing the motion of the baseball relative to the non-rotating reference frame, we can (and should) explain everything we see without any contribution from the centrifugal field.
A centrifugal field exists in a rotating frame and not otherwise.
The centrifugal field is not necessarily associated with the rotation of any material object. (A reference frame is something of an abstraction, and we can have a reference frame that is not tied to any material object.)
According to Einstein’s principle of equivalence, at each point in space, the gravitational field is indistinguishable from an acceleration of the reference frame. Einstein explained this in terms of the famous “elevator” argument. Please don’t try to tell me that Einstein’s argument is valid for elevators but not valid for merry-go-rounds; it’s the same physics in both cases. We conclude from this that the centrifugal field is as real as the gravitational field. Forsooth, at each point, the centrifugal field is indistinguishable from the gravitational field.
The centrifugal field is as real as the gravitational field.
As we shall see in section 1.2 and quantify in section 2, the equations of motion in the rotating frame contain additional terms, notably the Coriolis term (not just the centrifugal term).
In an introductory physics class, the teacher might decide that a formal analysis of rotating frames is “beyond the scope of the course”. That’s OK with me. Actually I recommend going one small step farther than that, and explaining that centrifugal field exists in a rotating frame and not otherwise. That allows the students to see the boundary between what they can handle and what they can’t. It would be completely unreasonable to say that rotating frames do not exist or that the centrifugal field does not exist. The physics of rotating frames has been understood for more than 180 years.
It must be emphasized that everybody, in physics class and elsewhere, uses rotating frames routinely. In particular, the usual “laboratory frame” is a rotating reference frame, co-rotating with the earth. If you measure the magnitude and direction of the gravitational acceleration g, the result is different from what it would be in a nearby non-rotating frame. It’s only a small percentage difference, but it’s huuuuge compared to the precision of the measurement, even using high-school grade equipment and procedures.
There is no need to formally analyze the rotation of the lab frame. Just patch it up by fudging the direction and magnitude of g, and leave it at that. It is still possible to observe the rotation, using e.g. a Foucault pendulum or a decent gyroscope, but the introductory class mostly refrains from measuring things (other than g) that are sensitive to the rotation.
It is purely a matter of interpretation whether we should say the centrifugal acceleration corresponds to a centrifugal force, or whether we should call it an unforced acceleration. De gustibus non disputandum. That is to say, there is no point in debating the matter, because the equation of motion is the same no matter which way we interpret this point. See reference 1 for details on this.
If somebody asks about centrifugal force, it is usually best to answer in terms of centrifugal field and centrifugal acceleration. If the answer cannot be expressed in those terms, there is probably something wrong with the question.
The tricky question of how and why you feel the effects of gravity is discussed in reference 2.
## * Contents
### 1.2 Qualitative Preview of the Coriolis Effect
We can get a qualitative appreciation for the physics involved in the Coriolis effect by considering the situation shown in figure 2. We have a puck (shown in red) that is initially at rest in the rotating reference frame. It is held in place by a string (not shown) running from the pivot to the puck. The string serves to counter the centrifugal force.
Figure 2: Velocities That Would Conserve Angular Momentum
We now consider what happens when we pull on the string, pulling the puck toward the pivot. By conservation of angular momentum, the puck’s velocity must increase, in inverse proportion to its distance from the pivot. These velocity vectors are shown in red in figure 2. Note that these are velocity vectors as measured in the nonrotating frame! As a reminder of this point, look at the initial velocity vector, which would be zero in the rotating frame, since the puck is initially at rest in the rotating frame.
Now let us consider a slightly different scenario, the scenario we really care about, where the puck is guided by a rail in the rotating frame, so that when we pull it toward the pivot it must move directly toward the pivot.
Figure 3: Velocities Required to Match the Turntable
The velocity that the puck must have at each point as it moves along the rail is shown in figure 3. The velocity varies in direct proportion to the distance from the pivot, which is the familiar rule for the rotation of a rigid body such as our merry-go-round.
The rail must exert a force on the puck to keep it moving directly toward the pivot. There are two terms that contribute to this force (plus a centrifugal force that we are ignoring for the moment).
1. The rail must undo the acceleration that would have happened according to figure 2.
2. The rail must do the deceleration that should happen according to figure 3.
3. We are ignoring the centrifugal force for the moment. It is in the radial direction.
Each of the two Coriolis terms is a product, proportional to how fast the frame is rotating multiplied by how fast the particle is moving relative to the rotating frame (i.e. how fast we pull on the string).
In the non-rotating frame, this force requires no special explanation; it is just the force required to make the puck follow a non-straight path as it spirals inward. In contrast, in the rotating frame, the puck is following a straight path, and we wish to preserve the spirit of the first law of motion, namely that a particle should move in a straight line if and only if it is subjected to no net force.1 So we say that there is a Coriolis force (to the right in this case) and the mechanical force exerted by the rail serves to counter the Coriolis force.
Beware that it is extremely common to find wrong explanations of the Coriolis effect. Most of the hand-wavy explanations are off by a factor of two. In effect they blow off the factor of 2 in front of the Coriolis term in equation 12 and all similar equations.
We can guess where this mistake comes from by comparing figure 2 with figure 3. The former says that as the puck is pulled inwards, the velocity “would” increase if the puck were not constrained by the rail, while the latter says that the velocity “should” decrease if it is to match the rigid-body rotation. So there are two contributions to the physics ... and they turn out to be numerically equal. Anybody who notices half the physics but overlooks the other half will get the wrong answer by a factor of 2.
Another moderately common mistake is to suggest that the force exerted by the rail is the Coriolis force. This is backwards. In our scenario the Coriolis force pushes to the right; the rail counters the Coriolis force by pushing to the left. The Coriolis effect exists whenever an object is moving relative to the rotating frame (whether or not it is opposed by a rail or anything else).
Let us now try to get a qualitative understanding of the Coriolis effect in the case where the puck is moving tangentially, as shown in figure 4. As before, the puck is shown in red, and the velocity vectors are measured in the nonrotating frame. A string holds the puck at a constant distance from the pivot. The puck is moving relative to the rotating frame, swinging like a tetherball, moving in the locally tangential direction.
• Relative to the rotating frame, at each point, the velocity of the puck is given by the blue vector in the diagram.
• Relative to the nonrotating frame, the velocity of the puck is given by the sum of the red vector and the blue vector. The red vector accounts for the rotation of the frame itself.
Figure 4: Tangential Motion
The figure shows three snapshots of the puck at three successive times. As the puck moves from place to place relative to the rotating frame, not only does its position change, but its velocity changes also. The string must apply a leftward force to the puck to impart a change in momentum, to turn the velocity vector to the left. Some of the tension in the string is needed to oppose the centrifugal force – which is independent of the blue velocity – but there is another contribution which is proportional to the blue velocity, which is needed to oppose the Coriolis force.
When we analyze the situation in the nonrotating frame, we again find two Coriolis terms (plus a centrifugal term), each of which is a product:
1. We need a force to explain the moment-to-moment rotation of the blue vector. This force will be proportional to the magnitude of the blue vector multiplied by the rate of rotation of the frame.
2. We must also account for the fact that motion along the blue vector brings the puck to a new location on the merry-go-round. At this new location, the local velocity of the rotating frame is different. That is, the new red vector is different. Therefore if we want the puck to maintain constant motion relative to the rotating frame, it cannot keep its velocity and momentum the way a free particle would; we need to rotate the blue vector to match the direction of the new local red vector. Again the force required to do this is proportional to the magnitude of the blue vector multiplied by the rate of rotation of the frame.
3. We also need to explain the rotation of the red contribution to the puck’s velocity. This is proportional to the magnitude of the red vector multiplied by the rate of rotation of the frame. It is independent of the blue vector, i.e. independent of how fast the puck is moving relative to the rotating frame. This is the familiar centrifugal force.
To summarize this section: Suppose the rotating frame is rotating counterclockwise. We have an observer moving relative to the rotating frame, facing forward along the direction of motion. We have shown that there will be a Coriolis force directed to the observer’s right. We have explained this in the case of northward motion and in the case of eastward motion. From there it is only a small leap to hypothesize a similar effect for all other directions of motion.
If you do the math – as in section 2 – you discover that in all generality, whenever an object has a velocity relative to some rotating frame, the Coriolis force is perpendicular to that velocity. You can find the direction of the Coriolis force by projecting the velocity onto the plane of rotation, then rotating the projection 90 degrees counter to the rotation of the frame.
## 2 The Equations of Motion
We now start building a quantitative understanding of motion in a rotating frame. The set of equations of motion we use in such a frame is different from the set of equations we would use in a non-rotating frame. The two sets of equations are consistent with each other, and indeed each can be derived from the other, as we now demonstrate.
### 2.1 Position
Suppose we have a pointlike object whose position is described by the abstract vector Q. This vector has components QJ relative to the nonrotating laboratory frame, and components QM relative to the rotating frame.
Let’s see if we can express QM in terms of QJ (and in terms of the other “givens”). We can get a good start just by turning the crank.
Let’s start with plain old position. To connect the position in one frame with the position in the other frame, we write
QJ = R{θ}(QM) (1)
where R{θ}(⋯) is the rotation operator that rotates something by an angle θ in our chosen plane of rotation.
We need to operate on this rotation operator. This includes being able to differentiate with respect to θ. This is easy to do if we know how to multiply vectors, as briefly defined in section 5.3 and more fully explained in reference 4. That allows us to express the rotation operator in terms of rotors, as briefly defined in section 5.4 and more fully explained in reference 5. Using that representation, we have:
QJ = r∼(θ/2) QM r(θ/2) = [cos(θ/2) + γ2 γ1 sin(θ/2)] QM [cos(θ/2) + γ1 γ2 sin(θ/2)]
(2)
where γ1 and γ2 are any two orthonormal vectors in the plane of rotation, and θ is the angle of rotation of the rotating frame relative to the laboratory frame, and r() is a rotor that generates a rotation in this plane, as discussed in section 5.4.
### 2.2 First Derivative
Now we can calculate the velocity, based on the position we just calculated. To do that, we differentiate both sides of equation 1. For simplicity we assume the plane of rotation is unchanging.
QJ• = (θ/2)• [−sin(θ/2) + γ2 γ1 cos(θ/2)] QM [cos(θ/2) + γ1 γ2 sin(θ/2)] + [cos(θ/2) + γ2 γ1 sin(θ/2)] QM [−sin(θ/2) + γ1 γ2 cos(θ/2)](θ/2)• + [cos(θ/2) + γ2 γ1 sin(θ/2)] QM• [cos(θ/2) + γ1 γ2 sin(θ/2)]
(3)
where the “dot” denotes differentiation with respect to time.
This can be put into a nicer form if we factor out γ1 γ2 in a couple of places:
QJ• = (θ/2)• γ2 γ1 [cos(θ/2) + γ2 γ1 sin(θ/2)] QM [cos(θ/2) + γ1 γ2 sin(θ/2)] + [cos(θ/2) + γ2 γ1 sin(θ/2)] QM [cos(θ/2) + γ1 γ2 sin(θ/2)] γ1 γ2 (θ/2)• + [cos(θ/2) + γ2 γ1 sin(θ/2)] QM• [cos(θ/2) + γ1 γ2 sin(θ/2)] = (θ/2)• γ2 γ1 r∼(θ/2) QM r(θ/2) + r∼(θ/2) QM r(θ/2) γ1 γ2 (θ/2)• + r∼(θ/2) QM• r(θ/2)
(4)
The first two terms on the RHS of equation 4 look tantalizingly similar, and indeed using equation 21 we can combine them to yield:
QJ• = θ• γ2 γ1 r∼(θ/2) ρM r(θ/2) + r∼(θ/2) QM• r(θ/2)
(5)
where ρM is defined to be the projection of QM onto the plane of rotation, that is:
ρM := (QM − γ2 γ1 QM γ1 γ2) / 2 (6)
Equation 5 can be made slightly more symmetric-looking by use of equation 24:
QJ• = θ• r∼(90∘/2) r∼(θ/2) ρM r(θ/2) r(90∘/2) tangential + r∼(θ/2) QM• r(θ/2) ordinary
(7)
Or, switching back to the more condensed “rotation operator” notation,
QJ• = θ• R{θ+90∘}(ρM) tangential + R{θ}(QM•) ordinary
(8)
The second term on the RHS of equation 8 can be interpreted as a rotated version of the “ordinary” velocity, namely the time-derivative of position, in close analogy to the LHS of the equation. Meanwhile, the first term on the RHS is an additional contribution, a “tangential velocity” term due solely to the rotation of the frame. It lies in the plane of rotation, and is 90 degrees “ahead” of ρM.
### 2.3 Second Derivative
Let’s continue to turn the crank. To form the second derivative, we differentiate both sides of equation 7.
QJ•• = θ•• r∼(90∘/2) r∼(θ/2) ρM r(θ/2) r(90∘/2) + θ• r∼(90∘/2) (θ/2)• γ2 γ1 r∼(θ/2) ρM r(θ/2) r(90∘/2) + θ• r∼(90∘/2) r∼(θ/2) ρM r(θ/2) γ1 γ2 (θ/2)• r(90∘/2) + θ• r∼(90∘/2) r∼(θ/2) ρM• r(θ/2) r(90∘/2) + (θ/2)• γ2 γ1 r∼(θ/2) QM• r(θ/2) + r∼(θ/2) QM• r(θ/2) γ1 γ2 (θ/2)• + r∼(θ/2) QM•• r(θ/2)
(9)
which can be simplified by collecting like terms, i.e. combining the 2nd and 3rd lines, and combining the 5th and 6th lines:
QJ•• = θ•• r∼(90∘/2) r∼(θ/2) ρM r(θ/2) r(90∘/2) + (θ•)2 r∼(180∘/2) r∼(θ/2) ρM r(θ/2) r(180∘/2) + θ• r∼(90∘/2) r∼(θ/2) ρM• r(θ/2) r(90∘/2) + θ• r∼(90∘/2) r∼(θ/2) ρM• r(θ/2) r(90∘/2) + r∼(θ/2) QM•• r(θ/2)
(10)
and that can be further simplified by some simple algebra:
QJ•• = θ•• r∼(90∘/2) r∼(θ/2) ρM r(θ/2) r(90∘/2) swat − (θ•)2 r∼(θ/2) ρM r(θ/2) pre-centrifugal + 2 θ• r∼(90∘/2) r∼(θ/2) ρM• r(θ/2) r(90∘/2) pre-Coriolis + r∼(θ/2) QM•• r(θ/2) ordinary
(11)
and put into more compact form:
QJ•• = ω• R{θ+90∘}(ρM) swat − ω2 R{θ}(ρM) pre-centrifugal + 2 ω R{θ+90∘}(ρM•) pre-Coriolis + R{θ}(QM••) ordinary
(12)
where we have introduced ω := θ to denote the rate of rotation.
The interpretation of the various terms on the RHS of equation 12 will be discussed in a moment, in connection with equation 14, but we can already begin to see what’s going to happen: The last term is a duly rotated version of the “ordinary” acceleration. The third term is destined to give rise to the Coriolis effect. The second term is destined to give rise to the centrifugal effect. The first term accounts for changes in the rotation rate.
We call these terms pre-centrifugal and pre-Coriolis because we need to uphold the principle that the centrifugal field and the Coriolis effect exist in the rotating frame and not otherwise. Since equation 12 is nominally an expression for the acceleration in the nonrotating frame, as specified by its LHS, these are not really centrifugal or Coriolis terms. They don’t even have the correct signs. The correct terms can be found in equation 14.
### 2.4 Switching to the Rotating Frame
Equation 12 has the elegant property of having the QJ-related term on one side, and all the QM-related terms on the other side. This makes clear the role of these equations; they are simply the mapping from one coordinate system to another.
We now sacrifice elegance by rearranging the equation to isolate QM on one side:
QM•• = − ω• r∼(90∘/2) ρM r(90∘/2) swat + ω2 ρM centrifugal − 2 ω r∼(90∘/2) ρM• r(90∘/2) Coriolis + r∼(−θ/2) QJ•• r(−θ/2) ordinary
(13)
or, in more compact notation:
QM•• = − ω• R{90∘}(ρM) swat + ω2 ρM centrifugal − 2 ω R{90∘}(ρM•) Coriolis + R{−θ}(QJ••) ordinary
(14)
So, finally, we have an expression for the acceleration in the rotating frame.
The last term on the RHS is a duly rotated version of the “ordinary” acceleration, i.e. the second derivative of the position, in close analogy to the LHS of the equation. It is independent of position and independent of velocity.
The third term describes the Coriolis effect. It is proportional to ρM• (the velocity in the plane of the rotation) and also proportional to ω (the rate of rotation). It is independent of position. It lies in the plane of rotation, and is perpendicular to the velocity.
The second term describes the centrifugal field. It is proportional to the square of the rotation rate, and also proportional to ρM (the distance from the axis of rotation). It is independent of velocity.
The first term describes the effect of unsteady rotation. Any observer attached to the rotating frame will be “swatted” as the rotation rate increases or decreases. It is proportional to the distance from the axis of rotation, and proportional to the rate of change of the rotation rate of the frame. It does not depend on the puck’s velocity relative to the rotating frame (unlike the Coriolis effect). It vanishes in the case of steady rotation.
Remark: The word “centrifugal” means, literally, fleeing away from the center. The rearrangement leading to equation 14 changed the sign of the ω2 ρM term, so that now it more clearly deserves the name “centrifugal”, in the sense that the centrifugal term ω2 ρM is directed radially outward, making an outward contribution to the acceleration QM. Similarly the Coriolis term in equation 14 has the correct sign. Since −R{90}(⋯) = R{−90}(⋯), we can say that the Coriolis force is 90 degrees behind the velocity vector, where the notions of “ahead” and “behind” are defined by the direction of rotation of the rotating frame.
Note that ρM is directed radially outward, by definition, by construction.
There are various possible reasons why you might want to use equation 14. For example, suppose you already know QJ somehow, and you wish to solve for QM by timestepping the equation of motion. Equation 14 will tell you the current value of QM, which you can integrate to get the next value of QM, and so forth.
In particular, we might know QJ from the second law of motion, equation 17. In its usual form, this law is not valid in rotating frames, but for present purposes that’s no problem. We simply apply the law in the nonrotating frame. This gives us the acceleration-components QJ in terms of the force-components FJ.
Plugging in, we immediately obtain:
QM•• = − ω• r∼(90∘/2) ρM r(90∘/2) + ω2 ρM − 2 ω r∼(90∘/2) ρM• r(90∘/2) + r∼(−θ/2) (1/m)FJ r(−θ/2)
(15)
At this point we may, optionally, apply the distributive rule in reverse to pull a factor of (1/m) out of every term on the RHS. Doing so means that each of the terms within braces now has dimensions of force:
QM•• = (1/m) { − m ω• r∼(90∘/2) ρM r(90∘/2) (pseudo-force) + m ω2 ρM (pseudo-force) − 2 m ω r∼(90∘/2) ρM• r(90∘/2) (pseudo-force) + r∼(−θ/2) FJ r(−θ/2) } (real force)
(16)
### 2.5 Discussion
Here’s an amusing mnemonic: If we restrict attention to uniform rotation, so that the swat term vanishes, the remaining three terms on the RHS of equation 14 has the same structure as a binomial expansion, namely (a+b)2, namely a2 + 2ab + b2. In this case a is the frequency omega, while b is, roughly speaking, the time derivative operator, denoted by a dot in the equations. The centrifugal term has two copies of ω, the ordinary F=ma term has two copies of the time derivative, and the cross term, i.e. the Coriolis term, has one copy of omega, one copy of the derivative, and a factor of two out front. Don’t forget the factor of two.
Equation 12 and equation 14 are not the almost general expressions, because we have assumed that the plane of rotation is not varying. In a two-dimensional world, this is obviously OK, but in three dimensions it is perfectly possible to be rotating around an axis while the orientation of the axis is changing. I’ve never actually seen the general expression. I suspect it would be quite a bit messier than equation 12.
Textbooks commonly present an expression that is even less general than equation 12, because they assume steady rotation, so that the ω term (the swat term) does not appear.
The key equations – equation 1, equation 8, and equation 12 – allow us to construct a complete description of the position, velocity, and acceleration of the object in the rotating frame, in terms of the position, velocity, and acceleration in the lab frame. Of course they also depend on the plane of rotation and rate of rotation, and other “givens” of the situation.
It is worth remarking, and perhaps worth emphasizing, that (except for the optional and digressive equation 16) these are not force equations. There are no forces appearing in any of these equations, and no masses. These equations are in no way dependent on the first, second, or third law of motion. There is a Coriolis term in equation 12, but it cannot be called a Coriolis force. Similarly the centrifugal term in equation 12 cannot be called a centrifugal force.
## 3 Conservation of Momentum
Momentum is conserved.
Momentum is conserved, no matter whether it is being observed from the laboratory frame, observed from a rotating frame, or not observed at all.
In the laboratory frame, momentum is equal to mass times velocity (i.e. velocity relative to the lab frame).
In the rotating frame, momentum is not equal to mass times velocity (i.e. velocity relative to the rotating frame). This is obvious if you consider a free particle moving slowly through the rotating frame. Its momentum will change direction completely, again and again, as the rotating frame goes around and around.
In the rotating frame, momentum is conserved, but MV is not (in general) conserved. The general expression for momentum is more complicated than MV.
## 4 Rotating Frames in the Real World
### 4.1 Gravity and Centrifugity
There is a profound analogy between gravitational acceleration and centrifugal acceleration. Both correspond to an acceleration of the local reference frame. If you are far enough away from the source of the gravitational field, and you look at a small-enough region, you can approximate the gravitational field as a uniform acceleration. Similarly, if you are far enough away from the center of rotation, and you look at a small-enough region, you can approximate the centrifugal field as a uniform acceleration.
As discussed in reference 6, the thing we think of as “gravity” in ordinary real-world situations includes centrifugal contributions. These contributions are smallish but definitely nontrivial at temperate latitudes. An architect who aligned his notion of “vertical” with the center of the earth (without regard for the centrifugal field) would be ridiculed.
### 4.2 Other Examples
In the real world, rotating frames are sometimes required ... not required by the laws of physics, but by man-made laws. For starters, the law requires you to stop at a stop sign. Relativity requires us to ask, stopped in what reference frame? Answer: the reference frame comoving and corotating with the earth. How would you like to get a ticket for speeding in a school zone, for doing 700 mph relative to the ECNR (earth centered nonrotating) reference frame? Similarly, how would you like to get a ticket for doing 67,000 mph relative to the solar system barycenter?
If you are in an airplane in a high-G turn, it is entirely appropriate to use a reference frame comoving and corotating with the airplane. Asking the pilot to do otherwise would be perverse.
Even closer to home, you can feel quite significant G-forces on a playground swing set, if you swing with a large amplitude.
Rotating frames may be convenient in some situations and inconvenient in others, but the physics is the same either way. It’s not an issue of principle.
## 5 Basic Laws + Lemmas
In this section we review some basic laws and lemmas, partly to establish notation, and partly to make this document more self-contained.
### 5.1 The First and Second Laws of Motion
The second law of motion may be written as:
Fimp = m a (17)
where m is the mass of the object, a is the acceleration of the object, and Fimp is the force impressed upon the object by the surroundings. It must be emphasized that this equation is valid only in Newtonian “inertial” frames, not in rotating frames.
Usually equation 17 is usually written as simply F=ma, where it is understood that F is shorthand for Fimp. However, for present purposes it is worth being explicit about the distinction:
Fimp := force impressed upon the object by its surroundings Fexp := force expressed by the object upon its surroundings
(18)
The force Fexp is a perfectly ordinary force, it just doesn’t happen to be particularly suitable for use in equation 17.
In principle it would be straightforward to reformulate the laws of motion in terms of Fexp instead of Fimp, but this would be highly unconventional. Non-experts would be well advised to adhere to the established convention.
The first law of motion can (in retrospect) be considered a simple corollary of the second law, and need not be discussed here.
### 5.2 The Third Law of Motion
The third law of motion may be stated in various ways. Perhaps the most general and most powerful way is to say that momentum is conserved. That is, momentum obeys a strict local conservation law.
The following is another (essentially equivalent) way of expressing the third law:
Fimp = − Fexp (19)
This is consistent with conservation of momentum because force is momentum per unit time.
### 5.3 Multiplying by γ1γ2
It is nice to be able to multiply vectors. The details of this are explained in reference 4, but the basic rules of multiplication can be briefly stated here:
In three dimensions, any vector QM can be expanded in terms of γ1, γ2, and γ3, such that
QM := aγ1 + bγ2 + cγ3 (20)
Then we find that
QM γ1 γ2 = (aγ1 + bγ2 + cγ3) γ1 γ2 = (aγ2 − bγ1 + cγ3 γ1 γ2) = γ1 γ2 (−aγ1 − bγ2 + cγ3)
(21)
which tells us that γ1 γ2 commutes with any vector entirely perpendicular to the γ1 γ2 plane, but anticommutes with any vector lying entirely in the plane.
In two dimensions, the foregoing is trivial. The generalization from three dimensions to higher dimensions is straightforward.
### 5.4 Rotors
Rotations can be expressed by multiplying something on one side by a rotor [denoted r()] and on the other side by the reverse of the rotor [denoted r()]. In the case of a rotation in the γ1 γ2 plane, we have
r(θ/2) = cos(θ/2) + γ1 γ2 sin(θ/2) r∼(θ/2) = cos(θ/2) + γ2 γ1 sin(θ/2)
(22)
The rotor angle θ/2 is half of the rotation angle θ. For details, see reference 5.
### 5.5 90 Degree Rotations in the Plane
Given any vector ρ lying in the γ1 γ2 plane:
ρ := aγ1 + bγ2 (23)
we can represent a 90 degree rotation in any of several ways:
aγ2 − bγ1 = (1/2) [1 + γ2 γ1] ρ [1 + γ1 γ2] = γ2 γ1 ρ = ρ γ1 γ2
(24)
### 5.6 Frequency Modulation
We are accustomed to seeing periodic functions written in the form f(ω t). If ω is constant, that’s fine ... but otherwise it’s a trap for the unwary.
If you consider ω to be a function of time, and differentiate f(ω t) with respect to time, you get two terms:
f(ω t)• = ω f′(ω t) + ω• t f′(ω t) (25)
where the second term is unphysical. We define f() to be the derivative of f() with respect to its argument.
The better approach would be to start over, and write f(θ) instead of f(ω,t). Then the derivative is:
f(θ)• = θ• f′(ω t) = ω f′(ω t)
(26)
where we have simply defined ω to be the derivative of θ.
Note that the RHS of equation 26 is the same as the RHS of equation 25, without the nasty (ω t) term.
Running the same argument backwards, we find
θ = ω dt (27)
which is equal to (ω t) if ω is constant, but generally not otherwise.
The physical significance of this can be appreciated with the help of figure 5, which compares ∫ ω dt to ω t. In both plots, ω=1 at early times, and ω=2 at later times. You can see that changing ω in the expression ω t produces a large unwanted “leverage” effect, with a lever-arm of length t.
Figure 5: Leverage: ω t
The same conceptual issue arose in section 1.1. The same issue arises in many other contexts, including the frequency modulation that one encounters in radio electronics and other signal-processing applications.
## 6 Acknowledgments
This document incorporates many insights and suggestions from the members of the phys-l discussion list.
## 7 References
1.
John Denker, “Fictitious Force? Pseudoforce” www.av8n.com/physics/fictitious-force.htm
2.
John Denker, “Can You Feel Gravity?” www.av8n.com/physics/gravity-perception.htm
3.
John Denker, “The Laws of Motion” [Chapter of See How It Flies] ../how/htm/motion.html
4.
John Denker, “Introduction to Clifford Algebra” www.av8n.com/physics/clifford-intro.htm
5.
John Denker, “Multi-Dimensional Rotations, Including Boosts” www.av8n.com/physics/rotations.htm
6.
John Denker,
“Weight, Gravitational Force, Gravity, g, Latitude, et cetera”
www.av8n.com/physics/weight.htm
1
This is often called “Newton’s” first law of motion, even though the idea was clearly set forth by Galileo, many decades before Newton came on the scene. Newton was the first to set this law atop a numbered list, but he did not originate the law itself.
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Lemma 100.4.12. Let $f : \mathcal{X} \to \mathcal{Y}$ and $g : \mathcal{Y} \to \mathcal{Z}$ be morphisms of algebraic stacks.
1. If $g \circ f$ is DM then so is $f$.
2. If $g \circ f$ is quasi-DM then so is $f$.
3. If $g \circ f$ is separated and $\Delta _ g$ is separated, then $f$ is separated.
4. If $g \circ f$ is quasi-separated and $\Delta _ g$ is quasi-separated, then $f$ is quasi-separated.
Proof. Consider the factorization
$\mathcal{X} \to \mathcal{X} \times _\mathcal {Y} \mathcal{X} \to \mathcal{X} \times _\mathcal {Z} \mathcal{X}$
of the diagonal morphism of $g \circ f$. Both morphisms are representable by algebraic spaces, see Lemmas 100.3.3 and 100.4.7. Hence for any scheme $T$ and morphism $T \to \mathcal{X} \times _\mathcal {Y} \mathcal{X}$ we get morphisms of algebraic spaces
$A = \mathcal{X} \times _{(\mathcal{X} \times _\mathcal {Z} \mathcal{X})} T \longrightarrow B = (\mathcal{X} \times _\mathcal {Y} \mathcal{X}) \times _{(\mathcal{X} \times _\mathcal {Z} \mathcal{X})} T \longrightarrow T.$
If $g \circ f$ is DM (resp. quasi-DM), then the composition $A \to T$ is unramified (resp. locally quasi-finite). Hence (1) (resp. (2)) follows on applying Morphisms of Spaces, Lemma 66.38.11 (resp. Morphisms of Spaces, Lemma 66.27.8). This proves (1) and (2).
Proof of (4). Assume $g \circ f$ is quasi-separated and $\Delta _ g$ is quasi-separated. Consider the factorization
$\mathcal{X} \to \mathcal{X} \times _\mathcal {Y} \mathcal{X} \to \mathcal{X} \times _\mathcal {Z} \mathcal{X}$
of the diagonal morphism of $g \circ f$. Both morphisms are representable by algebraic spaces and the second one is quasi-separated, see Lemmas 100.3.3 and 100.4.7. Hence for any scheme $T$ and morphism $T \to \mathcal{X} \times _\mathcal {Y} \mathcal{X}$ we get morphisms of algebraic spaces
$A = \mathcal{X} \times _{(\mathcal{X} \times _\mathcal {Z} \mathcal{X})} T \longrightarrow B = (\mathcal{X} \times _\mathcal {Y} \mathcal{X}) \times _{(\mathcal{X} \times _\mathcal {Z} \mathcal{X})} T \longrightarrow T$
such that $B \to T$ is quasi-separated. The composition $A \to T$ is quasi-compact and quasi-separated as we have assumed that $g \circ f$ is quasi-separated. Hence $A \to B$ is quasi-separated by Morphisms of Spaces, Lemma 66.4.10. And $A \to B$ is quasi-compact by Morphisms of Spaces, Lemma 66.8.9. Thus $f$ is quasi-separated.
Proof of (3). Assume $g \circ f$ is separated and $\Delta _ g$ is separated. Consider the factorization
$\mathcal{X} \to \mathcal{X} \times _\mathcal {Y} \mathcal{X} \to \mathcal{X} \times _\mathcal {Z} \mathcal{X}$
of the diagonal morphism of $g \circ f$. Both morphisms are representable by algebraic spaces and the second one is separated, see Lemmas 100.3.3 and 100.4.7. Hence for any scheme $T$ and morphism $T \to \mathcal{X} \times _\mathcal {Y} \mathcal{X}$ we get morphisms of algebraic spaces
$A = \mathcal{X} \times _{(\mathcal{X} \times _\mathcal {Z} \mathcal{X})} T \longrightarrow B = (\mathcal{X} \times _\mathcal {Y} \mathcal{X}) \times _{(\mathcal{X} \times _\mathcal {Z} \mathcal{X})} T \longrightarrow T$
such that $B \to T$ is separated. The composition $A \to T$ is proper as we have assumed that $g \circ f$ is quasi-separated. Hence $A \to B$ is proper by Morphisms of Spaces, Lemma 66.40.6 which means that $f$ is separated. $\square$
Comment #633 by Kestutis Cesnavicius on
The proofs of (3) and (4) are mixed up.
There are also:
• 2 comment(s) on Section 100.4: Separation axioms
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# ComplexIsFractional
#### implicit object ComplexIsFractional extends ComplexIsFractional with ComplexOrdering
Implicit object providing `scala.math.Fractional` capabilities. Although complex numbers have no natural ordering, some kind of `Ordering` is required because `Numeric` extends `Ordering`. Hence, an ordering based upon the real then imaginary components is used.
Linear Supertypes
ComplexOrdering, ComplexIsFractional, Fractional[Complex], ComplexIsConflicted, Numeric[Complex], Ordering[Complex], PartialOrdering[Complex], Equiv[Complex], Serializable, Serializable, Comparator[Complex], AnyRef, Any
Ordering
1. Alphabetic
2. By inheritance
Inherited
1. ComplexIsFractional
2. ComplexOrdering
3. ComplexIsFractional
4. Fractional
5. ComplexIsConflicted
6. Numeric
7. Ordering
8. PartialOrdering
9. Equiv
10. Serializable
11. Serializable
12. Comparator
13. AnyRef
14. Any
1. Hide All
2. Show all
Visibility
1. Public
2. All
### Type Members
1. #### class FractionalOps extends scala.math.Fractional.Ops
Definition Classes
Fractional
2. #### class Ops extends AnyRef
Definition Classes
Numeric → Ordering
### Value Members
1. #### final def !=(arg0: AnyRef): Boolean
Definition Classes
AnyRef
2. #### final def !=(arg0: Any): Boolean
Definition Classes
Any
3. #### final def ##(): Int
Definition Classes
AnyRef → Any
4. #### final def ==(arg0: AnyRef): Boolean
Definition Classes
AnyRef
5. #### final def ==(arg0: Any): Boolean
Definition Classes
Any
6. #### def abs(x: Complex): Complex
Definition Classes
Numeric
7. #### final def asInstanceOf[T0]: T0
Definition Classes
Any
8. #### def clone(): AnyRef
Attributes
protected[java.lang]
Definition Classes
AnyRef
Annotations
@throws( ... )
9. #### def compare(a: Complex, b: Complex): Int
Definition Classes
ComplexOrdering → Ordering → Comparator
10. #### def div(x: Complex, y: Complex): Complex
Definition Classes
ComplexIsFractional → Fractional
11. #### final def eq(arg0: AnyRef): Boolean
Definition Classes
AnyRef
12. #### def equals(arg0: Any): Boolean
Definition Classes
AnyRef → Any
13. #### def equiv(x: Complex, y: Complex): Boolean
Definition Classes
Ordering → PartialOrdering → Equiv
14. #### def finalize(): Unit
Attributes
protected[java.lang]
Definition Classes
AnyRef
Annotations
@throws( classOf[java.lang.Throwable] )
15. #### def fromInt(x: Int): Complex
Definition Classes
ComplexIsConflicted → Numeric
16. #### final def getClass(): Class[_]
Definition Classes
AnyRef → Any
17. #### def gt(x: Complex, y: Complex): Boolean
Definition Classes
Ordering → PartialOrdering
18. #### def gteq(x: Complex, y: Complex): Boolean
Definition Classes
Ordering → PartialOrdering
19. #### def hashCode(): Int
Definition Classes
AnyRef → Any
20. #### final def isInstanceOf[T0]: Boolean
Definition Classes
Any
21. #### def lt(x: Complex, y: Complex): Boolean
Definition Classes
Ordering → PartialOrdering
22. #### def lteq(x: Complex, y: Complex): Boolean
Definition Classes
Ordering → PartialOrdering
23. #### def max(x: Complex, y: Complex): Complex
Definition Classes
Ordering
24. #### def min(x: Complex, y: Complex): Complex
Definition Classes
Ordering
25. #### def minus(x: Complex, y: Complex): Complex
Definition Classes
ComplexIsConflicted → Numeric
26. #### implicit def mkNumericOps(lhs: Complex): FractionalOps
Definition Classes
Fractional → Numeric
27. #### implicit def mkOrderingOps(lhs: Complex): ComplexIsFractional.Ops
Definition Classes
Ordering
28. #### final def ne(arg0: AnyRef): Boolean
Definition Classes
AnyRef
29. #### def negate(x: Complex): Complex
Definition Classes
ComplexIsConflicted → Numeric
30. #### final def notify(): Unit
Definition Classes
AnyRef
31. #### final def notifyAll(): Unit
Definition Classes
AnyRef
32. #### def on[U](f: (U) ⇒ Complex): Ordering[U]
Definition Classes
Ordering
33. #### def one: Complex
Definition Classes
Numeric
34. #### def plus(x: Complex, y: Complex): Complex
Definition Classes
ComplexIsConflicted → Numeric
35. #### def reverse: Ordering[Complex]
Definition Classes
Ordering → PartialOrdering
36. #### def signum(x: Complex): Int
Definition Classes
Numeric
37. #### final def synchronized[T0](arg0: ⇒ T0): T0
Definition Classes
AnyRef
38. #### def times(x: Complex, y: Complex): Complex
Definition Classes
ComplexIsConflicted → Numeric
39. #### def toDouble(x: Complex): Double
Definition Classes
ComplexIsConflicted → Numeric
40. #### def toFloat(x: Complex): Float
Definition Classes
ComplexIsConflicted → Numeric
41. #### def toInt(x: Complex): Int
Definition Classes
ComplexIsConflicted → Numeric
42. #### def toLong(x: Complex): Long
Definition Classes
ComplexIsConflicted → Numeric
43. #### def toString(): String
Definition Classes
AnyRef → Any
44. #### def tryCompare(x: Complex, y: Complex): Some[Int]
Definition Classes
Ordering → PartialOrdering
45. #### final def wait(): Unit
Definition Classes
AnyRef
Annotations
@throws( ... )
46. #### final def wait(arg0: Long, arg1: Int): Unit
Definition Classes
AnyRef
Annotations
@throws( ... )
47. #### final def wait(arg0: Long): Unit
Definition Classes
AnyRef
Annotations
@throws( ... )
48. #### def zero: Complex
Definition Classes
Numeric
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https://www.education.com/common-core/CCSS.MATH.CONTENT.1.OA.B.3/worksheets/
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1.OA.B.3 Worksheets
CCSS.MATH.CONTENT.1.OA.B.3
:
"Apply properties of operations as strategies to add and subtract."
These worksheets can help students practice this Common Core State Standards skill.
Worksheets
Fact Families
Worksheet
Fact Families
Looking for a worksheet that practices addition and subtraction skills? This printable works with fact families and common denominators.
Math
Worksheet
Math Fact: Relationship Between Addition & Subtraction
Worksheet
Math Fact: Relationship Between Addition & Subtraction
Help children make the connection between addition and subtraction with this appealing worksheet!
Math
Worksheet
Subtraction Fact Families
Worksheet
Subtraction Fact Families
This solve-and-check worksheet encourages kids to solve subtraction problems, and then check their answers with addition.
Math
Worksheet
Complete the Inverse Operation
Worksheet
Complete the Inverse Operation
Your child will develop the skills necessary to recognize patterns by completing inverse operations as well make a connection between addition and subtraction.
Math
Worksheet
Inverse Operations Practice
Worksheet
Inverse Operations Practice
Math
Worksheet
Worksheet
Math
Worksheet
Practice with Ten Frames
Worksheet
Practice with Ten Frames
Each ten frame has ten boxes. Some boxes have red chips and others don't. If the top half of the ten frame is filled with red chips, how many are left? 10-5=5.
Math
Worksheet
Practice Adding and Subtracting with Ten Frames
Worksheet
Practice Adding and Subtracting with Ten Frames
Help your little one practice her adding subtracting skills with this game of ten frames! Each ten frame has a set of ten boxes.
Math
Worksheet
Worksheet
Teach your students to solve word problems with three addends. Plenty of space is provided for students to show their thinking!
Math
Worksheet
Number Roll Game
Worksheet
Number Roll Game
Looking for a fun way to engage students in addition practice? Your young mathematicians will love this entertaining game that involves rolling three dice and adding the results in any order to find the sum!
Math
Worksheet
Three Part Cube Trains
Worksheet
Three Part Cube Trains
Use the worksheet to help students visualize addition number sentences with three addends. Kids build 10 cube trains using three different color cubes, color their trains, and then write out the equations.
Math
Worksheet
Double-9 Domino Dash
Worksheet
Double-9 Domino Dash
Students will practice counting, adding, and comparing numbers with this engaging game. Each player chooses a domino, and the player with the greater number of dots takes both dominoes!
Math
Worksheet
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https://www.numbersaplenty.com/101622509
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Search a number
101622509 is a prime number
BaseRepresentation
bin1100000111010…
…10001011101101
321002012221212122
412003222023231
5202003410014
614030042325
72342530352
oct603521355
9232187778
10101622509
11523aa54a
122a0493a5
1318091181
14d6d4629
158dc558e
hex60ea2ed
101622509 has 2 divisors, whose sum is σ = 101622510. Its totient is φ = 101622508.
The previous prime is 101622503. The next prime is 101622517. The reversal of 101622509 is 905226101.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 100060009 + 1562500 = 10003^2 + 1250^2 .
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-101622509 is a prime.
It is a Chen prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (101622503) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 50811254 + 50811255.
It is an arithmetic number, because the mean of its divisors is an integer number (50811255).
Almost surely, 2101622509 is an apocalyptic number.
It is an amenable number.
101622509 is a deficient number, since it is larger than the sum of its proper divisors (1).
101622509 is an equidigital number, since it uses as much as digits as its factorization.
101622509 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 1080, while the sum is 26.
The square root of 101622509 is about 10080.7990258709. The cubic root of 101622509 is about 466.6557674892.
The spelling of 101622509 in words is "one hundred one million, six hundred twenty-two thousand, five hundred nine".
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http://math.stackexchange.com/questions/188195/inverse-of-binary-entropy-function-for-0-le-x-le-frac12?answertab=active
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# Inverse of binary entropy function for $0 \le x \le \frac{1}{2}$
I'm trying to find the inverse of $H_2(x) = -x \log_2 x - (1-x) \log_2 (1-x)$[1] subject to $0 \le x \le \frac{1}{2}$. This is for a computation, so an approximation is good enough.
My approach was to take the Taylor series at $x=\frac{1}{4}$, cut it off as a quadratic, then find the inverse of that. That yields
$$H_2^{-1}(x) \approx -\frac{1}{16} \, \sqrt{-96 \, x \log\left(2\right) + 9 \, \log\left(3\right)^{2} - 72 \, \log\left(3\right) + 96 \, \log\left(4\right)} + \frac{3}{16} \, \log\left(3\right) + \frac{1}{4}$$
Unfortunately, that's a pretty bad approximation and it's complex at $H_2^{-1}(1)$. What other approaches can I take?
[1] I originally forgot to write the base 2 subscript (I added that in a later edit)
-
## 3 Answers
I suggest you use the remez command in Chebfun (http://www.chebfun.org/) to compute a polynomial that approximates this function very well. Your question is closely connected to approximation theory, and this very user-friendly software should be able to give you anything you want. :)
-
A nice approximation I found in this thesis:
$\frac{x}{2 \log_2 (6/x)} \leq H_2^{-1}(x) \leq \frac{x}{ \log_2 (1/x)}$
-
A rather coarse approximation of $H_2(x)$ on the said interval is $4 \log(2) x(1-x)$. Hence a crude approximation for the inverse is: $$H_2^{-1}(y) \approx \frac{1}{2} \left(1- \sqrt{1-\frac{y}{\log(2)}}\right) = \frac12\frac{y}{\log(2) + \sqrt{\log(2)\left(\log(2)-y\right)}}$$ This initial approximation should be refined with Newton-Raphson method.
-
Thanks! How did you come up with $H_2(x) \approx 4 \log(2) x (1-x)$? – Red Aug 29 '12 at 3:17
The entropy is symmetric function of $x$, suggesting to approximate it by a polynomial in $x(1-x)$. – Sasha Aug 29 '12 at 3:19
@Red I meant symmetric around $x=1/2$. – Sasha Aug 29 '12 at 3:27
+1. A slightly better approximation would be to choose $a$ such that $$\int_0^1 \left(H_2(x) - ax(1-x) \right)^2 dx$$ is minimized. This gives us $a= \dfrac{35}{12} \approx 2.91667$ as opposed to $a = 4 \log(2) \approx 2.7726$. Sasha's $4 \log 2$ ensures that the approximation is always $\leq H_2(x)$, whereas the above approximation optimizes the choice over the entire domain. – user17762 Nov 28 '12 at 23:53
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https://courses.lumenlearning.com/cuny-hunter-collegealgebra/chapter/introduction-5/
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## Why learn about functions and function notation?
It is Joan’s birthday, and she has decided to spend it with her friend Hazel. To celebrate, Hazel and Joan are heading to the city to see a new art exhibit of paintings by artists from the Northwest School. Joan is really excited to check out some pieces by Mark Tobey, her favorite artist.
As Joan and Hazel are driving to the city, Joan wonders how many other people in the world share her birthday. Joan remembers following some click-bait on the internet about birthdays and reading that September 16 is the most common birthday in the U.S., while December 25 and February 29 are the least common. [1].
She considers how interesting it is that each person only has one unique birthday, while many people might share the same day of birth, and doodles the following drawing on the back of an envelope.
Without realizing it, Joan has discovered the definition of a mathematical function. Think of each individual person on the earth as a variable, p. Now imagine that all the birthdays are a function, B. For each individual person you place in the Birthday function, you will get out one unique birthday for that person. If you were to go backward, though, you could have many people with the same birthday.
In this module we will introduce the definition of a function and the formal mathematical notation that is used to express functions. You will see that there are many mathematical relationships that you are already familiar with that fit the definition of a function. Probably the most familiar of these relationships are linear equations for straight lines.
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Welcome to the Monastery PerlMonks
### Re^2: Quick Regex Question
on Oct 08, 2012 at 14:31 UTC ( #997824=note: print w/replies, xml ) Need Help??
in reply to Re: Quick Regex Question
Thank you for the quick reply. I must be doing something wrong as now it matches 7,8,9,10, etc numbers. I am looking to only match exactly 7?
Replies are listed 'Best First'.
Re^3: Quick Regex Question
by Athanasius (Chancellor) on Oct 08, 2012 at 14:37 UTC
OK, change the regex to:
```/INC\d{7}(?:[^\d]|\$)/
which should match only exactly 7 digits. (INC followed by 7 digits followed by either a non-digit or the end of the string.)
Update: Better solutions:
```/INC\d{7}(?!\d)/
using a zero-width negative look-ahead assertion; or:
```/INC\d{7}(?:(?=\D)|\$)/ # Corrected: See post by AnomalousMonk, belo
+w.
using a zero-width positive look-ahead assertion together with \D as per GrandFather’s suggestion. See Extended Patterns.
Athanasius <°(((><contra mundum
You can use \D for "not a digit". In like fashion you can use \W and \S to complement the common word and space matches.
True laziness is hard work
```... using a zero-width negative look-ahead assertion; or ... using a z
+ero-width positive look-ahead assertion together with \D ...
Note that (?!\d) and (?=\D) are not complementaryequivalent. The (?!\d) assertion is satisfied by having any non-digit follow it or by nothing, i.e., by the end of the string, nothing being not-a-digit. The (?=\D) assertion must be followed by a non-digit character. And similarly with the (?<!\d) and (?<=\D) look-behinds.
```>perl -wMstrict -le
"my \$s = 'xyzINC1234567';
;;
print 'negative assertion true' if \$s =~ /INC\d{7}(?!\d)/;
print 'positive assertion true' if \$s =~ /INC\d{7}(?=\D)/;
"
negative assertion true
In this behavior, (?!\d) and, in general, (?!character-or-character-class) and its negative look-behind cousin are similar to (but not exactly the same as) the \b assertion, which may be true at the beginning or end of a string.
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https://wowlazymacros.com/t/please-explain-cast-vs-castsequence-and-reset-time-gs/1496
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# Please explain: Cast vs Castsequence and Reset Time GS
Let’s start with my question about Castsequence and Reset Time: I assume that reset time is used when a spell has a duration of significant length (e.g., 15sec, 20sec, etc.) and you don’t want to recast it before that time is almost up. Theoretically, the spells that follow the initial spell in the castsequence are those that make up your typical rotation AND it seems, are those that you’d cast before your initial spell is recast AND the sum of whose CDs and GCDs is about the duration of the initial spell. Is all this correct so far?
Does that mean that if I don’t fill the cast sequence with enough spells to equal the reset time, that no spells in that sequence will proc until the reset? I assume the reset timer starts when the initial spell is activated. How’m I doing?
OK: Then how do the spells you place within the castsequence vary from those you place on separate lines within the macro either before of after the castsequence? How do macros that contain more than one castsequences work? I’ve seen macros where spells are contained inside the castsequence and then again outside on separate cast lines.
Sorry, that is a common misconception. Reset time is based on the last time the macro tried to activate that line. Further the way GS works reset target doesn’t work correctly so we drop in reset time but you must stop pressing the macro button for the duration of the time for it to have any effect. So, if reset time actually worked, correctly inside of GS what would happen is that the first time that cast sequence was activated it would start a time and when it reached that time limit it would reset the pointer back to the first position, this is not the behaviour in GS. Instead it resets after you stop pressing the macro for that length of time
[quote]Does that mean that if I don’t fill the cast sequence with enough spells to equal the reset time, that no spells in that sequence will proc until the reset? I assume the reset timer starts when the initial spell is activated. How’m I doing?[/quote] See above
[quote]OK: Then how do the spells you place within the castsequence vary from those you place on separate lines within the macro either before of after the castsequence? How do macros that contain more than one castsequences work? I’ve seen macros where spells are contained inside the castsequence and then again outside on separate cast lines.
[/quote]
Cast sequences move forward one unit each activation attempt then move mack to the beginning or reset based on timeout or a conditional such as target switch (not functioning presently under GS)
so castsequence=a,b,c,d
castsequence=x,y,z
cast sequence=p,d,q
Would result in attempts to cast a,x,p,b,y,d,c,z,q,d,x,p,a,y,d,b,z,q,c,x,p and so on
It would be nice if blizzard allowed us to cast a every 15s, but it does not. instead if you want to do that you create a cast sequence as long as your longest duration spell and create a timeline and generate a rotation accordingly.
The beauty of GS is that it allows a fall through of casts/castsequence but the pain of that same system is that it allows a fall through of cast/castsequence so where before a macro without a reset conditional might hang on a spell with a long CD because the spell was not ready, now the macro continues.
Thanks, Robert, that was an excellent tutorial and response to my question. I appreciate the time it took to clearly explain. So, if I parse each line in the following GS which I’m trying to use, I see that the macro will run through all of the spells in single item castsequences (i.e., Ice Barrier, Ice Ward, Comet Storm) and then it will ignore the reset=combat/target item, but proc the spells that follow it (i.e., Frostbolt, Frostbolt, etc.). Continuing beyond that huge castsequence, it will proc Frozen Orb, Ice Nova and then into the last castsequence before the Post-macro casting Deep Freeze, Ice Lance, etc. It will go into the Post Macro and attempt the 4 casts there. Right so far? Now here’s where it’s tricky. It now attempts to go back to the beginning, but since I’ve been continually mashing the key, it’ll skip all the lines with “reset=[time]” in them … right? Will it skip the line with reset=combat/target or not? If it skips it, it will next cast Frozen Orb and Ice Nova and move into the Post macro. So if a fight lasts more than 30 seconds, then all of the casts following reset= will never proc? And further, if the next target is less than 20 seconds after the last one dies, then again most of this GS will not proc? If most of what I’ve said is true, how would you re-write this GS to work?
Sequences[‘Frosty’] = {
PreMacro = [[
]],
[[/castsequence reset=25, Ice Barrier]],d
[[/castsequence reset=20, Ice Ward]],
[[/castsequence reset=30, Comet Storm]],
[[/castsequence reset=combat/target Frostbolt,Frostbolt,Frostbolt,Ice Lance,Ice Lance,Frostbolt,Frostbolt,Frostbolt,Frostbolt,Ice Lance,Ice Lance,Frostbolt,Frostbolt,Frostbolt,Frostbolt,Ice Lance,Ice Lance,Frostbolt,Frostbolt,Frostbolt,Ice Lance,Ice Lance]],
[[/cast Frozen Orb]],
[[/cast Ice Nova]],
[[/castsequence reset=30 Deep Freeze,Ice Lance,Ice Lance]],
PostMacro = [[
/startattack
/use [combat]Ice Floes
/use [combat]Icy Veins
/use [combat]Mirror Image
/use [combat]Frozen Orb
]],
}
GS does not honor the reset conditional, so if the command is ‘/castsequence reset=not a better love story than twilight, Frostbolt, Frostbolt Frostbolt, Ice lance, Ice lance’, then you might as well have written ‘/castsequence Frostbolt, Frostbolt Frostbolt, Ice lance, Ice lance’,
Well, I would lose the ‘d’ on the end of line #5 and use " instead of [[ ]] but the 2nd part is stylistic.
In all honesty I do not play my mage enough to know. If the CD on the first 3 is close to the reset time, there is no reason to change them as they would only cast when they are off CD. OK so I looked on wowhead and it says: Comet Storm is Instant with a 30sec cooldown so you just want it to fire every time it is off CD same with the other 2. I would simply put them in the premacro area or as priorities 1,2 & 3 as “/cast Ice Barrier” etc. Are they on the GCD? if not definitely put them in premacro so the script tries to pop them at every iteration.
It appears as though you want to cast frostbolt 3x -> then ice lance 2x, then frost bolt 4x -> ice lance, frost bolt 4x -> ice lance 2x, frostbolt 3x -> then ice lance 2x
So you are building up charges then releasing them so that is a great use for the cast sequence part
Let me know if I am making sense, I had a doc appt today and got some weird news so I have been processing that rather than WoW today. Sorry.
Thanks, again! Hope all is well for you!
I don’t know if im writing this in the right place but i hope some one can help, been trying to find a good Affliction macro but the isnt any great ones so i decided to do my own, i kinda know what im doing but not a 100% sure. This is what I got
Sequences[‘AfflictionTest’] = {
StepFunction = [[
limit = limit or 1
if step == limit then
limit = limit % #macros + 1
step = 1
else
step = step % #macros + 1
end
]],
PreMacro = [[
/console Sound_EnableSFX 0
/petattack [@target,harm]
]],
“/castsequence reset=target Unstable Affliction,Corruption,Agony”,
“/cast [combat,nochanneling]Seed of Corruption”,
‘/cast Drain Soul’,
‘/cast [combat,nochanneling] Dark Soul: Misery’,
‘/cast [combat,nochanneling] Command Demon’,
‘/cast [mod:shift] Life Tap’,
‘/cast [mod:ctrl] Haunt’,
PostMacro = [[
/startattack
/use [combat]13
/use [combat]14
/script UIErrorsFrame:Hide();
/console Sound_EnableSFX 1
]],
}
It fires off random Drain Souls and doesn’t do my Agony,Corruption or Unstable in the right order. Also is the any way to make it so my dots will only fire off a few secs before their duration is up?
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2021.10.05 13:05
# One Day Workshop on the Arithmetic Theory of Quadratic Forms
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일정시작 2021-10-09 2021-10-09 #FF5733
One Day Workshop on the Arithmetic Theory of Quadratic Forms
Date: 10월 9일 (토) 10:00 ~ 11:50, 14:00 ~ 15:50, 16:00 ~ 17:50
Place: 129-406 (SNU)
Talk 1: 10:00 ~ 11:50
Title: Universal sums of generalized polygonal numbers
Speaker: Jangwon Ju (Ulsan University)
Abstract:
The sum of generalized polygonal numbers is said to be universal if it represents all nonnegative integers. In this talk, we introduce some arithmetic method on studying representations of sums of generalized polygonal numbers. We provide effective criteria on the universalities of sums of generalized polygonal numbers with sime small order. These might be considered as a generalization of the "15-Theorem" of Conway and Schneeberger.
Talk 2: 14:00 ~ 15:50
Title: The use of modular form theory in studying quadratic forms
Speaker: Kyoungmin Kim (Hannam University)
Abstract:
In this talk, we introduce some modular form theory used in studying the number of representations of integers by quadratic forms.
Talk 3: 16:00 ~ 17:50
Speaker: Mingyu Kim (Sungkyunkwan University)
Abstract:
For a positive integer $n$, let $\mathcal{T}(n)$ be the set of all integers greater than or equal to $n$. An integral quadratic form $f$ is called tight $\mathcal{T}(n)$-universal if the set of nonzero integers that are represented by $f$ is exactly $\mathcal{T}(n)$. The smallest possible rank over all tight $\mathcal{T}(n)$-universal quadratic forms is denoted by $t(n)$. In this talk, we prove that $t(n) \in \Omega(\log_2(n)) \cap \mathcal{O}(\sqrt{n})$. Explicit lower and upper bounds for $t(n)$ will be provided for some small integer $n$. We also consider the classification of all tight $\mathcal{T}(n)$-universal diagonal quadratic forms.
This is a joint work with Byeong-Kweon Oh.
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19 Mini-workshop on deformed W-algebras and q-characters II 2022.06.03 475
18 Mini-workshop on deformed W-algebras and q-characters I 2022.05.15 502
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16 QSMS winter school on symplectic geometry and mirror symmetry 2022.01.07 533
15 [QSMS 2022 Winter School] Workshop on Representation Theory 2022.01.07 679
14 2021 Winter School on Number Theory 2021.12.13 567
» One Day Workshop on the Arithmetic Theory of Quadratic Forms 2021.10.05 403
12 QUANTUM GROUPS AND CLUSTER ALGEBRAS 2021.10.04 825
11 CONFERENCE ON ALGEBRAIC REPRESENTATION THEORY 2021 2021.10.04 658
10 QSMS 2021 위상기하 가을 워크숍 프로그램 2021.09.28 694
9 QSMS 2021 Summer Workshop on Representation theory (Week2) 2021.08.02 512
8 QSMS 2021 Summer Workshop on Representation theory (Week1) 2021.08.02 565
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6 QSMS 20/21 Winter School on Representation Theory 2021.01.13 1061
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4 QSMS 20/21 Winter mini-school on Mirror symmetry and related topics (Part 2) 2021.01.13 1004
3 QSMS 20/21 Winter mini-school on Mirror symmetry and related topics (Part 1) 2021.01.13 14157
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Free Version
Easy
# Data Sufficiency: Assorted Practice $a=j$
GMAT-YWGJZ4
INSTRUCTIONS
This data sufficiency problem consists of a question and two statements, labeled (1) and (2), in which certain data are given. You have to decide whether the data given in the statements are sufficient for answering the question. Using the data given in the statements, plus your knowledge of mathematics and everyday facts, you must indicate whether:
• Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked.
• Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked.
• BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.
• Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.
--
If $w=1$, can we verify that $a=j$?
(1) $a=\frac{j}{|w|}$
(2) $a=\frac{|j|}{w}$
A
Statement (1) ALONE is sufficient to answer the question, but statement (2) ALONE is not.
B
Statement (2) ALONE is sufficient to answer the question, but statement (1) ALONE is not.
C
Statement (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient.
D
EITHER statement ALONE is sufficient to answer the question.
E
Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked - more relevant data is required.
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# Search by Topic
#### Resources tagged with Divisibility similar to Remainders:
Filter by: Content type:
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### There are 40 results
Broad Topics > Numbers and the Number System > Divisibility
### Eminit
##### Age 11 to 14 Challenge Level:
The number 8888...88M9999...99 is divisible by 7 and it starts with the digit 8 repeated 50 times and ends with the digit 9 repeated 50 times. What is the value of the digit M?
### Remainders
##### Age 7 to 14 Challenge Level:
I'm thinking of a number. My number is both a multiple of 5 and a multiple of 6. What could my number be?
### Ewa's Eggs
##### Age 11 to 14 Challenge Level:
I put eggs into a basket in groups of 7 and noticed that I could easily have divided them into piles of 2, 3, 4, 5 or 6 and always have one left over. How many eggs were in the basket?
### Oh! Hidden Inside?
##### Age 11 to 14 Challenge Level:
Find the number which has 8 divisors, such that the product of the divisors is 331776.
### Digat
##### Age 11 to 14 Challenge Level:
What is the value of the digit A in the sum below: [3(230 + A)]^2 = 49280A
### Gaxinta
##### Age 11 to 14 Challenge Level:
A number N is divisible by 10, 90, 98 and 882 but it is NOT divisible by 50 or 270 or 686 or 1764. It is also known that N is a factor of 9261000. What is N?
### Remainder
##### Age 11 to 14 Challenge Level:
What is the remainder when 2^2002 is divided by 7? What happens with different powers of 2?
### Divisively So
##### Age 11 to 14 Challenge Level:
How many numbers less than 1000 are NOT divisible by either: a) 2 or 5; or b) 2, 5 or 7?
### Dozens
##### Age 7 to 14 Challenge Level:
Do you know a quick way to check if a number is a multiple of two? How about three, four or six?
### AB Search
##### Age 11 to 14 Challenge Level:
The five digit number A679B, in base ten, is divisible by 72. What are the values of A and B?
### Powerful Factorial
##### Age 11 to 14 Challenge Level:
6! = 6 x 5 x 4 x 3 x 2 x 1. The highest power of 2 that divides exactly into 6! is 4 since (6!) / (2^4 ) = 45. What is the highest power of two that divides exactly into 100!?
### Factoring Factorials
##### Age 11 to 14 Challenge Level:
Find the highest power of 11 that will divide into 1000! exactly.
### Square Routes
##### Age 11 to 14 Challenge Level:
How many four digit square numbers are composed of even numerals? What four digit square numbers can be reversed and become the square of another number?
### Counting Factors
##### Age 11 to 14 Challenge Level:
Is there an efficient way to work out how many factors a large number has?
### Elevenses
##### Age 11 to 14 Challenge Level:
How many pairs of numbers can you find that add up to a multiple of 11? Do you notice anything interesting about your results?
### Flow Chart
##### Age 11 to 14 Challenge Level:
The flow chart requires two numbers, M and N. Select several values for M and try to establish what the flow chart does.
##### Age 11 to 14 Challenge Level:
Powers of numbers behave in surprising ways. Take a look at some of these and try to explain why they are true.
### Going Round in Circles
##### Age 11 to 14 Challenge Level:
Mathematicians are always looking for efficient methods for solving problems. How efficient can you be?
### Differences
##### Age 11 to 14 Challenge Level:
Can you guarantee that, for any three numbers you choose, the product of their differences will always be an even number?
### What Numbers Can We Make?
##### Age 11 to 14 Challenge Level:
Imagine we have four bags containing a large number of 1s, 4s, 7s and 10s. What numbers can we make?
### What Numbers Can We Make Now?
##### Age 11 to 14 Challenge Level:
Imagine we have four bags containing numbers from a sequence. What numbers can we make now?
### Digital Roots
##### Age 7 to 14
In this article for teachers, Bernard Bagnall describes how to find digital roots and suggests that they can be worth exploring when confronted by a sequence of numbers.
### The Remainders Game
##### Age 7 to 14 Challenge Level:
Play this game and see if you can figure out the computer's chosen number.
### What an Odd Fact(or)
##### Age 11 to 14 Challenge Level:
Can you show that 1^99 + 2^99 + 3^99 + 4^99 + 5^99 is divisible by 5?
### Three Times Seven
##### Age 11 to 14 Challenge Level:
A three digit number abc is always divisible by 7 when 2a+3b+c is divisible by 7. Why?
### Just Repeat
##### Age 11 to 14 Challenge Level:
Think of any three-digit number. Repeat the digits. The 6-digit number that you end up with is divisible by 91. Is this a coincidence?
### Skeleton
##### Age 11 to 14 Challenge Level:
Amazing as it may seem the three fives remaining in the following `skeleton' are sufficient to reconstruct the entire long division sum.
### American Billions
##### Age 11 to 14 Challenge Level:
Play the divisibility game to create numbers in which the first two digits make a number divisible by 2, the first three digits make a number divisible by 3...
##### Age 11 to 14 Challenge Level:
List any 3 numbers. It is always possible to find a subset of adjacent numbers that add up to a multiple of 3. Can you explain why and prove it?
### Book Codes
##### Age 7 to 11 Challenge Level:
Look on the back of any modern book and you will find an ISBN code. Take this code and calculate this sum in the way shown. Can you see what the answers always have in common?
### Legs Eleven
##### Age 11 to 14 Challenge Level:
Take any four digit number. Move the first digit to the end and move the rest along. Now add your two numbers. Did you get a multiple of 11?
### Repeaters
##### Age 11 to 14 Challenge Level:
Choose any 3 digits and make a 6 digit number by repeating the 3 digits in the same order (e.g. 594594). Explain why whatever digits you choose the number will always be divisible by 7, 11 and 13.
### Ben's Game
##### Age 11 to 14 Challenge Level:
Ben passed a third of his counters to Jack, Jack passed a quarter of his counters to Emma and Emma passed a fifth of her counters to Ben. After this they all had the same number of counters.
### Big Powers
##### Age 11 to 16 Challenge Level:
Three people chose this as a favourite problem. It is the sort of problem that needs thinking time - but once the connection is made it gives access to many similar ideas.
### Peaches Today, Peaches Tomorrow...
##### Age 11 to 14 Challenge Level:
Whenever a monkey has peaches, he always keeps a fraction of them each day, gives the rest away, and then eats one. How long could he make his peaches last for?
### Neighbours
##### Age 7 to 11 Challenge Level:
In a square in which the houses are evenly spaced, numbers 3 and 10 are opposite each other. What is the smallest and what is the largest possible number of houses in the square?
### Gran, How Old Are You?
##### Age 7 to 11 Challenge Level:
When Charlie asked his grandmother how old she is, he didn't get a straightforward reply! Can you work out how old she is?
### Division Rules
##### Age 7 to 11 Challenge Level:
This challenge encourages you to explore dividing a three-digit number by a single-digit number.
### Curious Number
##### Age 7 to 11 Challenge Level:
Can you order the digits from 1-3 to make a number which is divisible by 3 so when the last digit is removed it becomes a 2-figure number divisible by 2, and so on?
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# The Lambda Calculus
The revised edition contains a new chapter which provides an elegant description of the semantics. The various classes of lambda calculus models are described in a uniform manner. Some didactical improvements have been made to this edition.
Author: Hendrik Pieter Barendregt
Publisher: Elsevier Science Limited
ISBN: UCSD:31822000132415
Category: Mathematics
Page: 621
View: 481
The revised edition contains a new chapter which provides an elegant description of the semantics. The various classes of lambda calculus models are described in a uniform manner. Some didactical improvements have been made to this edition. An example of a simple model is given and then the general theory (of categorical models) is developed. Indications are given of those parts of the book which can be used to form a coherent course.
Categories: Mathematics
# The Lambda Calculus
The Lambda Calculus, treated in this book mainly in its untyped version, consists of a collection of expressions, called lambda terms, together with ways how to rewrite and identify these.
Author: Henk Barendregt
Publisher:
ISBN: 184890066X
Category: Computers
Page: 656
View: 468
The Lambda Calculus, treated in this book mainly in its untyped version, consists of a collection of expressions, called lambda terms, together with ways how to rewrite and identify these. In the parts conversion, reduction, theories, and models the view is respectively 'algebraic', computational, with more ('coinductive') identifications, and finally set-theoretic. The lambda terms are built up from variables, using application and abstraction. Applying a term F to M has as intention that F is a function, M its argument, and FM the result of the application. This is only the intention: to actually obtain the result one has to rewrite the expression FM according to the reduction rules. Abstraction provides a way to create functions according to the effect when applying them. The power of the theory comes from the fact that computations, both terminating and infinite, can be expressed by lambda terms at a 'comfortable' level of abstraction.
Categories: Computers
# Semantics of the Probabilistic Typed Lambda Calculus
Markov Chain Semantics, Termination Behavior, and Denotational Semantics Dirk Draheim. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. ... The Lambda CalculusIts Syntax and Semantics. North Holland, Amsterdam, 1984.
Author: Dirk Draheim
Publisher: Springer
ISBN: 9783642551987
Category: Computers
Page: 218
View: 669
This book takes a foundational approach to the semantics of probabilistic programming. It elaborates a rigorous Markov chain semantics for the probabilistic typed lambda calculus, which is the typed lambda calculus with recursion plus probabilistic choice. The book starts with a recapitulation of the basic mathematical tools needed throughout the book, in particular Markov chains, graph theory and domain theory, and also explores the topic of inductive definitions. It then defines the syntax and establishes the Markov chain semantics of the probabilistic lambda calculus and, furthermore, both a graph and a tree semantics. Based on that, it investigates the termination behavior of probabilistic programs. It introduces the notions of termination degree, bounded termination and path stoppability and investigates their mutual relationships. Lastly, it defines a denotational semantics of the probabilistic lambda calculus, based on continuous functions over probability distributions as domains. The work mostly appeals to researchers in theoretical computer science focusing on probabilistic programming, randomized algorithms, or programming language theory.
Categories: Computers
# Type Assignment in the Lambda calculus
Author: C. B. Ben-Yelles
Publisher:
ISBN: OCLC:59728610
Category:
Page:
View: 509
Categories:
# From Lambda Calculus to Cybersecurity Through Program Analysis
Barendregt, H.: The Lambda Calculus, Its Syntax and Semantics, Studies in Logic and the Foundations of Mathematics, vol. 103. Elsevier (1984) 7. Cardelli, L., Wegner, P.: On understanding types, data abstraction, and polymorphism.
Author: Alessandra Di Pierro
Publisher: Springer Nature
ISBN: 9783030411039
Category: Mathematics
Page: 293
View: 987
This Festschrift is in honor of Chris Hankin, Professor at the Imperial College in London, UK, on the Occasion of His 65th Birthday. Chris Hankin is a Fellow of the Institute for Security Science and Technology and a Professor of Computing Science. His research is in cyber security, data analytics and semantics-based program analysis. He leads multidisciplinary projects focused on developing advanced visual analytics and providing better decision support to defend against cyber attacks. This Festschrift is a collection of scientific contributions related to the topics that have marked the research career of Professor Chris Hankin. The contributions have been written to honour Chris' career and on the occasion of his retirement.
Categories: Mathematics
# Lambda Calculus with Types
Barendregt, H. P. 1984. The Lambda Calculus: its Syntax and Semantics. Revised edition. North—Holland. Barendregt, H. P. 1991. Self—interpretation in lambda calculus. Journal of Functional Programming, 1(2), 229*233.
Author: Henk Barendregt
Publisher: Cambridge University Press
ISBN: 9780521766142
Category: Mathematics
Page: 833
View: 372
This handbook with exercises reveals the mathematical beauty of formalisms hitherto mostly used for software and hardware design and verification.
Categories: Mathematics
# The Parametric Lambda Calculus
The Lambda Calculus: Its Syntax and Semantics (2nd edition). North-Holland, Amsterdam, 1984. Henk Barendregt, Mario Coppo, and Mariangiola Dezani-Ciancaglini. A filter lambda model and the completeness of type assignment.
Author: Simona Ronchi Della Rocca
Publisher: Springer Science & Business Media
ISBN: 9783662103944
Category: Mathematics
Page: 248
View: 998
The book contains a completely new presentation of classical results in the field of Lambda Calculus, together with new results. The text is unique in that it presents a new calculus (Parametric Lambda Calculus) which can be instantiated to obtain already known lambda-calculi. Some properties, which in the literature have been proved separately for different calculi, can be proved once for the Parametric one. The lambda calculi are presented from a Computer Science point of view, with a particular emphasis on their semantics, both operational and denotational.
Categories: Mathematics
# Computer Science Logic
Barendregt, H.P.: The lambda calculus: Its syntax and semantics. North-Holland Publishing Co, Amsterdam (1984) 5. Berline, C.: From computation to foundations via functions and application: The λ- calculus and its webbed models.
Author: Jacques Duparc
Publisher: Springer Science & Business Media
ISBN: 9783540749141
Category: Computers
Page: 600
View: 238
This book constitutes the refereed proceedings of the 21st International Workshop on Computer Science Logic, CSL 2007, held as the 16th Annual Conference of the EACSL in Lausanne, Switzerland. The 36 revised full papers presented together with the abstracts of six invited lectures are organized in topical sections on logic and games, expressiveness, games and trees, logic and deduction, lambda calculus, finite model theory, linear logic, proof theory, and game semantics.
Categories: Computers
# Typed Lambda Calculi and Applications
Thispaperhasraised many questions, mainly concerninga possible implementation of lambda calculus with constructors.The first one is about recursively defined data types, ... Barendregt, H.: The Lambda Calculus: Its Syntax and Semantics.
Author: Pierre-Louis Curien
Publisher: Springer Science & Business Media
ISBN: 9783642022722
Category: Computers
Page: 417
View: 366
This book constitutes the refereed proceedings of the 9th International Conference on Typed Lambda Calculi and Applications, TLCA 2009, held in Brasilia, Brazil in July 2008 in conjunction with RTA 2007, the 19th International Conference on Rewriting Techniques and Applications as part of RDP 2009, the 5th International Conference on Rewriting, Deduction, and Programming. The 27 revised full papers presented together with 2 invited talks were carefully reviewed and selected from 53 submissions. The papers present original research results that are broadly relevant to the theory and applications of typed calculi and address a wide variety of topics such as proof-theory, semantics, implementation, types, and programming.
Categories: Computers
# Typed Lambda Calculi and Applications
Barendregt, H.: The Lambda Calculus: its Syntax and Semantics. North-Holland, Amsterdam (1984) 8. Barendregt, H., Coppo, M., Dezani-Ciancaglini, M.: A filter lambda model and the completeness of type assignment.
Author: Luke Ong
Publisher: Springer Science & Business Media
ISBN: 9783642216909
Category: Computers
Page: 245
View: 938
This book constitutes the refereed proceedings of the 10th International Conference on Typed Lambda Calculi and Applications, TLCA 2011, held in Novi Sad, Serbia, in June 2011 as part of RDP 2011, the 6th Federated Conference on Rewriting, Deduction, and Programming. The 15 revised full papers presented were carefully reviewed and selected from 44 submissions. The papers provide prevailing research results on all current aspects of typed lambda calculi, ranging from theoretical and methodological issues to applications in various contexts addressing a wide variety of topics such as proof-theory, semantics, implementation, types, and programming.
Categories: Computers
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Friday, 21 March 2014
Speeling Test for Class 6B / 6A
Assalamu alaikum,
This is a reminder to Grade 6B and 6A students that the upcoming Speeling test will be on
Tuesday, March 25,2014
You find below the List 1 .
le/ un bistro
l' / un éléphant
la/ une souris
l'/un alligator
se bat
vieux
vieil/vieille
l'énergie
le / un câble
secoue
l'air
apprécie
chez
la/une cuisine
le/un bol
a l'air de
terrible
le/un café
allons-y
étudie
la/une boule
bâtit
Tr. Said
Jazakoum Allah khairan
Wednesday, 19 March 2014
Alsalamo Alikom Dear Parents
Alhamdullillah we are done with chapter 12. It was a very important chapter about Fractions, Decimal, Ratios, and Percentages.
Alhamdullillah students now are able to relate and compare fractions, decimals, and percents.
They are able to identify, model, and apply ratios in various situations.
They are able to represent relationships using unit rates and solve problems using a guess and test strategy.
Inshaa'Allah we will have our test on chapter 12 next Monday Mrach 31st.
Jazakom Allah Khairan for your cooperation.
Tuesday, 18 March 2014
Assalam alaikoum Dear parent,
Student will have a French test this Thursday,March 20,2014 ( possessive adjectives)
The possessive adjectives (les adjectifs possessifs). In French, they agree with the following noun. But in the plural, there is no difference between masculine and feminine.
Masculine Feminine Plural my mon ma mes your ton ta tes his, her, its son sa ses our notre notre nos your votre votre vos their leur leur leurs
Examples:
• C'est le chat de Marie ; c'est son chat.
• Mon père travaille dur.
• Nous avons trouvé tes chaussures.
• Nous gardons leurs enfants.
• C'est notre père.
Jazakoum Allah khairan.
Tr.Said
French Teacher
Assalamu Alaikum Dear Parents,
You find below what we did in the class Fev 24 to Fev 28. :
1- We introduced the characters of story( Le Bistro des animaux) :
* Le grand Éléphant
* Mme Souris
* Beau Lion
* Louis la Grenouille
* Alice l'Alligateur
2- We started a new gestured vocabulary
le/ un bistro
l' / un éléphant
la/ une souris
l'/un alligator
se bat
vieux
vieil/vieille
l'énergie
le / un câble
secoue
l'air
apprécie
chez
la/une cuisine
le/un bol
a l'air de
terrible
le/un café
allons-y
étudie
la/une boule
bâtit
3- We started to read the story.
4- play rehearsal
viewing play on video/DVD and questions
Jazakoum allah khairan
Tr.Said
French Teacher.
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# Quants Questions : Profit and Loss Set 4
Hello Aspirants. Welcome to Online Quantitative Aptitude section in AffairsCloud.com. Here we are creating question sample in Profit and loss , which is common for all the competitive exams. We have included Some questions that are repeatedly asked in exams !!
1. The marked price of a sofa is 11,500 . The shopkeeper sold it by allowing 18% discount on the market price and earned 15% profit. What is the cost price of the sofa ?
A)8000
B)8100
C)8200
D)8400
Explanation :
11500×(82/100)×(100/115) = 8200
2. The ratio of the cost price to the marked price of the furniture is 3:5 and ratio of percentage profit to the percentage discount is 5:3.What is the discount percentage ?
A)3.33
B)6.66
C)15.15
D)9.99
Explanation :
CP:MP = 3: 5
P:D = 5:3
Let CP = 300; MP = 500
(5X/100)×300 + (3X/100)×500 = 100
15X+15X = 100
X=(100/30) = 3.33
Discount = 3×3.33 = 9.99
3. A shopkeeper expect the gain of 19% on his cost price.If in a week his sale was of Rs.462 , What was his profit ?
A)73
B)73.77
C)73.50
D)74
Explanation :
CP=(100/119)×462 = 388. 23
P = 462 – 388.23 = 73.77
4. Prakash bought a bike at 20% discount on its original price.He sold it with 30% increase on the price he bought it.The new sale price is by what percent more than the original price ?
A)4%
B)5%
C)10%
D)22%
Explanation :
Let original price = Rs.100
20% discount = Rs.20
CP = 80
SP = (130/100)×80 = 104
Percentage = (104 – 100 )% = 4%
5. Shopkeeper marked the selling price of an article at 12% above the cost price . At the time of selling , he allow certain discount and suffer a loss of 2%.Find the discount percentage
A)12%
B)12.25%
C)12.5%
D)13%
Explanation :
Let CP = 100
MP = 112
SP = 98
Discount percentage = ([112 – 98]/112)×100 = (14/112)×100 = 12.5%
6. A Grinder is listed at Rs.1200 and a discount of 25%is offered on the list price. What additional discount must be offered to the customer to bring the net price to Rs. 990.
A)10%
B)12%
C)15%
D)20%
Explanation :
SP after 1st discount = [(75/100)×1200] = Rs.900
Net SP = Rs.990
Discount on Rs.900 = 990 – 900 = Rs.90
Required discount = [(90/900)×100] = 10%
7. A sales man sells 2 watches for Rs.460 each.On one he gets 15% profit and on the other 15% loss.His profit or loss in the entire transaction was
A)2(4/6)%
B)2(1/4)%
C)2(3/4)%
D)2(7/4)%
Explanation :Loss = (15/10)^2 = 225/100 = 9/4 = 2(1/4)%
8. A shopkeeper takes 12% profit on his goods.He lost 22% goods during theft.His loss percentage is
A)12.56
B)12.55
C)12.64
D)12.60
Explanation :
Let no of goods = 100
CP of each item Rs.1 then total cost = Rs.100
No of item theft = 22
Remaining item = 78
Profit 12%, SP = 1.12 each item
Total sale = 78 ×1.12 = 87.36
Loss = 100 – 87.36 = 12.64
9. The difference the cost price and selling price of the book is Rs.260 .If the profit is 25%, then the selling price is
A)1100
B)1200
C)1250
D)1300
Explanation :
SP = 125% of Rs.x = (x×125/100) = 5x/4
(5x/4) – x = 260
X = (260 × 4) = 1040
SP = (1040×5)/4 = 1300
10. The CP of 18 articles is equal to SP 15 articles .Find the gain or loss
A)15%
B)18%
C)20%
D)22%
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# math
posted by on .
OMG I CAN NOT get this !!!!
Purchase price of article = \$495
Down payment = \$50
Number of payments = 36
True annual interest rate = 18%
Monthly payment amount = \$
The formula is I= 2YC
________
M(N+1)
y = the number of payments made in a year.
m = the amount financed or how much money was loaned.
c = the total amount of interest charges.
n = the number of payments for the whole loan.
I = interest
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https://arabic.oercommons.org/EN/browse?f.alignment=CCSS.Math.Content.HSF-IF.B.5
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# 4 Results
View
Selected filters:
• CCSS.Math.Content.HSF-IF.B.5
Unrestricted Use
CC BY
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In this real world problem students solve questions based on the relationship between production costs and price.
Material Type:
Activity/Lab
Provider:
Illustrative Mathematics
Provider Set:
Illustrative Mathematics
Author:
Illustrative Mathematics
05/01/2012
Only Sharing Permitted
CC BY-NC-ND
Rating
This lesson unit is intended to help teachers assess how well students are able to translate between graphs and algebraic representations of polynomials. In particular, this unit aims to help you identify and assist students who have difficulties in: recognizing the connection between the zeros of polynomials when suitable factorizations are available, and graphs of the functions defined by polynomials; and recognizing the connection between transformations of the graphs and transformations of the functions obtained by replacing f(x) by f(x + k), f(x) + k, -f(x), f(-x).
Material Type:
Assessment
Lesson Plan
Provider:
Shell Center for Mathematical Education
U.C. Berkeley
Provider Set:
Mathematics Assessment Project (MAP)
04/26/2013
Only Sharing Permitted
CC BY-NC-ND
Rating
This lesson unit is intended to help teachers assess how well students are able to understand what the different algebraic forms of a quadratic function reveal about the properties of its graphical representation. In particular, the lesson will help teachers identify and help students who have the following difficulties: understanding how the factored form of the function can identify a graphŐs roots; understanding how the completed square form of the function can identify a graphŐs maximum or minimum point; and understanding how the standard form of the function can identify a graphŐs intercept.
Material Type:
Assessment
Lesson Plan
Provider:
Shell Center for Mathematical Education
U.C. Berkeley
Provider Set:
Mathematics Assessment Project (MAP)
04/26/2013
Only Sharing Permitted
CC BY-NC-ND
Rating
This lesson unit is intended to help teachers assess how well students are able to: articulate verbally the relationships between variables arising in everyday contexts; translate between everyday situations and sketch graphs of relationships between variables; interpret algebraic functions in terms of the contexts in which they arise; and reflect on the domains of everyday functions and in particular whether they should be discrete or continuous.
Material Type:
Assessment
Lesson Plan
Provider:
Shell Center for Mathematical Education
U.C. Berkeley
Provider Set:
Mathematics Assessment Project (MAP)
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The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A307223 Irregular table T(n, k) read by rows: n-th row gives number of subsets of the divisors of n which sum to k for 1 <= k <= sigma(n). 2
%I
%S 1,1,1,1,1,0,1,1,1,1,1,1,1,1,1,1,0,0,0,1,1,1,1,2,1,1,2,1,1,2,1,1,1,1,
%T 0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1,1,0,0,0,0,1,1,0,1,
%U 1,1,1,1,0,1,1,1,1,0,1,1,1,1,0,1,1,1,1
%N Irregular table T(n, k) read by rows: n-th row gives number of subsets of the divisors of n which sum to k for 1 <= k <= sigma(n).
%C T(n, k) > 0 for all values of k iff n is practical (A005153).
%F T(n, n) = A033630(n).
%F T(n, A030057(n)) = 0 if there is a 0 in the n-th row, i.e. A030057(n) <= sigma(n) or n is not practical.
%e Table begins as:
%e 1
%e 1,1,1
%e 1,0,1,1
%e 1,1,1,1,1,1,1
%e 1,0,0,0,1,1
%e 1,1,2,1,1,2,1,1,2,1,1,1
%e 1,0,0,0,0,0,1,1
%e 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1
%e 1,0,1,1,0,0,0,0,1,1,0,1,1
%e 1,1,1,0,1,1,1,1,0,1,1,1,1,0,1,1,1,1
%t T[n_,k_] := Module[{d = Divisors[n]}, SeriesCoefficient[Series[Product[1 + x^d[[i]], {i, Length[d]}], {x, 0, k}], k]]; Table[T[n, k], {n,1,10}, {k, 1, DivisorSigma[1,n]}] // Flatten
%Y Cf. A005153, A027750, A030057, A033630, A119348, A225561, A237287, A322860.
%K nonn,tabf
%O 1,24
%A _Amiram Eldar_, Mar 29 2019
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http://www.chegg.com/homework-help/questions-and-answers/force-necessary-start-crate-moving-given-mass-crate-32-rm-kg-coefficient-static-friction-c-q972896
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# To move a large crate across a rough floor, you push on it with a force F at an angle of ...
0 pts ended
Find the force necessary to start the crate moving, given that the mass of the crate is 32 {\rm kg} and the coefficient of static friction between the crate and the floor is 0.55.
| 71
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http://www.java-forums.org/new-java/70265-finding-hardest-hole-mini-golf-game.html
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# Thread: Finding hardest hole in mini-golf game
1. Member
Join Date
Nov 2012
Posts
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0
## Finding hardest hole in mini-golf game
So What I need to have is to find the hardest hole in a minigolf game, and that is the highest number. Now the problem is, the program starts counting for the hardest hole from the end (hole 18), but I need to get it to do it from (hole 1). Scoreboard contains the method findHardestHole. Other files are also included. 2nd file is panel. 3rd is driver.
Java Code:
``` //Name______________________________ Date_____________
import javax.swing.*;
import java.awt.*;
public class ScoreCard09 extends JPanel
{
private JTextField[] input;
public ScoreCard09()
{
setLayout(new GridLayout(2, 18));
for(int x = 1; x <= 18; x++)
{
add(new JLabel("" + x, SwingConstants.CENTER));
}
input = new JTextField[18];
for(int x = 0; x < input.length; x++)
{
input[x] = new JTextField();
add(input[x]);
}
}
public void randomize()
{
for (int i = 0; i < input.length; i++)
input[i].setText("" + (int)(Math.random() * 4 + 1));
}
public int findTotal()
{
int total = 0;
for (int i = 0; i < input.length; i++)
total += Integer.parseInt(input[i].getText());
return total;
}
public int findAces()
{
int count = 0;
for (int i = 0; i < input.length; i++)
if (Integer.parseInt(input[i].getText()) == 1)
count++;
return count;
}
public int findHardestHole()
{
int hardest = 1;
int strokes = Integer.parseInt(input[0].getText());
for (int i = 0; i < input.length; i++)
if (Integer.parseInt(input[i].getText()) > strokes)
hardest = i+1;
return hardest;
}
}```
PANEL:
Java Code:
``` //Torbert, e-mail: mr@torbert.com, website: www.mr.torbert.com
//version 7.14.2003
import javax.swing.*;
import java.awt.*;
import java.awt.event.*;
public class Panel09 extends JPanel
{
ScoreCard09 scorecard;
JLabel output;
public Panel09()
{
setLayout(new BorderLayout());
output = new JLabel("------", SwingConstants.CENTER);
add(output, BorderLayout.NORTH);
scorecard = new ScoreCard09();
add(scorecard, BorderLayout.CENTER);
JPanel panel = new JPanel();
panel.setLayout(new FlowLayout());
addButton(panel, "Randomize", new Listener1());
addButton(panel, "Total Score", new Listener2());
addButton(panel, "Holes in One", new Listener3());
addButton(panel, "Hardest Hole", new Listener4());
add(panel, BorderLayout.SOUTH);
}
private void addButton(JPanel p, String s, ActionListener a)
{
JButton b = new JButton(s);
b.addActionListener(a);
p.add(b);
}
private class Listener1 implements ActionListener
{
public void actionPerformed(ActionEvent e)
{
scorecard.randomize();
output.setText("------");
}
}
private class Listener2 implements ActionListener
{
public void actionPerformed(ActionEvent e)
{
output.setText("Total Score: " + scorecard.findTotal());
}
}
private class Listener3 implements ActionListener
{
public void actionPerformed(ActionEvent e)
{
output.setText("Holes in One: " + scorecard.findAces());
}
}
private class Listener4 implements ActionListener
{
public void actionPerformed(ActionEvent e)
{
output.setText("Hardest Hole: " + scorecard.findHardestHole());
}
}
}```
DRIVER:
Java Code:
``` import javax.swing.JFrame;
public class Driver09
{
public static void main(String[] args)
{
JFrame frame = new JFrame("Unit4, Lab09: Miniature Golf");
frame.setSize(500, 150);
frame.setLocation(150, 100);
frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
frame.setContentPane(new Panel09());
frame.setVisible(true);
}
}```
2. Member
Join Date
Nov 2012
Posts
29
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0
Bump
3. ## Re: Finding hardest hole in mini-golf game
What is wrong?
kind regards,
Jos
4. Member
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Nov 2012
Posts
29
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0
## Re: Finding hardest hole in mini-golf game
Originally Posted by JosAH
What is wrong?
kind regards,
Jos
The problem is, the program starts counting from the last hole (hole 18), but I need it to count from the first hole to look for the hardest hole.
5. Senior Member
Join Date
Oct 2010
Posts
317
Rep Power
4
## Re: Finding hardest hole in mini-golf game
Hi Wizard0860,
There is a problem with your implementation of the findHardestHole() method.
At the moment 'strokes' is set to the value of the first element in 'input'. Within the loop the value of each element is compared to 'strokes' but 'strokes' never changes so you will always be comparing the value of the current element with the first.
Regards.
#### Posting Permissions
• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
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# display
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3. ### Data labels color in stacked bar data chart
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Is it possible to display a custom userform message at two different times on the same day using 1) one userfrom 2) not being available to close the earlier form ? The earlier userform time to display blocks the later userform display from displaying. What I want is to be able to show BOTH...
6. ### Number formating
I have a numeric cell, that I would like to be able to enter "-123" and the cell displays "-1.23" or if the user enters "-1.23" the cell will display "-1.23" is this even possible? Thanks
7. ### Custom number formatting in Excel
Hi there, Im trying to display some tricky % number formatting and have hit a dead end. Just wondering if it's possible to display % under the following criteria Over 100%: +xxx% 0-100%: +xx.x% -100%-0%: [red]-xx.x% Under -100%:[red]-xxx% The key is basically to retain the +/-, the...
8. ### RE: Multiple display of timed userforms ...
RE: Multiple display of timed userforms ... Action1: it's 8:00 AM. Open a new Userform1(blank textbox1): Enter data: Userform.DTPicker1.value = Time 10:00 AM (display time) Userform1.Textbox1.value = Buy oranges close userform Action2: 5 minutes later: open a new Userform1(blank...
9. ### Checkbox issue
Hi All, This may be easy for some but I have looked around and can't see an answer to my issue. I have created a form with a number of checkboxes and when I pull the data from access into excel in a table format all the details from the check boxes show ether False or True. Is there anyway...
10. ### Getting value on a different tab (same spreadsheet)
Hello, I’m trying to display the latest value entered on adifferent tab (pretty much a Vlookup example). In the column on “Tab_2” an employee enters the new amountof mileage driven in the rented car once it’s returned (see example below. Date and Mileage are two different columns)...
11. ### Vlookup yet again
Hi, I have a spreadsheet with 5000 rows and 15. The 1st Column either has a Y or N in it and the 12 column either has a 'South' or "West" in it. What I would like to do o a different sheet is display only the rows that have a Y in column 1 and South in column 12. I would also like to only...
12. ### TextBox value as Formula Overwrites
Hello The below code somehow overwrites the formula typed in rangeE1 What i am trying to do is that if i check value as formula then the txtValue should display the Formula if unchecked then it should display the value in textbox this is like Formula Bar and Cell Value. Second thing if i type...
13. ### ROI value for an investment portfolio.
Could I ask for some help from you experts with a formula please. I am having an issue with trying to get the correct result for an investment portfolio that I am building. My isue is obtaining the correct RIO value in cell O27 for the month of July. Here are the cells that are being...
14. ### Comma Seperation issues: Numeric values are completely displayed differently
Hello OMG What i face now is when values displayed in formula bar as 200600 it displays as 200,600 in cell instead of 2,00,600 ie incorrect comma placement Following values 4,14,180 to display as 4,14,180 it displays as 4104,180 3,18,600 to display as 3,18,600 it dispalys as 3108,600 17676.40...
15. ### Custom Format of Numbers
I am trying to figure out how to create a custom format for numbers. Is it possible to do this? Positive Number Display (in color black) 145.00 Negative Number Display (in color red) (-145.00)
16. ### Cell Formatting
I don't know why I cannot remember how to do this...but... I have a time cell and I just want to type "1753" in the cell and for it to display as "17:53". How do I have it automatically add the ":" to that cell?
I have a column of unique text values that are URLs (web addresses). I need to convert all of them to hyperlinks that point to the unique URL but want all of the rows to display the same text. My actual dataset is hundreds of rows. Data www.1234.com www.5678.com www.9101.com Desired output...
18. ### Add Numerical Tab Names in Combo box
My workbook sheet tabs would be named by Year (i.e. 2018, 2019, etc). Currently I have one sheet (i.e. 2019). I would like to display in a combo box the previous 2yrs (i.e. 2018 & 2017) and also the following year (i.e. 2020). Maximum years displayed should be limited to 4. (i.e. 2017 -...
19. ### IF formula
Can someone explain how to write a IF formula in excel? If cell 10% of B9 is greater than 35, display the value, otherwise display 35.
20. ### VBA CheckBox If True, Display Date / Time
Hi newbie question please. I'm trying to create vba for checkbox when checkbox = true, display date/time else display blank/null Thanks in advance:)
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https://aviation.stackexchange.com/questions/87286/why-is-it-safe-flying-in-a-tailwind-when-taking-off-is-not?noredirect=1
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# Why is it safe flying in a tailwind when taking off is not? [duplicate]
If taking off in q tailwind is unsafe (due to loss of lift from reduced air speed), then why is it safe flying in tailwind, such as in jet stream?
• Does this answer your question? Why is tailwind during final approach and landing so dangerous? and see also: How do jet streams affect the fuel consumption? May 24, 2021 at 1:23
• Do any of these questions help? Here, here, here May 24, 2021 at 3:28
• Cruising speeds are also generally quite a lot higher than landing / take-off speeds - so the effective loss of lift is not relevant - 100% - 2% is still 98%... At landing speed 100% - 40% - well you get the idea ....
– Mr R
May 24, 2021 at 4:02
• BTW, I downvoted because the statement concerning loss of lift from reduced airspeed is not clear. Yes, you will have reduced lift at lower airspeeds, but with a tailwind you will simply take off at a higher groundspeed further down the runway. Hint: Airplanes don't rotate after a fixed ground roll distance, they rotate at a predetermined airspeed. Always... May 24, 2021 at 19:40
• Another hint: If you are flying in the jet stream at 300 knots of indicated airspeed you will have the same IAS no matter whether you have a tailwind or a headwind. The ONLY thing that changes based on direction of flight is ground speed. May 24, 2021 at 20:15
I once had a flight instructor tell me, "As soon as you get airborne, there's no wind." While this is not strictly true in all cases (wind shear is still a thing), it does hold true for flying with the wind. The aircraft is moving in the moving air mass, so the fact that the wind is "coming from" behind you only helps you get to your destination faster. Aircraft can fly at the same airspeed no matter which direction the air mass is moving.
• Or how fast the air mass is moving. May 24, 2021 at 3:57
• What about a tailwind matching the max speed IAS an aircraft can gain, and thus reducing IAS to 0, with no the wings generating no lift? May 24, 2021 at 5:46
• With tailwind you fly at same IAS but increase ground speed. You look at IAS and ignore ground speed. May 24, 2021 at 6:07
• @HarshilSharma In that case, the IAS of the airplane wouldn't change, but its ground speed would double. May 24, 2021 at 13:21
“Wind” is simply an air mass moving over the ground. When flying, planes do not experience wind per se because they are flying at a (mostly) fixed speed relative to that air mass.
We speak of a “headwind” or “tailwind”because the movement of the air either reduces or increases the plane’s speed relative to the ground.
We want to takeoff and land with a headwind because the reduced ground speed means we need less runway to achieve the same speed relative to the air, whereas a tailwind means we need more runway.
• From what I think, more important than runway length required, a tailwind will require more engine power to maintain sufficient airspeed to prevent a stall. Which is what makes me think that a jetstream can theoretically reduce an aircraft's airspeed and may remove all lift. It's not a problem if jetstream is at 300 kph and the aircraft is moving at 800 kph IAS. Can, theoretically, would a 800 kph tailwind possibly stall the aircraft? May 24, 2021 at 5:44
• @HarshilSharma sudden increase of tailwind can stall an aircraft, see e.g. Delta Air Lines flight 191. The jet stream is, however, hundreds of kilometers wide and kilometers high, so even though the wind speed can be high, the change when entering or leaving it is gradual, so the aircraft will catch up it easily, the airspeed remaining constant and just change in ground speed indicating the presence of a jet stream. May 24, 2021 at 8:31
• @HarshilSharma Two side notes: 1. transport aircraft never move at 800 km/h indicated. 800 km/h true airspeed is rather low, but due to the low air density up in the cruise flight level the indicated airspeed is only around 250 knots (460 km/h). 2. in cruise with flaps retracted the margin to stall isn't that large, so a 300 km/h wind shear would almost certainly stall the aircraft—fortunately the jet stream is never sudden. May 24, 2021 at 8:35
• @HarshilSharma, no, tail wind does not require more engine power to maintain sufficient airspeed. The aircraft flies relative to the air, so the power to airspeed is always the same. The only problem with tail-wind is that the runway is moving aft under the air mass while the aircraft is taking off or landing, so more of it is needed. May 24, 2021 at 8:37
• @HarshilSharma Incorrect. I normally cruise at 75% power, which results in an airspeed of 110kt. If there is a 10kt tailwind, that means a groundspeed of 120kt. If there is a 10kt headwind, that means a groundspeed of 100kt. My plane stalls at an airspeed of about 45kt; my ground speed is irrelevant. May 24, 2021 at 12:41
If taking off in q tailwind is unsafe (due to loss of lift from reduced air speed)
This is incorrect. The airspeed required to take off is the same, no matter the speed of the wind relative to the ground.
Taking off in a tailwind is potentially unsafe because the required ground speed is higher. If you take off in a tailwind, you may run out of runway before your airspeed is high enough.
You cannot run out of runway while in flight.
• Why was this downvoted?! May 24, 2021 at 14:45
• @MichaelHall -- Maybe someone thought the quote was correct-- in the sense that at any given point in the take-off run (specified either in terms of elapsed distance, or elapsed time), the airspeed will be less when there is a tailwind. I'd say the answer would be improved by deleting the very first sentence after the quote. May 24, 2021 at 16:55
• But the airspeed will NOT be less, the ground speed will just be greater. (Unless you say airspeed compared to ____.) This will affect time to accelerate, ground roll distance, tires speed, etc. May 24, 2021 at 18:48
• The word "less" always implies a comparison, so I stand by my comment. Of course, the airspeed at the moment of liftoff will not be less, normally. The quoted statement is vague enough that you one can argue either way, that it is correct or not correct. May 24, 2021 at 19:17
• No, it doesn't imply any comparison! If you rotate at 80 knots you should ALWAYS rotate at 80 knots whether you have a headwind, tailwind, or no wind. Period. Therefore airspeed will NOT be less - instead ground roll, ground speed, and time to accelerate will all be increased with a tailwind. This is the fundamental misconception that non-flyers can't seem go get over, and it baffles me because it really isn't complicated... May 24, 2021 at 19:33
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# Java Make Numbers With Code Examples
Java Make Numbers With Code Examples
In this session, we will try our hand at solving the Java Make Numbers puzzle by using the computer language. The following piece of code will demonstrate this point.
```public class Main {
public int number = 12; // 'int' is a number in Java. but without digits.
public double digitNumber = 12.25; // 'double' is a number in Java, but with digits.
}
```
By investigating a variety of use scenarios, we were able to demonstrate how to solve the Java Make Numbers problem that was present.
## How do you create a number in Java?
In Java, there is three-way to generate random numbers using the method and classes.Using the Random Class
• First, import the class java.lang.Random.
• Create an object of the Random class.
• Invoke any of the following methods:
• nextInt(int bound)
• nextInt()
• nextFloat()
• nextDouble()
• nextLong()
## What is %s and %D in Java?
Answer 548f5741937676d496009b7d. 0 votes. %s refers to a string data type, %f refers to a float data type, and %d refers to a double data type.
## What does %d do in Java?
%d: Specifies Decimal integer. %c: Specifies character. %T or %t: Specifies Time and date. %n: Inserts newline character.20-Mar-2017
## How do you do 2 decimal places in Java?
Just use %. 2f as the format specifier. This will make the Java printf format a double to two decimal places.13-Jun-2022
## How do you generate a 3 digit random number in Java?
You should create a Random object instead. Random random = new Random(); int randomNumber = random. nextInt(900) + 100; Now randomNumber must be three digit.12-Sept-2015
## How do you generate a random 10 digit number in Java?
To generate a random integer in this range, you can do this:
• import java. util. Random;
• Random rand = new Random();
• int num = rand. nextInt(9000000) + 1000000;
for string
## What is %s in Java example?
Java String printf format examples
## What is %s in a string Java?
Example 1: Java String format() %s in the format string is replaced with the content of language . %s is a format specifier. Similarly, %x is replaced with the hexadecimal value of number in String.
## What is printf () in Java?
The printf() method of Java PrintStream class is a convenience method which is used to write a String which is formatted to this output Stream. It uses the specified format string and arguments to write the string.
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Home » How to Use a Voltmeter
# How to Use a Voltmeter
The voltmeter is a device used to measure the voltage of electricity that is present in outlets and other electrical devices and access points. Voltmeters are quite common and can be found in most retail stores that carry electrical wiring and products.
However, most meters available today are called multimeters because they measure voltage, ohms, and amperage. They can be found using either digital or analog displays which show you how much electricity is present. Today, more voltmeters are made with digital displays, but the old analog ones work just as well.
Technically, a voltmeter registers the difference between the electricity present in two nodes of the circuit. And while that may sound complicated, all you really need to know is how to read it and what it means in terms of the electricity that is present.
## Set the Meter
Since multimeters have different settings, the first thing you will have to learn is where to set the meter to get the reading that you want. This is usually found on a large dial in the center of the voltmeter that can be switched to different settings depending on where it is turned.
• V~: AC
• V-: DC
AC stands for alternating current. This is the current used in the household and is normally accessed through the outlets DC stands for direct current and is used in batteries and portable electronic devices. The reason why there is a difference is that DC is simpler and easier to use for small devices, but it cannot travel very far. AC is a little more complicated but has considerable range which is why it is used for the transmission of power over long distances.
Once you have chosen the AC or DC setting depending on what you are trying to read, you now need to set the range of the voltmeter to just above what you expect to find. Some multimeters have no range setting which means they will detect automatically the voltage that is present. Otherwise, you will need to set the meter to the voltage just above what you expect to find.
For example, if you are checking a 120V outlet, set the meter to 200 V~. If you are checking power tools, they should have their voltage marked on the device. Household batteries tend to be at 9 volts while a car battery is usually around 12.6 volts. If you are not sure, just set the device to the maximum setting.
You will see one red and one black wire with a probe on one end and a jack on the other. Put the jacks according to their color into their respective slots on the voltmeter. You will put the blackjack into the port which is labeled “COM” or common. The red jack will be put into the hole marked with a V. If you do not have such a hole on the device, place it in the one marked with the lowest number or go with mA.
Now you are ready to check the voltage. Keep in mind that the test leads are protected, but do not touch the metal on the probes when inserting them into an outlet or other electrical device. Do not allow the probes to touch each other as well.
## Insert Probes
The probes or leads are what you will use to measure the voltage that is present.
### Outlet
If you are using the voltmeter to check to see if electricity is present in an outlet, you set the meter to the appropriate voltage. Then, insert the black or negative probe into the larger slot on the outlet which is usually the one on the left or upper left if there is a ground. Because most black test leads have a bump on them, you can insert it into the outlet and let it go. The lead will stay in place.
Now, take the red lead and insert it into the smaller vertical slot. If you look at the display on your device, you should be getting a 120V reading. However, if you are getting an “OL”, you are overloading the voltmeter and it will need to be removed quickly before it becomes damaged. Raise the range on the voltmeter if you get an OL reading and reinsert the red lead.
### Batteries
This is simpler compared to an outlet because the positive and negative poles on the batteries are marked. Set the multimeter to read for DC or V- and place the black lead on the negative terminal and the red lead on the positive one. You may have a DC+ or DC- position on the voltmeter. If so, change it to the other setting if you get no reading. If that does not work, exchange the positions of the red and black leads.
If that does not work, drop the setting on the voltmeter and go through the process again. You may have to check it with a source that you know is good first to ensure that the meter is working properly.
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# Calculate Fractions – What is 1/44 of 44?
In mathematical terms, a fraction of the number one is equal to a quarter. In this article, we’ll learn how to convert 1/44 to a decimal, as well as how to work out 1/5 of eight or 3/12 of 48. These fractions are easy to understand and can be useful in solving everyday problems. The following table shows some examples. Read on to learn how to calculate one-fourth percent and other common fractions.
## How to convert 1/44 to a decimal
How to convert 1/44 to a decimals? Firstly, we need to identify whether 1/44 is a fraction or a whole number. A fraction is an amount that is not a whole number, while a decimal is an integer without a fraction. Regardless, we can easily convert 1/44 to a decimal by rounding it to the nearest whole number. As a result, 1/44 is 0.0227272727273.
When converting a fraction, we can use a calculator to find the decimal equivalent of the fraction. The formula for doing so is the same as that for converting a whole number, but this method is a bit more convenient. Moreover, most fraction calculators also provide graphical representations of the fraction. As such, a fraction calculator will not only show the decimal equivalent of 1/44, but will also allow students to understand how fractions are calculated.
In case you don’t have a calculator, you can also use an online calculator to convert the fraction. This will automatically round the number to the nearest whole number and allow you to determine the number of digits. You can also choose the level of precision to ensure that the number is as precise as possible. By default, we use a precision point of five. Alternatively, you can choose to round half up, which will round the last trailing digit up.
## How to work out 1/5 of 8
How to work out 1/5 of an odd number? There are two ways to express the fraction. First, it is important to note that one fifth equals four and the other half equals five. However, it can also be written as 1/5 x 5/8. If you’re unsure, simply use the following formula to get a quick estimate. In addition, you can also use this method to calculate fractions in a wide variety of situations.
## How to work out 3/12 of 48
Three quarters of a dozen equals nine. A third quarter equals nine times eleven, which is forty-four. Then, four times eleven equals forty-four, and so on. So, if Orla spent five eighths of her money on clothes, she would have three-eighths of a dozen left over. How do you work out three-eighths of a dozen?
#### Asim Boss
Muhammad Asim is a Professional Blogger, Writer, SEO Expert. With over 5 years of experience, he handles clients globally & also educates others with different digital marketing tactics.
Asim Boss has 3444 posts and counting. See all posts by Asim Boss
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Cody
# Problem 43968. Concatenated roots
Solution 1100664
Submitted on 9 Jan 2017 by Binbin Qi
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
x = 1; y = 1; assert(abs(infinteroots(x)-y)<1e-11)
2 Pass
x = 10; y = 52.2735299670437; assert(abs(infinteroots(x)-y)<1e-11)
3 Pass
x=5; y=15.8864718332426; assert(abs(infinteroots(x)-y)<1e-11)
4 Pass
x=6; y=21.7311722059576; assert(abs(infinteroots(x)-y)<1e-11)
5 Pass
x=4; y=10.827015106694; assert(abs(infinteroots(x)-y)<1e-11)
6 Pass
x=3.2; y=7.37887287693964; assert(abs(infinteroots(x)-y)<1e-11)
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You are on page 1of 9
# UNIT G481
Module 1
1.1.4
Linear Motion
Candidates should be able to :
Derive the equations of motion for constant acceleration in a
straight line from a velocity-time graph.
Select and use the equations of motion for constant
acceleration in a straight line :
v = u + at s = (u + v)t s = ut + at
2
v
2
= u
2
+ 2as
Apply the equations for constant acceleration in a straight line,
including the motion of bodies falling in the Earths uniform
gravitational field without air resistance.
Explain how experiments carried out by Galileo overturned
Aristotles ideas of motion.
Describe an experiment to determine the acceleration of free
fall (g) using a falling body.
Apply the equations of constant acceleration to describe and
explain the motion of an object due to a uniform velocity in
one direction and a constant acceleration in a perpendicular
direction.
DERIVING THE EQUATIONS OF MOTION 1
Consider an object moving with an
initial velocity (u) which accelerates
with a constant acceleration (a) to
reach a final velocity (v) after a
time (t). The distance moved in
this time is (s).
This is the velocity-time graph
for the motion.
2008 FXA
EQUATION 1
Acceleration = gradient of v/t graph
a = (v - u)
t
at = v - u
v = u + at
EQUATION 2
Total displacement = average velocity x time
s = (u + v)t
UNIT G481
Module 1
1.1.4
Linear Motion
EQUATIONS OF MOTION - SUMMARY 2
These equations can also be used for objects falling without air
resistance in the Earths uniform gravitational field. Then :
a = g = acceleration due to gravity = 9.81 m s
-2
.
g is positive for objects which are initially falling.
g is negative for objects which are projected upwards.
h = vertical displacement.
PRACTICE QUESTIONS (1)
1 An aircraft accelerates uniformly from rest at 3.1 m s
-2
to reach its
take-off velocity of 100 m s
-1
.
(a) How long does it take for the aircraft to leave the ground ?
(b) How far does it travel during take-off ?
2 A motorway designer assumes that cars approaching a motorway enter
a slip road with a velocity of 8 m s
-1
and need to reach a velocity of
25 m s
-1
.
Assuming that vehicles have an acceleration of 5 m s
-2
, calculate the
minimum length for a slip road.
2008 FXA
EQUATION 3
v = u + at . (1) and s = (u + v) t .. (2)
2
Substituting (u +at) for (v) in equation (2) gives :
s = (u + u + at) t = 2ut + at
2
2 2
So,
s = ut + at
2
EQUATION 4
v = u + at , so t = (v - u)
a
Substituting for (t) in s = (u + v) t gives :
2
s = (u + v) x (v - u) = (uv - uv + v
2
- u
2
)
2 a 2a
2as = v
2
- u
2
So,
v
2
= u
2
+ 2as
v = u + at s = (u + v)t
s = ut + at
2
v
2
= u
2
+ 2as
UNIT G481
Module 1
1.1.4
Linear Motion
3 The cyclist shown in the
diagram opposite is
travelling at 15 m s
-1
.
She brakes so as to avoid
colliding with the wall.
Calculate the deceleration
needed in order to come
to rest in 18 m.
4
The car shown in the diagram above is travelling along a straight road
at 8 m s
-1
. If it accelerates uniformly at 1.5 m s
-2
over a distance
of 18 m, calculate its final velocity.
5 A lorry accelerates from rest at a steady rate of 1.2 m s
-2
.
Calculate :
(a) The time taken to reach a velocity of 15 m s
-1
.
(b) The distance travelled during this time.
(c) The velocity of the lorry when it is 100 m from the
start.
6 3
A train travelling at 20 m s
-1
accelerates uniformly at 0.75 m s
-2
for
25 s. Calculate the distance travelled by the train in this time.
7 The diagram opposite shows a
velocity-time graph for two
cars, A and B, which are moving
in the same direction over a
40 s time period.
Car A, travelling at a constant
velocity of 40 m s
-1
, overtakes
car B at time, t = 0.
In order to catch up with car A, car B immediately accelerates
uniformly for 20 s to reach a constant velocity of 50 m s
-1
.
Calculate :
(a) The distance travelled by car A during the first 20 s.
(b) Car Bs acceleration during the first 20 s.
(c) The distance travelled by car B during the first 20 s.
(d) The additional time taken for car B to catch up with
car A.
(e) The total distance each car will have
travelled.
2008 FXA
UNIT G481
Module 1
1.1.4
Linear Motion
8 A coin is dropped down a mine shaft and falls through a height of
55 m before hitting the bottom of the shaft. Assuming negligible
air resistance and taking the acceleration due to gravity (g)
as 9.81 m s
-2
, calculate :
(a) The time taken for the coin to hit the bottom.
(b) The velocity of the coin on impact.
9 A stone is projected vertically upwards from the ground with an
initial velocity of 30 m s
-1
. Assuming air resistance is negligible
and taking the acceleration due to gravity (g) as 9.81 m s
-2
,
Calculate :
(a) The maximum height reached by the stone.
10 The diagram opposite shows a
velocity-time graph for a vehicle
for a time of 30 s.
(a) Describe the motion of the
vehicle.
(b) Use the graph to determine
the acceleration of the
vehicle over the 30 s period.
(c) Use the graph to determine the displacement of the vehicle over
the 30 s period.
using a suitable equation of motion.
NOTE ON NON-UNIFORM ACCELERATION 4
The EQUATIONS OF MOTION only apply to objects moving with
CONSTANT or UNIFORM ACCELERATION.
The v/t graph opposite shows
the motion of an object which
is moving with non-uniform
acceleration.
The acceleration at any time
is given by the gradient of
the v/t graph at that time.
To find the acceleration at
Any given time :
At the time in question, mark a point on the v/t graph.
Draw a tangent to the curve at that point.
Construct a large right-angled triangle and use it to
Use the procedure outlined above to calculate the acceleration at
Point P in the v/t graph shown below.
Acceleration = = m s
-2
2008 FXA
UNIT G481
Module 1
1.1.4
Linear Motion
HOW GALILEOS EXPERIMENTS OVERTURNED
ARISTOTLES IDEAS OF MOTION
The Greek philosopher ARISTOTLE thought that a force must act all the time in order
to keep an object moving. Our own experience seems to support this idea in that a car,
for example, will slow down and eventually stop when the engine is switched off. But,
does this happen because the driving force has been switched off ? The reality is that
the slowing down and stopping occurs because there is an opposing FRICTIONAL
FORCE.
Experiments carried out by GALILEO (about 1600 years after Aristotle) showed that
Constant force is not needed to maintain motion, but force is needed to :
Start and stop motion,
Change the speed of an object,
Change the direction of an object.
GALILEOS EXPERIMENTS
Galileo simultaneously dropped two objects of different weight from the top of
the leaning Tower of Pisa and found that they hit the ground at the same time.
He concluded that any two objects will fall at the same rate, regardless of
their relative weights.
Using a dripping water clock which counted the volume of water drips as a
measure of time, Galileo was able to measure the time taken by a ball to travel
equal distances down a slope from rest. His results showed that the ball
accelerates as it rolls down the slope and that the greater the slope, the
greater is the acceleration. From this, he concluded that an object falling
vertically will accelerate.
DETERMINATION OF THE ACCELERATION OF FREE FALL (g) 5
A steel ball-bearing is held
by an electromagnet. When
the current to the magnet
is switched off, the ball is
released and the timer is
started. The ball strikes and
opens a trapdoor which then
stops the timer. The time
taken (t) for the ball to fall
through a given height (h) is
recorded. The timing is
repeated and an average
t-value is calculated.
The procedure is repeated for several different h-values and the
results are recorded in the table below :
h = ut + gt
2
And since u = 0, ut = 0
So h = g t
2
Compare y = m x + c
with
Therefore, plotting a
graph of (h) against (t
2
)
gives a best-fit, straight
line through the origin,
From which, acceleration of free fall, g = 2 x gradient of h/t
2
graph
= 2 x
= m s
-2
2008 FXA
h/m t
1
/s t
2
/s t
3
/s tav /s t
2
/s
2
UNIT G481
Module 1
1.1.4
Linear Motion
MOTION DUE TO CONSTANT VELOCITY IN ONE DIRECTION AND
A CONSTANT ACCELERATION IN A PERPENDICULAR DIRECTION
The motion of objects projected at an angle or horizontally from
some height above the ground is called PROJECTILE MOTION.
All PROJECTLIES have the following in common :
They are given some initial velocity (by kicking, firing etc.)
Throughout their flight, the only force acting (neglecting
air resistance) is their WEIGHT due to gravity, which
exerts a constant force acting vertically downwards .
This gives the projectile :
CONSTANT DOWNWARD ACCELERATION = g (9.81 m s
-2
)
Neglecting air resistance, the HORIZONTAL COMPONENT
OF VELOCITY REMAINS CONSTANT THROUGHOUT THE
MOTION and the path followed is a PARABOLA.
6
SUMMARY FOR SOLVING PROJECTILE PROBLEMS
2008 FXA
max. horizontal
displacement
maximum height
vsin
v
vcos
PARABOLA
v
Resolve the initial velocity into :
HORIZONTAL and VERTICAL COMPONENTS.
Assume NEGLIGIBLE AIR RESISTANCE
For the Horizontal Direction :
The HORIZONTAL COMPONENT of velocity remains
constant throughout the flight.
horizontal displacement = horizontal velocity x flight time
For the Vertical Direction :
The acceleration is constant, equal to g.
The equations of motion apply.
At the maximum height, vertical velocity = 0.
When the projectile has returned to the level of its launch
point :
Vertical displacement = 0
Final vertical velocity is equal, but oppositely
Directed to the initial vertical velocity.
v = u + gt h = (u + v)t
h = ut + gt
2
v
2
= u
2
+ 2gh
UNIT G481
Module 1
1.1.4
Linear Motion
Practice Questions (2)
1 A helicopter is flying in a straight line at a speed of 20 m s
-1
and at
a constant height of 180 m. A small object is released from the
helicopter and falls to the ground. Assuming air resistance is
negligible, calculate :
(a) The time taken for the object to reach the ground.
(b) The vertical component of velocity of the object when it
hits the ground.
(c) The horizontal component of velocity of the object when it
hits the ground.
(d) The horizontal displacement of the object in the time taken
to reach the ground.
2 During a European Champions League match, a free kick was taken
by Steven Gerrard and the ball was projected with a velocity of
20 m s
-1
at an angle of 35 to the pitch. Assuming that air
resistance is negligible, calculate :
(a) The initial vertical and horizontal components of velocity.
(b) The time taken for the ball to reach its maximum height.
(c) The maximum height reached by the ball.
(d) The horizontal displacement of the ball in the time taken
3 7
The diagram above shows the path of a ball that is thrown from point A
to point B. The ball reaches its maximum height at point H. The ball
is thrown with an initial velocity of 25.0 m s
-1
at 60 to horizontal.
Assume that there is no air resistance.
(a) (i) Show that the vertical component of the initial velocity is
21.7 m s
-1
.
(ii) Calculate the time taken for the ball to reach point H.
(iii) Calculate the displacement from A to B.
(b) For the path of the ball shown in the diagram, draw sketch graphs,
with labelled axes but without numerical values, to show the
variation of :
(i) The vertical component of the balls velocity against time.
(ii) The distance travelled along its path against time.
(OCR Physics AS - Module 2821 - May 2008)
2008 FXA
UNIT G481
Module 1
1.1.4
Linear Motion
HOMEWORK QUESTIONS
1 When the brakes are applied in a car which is moving at 40 m s
-1
,
the velocity is reduced to 25 m s
-1
over a distance of 140 m. If
the deceleration remains constant, what further distance will the
car travel before coming to rest ?
Sketch a velocity-time graph for the whole motion showing numerical
values on the axes.
2 Calculate the deceleration of a bullet initially travelling at 400 m s
-1
,
if it is brought to rest after travelling 10 cm through a wooden
block.
3 A hawk is hovering above a field at a height of 50 m. It sees a
Mouse directly below it and dives vertically with an acceleration of
12 m s
-2
. Calculate :
(a) The hawks velocity at the instant it reaches the mouse.
(b) The time taken to reach the mouse.
4 A train accelerates steadily from 4.0 m s
-1
to 24.0 m s
-1
in a time
of 180 s. Calculate :
(a) The trains acceleration.
(b) The average velocity of the train.
(c) The distance travelled by the train during the acceleration.
5 Traffic police investigators use the length of skid marks left on the 8
road by a decelerating vehicle in order to determine whether or not
the speed limit has been exceeded.
If the skid marks at the scene of an accident are 52m long and other
tests on the road surface show that the skidding car was decelerating
at 6.5 m s
-2
, was the car breaking the speed limit of 30 m s
-1
?
6 A sandbag is dropped from a height of 180 m, from a helicopter that
is moving vertically upwards with a velocity of 6 m s
-1
. If air
resistance is neglected, calculate :
(a) The initial velocity of the sandbag.
(b) The final velocity of the sandbag.
(c) The time taken for the sandbag to reach the ground.
7 An aid parcel is released from a plane flying horizontally at 65 m s
-1
,
at a height of 800 m above the ground.
(a) Calculate the horizontal and vertical components of the parcels
initial velocity.
(b) How long does it take for the parcel to reach the ground ?
(c) At what horizontal distance from the target should the plane be
when the parcel is released ?
2008 FXA
UNIT G481
Module 1
1.1.4
Linear Motion
8
The diagram above shows the path of a tennis ball after passing over
the net.
As it passes over the net, it is travelling at a height of 1.20 m. The
ball strikes the ground on a line which is 11.9 m from the net.
(a) Assuming air resistance to be negligible,
(i) Show that the time taken for the ball to reach the line
After passing over the net is 0.495 s.
(ii) At the instant the ball strikes the line, calculate :
1. The horizontal component of its velocity.
2. The vertical component of its velocity.
(OCR Physics AS - Module 2821 - June 2007)
9
2008 FXA
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# Introduction to Computing and Programming in Python:
## Introduction to Computing and Programming in Python:
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## Presentation on theme: "Introduction to Computing and Programming in Python:"— Presentation transcript:
Slide1
Introduction to Computing and Programming in Python: A Multimedia Approach
Slide2
Chapter Objectives
Slide3
Introducing IF
if (some test):
#These Python statements will execute if true
print
This is true!
print
This will print, either way.
Computers can make choices.
They can perform tests.
Then execute different statements depending on the result of the test.
Slide4
An
if
statement uses a
test
and a
block
Block is executed
only if
the test is true
“some test” (i.e. the test) can be any kind of logical expression.
E.g
:
1<2 is a logical expression that always return true
a<2 is a logical expression that return true or false depending on the value of a.
Possible tests: <, <=(less than or equal), >, >=, == (equal), <>(not equal)
Slide5
Computing a distance between colors
What does it mean for a color to be “close” to another color?We will compute a color distance, based on the formula for the distance between two points.
Slide6
How to replace brown color in Barbara’s hair with red?
Use JES picture tool to figure out roughly what RGB values are on Katie’s brown hair.
Use JES picture tool figure out
x,y
coordinate ranges where you want to change (e.g. hair near Barbara’s forehead)
Then write a program to look for colors close to that color and increase the redness of those pixels.
Slide7
x in range(70,168)
y in range(56,190)
Slide8
Replacing colorsin a range
def
turnRedInRange(): brown = makeColor(57,16,8) file=“c:/Users/guzdial/mediasources/barbara.jpg" picture=makePicture(file) for x in range(70,168): for y in range(56,190): px=getPixel(picture,x,y) color = getColor(px) if distance(color,brown)<50.0: redness=getRed(px)*1.5 setRed(px,redness) show(picture) return(picture)
Get the range using MediaTools
Slide9
Walking this code
Like last time: Don’t need input, same color we want to change, same file, make a picture
def turnRedInRange():
brown = makeColor(57,16,8)
file=
/Users/guzdial/mediasources/barbara.jpg"
picture=makePicture(file)
for x in range(70,168):
for y in range(56,190):
px=getPixel(picture,x,y)
color = getColor(px)
if distance(color,brown)<50.0:
redness=getRed(px)*1.5
setRed(px,redness)
show(picture)
return(picture)
Slide10
The nested loop
I used MediaTools to find the rectangle where most of the hair is that I want to change
def turnRedInRange():
brown = makeColor(57,16,8)
file=
/Users/guzdial/mediasources/barbara.jpg
picture=makePicture(file)
for x in range(70,168):
for y in range(56,190):
px=getPixel(picture,x,y)
color = getColor(px)
if distance(color,brown)<50.0:
redness=getRed(px)*1.5
setRed(px,redness)
show(picture)
return(picture)
Slide11
Same thing as last time (could raise threshold now)
Then we’re looking for a close-match on hair color, and increasing the redness
def
turnRedInRange
():
brown =
makeColor
(57,16,8)
file=
/Users/
guzdial
/
mediasources
/barbara.jpg
picture=
makePicture
(file)
for x in range(70,168):
for y in range(56,190):
px
=
getPixel
(
picture,x,y
)
color =
getColor
(
px
)
if distance(
color,brown
)<50.0:
redness=
getRed
(
px
)*1.5
setRed
(
px,redness
)
show(picture)
return(picture)
Slide12
QUESTION
Could we write
turnRedInRange
without
the nested
loops?
Slide13
Yes, but complicated IF
def turnRedInRange2():
brown =
makeColor
(57,16,8)
file=
/Users/
guzdial
/
mediasources
/barbara.jpg
picture=
makePicture
(file)
for p in
getPixels
(picture):
x =
getX
(p)
y =
getY
(p)
if x >= 70 and x < 168:
if y >=56 and y < 190:
color =
getColor
(p)
if distance(
color,brown
)<100.0:
redness=
getRed
(p)*2.0
setRed
(
p,redness
)
show(picture)
return picture
Slide14
Working on Katie’s Hair
def turnRed(): brown = makeColor(42,25,15) file="C:/ip-book/mediasources/katieFancy.jpg" picture=makePicture(file) for px in getPixels(picture): color = getColor(px) if distance(color,brown)<50.0: redness=int(getRed(px)*2) blueness=getBlue(px) greenness=getGreen(px) setColor(px,makeColor(redness,blueness,greenness)) show(picture) return(picture)
This version doubles all
close
reds. Notice the couch.
Slide15
Working on Katie’s hair, in a range
def turnRedInRange(): brown = makeColor(42,25,15) file="C:/ip-book/mediasources/katieFancy.jpg" picture=makePicture(file) for x in range(63,125): for y in range(6,76): px=getPixel(picture,x,y) color = getColor(px) if distance(color,brown)<50.0: redness=int(getRed(px)*2) blueness=getBlue(px) greenness=getGreen(px) setColor(px,makeColor(redness,blueness,greenness)) show(picture) return(picture)
Left is one we did with all
close
browns.
Right is same, but only in
rectangle
Slide16
Removing “Red Eye”
When the flash of the camera catches the eye just right (especially with light colored eyes), we get bounce back from the back of the retina.This results in “red eye”We can replace the “red” with a color of our choosing.First, we figure out where the eyes are (x,y) using MediaTools
Slide17
Removing Red Eye
def removeRedEye(pic,startX,startY,endX,endY,replacementcolor): red = makeColor(255,0,0) for x in range(startX,endX): for y in range(startY,endY): currentPixel = getPixel(pic,x,y) if (distance(red,getColor(currentPixel)) < 165): setColor(currentPixel,replacementcolor)
What we
re doing here:
Within the rectangle of pixels (startX,startY) to (endX, endY)
Find pixels close to red, then replace them with a new color
Slide18
Clicker: What would happen if we just did getPixels() here?
Why not process all pixels the same to remove redeye?
We would remove the red in her dressThe whole picture would go redThe whole picture would go blackWe would probably miss her eyes
Why use a range? Because we don
t want to replace her red dress!
Slide19
“Fixing” it: Changing red to black
removeRedEye(jenny, 109, 91, 202, 107, makeColor(0,0,0))Jenny’s eyes are actually not black—could fix thatEye are also not mono-colorA better function would handle gradations of red and replace with gradations of the right eye color
Slide20
Replacing colors using IF
We don
t have to do one-to-one changes or replacements of color
We can use
if
to decide if we want to make a change.
We could look for a
range
of colors, or
one
specific color.
We could use an operation (like multiplication) to set the new color, or we can set it to a specific value.
It all depends on the effect that we want.
Slide21
Posterizing: Reducing range of colors
Slide22
Slide23
Posterizing: How we do it
We look for a
range
of colors, then map them to a
single
color.
E.g.:
If red is less than 64, set it to 31
If red is between 63 and 128, set it to 95
If red is between 127 and 192, set it to 159
If red is between 191 and 256, set it to 223
Do similarly for green and blue
It
requires a lot of if statements
, but it
s really pretty
simple
.
The end result is that
a
bunch
of different colors, get set to a
few
colors
.
Slide24
Posterizing function
def posterize(picture): #loop through the pixels for p in getPixels(picture): #get the RGB values red = getRed(p) green = getGreen(p) blue = getBlue(p) #check and set red values if(red < 64): setRed(p, 31) if(red > 63 and red < 128): setRed(p, 95) if(red > 127 and red < 192): setRed(p, 159) if(red > 191 and red < 256): setRed(p, 223)
#check and set green values
if(green < 64):
setGreen
(p, 31)
if(green > 63 and green < 128):
setGreen
(p, 95)
if(green > 127 and green < 192):
setGreen
(p, 159)
if(green > 191 and green < 256):
setGreen
(p, 223)
#check and set blue values
if(blue < 64):
setBlue
(p, 31)
if(blue > 63 and blue < 128):
setBlue
(p, 95)
if(blue > 127 and blue < 192):
setBlue
(p, 159)
if(blue > 191 and blue < 256):
setBlue
(p, 223)
Slide25
What’s with this “#” stuff?
Any line that starts with a
#
is
ignored
by Python.
This allows you to insert
: Notes to yourself (or another programmer) that explains what
s going on here.
When programs get longer, there are lots of pieces to them, and it
s hard to figure out what each piece does.
Slide26
QUESTION
grayPosterize
(
pic
):
How do you
posterize
to produce black/white picture?
Slide27
Posterizing to b/w levels
def grayPosterize(pic): for p in getPixels(pic): r = getRed(p) g = getGreen(p) b = getBlue(p) luminance = (r+g+b)/3 if luminance < 64: setColor(p,black) if luminance >= 64: setColor(p,white)
We check luminance on each pixel.
If it
s low enough, it
s black, and
Otherwise, it
s white
Slide28
Another way to do that
We don’t have to do both tests. We can just say elseIt’s easier to write, and it’s more efficient.But it can be harder to read, especially for new programmers.
def
grayPosterize
(pic):
for p in
getPixels
(pic):
r =
getRed
(p)
g =
getGreen
(p)
b =
getBlue
(p)
luminance = (
r+g+b
)/3
if luminance < 64:
setColor
(
p,black
)
else:
setColor
(
p,white
)
Slide29
Generating sepia-toned prints
Pictures that are
sepia-toned
have a
yellowish tint
to them that we associate with
older pictures
.
It
s
not directly a matter of simply increasing the yellow
in the picture, because it
s not a one-to-one correspondence.
colors in different ranges get mapped to other colors
.
We can create such a mapping using IF
Slide30
Example of sepia-toned prints
Slide31
Slide32
Here’s how we do it
def sepiaTint(picture): #Convert image to greyscale greyScaleNew(picture) #loop through picture to tint pixels for p in getPixels(picture): red = getRed(p) blue = getBlue(p) #tint shadows if (red < 63): red = red*1.1 blue = blue*0.9
#tint
midtones
if (red > 62 and red < 192):
red = red*1.15
blue = blue*0.85
#tint highlights
if (red > 191):
red = red*1.08
if (red > 255):
red = 255
blue = blue*0.93
#set the new color values
setBlue
(p, blue)
setRed
(p, red)
Slide33
What’s going on here?
First, we
re calling
greyScaleNew
(the one with weights).
It
s perfectly okay to have one function calling another.
We then manipulate the red (increasing) and the blue (decreasing) channels to bring out more yellows and oranges.
Why are we doing the comparisons on the red? Why
not
? After
greyscale
conversion, all channels are the same!
Why these values?
Trial-and-error
: Twiddling the values until it looks the way that you want.
Slide34
What to do?
How do we clear up the degradation of scaling up?
Variety of techniques, but mostly following the same basic idea:
Use the pixels around to figure out what color a new pixel
should
be, then somehow (e.g., by averaging) compute the right color.
Different techniques look at different pixels and compute different averages in different ways.
Slide35
Blurring out the pixelation
Slide36
Blurring
When we scale up pictures (make them bigger), we get sharp lines and boxes: pixelation.Can reduce that by purposefully blurring the image.One simple algorithm: Take the pixels left, right, bottom, and top of yours. Average the colors.
Pre-Blur
Post-Blur
Slide37
Average from four sides, to compute new color
Slide38
Blurring code
def blur(filename): source=makePicture(filename) target=makePicture(filename) for x in range(0, getWidth(source)-1): for y in range(0, getHeight(source)-1): top = getPixel(source,x,y-1) left = getPixel(source,x-1,y) bottom = getPixel(source,x,y+1) right = getPixel(source,x+1,y) center = getPixel(target,x,y) newRed=(getRed(top)+ getRed(left)+ getRed(bottom)+getRed(right)+ getRed(center))/5 newGreen=(getGreen(top)+ getGreen(left)+getGreen(bottom)+getGreen(right)+getGreen(center))/5 newBlue=(getBlue(top)+ getBlue(left)+ getBlue(bottom)+getBlue(right)+ getBlue(center))/5 setColor(center, makeColor(newRed, newGreen, newBlue)) return target
We make two copies of the picture.
We read pixel colors from one, and set them in the other.
Slide39
Same thing
def blur(pic,size): for pixel in getPixels(pic): currentX = getX(pixel) currentY = getY(pixel) r = 0 g = 0 b = 0 count = 0 for x in range(currentX - size,currentX + size): for y in range(currentY - size, currentY + size): if(x<0) or (y<0) or (x >= getWidth(pic)) or (y >=getHeight(pic)): pass # Skip if we go off the edge else: r = r + getRed(getPixel(pic,x,y)) g = g + getGreen(getPixel(pic,x,y)) b = b + getBlue(getPixel(pic,x,y)) count = count + 1 newColor = makeColor(r/count,g/count,b/count) setColor(pixel,newColor)
Here
s a version with
pass
and
else
which we
ll see
later.
Slide40
Edge Detection
Blurring is averaging across pixels.
Edge detection is looking for
differences
between pixels.
We
draw lines
that our eyes see—where the luminance changes.
If the pixel changes
left-to-right
,
up and down
, then we make our pixel
black
. Else
white
.
Slide41
Edge Detection
def lineDetect(filename): #make two picture objects of image file orig = makePicture(filename) makeBw = makePicture(filename) for x in range(0,getWidth(orig)-1): for y in range(0,getHeight(orig)-1): #get this pixel, down pixel and right pixel here=getPixel(makeBw,x,y) down=getPixel(orig,x,y+1) right=getPixel(orig,x+1,y) #get luminance of this pixel, down pixel and right pixel hereL=(getRed(here)+getGreen(here)+getBlue(here))/3 downL=(getRed(down)+getGreen(down)+getBlue(down))/3 rightL=(getRed(right)+getGreen(right)+getBlue(right))/3 #detect edges if abs(hereL-downL)>10 and abs(hereL-rightL)>10: setColor(here,black) if abs(hereL-downL)<=10 and abs(hereL-rightL)<=10: setColor(here,white) return makeBw
Notice the use of absolute value (abs) here. We don
t care which is larger. We care about a large difference.
Slide42
Blending pictures
How do we get
part of one picture and part of another to blur together
, so that we see some of each?
It
s about making one a bit
transparent
.
Video cards sometimes support this transparency
in hardware, called an
alpha level
to each pixel.
We do it as a
weighted sum
If it
s
50-50
, we
take 50% of red of picture1
s pixels + 50% of red of picture2
s pixels, and so on
for green and blue, across all overlapping pixels.
Slide43
Example blended picture
Blended here
Slide44
bwidth
kwidth
X=0
X=150
X=0
Overlap = bwidth-150
barbara
katie
canvas
X:[0,150]
X:[150, 150+overlap]
X:[150+overlap, 150+
kwidth
]
Slide45
Blending code (1 of 3)
def blendPictures(): barb = makePicture(getMediaPath("barbara.jpg")) katie = makePicture(getMediaPath("Katie-smaller.jpg")) canvas = makePicture(getMediaPath("640x480.jpg")) #Copy first 150 columns of Barb sourceX=0 for targetX in range(0,150): sourceY=0 for targetY in range(0,getHeight(barb)): color = getColor(getPixel(barb,sourceX,sourceY)) setColor(getPixel(canvas,targetX,targetY),color) sourceY = sourceY + 1 sourceX = sourceX + 1
Straightforward copy of 150 column
s of Barb
s picture
Slide46
Blending code (2 of 3)
#Now, grab the rest of Barb and part of Katie # at 50% Barb and 50% Katie overlap = getWidth(barb)-150 sourceX=0 for targetX in range(150,getWidth(barb)): sourceY=0 for targetY in range(0,getHeight(katie)): bPixel = getPixel(barb,sourceX+150,sourceY) kPixel = getPixel(katie,sourceX,sourceY) newRed= 0.50*getRed(bPixel)+0.50*getRed(kPixel) newGreen=0.50*getGreen(bPixel)+0.50*getGreen(kPixel) newBlue = 0.50*getBlue(bPixel)+0.50*getBlue(kPixel) color = makeColor(newRed,newGreen,newBlue) setColor(getPixel(canvas,targetX,targetY),color) sourceY = sourceY + 1 sourceX = sourceX + 1
Here
s the trick. For each pixel, grab 50% of each red, green and blue
Slide47
Blending code (3 of 3)
# Last columns of Katie
sourceX
=overlap
for
targetX
in range(150+overlap,150+getWidth(
katie
)):
sourceY
=0
for
targetY
in range(0,getHeight(
katie
)):
color =
getColor
(
getPixel
(
katie,sourceX,sourceY
))
setColor
(
getPixel
(
canvas,targetX,targetY
),color)
sourceY
=
sourceY
+ 1
sourceX
=
sourceX
+ 1
show(canvas)
return canvas
Slide48
Background subtraction
Let
s say that you have a
picture of someone
,
and a picture of the same place
(same background) without the someone there,
could you subtract out the background and leave the picture of the person?
Maybe even change the background?
Let
s take that as our problem!
Slide49
Person (Katie) and Background
Let
s put Katie on the moon!
Slide50
Where do we start?
What we most need to do is to
figure out whether the pixel in the Person shot is the same as the in the Background shot
.
Will they be the EXACT same color? Probably not.
So, we
ll
need some way of
figuring out if two colors are close
Slide51
Remember this?
def
turnRed
():
brown = makeColor(57,16,8) file = r"C:\Documents and Settings\Mark Guzdial\My Documents\mediasources\barbara.jpg" picture=makePicture(file) for px in getPixels(picture): color = getColor(px) if distance(color,brown)<50.0: redness=getRed(px)*1.5 setRed(px,redness) show(picture) return(picture)
Original:
Slide52
Using distance
So we know that we want to ask:
if
distance(
personColor,bgColor
) >
someValue
And what do we then?
We want to
grab the color from another background (a new background) at the same point
.
Do we have any examples of doing that?
Slide53
Copying Barb to a canvas
def
copyBarb(): # Set up the source and target pictures barbf=getMediaPath("barbara.jpg") barb = makePicture(barbf) canvasf = getMediaPath("7inX95in.jpg") canvas = makePicture(canvasf) # Now, do the actual copying targetX = 1 for sourceX in range(1,getWidth(barb)): targetY = 1 for sourceY in range(1,getHeight(barb)): color = getColor(getPixel(barb,sourceX,sourceY)) setColor(getPixel(canvas,targetX,targetY), color) targetY = targetY + 1 targetX = targetX + 1 show(barb) show(canvas) return canvas
Slide54
Where we are so far:
if distance(
personColor,bgColor
) >
someValue
:
bgcolor
=
getColor
(
getPixel
(
newBg,x,y
))
setColor
(
getPixel
(
person,x,y
),
bgcolor
)
What else do we need?
We need to get all these variables set up
We need to input a person picture, a background (background without person), and a new background.
We need a loop where x and y are the right values
We have to figure out
personColor
and
bgColor
Slide55
Swap a background using background subtraction
def
swapbg
(
person,
bg
,
newbg
):
for x in range(1,getWidth(person)):
for y in range(1,getHeight(person)):
personPixel
=
getPixel
(
person,x,y
)
bgpx
=
getPixel
(
bg,x,y
)
personColor
=
getColor
(
personPixel
)
bgColor
=
getColor
(
bgpx
)
if distance(
personColor,bgColor
) >
someValue
:
bgcolor
=
getColor
(
getPixel
(
newbg,x,y
))
setColor
(
getPixel
(
person,x,y
),
bgcolor
)
Slide56
Simplifying a little, and specifying a little
def swapbg(person, bg, newbg): for x in range(1,getWidth(person)): for y in range(1,getHeight(person)): personPixel = getPixel(person,x,y) bgpx = getPixel(bg,x,y) personColor= getColor(personPixel) bgColor = getColor(bgpx) if distance(personColor,bgColor) > 10: bgcolor = getColor(getPixel(newbg,x,y)) setColor(personPixel, bgcolor)
Specifying a threshold.
Using a variable for the person pixel
Slide57
Trying it with a jungle background
Slide58
What happened?
It looks like we reversed the swap
If the distance is great, we want to KEEP the pixel.
If the distance is small (it
s basically the same thing), we want to get the NEW pixel.
Slide59
Reversing the swap
def
swapbg
(person,
bg
,
newbg
):
for x in range(1,getWidth(person)):
for y in range(1,getHeight(person)):
personPixel
=
getPixel
(
person,x,y
)
bgpx
=
getPixel
(
bg,x,y
)
personColor
=
getColor
(
personPixel
)
bgColor
=
getColor
(
bgpx
)
if distance(
personColor,bgColor
)
<
10:
bgcolor
=
getColor
(
getPixel
(
newbg,x,y
))
setColor
(
personPixel
,
bgcolor
)
Slide60
Better!
Slide61
But why isn’t it alot better?
We’ve got places where we got pixels swapped that we didn’t want to swapSee Katie’s shirt stripesWe’ve got places where we want pixels swapped, but didn’t get them swappedSee where Katie made a shadow
Slide62
How could we make it better?
What could we change in the program?
We could
change the threshold
someValue
If we
increase
it, we get
fewer
pixels matching
That won
If we
decrease
it, we get
more
pixels matching
That won
t help with the stripe
What could we change in the pictures?
Take them in better light, less shadow
Make sure that the person
isn
t wearing clothes near the background colors.
Slide63
Another way: Chromakey
Have a background of a known colorSome color that won’t be on the person you want to mask outPure green or pure blue is most often usedI used my son’s blue bedsheetThis is how the weather people seem to be in front of a map—they’re actually in front of a blue sheet.
Slide64
Chromakey recipe
def
chromakey
(
source,bg
):
# source should have something in front of blue,
bg
is the new background
for x in range(0,getWidth(source)):
for y in range(0,getHeight(source)):
p =
getPixel
(
source,x,y
)
# My definition of blue: If the redness + greenness < blueness
if (
getRed
(p) +
getGreen
(p) <
getBlue
(p)):
#Then, grab the color at the same spot from the new background
setColor
(
p,getColor
(
getPixel
(
bg,x,y
)))
Slide65
Can also do this with getPixels()
def chromakey2(
source,bg
):
# source should have something in front of blue,
#
bg
is the new background
for p in
getPixels
(source):
# My definition of blue: If the redness + greenness < blueness
if (
getRed
(p) +
getGreen
(p) <
getBlue
(p)):
#Then, grab the color at the same spot from the new background
setColor
(
p,getColor
(
getPixel
(
bg,getX
(p),
getY
(p))))
Slide66
Example results
Slide67
Just trying the obvious thing for Red
def chromakey2(
source,bg
):
# source should have something in front of red,
bg
is the new background
for p in
getPixels
(source):
if
getRed
(p) > (
getGreen
(p) +
getBlue
(p)):
#Then, grab the color at the same spot from the new background
setColor
(
p,getColor
(
getPixel
(
bg,getX
(p),
getY
(p))))
Slide68
Doesn’t always work as you expect
Slide69
Let’s try that with green
def
chromakeyGreen
(
source,bg
):
# source should have something in front of green,
bg
is the new background
for x in range(1,getWidth(source)):
for y in range(1,getHeight(source)):
p =
getPixel
(
source,x,y
)
# My definition of green: If the greenness > redness + blueness
if
getGreen
(p) >
getBlue
(p) +
getRed
(p):
#Then, grab the color at the same spot from the new background
setColor(
p,getColor
(
getPixel
(
bg,x,y
)))
Slide70
The same definition of green doesn’t work
Changes only a few pixels
Slide71
What happened?
The towel isn
t
just
green
The green of the towel has lots of blue and red in it.
Use MediaTools to figure out a new rule that makes sense.
Slide72
Tweaking Chromakey
def
chromakeyGreen
(
source,bg
):
# source should have something in front of green,
bg
is the new background
for x in range(1,getWidth(source)):
for y in range(1,getHeight(source)):
p =
getPixel
(
source,x,y
)
# My definition of green: If the greenness > redness AND blueness
if
getGreen
(p) >
getBlue
(p) and
getGreen
(p) >
getRed
(p):
#Then, grab the color at the same spot from the new background
setColor
(
p,getColor
(
getPixel
(
bg,x,y
)))
Slide73
That looks better
Slide74
Drawing on images
Sometimes you want to draw on pictures,
to add something to the pictures.
Lines
Text
Circles and boxes.
We can do that pixel by pixel, setting black and white pixels
Slide75
Drawing lines on Carolina
def
lineExample
():
img
=
makePicture
(
getMediaPath
("barbara.jpg"))
verticalLines
(
img
)
horizontalLines
(
img
)
show(
img
)
return
img
Slide76
Another Way: Does the same
def
lineExample
():
img
=
makePicture
(
getMediaPath
("barbara.jpg"))
verticalLines
(
img
)
horizontalLines
(
img
)
show(
img
)
return
img
def
horizontalLines
(
src
):
for y in range(0,getHeight(
src
),5):
for x in range(0,getWidth(
src
)):
setColor(
getPixel
(
src,x,y
),
black
)
def
verticalLines
(
src
):
for x in range(0,getWidth(
src
),5):
for y in range(0,getHeight(
src
)):
setColor(
getPixel
(
src,x,y
),
black
)
Slide77
Yes, some colors are already defined
black, white, blue, red, green, gray,
lightGray
,
darkGray
, yellow, orange, pink, magenta, and cyan
Slide78
That’s tedious
That
s slow and tedious to set every pixel you want to make lines and text, etc.
What you really want to do is to think in terms of your desired effect (think about
requirements
and
design
)
Slide79
New functions
(
pict,x,y,string
)
puts the string starting at position (
x,y
) in the picture
(picture,x1,y1,x2,y2)
draws a line from position (x1,y1) to (x2,y2)
(pict,x1,y1,w,h)
draws a black rectangle (unfilled) with the upper left hand corner of (x1,y1) and a width of w and height of h
(pict,x1,y1,w,h,color)
draws a rectangle filled with the color of your choice with the upper left hand corner of (x1,y1) and a width of w and height of h
Slide80
The mysterious red box on the beach
Slide81
Example picture
def
littlepicture
():
canvas=
makePicture
(
getMediaPath
("640x480.jpg"))
(canvas,10,50,"This is not a picture")
(canvas,10,20,300,50)
(canvas,0,200,300,500,yellow)
(canvas,10,210,290,490)
return canvas
Slide82
A thought experiment
Look at that previous page: Which has a fewer number of bytes?
The program that drew the picture
The pixels in the picture itself.
It
s a no-brainer
The program is less than 100 characters (100 bytes)
The picture is stored on disk at about 15,000 bytes
Slide83
Vector-based vs. Bitmap Graphical representations
Vector-based graphical representations
are basically
executable programs
that generate the picture on demand.
Postscript, Flash, and AutoCAD use vector-based representations
Bitmap graphical representations
(like JPEG, BMP, GIF) store individual pixels or representations of those pixels.
JPEG and GIF are actually
compressed
representations
Slide84
Vector-based representations can be smaller
Vector-based representations can be
much smaller
than bit-mapped representations
Smaller means faster transmission (Flash and Postscript)
If you want all the detail of a complex picture, no, it
s not.
Slide85
But vector-based has more value than that
Imagine that you
re editing a
picture with lines
on it.
If you edit a
bitmap image
and
extend a line
, it
s just
more bits
.
There
s no way to really realize that you
ve
extended
or
shrunk
the line.
If you edit a
vector-based image
, it
s possible to just
change the specification
Change the numbers saying where the line is
Then it
really is
the same line
That
s important when the picture drives the creation of the product, like in automatic cutting machines
Slide86
How are images compressed?
Sometimes
lossless
using techniques
like
run length encoding (RLE)
B
B
Y
Y
Y
Y
Y
Y
Y
Y
Y
B
B
We could say
9 Y
s
like this:
B
B
9 Y B
B
Lossy
compression
(like JPEG and GIF) loses detail, some of which is invisible to the eye.
Slide87
When changing the picture means changing a program…
In a
vector-based drawing
package, changing the drawing is
changing a
program
.
How could we reach in and change the actual program?
We can using
string manipulation
The program is just a string of characters
We want to manipulate those characters, in order to manipulate the program
Slide88
Example programmed graphic
If I did this right, we perceive the left half as lighter than the right halfIn reality, the end quarters are actually the same colors.
Slide89
640x480.jpg
X=0
X=100
X=200
X=300
X=400
Y
=0
Y=100
Grey
R:100
G:100
B:100
increase
greyness
By 1 for every
c
olumn from
current grey level
increase greynessBy 1 for every column from grey Level = 0 i.e.(R:0,G:0,B:0)
Grey
R:100
G:100
B:100
Slide90
Building a programmed graphic
def greyEffect(): file = getMediaPath("640x480.jpg") pic = makePicture(file) # First, 100 columns of 100-grey grey = makeColor(100,100,100) for x in range(1,100): for y in range(1,100): setColor(getPixel(pic,x,y),grey) # Second, 100 columns of increasing greyness greyLevel = 100 for x in range(100,200): grey = makeColor(greyLevel, greyLevel, greyLevel) for y in range(1,100): setColor(getPixel(pic,x,y),grey) greyLevel = greyLevel + 1
# Third, 100 colums of increasing greyness, from 0 greyLevel = 0 for x in range(200,300): grey = makeColor(greyLevel, greyLevel, greyLevel) for y in range(1,100): setColor(getPixel(pic,x,y),grey) greyLevel = greyLevel + 1 # Finally, 100 columns of 100-grey grey = makeColor(100,100,100) for x in range(300,400): for y in range(1,100): setColor(getPixel(pic,x,y),grey) return pic
Slide91
Slide92
Another Programmed Graphic
def coolpic(): canvas=makePicture(getMediaPath("640x480.jpg")) for index in range(25,1,-1): color = makeColor(index*10,index*5,index) addRectFilled(canvas,0,0,index*10,index*10,color) show(canvas) return canvas
Slide93
And another
def coolpic2(): canvas=makePicture(getMediaPath("640x480.jpg")) for index in range(25,1,-1): addRect(canvas,index,index,index*3,index*4) addRect(canvas,100+index*4,100+index*3,index*8,index*10) show(canvas) return canvas
Slide94
Why do we write programs?
Could we do this in Photoshop?
Maybe
I
m sure that you can, but you need to know how.
Could I teach you to do this in Photoshop?
Maybe
Might take a lot of demonstration
But this program is an
exact
definition of the process of generating this picture
It works for anyone who can run the program, without knowing Photoshop
Slide95
We write programs to encapsulate and communicate process
If you can do it by hand, do it.
If you need to teach someone else to do it, consider a program.
If you need to explain to lots of people how to do it, definitely use a program.
If you want lots of people to do it without having to teach them something first, definitely use a program.
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90540 (number)
90,540 (ninety thousand five hundred forty) is an even five-digits composite number following 90539 and preceding 90541. In scientific notation, it is written as 9.054 × 104. The sum of its digits is 18. It has a total of 6 prime factors and 36 positive divisors. There are 24,096 positive integers (up to 90540) that are relatively prime to 90540.
Basic properties
• Is Prime? No
• Number parity Even
• Number length 5
• Sum of Digits 18
• Digital Root 9
Name
Short name 90 thousand 540 ninety thousand five hundred forty
Notation
Scientific notation 9.054 × 104 90.54 × 103
Prime Factorization of 90540
Prime Factorization 22 × 32 × 5 × 503
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 6 Total number of prime factors rad(n) 15090 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 0 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 90,540 is 22 × 32 × 5 × 503. Since it has a total of 6 prime factors, 90,540 is a composite number.
Divisors of 90540
1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45, 60, 90, 180, 503, 1006, 1509, 2012, 2515, 3018, 4527, 5030, 6036, 7545, 9054, 10060, 15090, 18108, 22635, 30180, 45270, 90540
36 divisors
Even divisors 24 12 6 6
Total Divisors Sum of Divisors Aliquot Sum τ(n) 36 Total number of the positive divisors of n σ(n) 275184 Sum of all the positive divisors of n s(n) 184644 Sum of the proper positive divisors of n A(n) 7644 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 300.899 Returns the nth root of the product of n divisors H(n) 11.8446 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 90,540 can be divided by 36 positive divisors (out of which 24 are even, and 12 are odd). The sum of these divisors (counting 90,540) is 275,184, the average is 7,644.
Other Arithmetic Functions (n = 90540)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 24096 Total number of positive integers not greater than n that are coprime to n λ(n) 3012 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 8758 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 24,096 positive integers (less than 90,540) that are coprime with 90,540. And there are approximately 8,758 prime numbers less than or equal to 90,540.
Divisibility of 90540
m n mod m 2 3 4 5 6 7 8 9 0 0 0 0 0 2 4 0
The number 90,540 is divisible by 2, 3, 4, 5, 6 and 9.
• Arithmetic
• Refactorable
• Abundant
• Polite
• Practical
Base conversion (90540)
Base System Value
2 Binary 10110000110101100
3 Ternary 11121012100
4 Quaternary 112012230
5 Quinary 10344130
6 Senary 1535100
8 Octal 260654
10 Decimal 90540
12 Duodecimal 44490
20 Vigesimal b670
36 Base36 1xv0
Basic calculations (n = 90540)
Multiplication
n×y
n×2 181080 271620 362160 452700
Division
n÷y
n÷2 45270 30180 22635 18108
Exponentiation
ny
n2 8197491600 742200889464000 67198868532070560000 6084185556893668502400000
Nth Root
y√n
2√n 300.899 44.9035 17.3464 9.80321
90540 as geometric shapes
Circle
Diameter 181080 568880 2.57532e+10
Sphere
Volume 3.10892e+15 1.03013e+11 568880
Square
Length = n
Perimeter 362160 8.19749e+09 128043
Cube
Length = n
Surface area 4.91849e+10 7.42201e+14 156820
Equilateral Triangle
Length = n
Perimeter 271620 3.54962e+09 78409.9
Triangular Pyramid
Length = n
Surface area 1.41985e+10 8.74692e+13 73925.6
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# k-Nearest Neighbors Classification (k-NN)¶
$$k$$-NN classification algorithm infers the class for the new feature vector by computing majority vote of the $$k$$ nearest observations from the training set.
Operation Computational methods Programming Interface Training Brute-force k-d tree train(…) train_input train_result Inference Brute-force k-d tree infer(…) infer_input infer_result
## Mathematical formulation¶
### Training¶
Let $$X = \{ x_1, \ldots, x_n \}$$ be the training set of $$p$$-dimensional feature vectors, let $$Y = \{ y_1, \ldots, y_n \}$$ be the set of class labels, where $$y_i \in \{ 0, \ldots, c-1 \}$$, $$1 \leq i \leq n$$. Given $$X$$, $$Y$$ and the number of nearest neighbors $$k$$, the problem is to build a model that allows distance computation between the feature vectors in training and inference sets at the inference stage.
#### Training method: brute-force¶
The training operation produces the model that stores all the feature vectors from the initial training set $$X$$.
#### Training method: k-d tree¶
The training operation builds a $$k$$-$$d$$ tree that partitions the training set $$X$$ (for more details, see k-d Tree).
### Inference¶
Let $$X' = \{ x_1', \ldots, x_m' \}$$ be the inference set of $$p$$-dimensional feature vectors. Given $$X'$$, the model produced at the training stage and the number of nearest neighbors $$k$$, the problem is to predict the label $$y_j'$$ for each $$x_j'$$, $$1 \leq j \leq m$$, by performing the following steps:
1. Identify the set $$N(x_j') \subseteq X$$ of the $$k$$ feature vectors in the training set that are nearest to $$x_j'$$ with respect to the Euclidean distance.
2. Estimate the conditional probability for the $$l$$-th class as the fraction of vectors in $$N(x_j')$$ whose labels $$y_j$$ are equal to $$l$$:
(1)$P_{jl} = \frac{1}{| N(x_j') |} \Big| \big\{ x_r \in N(x_j') : y_r = l \big\} \Big|, \quad 1 \leq j \leq m, \; 0 \leq l < c.$
3. Predict the class that has the highest probability for the feature vector $$x_j'$$:
(2)$y_j' = \mathrm{arg}\max_{0 \leq l < c} P_{jl}, \quad 1 \leq j \leq m.$
#### Inference method: brute-force¶
Brute-force inference method determines the set $$N(x_j')$$ of the nearest feature vectors by iterating over all the pairs $$(x_j', x_i)$$ in the implementation defined order, $$1 \leq i \leq n$$, $$1 \leq j \leq m$$. The final prediction is computed according to the equations (1) and (2).
#### Inference method: k-d tree¶
K-d tree inference method traverses the $$k$$-$$d$$ tree to find feature vectors associated with a leaf node that are closest to $$x_j'$$, $$1 \leq j \leq m$$. The set $$\tilde{n}(x_j')$$ of the currently-known nearest $$k$$-th neighbors is progressively updated during tree traversal. The search algorithm limits exploration of the nodes for which the distance between the $$x_j'$$ and respective part of the feature space is not less than the distance between $$x_j'$$ and the most distant feature vector from $$\tilde{n}(x_j')$$. Once tree traversal is finished, $$\tilde{n}(x_j') \equiv N(x_j')$$. The final prediction is computed according to the equations (1) and (2).
## Usage example¶
### Training¶
knn::model<> run_training(const table& data,
const table& labels) {
const std::int64_t class_count = 10;
const std::int64_t neighbor_count = 5;
const auto knn_desc = knn::descriptor<float>{class_count, neighbor_count};
const auto result = train(knn_desc, data, labels);
return result.get_model();
}
### Inference¶
table run_inference(const knn::model<>& model,
const table& new_data) {
const std::int64_t class_count = 10;
const std::int64_t neighbor_count = 5;
const auto knn_desc = knn::descriptor<float>{class_count, neighbor_count};
const auto result = infer(knn_desc, model, new_data);
print_table("labels", result.get_labels());
}
## Examples¶
Batch Processing:
Batch Processing:
Batch Processing:
## Programming Interface¶
All types and functions in this section are declared in the oneapi::dal::knn namespace and be available via inclusion of the oneapi/dal/algo/knn.hpp header file.
### Descriptor¶
template<typename Float = detail::descriptor_base<>::float_t, typename Method = detail::descriptor_base<>::method_t, typename Task = detail::descriptor_base<>::task_t>
class descriptor
Template Parameters
• Float – The floating-point type that the algorithm uses for intermediate computations. Can be float or double.
• Method – Tag-type that specifies an implementation of algorithm. Can be method::v1::brute_force or method::v1::kd_tree.
• Task – Tag-type that specifies type of the problem to solve. Can be task::v1::classification.
Constructors
descriptor(std::int64_t class_count, std::int64_t neighbor_count)
Creates a new instance of the class with the given class_count and neighbor_count property values.
Public Methods
auto &set_class_count(std::int64_t value)
auto &set_neighbor_count(std::int64_t value)
#### Method tags¶
struct brute_force
Tag-type that denotes brute-force computational method.
struct kd_tree
Tag-type that denotes k-d tree computational method.
using by_default = brute_force
Alias tag-type for brute-force computational method.
struct classification
Tag-type that parameterizes entities used for solving classification problem.
using by_default = classification
Alias tag-type for classification task.
### Model¶
template<typename Task = task::by_default>
class model
Template Parameters
Task – Tag-type that specifies type of the problem to solve. Can be task::v1::classification.
Constructors
model()
Creates a new instance of the class with the default property values.
### Training train(...)¶
#### Input¶
template<typename Task = task::by_default>
class train_input
Template Parameters
Task – Tag-type that specifies type of the problem to solve. Can be task::v1::classification.
Constructors
train_input(const table &data, const table &labels)
Creates a new instance of the class with the given data and labels property values.
Properties
const table &data = table{}
The training set X.
Getter & Setter
const table & get_data() const
auto & set_data(const table &data)
const table &labels = table{}
Vector of labels y for the training set X.
Getter & Setter
const table & get_labels() const
auto & set_labels(const table &labels)
#### Result¶
template<typename Task = task::by_default>
class train_result
Template Parameters
Task – Tag-type that specifies type of the problem to solve. Can be task::v1::classification.
Constructors
train_result()
Creates a new instance of the class with the default property values.
Properties
const model<Task> &model = model<Task>{}
The trained k-NN model.
Getter & Setter
const model< Task > & get_model() const
auto & set_model(const model< Task > &value)
#### Operation¶
template<typename Descriptor>
knn::train_result train(const Descriptor &desc, const knn::train_input &input)
Template Parameters
• desc – k-NN algorithm descriptor knn::desc
• input – Input data for the training operation
Preconditions
input.data.has_data == true
input.labels.has_data == true
input.data.row_count == input.labels.row_count
input.labels.column_count == 1
input.labels[i] >= 0
input.labels[i] < desc.class_count
### Inference infer(...)¶
#### Input¶
template<typename Task = task::by_default>
class infer_input
Template Parameters
Task – Tag-type that specifies type of the problem to solve. Can be task::v1::classification.
Constructors
infer_input(const table &data, const model<Task> &model)
Creates a new instance of the class with the given model and data property values.
Properties
const table &data = table{}
The dataset for inference $$X'$$.
Getter & Setter
const table & get_data() const
auto & set_data(const table &data)
const model<Task> &model = model<Task>{}
The trained k-NN model.
Getter & Setter
const model< Task > & get_model() const
auto & set_model(const model< Task > &m)
#### Result¶
template<typename Task = task::by_default>
class infer_result
Template Parameters
Task – Tag-type that specifies type of the problem to solve. Can be task::v1::classification.
Constructors
infer_result()
Creates a new instance of the class with the default property values.
Properties
const table &labels = table{}
The predicted labels.
Getter & Setter
const table & get_labels() const
auto & set_labels(const table &value)
#### Operation¶
template<typename Descriptor>
knn::infer_result infer(const Descriptor &desc, const knn::infer_input &input)
Template Parameters
• desc – k-NN algorithm descriptor knn::desc
• input – Input data for the inference operation
Preconditions
input.data.has_data == true
Postconditions
result.labels.row_count == input.data.row_count
result.labels.column_count == 1
result.labels[i] >= 0
result.labels[i] < desc.class_count
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Sum of series with triangular numbers
Can you please tell me the sum of the seires
$\frac {1}{10} + \frac {3}{100} + \frac {6}{1000} + \frac {10}{10000} + \frac {15}{100000} + \cdots$
where the numerator is the series of triangular numbers?
Is there a simple way to find the sum?
Thank you.
-
Fixed $\LaTeX$ and problem statement. – Ahaan S. Rungta Jul 16 '14 at 17:39
@AhaanRungta you do not need to comment that here. – picakhu Jul 16 '14 at 21:44
$$S={1\over10}+{3\over100}+{6\over1000}+{10\over10000}+\cdots$$ $${S\over10}={1\over100}+{3\over1000}+{6\over10000}+\cdots$$ Subtracting, $${9S\over10}={1\over10}+{2\over100}+{3\over1000}+{4\over10000}+\cdots$$ Now do the same thing again, that is, divide by $10$ and subtract, to get $${81S\over100}={1\over10}+{1\over100}+{1\over1000}+\cdots={1\over9}$$
-
Thanks for the fast response. Your manipulation is quite nice and makes the question too easy to solve. – Mustafa Saad Jul 16 '14 at 13:46
Indeed, WolframAlpha agrees – Ryan Jul 16 '14 at 15:50
Your expression is equal to $g(1/10)$, where $$g(x)=\frac{x}{2}\left((2)(1)+(3)(2)x+(4)(3)x^2+(5)(4)x^3+\cdots\right)$$
Take the power series $1+x+x^2+x^3+\cdots$ for $\frac{1}{1-x}$ and differentiate twice. We get $(2)(1)+(3)(2)x+(4)(3)x^2+\cdots$ if we do it term by term, and $\frac{2!}{(1-x)^3}$ if we do it the usual way. Thus $$g(x)=\frac{x}{2}\cdot\frac{2!}{(1-x)^3}$$ (when $|x|\lt 1$, and in particular at $x=1/10$).
Remark: The idea generalizes. The $n$-th triangular nunber is $\binom{n}{2}$. The same idea can be used to calculate $\sum \binom{n}{k}x^n$ for $|x|\lt 1$ and fixed positive integer $k$.
-
Good answer, but one mistake. Differentiating $\frac1{1-x}$ twice gives you $\frac2{(1-x)^3}$ or $g(x)=\frac x{(1-x)^3}$. Then plugging in $\frac1{10}$ matches Gerry's answer. – Mike Jul 16 '14 at 13:49
Thanks, fixed! I tend to forget about constants. – André Nicolas Jul 16 '14 at 13:58
It was more than just the constant. I think you only differentiated once. The exponent is still wrong. – Mike Jul 16 '14 at 19:57
Thank you, exponent also fixed. Bad day! – André Nicolas Jul 16 '14 at 20:16
I thought I might add another derivation (devised by me). This one is long and involves dissecting the sequence into its simplest terms.
$1/10 + 3/100 + 6/1000 + \ldots$
$= 1/10 + (1+2)/100 + (1+2+3)/1000 + \ldots$ (from the definition of triangular numbers.)
$= 1/10 + 1/100 + 2/100 + 1/1000 + 2/1000 + 3/1000 + \ldots$
(by grouping terms with similar numerator together)
$= (1/10 + 1/100 + 1/1000 + \ldots) + (2/100 + 2/1000 + \ldots) + (3/1000 + \ldots) + \ldots$ $= 1/9 + 2/90 + 3/900 + \ldots$
($1/9$ is a common factor)
$= 1/9 [ 1 + 2/10 + 3/100 + \ldots]$ $= 1/9 [ 1 + 1/10 + 1/10 + 1/100 + 1/100 + 1/100 + \ldots ]$
(after rearranging the terms)
$= 1/9 [ 1 + (1/10 + 1/100 + 1/100 + \ldots) + (1/10 + 1/100 + 1/100 + \ldots) + (1/100 + \ldots) + \ldots ]$ $= 1/9 [ 1 + 1/9 + (1/9 + 1/90 + 1/900 + 1/900 + \ldots) ]$
(the terms between the parentheses represent a geometric series whose sum is $10/81$)
$= 1/9 [ 1 + 1/9 + 10/81 ]$ $= 1/9 \times 100/81$ $= 100/729$
-
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# Uniqueness of projective covers
I want to show that if projective covers exist then they are unique up to isomorphism.
More precisely let $f: P \rightarrow M$ and $g: Q \rightarrow M$ be projective covers of an $R$-module $M$.
Using the fact that $g$ is surjective and that $P$ is projective we can find an $R$-map $h: P \rightarrow Q$ such that $g \circ h=f$.
Note then that $\operatorname{Im}(h)+\operatorname{ker}(g)=Q$. Since $ker(g)$ is superfluous this implies that $Q=Im(h)$ so that $h$ is surjective.
But how do we conclude that $h$ is injective?
-
I'm not sure what category you're working in, but I'm pretty certain it's not true if you're working in the category of $R$-modules. For example, take $Q = P \oplus P$. – Zhen Lin Apr 19 '12 at 7:20
@zhen, if P is a cover then that Q won't be. – Mariano Suárez-Alvarez Apr 19 '12 at 8:34
Hm? Why would $g$ be a monomorphism? (if it were, since it is also surjective, $M$ would be projective...) – Mariano Suárez-Alvarez Apr 19 '12 at 18:25
Repeat your argument. Using the fact that $f$ is surjective and $Q$ is projective there exists a map $h'$ such that $f \circ h' = g$. As above, you conclude that $h'$ is surjective, and you have $g \circ h \circ h' =g$.
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» Home » Resources & support » FAQs » Producing adjusted means after ANOVA
Note: This FAQ is for Stata 10 and older versions of Stata.
Stata 11 introduced the margins command, which superseded adjust.
## How can I produce adjusted means after ANOVA?
Title Producing adjusted means after ANOVA Author Kenneth Higbee, StataCorp
### Question:
Someone posed the following question:
I am running some simple ANOVAs and wanted also to produce the adjusted means. The command is a 3-way ANOVA with a single 2-way interaction. All the predictors are dichotomous (0/1) variables. There were a few problems with the output.
First, when I run
. anova opportu2 volsex frcsex volsex*frcsex q3
and then
. adjust q3 if e(sample), by(volsex frcsex) se ci
I get a table with all cells missing.
I then decided to run
. anova opportu2 volsex frcsex volsex*frcsex q3, cont(q3)
. adjust q3 if e(sample), by(volsex frcsex) se ci
This did work! What is going on?
The result you show, comparing when q3 was used as a categorical variable and when it was specified to be a continuous variable in the ANOVA, does not surprise me. Let me explain why using the auto data.
. sysuse auto
(1978 Automobile Data)
. gen z = trunk < 14
. anova wei rep for rep*for z
Number of obs = 69 R-squared = 0.6079
Root MSE = 528.54 Adj R-squared = 0.5556
Source | Partial SS df MS F Prob > F
--------------+----------------------------------------------------
Model | 25984464.5 8 3248058.07 11.63 0.0000
|
rep78 | 1524294.09 4 381073.522 1.36 0.2571
foreign | 3521325.49 1 3521325.49 12.61 0.0008
rep78*foreign | 2300624.62 2 1150312.31 4.12 0.0211
z | 3248513.88 1 3248513.88 11.63 0.0012
|
Residual | 16761251.4 60 279354.19
--------------+----------------------------------------------------
Total | 42745715.9 68 628613.47
I used the 0/1 variable z as a categorical variable in the anova above.
Now, just like you experienced, when I use adjust to adjust to the MEAN of z, I get nothing useful.
. adjust z if e(sample), by(for rep) se ci
----------------------------------------------------------------------------
Dependent variable: weight Command: anova
Covariate set to mean: z = .43478259
----------------------------------------------------------------------------
(8 missing values generated)
(8 missing values generated)
----------------------------------------
| Repair Record 1978
Car type | 1 2 3 4 5
----------+-----------------------------
Domestic |
|
|
|
Foreign |
|
|
----------------------------------------
Key: Linear Prediction
(Standard Error)
[95% Confidence Interval]
Notice the messages about missing values generated.
Think about the parameterization used by ANOVA models. Categorical variables enter the design matrix as a set of indicator (also called dummy) variables. For instance, rep78, which has five levels, becomes 5 columns in the ANOVA design matrix (one of these levels will later be dropped in the estimation process since there are only 4 degrees of freedom for the 5 levels. foreign becomes 2 columns in the design matrix, and one of them will be dropped later in the estimation process. The same is true for the z variable.
Here is a look at the underlying regression for the ANOVA above:
. regress
Source | SS df MS Number of obs = 69
-------------+------------------------------ F( 8, 60) = 11.63
Model | 25984464.5 8 3248058.07 Prob > F = 0.0000
Residual | 16761251.4 60 279354.19 R-squared = 0.6079
-------------+------------------------------ Adj R-squared = 0.5556
Total | 42745715.9 68 628613.47 Root MSE = 528.54
------------------------------------------------------------------------------
weight Coef. Std. Err. t P>|t| [95% Conf. Interval]
------------------------------------------------------------------------------
_cons 2182.261 187.7289 11.62 0.000 1806.748 2557.775
rep78
1 1140 528.5397 2.16 0.035 82.76324 2197.237
2 1020.691 431.9311 2.36 0.021 156.7003 1884.682
3 -338.0654 352.7323 -0.96 0.342 -1043.635 367.5043
4 -85.01959 251.2557 -0.34 0.736 -587.6057 417.5666
5 (dropped)
foreign
1 -222.2614 418.2335 -0.53 0.597 -1058.853 614.3301
2 (dropped)
z
1 497.4118 145.8651 3.41 0.001 205.6382 789.1855
2 (dropped)
rep78*foreign
1 1 (dropped)
2 1 (dropped)
3 1 1470.257 536.9423 2.74 0.008 396.2125 2544.301
3 2 (dropped)
4 1 1270.366 504.0553 2.52 0.014 262.1051 2278.627
4 2 (dropped)
5 1 (dropped)
5 2 (dropped)
------------------------------------------------------------------------------
predict produces missing values when asked to produce predictions for these 10 points. It does this because since z entered the ANOVA model as a categorical variable with 0 and 1 as the valid values of z, having z = .43478259 doesn’t correspond to either 0 or 1.
predict would, for instance, also produce a missing value if you asked for a prediction when rep78 = 3.257, rep78 = 12, etc. After anova, the only valid values for categorical variables for predict are those values present in the ANOVA.
Now watch what happens when I do the following adjust:
. adjust z=0 if e(sample), by(rep for) se ci
----------------------------------------------------------------------------
Dependent variable: weight Command: anova
Covariate set to value: z = 0
----------------------------------------------------------------------------
------------------------------------------------
Repair |
Record | Car type
1978 | Domestic Foreign
----------+-------------------------------------
1 | 3597.41
| (401.19)
| [2794.91,4399.91]
|
2 | 3478.1
| (190.392)
| [3097.26,3858.94]
|
3 | 3589.6 2341.61
| (110.519) (320.272)
| [3368.53,3810.67] [1700.97,2982.25]
|
4 | 3642.76 2594.65
| (179.137) (209.548)
| [3284.43,4001.09] [2175.5,3013.81]
|
5 | 2457.41 2679.67
| (401.19) (193.923)
| [1654.91,3259.91] [2291.77,3067.58]
------------------------------------------------
Key: Linear Prediction
(Standard Error)
[95% Confidence Interval]
The answers I got are the adjusted predictions when z is 0. I could also get predictions when z is 1 with
. adjust z=1 if e(sample), by(rep for) se ci
output omitted
If I ask for predictions at any other values of z besides 0 or 1, I will get missing values from the predictions.
Now, instead of having z enter the anova model as a categorical variable, you instead send it in as a continuous variable (a covariate in an ANCOVA).
. anova wei rep for rep*for z, cont(z)
Number of obs = 69 R-squared = 0.6079
Root MSE = 528.54 Adj R-squared = 0.5556
Source | Partial SS df MS F Prob > F
--------------+----------------------------------------------------
Model | 25984464.5 8 3248058.07 11.63 0.0000
|
rep78 | 1524294.09 4 381073.522 1.36 0.2571
foreign | 3521325.49 1 3521325.49 12.61 0.0008
rep78*foreign | 2300624.62 2 1150312.31 4.12 0.0211
z | 3248513.88 1 3248513.88 11.63 0.0012
|
Residual | 16761251.4 60 279354.19
--------------+----------------------------------------------------
Total | 42745715.9 68 628613.47
The ANOVA table looks the same, but the underlying representation is different. There is one less column in the design matrix. z only has one column instead of two (corresponding to z=0 and z=1). Instead, since z is “continuous”, the ANOVA is happy with whatever values might happen to be in z. Since z had only two levels (0 and 1), the resulting ANOVA table is identical. This would not be true if z had 3 or more levels. Then the first anova would have had more degrees of freedom for the z, while the second anova would continue to have only 1 degree of freedom.
Here is the underlying regression for the ANOVA above:
. regress
Source | SS df MS Number of obs = 69
-------------+------------------------------ F( 8, 60) = 11.63
Model | 25984464.5 8 3248058.07 Prob > F = 0.0000
Residual | 16761251.4 60 279354.19 R-squared = 0.6079
-------------+------------------------------ Adj R-squared = 0.5556
Total | 42745715.9 68 628613.47 Root MSE = 528.54
------------------------------------------------------------------------------
weight Coef. Std. Err. t P>|t| [95% Conf. Interval]
------------------------------------------------------------------------------
_cons 2679.673 193.9232 13.82 0.000 2291.769 3067.577
rep78
1 1140 528.5397 2.16 0.035 82.76324 2197.237
2 1020.691 431.9311 2.36 0.021 156.7003 1884.682
3 -338.0654 352.7323 -0.96 0.342 -1043.635 367.5043
4 -85.01959 251.2557 -0.34 0.736 -587.6057 417.5666
5 (dropped)
foreign
1 -222.2614 418.2335 -0.53 0.597 -1058.853 614.3301
2 (dropped)
z -497.4118 145.8651 -3.41 0.001 -789.1855 -205.6382
rep78*foreign
1 1 (dropped)
2 1 (dropped)
3 1 1470.257 536.9423 2.74 0.008 396.2125 2544.301
3 2 (dropped)
4 1 1270.366 504.0553 2.52 0.014 262.1051 2278.627
4 2 (dropped)
5 1 (dropped)
5 2 (dropped)
------------------------------------------------------------------------------
Unlike the first anova, the z here has only one row in the output. The underlying representation within anova is different.
Here it makes sense for predict after anova to ask for predictions when z is .43478259. As far as anova and predict are concerned, the z variable is continuous and can take on any value.
. adjust z if e(sample), by(rep for) se ci
----------------------------------------------------------------------------
Dependent variable: weight Command: anova
Covariate set to mean: z = .43478259
----------------------------------------------------------------------------
------------------------------------------------
Repair |
Record | Car type
1978 | Domestic Foreign
----------+-------------------------------------
1 | 3381.15
| (382.72)
| [2615.59,4146.7]
|
2 | 3261.84
| (188.801)
| [2884.18,3639.49]
|
3 | 3373.34 2125.34
| (103.704) (307.021)
| [3165.9,3580.78] [1511.21,2739.48]
|
4 | 3426.49 2378.39
| (178.887) (183.146)
| [3068.66,3784.32] [2012.04,2744.73]
|
5 | 2241.15 2463.41
| (382.72) (177.058)
| [1475.59,3006.7] [2109.24,2817.58]
------------------------------------------------
Key: Linear Prediction
(Standard Error)
[95% Confidence Interval]
| 3,429
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