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Guest Dec 12, 2014
#3
+91479
+10
Welcome to web2.0calc forum Catboy13
Umm - Interesting question.
$$\begin{array}{rll} 3x^2-2x+5&=&y\\\\ 3x^2-2x&=&y-5\\\\ x^2-\frac{2}{3}x&=&\frac{y-5}{3}\\\\ x^2-\frac{2}{3}x+\left(\frac{2}{6}\right)^2&=&\frac{y-5}{3}+\left(\frac{2}{6}\right)^2\\\\ x^2-\frac{2}{3}x+\left(\frac{1}{3}\right)^2&=&\frac{y-5}{3}+\left(\frac{1}{3}\right)^2\\\\ x^2-\frac{2}{3}x+\left(\frac{1}{3}\right)^2&=&\frac{3(y-5)}{9}+\frac{1}{9}\\\\ \left(x-\frac{1}{3}\right)^2&=&\frac{3y-15+1}{9}\\\\ \left(x-\frac{1}{3}\right)^2&=&\frac{3y-14}{9}\\\\ x-\frac{1}{3}&=&\pm\sqrt{\frac{3y-14}{9}}\\\\ x&=&\frac{1}{3}\pm\sqrt{\frac{3y-14}{9}}\\\\ x&=&\frac{1\pm\sqrt{3y-14}}{3}}\\ \end{array}$$
Melody Dec 13, 2014
#4
+91479
0
I have added this to the Sticky Topic - "Great Answers to Learn From"
Melody Dec 14, 2014
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# Calculation puzzle 010
Find the missing number in the sequence.
2 9 28 65 ? 217
Source: This question is taken from YTU YOS 2018 exam. I have mentioned them in other posts.
It's simply just
$$n^3 + 1$$
So
$$1^3 + 1 = 1 + 1 = 2$$
$$2^3 + 1 = 8 + 1 = 9$$
$$3^3 + 1 = 27 + 1 = 28$$
$$4^3 + 1 = 64 + 1 = 65$$
$$5^3 + 1 = 125 + 1 = 126$$
$$6^3 + 1 = 216 + 1 = 217$$
Therefore the missing number is
$$126$$
• Gz mate nice one Jul 27, 2020 at 12:51
The answer is 126 because of the pattern 1cubed +1, 2cubed+1, 3 cubed +1,4cubed +1, 5cubed+1,6cubed+1
• By the way I did not copy this answer Jul 27, 2020 at 12:47
• There was no answer as I was typing it Jul 27, 2020 at 12:47
• Don't worry, people will be able to see that you didnt from the timestamp, as you can see we both answered at almost the same time Jul 27, 2020 at 12:52
• Yesterday I answered the question 3 minutes before another guy and he still got the credit😂 I don't have good luck here Jul 27, 2020 at 12:54
• Sometimes a later answer will be accepted if the OP thinks that it's more correct or explains it better, it's not always first to get it is accepted :) Jul 27, 2020 at 12:55 | {
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# If $n$ is a positive integer, does $n^3-1$ always have a prime factor that's 1 more than a multiple of 3?
It appears to be true for all $n$ from 1 to 100. Can anyone help me find a proof or a counterexample?
If it's true, my guess is that it follows from known classical results, but I'm having trouble seeing it.
In some cases, the prime factors congruent to 1 mod 3 are relatively large, so it's not as simple as "they're all divisible by 7" or anything like that.
It's interesting if one can prove that an integer of a certain form must have a prime factor of a certain form without necessarily being able to find it explicitly.
EDITED TO ADD: It appears that there might be more going on here!
$n^2-1$ usually has a prime factor congruent to 1 mod 2 (not if n=3, though!)
$n^3-1$ always has a prime factor congruent to 1 mod 3
$n^4-1$ always has a prime factor congruent to 1 mod 4
$n^5-1$ appears to always have a prime factor congruent to 1 mod 5.
Regarding $n^2-1$: If $n>3$, then $n^2-1=(n-1)(n+1)$ is a product of two numbers that differ by 2, which cannot both be powers of 2 if they are bigger than 2 and 4. Therefore at least one of $n-1,n+1$ is divisible by an odd prime.
Regarding $n^4-1$: If $n>1$, we factor $n^4-1$ as $(n+1)(n-1)(n^2+1)$. We claim that in fact, every prime factor of $n^2+1$ is either 2 or is congruent to 1 mod 4. If $p$ is an odd prime that divides $n^2+1$, then $-1$ is a square mod $p$, but the odd primes for which $-1$ is a square mod $p$ are precisely the primes congruent to 1 mod 4. It remains just to show that $n^2+1$ cannot be a power of 2. If $n$ is even this is obvious, and if $n=2k+1$ is odd, then $n^2+1=(2k+1)^2+1=4k^2+4k+2$ is 2 more than a multiple of 4.
Regarding $n^5-1$, I don't have a proof, but based on experimenting with a few dozen numbers, I conjecture that in fact, every prime factor of $n^4+n^3+n^2+n+1$ is either 5 or is 1 more than a multiple of 5. | {
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• From the factorization you can get a factor that's $\equiv 1 \mod 3$, but not necessarily a prime one – MCT Apr 1 '16 at 15:38
• Ran a code for the for all such numbers less than $10^7$. Seems ok. I am addin a list of the first few numbers and their prime factors. (7, 7) (26, 13) (63, 7) (124, 31) (215, 43) (342, 19) (511, 7) (728, 7) (999, 37) (1330, 7) (1727, 157) (2196, 61) (2743, 13) (3374, 7) (4095, 7) (4912, 307) (5831, 7) (6858, 127) (7999, 19) (9260, 463) (10647, 7) (12166, 7) (13823, 601) (15624, 7) (17575, 19) (19682, 13) (21951, 271) (24388, 7) (26999, 7) (29790, 331) (32767, 7) (35936, 1123) (39303, 397) (42874, 13) (46655, 7) (50652, 7) (54871, 37) (59318, 7) (63999, 13) (68920, 1723) (74087, 13) (79506, 7) – Banach Tarski Apr 1 '16 at 15:49
• Ran the code again. This statement seems to be true for all such numbers smaller than $2 * 10^8$ – Banach Tarski Apr 1 '16 at 15:56
• @idmercer: For prime powers $p>2$, you can generalize it neatly. Define $$F(n) = \frac{n^p-1}{n-1}$$ Conjecture: "If $F(n)$ is not divisible by $p$, then every odd prime factor of $F(n)$ is $1\mod 2p$." I'm not sure this is a proven result though, because otherwise the guys who answered would have used it. Maybe you can ask the general case for primes separately. – Tito Piezas III Apr 1 '16 at 22:01
The case $n=1$ is uninteresting, so let $n\gt 1$. We show that $n^2+n+1$ has a prime factor congruent to $1$ modulo $3$, by showing that $4(n^2+n+1)$ has such a prime factor.
Note that $n^2+n+1$ is odd. We first show that $3$ is not the only odd prime that divides $n^2+n+1$. For suppose to the contrary that $(2n+1)^2+3=4\cdot 3^k$ where $k\gt 1$. Then $2n+1$ is divisible by $3$, so $(2n+1)^2+3\equiv 3\pmod{9}$, contradiction.
Now let $p\gt 3$ be a prime divisor of $(2n+1)^2+3$. Then $-3$ is a quadratic residue of $p$. A straightforward quadratic reciprocity argument shows that $p$ cannot be congruent to $2$ modulo $3$. | {
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Remark: By considering $n$ of the form $q!$ for possibly large $q$, we can use the above result to show that there are infinitely many primes of the form $6k+1$.
• Is there a general result for the odd prime factors of $\displaystyle F(n)=\frac{n^p-1}{n-1}$ for prime $p$? (The OP was just focusing on the case $p=3$.) In other words, is every odd prime factor of $F(n)$ either $p$ or $1\mod p$? – Tito Piezas III Apr 2 '16 at 18:51
• Analysis is pretty simple (Fermat's Theorem) for $n$ and $p$ such that $n$ is not congruent to $1$ mod $p$. But there can be funny prime divisors otherwise. There should be a complete easy answer in general, but I looked and am missing something. That's why the appeal to reciprocity in the answer. – André Nicolas Apr 2 '16 at 18:54
• Yes, Fermat's little theorem occurred to me as well. I may ask the general prime case in the forum. Do you think you'll figure out the answer to the general case? :) – Tito Piezas III Apr 2 '16 at 18:57
• @TitoPiezasIII: Someone will. Conceivably me, but probably not. Would likely not be in any case, but also today is a massive cooking day. – André Nicolas Apr 2 '16 at 19:00
• Ok, Bon appetit. (P.S. The question is here.) – Tito Piezas III Apr 2 '16 at 19:20
For $p\equiv 1\pmod 3$, there exist three distinct solutions of $x^3\equiv 1\pmod p$, and if $n$ is congruent modulo $p$ to such $x$ then $n^3-1$ is a multiple of $p$. Heuristically, this solves the problem affirmatively for a fraction $\frac 3p$ of all possible $n$. | {
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Let's make this a bit less hand-wavy: Let $p_1=7, p_2=13, p_3=19,\ldots$ denote the sequence of prime $\equiv 1\pmod 3$. Then of the $M:=p_1p_2\cdots p_m$ residue classes modulo $M$, there are only $(1-\frac3{p_1})(1-\frac3{p_2})\cdots(1-\frac3{p_m})M=(p_1-3)(p_2-3)\cdots(p_m-3)$ residue classes $x$ for which $n\equiv x\pmod M$ does not imply that $n^3-1$ has a prime factor $\in\{p_1,\ldots,p_m\}$. As we let $m\to \infty$, the product $(1-\frac3{p_1})(1-\frac3{p_2})\cdots(1-\frac3{p_m})$ can be shown to tend to $0$. So at least the density of $n$ that do not work must be zero. | {
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# Find a simple formula for $\binom{n}{0}\binom{n}{1}+\binom{n}{1}\binom{n}{2}+…+\binom{n}{n-1}\binom{n}{n}$
$$\binom{n}{0}\binom{n}{1}+\binom{n}{1}\binom{n}{2}+...+\binom{n}{n-1}\binom{n}{n}$$
All I could think of so far is to turn this expression into a sum. But that does not necessarily simplify the expression. Please, I need your help.
First note that $\dbinom{n}k = \dbinom{n}{n-k}$. Hence, your sum can be written as $$\sum_{k=0}^{n-1} \dbinom{n}k \dbinom{n}{k+1} = \sum_{k=0}^{n-1} \dbinom{n}k \dbinom{n}{n-k-1}$$ Now consider a bag with $n$ red balls and $n$ blue balls. We want to choose a total of $n-1$ bals. The total number of ways of doing this is given by $$\dbinom{2n}{n-1} \tag{\star}$$ However, we can also count this differently. Any choice of $n-1$ balls will involve choosing $k$ blue balls and $n-k-1$ red balls. Hence, the number of ways of choose $n-1$ balls with $k$ blue balls is $$\color{blue}{\dbinom{n}k} \color{red}{\dbinom{n}{n-1-k}}$$ Now to count all possible ways of choosing $n-1$ balls, we need to let our $k$ run from $0$ to $n-1$. Hence, the total number of ways is $$\sum_{k=0}^{n-1} \color{blue}{\dbinom{n}k} \color{red}{\dbinom{n}{n-k-1}} \tag{\dagger}$$ Now since $(\star)$ and $(\dagger)$ count the same thing, they must be equal and hence, we get that $$\sum_{k=0}^{n-1} \color{blue}{\dbinom{n}k} \color{red}{\dbinom{n}{n-k-1}} = \dbinom{2n}{n-1}$$
EDIT As Brian points out in the comments, the above is a special case of the more general Vandermonde's identity. $$\sum_{k=0}^r \dbinom{m}k \dbinom{n}{r-k} = \dbinom{m+n}r$$ The proof for this is essentially the same as above. Consider a bag with $m$ blue balls and $n$ red balls and count the number of ways of choosing $r$ balls from these $m+n$ balls in two different ways as discussed above. | {
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• Love the formatting! +1 – Macavity Apr 4 '13 at 18:32
• (+1) Nicely explained. You might want to mention that it’s a case of Vandermonde’s identity, which has essentially the same proof. – Brian M. Scott Apr 4 '13 at 18:43
• Thanks so much!!!! Your explanation was great!!!! – Dome Apr 4 '13 at 18:49
We have $$\sum_{k=0}^{n-1}\dbinom{n}{k}\dbinom{n}{k+1}=\sum_{k=0}^{n}\dbinom{n}{k}\dbinom{n}{k+1}=\sum_{k=0}^{n}\dbinom{n}{k}\dbinom{n}{n+1-k}=\color{red}{\dbinom{2n}{n+1}}$$ by the Chu-Vandermonde identity and since $\dbinom{n}{n}\dbinom{n}{n+1}=0$. Another way is to use the integral representation of the binomial coefficient $$\dbinom{n}{k}=\frac{1}{2\pi i}\oint_{\left|z\right|=1}\frac{\left(1+z\right)^{n}}{z^{k+1}}dz$$ and get $$\sum_{k=0}^{n}\dbinom{n}{k}\dbinom{n}{k+1}=\frac{1}{2\pi i}\oint_{\left|z\right|=1}\frac{\left(1+z\right)^{n}}{z^{2}}\sum_{k=0}^{n}\dbinom{n}{k}z^{-k}dz$$ $$=\frac{1}{2\pi i}\oint_{\left|z\right|=1}\frac{\left(1+z\right)^{n}}{z^{2}}\left(1+\frac{1}{z}\right)^{n}dz=\frac{1}{2\pi i}\oint_{\left|z\right|=1}\frac{\left(1+z\right)^{2n}}{z^{n+2}}dz=\color{blue}{\dbinom{2n}{n+1}}.$$
There is a simple combinatorial interpretation. Assume that you have a parliament with $n$ politicians in the left wing and $n$ politicians in the right wing. If you want to select a committee made by $n-1$ politicians, you obviously have $\binom{2n}{n-1}=\binom{2n}{n+1}$ ways for doing it. On the other hand, by classifying the possible committees according to the number of politicians of the left wing in them, you have: $$\binom{2n}{n-1}=\sum_{l=0}^{n-1}\binom{n}{l}\binom{n}{n-1-l} = \sum_{l=0}^{n-1}\binom{n}{l}\binom{n}{l+1}$$ as wanted.
Hint: it's the coefficient of $T$ in the binomial expansion of $(1+T)^n(1+T^{-1})^n$, which is equivalent to saying that it's the coefficient of $T^{n+1}$ in the expansion of $(1+T)^n(1+T^{-1})^nT^n=(1+T)^{2n}$.
Note that $\ds{{n \choose k} + {n \choose k + 1} = {n + 1 \choose k + 1}}$ | {
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Note that $\ds{{n \choose k} + {n \choose k + 1} = {n + 1 \choose k + 1}}$
\begin{align} \sum_{k = 0}^{n - 1}{n \choose k}{n \choose k + 1} & = \sum_{k = 0}^{n - 1} {\bracks{{n \choose k} + {n \choose k + 1}}^{\,2} - {n \choose k}^{2} - {n \choose k + 1}^{2}\over 2} \\[5mm] & = {1 \over 2}\sum_{k = 0}^{n - 1}{n + 1 \choose k + 1}^{2} - {1 \over 2}\sum_{k = 0}^{n - 1}{n \choose k}^{2} - {1 \over 2}\sum_{k = 0}^{n - 1}{n \choose k + 1}^{2} \\[5mm] & = {1 \over 2}\sum_{k = 1}^{n}{n + 1 \choose k}^{2} - {1 \over 2}\sum_{k = 0}^{n - 1}{n \choose k}^{2} - {1 \over 2}\sum_{k = 1}^{n}{n \choose k}^{2} \\[5mm] & = {1 \over 2}\bracks{\sum_{k = 0}^{n + 1}{n + 1 \choose k}^{2} -2} - {1 \over 2}\bracks{\sum_{k = 0}^{n}{n \choose k}^{2} - 1} - {1 \over 2}\bracks{\sum_{k = 0}^{n}{n \choose k}^{2} - 1} \\[5mm] & = {1 \over 2}{2n + 2 \choose n + 1} - {2n \choose n} \end{align} where I used the well known result $\bbx{\ds{\quad\sum_{i = 0}^{m}{m \choose i}^{2} = {2m \choose m}}}$. Moreover, \begin{align} \sum_{k = 0}^{n - 1}{n \choose k}{n \choose k + 1} & = {1 \over 2}\,{\pars{2n + 2}\pars{2n + 1}\pars{2n}! \over \pars{n + 1}n!\pars{n + 1}n!} - {2n \choose n} = \pars{{2n + 1 \over n + 1} - 1}{2n \choose n} \\[5mm] & = {n \over n + 1}{2n \choose n} = {n \over n + 1}{\pars{2n}! \over n!\,n!} = {\pars{2n}! \over \pars{n + 1}!\pars{n - 1}!} = \bbx{\ds{2n \choose n + 1}} \end{align}
• It is not easy to add a new perspective to a popular sum like this one. Upvoted for originality of the approach. (+1): – Marko Riedel Feb 1 '17 at 23:24
• @MarkoRiedel Thanks. Cross terms always suggest this approach. It's useful too for products of $\log$'s, etc... – Felix Marin Feb 2 '17 at 0:14
Rewrite the sum as $\displaystyle\sum_0^{n-1}\binom{n}{k}\binom{n}{k+1}$, then by combinatorial interpretation, this is: $$\sum_{0}^{n-1}\binom{n}{k}\binom{n}{k+1}=\binom{2n}{2k+1}$$ | {
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The right hand side $\binom{2n}{2k+1}$ is the number of ways to choose $2k+1$ from a total of $2n$. For the left hand side, divide $2n$ into two groups of size $n$ of each, then if choose $k$ from one group, then must choose $2k+1-k=k+1$ from another group, sum over all possible $k$, then you get $\sum_0^{n-1}\binom{n}{k}\binom{n}{k+1}$
• What is $k$ in the rhs? – Julien Apr 4 '13 at 18:28
• I have downvoted this answer for being (mathematically) unintelligible. – anon Apr 5 '13 at 0:51
Using the estimate derived in this answer, we get \begin{align} \sum^{n-1}_{j=0}\binom{n}{j}\binom{n}{j+1} &=\sum^{n-1}_{j=0}\binom{n}{n-j}\binom{n}{j+1}\\ &=\binom{2n}{n+1}\\ &=\binom{2n}{n}\frac{n}{n+1}\\[3pt] &\sim\frac{4^n}{\sqrt{\pi n}}\left(1-\frac9{8n}\right) \end{align} | {
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# Orthonormal Eigenbasis
I am a little apprehensive to ask this question because I have a feeling it's a "duh" question but I guess that's the beauty of sites like this (anonymity):
I need to find an orthonormal eigenbasis for the $2 \times 2$ matrix $\left(\begin{array}{cc}1&1\\ 1&1\end{array}\right)$. I calculated that the eigenvalues were $x=0$ and $x=2$ and the corresponding eigenvectors were $E(0) = \mathrm{span}\left(\begin{array}{r}-1\\1\end{array}\right)$ and $E(2) = \mathrm{span}\left(\begin{array}{c}1\\1\end{array}\right)$. Therefore, an orthonormal eigenbasis would be: $$\frac{1}{\sqrt{2}}\left(\begin{array}{r}-1\\1\end{array}\right), \frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\1\end{array}\right).$$
Here my question: Could the eigenvalues for $E(0)$ been $\mathrm{span}\left(\begin{array}{r}1\\-1\end{array}\right)$?? This would make the final answer $\frac{1}{\sqrt{2}}\left(\begin{array}{r}1\\-1\end{array}\right), \frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\1\end{array}\right)$. Is one answer more correct than the other (or are they both wrong)?
Thanks! | {
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Thanks!
-
I don't quite understand the question as written, but if it's what I think it is, then yes, orthonormal eigenbases are not unique; you can multiply any eigenvector by a complex number of absolute value 1. – Qiaochu Yuan Jan 6 '11 at 13:28
Or by a real number or whatever the field is that you are working over. One more thing: professional mathematicians have "duh" moments all the time. It's part of the job description, so to speak. There is absolutely no need to hide behind anonymity. On the contrary, it is important for a mathematician to learn to live with those "duh" moments. Otherwise he will never be able to freely talk with his colleagues, which would greatly hinder his progress. – Alex B. Jan 6 '11 at 13:32
@Alex: the absolute value 1 condition is important for orthonormality, which doesn't make sense over an arbitrary field. – Qiaochu Yuan Jan 6 '11 at 13:55
@Qiaochu You are right. Read that as "real or complex, depending on..." – Alex B. Jan 6 '11 at 14:38
0 and 2 are the correct eigenvalues to your matrix; (1, -1) is one eigenvector, (1, 1) the other. Your solution is correct.
Span is the set of all linear combinations, so if you consider a vector space over $\mathbb{R}$, it absolutely doesn't matter what scalar in $\mathbb{R}$ you multiply your vectors with inside Span. This does not affect the set Span at all. So the solution is the same.
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There is no such thing as the eigenvector of a matrix, or the orthonormal basis of eigenvectors. There are usually many choices. | {
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Remember that an eigenvector $\mathbf{v}$ of eigenvalue $\lambda$ is a nonzero vector $\mathbf{v}$ such that $T\mathbf{v}=\lambda\mathbf{v}$. That means that if you take any nonzero multiple of $\mathbf{v}$, say $\alpha\mathbf{v}$, then we will have $$T(\alpha\mathbf{v}) = \alpha T\mathbf{v} = \alpha(\lambda\mathbf{v}) = \alpha\lambda\mathbf{v}=\lambda(\alpha\mathbf{v}),$$ so $\alpha\mathbf{v}$ is also an eigenvector corresponding to $\lambda$. More generally, if $\mathbf{v}_1,\ldots,\mathbf{v}_k$ are all eigenvectors of $\lambda$, then any nonzero linear combination $\alpha_1\mathbf{v}_1+\cdots+\alpha_k\mathbf{v}_k\neq \mathbf{0}$ is also an eigenvector corresponding to $\lambda$.
So, of course, since $\left(\begin{array}{r}-1\\1\end{array}\right)$ is an eigenvector (corresponding to $x=0$), then so is $\alpha\left(\begin{array}{r}-1\\1\end{array}\right)$ for any $\alpha\neq 0$, in particular, for $\alpha=-1$ as you take.
Now, a set of vectors $\mathbf{w}_1,\ldots,\mathbf{w}_k$ is orthogonal if and only if $\langle \mathbf{w}_i,\mathbf{w}_j\rangle = 0$ if $i\neq j$. If you have an orthogonal set, and you replace, say, $\mathbf{w}_i$ by $\alpha\mathbf{w}_i$ with $\alpha$ any scalar, then the result is still an orthogonal set: because $\langle\mathbf{w}_k,\mathbf{w}_j\rangle=0$ if $k\neq j$ and neither is equal to $i$, and for $j\neq i$, we have $$\langle \alpha\mathbf{w}_i,\mathbf{w}_j\rangle = \alpha\langle\mathbf{w}_i,\mathbf{w}_j\rangle = \alpha 0 = 0$$ by the properties of the inner product. As a consequence, if you take an orthogonal set, and you take any scalars $\alpha_1,\ldots,\alpha_k$, then $\alpha_1\mathbf{w}_1,\ldots,\alpha_k\mathbf{w}_k$ is also an orthogonal set. | {
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A vector $\mathbf{n}$ is normal if $||\mathbf{n}||=1$. If $\alpha$ is any scalar, then $||\alpha\mathbf{n}|| = |\alpha|\,||\mathbf{n}|| = |\alpha|$. So if you multiply any normal vector $\mathbf{n}$ by a scalar $\alpha$ of absolute value $1$ (or of complex norm $1$), then the vector $\alpha\mathbf{n}$ is also a normal vector.
A set of vectors is orthonormal if it is both orthogonal, and every vector is normal. By the above, if you have a set of orthonormal vectors, and you multiply each vector by a scalar of absolute value $1$, then the resulting set is also orthonormal.
In summary: you have an orthonormal set of two eigenvectors. You multiply one of them by $-1$; this does not affect the fact that the two are eigenvectors. The set was orthogonal, so multiplying one of them by a scalar does not affect the fact that the set is orthogonal. And the vectors were normal, and you multiplied one by a scalar of absolute value $1$, so the resulting vectors are still normal. So you still have an orthonormal set of two eigenvectors. I leave it to you to verify that if you have a linearly independent set, and you multiply each vector by a nonzero scalar, the result is still linearly independent.
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Dijkstra's algorithm, published in 1959 and named after its creator Dutch computer scientist Edsger Dijkstra, can be applied on a weighted graph. Very nice article. Graph Algorithms I 12. We use n = jV(G)jand m = jE(G)jto denote the number of ver-. Shortest Path Queries A shortest path query on a(n) (undirected) graph finds the shortest path for the given source and target vertices in the graph. The graph is not weighted. Fast Paths allows a massive speed-up when calculating shortest paths on a weighted directed graph compared to the standard Dijkstra algorithm. it, daniele. In the first part of the paper, we reexamine the all-pairsshortest paths (APSP) problem and present a newalgorithm with running time approaching O(n3/log2n), which improves all known algorithms for general real-weighted dense graphs andis perhaps close to the best result possible without using fast matrix multiplication, modulo a few log log n factors. Since the edges in the center of the graph have large weights, the shortest path between nodes 3 and 8 goes around the boundary of the graph where the edge weights are smallest. ! But what if edges have different 'costs'? s v δ(, ) 3sv = δ(, ) 12sv = 2 s v 2 5 1 7. Consumes a graph and two vertices, and returns the shortest path (in terms of number of vertices) between the two vertices. When There Is An Edge Between I And J, Then G[i][j] Denotes The Weight Of The Edge. Weighted graphs are commonly used in determining the most optimal path, most expedient, or the. I maintain a count for the number of shortest paths; I would like to use BFS from v first and also maintain a global level. Weighted Graph ( भारित ग्राफ ) Discrete Mathematics Shortest Path || Dijkstra Algorithm #weightedgraph #grewalpinky B. Shortest Path (Unweighted Graph) Goal: find the shortest route to go from one node to another in a graph. Exercise 3 [Modeling a problem as a shortest path problem in graphs] Four imprudent walkers are caught in the storm and nights. It maintains a | {
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path problem in graphs] Four imprudent walkers are caught in the storm and nights. It maintains a set of nodes for which the shortest paths are known. 7 s: source 8 ’’’ 9 result = ShortestPathResult() 10 num_vertices = graph. the first assumes that the graph is weighted, which means that each edge has a cost to traverse it. BFS always visits nodes in increasing order of their distance from the source. An interesting problem is how to find shortest paths in a weighted graph; i. The previous state of the art for this problem was total update time O ̃(n 2 √m/ε) for directed, unweighted graphs [2], and Õ(mn=ε) for undirected, unweighted graphs [12]. The result is a list of vertices, or #f if there is no path. G Is A Weighted Graph With Vertex Set {0,1,2,,1-1} And Integer Weights Represented As The Following Adjacency Matrix Form. Assignment: Given any connected, weighted graph G, use Dijkstras algorithm to compute the shortest (or smallest weight) path from any vertex a to any other vertex b in the graph G. Our main goal is to characterize exactly which sets of node sequences, which we call path systems, can appear as unique shortest paths in a graph with arbitrary real edge weights. If There Is An Edge Between I And 1, Then G[16] > 0, Otherwise G[i,j]=-1. I'm using the networkx package in Python 2. If the graph is weighted (that is, G. Shortest paths. Here a, b, c. A cycle is a path where the first and last vertices are the same. We wish to determine a shortest path from v 0 to v n Dijkstra's Algorithm Dijkstra's algorithm is a common algorithm used to determine shortest path from a to z in a graph. Dijkstra's Algorithm. Counting the number of shortest paths in various graphs is an important and interesting combinatorial problem, especially in weighted graphs with various applications. A few months ago, mathematicians Andrew Beveridge and Jie Shan published Network of Thrones in Math Horizon Magazine where they analyzed a network of character interactions from the | {
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of Thrones in Math Horizon Magazine where they analyzed a network of character interactions from the novel “A Storm of Swords”, the third book in the popular “A Song of Ice and Fire” and the basis for the Game of Thrones TV series. numPaths initialized to 1). Combinatorics is the branch of mathematics concerned with selecting, arranging, and listing or counting collections of objects. Newman Santa Fe Institute, 1399 Hyde Park Road, Santa Fe, New Mexico 87501 and Center for Applied Mathematics, Cornell University, Rhodes Hall, Ithaca, New York 14853 ~Received 1 February 2001; published 28 June 2001!. Shortest Path In A Weighted Directed Graph With Dijkstra's Algorithm - posted in C and C++: Well, I encountered an interesting problem. We start with undirected graphs. The main algorithms that fall under this definition are Breadth-First Search (BFS) and Dijkstra's algorithms. A logical scalar. Weighted Graphs Data Structures & Algorithms 2 [email protected] ©2000-2009 McQuain Shortest Paths (SSAD) Given a weighted graph, and a designated node S, we would like to find a path of least total weight from S to each of the other vertices in the graph. The graph given in the test case is shown as : The shortest paths for the 3 queries are :: The direct Path is shortest with weight 5: There is no way of reaching node 1 from node 3. It is used to identify optimal driving directions or degree of separation between two people on a social network for example. The gist of Dijkstra's single source shortest path algorithm is as below : Dijkstra's algorithm finds the shortest path in a weighted graph containing only positive edge weights from a single source. The shortest path between two points in a weighted graph can be found with Dijkstra’s algorithm. Now we are going to find the shortest path between source (a) and remaining vertices. Find the shortest distance from C to D and if it is impossible to reach node D from C then return -1. The algorithm exists in many variants. This | {
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it is impossible to reach node D from C then return -1. The algorithm exists in many variants. This means that, given a weighted graph, this algorithm will output the shortest distance from a selected node to all other nodes. shortest path. In this post, I explain the single-source shortest paths problems out of the shortest paths problems, in which we need to find all the paths from one starting vertex to all other vertices. Changing to its dual, the triangular grid, paths between triangle pixels (we abbreviate this term to trixels) are counted. Question: Shortest Paths(int[0). The total cost of a path in a graph is equal to the sum of the weights of the edges that connect the vertices in the path. implementations of the shortest-path search algorithms on graphs on the number of graph vertices experimentally. In the first part of the paper, we reexamine the all-pairsshortest paths (APSP) problem and present a newalgorithm with running time approaching O(n3/log2n), which improves all known algorithms for general real-weighted dense graphs andis perhaps close to the best result possible without using fast matrix multiplication, modulo a few log log n factors. PRACTICE PROBLEM BASED ON FLOYD WARSHALL ALGORITHM- Problem- Consider the following directed weighted graph- Using Floyd Warshall Algorithm, find the shortest path distance between every pair of vertices. The shortest path from 0 to 4 uses the shortest path from 0 to 1 and the edge 1–4. Our algorithm is deterministic and has a running time of O(k(m√n + n 3/2 log n)) where m is the number of edges in the graph and n is the number of vertices. In this graph, vertex A and C are connected by two parallel edges having weight 10 and 12 respectively. Abstract: In this paper we evaluate our presented Quantum Approach for finding the Estimation of the Length of the Shortest Path in a Connected Weighted Graph which is achieved with a polynomial time complexity about O(n) and as a result of evaluation we show that the | {
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achieved with a polynomial time complexity about O(n) and as a result of evaluation we show that the Probability of Success of our presented Quantum Approach is increased if the Standard Deviation of the Length of all possible. An assigned number is called the weight of the edge, and the collection of all weights is called a weighting of the graph Γ. A path in a graph is a sequence of adjacent vertices. In any graph G, the shortest path from a source vertex to a destination vertex can be calculated using Dijkstra Algorithm. Path Graphs. Only edges with non-negative costs are included. Discuss an efficient algorithm to compute a shortest path from node s to node t in a weighted directed graph G such that the path is of minimum cardinality among all shortest s - t paths in G graph-theory hamiltonian-path path-connected. This means it finds a shortest paths between nodes in a graph, which may represent, for example, road networks; For a given source node in the graph, the algorithm finds the shortest path between source node and every other node. dijkstra_path¶ dijkstra_path (G, source, target, weight='weight') [source] ¶. 0 = no epidemic 1 = epidemic NW N SW C E Graph 3: Vertex coloring of weighted graph for the county. In this category, Dijkstra’s algorithm is the most well known. You are expected to do it in Time Complexity of O(A + M). shortest path functions use it as the cost of the path; community finding methods use it as the strength of the relationship between two vertices, etc. Shortest path algorithms are a family of algorithms designed to solve the shortest path problem. The shortest path between node 222 and node 444 is 222 -> 555 -> 666 -> 777 -> 444, which has a weighted distance 1. The number of diagonal steps in a shortest path of the chessboard distance is min{w 1,w 2}, and the number of cityblock steps (i. A cycle is a path where the first and last vertices are the same. Given an undirected, weighted graph, find the minimum number of edges to travel | {
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are the same. Given an undirected, weighted graph, find the minimum number of edges to travel from node 1 to every other node. (Consider what this means in terms of the graph shown above right. The Edge can have weight or cost associate with it. Simple path is a path with no repeated vertices. Hence, parallel computing must be applied. In this post I will be discussing two ways of finding all paths between a source node and a destination node in a graph: Using DFS: The idea is to do Depth First Traversal of given directed graph. , capacity, cost, demand, traffic frequency, time, etc. There are already comprehensive network analysis packages in R, notably igraph and its tidyverse-compatible interface tidygraph. G Is A Weighted Graph With Vertex Set {0,1,2,,1-1} And Integer Weights Represented As The Following Adjacency Matrix Form. Geodesic paths are not necessarily unique, but the geodesic distance is well-defined since all geodesic paths have. ple, Figure 1a illustrates a graph G, and Figure 1e shows an aug-mented graph G∗ constructed from G. The latter only works if the edge weights are non-negative. 4 5 Args: 6 graph: weighted graph with no negative cycles. Weighted directed graphs may be used to model communication networks, and shortest distances (shortest-path weights) between nodes may be used to suggest routes for messages. Chan⁄ September 30, 2009 Abstract Intheflrstpartofthepaper,wereexaminetheall-pairs shortest paths (APSP)problemand present a new algorithm with running time O(n3 log3 logn=log2 n), which improves all known algorithmsforgeneralreal-weighteddensegraphs. This module covers weighted graphs, where each edge has an associated weight or number. The problem is to find k directed paths starting at s, such that every node of G lies on at least one of those paths, and such that the sum of the weights of all the edges in the paths is minimized. The one-to-all shortest path problem is the problem of determining the shortest path from node s to all the | {
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shortest path problem is the problem of determining the shortest path from node s to all the other nodes in the. Finding shortest paths in weighted graphs In the past two weeks, you've developed a strong understanding of how to design classes to represent a graph and how to use a graph to represent a map. Exercise 3 [Modeling a problem as a shortest path problem in graphs] Four imprudent walkers are caught in the storm and nights. How to use BFS for Weighted Graph to find shortest paths ? If your graph is weighted, then BFS may not yield the shortest weight paths. You are given a weighted directed acyclic graph G, and a start node s in G. It finds a shortest-path tree for a weighted undirected graph. Journal of the ACM 65 :6, 1-40. Is there a cycle that uses each vertex. Mizrahi et al. The number dist[w] equals the length of a shortest path from v to w, or is -1 if w cannot be reached. Dijkstra's Algorithm is useful for finding the shortest path in a weighted graph. Given a directed weighted graph G= (V;E;w) with non-negative weights w: E!R+ and a vertex s2V, the single-source shortest paths is the family of shortest paths s vfor every vertex v2V. If finds only the lengths not the path. If There Is An Edge Between I And 1, Then G[16] > 0, Otherwise G[i,j]=-1. A logical scalar. We use n = jV(G)jand m = jE(G)jto denote the number of ver-. The order of a graph is the number of nodes. When There Is An Edge Between I And J, Then G[i][j] Denotes The Weight Of The Edge. The Dijkstra’s algorithm is provided in Algorithm 2. For example, the length of v8,v9 equals 2, which is identical to the length of the. As such, we say that the weight of a path is the sum of the weights of the edges it contains. The shortest-path from vertex u to vertex v in a weighted graph is a path with minimum sum-weight B All-Pairs Shortest Paths For a weighted graph, the all-pairs shortest paths problem is to nd the shortest path between all pairs of vertices. 2 Representing Weighted Graphs Often it | {
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is to nd the shortest path between all pairs of vertices. 2 Representing Weighted Graphs Often it is desirable to use a priority queue to store weighted edges. Question: Shortest Paths(int[0). We first propose an exact (and. A single execution of the algorithm will. shortest path. Consider a shortest path p from vertex i to vertex j, and suppose that p contains at most m edges. In the weighted matching [G85b, GT89, GT91] and maxi-mum flow problems [GR98], for instance, the best algorithms for real- and integer-weighted graphs have running times differing by a polynomial factor. Slight Adjustment of Dijkstra's Algorithm to Solve Shortest Path Problem… 783 The i-number weight of an edge is also called the i-distance between the corresponding two nodes. The distance between two vertices u and v, denoted d(u,v), is the length of a shortest. Suppose you are given a directed graph G = (V, E), with costs on the edges; the costs may be. A path in a graph is a sequence of adjacent vertices. The previous state of the art for this problem was total update time O ̃(n 2 √m/ε) for directed, unweighted graphs [2], and Õ(mn=ε) for undirected, unweighted graphs [12]. Assignment: Given any connected, weighted graph G, use Dijkstras algorithm to compute the shortest (or smallest weight) path from any vertex a to any other vertex b in the graph G. Shortest Path. An example of a weighted graph would be the distance between the capitals of a set of countries. You are given a weighted undirected graph. Design a linear time algorithm to find the number of different shortest paths (not necessarily vertex disjoint) between v and w. We present an improved algorithm for maintaining all-pairs (1 + ε) approximate shortest paths under deletions and weight-increases. Consider the graph above. Your task is to find the shortest path between the vertex 1 and the vertex n. In weighted graphs, where we assume that the edge weights do not represent the communication speed, a straightforward distributed | {
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assume that the edge weights do not represent the communication speed, a straightforward distributed variant of the Bellman-Ford algorithm [2], [3], [4] computes. Shortest paths problems are among the most fundamental algorithmic graph problems. The actual shortest paths can also be constructed by modifying the above algorithm. Question: Shortest Paths(int[0). If There Is An Edge Between I And 1, Then G[16] > 0, Otherwise G[i,j]=-1. To find the shortest path on a weighted graph, just doing a breadth-first search isn't enough - the BFS is only a measure of the shortest path based on number of edges. A comparison of the data obtained as a result of the study was carried out to find the best applications of implementations of the shortest path search algorithms in the Postgre SQL DBMS. Single source shortest path for undirected graph is basically the breadth first traversal of the graph. Shortest Path Algorithms?(a) Breadth-first search (BFS) can be used to perform single source short- est paths on any graph where all edges have the same costs 1. A cycle is a path where the first and last vertices are the same. Let s denote the number of edges of H. Shortest-Path Problems (cont'd) Single-source shortest path problem Given a weighted graph G = (V, E), and a distinguished start vertex, s, find the minimum weighted path from s to every other vertex in G The shortest weighted path from v 1 to v 6 has a cost of 6 and v 1 v 4 v 7 v 6. Shortest Path Problem. Dijkstra's algorithm (or Dijkstra's Shortest Path First algorithm, SPF algorithm) is an algorithm for finding the shortest paths between nodes in a graph, which may represent, for example, road networks. Node is a vertex in the graph at a position. Assignment: Given any connected, weighted graph G, use Dijkstras algorithm to compute the shortest (or smallest weight) path from any vertex a to any other vertex b in the graph G. Using similar ideas, we can construct a (1+epsilon)-approximate distance oracle for weighted | {
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G. Using similar ideas, we can construct a (1+epsilon)-approximate distance oracle for weighted unit-disk graphs with O(1) query. Finding the shortest path, with a little help from Dijkstra! If you spend enough time reading about programming or computer science, there’s a good chance that you’ll encounter the same ideas. in Proceedings - 25th IEEE International Conference on High Performance Computing, HiPC 2018. Shortest Path Problem. Graph Algorithms I 12. We let k denote the number of source. Given a weighted line-graph (undirected connected graph, all vertices of degree 2, except two endpoints which have degree 1), devise an algorithm that preprocesses the graph in linear time and can return the distance of the shortest path between any two vertices in constant time. The All-Pairs Shortest Paths (APSP) problem seeks the shortest path distances between all pairs of vertices, and is one of the most fundamental graph problems. Once we have reached our destination, we continue searching until all possible paths are greater than 11; at that point we are certain that the shortest path is 11. The vertices V are connected to each other by these edges E. Key Graph Based Shortest Path Algorithms With Illustrations - Part 1: Dijkstra's And Bellman-Ford Algorithms Bellman-Ford algorithm is used to find the shortest paths from a source vertex to all other vertices in a weighted graph. Distributed Exact Weighted All-Pairs Shortest Paths in O˜(n5/4) network is modeled by a weighted n-node m-edge graph G. There are already comprehensive network analysis packages in R, notably igraph and its tidyverse-compatible interface tidygraph. We revisit a classical graph-theoretic problem, the single-source shortest-path (SSSP) problem, in weighted unit-disk graphs. An interesting side-effect of traversing a graph in BFS order is the fact that, when we visit a particular node, we can easily find a path from the source node to the newly visited node with the least number of edges. The | {
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find a path from the source node to the newly visited node with the least number of edges. The Symmetric Shortest-Path Table Routing Conjecture Thomas L. Question: Shortest Paths(int[0). Mizrahi et al. C++ Program to Generate a Random UnDirected Graph for a Given Number of Edges; Shortest Path in a Directed Acyclic Graph. Shortest Distance in a graph with two different weights : Given a weighted undirected graph having A nodes, a source node C and destination node D. Exercise 7 Consider the following modification of the Dijkstra’s algorithm to work with negative weights:. The algorithm creates a tree of shortest paths from the starting vertex, the source, to all other points in the graph. The previous state of the art for this problem was total update time O ̃(n 2 √m/ε) for directed, unweighted graphs [2], and Õ(mn=ε) for undirected, unweighted graphs [12]. I will cover several other graph based shortest path algorithms with concrete illustrations. Shortest path in a graph with weighted edges and vertices. If there are multiple paths from node 1 to a node that have the same minimum number of. Hi I have already posted a similar question but there was a misunderstanding of the problem by my side so here I post it again. There is one shortest path vertex 0 to vertex 0 (from each vertex there is a single shortest path to itself), one shortest path between vertex 0 to vertex 2 (0->2), and there are 4 different shortest paths from vertex 0 to vertex 6: 1. A shortest path between two nodes u and v in a graph is a path that starts at u and ends at v and has the lowest total link weight. Given an undirected, weighted graph, find the minimum number of edges to travel from node 1 to every other node. Row i of the predecessor matrix contains information on the shortest paths from point i: each entry predecessors[i, j] gives the index of the previous node in the path from point i to point j. You are expected to do it in Time Complexity of O(A + M). For example, the length of | {
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i to point j. You are expected to do it in Time Complexity of O(A + M). For example, the length of v8,v9 equals 2, which is identical to the length of the. A shortest path (with the lowest weight) from $$u$$ to $$v$$; The weight of the path is the sum of the weights of its edges. Both algorithms were randomized and had constant query time. It was conceived by computer scientist Edsger W. Deterministic Partially Dynamic Single Source Shortest Paths in Weighted Graphs Aaron Bernstein May 30, 2017 Abstract In this paper we consider the decremental single-source shortest paths (SSSP) problem, where given a graph Gand a source node sthe goal is to maintain shortest distances between sand all other nodes. In this category, Dijkstra's algorithm is the most well known. An undirected graph that has a path from every vertex to every other vertex in the graph. As noted earlier, mapping software like Google or Apple maps makes use of shortest path algorithms. C++ Program to Generate a Random UnDirected Graph for a Given Number of Edges; Shortest Path in a Directed Acyclic Graph. If you think carefully, it's easy to see that there can be many graphs such that the. A new approach to all-pairs shortest paths on real-weighted graphs Seth Pettie1 Department of Computer Sciences, The University of Texas at Austin, Austin, TX 78712, USA Abstract We present a new all-pairs shortest path algorithm that works with real-weighted graphs in the traditional comparison-additionmodel. The gist of Dijkstra's single source shortest path algorithm is as below : Dijkstra's algorithm finds the shortest path in a weighted graph containing only positive edge weights from a single source. How to use BFS for Weighted Graph to find shortest paths ? If your graph is weighted, then BFS may not yield the shortest weight paths. Chan⁄ September 30, 2009 Abstract Intheflrstpartofthepaper,wereexaminetheall-pairs shortest paths (APSP)problemand present a new algorithm with running time O(n3 log3 logn=log2 n), | {
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shortest paths (APSP)problemand present a new algorithm with running time O(n3 log3 logn=log2 n), which improves all known algorithmsforgeneralreal-weighteddensegraphs. shortest path algorithm. In this tutorial, we will present a general explanation of both algorithms. A cycle is a path where the first and last vertices are the same. techniques to speed the computation of shortest paths in the discretization graph [3,21]. ; It uses a priority based dictionary or a queue to select a node / vertex nearest to the source that has not been edge relaxed. This turns out to be a problem that can be solved efficiently, subject to some restrictions on the edge costs. An observed image is composed of multiple components based on optical phenomena, such as light reflection and scattering. The multiplicity of a path is the maximum number of times that an edge appears in it. The latter only works if the edge weights are non-negative. It maintains a set S of vertices whose final shortest path from the source has already been determined and it repeatedly selects the left vertices with the minimum shortest-path estimate, inserts them. Fast Paths allows a massive speed-up when calculating shortest paths on a weighted directed graph compared to the standard Dijkstra algorithm. Directed and undirected graphs may both be weighted. I maintain a count for the number of shortest paths; I would like to use BFS from v first and also maintain a global level. Single Source Shortest Path. This section discusses three algorithms for this problem: breadth-first search for unweighted graphs, Dijkstra’s algorithm for weighted graphs, and the Floyd-Warshall algorithm for computing distances between all pairs of vertices. And here is some test code: test_graph. Weighted directed graphs may be used to model communication networks, and shortest distances (shortest-path weights) between nodes may be used to suggest routes for messages. An undirected graph that has a path from every vertex to every other | {
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to suggest routes for messages. An undirected graph that has a path from every vertex to every other vertex in the graph. shortest path problem. shortest_paths calculates a single shortest path (i. We will see how simple algorithms like depth-first-search can be used in clever ways (for a problem known as topological sorting) and will see how Dynamic Programming can be used to solve problems of finding shortest paths. yenpathy: An R Package to Quickly Find K Shortest Paths Through a Weighted Graph Submitted 05 September 2019 This paper is under review which means review has begun. ! Here, the length of a path is simply the number of edges on the path. One of these challenges is the need to generate a limited number of test cases of a given regression test suite in a manner that does not compromise its defect detection ability. For example finding the ‘shortest path’ between two nodes, e. We would then assign weights to vertices, not edges. Python – Get the shortest path in a weighted graph – Dijkstra Posted on July 22, 2015 by Vitosh Posted in VBA \ Excel Today, I will take a look at a problem, similar to the one here. FindShortestPath[g, s, All] generates a ShortestPathFunction[] that can be applied repeatedly to different t. [email protected] ple, Figure 1a illustrates a graph G, and Figure 1e shows an aug-mented graph G∗ constructed from G. According to [11], [12], [13], since 1959 Dijkstra's algorithm has been recognized as the best algorithm and used as method to find the shortest path. nodes in a given directed graph is a very common problem. You are also given a positive integer k. A subgraph is a subset of a graph’s edges (with associated vertices) that form a graph. ” Dijkstra’s algorithm is an iterative algorithm that provides us with the shortest path from one particular starting node to all other nodes in the graph. The shortest path function can also be used to compute a transitive closure or for arbitrary length traversals. the number of pairs of vertices | {
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to compute a transitive closure or for arbitrary length traversals. the number of pairs of vertices not including v, which for directed graphs is and for undirected graphs is. The all-pairs shortest paths problem for unweighted directed graphs was introduced by Shimbel (1953), who observed that it could be solved by a linear number of matrix multiplications that takes a total time of O(V 4). There are two types of weighted graphs: vertex weighted and edge weighted. Shortest path problem (SPP) is a fundamental and well-known combinatorial optimization problem in the area of graph theory. A weighted graph is a one which consists of a set of vertices V and a set of edges E. The shortest path problem is about finding a path between $$2$$ vertices in a graph such that the total sum of the edges weights is minimum. Dijkstra's Shortest Path Algorithm in Java. The single pair shortest path problem seeks to compute (u;v) and construct a shortest path from. Consumes a graph and two vertices, and returns the shortest path (in terms of number of vertices) between the two vertices. A subgraph is a subset of a graph’s edges (with associated vertices) that form a graph. If There Is An Edge Between I And 1, Then G[16] > 0, Otherwise G[i,j]=-1. the sum of the weights of the edges in the paths is minimized. shortest path algorithm. Variations of the Shortest Path Problem. We present a new all-pairs shortest path algorithm that works with real-weighted graphs in the traditional comparison-addition model. I Therefore, the numbers d 1;d 2; ;d n must include an even number of odd numbers. Your solution should be complete in that it shows the shortest path from all starting vertices to all other vertices. All-Pairs Shortest Paths Problem To find the shortest path between all verticesv 2 V for a graph G =(V,E). This algorithm has numerous applications in network analysis, such as transportation planning. (2018) Decremental Single-Source Shortest Paths on Undirected Graphs in Near-Linear | {
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planning. (2018) Decremental Single-Source Shortest Paths on Undirected Graphs in Near-Linear Total Update Time. Dijkstra’s Shortest Path Algorithm in Java. Fast Paths allows a massive speed-up when calculating shortest paths on a weighted directed graph compared to the standard Dijkstra algorithm. It was conceived by computer scientist Edsger W. Shortest Path in a weighted Graph where weight of an edge is 1 or 2 Given a directed graph where every edge has weight as either 1 or 2, find the shortest path from a given source vertex ‘s’ to a given destination vertex ‘t’. Given an undirected, weighted graph, find the minimum number of edges to travel from node 1 to every other node. be contained in shortest augmenting paths, and the lay-ered network contains all augmenting paths of shortest length. Dating back some 3000 years, and initially consisting mainly of the study of permutations and combinations, its scope has broadened to include topics such as graph theory, partitions of numbers, block designs, design of codes, and latin squares. Simple path is a path with no repeated vertices. Find a TSP solution using state-of-the-art software, and then remove that dummy node (subtracting 2 from the total weight). C… Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. This module covers weighted graphs, where each edge has an associated weight or number. Discuss an efficient algorithm to compute a shortest path from node s to node t in a weighted directed graph G such that the path is of minimum cardinality among all shortest s - t paths in G graph-theory hamiltonian-path path-connected. Dijkstra’s Algorithms describes how to find the shortest path from one node to another node in a directed weighted graph. Print the number of shortest paths from a given vertex to each of the vertices. Given a weighted line-graph (undirected connected graph, all vertices of degree 2, except two endpoints which have degree 1), devise | {
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connected graph, all vertices of degree 2, except two endpoints which have degree 1), devise an algorithm that preprocesses the graph in linear time and can return the distance of the shortest path between any two vertices in constant time. Dijkstra's Algorithm: Finds the shortest path from one node to all other nodes in a weighted graph. It finds a shortest-path tree for a weighted undirected graph. The SQL Server graph extensions are amazing. Introduction. In this week, you'll add a key feature of map data to our graph representation -- distances -- by adding weights to your edges to produce a "weighted. Must Read: C Program. The Degree of a vertex is the number of edges incident on it. I'm using the networkx package in Python 2. adjacent b. Shortest Paths: Problem Statement Given a weighted graph and two vertices u and v, we want to find a path of minimum total weight between u and v Length (or distance) of a path is the sum of the weights of its edgesLength (or distance) of a path is the sum of the weights of its edges. Floyd-Warshall algorithm is an algorithm for finding shortest paths in a weighted graph with positive or negative edge weights (but with no negative cycles). The Degree of a vertex is the number of edges incident on it. for unweighted. If the graph is unweighed, then finding the shortest path is easy: we can use the breadth-first search algorithm. A path graph is a graph consisting of a single path. So, the shortest path would be of length 1 and BFS would correctly find this for us. Slight Adjustment of Dijkstra's Algorithm to Solve Shortest Path Problem… 783 The i-number weight of an edge is also called the i-distance between the corresponding two nodes. We know that breadth-first search can be used to find shortest path in an unweighted graph or in weighted graph having same cost of all its edges. Shortest path in a graph with weighted edges and vertices. PY - 2007/10/30. If There Is An Edge Between I And 1, Then G[16] > 0, Otherwise G[i,j]=-1. | {
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vertices. PY - 2007/10/30. If There Is An Edge Between I And 1, Then G[16] > 0, Otherwise G[i,j]=-1. Variations of the Shortest Path Problem. We start with a directed and weighted graph $$G = (V, E)$$ with a weights function $$w: E \rightarrow R$$ that maps edges to weights with real values; Find for each pair of vertices $$u$$, $$v \in V$$. Shortest paths&Weighted graphs. An undirected graph that has a path from every vertex to every other vertex in the graph. Consider any node that is not the root: its possible distances from the root are all possible distances of its neighbors plus the weight of the connecting edges. If an edge is missing a special value, perhaps a negative value, zero or a large value to represent "infinity", indicates this fact. In a weighted graph with start node s, there are often multiple shortest paths from s to any other node. 1 def shortest_path_cycle(graph, s): 2 ’’’Single source shortest paths using DP on a graph with cycles but no 3 negative cycles. Dijkstra’s algorithm is similar to Prim’s algorithm. 4 5 Args: 6 graph: weighted graph with no negative cycles. b) Weighted graph matching: Mulmuley & Shah ob-served that their result for the shortest path problem yields the same lower bound for the WEIGHTED GRAPH MATCHING problem [3, Corollary 1. In any graph G, the shortest path from a source vertex to a destination vertex can be calculated using this algorithm. 1 Shortest paths and matrix multiplication 25. This means it finds a shortest paths between nodes in a graph, which may represent, for example, road networks; For a given source node in the graph, the algorithm finds the shortest path between source node and every other node. We know that breadth-first search can be used to find shortest path in an unweighted graph or in weighted graph having same cost of all its edges. Suppose we have to following graph: We may want to find out what the shortest way is to get from node A to node F. Graphs: Finding shortest paths Definition: | {
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what the shortest way is to get from node A to node F. Graphs: Finding shortest paths Definition: weight of a path 4 Tecniche di programmazione A. Weighted Graph ( भारित ग्राफ ) Discrete Mathematics Shortest Path || Dijkstra Algorithm #weightedgraph #grewalpinky B. P = shortestpath(G,s,t) computes the shortest path starting at source node s and ending at target node t. 18 between shortest paths in networks and Hamilton paths in graphs ties in with our observation that finding paths of low weight (which we have been calling "short") is tantamount to finding paths with a high number of edges (which we might consider to be "long"). Exercise 7 Consider the following modification of the Dijkstra’s algorithm to work with negative weights:. If you think carefully, it's easy to see that there can be many graphs such that the. A path in a graph is a sequence of adjacent vertices. As there are a number of different shortest path algorithms, we've gathered the most important to help you understand how they work and which is the best. This problem also is known as "Print all paths between two nodes". The shortest paths followed for the three nodes 2, 3 and 4 are as follows : 1/S->2 - Shortest Path Value : 1/S->3 - Shortest Path Value : 1/S->3->4 - Shortest Path Value :. Data Structure Graph Algorithms Algorithms. It runs in O (mn+n 2 log log n) time, improving on the long-standing bound of O (mn+n 2 log n) derived from an implementation of Dijkstra's algorithm with Fibonacci heaps. The total cost of a path in a graph is equal to the sum of the weights of the edges that connect the vertices in the path. By contrast, the graph you might create to specify the shortest path to hike every trail could be a directed graph, where the order and direction of edges matters. 6 2, 6(a), 6(c), 18 In Exercises 2-4 find the length of a shortest path between a and z in the given weighted graph. Shortest Paths, and Dijkstra's Algorithm: Overview Graphs with lengths/weights/costs on edges. | {
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Shortest Paths, and Dijkstra's Algorithm: Overview Graphs with lengths/weights/costs on edges. Scientific collaboration networks. 2 You can use the tool to create a weighted graph with mouse gestures and show the MST and shortest paths. If say we were to find the shortest path from the node A to B in the undirected version of the graph, then the shortest path would be the direct link between A and B. shortest_paths calculates a single shortest path (i. Weighted graphs and path length Weighted graphs A weighted graph is a graph whose edges have weights. Topics in this lecture include:. You may print the results of this algorithm to the screen or to a log file. A B C F D E 7 14 9 8 15 2 9 4 6 Figure 3. Dijkstra’s algorithm is similar to Prim’s algorithm. This module covers weighted graphs, where each edge has an associated weight or number. I'm restricting myself to Unweighted Graph only. The relationship shown in the proof of Property 21. Shortest Path (Unweighted Graph) Goal: find the shortest route to go from one node to another in a graph. Undirected graph. Shortest paths in an edge-weighted digraph 4->5 0. This means it finds the shortest paths between nodes in a graph, which may represent, for example, road networks; For a given source node in the graph, the algorithm finds the shortest path between the source node and every other node. If the graph is weighted, it is a path with the minimum sum of edge weights. Algorithms to find shortest paths in a graph are given later. Breadth First Search, BFS, can find the shortest path in a non-weighted graphs or in a weighted graph if all edges have the same non-negative weight. More Algorithms for All-Pairs Shortest Paths in Weighted Graphs Timothy M. world as a directed graph, the problem of directing the robot was transformed into a shortest-path problem. As noted earlier, mapping software like Google or Apple maps makes use of shortest path algorithms. Variations of the Shortest Path Problem. If an edge is missing a | {
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use of shortest path algorithms. Variations of the Shortest Path Problem. If an edge is missing a special value, perhaps a negative value, zero or a large value to represent "infinity", indicates this fact. Print the number of shortest paths from a given vertex to each of the vertices. Consider the graph above. The best algorithm (modulo small polylogarithmic improvements) for this problem runs in cubic time, a. By contrast, the graph you might create to specify the shortest path to hike every trail could be a directed graph, where the order and direction of edges matters. Your solution should be complete in that it …. Representing a Graph. For example, the two paths we mentioned in our example are C, B and C, A, B. Thanks for pointing to Gephi. In order to solve the load-balancing problem for coarse-grained parallelization, the relationship between the computing time of a single-source shortest-path length of node and the features of node is studied. The all-pairs shortest paths problem for unweighted directed graphs was introduced by Shimbel (1953), who observed that it could be solved by a linear number of matrix multiplications that takes a total time of O(V 4). CS 340 Programming Assignment VII: Single-Source Shortest Paths in generally Weighted graphs Description: You are to implement both the DAG SP algorithm and the Bellman-Ford algorithm for single-source shortest paths based on whether or not you detect a cycle. a i g f e d c b h 25 15 10 5 10. Hart, Nilsson, and Raphael [12] discovered how to use this information to improve the efficiency of computing the shortest path. When There Is An Edge Between I And J, Then G[i][j] Denotes The Weight Of The Edge. The number parent[w] is the predecessor of w on a shortest path from v to w, or -1 if none exists. minimum path sizeis the shortest distance measured in the number of edges traversed. The algorithm can be implemented with O(m+nlog(n)) time; therefore, it is efficient in sparsely connected networks. The | {
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implemented with O(m+nlog(n)) time; therefore, it is efficient in sparsely connected networks. The Bellman-Ford algorithm is a single-source shortest path algorithm. Michael Quinn, Parallel Programming in C with MPI and OpenMP,. Decomposing the observed image into individual components is an important process for various computer vision tasks. Description and notations-building the graph Given a weighted graph G = (V, E), where V. Given an undirected, weighted graph, find the minimum number of edges to travel from node 1 to every other node. By comparison, if the graph is permitted. The order of a graph is the number of nodes. P( s,t ) is the shortest path between the given vertices and containing the least sum of edge weights on the path from to. Bellman-Ford Algorithm is computes the shortest paths from a single source vertex to all of the other vertices in a weighted digraph. Shortest paths&Weighted graphs. It was conceived by computer scientist Edsger W. Finding the shortest path, with a little help from Dijkstra! If you spend enough time reading about programming or computer science, there’s a good chance that you’ll encounter the same ideas. This technique is applied to a minute level bid/ask quote dataset consisting of rates constructed from all G10 currency pairs. • In addition, the first time we encounter a vertex may, we may not have found the shortest path to it, so we need to delay committing to that path. 1 Shortest Path Queries Let G= (V;E;˚) be a road network (i. Start by setting the distance of all notes to infinity and the source's distance to 0. The previous state of the art for this problem was total update time O ̃(n 2 √m/ε) for directed, unweighted graphs [2], and Õ(mn=ε) for undirected, unweighted graphs [12]. A path in a graph is a sequence of adjacent vertices. shortest s-tpath in G n. Algorithms to find shortest paths in a graph are given later. These weights represent the cost to traverse the edge. A cycle is a path where the first and last | {
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These weights represent the cost to traverse the edge. A cycle is a path where the first and last vertices are the same. In any graph G, the shortest path from a source vertex to a destination vertex can be calculated using Dijkstra Algorithm. It turns out that it is as easy to compute the shortest paths from s to every node in G (because if the shortest path from s to t is s = v0, v1, v2, , vk = t, then the path v0,v1 is the shortest path from s to v1, the path v0,v1,v2 is the shortest path from s to v2, the path v0,v1,v2,v3 is the shortest path from s to v3, etc. Assignment: Given any connected, weighted graph G, use Dijkstras algorithm to compute the shortest (or smallest weight) path from any vertex a to any other vertex b in the graph G. Essentially, you replace the stack used by DFS with a queue. I'm looking for an algorithm that will perform as in the title. Again this is similar to the results of a breadth first search. It can be used in numerous fields such as graph theory, game theory, and network. Dijkstra's Shortest Path Algorithm in Java. linear-time. So there is a weighted graph, and the shortest paths of all pair of n. dijkstra_path_length¶ dijkstra_path_length (G, source, target, weight='weight') [source] ¶. What is the longest simple path between s and t? Cycle. Graphs can be weighted (edges carry values) and directional (edges have direction). Let u and v be two vertices in G, and let P be a path in G from u to v. We are also given a starting node s ∈ V. Dijkstra's Single Source Shortest Path. Exercise 7 Consider the following modification of the Dijkstra’s algorithm to work with negative weights:. Among the various shortest path algorithms, Dijkstra’s shortest path algorithm [1] is said to have better performance with regard to run time than the other algorithms. Finally, at k = 4, all shortest paths are found. Shortest Path in a weighted Graph where weight of an edge is 1 or 2 Given a directed graph where every edge has weight as either 1 or 2, | {
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weight of an edge is 1 or 2 Given a directed graph where every edge has weight as either 1 or 2, find the shortest path from a given source vertex ‘s’ to a given destination vertex ‘t’. However, the robot had a means for estimating Euclidean distances in his world. Weighted graphs and path length Weighted graphs A weighted graph is a graph whose edges have weights. The Floyd-Warshall algorithm is an example of dynamic programming. 2 commits 1 branch. The previous state of the art for this problem was total update time O ̃(n 2 √m/ε) for directed, unweighted graphs [2], and Õ(mn=ε) for undirected, unweighted graphs [12]. The Floyd Warshall Algorithm is for solving the All Pairs Shortest Path problem. However, most communication networks are dynamic, i. shortest path problem. Referred to as the shortest path between vertices For weighted graphs this is the path that has the smallest sum of its edge weights ijkstra’salgorithm finds the shortest path between one vertex and all other vertices The algorithm is named after its discoverer, Edgser Dijkstra 24 The shortest path between B and G is: 1 4 3 5 8 2 2 1 5 1 B A. Design a linear time algorithm to find the number of different shortest paths (not necessarily vertex disjoint) between v and w. Graph Characteristics-Undirected-Weighted-Journey: (1, 7)-Shortest path: 1 – 4 – 6 – 7 (in purple)-Total cost: 6. We first propose an exact (and. The graph has eight nodes. The Degree of a vertex is the number of edges incident on it. For the shortest path problem on positively weighted graphs the integer/real gap is only logarith-mic. Fine the shortest weighted path from a vertex, s, to every other vertex in the graph. The weights of the edges can be positive or negative. A near linear shortest path algorithm for weighted undirected graphs Abstract: This paper presents an algorithm for Shortest Path Tree (SPT) problem. Here the graph we consider is unweighted and hence the shortest path would be the number of edges it takes to go | {
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we consider is unweighted and hence the shortest path would be the number of edges it takes to go from source to destination. The longest path is based on the number of edges in the path if weighted == false and the unweighted shortest path algorithm is being used. Your solution should be complete in that it shows the shortest path from all starting vertices to all other vertices. Suppose G be a weighted directed graph where a minimum labeled w(u, v) associated with each edge (u, v) in E, called weight of edge (u, v). We introduce a metric on the. It asks for the number of different shortest paths. Day 3: Weighted Graphs. This problem also is known as "Print all paths between two nodes". So there is a weighted graph, and the shortest paths of all pair of n. texens | Hello World ! | Page 3 Hello World !. In this tutorial, we will present a general explanation of both algorithms. The graph given in the test case is shown as : * The lines are weighted edges where weight denotes the length of the edge. Graphs can be weighted (edges carry values) and directional (edges have direction). We present an improved algorithm for maintaining all-pairs (1 + ε) approximate shortest paths under deletions and weight-increases. Give an efficient algorithm to solve the single-destination shortest paths problem. Thus, the shortest path between any two nodes is the path between the two nodes with the lowest total length. The weighted path length is given by ∑C(vi,vi+1) i=1 k-1 The general problem Given an edge-weighted graph G = (V,E) and two vertices, vs ∈V and vd ∈V, find the path that starts at vs and ends at vd that has the smallest weighted path length Single-source shortest. Uses Dijkstra's Method to compute the shortest weighted path between two nodes in a graph. It finds a shortest-path tree for a weighted undirected graph. Crossref, ISI, Google Scholar; 19. When There Is An Edge Between I And J, Then G[i][j] Denotes The Weight Of The Edge. The shortest pair edge disjoint paths | {
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Between I And J, Then G[i][j] Denotes The Weight Of The Edge. The shortest pair edge disjoint paths in the new graph corresponds to the required solution in the original graph. the path itself, not just its length) between the source vertex given in from, to the target vertices given in to. A subgraph is a subset of a graph’s edges (with associated vertices) that form a graph. Length of a path is the sum of the weights of its edges. Let G = (V, E) be a directed graph and let v and w be two vertices in G. We present an improved algorithm for maintaining all-pairs (1 + ε) approximate shortest paths under deletions and weight-increases. weighted › cyclic vs. num_vertices() 11 for i in range(num_vertices): 12 result. Finding paths. The Bellman-Ford algorithm is a single-source shortest path algorithm. Topological Sort: Arranges the nodes in a directed, acyclic graph in a special order based on incoming edges. We will see how simple algorithms like depth-first-search can be used in clever ways (for a problem known as topological sorting) and will see how Dynamic Programming can be used to solve problems of finding shortest paths. Asked in Graphs , C Programming. Dijkstra’s Algorithm for Finding the Shortest Path Through a Weighted Graph E. Simple path is a path with no repeated vertices. This algorithm aims to find the shortest-path in a directed or undirected graph with non-negative edge weights. BFS runs in O(E+V) time where E is the number of edges and V is number of vertices in the graph. Those times are the weights of those paths. A path in a graph is a sequence of adjacent vertices. We revisit a classical graph-theoretic problem, the single-source shortest-path (SSSP) problem, in weighted unit-disk graphs. Floyd-Warshall algorithm is an algorithm for finding shortest paths in a weighted graph with positive or negative edge weights (but with no negative cycles). Hi I have already posted a similar question but there was a misunderstanding of the problem by my side so | {
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have already posted a similar question but there was a misunderstanding of the problem by my side so here I post it again. Mark Dolan CIS 2166 10. weighted › cyclic vs. The N x N matrix of predecessors, which can be used to reconstruct the shortest paths. MINIMUM-WEIGHT SPANNING TREE 49 4. Year 2001 Predicted year 2005 Graph 2: Evolution of flu epidemic behavior. CS 312 Lecture 26 Finding Shortest Paths Finding Shortest Paths. C++ Program to Generate a Random UnDirected Graph for a Given Number of Edges; Shortest Path in a Directed Acyclic Graph. The weight of the path P is the sum of the weights of all the _____ on the path P. This problem could be solved easily using (BFS) if all edge weights were ($$1$$), but here weights can take any value. The length of a path is the sum of the lengths of all component edges. The weighted inverse distance is the total number of shortest paths from node \(s G From the Betweenness Centrality in Street Networks to Structural Invariants in Random Planar Graphs. In order to solve the load-balancing problem for coarse-grained parallelization, the relationship between the computing time of a single-source shortest-path length of node and the features of node is studied. So, the shortest path would be of length 1 and BFS would correctly find this for us. The total length of a path is the sum of the lengths of its component edges. In the notes below I am going to describe the Dijkstra algorithm, which is a widely-used algorithm for finding shortest paths in weighted, directed graphs. 2 commits 1 branch. In this post, I explain the single-source shortest paths problems out of the shortest paths problems, in which we need to find all the paths from one starting vertex to all other vertices. 7 (Single-Source Shortest Paths). The Degree of a vertex is the number of edges incident on it. Single-Source Shortest Path on Weighted Graphs. On weighted graphs Weighted Shortest Paths The shortest path from a vertex u to a vertex v in a graph is a | {
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graphs Weighted Shortest Paths The shortest path from a vertex u to a vertex v in a graph is a path w1 = u, w2,…,wn= v, where the sum: Weight(w1,w2)+…+Weight(wn-1,wn) attains its minimal value among all paths that start at u and end at v The length of a path of n vertices is n-1 (the number of edges) If a graph is connected, and the weights are all non-negative, shortest paths exist for any pair of vertices Similarly for strongly connected digraphs with non-negative weights. Write an algorithm to print all possible paths between source and destination. Floyd-Warshall algorithm is used to find all pair shortest path problem from a given weighted graph. In real-life scenarios, the arc weighs in a shortest path of a network/graph have the several parameters which are very hard to define exactly (i. $\begingroup$ Shortest Path on an Undirected Graph? might be interesting. Weighted graphs and path length Weighted graphs A weighted graph is a graph whose edges have weights. Cover Photo by Thor Alvis on Unsplash. The path graph with n vertices is denoted by P n. In this tutorial, we will present a general explanation of both algorithms. Implementation of Dijkstra's algorithm in C++ which finds the shortest path from a start node to every other node in a weighted graph. There is a simple tweak to get from DFS to an algorithm that will find the shortest paths on an unweighted graph. the path itself, not just its length) between the source vertex given in from, to the target vertices given in to. (Consider what this means in terms of the graph shown above right. Furthermore, if we perform relaxation on the set of edges once, then we will at least have determined all the one-edged shortest paths; if we traverse the set of edges twice, we will have solved at least all the two-edged shortest paths; ergo, after the V-1 iteration. However, most communication networks are dynamic, i. shortest path functions use it as the cost of the path; community finding methods use it as the | {
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i. shortest path functions use it as the cost of the path; community finding methods use it as the strength of the relationship between two vertices, etc. More Algorithms for All-Pairs Shortest Paths in Weighted Graphs Timothy M. Shortest paths in an edge-weighted digraph 4->5 0. Graphs can be weighted (edges carry values) and directional (edges have direction). The Degree of a vertex is the number of edges incident on it. In the weighted matching [G85b, GT89, GT91] and maxi-mum flow problems [GR98], for instance, the best algorithms for real- and integer-weighted graphs have running times differing by a polynomial factor. Implementation of Dijkstra's algorithm in C++ which finds the shortest path from a start node to every other node in a weighted graph. $\begingroup$ Shortest Path on an Undirected Graph? might be interesting. Simple path is a path with no repeated vertices. Is there a cycle in the graph? Euler tour. So, the shortest path would be of length 1 and BFS would correctly find this for us. First, the paths should be shortest, then there might be more than one such shortest paths whose length are the same. A path in a graph is a sequence of adjacent vertices. Mark Dolan CIS 2166 10. goldberg_radzik (G, source[, weight]) Compute shortest path lengths and predecessors on shortest paths in weighted graphs. Increasingly, there is interest in using asymmetric structure of data derived from Markov chains and directed graphs, but few metrics are specifically adapted to this task. We may also want to associate some cost or weight to the traversal of an edge. It is a real time graph algorithm, and can be used as part of the normal user flow in a web or mobile application. You may print the results of this algorithm to the screen or to a log file. shortest_paths calculates a single shortest path (i. However, the robot had a means for estimating Euclidean distances in his world. PY - 2007/10/30. More Algorithms for All-Pairs Shortest Paths in Weighted Graphs Timothy | {
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his world. PY - 2007/10/30. More Algorithms for All-Pairs Shortest Paths in Weighted Graphs Timothy M. Based on Data Structures, Algorithms & Software Principles in CT. Finding shortest paths in weighted graphs In the past two weeks, you've developed a strong understanding of how to design classes to represent a graph and how to use a graph to represent a map. We ˙rst propose an exact (and deterministic) al-. In any graph G, the shortest path from a source vertex to a destination vertex can be calculated using Dijkstra Algorithm. , capacity, cost, demand, traffic frequency, time, etc. A subgraph is a subset of a graph’s edges (with associated vertices) that form a graph. Path Graphs. dijkstra_path¶ dijkstra_path (G, source, target, weight='weight') [source] ¶. This article presents a Java implementation of this algorithm. The previous state of the art for this problem was total update time O ̃(n 2 √m/ε) for directed, unweighted graphs [2], and Õ(mn=ε) for undirected, unweighted graphs [12]. Weighted Graphs Data Structures & Algorithms 2 [email protected] ©2000-2009 McQuain Shortest Paths (SSAD) Given a weighted graph, and a designated node S, we would like to find a path of least total weight from S to each of the other vertices in the graph. Chapter 4 Algorithms in edge-weighted graphs Recall that anedge-weighted graphis a pair(G,w)whereG=(V,E)is a graph andw:E →IR number of edges in a shortest path. , capacity, cost, demand, traffic frequency, time, etc. So, for this purpose we use the retroactive priority queue which allows to perform opeartions at any point of time. The Degree of a vertex is the number of edges incident on it. Shortest path algorithms are a family of algorithms designed to solve the shortest path problem. BFS always visits nodes in increasing order of their distance from the source. These values become important when calculating the. weighted graphs: find a path from a given source to a given target such that the consecutive weights on the path | {
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find a path from a given source to a given target such that the consecutive weights on the path are nondecreasing and the last weight on the path is minimized. A single execution of the algorithm will. Analyze your algorithm. ) B C A E D F 0 7 2 3 5 8 8 4 7 1 2 5 2. , its number of edges. Finding the Shortest Path in Weighted Directed Acyclic Graph For the graph above, starting with vertex 1, what're the shortest paths(the path which edges weight summation is minimal) to vertex 2. The Classical Dijkstra’s algorithm [16] solves the single-source shortest path problems in a simple graph. Shortest Distance in a graph with two different weights : Given a weighted undirected graph having A nodes, a source node C and destination node D. If G is a weighted graph, the length/weight of a path is the sum of the weights of the edges that compose the path. Dimitrios Skrepetos, PhD candidate David R. Check the manual pages of the functions working with weighted graphs for details. Another source vertex is also provided. A cycle is a path where the first and last vertices are the same. A subgraph is a subset of a graph’s edges (with associated vertices) that form a graph. Print the number of shortest paths from a given vertex to each of the vertices. However, the resulting algorithm is no longer called DFS. If There Is An Edge Between I And 1, Then G[16] > 0, Otherwise G[i,j]=-1. The weight of the path P is the sum of the weights of all the _____ on the path P. We will see how simple algorithms like depth-first-search can be used in clever ways (for a problem known as topological sorting) and will see how Dynamic Programming can be used to solve problems of finding shortest paths. Dijkstra's Algorithm. ; It uses a priority based set or a queue to select a node / vertex nearest to the source that has not been edge relaxed. But how should we evaluate a path in a weighted graph? Usually, a path is assessed by the sum of weights of its edges and, based on this assumption, many authors | {
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a path is assessed by the sum of weights of its edges and, based on this assumption, many authors proposed their. A graph is a series of nodes connected by edges. (2018) A Faster Distributed Single-Source Shortest Paths Algorithm. One algorithm for finding the shortest path from a starting node to a target node in a weighted graph is Dijkstra's algorithm. Count the number of updates At each step we compute the shortest path through a subset of vertices. When There Is An Edge Between I And J, Then G[i][j] Denotes The Weight Of The Edge. In this tutorial, we will present a general explanation of both algorithms. Single source shortest path for undirected graph is basically the breadth first traversal of the graph. Assignment: Given any connected, weighted graph G, use Dijkstras algorithm to compute the shortest (or smallest weight) path from any vertex a to any other vertex b in the graph G. If There Is An Edge Between I And 1, Then G[16] > 0, Otherwise G[i,j]=-1. G∗ contains threeshortcuts: v8,v9, v9,v7,and v9,v10. One problem might be the shortest path in a given undirected, weighted graph. GUVEW ,, eE , where (non-negative real number) is a weight function, by which each edge is associ-ated with a weight The weight of a matching M is. 1 def shortest_path_cycle(graph, s): 2 '''Single source shortest paths using DP on a graph with cycles but no 3 negative cycles. num_vertices() 11 for i in range(num_vertices): 12 result. We consider a specific infinite graph here, namely the honeycomb grid. Fine the shortest weighted path from a vertex, s, to every other vertex in the graph. We introduce a metric on the. The N x N matrix of predecessors, which can be used to reconstruct the shortest paths. There are already comprehensive network analysis packages in R, notably igraph and its tidyverse-compatible interface tidygraph. None of these algorithms for the WRP permit one to bound the number of links/turns in the produced path. | {
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# How to prove $n!>(\frac{n}{e})^{n}$
Prove that $n!>\left(\dfrac{n}{e}\right)^{n}$.
I used induction principle but cannot solve it for the $(m+1)$-th term after taking the $m$th term to be true.
Here the key is to use the appropriate definition of $e^x$, namely:
$$e^x = \sum_{k=0}^{\infty}\frac{1}{k!}x^k$$
Plugging in $x = n$ we get
$$e^n = \sum_{k=0}^{\infty} \frac{1}{k!}n^k$$
and hence, breaking this sum up a little we get our inequality: $$n! e^n = n^n + \sum_{k\ne n} \frac{n!}{k!}n^k > n^n$$
Hint: Show that $$\ln(n!)=\sum_{k=1}^n \ln k >\int_1^n\ln x\, dx.$$
• That's a clever approach! Nice one! Aug 25 '13 at 18:02
Inductively, if $n!>\frac{n^n}{e^n}$ and you multiply both sides by $n+1$, then you have that $(n+1)!>(n+1)\frac{n^n}{e^n}$, so it suffices to prove that $(n+1)\frac{n^n}{e^n}>\frac{(n+1)^{n+1}}{e^{n+1}}$. Can you continue from here?
• +1 This requires the least amount of machinery of the presented solutions, I think. Aug 25 '13 at 17:11
Hint: write out the series for $e^n$ and pick out a relevant term amongst the positive terms which make up the sum.
• You posted a hint and Deven independently posted a solution using the same idea => 10 vote difference. Isn't that always so? +1 to both of you from me has been there from the beginning, of course. Aug 25 '13 at 18:10
I had originally written this up for another question but it seems fitting here as well. Maybe this can help someone.
Depending on how you introduced $e$, you might be able to use the fact that there are two sequences $(a_n)_{n \in \mathbb{N}}$, $(b_n)_{n \in \mathbb{N}}$ with
\begin{align} a_n ~~~&:=~~~ \left ( 1 + \frac{1}{n} \right ) ^n \\ ~ \\ b_n ~~~&:=~~~ \left ( 1 - \frac{1}{n} \right ) ^{-n} \end{align}
and
$$\underset{n \rightarrow \infty}{\lim} a_n ~~~=~~~ \underset{n \rightarrow \infty}{\lim} b_n ~~~=~~~ e \\ ~ \\$$
While both sequences converge to the same limit, $a_n$ approaches from the bottom and $b_n$ approaches from the top: | {
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import numpy as np
import matplotlib.pyplot as plt
from matplotlib import rcParams
rcParams.update({'figure.autolayout': True})
pts = np.arange(0, 20, 1)
a_n = lambda n: (1+1/n)**n
b_n = lambda n: (1-1/n)**(-n)
plt.errorbar(x = pts, xerr = None, y = a_n(pts), yerr = None, fmt = "bx", markersize = "5", markeredgewidth = "2", label = "$a_n$")
plt.errorbar(x = pts, xerr = None, y = b_n(pts), yerr = None, fmt = "rx", markersize = "5", markeredgewidth = "2", label = "$b_n$")
plt.plot(pts, [np.exp(1)]*len(pts), color = "black", linewidth = 2, label = "$e$")
plt.xlim(1.5, 14.5)
plt.ylim(2.0, 3.5)
plt.legend(loc = "best")
plt.setp(plt.gca().get_legend().get_texts(), fontsize = "22")
plt.show()
So we're going to use the following inequality:
$$\forall n \in \mathbb{N} ~ : ~~~~~ \left ( 1 + \frac{1}{n} \right ) ^n ~~~~<~~~~ e ~~~~<~~~~ \left ( 1 - \frac{1}{n} \right ) ^{-n} \tag*{\circledast} \\ ~ \\$$
Thesis
$$\forall n \in \mathbb{N}, ~ n \geq 2 ~ : ~~~~~ e \cdot \left ( \frac{n}{e} \right )^n ~~~~<~~~~ n! ~~~~<~~~~ n \cdot e \cdot \left ( \frac{n}{e} \right )^n \\ ~ \\$$
Proof By Induction
Base Case
We begin with $n = 2$ and get
\begin{align} & ~ && e \cdot \left ( \frac{2}{e} \right )^2 ~~~~&&<~~~~ 2! ~~~~&&<~~~~ 2 \cdot e \cdot \left ( \frac{2}{e} \right )^2 \\ ~ \\ & \Leftrightarrow && e \cdot \frac{4}{e^2} ~~~~&&<~~~~ 1 \cdot 2 ~~~~&&<~~~~ 2 \cdot e \cdot \frac{4}{e^2} \\ ~ \\ & \Leftrightarrow && \frac{4}{e} ~~~~&&<~~~~ 2 ~~~~&&<~~~~ \frac{8}{e} \\ ~ \\ &\Leftrightarrow && 2 ~~~~&&<~~~~ e ~~~~&&<~~~~ 4 ~~~~ \\ \end{align}
Which is a true statement.
Inductive Hypothesis
Therefore the statement holds for some $n$. $\tag*{$\text{I.H.}$}$
Inductive Step | {
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Therefore the statement holds for some $n$. $\tag*{$\text{I.H.}$}$
Inductive Step
\begin{align} & ~ && e \cdot \left ( \frac{n+1}{e} \right )^{n+1} \\ ~ \\ & = && (n+1) \cdot \frac{1}{e} \cdot e \cdot \left ( \frac{n+1}{e} \right )^n\\ ~ \\ & = && (n+1) \cdot \left ( \frac{n}{e} \right )^n \cdot \left ( \frac{n+1}{n} \right )^n\\ ~ \\ & = && (n+1) \cdot \left ( \frac{n}{e} \right )^n \cdot \left ( 1 + \frac{1}{n} \right )^n\\ ~ \\ & \overset{\circledast}{<} && (n+1) \cdot \left ( \frac{n}{e} \right )^n \cdot e\\ ~ \\ & \overset{\text{I.H.}}{<} && (n+1) \cdot n!\\ ~ \\ & = && (n+1)!\\ ~ \\ & = && (n+1) \cdot n!\\ ~ \\ & \overset{\text{I.H.}}{<} && (n+1) \cdot n \cdot e \cdot \left ( \frac{n}{e} \right )^n\\ ~ \\ & = && (n+1) \cdot e \cdot \left ( \frac{n}{e} \right )^{n+1} \cdot e \\ ~ \\ & = && (n+1) \cdot e \cdot \left ( \frac{n+1}{e} \right )^{n+1} \cdot \left ( \frac{n}{n+1} \right )^{n+1} \cdot e \\ ~ \\ & = && (n+1) \cdot e \cdot \left ( \frac{n+1}{e} \right )^{n+1} \cdot \left ( 1 - \frac{1}{n+1} \right )^{n+1} \cdot e \\ ~ \\ & \overset{\circledast}{<} && (n+1) \cdot e \cdot \left ( \frac{n+1}{e} \right )^{n+1} \cdot \left ( 1 - \frac{1}{n+1} \right )^{n+1} \cdot \left ( 1 - \frac{1}{n+1} \right )^{-(n+1)} \\ ~ \\ & = && (n+1) \cdot e \cdot \left ( \frac{n+1}{e} \right )^{n+1} \\ ~ \\ \end{align}
Conclusion
Therefore the statement holds $\forall n \in \mathbb{N}, ~ n \geq 2$. $$\tag*{\square}$$ | {
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# Probability of drawing cards on specific draw counts
There is a deck of $30$ cards, each card labeled a number from $1$ to $15$, with exactly $2$ copies of a card for each number. You draw $8$ cards. What is the probability that you draw the number '$1$' card by the $5$th draw (on the $5$th draw or before that), AND also drawing the number '$2$' card on or before the $8$th draw?
I know how to compute the probability of drawing both the cards on or before the $5$th draw:
$$\frac{\binom{2}{1}\cdot \binom{2}{1} \cdot \binom{26}{3}}{\binom{30}{5}}$$
Since there's $2$ ways to choose from each of the '$1$' and '$2$' cards, and then there's $26$ cards left after those $4$ cards so the other $3$ cards can be any of those $26$, and the total number of combinations you can draw $5$ cards from $30$.
But we want to expand this search to $8$ draws, and also at the same time want to have assumed that we have already drawn the '$1$' card on or before the $5$th draw (if we don't get the '$2$' card by the $5$th draw. How can I combine these ideas? Thanks
• number 2 before 8-th draw... Then the 8th draw is irrelevant? Don't you mean (again) "on or before"? – drhab Sep 3 '17 at 9:07
• @user152294 Just to check my try, have you the result of this exercise? – Robert Z Sep 3 '17 at 9:46
• @drhab Yes, on or before – user152294 Sep 3 '17 at 17:29
• @RobertZ No, I don't have the solution unfortunately – user152294 Sep 3 '17 at 17:30
• @user152294 "before 8-th draw" means "on or before 8-th draw"? In case I have to modify my solution. P.S. Where does his exercise come from? – Robert Z Sep 3 '17 at 17:33
I think it is more convenient to evaluate the probability of the complementary event: draw NO number '1' card by the $5$th draw, OR draw NO number '2' card. Here we consider the 2 copies of a card with the same number distinguishable (for example assume that one is red and the other is blue). Let $n^{\underline{k}}:=n(n-1)\cdots(n-k+1)$. | {
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1) If we draw NO number '1' card by the 5th draw then we can have zero, one or two '1's in the $6$th, $7$th or $8$th draw $$p_1=\frac{1}{30^{\underline{8}}}\left(28^{\underline{8}}+(3+3)\cdot 28^{\underline{7}}+3\cdot 2\cdot 28^{\underline{6}}\right)$$
2) If we draw NO number '2' card then $$p_2=\frac{28^{\underline{8}}}{30^{\underline{8}}}$$
3) If we draw NO number '1' card by the $5$th draw AND NO number '2' card then, similarly to case 1), $$p_3=\frac{1}{30^{\underline{8}}}\left(26^{\underline{8}}+(3+3)\cdot 26^{\underline{7}}+3\cdot 2\cdot 26^{\underline{6}}\right)$$
Hence, the desired probability is $$p=1-(p_1+p_2-p_3)=211/1566\approx 0.134738.$$
• Is it also possible to just do these two cases? 1) Draw a '1' and '2' before the 5th draw, or 2) Draw a '1' before the 5th draw AND draw a '2' before the 8th draw? I'm not sure how to express 2) but would it be more cumbersome than the complementary events? – user152294 Sep 3 '17 at 17:34
• @user152294 Yes you can consider two cases, but I think that with 3 cases is simpler. – Robert Z Sep 3 '17 at 17:36
• How come in this solution, we don't have to use binomial coefficients? – user152294 Sep 3 '17 at 17:43
• @user152294 Let me know if the official solution is the one that I have found. – Robert Z Sep 3 '17 at 17:44
• @user152294 $3$ is the number of ways to place the red $1$ (position 6,7, 8), $3$ is the number of ways to place the blue $1$ (position 6,7, 8) and $3\cdot 2$ is the number of ways to place the red $1$ and blue $1$ (positions (6,7), (6,8), (7,8), (7,6), (8,6), (8,7)). $28^{\underline{8}}$, $28^{\underline{7}}$ and $28^{\underline{6}}$ are the number of ways to fill the remaining positions (cards different from 1). – Robert Z Sep 3 '17 at 18:21 | {
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# What does $f|_A$ mean? [duplicate]
If $f$ a is a function and $A$ is a set, what could the notation
$$f|_A$$
mean? Is it perhaps "restricted to set $A$"?
• Yes, if $f$ is a function with domain $D$, then for a set $A \subseteq D$, $f|_A$ is used for the restriction of $f$ to $A$. Feb 2, 2016 at 15:43
• That's what I would assume. You might have two functions $f$ and $g$ for which $f \neq g$ but $f|_A = g|_A$ for some $A$. Feb 2, 2016 at 15:43
It means that I am constricting the domain of the function $f$. If $f:X\to Y$, then $g=f|_A$ means that $g:A\to Y$ where $A\subseteq X$.
Intuitively speaking, a function $f$ is constituted of three ingredients:
• a domain;
• a codomain;
• a rule (that, for each element in the domain, assigns a unique element in the codomain).
If we change any of these three ingredients, we obtain a different function. In particular, if we change the domain by a subset $A$ of the original domain (keeping the codomain and the rule), we get a new function which is represented by $f|_A$.
In other words: given a function $f:X\to Y$ and a set $A\subset X$, the notation $f|_A$ denotes the function $g:A\to Y$ given by $$g(x)=f(x),\quad \forall \ x\in A.$$
This is the usual meaning but, maybe, there are different meanings in other contexts.
That's correct.
Suppose we have a function $$f : Y \leftarrow X,$$ and a subset $A$ of $X$.
Approach 0. Then $f \restriction_A$ is defined as the unique function $Y \leftarrow A$ that agrees with $f$ on $A$. That is: $$\mathop{\forall}_{a \in A} ((f \restriction_A)(a) = f(a))$$
However, there's a cleaner way of formalizing this.
Approach 1. Write $$\mathrm{incl}_A : X \leftarrow A$$ for the inclusion of $A$ into $X$. Then we can form the composite $$f \circ \mathrm{incl}_A : Y \leftarrow A.$$ Write $f\restriction_A$ as a shorthand for this composite.
The nice thing about Approach 1 is that it makes proving the basic properties of the restriction operator trivial. In particular: | {
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Claim. $(g \circ f)\restriction_A = g \circ (f \restriction_A)$
Proof.
$$(g \circ f)\restriction_A = (g \circ f) \circ \mathrm{incl}_A = g \circ (f \circ \mathrm{incl}_A) = g \circ (f \restriction_A)$$
Approach 1 is especially appealing from a category-theory perspective. In that context:
• $X$ is an object of some category
• a subobject of $X$ is by definition an object $\underline{A}$ together with a monomorphism $\mathrm{incl}_A : X \leftarrow \underline{A}$.
• a partial morphism $Y \leftarrow X$ consists of a subobject $A$ of $X$ together with a morphism $Y \leftarrow \underline{A}$.
Hence, if we're given a morphism $f : Y \leftarrow X$ and a subobject $A$ of $X$, then we get a partial morphism $f \restriction_A : Y \leftarrow X$ by forming the obvious composite.
The notation $f|_A$ is probably best understood via a meaningful example. Before giving one (I hope it will be useful, anyway), it would probably be good to consult two decent references:
2) Abstract Algebra by Dummit and Foote (p. 3, 3rd Ed.).
The relevant portion from the Wiki blurb:
Let $f\colon E\to F$ be a function from a set $E$ to a set $F$, so that the domain of $f$ is in $E$ (i.e., $\operatorname{dom}f\subseteq E$). If $A\subseteq E$, then the restriction of $f$ to $A$ is the function $f|_A\colon A\to F$.
Informally, the restriction of $f$ to $A$ is the same function as $f$, but is only defined on $A\cap\operatorname{dom} f$.
Wiki's "informal" remark is the key part in my opinion. The following excerpt from Dummit and Foote's Abstract Algebra may be slightly more abstract, but I think a meaningful example will clear everything up.
If $A\subseteq B$, and $f\colon B\to C$, we denote the restriction of $f$ to $A$ by $f|_A$. When the domain we are considering is understood we shall occasionally denote $f|_A$ again simply as $f$ even though these are formally different functions (their domains are different). | {
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If $A\subseteq B$ and $g\colon A\to C$ and there is a function $f\colon B\to C$ such that $f|_A=g$, we shall say $f$ is an extension of $g$ to $B$ (such a map $f$ need not exist nor be unique).
Example: Let $g\colon\mathbb{Z}^+\to\{1\}$ be defined by $g(x)=1$ and let $f\colon\mathbb{Z}\setminus\{0\}\to\{1\}$ be defined by $f(x)=\dfrac{|x|}{x}$. Using the notation from the second paragraph above, we have $g\colon A\to C$ and $f\colon B\to C$, where
• $A = \mathbb{Z^+}$
• $B=\mathbb{Z}\setminus\{0\}$
• $C=\{1\}$
and, clearly, $A\subseteq B$. Thus, we have the following: \begin{align} f|_A &\equiv f\colon\mathbb{Z^+}\to\{1\}\tag{by definition}\\[0.5em] &= \frac{|x|}{x}\tag{by definition}\\[0.5em] &= \frac{x}{x}\tag{if $x\in\mathbb{Z^+}$, then $|x|=x$ }\\[0.5em] &= 1\tag{simplify}\\[0.5em] &\equiv g\colon\mathbb{Z^+}\to\{1\}\tag{by definition}\\[0.5em] &= g. \end{align} Apart from some slight notational abuse, perhaps, the above example shows that $f$ is an extension of $g$ to $B$ since $f|_A=g$. | {
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Mean and Variance of "Piecewise" Normal Distribution
Note - I put piecewise in quotes because I don't think it's the right term to use (I can't figure out what to call it).
I am building a program to model the load that a user places on a server. The load a user produces follows a normal distribution. Depending on which application the user is using, however, the mean and variance of that normal distribution will be different. What I am trying to do is calculate the overall mean and variance for a user's activity given the proportions of time they use each application.
For example, Application A follows $\mathcal{N}(100, 50)$ and Application B follows $\mathcal{N}(500, 20)$. If a user uses A 50% of the time and B the other 50%, what is the mean and variance of the data that the user would produce during a day?
I'm able to simulate this by selecting a number from a uniform distribution between 0 and 1 and then generating a value from the appropriate distribution. Something like this:
$f(x) = \begin{cases} \mathcal{N}(100, 50), &0 \le x \lt 0.5\\ \mathcal{N}(500, 20), &0.5 \le x \lt 1\\ \end{cases}$
When I simulate a large number of these values and measure the results, it looks like the mean is just
$\sum\limits_{i=1}^n\mu_ip$
where $p$ is the percentage of the day a user is using each application.
I can't figure out what pattern the variance follows or what the formula might be to determine it without measuring a bunch of simulated values (When I simulate the above example, the variance looks to be something close to 41500).
I'd appreciate confirmation that how I'm calculating the combined mean is correct and some help in figuring out how to determine the variance of the overall distribution.
Let the two normal random variables be $X$ and $Y$, where $X$ is chosen with probability $p$, and $Y$ is chosen with probability $q=1-p$.
If $W$ is the resulting random variable, then $\Pr(W\le w)=p\Pr(X\le w)+q\Pr(Y\le w)$. | {
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If $W$ is the resulting random variable, then $\Pr(W\le w)=p\Pr(X\le w)+q\Pr(Y\le w)$.
Differentiate. We get $f_W(w)=pf_X(w)+qf_Y(w)$.
The mean of $W$ is $\int_{-\infty}^\infty wf_W(w)$. Calculate. We get $$\int_{-\infty}^\infty w(pf_X(w)+qf_Y(w))\,dw.$$ This is $pE(X)+qE(Y)$, confirming your observation.
For the variance, we want $E(W^2)-(E(W))^2$. For $E(W^2)$, we calculate $\int_{-\infty}^{\infty} w^2(pf_X(w)+qf_Y(w))\,dw$. This is $pE(X^2)+qE(Y^2)$.
But $pE(X^2)= p(\text{Var}(X)+(E(X))^2)$ and $qE(Y^2)= q(\text{Var}(Y)+(E(Y))^2)$
Putting things together we get $$\text{Var}(W)=p\text{Var}(X)+q\text{Var}(Y)+ p(E(X))^2+q(E(Y))^2- (pE(X)+qE(Y))^2.$$
Remark: For a longer discussion, please look for Mixture Distributions.
• So, to put it more generally: $\text{Var}(W) = \sum\limits_{i=1}^np_i(\text{Var}(X_i) + E(X_i)^2) - (\sum\limits_{i=1}^np_iE(X_i))^2$ Aug 5, 2014 at 17:57
• Yes, the same argument works for $X_i$ with probability $p_i$. One could generalize further, to sums to $\infty$, also to "continuous" mixtures. Mixtures happen a lot in Statistics, since populations can often be stratified. Aug 5, 2014 at 18:15
• Perfect. Thank you so much for your help! I already ran some simulations with a few different combinations and it works beautifully. Aug 5, 2014 at 18:47
• You are welcome. Fast work. Aug 5, 2014 at 19:49 | {
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What you are wanting are the distributional quantities of the mixture distribution. There are some convenient formulas to do this: let $T$ be the unconditional load a user places on the system, and let $X$ be a Bernoulli random variable that indicates whether a user is on system $A$ or $B$. So $A = (T \mid X = 0) \sim \mathrm{Normal}(\mu_A = 100, \sigma_A^2 = 50)$, and $B = (T \mid X = 1) \sim \mathrm{Normal}(\mu_B = 500, \sigma_B^2 = 20)$. That is to say, \begin{align*} \mathrm{E}[T \mid X = 0] &= \mu_A, \\ \mathrm{E}[T \mid X = 1] &= \mu_B, \\ \mathrm{Var}[T \mid X = 0] &= \sigma_A^2, \\ \mathrm{Var}[T \mid X = 1] &= \sigma_B^2. \end{align*} | {
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Then by the law of total expectation, \begin{align*} \mathrm{E}[T] &= \mathrm{E}[\mathrm{E}[T \mid X]] \\ &= \mathrm{E}[T \mid X = 0]\Pr[X = 0] + \mathrm{E}[T \mid X = 1]\Pr[X = 1] \\ &= \mu_A (1-p) + \mu_B p,\end{align*} where $p$ is the probability that a user is on system $B$. The variance is calculated by \begin{align*} \mathrm{Var}[T] &= \mathrm{E}[\mathrm{Var}[T \mid X]] + \mathrm{Var}[\mathrm{E}[T \mid X]] \\ &= \mathrm{Var}[T \mid X = 0]\Pr[X = 0] + \mathrm{Var}[T \mid X = 1]\Pr[X = 1] + \mathrm{Var}[\mathrm{E}[T \mid X]] \\ &= \sigma_A^2 (1-p) \sigma_B^2 p + \mathrm{Var}[\mathrm{E}[T \mid X]]. \end{align*} The last term requires a little subtlety to understand. The variable $\mathrm{E}[T \mid X]$ is a generalized Bernoulli, which takes on the value $\mu_A$ with probability $1-p$ and $\mu_B$ with probability $p$ (rather than $0$ and $1$). So we may write this as $$\mathrm{E}[T \mid X] = X(\mu_B - \mu_A) + \mu_A,$$ where $X \sim \mathrm{Bernoulli}(p)$. Therefore, $$\mathrm{Var}[\mathrm{E}[T \mid X]] = \mathrm{Var}[(\mu_B - \mu_A)X + \mu_A] = (\mu_B - \mu_A)^2 \mathrm{Var}[X] = (\mu_B - \mu_A)^2 p(1-p).$$ Hence $$\mathrm{Var}[T] = \sigma_A^2 (1-p) + \sigma_B^2 p + (\mu_B - \mu_A)^2 p(1-p).$$ The general case with $n$ systems $S_i \sim \mathrm{Normal}(\mu_i, \sigma_i^2)$, $i = 1, 2, \ldots, n$, where the user is on system $i$ with a probability of $p_i$, with $\sum_{i=1}^n p_i = 1$, involves a categorical distribution rather than a Bernoulli. | {
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Recall that $V[X]=E[X^2]-E[X]^2$. Hence if you are given $E$ and $V$ of $X_1,X_2$ and combine them as described to a new random variable $Y$ by using a 0-1-random variable $Z$, i.e. picking $X_1$ if $Z=1$ and picking $X_2$ if $Z=0$, we find $$E[Y]=P(Z=1)\cdot E[Y|Z=1]+P(Z=0)\cdot E[Y|Z=0]=pE[X_1]+(1-p)E[X_2].$$ By the same argument we find $$E[Y^2] = pE[X_1^2]+(1-p)E[X_2^2]$$ Substituting $E[X_i^2]=V[X_i]+E[X_i]^2$, we obtain \begin{align}V[Y]&=E[Y^2]-E[Y]^2 \\&= p(V[X_1]+E[X_1]^2) + (1-p)(V[X_2]+E[X_2]^2)-(pE[X_1]+(1-p)E[X_2])^2\\ &=pV[X_1]+(1-p)V[X_2]+p(1-p)(E[X_1]^2+E[X_2]^2)-2p(1-p)E[X_1]E[X_2]\\ &=pV[X_1]+(1-p)V[X_2]+p(1-p)(E[X_1]-E[X_2])^2.\end{align} | {
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# Linear Equations
Module by: Rupinder Sekhon. E-mail the author
Summary: This chapter covers principles of linear equations. After completing this chapter students should be able to: graph a linear equation; find the slope of a line; determine an equation of a line; solve linear systems; and complete application problems using linear equations.
## Chapter Overview
In this chapter, you will learn to:
1. Graph a linear equation.
2. Find the slope of a line.
3. Determine an equation of a line.
4. Solve linear systems.
5. Do application problems using linear equations.
## Graphing a Linear Equation
Equations whose graphs are straight lines are called linear equations. The following are some examples of linear equations:
2x3y=62x3y=6 size 12{2x - 3y=6} {}, 3x=4y73x=4y7 size 12{3x=4y - 7} {}, y=2x5y=2x5 size 12{y=2x - 5} {}, 2y=32y=3 size 12{2y=3} {}, and x2=0x2=0 size 12{x - 2=0} {}.
A line is completely determined by two points, therefore, to graph a linear equation, we need to find the coordinates of two points. This can be accomplished by choosing an arbitrary value for xx size 12{x} {} or yy size 12{y} {} and then solving for the other variable.
### Example 1
#### Problem 1 | {
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### Example 1
#### Problem 1
Graph the line: y=3x+2y=3x+2 size 12{y=3x+2} {}
### Example 2
#### Problem 1
Graph the line: 2x+y=42x+y=4 size 12{2x+y=4} {}
The points at which a line crosses the coordinate axes are called the intercepts. When graphing a line, intercepts are preferred because they are easy to find. In order to find the x-intercept, we let y=0y=0 size 12{y=0} {}, and to find the y-intercept, we let x=0x=0 size 12{x=0} {}.
### Example 3
#### Problem 1
Find the intercepts of the line: 2x3y=62x3y=6 size 12{2x - 3y=6} {}, and graph.
### Example 4
#### Problem 1
Graph the line given by the parametric equations: x=3+2tx=3+2t size 12{x=3+2t} {}, y=1+ty=1+t size 12{y=1+t} {}
### Horizontal and Vertical Lines
When an equation of a line has only one variable, the resulting graph is a horizontal or a vertical line.
The graph of the line x=ax=a size 12{x=a} {}, where aa size 12{a} {} is a constant, is a vertical line that passes through the point ( aa size 12{a} {}, 0). Every point on this line has the x-coordinate aa size 12{a} {}, regardless of the y-coordinate.
The graph of the line y=by=b size 12{y=b} {}, where bb size 12{b} {} is a constant, is a horizontal line that passes through the point (0, bb size 12{b} {}). Every point on this line has the y-coordinate bb size 12{b} {}, regardless of the x-coordinate.
#### Example 5
##### Problem 1
Graph the lines: x=2x=2 size 12{x= - 2} {} , and y=3y=3 size 12{y=3} {}.
## Slope of a Line
### Section Overview
In this section, you will learn to:
1. Find the slope of a line if two points are given.
2. Graph the line if a point and the slope are given.
3. Find the slope of the line that is written in the form y=mx+by=mx+b size 12{y= ital "mx"+b} {}.
4. Find the slope of the line that is written in the form Ax+By=cAx+By=c size 12{ ital "Ax"+ ital "By"=c} {}. | {
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In the last section, we learned to graph a line by choosing two points on the line. A graph of a line can also be determined if one point and the "steepness" of the line is known. The precise number that refers to the steepness or inclination of a line is called the slope of the line.
From previous math courses, many of you remember slope as the "rise over run," or "the vertical change over the horizontal change" and have often seen it expressed as:
riserun, vertical changehorizontal change, ΔyΔxetc.riserun size 12{ { {"rise"} over {"run"} } } {}, vertical changehorizontal change size 12{ { {"vertical change"} over {"horizontal change"} } } {}, ΔyΔx size 12{ { {Δy} over {Δx} } } {}etc.
(5)
We give a precise definition.
Definition 1:
If ( x1x1 size 12{x rSub { size 8{1} } } {}, y1y1 size 12{y rSub { size 8{1} } } {}) and ( x2x2 size 12{x rSub { size 8{2} } } {}, y2y2 size 12{y rSub { size 8{2} } } {}) are two different points on a line, then the slope of the line is
Slope = m = y 2 y 1 x 2 x 1 Slope = m = y 2 y 1 x 2 x 1 size 12{"Slope"=m= { {y rSub { size 8{2} } - y rSub { size 8{1} } } over {x rSub { size 8{2} } - x rSub { size 8{1} } } } } {}
### Example 6
#### Problem 1
Find the slope of the line that passes through the points (-2, 3) and (4, -1), and graph the line.
### Example 7
#### Problem 1
Find the slope of the line that passes through the points (2, 3) and (2, -1), and graph.
### Example 8
#### Problem 1
Graph the line that passes through the point (1, 2) and has slope 3434 size 12{ - { {3} over {4} } } {} .
### Example 9
#### Problem 1
Find the slope of the line 2x+3y=62x+3y=6 size 12{2x+3y=6} {}.
### Example 10
#### Problem 1
Find the slope of the line y=3x+2y=3x+2 size 12{y=3x+2} {}.
### Example 11
#### Problem 1
Determine the slope and y-intercept of the line 2x+3y=62x+3y=6 size 12{2x+3y=6} {}.
## Determining the Equation of a Line
### Section Overview
In this section, you will learn to: | {
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## Determining the Equation of a Line
### Section Overview
In this section, you will learn to:
1. Find an equation of a line if a point and the slope are given.
2. Find an equation of a line if two points are given.
So far, we were given an equation of a line and were asked to give information about it. For example, we were asked to find points on it, find its slope and even find intercepts. Now we are going to reverse the process. That is, we will be given either two points, or a point and the slope of a line, and we will be asked to find its equation.
An equation of a line can be written in two forms, the slope-intercept form or the standard form.
The Slope-Intercept Form of a Line: y = mx + b y = mx + b size 12{y= ital "mx"+b} {}
A line is completely determined by two points, or a point and slope. So it makes sense to ask to find the equation of a line if one of these two situations is given.
### Example 12
#### Problem 1
Find an equation of a line whose slope is 5, and y-intercept is 3.
### Example 13
#### Problem 1
Find the equation of the line that passes through the point (2, 7) and has slope 3.
### Example 14
#### Problem 1
Find an equation of the line that passes through the points (–1, 2), and (1, 8).
### Example 15
#### Problem 1
Find an equation of the line that has x-intercept 3, and y-intercept 4.
The Standard form of a Line: Ax + By = C Ax + By = C size 12{ ital "Ax"+ ital "By"=C} {}
Another useful form of the equation of a line is the Standard form.
Let LL size 12{L} {} be a line with slope mm size 12{m} {}, and containing a point (x1,y1)(x1,y1) size 12{ $$x rSub { size 8{1} } ,y rSub { size 8{1} }$$ } {}. If (x,y)(x,y) size 12{ $$x,y$$ } {} is any other point on the line LL size 12{L} {}, then by the definition of a slope, we get | {
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m = y y 1 x x 1 m = y y 1 x x 1 size 12{m= { {y - y rSub { size 8{1} } } over {x - x rSub { size 8{1} } } } } {}
(21)
y y 1 = m ( x x 1 ) y y 1 = m ( x x 1 ) size 12{y - y rSub { size 8{1} } =m $$x - x rSub { size 8{1} }$$ } {}
(22)
The last result is referred to as the point-slope form or point-slope formula. If we simplify this formula, we get the equation of the line in the standard form, Ax+By=CAx+By=C size 12{ ital "Ax"+ ital "By"=C} {}.
### Example 16
#### Problem 1
Using the point-slope formula, find the standard form of an equation of the line that passes through the point (2, 3) and has slope –3/5.
### Example 17
#### Problem 1
Find the standard form of the line that passes through the points (1, -2), and (4, 0).
### Example 18
#### Problem 1
Write the equation y=2/3x+3y=2/3x+3 size 12{y= - 2/3x+3} {} in the standard form.
### Example 19
#### Problem 1
Write the equation 3x4y=103x4y=10 size 12{3x - 4y="10"} {} in the slope-intercept form.
Finally, we learn a very quick and easy way to write an equation of a line in the standard form. But first we must learn to find the slope of a line in the standard form by inspection.
By solving for yy size 12{y} {}, it can easily be shown that the slope of the line Ax+By=CAx+By=C size 12{ ital "Ax"+ ital "By"=C} {} is A/BA/B size 12{ - A/B} {}. The reader should verify.
### Example 20
#### Problem 1
Find the slope of the following lines, by inspection.
1. 3x5y=103x5y=10 size 12{3x - 5y="10"} {}
2. 2x+7y=202x+7y=20 size 12{2x+7y="20"} {}
3. 4x3y=84x3y=8 size 12{4x - 3y=8} {}
Now that we know how to find the slope of a line in the standard form by inspection, our job in finding the equation of a line is going to be very easy.
### Example 21
#### Problem 1
Find an equation of the line that passes through (2, 3) and has slope 4/54/5 size 12{ - 4/5} {}.
If you use this method often enough, you can do these problems very quickly.
## Applications | {
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If you use this method often enough, you can do these problems very quickly.
## Applications
Now that we have learned to determine equations of lines, we get to apply these ideas in real-life equations.
### Example 22
#### Problem 1
A taxi service charges $0.50 per mile plus a$5 flat fee. What will be the cost of traveling 20 miles? What will be cost of traveling xx size 12{x} {} miles?
### Example 23
#### Problem 1
The variable cost to manufacture a product is $10 and the fixed cost$2500. If xx size 12{x} {} represents the number of items manufactured and yy size 12{y} {} the total cost, write the cost function.
### Example 24
#### Problem 1
It costs $750 to manufacture 25 items, and$1000 to manufacture 50 items. Assuming a linear relationship holds, find the cost equation, and use this function to predict the cost of 100 items.
### Example 25
#### Problem 1
The freezing temperature of water in Celsius is 0 degrees and in Fahrenheit 32 degrees. And the boiling temperatures of water in Celsius, and Fahrenheit are 100 degrees, and 212 degrees, respectively. Write a conversion equation from Celsius to Fahrenheit and use this equation to convert 30 degrees Celsius into Fahrenheit.
### Example 26
#### Problem 1
The population of Canada in the year 1970 was 18 million, and in 1986 it was 26 million. Assuming the population growth is linear, and x represents the year and y the population, write the function that gives a relationship between the time and the population. Use this equation to predict the population of Canada in 2010.
## More Applications
### Section Overview
In this section, you will learn to:
1. Solve a linear system in two variables.
2. Find the equilibrium point when a demand and a supply equation are given.
3. Find the break-even point when the revenue and the cost functions are given. | {
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In this section, we will do application problems that involve the intersection of lines. Therefore, before we proceed any further, we will first learn how to find the intersection of two lines.
### Example 27
#### Problem 1
Find the intersection of the line y=3x1y=3x1 size 12{y=3x - 1} {} and the line y=x+7y=x+7 size 12{y= - x+7} {}.
### Example 28
#### Problem 1
Find the intersection of the lines 2x+y=72x+y=7 size 12{2x+y=7} {} and 3xy=33xy=3 size 12{3x - y=3} {} by the elimination method.
### Example 29
#### Problem 1
Solve the system of equations x+2y=3x+2y=3 size 12{x+2y=3} {} and 2x+3y=42x+3y=4 size 12{2x+3y=4} {} by the elimination method.
### Example 30
#### Problem 1
Solve the system of equations 3x4y=53x4y=5 size 12{3x - 4y=5} {} and 4x5y=64x5y=6 size 12{4x - 5y=6} {}.
### Supply, Demand and the Equilibrium Market Price
In a free market economy the supply curve for a commodity is the number of items of a product that can be made available at different prices, and the demand curve is the number of items the consumer will buy at different prices. As the price of a product increases, its demand decreases and supply increases. On the other hand, as the price decreases the demand increases and supply decreases. The equilibrium price is reached when the demand equals the supply.
### Example 31
#### Problem 1
The supply curve for a product is y=1.5x+10y=1.5x+10 size 12{y=1 "." 5x+"10"} {} and the demand curve for the same product is y=2.5x+34y=2.5x+34 size 12{y= - 2 "." 5x+"34"} {}, where xx size 12{x} {} is the price and y the number of items produced. Find the following.
1. How many items will be supplied at a price of $10? 2. How many items will be demanded at a price of$10?
3. Determine the equilibrium price.
4. How many items will be produced at the equilibrium price?
### Break-Even Point | {
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### Break-Even Point
In a business, the profit is generated by selling products. If a company sells xx size 12{x} {} number of items at a price PP size 12{P} {}, then the revenue RR size 12{R} {} is PP size 12{P} {} times xx size 12{x} {} , i.e., R=PxR=Px size 12{R=P cdot x} {}. The production costs are the sum of the variable costs and the fixed costs, and are often written as C=mx+bC=mx+b size 12{C= ital "mx"+b} {}, where xx size 12{x} {} is the number of items manufactured.
A company makes a profit if the revenue is greater than the cost, and it shows a loss if the cost is greater than the revenue. The point on the graph where the revenue equals the cost is called the Break-even point.
### Example 32
#### Problem 1
If the revenue function of a product is R=5xR=5x size 12{R=5x} {} and the cost function is y=3x+12y=3x+12 size 12{y=3x+"12"} {}, find the following.
1. If 4 items are produced, what will the revenue be?
2. What is the cost of producing 4 items?
3. How many items should be produced to break-even?
4. What will be the revenue and the cost at the break-even point?
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# Product of concave functions and harmonic mean
I discovered something interesting, and I would like to know whether it is a known result or not. Say that a function $$f: \Omega \subset \mathbb{R} \rightarrow \mathbb{R_+^*}$$ is $$\alpha$$-concave if $$f^\alpha$$ is concave.
Let $$f$$ a $$\alpha$$-concave function, $$g$$ a $$\beta$$-concave function. Let $$\gamma$$ be the half of the harmonic mean of $$\alpha$$ and $$\beta$$, ie. $$\begin{equation*}\frac{1}{\gamma} = \frac{1}{\alpha} + \frac{1}{\beta}.\end{equation*}$$ Then $$fg$$ is a $$\gamma$$-concave function.
This can be proved by computing the hessian of $$fg$$. Have you already seen this result in the literature?
• A small remark: $\gamma$ is not the harmonic mean, but half of it. – Ivan Izmestiev Nov 24 '18 at 6:59
• You're perfectly right, thanks! I've just corrected it. – LacXav Nov 26 '18 at 15:51
We want to prove that $$(f(ax+by)g(ax+by))^\g \ge a(f(x)g(x))^\g + b (f(y)g(y))^\g$$ for every $$x, y$$ and $$a+b = 1$$. Since we know that $$f(ax+by)^\a \ge af(x)^\a + bf(y)^\a$$ and $$g(ax+by)^\b \ge ag(x)^\b + bg(y)^\b$$ it is enough to prove that
$$(af(x)^\a + bf(y)^\a)^{\frac{\g}{\a}} (ag(x)^\b + bg(y)^\b)^{\frac{\g}{\b}}\ge af(x)^\g g(x)^\g + b f(y)^\g g(y)^\g.$$
Consider set $$\{x, y\}$$ as measurable space with $$m(x) = a$$, $$m(y) = b$$. Then our desired inequality (after rising to the power $$\frac{1}{\g}$$) is nothing but Holder inequality with parameters $$\a, \b$$.
Unfortunatly, I too do not have any reference for this cute result, but I'm sure it is known.
This is nice, and I do not know a reference for this.
The notion of $$p$$-concavity is mentioned, for example, in Section 9 of
Gardner, R. J., The Brunn-Minkowski inequality, Bull. Am. Math. Soc., New Ser. 39, No. 3, 355-405 (2002). ZBL1019.26008.
Note that a natural interpretation of $$0$$-concavity is log-concavity. Your result generalizes the fact that the product of log-concave functions is log-concave. | {
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Also your result is equivalent to the following.
Theorem. If functions $$F$$ and $$G$$ are concave, then so is $$F^tG^{1-t}$$ for all $$t \in [0,1]$$.
Indeed, substitute $$f^\alpha = F$$, $$g^\beta = G$$, $$t = \frac{\beta}{\alpha+\beta}$$.
The concavity of $$F^tG^{1-t}$$ can be proved as follows. For every $$\lambda \in (0,1)$$ and $$\mu = 1-\lambda$$ the concavity of $$F$$ and $$G$$ imply $$\begin{multline*} F^tG^{1-t}(\lambda x + \mu y) = F^t(\lambda x + \mu y) G^{1-t}(\lambda x + \mu y)\\ \ge (\lambda F(x) + \mu F(y))^t (\lambda G(x) + \mu G(y))^{1-t}\\ \ge (\lambda F(x))^t(\lambda G(x))^{1-t} + (\mu F(y))^t(\mu G(y))^{1-t}\\ = \lambda F^tG^{1-t}(x) + \mu F^tG^{1-t}(y) \end{multline*}$$ (In the middle we have used the inequality $$(1+u)^t(1+v)^{1-t} \ge 1+u^t v^{1-t}$$, for which I have only a nasty proof.)
One can prove the theorem also by computing the second derivative of $$F^tG^{1-t}$$ (which is fun), but the above argument is more general because it does not require differentiability.
I think these results should be known, but have no reference.
EDIT: As Alexei Kulikov pointed out, one can prove the Theorem by first proving the special case $$t=\frac12$$ as a lemma (in this case the nasty inequality becomes $$\sqrt{(1+u)(1+v)} \ge 1 + \sqrt{uv}$$, which follows from Cauchy-Schwarz). From the special case one infers by induction all $$t = p/2^n$$ cases, which implies the general case by a limit argument.
EDIT: A reformulation of the theorem: the space of exp-concave functions is convex. | {
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EDIT: A reformulation of the theorem: the space of exp-concave functions is convex.
• It seems to me that your inequality(and the initial problem as well) can be redused to the case $t=\frac{1}{2}$ via mimicking proof of the fact that mid-point concavity implies concavity for continuous functions. And for $t=\frac{1}{2}$ it is just C-S. – Aleksei Kulikov Nov 23 '18 at 23:33
• And moreover it is actually Holder inequality on $2$-point measurable space with appropriate functions. – Aleksei Kulikov Nov 24 '18 at 0:37
• @AlekseiKulikov: I will add the $t=1/2$ approach to my answer. What do you mean by Hoelder inequality on $2$-point measurable space? Could you write this up as an answer? – Ivan Izmestiev Nov 24 '18 at 6:52
• Thanks for the reference, and having put in perspective. – LacXav Nov 26 '18 at 16:35 | {
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# Permutations expressed as product of transpositions
There is a theorem that states that all permutations can be expressed as a product of transpositions. I have a couple of questions about this theorem:
1. Does the product which is equal to the permutation always start from the identity permutation?
In the proof for this theorem our professor has argued that every permutation can be transformed into the identity permutation by applying a certain number of transpositions, e.g. if $\sigma$ is a permutation not equal to the identity permutation then you can apply, say l transpositions, so that you get: $\tau_l \circ \tau_{l-1} \circ .... \circ \tau_1 \circ \sigma = id \Rightarrow \tau_1^{-1} \circ \tau_2^{-1} \circ ... \circ \tau_l^{-1}=\tau_1 \circ \tau_2 \circ .... \circ \tau_l$.
Is this product of transpositions always unique, or can you start from any arbitrary permutation and perform the required number of transpositions to get your permutation?
2 If I form the composition of two permutations, say $\sigma_1$ and $\sigma_2$ given by: $$\sigma_1 = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 3 & 4 &5 &6 &1 &2 \\ \end{pmatrix}$$
$$\sigma_2 = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 2 & 4 &6 &1 &5 &3 \\ \end{pmatrix}$$
I can express them in terms of the following transpositions $$\sigma_1 = (1,5)\circ (2,6) \circ (3,5) \circ (4,6)$$ $$\sigma_2 = (1,4) \circ (2,4) \circ (3,6)$$
When I form the composition $\sigma_1 \circ \sigma_2$ I get:
$$\sigma_1 \circ \sigma_2 = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 4 & 6 &2 &3 &1 &5 \\ \end{pmatrix}$$
But since the product of the transpositions is equal to the permutations, I should get the same result when I use:
$$\sigma_1 \circ \sigma_2 = (1,5)\circ (2,6) \circ (3,5) \circ (4,6) \circ (1,4) \circ (2,4) \circ (3,6)$$
but I get:
$$\sigma_1 \circ \sigma_2 = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 6 & 1 &5 &3 &2 &4 \\ \end{pmatrix}$$
Why doesn't this work? | {
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Why doesn't this work?
• I'm not sure I understand your first question - but for the second one, you're just reading the transpositions in the wrong order. Your convention for $\sigma_1\circ\sigma_2$ is that $\sigma_2$ is applied first, so with the transpositions you need to start from $(3,6)$ and read from right to left. So $1\mapsto 4\mapsto 6\mapsto 2$ etc. – mdp Apr 22 '14 at 13:49
• Thanks for your comment, but I did start from the right side: $\begin{pmatrix} 1 & 2 & 3&4&5&6\end{pmatrix} \Rightarrow \begin{pmatrix} 1 & 2 & 6&4&5&3\end{pmatrix} \Rightarrow \begin{pmatrix} 1 & 4 & 6&2&5&3\end{pmatrix} \Rightarrow \begin{pmatrix} 2 & 4 & 6&1&5&3\end{pmatrix} \Rightarrow \begin{pmatrix} 2 & 4 & 6&3&5&1\end{pmatrix} \Rightarrow \begin{pmatrix} 2 & 4 & 5&3&6&1\end{pmatrix} \Rightarrow \begin{pmatrix} 2 & 1 & 5&3&6&4\end{pmatrix} \Rightarrow \begin{pmatrix} 6 & 1 & 5&3&2&4\end{pmatrix}$ I'm not sure what you mean by $1 \rightarrow 4 \rightarrow 6 \rightarrow 2...$ – eager2learn Apr 22 '14 at 13:58
• Ah, you made a different mistake that gives the same result as doing the transpositions backwards - on the third arrow you should be swapping the numbers $1$ and $4$ over, wherever they appear, not the first position with the fourth. (I was drawing my way of doing this calculation; put $1$ in on the right and see where it goes - it first gets mapped to $4$ (by $(1,4)$), then $4$ is mapped to $6$ by $(4,6)$, and finally $6$ is mapped to $2$ by $(2,6)$). – mdp Apr 22 '14 at 14:04
• Yes that was the mistake I made. Thanks a lot for your help. – eager2learn Apr 22 '14 at 14:33 | {
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Note that the product of transpositions that you used to express $\sigma_1$, when composed, does not again yield $\sigma_1$; I.e., you did not correctly express $\sigma_1$ as a product of transpositions.
Rather, $\sigma_1$ can be written $(1,5)\circ (1, 3)\circ (2, 6)\circ (2, 4)$, or $(1, 3)\circ (3, 5)\circ (2, 4)\circ(4, 6)$...
Similarly, $\sigma_2$ is incorrectly decomposed. Two correct decompositions include $(1, 4)\circ (1, 2)\circ (3, 6)$ and $(1, 2)\circ (2, 4) \circ (3, 6)$...
...which answers your question about uniqueness. When writing a permutation as a product of transpositions, there are many such ways to do this. What does not vary is the parity: an "odd" permutation is one that can only be decomposed to a product of an odd number of transpositions, and "even" permutations can only be decomposed into a product of an even number of transpositions. So, for example, $\sigma_1$ is even, and $\sigma_2$ is odd.
• I don't understand this. If I apply $(1,5)\circ (1,3) \circ (2,6) \circ (2,4)$ on the id permutation I get the permutation $\begin{pmatrix} 5 &6&1&2&3&4 \end{pmatrix}$ Why is this a transposition for $\sigma_1$? I think this goes back to my original question, from which permutation do I start when applying those transpositions? – eager2learn Apr 22 '14 at 14:26
• You start from the rightmost permutation. Where does it send $1$?. If it sends $1$ to $a$, then you move to the next transposition to its left to see where it sends $a$. Etc. – Namaste Apr 22 '14 at 14:30 | {
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By default, glmnet will do two things that you should be aware of: Since regularized methods apply a penalty to the coefficients, we need to ensure our coefficients are on a common scale. Shows the effect of collinearity in the coefficients of an estimator. Ridge regression is a method by which we add a degree of bias to the regression estimates. The second line fits the model to the training data. Keep in mind, ridge is a regression … 11. Ridge, LASSO and Elastic net algorithms work on same principle. In scikit-learn, a ridge regression model is constructed by using the Ridge class. Lasso is great for feature selection, but when building regression models, Ridge regression should be your first choice. In R, the glmnet package contains all you need to implement ridge regression. Active 2 years, 8 months ago. Let us first implement it on our above problem and check our results that whether it performs better than our linear regression model. They all try to penalize the Beta coefficients so that we can get the important variables (all in case of Ridge and few in case of LASSO). Ridge Regression is the estimator used in this example. Linear regression is the standard algorithm for regression that assumes a linear relationship between inputs and the target variable. The value of alpha is 0.5 in our case. Ridge Regression. Elastic net regression combines the properties of ridge and lasso regression. Ridge Regression is a neat little way to ensure you don't overfit your training data - essentially, you are desensitizing your model to the training data. Use the below code for the same. This is also known as $$L1$$ regularization because the regularization term is the $$L1$$ norm of the coefficients. Ridge regression - introduction¶. Yes simply it is because they are good biased. Important things to know: Rather than accepting a formula and data frame, it requires a vector input and matrix of predictors. regression_model = LinearRegression() regression_model.fit(X_train, | {
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input and matrix of predictors. regression_model = LinearRegression() regression_model.fit(X_train, y_train) ridge = Ridge(alpha=.3) For example, to conduct ridge regression you may use the sklearn.linear_model.Ridge regression model. scikit-learn provides regression models that have regularization built-in. It works by penalizing the model using both the 1l2-norm1 and the 1l1-norm1. Generally speaking, alpha increases the affect of regularization, e.g. The Alpha Selection Visualizer demonstrates how different values of alpha influence model selection during the regularization of linear models. Ridge Regression is a technique for analyzing multiple regression data that suffer from multicollinearity. There are two methods namely fit() and score() used to fit this model and calculate the score respectively. Here, we are using Ridge Regression as a Machine Learning model to use GridSearchCV. One commonly used method for determining a proper Γ \boldsymbol{\Gamma} Γ value is cross validation. Note that scikit-learn models call the regularization parameter alpha instead of $$\lambda$$. The alpha parameter tells glmnet to perform a ridge (alpha = 0), lasso (alpha = 1), or elastic net (0 < alpha < 1) model. ridgeReg = Ridge(alpha=0.05, normalize=True) ridgeReg.fit(x_train,y_train) pred = ridgeReg.predict(x_cv) calculating mse When multicollinearity occurs, least squares estimates are unbiased, but their variances are large so they may be far from the true value. This notebook is the first of a series exploring regularization for linear regression, and in particular ridge and lasso regression.. We will focus here on ridge regression with some notes on the background theory and mathematical derivations that are useful to understand the concepts.. Then, the algorithm is implemented in Python numpy | {
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## ridge regression alpha
Facebook Message Says Seen But No Time, Which Is Worse Agnostic Or Atheist, Activity Diagram For Online Shopping, Sprinkler Cad Block, Amaryllis Name Meaning, Easton Beast Speed, Meal Village Menu, | {
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# Lecture 17: Rapidly Decreasing Singular Values
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## Description
Professor Alex Townsend gives this guest lecture answering the question “Why are there so many low rank matrices that appear in computational math?” Working effectively with low rank matrices is critical in image compression applications.
## Summary
Professor Alex Townsend's lecture
Why do so many matrices have low effective rank?
Sylvester test for rapid decay of singular values
Image compression: Rank $$k$$ needs only $$2kn$$ numbers.
Flags give many examples / diagonal lines give high rank.
Related section in textbook: III.3
Instructor: Prof. Alex Townsend
GILBERT STRANG: So let me use the mic to introduce Alex Townsend, who taught here at MIT-- taught Linear Algebra 18.06 very successfully.
And then now he's at Cornell on the faculty, still teaching very successfully. And he was invited here yesterday for a big event over in Engineering.
And he agreed to give a talk about a section of the book-- section 4.3-- which, if you look at it, you'll see is all about his work. And now you get to hear from the creator himself. OK.
ALEX TOWNSEND: OK. Thanks. Thank you, Gil. Thank you for inviting me here. I hope you're enjoying the course.
Today I want to tell you a little about why there so many matrices that are low rank in the world. So as computational mathematicians-- Gil and myself-- we come across low-rank matrices all the time. And we started wondering, as a community, why?
What is it about the problems that we are looking at? What makes low-rank matrices appear? And today I want to give you that story-- or at least an overview of that story.
So for this class, x is going to be n by n real matrix. So nice and square. And you already know, or are very comfortable with, the singular values of a matrix. | {
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So the singular values of a matrix, as you know, are a sequence of numbers that are monotonically non-increasing that tell us all kinds of things about the matrix x.
For example, the number of nonzero singular values tell us the rank of the matrix x. And they also, you probably know, tell us how well a matrix x can be approximated by a low-rank matrix.
So let me just write two facts down that you already are familiar with. So here's a fact-- that, if I look at the number of non-zero singular values in x-- so I'm imagining there's going to be k non-zero singular values-- then we can say a few things about x.
For example, the rank of x, as we know, is k-- the number of non-zero singular values. But we also know from the SVD that we can decompose x into a sum of rank 1 matrices-- in fact, the sum of k of them.
So because x is rank k, we can write down a low-rank representation for x, and it involves k terms, like this.
Each one of these vectors here is a column vector. So if I draw this pictorially, this guy looks like this, right? And we have k of them. So because x is rank k, we can write x as a sum of k rank 1 matrices.
And we also have an initial fact that we already know-- that the dimension of the column space of x is equal to k, and the same with the row space. So the column space of x equals the row space of x-- the dimension-- and they all equal k.
And so there are three facts we can determine from looking at this sequence of singular values of a matrix x. Of course, the singular value sequence is unique. X defines its own singular values.
What we're interested in here is, what makes x? What are the properties of x that make sure that the singular values have a lot of zeros in that sequence? Can we try to understand what kind of x makes that happen? | {
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And we really like matrices that have a lot of zeros here, for the following reason-- we say x is low rank if the following holds, right? Because if we wanted to send x to our friend-- we're imagining x as picture where each entry is a pixel of that image.
If that matrix-- that image-- was low rank, we could send the picture to our friend in two ways. We could send one every single entry of x. And for us to do that, we would have to send n squared pieces of information, because we'd have to send every entry.
But if x is sufficiently low rank, we could also send our friend the vectors-- u, u1, v1, uk, up to vk. And how much pieces of data would we have to send our friend to get x to them if we sent in the low-rank form?
Well, there's 2n here, 2n here numbers. There's k of them. So we'd have to send 2kn numbers. And we strictly say a matrix is low rank if it's more efficient to send x to our friend in low-rank form then in full-rank form.
So this, of course, by a little calculation, just shows us that, provided the rank is less than half the size of the matrix, we are calling the matrix low rank.
Now, often, in practice, we demand more. We demand that k is much smaller than this number, so that it's far more efficient to send our friend the matrix x in low-rank form than in full-rank form. So the colloquial use of the word low rank is kind of this situation. But this is the strict definition of it.
So what do low-rank matrices look like? And to do that, I have some pictures for you. I have some flags-- the world flags. So these are all matrices x-- these examples-- because their flags happen to not be square. I hope you can all see this. | {
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But the top row here are all matrices that are extremely low rank. For example, the Austria flag-- if you want to send that to your friend, that matrix is of rank 1. So all you have to do is send your friend two vectors. You have to tell your friend the column space and the row space. And there's only the dimensions of one of both.
For the English flag, you need to send them two column vectors and two row vectors-- u1, v1, u2 and v2. And as we go down this row, they get slowly fuller and fuller rank.
So the Japanese flag, for example, is low rank but not that small. The Scottish flag is essentially full rank. So it's very inefficient to send your friend the Scottish flag in low-rank form. You're better off sending almost every single entry.
So what do low-rank matrices look like? Well, if the matrix is extremely low rank, like rank 1, then when you look at that matrix-- like here, like the flag-- it's highly aligned with the coordinates-- with the rows and columns.
So if it's rank 1, the matrix is highly aligned-- like the Austria flag. And of course, as we add in more and more rank here, the situation gets a bit blurry. For example, once we get into the medium rank situation, which is a circle, it's very hard to see that the circle is actually, in fact, low rank.
But what I'm going to do was try to understand why the Scottish flag or diagonal patterns-- particularly a bad example for low rank. So I'm going to take the triangular flag to examine that more carefully. So the triangular flag looks like-- I'll take a square matrix and I'll color in the bottom half.
So this matrix is the matrix of ones below the diagonal. And I'm interested in this matrix and, in particular, its singular values, to try to understand why diagonal patterns are not particularly useful for low-rank compression. And this matrix of all ones has a really nice property that, if I take its inverse, it looks a lot like-- getting close to Gil's favorite matrix. | {
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So if I take the inverse of this matrix-- it has an inverse because it's got ones on the diagonal-- then its inverse is the following matrix, which people familiar with finite difference schemes will notice the familiarity between that and the first order finite difference approximation.
In particular, if I go a bit further and times two of these together, and do this, then this is essentially Gil's favorite matrix, except one entry happens to be different-- ends up being this matrix, which is very close to the second order, central, finite difference matrix.
And people have very well studied that matrix and know its eigenvalues, its singular values-- they know everything about that matrix. And you'll remember that if we know the eigenvalues of a matrix, like x transpose x, we know the singular values of x. So this allows us to show, by the fact that we know that, that the singular values of this matrix are not very amenable to low rank. They're all non-zero, and they don't even decay.
So I'm getting this from-- I rang up Gil, and Gil tells me these numbers. That allows us to work out exactly what the singular values of this matrix are, from the connection to finite differences.
And so we can understand why this is not good by looking at the singular values. So the first singular value of x from this expression is going to be approximately 2n over pi. And from this expression, again, for the last guy-- the last singular value of x is going to be approximately a half.
So these singular values are all large. They're not getting close to zero. If I plotted these singular values on a graph-- so here's the first singular value, the second, and the n-th-- then what would the graph look like?
Well, plot these numbers. Divide by this guy so that they all are bounded between 1 and 0 because of the normalization, because I divided by sigma 1 of x. | {
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And so we can plot them, and they will look like this kind of thing. This number happens to be here where they come to be pi over 4n, which is me dividing this number by this number, approximately.
So triangular patterns are extremely bad for low rank. We need things-- or we at least intuitively think that we need things-- aligned with the rows and columns, but the circle case happens to also be low rank. And so what happened to the Japanese flag? Why is the Japanese flag convenient for low rank?
Well it's the fact that it's a circle, and there's lots of symmetry in a circle. So if I try to look at the rank of a circle, the Japanese flag, then I can bound this rank by decomposing the Japanese flag into two things. So this is going to be less than or equal to the rank of sum of two matrices, and I'll do it so that the decomposition works out.
I have the circle. I'm going to cut out a rank one piece that lives in the middle of this circle. OK? And I'm going to cut out a square from the interior of that circle. OK? And I can figure out-- of course the rank is just bounded by the sum of those two ranks. This guy is bounded by rank one because it's highly aligned with the grid. So this guy is bounded by rank one. So this thing here plus 1.
And now I have to try to understand the rank of this piece. Now this piece has lots of symmetry. For example, we know that the rank of that matrix is the dimension of the column space and the dimension of the row space. So when we look at this matrix, because of symmetry, if I divide this matrix in half along the columns, all the columns on the left appear on the right. So for example, the rank of this matrix is the same as the rank of that matrix because I didn't change the column space. OK? | {
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Now I go again and divide along the rows, and now the row dimension of this matrix is the same as the top half, because as I wipe out those, I didn't change the dimension of the row space because the rows are the same top-bottom. And so this becomes the rank of that tiny little matrix there. And because it's small, it won't have too large a rank. So this is definitely less than-- if I divide that up, a little guy here looks like that plus the other guy that looks like that plus 1.
And so of course the row space of this matrix cannot be very high because this is a very thin matrix. There's lots of zeros in that matrix, only a few ones. And so you can go along and do a bit of trig to try to figure out how many rows are non-zero in this matrix.
And a bit of trig tells you-- well it depends on the radius of this original circle. So if I make the original radius r of this Japanese flag, then the bound that you end up getting will be, for this matrix, r 1 minus square root 2 over 2 for this guy. That's a bit of trig. I've got to make sure that's an integer. And then again, here it's the same but for the column space. So this is me just doing trig. OK?
And that's bound on the rank. It happens to be extremely good. And if you work out what that rank is and try to look back, you will find it's extremely efficient to send the Japanese flag to your friend in low rank form, because it's not full rank because these numbers are so small. So this comes out to be, like, approximately 1/2 r plus 1. So much smaller than what you would expect, because remember, a circle is almost the anti-version version of a line with the grid, but yet, it's still low rank. OK. | {
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Now most matrices that we come up with in computational math are not exactly of finite rank. They are of numerical rank. And so I'll just define that. So the numerical rank of a matrix is very similar to the rank, except we allow ourselves a little bit of wiggle room when we define it, and so that amount of wiggle room will be of parameter called tol called epsilon. That's a tolerance. I'm thinking of epsilon as a tolerance. That's the amount of wiggle room I'm going to give myself. OK.
And we say that the numerical rank-- I'll put an epsilon there to denote numerical rank-- is k. k is the first singular value, or the last singular value, above epsilon. In the following sense, I'm copying the definition above but with epsilons instead of zeros. If this singular value is less than epsilon, relatively, and the kth one was not below. So k plus 1 is the first singular value below epsilon in this relative sense.
So of course the rank of 0x, if that was defined, is the same as the rank of x. OK? So this is just allowing ourselves some wiggle room. But this is actually what we're interested more in practice. All right? I don't want to necessarily send my friend the flag to exact precision. I would actually be happy to send my friend the flag up to 16 digits of precision, for example. They're not going to tell the difference between those two flags. And if I can get away with compressing the matrix a lot more once I have a little bit of wiggle room, that would be a good thing.
So we know from the Eckart and Young that the singular values tell us how well we can approximate x by a low-rank matrix. In particular, we know that the k plus 1 singular value of x tells us how well x can be approximated by a rank k matrix. OK? For example, when the rank was exactly k, the sigma k plus 1 was 0, and then this came out to be 0 and we found that x was exactly a rank k matrix. | {
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Here, because we have the wiggle room, the epsilon, we get an approximation, not an exact. So this is telling us how well we can approximate x by a rank k matrix. OK? That's what the singular values are telling us. And so this allows us to try our best to compress matrices but use low-rank approximation rather than doing things exactly.
And of course, on a computer, when we're using floating point arithmetic, or on a computer because we always round numbers to the nearest 16-digit number, if epsilon was 16 digits, your computer wouldn't be able to tell the difference between x or x the rank k approximation if this number satisfied this expression. Your computer would think of x and xk as the same matrix because it would inevitably round both to epsilon, within epsilon. OK.
So what kind of matrices are numerically of low rank? Of course all low-rank matrices are numerically of low rank because the wiggle room can only help you but it's far more than that. There are many full-rank matrices-- matrices that don't have any singular values that are zero-- but the singular values decay rapidly to zero. That are full-rank matrices with low numerical rank because of the wiggle room.
So for example, here is the classic matrix that fits this regime. If I give you this, this is called the Hilbert matrix. This is a matrix that happens to have extremely low numerical rank but it's actually full rank, which means that I can approximate H by a rank k matrix where k is quite small very well, provided you give me some wiggle room, but it's not a low-rank matrix in the sense that if epsilon was zero here, you didn't allow me the wriggle room, all the singular values of this matrix are positive. So it's of low numerical rank but it's not a low-rank matrix. | {
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The other classical example which motivated a lot of the research in this area was the Vandermonde matrix. So here is the Vandermonde matrix. An n by n version of it. Think of the xi's as real. And this is Vandermonde. This is the matrix that comes up when you try to do polynomial interpolation at real points.
This is an extremely bad matrix to deal with because it's numerically low rank, and often, you actually want to solve a linear system with this matrix. And numerical low rank implies that it's extremely hard to invert, so numerical low rank is not always good for you. OK? Often, we want the inverse, which exists, but it's difficult because V has low numerical rank.
OK. So people have been trying to understand why these matrices are numerically of low rank for a number of years, and the classic reason why there are so many low-rank matrices is because the world is smooth, as people say. They say, the world is smooth. That's why matrices are of numerical low rank. And to illustrate that point, I will do an example.
So this is classically understood by a man called Reade in 1983, and this is what his reason was. I have a picture of John Reade. He's not very famous, so I try to make sure his picture gets around. He's playing the piano. It's, like, one of the only pictures I could find of him.
So what is in this reason? Why do people say this? Well here's an example that illustrates it. If I take a polynomial in two variables and I-- for example, this is a polynomial of two variables-- and my x matrix comes from sampling that polynomial integers-- for example, this matrix-- then that matrix happens to be of low rank-- mathematically of low rank, with epsilon equals zero. | {
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Why is that? Well if I write down x in terms of matrices, you could easily see it. So this is made up of a matrix of all ones plus a matrix of j-- so that's 1, 2, up to n, 1, 2, up to n, because every entry of that matrix just depends on the row index. And then this guy depends on both j and k. So this is a multiplication table, right? So this is n, 2, 4, up to 2n, n, 2n, n squared. OK.
Clearly, the matrix of all ones is a rank one matrix. The same with this guy. The column space is just of dimension one. And the last guy also happens to be of rank one because I can write this matrix in rank one form, which is a column vector times a row vector. OK. So this matrix x is of rank three. I guess at lowest rank three is what I've actually shown. OK.
Now of course this hasn't got to numerical low rank yet, so let's get ourselves there. So Reade knew this, and he said to himself, OK, well if I can approximate-- if x is actually coming from sampling a function, and I approximate that function by polynomial, then I'm going to get myself a low-rank approximation and get a bound on the numerical rank.
So in general, if I give you a polynomial of two variables, which can be written down-- it's degree n in both x and y. Let's just keep these indexes away from the matrix index. I give you this such polynomial, and I go away and I sample it and make a matrix X, then X, by looking at each term individually like I did there, will have low rank mathematically, with epsilon equals zero. This will have, at most, m squared rank, and if m is 3 or 4 or 10, it possibly could be low because this X could be a large matrix. OK.
So what Reade did for the Hilbert matrix was said, OK, well look at that guy. That guy looks like it's sampling a function. It looks like it's sampling the function 1 over x plus y minus 1. So he said to himself, well, that x, if I look at the Hilbert matrix, then that is sampling a function. It happens to not be a polynomial. It happens to be this function. | {
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But that's OK because sampling polynomials, integers, gives me low rank exactly. Maybe sampling smooth functions, functions like this, can be well approximated by polynomials and therefore have low numerical rank. And that's what he did in this case. So he tried to find a p, a polynomial approximation to f. In particular, he looked at exactly this kind of approximation. So he has some numbers here so that things get dissolved later. And he tried to find a p that did this kind of approximation. So this approximates f.
And then he would develop a low-rank approximation to X by sampling p. So he would say, OK, well if I let y be a sampling of p, then from the fact that f is a good approximation to p, y is a good approximation to X. And so this has finite rank. He wrote down that this must hold. And the epsilon comes out here because these factors were chosen just right. The divide by n was chosen so that the epsilon came out just there. OK?
So, for many years, that was kind of the canonical reason that people would give, that, well, if the matrix X is sampled from a smooth function, then we can approximate our function by a polynomial and get polynomial rank approximations. And therefore, the matrix X will be of low numerical rank. There's an issue with this reasoning, especially for the Hilbert matrix, that it doesn't actually work that well. | {
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So for example, if I take the 1,000 by 1,000 Hilbert matrix and I look at its rank-- OK, well I've already told you this is full rank. You'll get 1,000. All the singular values are positive. If I look at the numerical rank of this 1,000 by 1,000 Hilbert matrix and I compute it, I compute the SVD and I look at how many are above epsilon where epsilon is 10 to the minus 15, so that means I can approximate the 1,000 by 1,000 Hilbert matrix by a rank 28 matrix and only give up 15-- there will be exact 15 digits, which is a huge amount. So this is what we get in practice, but Reade's argument here shows that the rank of this matrix, the numerical rank, is at most. So it doesn't do a very good job on the Hilbert matrix for bounding the rank, right?
So Reade comes along, takes this function. He tries to find a polynomial that does this, where epsilon is 10 to the minus 15. He finds that the number of terms that he needs in this expression here is around 719, and therefore, that's the rank that he gets. The bound on the numerical rank. The trouble is that 719 tells us that this is not of low numerical rank, but we know it is, so it's an unsatisfactory reason.
So there's been several people trying to come up with more appropriate reasons that explain the 28 here. And so one reason that I've started to use is another slightly different way of looking at things, which is to say the world is Sylvester. Now Sylvester, what does that mean? What does the word "Sylvester" mean in this case? It means that the matrices satisfy a certain type of equation called the Sylvester equation, and so the reason is really, many of these matrices satisfy a Sylvester equation, and that takes the form-- for sum A, B, and C. | {
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OK. So X is your matrix of interest. You want to show X is of numerical low rank. And the task at hand is to find an A, B, and C so that X satisfies that equation. OK. For example, the two matrices I've had on the board satisfy a Sylvester equation-- a Sylvester matrix equation. There is an A, a B, and a C for which they do this.
For example, remember the Hilbert matrix, which we have there still, but I'll write it down again. Has these entries. So all we need to do is to try to figure out an A, a B, and then a C so that we can make it fit a Sylvester equation. There's many different ways of doing this. The one that I like is the following, where if I put 1/2 here and 3/2 here, all the way down to n minus 1/2, times this matrix-- so this is timesing the top of this matrix by 1/2 and then 3/2 and then 5/2. So we're basically timesing each entry of this matrix by j minus 1/2.
And then I do something on the right here, which I'm allowed to do because I've got the B freedom, and I choose this to be the same up to a minus sign. Then when you think about this, what is it doing? It's timing the jk entry-- this is-- by j minus 1/2. That's what this is doing. And what's this doing is timesing the jk entry by k minus 1/2. So this is, in total, timesing the jk entry by j plus k minus 1/2 minus 1/2, which is minus 1, so this is timesing the jk entry by j plus k minus 1. So it knocks out the denominator. And what we get from this equation is a bunch of ones. So in this case, A and B are diagonal, and C is the matrix of all ones. OK?
We can also do this for Vandermonde. So Vandermonde, you'll remember, looks like this. And then over here, we have this guy, the matrix that appears with polynomial interpolation. OK. So if I think about this, I could also come up with an A, B, and C, and for example, here's one that works. I can stick the x's on the diagonal. | {
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So if you imagine what that matrix on the left is doing, it's timesing each column by the vector x. OK? So the first column of this matrix becomes x, the vector x. The second becomes the vector x squared, where squared is done entry-wise. And then the third entry is now x cubed, and when we get to the last, it's x to the n. OK? So that's like, multiply each column by the vector x.
So if I want to try to come up with a matrix-- so what's left is of low rank, is like of this form. What I can do is shift the columns. So I've noticed that this product here, this diagonal matrix, has made the first column x. So if I want to kill off that column, I can take the second column and permute it to the first column. I could take the third column and permute it to the second, the last column and permute it to the penultimate column here. And that will actually kill off a lot of what I've created in this matrix right here.
So let me write that down. This is a circumshift matrix. This does that permutation. I've put a minus 1 there. I could have put any number there. It doesn't make any difference. But this is the one that works out extremely nicely. Now this zeros out lots of things because of the way I've done the multiplication by x and the circumshift of the columns.
And so the first column is zero because this first column is x, this first column is x, so I've got x minus x. This column was x squared minus x squared, so I got zero, and I just keep going along until that last column. That last column is a problem because the last column of this guy is x to the n, whereas I don't have x to the n in V, so there are some numbers here. OK. | {
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You'll notice that C in both cases happens to be a low-rank matrix. In these cases, it happens to be of rank one. And so people were wondering, maybe it's something to do with satisfying these kind of equations that makes these matrices that appear in practice numerically of low rank. And after a lot of work in this area, people have come up with a bound that demonstrates that these kind of equations are key to understanding numerical low rank.
So if X satisfies a Sylvester equation, like this, and A is normal, B is normal-- I don't really want to concentrate on those two conditions. It's a little bit academic. Then-- people have found a bound on the singular values of any matrix that satisfies this kind of expression, and they found this following bound. OK, so here, the rank of C is r. So that goes there. So in our cases, the two examples we have, r is 1, so we can forget about r.
This nasty guy here is called the Zolotarev number. E is a set that contains the eigenvalues of A, and F is a set that contains the eigenvalues of B. OK. Now it looks like we have gained absolutely nothing by this bound, because I've just told you singular values are bound by Zolotarev numbers. That doesn't mean anything to anyone. It means a little bit to me but not that much.
So the key to this bound-- the reason this is useful-- is that so many people have worked out what these Zolotarev numbers actually mean. OK? So these are two key people that worked out what this bound means. And we have gained a lot because people have been studying this number. This is, like, a number that people cared about from 1870 onwards to the present day, and people have studied this number extremely well. So we've gained something by turning it into a more abstract problem that people have thought about previously, and now we can go to the literature on Zolotarev numbers, whatever they are, and discover this whole literature of work on this Zolotarev number. | {
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And the key part-- I'll just tell you the key-- is that the sets E and F are separated. So for example, in the Hilbert matrix, the eigenvalues of A can be read off the diagonal. What are they? They are between minus 1/2 and n minus 1/2. And the eigenvalues of B lie in the set minus 1/2 minus n plus 1/2. And the key reason why the Hilbert matrix is of low numerical rank is the fact that these two sets are separated, and that makes this Zolotarev number gets small extremely quickly with k.
Now you might wonder why there is a question mark on Penzl's name. There is an unofficial curse that's been going on for a while. Both these men died while working on the Zolotarev problem. They both died at the age of 31. One died by being hit by a train, Zolotarev. It's unclear whether he was suicidal or it was accidental.
Penzl died at the age of 31 in the Canadian mountains by an avalanche. I am currently not yet 31 but going to be 31 very soon, and I'm scared that I may join this list. OK. But for the Hilbert matrix, what you get from this analysis, based on these two peoples' work, is a bound on the numerical rank. And the rank that you get is, let's say, a world record bound. For the Hilbert matrix is 34, which is not quite 28, not yet, but it's far more descriptive of 28 than 719.
And so this technique of bounding singular values by using these Zolotarev numbers is starting to gain popularity because we can finally answer to ourselves why there are so many low-rank matrices that appear in computational math. And it's all based on two 31-year-olds that died. And so if you ever wonder when you're doing computational science when a low rank appears and the smoothness argument does not work for you, you might like to think about Zolotarev and the curse. OK, thank you very much.
[APPLAUSE]
GILBERT STRANG: Thank you [INAUDIBLE] Excellent.
ALEX TOWNSEND: How does it work now?
GILBERT STRANG: We're good. Yeah. | {
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ALEX TOWNSEND: How does it work now?
GILBERT STRANG: We're good. Yeah.
ALEX TOWNSEND: I'm happy to take questions if we have a minute, if you have any questions.
GILBERT STRANG: How near of 31 are you?
ALEX TOWNSEND: [INAUDIBLE] I get a spotlight. I'm 31 in December.
GILBERT STRANG: Wow. OK.
ALEX TOWNSEND: So they died at the age of 31, so you know, next year is the scary year for me. So I'm not driving anywhere. I'm not leaving my house until I become 32.
GILBERT STRANG: Well, thank you [INAUDIBLE]
ALEX TOWNSEND: OK, thanks.
[APPLAUSE] | {
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# How many trailing zeroes in $52!$? [duplicate]
Problem
How many trailing zeroes are there in $52!$ ?
My thoughts
I believe I correctly solved it, but I'm not happy with the scalability of my method.
I figure the number of trailing zeroes is equal to the number of times $52!$ is divisible by 10. I wrote out every integer from 1-52 that is divisible by 2 or 5. The idea being that the number of 2 AND 5-factors equals the number of 10-factors.
I quickly noted that the number of 2-factors is greater than the number of 5-factors, so I figured finding the number of 5-factors will do. There were 12.
I'm not very happy with this, because if they now ask me to do the same for $152!$, I'll have to tell them to shove it. I'm not doing this again.
Question
Is there a better way to do this? Perhaps a method that scales better?
## marked as duplicate by Jyrki LahtonenAug 27 '17 at 16:35
• Your method looks good and should be quick now you have the basics. Undertaking the same task for $152!$ should rapidly give you some insight into how to solve the problem quicker. Look out for $125$. – Joffan Aug 27 '17 at 16:39
You get your $12$ by counting $\left\lfloor\frac{52}{5}\right\rfloor+\left\lfloor\frac{52}{25}\right\rfloor$. This counts each multiple of $5$ in the product $52!$, then adds on a count for each multiple of $25$. This can generalize to $152$. You'll need to go up to $125$ with divisors.
In general, the number of trailing $0$s in $n!$ will be $$\sum_{k=1}^{\infty}\left\lfloor\frac{n}{5^k}\right\rfloor$$ where for any $n$, only the first few terms in that infinite sum will be nonzero. | {
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# Which transformation needed to make variance independent of population parameter?
Suppose $$s^2$$ is the sample variance of a sample$$(\text{of size }n)$$ from a normal population with mean $$\mu$$ and variance $$\sigma^2. \text{Here }s^2=\frac{\sum_{i=1}^n(x_i-\overline{x})^2}{n-1}$$
As we know $$\frac{(n-1)s^2}{\sigma^2}\sim \chi^2_{(n-1)}$$ here . Let $$x\sim \chi^2_{(n-1)}$$ then $$E[x]=n-1$$ and $$Var[x]=2(n-1)$$ \begin{align*} E\left[\frac{(n-1)s^2}{\sigma^2}\right]=E[x]&=n-1\\ \implies \frac{(n-1)}{\sigma^2}E[s^2] &= n-1\\ \implies E[s^2]&=\sigma^2 \end{align*} Similarly, \begin{align*} Var\left[\frac{(n-1)s^2}{\sigma^2}\right]=Var[x]&=2(n-1)\\ \implies \frac{(n-1)^2}{\sigma^4}E[s^2] &=2(n-1)\\ \implies Var[s^2]&=\color{red}{\frac{2\sigma^4}{(n-1)} } \end{align*} Now we need a function of $$s^2$$ whose variance will be independent of $$\sigma^2.$$ Let $$f(s^2)$$ be the required transformation.
The required transformation is $$f(s^2)\stackrel{?}{=}\ln{s^2}$$
Question: How they get that transformation $$?$$
Similar things happen with Poisson variate and Binomial proportion with square root and $$\sin^{-1}$$ transformation. So I need a general approach to get that transformation which will make variance independent of population parameter.
• Search for Variance-stabilizing_transformation. – StubbornAtom Nov 6 '19 at 14:33
• "The aim behind the choice of a variance-stabilizing transformation is to find a simple function f to apply to values x in a data set to create new values y=f(x) such that the variability of the values y is not related to their mean value." But I need the variability of the values y is not related to their variance(population) value? @StubbornAtom Sir – emonhossain Nov 6 '19 at 14:45
• The wiki article might not be entirely clear. I gave you the keyword to look for the topic. (And I am no 'sir'.) – StubbornAtom Nov 6 '19 at 14:51 | {
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