PROBLEM LINK:Author: Trung Nguyen DIFFICULTY:MEDIUM PREREQUISITES:Trigonometry, Fast Matrix Exponentiation, Modular Inverse PROBLEM:A clock has a minute hand which goes $x$ angle per second. The minute hand has length $l$ and the center of clock has coordinate $(0,0)$. Initially, the other endpoint is at coordinate $(0,l)$, and after $1$ second its $y$coordinate becomes $d$. Given $l,d,t$($x$ is unknown), calculate the $y$coordinate of this endpoint after $t$ seconds. QUICK EXPLANATION:By the problem statement we have $\cos(x)=d/l$, and the final answer is $\cos(tx)\cdot l$. So we only need to compute $\cos(tx)$. There is a formula computing $\cos(nx)$: $$\cos(nx)=2\cos(x)\cos((n1)x)\cos((n2)x),$$ and we can use fast matrix exponentiation to compute $\cos(tx)$ in $O(\log t)$ time. EXPLANATION:A note to modular arithmeticSetter's SolutionFirst notice that $x$ is given by $\cos(x)=d/l$. To see this, consider the following figure which illustrates the problem statement: $OA$ and $OB$ are the position of minute hand initially and after $1$ seconds, respectively. And the answer after $t$ seconds is $\cos(tx)\cdot l$. Therefore, our actual task is: given $\cos(x)$ and integer $t$, compute $\cos(tx)$. Notice the following formula: $$\cos(nx)=2\cos(x)\cos((n1)x)\cos((n2)x)\ \ \ (*)$$ that means we can write $\cos(tx)$ in a matrixmultiplication form: $$\begin{pmatrix}\cos(tx)\\\cos((t1)x)\end{pmatrix}=\begin{pmatrix}2\cos(x)&1\\1&0\end{pmatrix}\begin{pmatrix}\cos((t1)x)\\\cos((t2)x)\end{pmatrix},$$ and therefore $$\begin{pmatrix}\cos(tx)\\\cos((t1)x)\end{pmatrix}=\begin{pmatrix}2\cos(x)&1\\1&0\end{pmatrix}^{t1}\begin{pmatrix}\cos(x)\\1\end{pmatrix}.$$ Thus we can compute $\cos(tx)$ by fast matrix exponentiation in $O(\log t)$ time. Tester's SolutionAgain, we're given $\cos(x)$ and integer $t$ and we want to compute $\cos(tx)$. But what if we don't know formula $(*)$? We can use the following (betterknown) formula: $$\begin{align*} \cos(a+b)=&\cos(a)\cos(b)\sin(a)\sin(b)\ \ \ (1)\\ \sin(a+b)=&\sin(a)\cos(b)+\cos(a)\sin(b)\ \ \ (2) \end{align*}$$ All numbers we come across will be in the form $a+b\sin(x)$. Although we don't know $\sin(x)$, we can represent such a number as $(a,b)$. Therefore we can define $+,,\times$ on these numbers: $$\begin{align*} (a,b)\pm(c,d)=&(a\pm c,b\pm d)\\ (a,b)\cdot(c,d)=&(a+b\sin(x))(c+d\sin(x))\\ =&ac+bd(1\cos^2(x))+(ad+bc)\sin(x)\\ =&(ac+bd(1\cos^2(x)),ad+bc). \end{align*}$$ We can write a recursive procedure $\mathrm{Solve}(t)$ for computing the pair $(\cos(tx),\sin(tx))$.
This procedure runs in $O(\log t)$ time since every two steps decrease $t$ by half. Suppose we've computed $\cos(tx)=(a,b)$. $b$ must be equal to $0$ since it's possible to represent $\cos(tx)$ by only $\cos(x)$ terms. Therefore $\cos(tx)=a$ and we're done. On formula $(*)$How to prove formula $(*)$? In fact I don't know, but one might look into this page for answer. The Chebyshev polynomial of the first kind is defined as $T_0(x)=1$, $T_1(x)=x$ and $T_{n+1}(x)=2xT_n(x)T_{n1}(x)$. It can be proved that $T_n(\cos(x))=\cos(nx)$, which directly implies formula $(*)$. ALTERNATIVE SOLUTION:Please feel free to share your approach :) AUTHOR'S AND TESTER'S SOLUTIONS:Author's solution can be found here. RELATED PROBLEMS:
This question is marked "community wiki".
asked 27 Jan '18, 14:05

I calculated cosnx by breaking n into a sum of powers of 2 and then calculated each power recursively using the formula cos2x=2cos^21. I used memoization to avoid repeating the same calculations. This required no more than log(n) operations. To combine the powers of 2, I used the formula cos(a+b)=2cos(a)cos(b)cos(ab). I also used memoisation in this approach by mapping the previously calculated values of cosx otherwise it would result in TLE. answered 13 Feb '18, 04:00

For formula proof: answered 12 Feb '18, 17:46

Really good problem... never used matrix exponentiation like fibonacci's anywhere else before answered 13 Feb '18, 01:28

Wow, I found (another) simple proof for the formula in @taran_1407's editorial! Link. answered 12 Feb '18, 17:50

Can someone help me understand the formula manipulation to get from the first matrix representation of the formula to the second one with the second 2x2 matrix raised to the (t1) power? answered 13 Feb '18, 03:31
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It is written in French but I guess you should be able to understand the basic idea by reading the equation. So look for chapter: "Suite récurrente d’ordre p" on page: https://fr.wikipedia.org/wiki/Suite_r%C3%A9currente_lin%C3%A9aire NB: I did not find the same explanation in English.
(13 Feb '18, 14:19)
It just follows from plain matrix multiplication. Any linear recurrence can solved in this manner using matrix exponentiation. You can find relevant tutorials on the internet.
(13 Feb '18, 21:43)
Okay thanks for the replies. I'm actually relatively new to competitive programming and have done 0 matrix exponentiation problems so far. I think I just need to learn the topic and then get some practice with it.
(17 Apr '18, 22:49)

need some help!
then i found the angle with respect to y axis after t seconds as this
now I found the quadrant where the final angle will lie and calculate the value of d(the final y coordinate) the i converted d into fraction using the function and just find the inverse and print it where i am wrong my solution my solution for BROCLK
link
This answer is marked "community wiki".
answered 13 Feb '18, 07:12

hey everybody, I found the x correctly. using cos(x) = d/l; now why does doesn't we get the final length by cos(x*t) this can +ve or ve depending on the quadrent. I believe cos will take care of everything as it changes the quadrent it switched the sign of value i.e from ve to +ve or vice versa. Please confirm.. my broken code is https://www.codechef.com/viewsolution/17355814 answered 13 Feb '18, 08:28

After finding x, and also given t, why can't we straight away find cos(tx)? answered 13 Feb '18, 13:42
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you wont get ans in p/q form, even if u try to use fractions in python, you'll have to round your cos(tx), in this case you'll get wrong ans because precision will be different.
(13 Feb '18, 16:00)
@siddharthp538 why not do mod 360 or 2*PI depending whether you are using degrees or radians, and then t will be a very small value and precision will be saved right ?
(13 Feb '18, 16:45)
Actually this happened with me during the contest, i was getting cos(60) as 0.4999999999999999 and using Fraction if u convert it in p/q form then p comes as 4503599627370495 and q comes as 9007199254740992. p*(q^1) comes out to be 879237148. It makes a huge diff to our ans. Actually if u do mod 360 still values will be rounded. It can give us wa.
(13 Feb '18, 17:12)

There is no need to use matrices or trigonometry at all. In my solution, all you need to know is how to deal with complex numbers. First, swap X and Y coords for convenience. Then, we have complex number Let's denote that nasty root
As we can see, field of numbers of that kind is closed under multiplication. So, all we need is to store complex number (a, b*R) as a pair of integers (a, b), write our own multiplication function, and use binpow. Code answered 13 Feb '18, 17:39
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This is identical to the tester's solution, only interpreted as complex numbers.
(13 Feb '18, 21:33)

When I use the normal iterative to compute the chebyshev and fermat's little theorem for modular inverse, I get wrong answer on the first subtask. The given test cases are computed right tho, and yes, I need something better for when t is large but could someone tell me what's wrong in this code, at least for t <= 3? answered 13 Feb '18, 21:58

There is other way to solve the question. Below is my approach: We know e^ix = cos(x) + isin(x) => e^inx = cos(nx) + isin(nx) We know cos(nx) + isin(nx) = (cos(x) + isin(x))^n After computing we would have cos(nx) = real((cos(x) + isin(x))^n) The multiplication can can be done in logn. Here is the code: https://www.codechef.com/viewsolution/17348197 answered 13 Feb '18, 23:13

@manjuransari at one place u used the line t_x = (xn_x)%mod  ((((y_xn_y_x)%mod)*diff)%mod)%mod; cant we write it as t_x = [(xn_x) ((((y_xn_y_x))*diff))]%mod; i.e after multiplying all the values at last we are calculating mod why using mod everytime as question clearly mentions you have to give FINAL result as ans%mod why using mod every time will not affect our answer answered 13 Feb '18, 23:23

i used binomial coefficients but iam getting wrong answer for some subtasks answered 14 Feb '18, 17:35

If t=1, we return cos(tx)=(cos(x),0)and sin(tx)=(0,1); Can anyone tell me why we are initializing sin(tx)=(0,1) ? answered 15 Feb '18, 12:04

Can anyone please explain one example of this problem along with modular inverse calculation. answered 17 Feb '18, 14:14

What is the problem if i do the following :
I find x as $x = t* \theta  (int)(t* \theta/ \arctan(1)*8)* \theta$ Is the problem due to floating point airthematic ? answered 18 Feb '18, 16:00

Hello 'd' is opposite side right?'l' is Hypotenuse right? answered 26 Feb '18, 16:41

came up with the exact formula of author . but didn't have prior knowledge of Mexp.. it was a very good question ..
What about simply using the 2 formulas
$cos (tx)= 2*{cos}^{2}(tx/2)1$ (even t)
and $cos(tx)+cos(x)=2cos((t+1)x/2)*cos((t1)x/2)$ (odd t)
No need for any complex as such formula if we use these.
yes.. this will be much time saver and also no dealing with irrational parts.