The equation is 7a^3=11b^5. It cannot be 0 and both must be a positive integer. I have tried substitution and everything in algebra I know (not including calculus). Trial and error doesn't work because a is bigger than 40 000. I don't mean that I actually tried all till 40 000, i know the answer. I mainly just want to know how to solve it and explain how I solved it.
a = 41503 and b = 539
Thank you in advance.
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This is a homework question. I'm asked to find the coefficients of the line of best fit for a given set of n dots(2D). The coefficients are a b c in: ax+by=c.Say there're n dots, use linear programming to find the coefficients that leads to the smallest 'maximum absolute error', defined as: max(|a*xi+b*yi-c|), i ranges from 1-n.
Here's my thought process:
Let M denote the maximum absolute error. The objective of linear programming is to minimize M. Since M is the biggest of all |a*xi+b*yi-c|, it must be bigger than every one of them. So (a*xi+b*yi-c)<= M, and (a*xi+b*yi-c)>= -M, for all i (the second expression is to account for the absolute sign).
I thought this is sufficient to define the problem. When i put the conditions into a solver, it returned a b c all equal to 0, but in reality it shouldn't. I think I'm missing some conditions here. Can someone point it out to me?
You should add one extra statement that is: either a or b should not be 0. If both values are 0 you have a valid solution to your system but there is no line with both a and b equal to 0.
EDIT: improving Rerito's suggestion. Any line has either a or b not equal to 0. Alo the lines (k*a)*x + (k*b)* y + (k*c) and (a)*x + (b)* y + (c) are the same for any non-zero k. So I would say you need to run the solver twice- once when specifying a is 1 and once when specifying b is 1 and after than select the better solution. You have to run the solver twice because it might be the case that the best solution has a=0 or b=0(but not both).
In today's edition of the Guardian (a newspaper in the UK), in the "Pyrgic puzzles" section on page 43 by Chris Maslanka, the following puzzle was given:
The 3 wise men ... went to Herrods to do their Christmas shopping. Caspar bought Gold, Melchior bought Frankincense, and Balthazar bought a copy of the Daily Myrrh. The cashier tapped in the number of euros of each of these things cost, and meant to add the three numbers, but multiplied them instead. ... the marvel of the thing was that the result was exactly the same: €65.52. What were the three sums [I assume he meant the three numbers]?
My interpretation is: Find an a, b and c such that a + b + c = abc = 65.52 (exactly) where a, b and c are positive decimal numbers with no more than two decimal places. It follows that a, b and c must also be less than 65.52 (approximately).
My approach is thus: I shall find all the candidate sets of a, b and c where a + b + c = 6552 and a, b and c are integers from {1 ... 6550} (Notionally I have multiplied all the operands by 100 for convenience). Then, for all the candidate sets, it is trivial to satisfy the other condition by dividing all the operands by 100 then multiplying them (with arbitrary-precision arithmetic).
This, as I see it, is an instance of the subset sum problem. So I implemented a dirty (exponential time) algorithm which found one distinct solution: a=0.52, b=2, c=63.
Ok, there are better algorithms for the subset sum problem, but don't you think this is getting a little out-of-reach for an average Guardian reader?
On page 40 the answer is listed:
This is easy, by trial and error. Guess 52p for the Daily Myrrh. But by multiplying by 0.52 is roughly halving, so we need one sum to be about 2; so try 2 X 63 X 0.52. Et voilà. Is this answer unique?
Well, we know that the answer is unique (disregarding the other permutations of 2, 63 and 0.52).
What I want to know is: How can this be "easy"? Am I right in characterising the puzzle as an instance of the subset sum problem? Have I overlooked some characteristic of the puzzle which can be utilized to simplify the solution? Was anyone able to adopt a similar "trial and error" approach and if so can they take me through it? Is Chris Maslanka simply undaunted by NP-complete problems?
No, it is not an instance of the subset sum problem, because:
The subset size is limited to 3, making it O(n^3) solution worst case with naive exhaustive search (and not exponential)
There is additional data in here, the product of the numbers.
You are not actually given a set, a set of all integers is just a subproblem of subset-sum, a much easier one.
The important thing to understand here is: if a problem can be solved by an NP-Hard problem - it doesn't mean it is NP-Hard as well, the other way around holds - if you have a problem, and you can solve some NP-Hard problem (polynomially) with it, then your problem is NP-Hard. It is called polynomial reduction1.
The approach is easy because all you have to do is "guess" (by iterating all candidates) a value for a, and from this you can derive what is the possible solution for b,c - (2 variables, two equations if a is known - and in each iteration - it is), thus the solution is even linear - not only sub exponential.
It might even be optimized to use a variation of binary search to get a sub-linear optimization, but I cannot think of that optimization at the moment.
(1) Note: this is some intuitive explanation, and not a formal definition.
to check the divisibility of an integer (say A) by another integer(say B) , i have tried an approach by factoring B and checking if A is divisible by all the prime factors of B . However i'm not sure if its correct ? Could you please suggest what can be done ? For e.g if we have a very large integer say 10^100 and we want to check if its divisible by another integer say 200 then i was trying if 10^100 is divisible by 2 and 5 (by noticing the last digit).
If A is small enough we could directly have checked if A%B==0 but i was trying this for larger numbers.
Thanks,
You have to count the number of times that the prime appears in the factorization of B, and ensure that it appears at least as many times in the factorization of A.
So, 200 = 23 * 52. Then A is divisible by 200 if and only if it is divisible by 23 and by 52.
Unless you somehow know the factorization already, factorizing A is far slower than just dividing it by B. The reason is that it will take many trial divisions (or equivalent work) to fully factorize A, whereas it only takes one trial division to check it for divisibility by B. After all, consider the case where B is prime: you've found all prime factors of A, when all you needed was to test whether or not B is one of them. This cannot possibly be less work.
This is the best idea about integer factorization, written here is to let more people know and participate.
A New Way of the integer factorization
1+2+3+4+……+k=Ny,(k
True gold fears fire, you can test 1+2+3+...+k=Ny(k<N/2).
How do I know "k" and "y"?
"P" is a factor of "N",GCD(k,N)=P.
The idea may be a more simple way faster than Fermat's factorization method(x^2-N=y^2)!
True gold fears fire, you can test 1+2+3+...+k=Ny(k<N/2).
More details of the process in my G+ and BLOG.
How can I determine the initial guess of the equation Ax+Bsin(x)=C in terms of A,B and C ?
I am trying to solve it using Newton Raphson. A,B and C will be given during runtime.
Is there any other method more efficient than Newton Raphson for this purpose ?
The optimal initial guess is the root itself, so finding an "optimal" guess isn't really valid.
Any guess will give you a valid solution eventually as long as f'(x0) != 0 for any step, which only occurs at the zeroes of cos(x), which are k*pi + pi/2 for any integer k.
I would try x0 = C * pi, just to see if it works.
Your biggest problem, however, would be the periodic nature of your function. Newton's method will be slow (if it even works) for your function as sin(x) will shift x0 back and forth over and over.
Precaution:
In Newton's method, do you notice how f'(xn) is in the denominator? f'(x) approaches 0 infinitely many times. If your f'(x) = 0.0001 (or anywhere close to zero, which has a chance of happening), your xn+1 gets thrown really far away from xn.
Worse yet, this can happen over and over due to f'(x) being a periodic function, which means that Newton's method might never even converge for an arbitrary x0.
The simplest "good" approximation is to just assume that sin(x) is approximately zero, and so set:
x0 = C/A
Well, if A,B and C are real and different from 0, then (B+C)/A is an upper quote to the highest root and (C-B)/A is a lower quote to the lowest root, as -1 <= sin(x) <= 1. You could start with those.
Newton method can work with any guess. the problem is simple,
if there is an equation and I guessed x0=100
and the best close solution for it is x0=2
and I know the answer is 2.34*
by using any guess in the world you will eventually get to 2.34*
the method says to choose a guess because without a valid guess it will take many solutions which aren't comfortable no one wants to repeat the method 20 times
and guessing a solution is not hard
you just find a critical point-
for example, 3 is too big and 2 is too small
so the answer is between 2 and 3
but if instead guessing 2 you guess 50
you will still get to the right solution.
like I said it will just take you much longer
I tested the method by myself
I guessed 1000 to a random equation
and I knew the best guess was 4
the answer was between 4 and 5
I chose 1000 it took me much time
but after a few hours, I got down from 1000 to 4.something
if you somehow can't find a critical point you can actually put a random number equals to x0 and then eventually you will get to the right solution
no matter what number you guessed.
This is the problem in question: Problem #78
This is driving me crazy. I've been working on this for a few hours now and I've been able to reduce the complexity of finding the number of ways to stack n coins to O(n/2), but even with those improvements and starting from an n for which p(n) is close to one-million, I still can't reach the answer in under a minute. Not at all, actually.
Are there any hints that could help me with this?
Keep in mind that I don't want a full solution and there shouldn't be any functional solutions posted here, so as not to spoil the problem for other people. This is why I haven't included any code either.
Wikipedia can help you here. I assume that the solution you already have is a recursion such as the one in the section "intermediate function". This can be used to find the solution to the Euler problem, but isn't fast.
A much better way is to use the recursion based on the pentagonal number theorem in the next section. The proof of this theorem isn't straight forward, so I don't think the authors of the problem expect that you come up with the theorem by yourself. Rather it is one of the problems, where they expect some literature search.
This problem is really asking to find the first term in the sequence of integer partitions that’s divisible by 1,000,000.
A partition of an integer, n, is one way of describing how many ways the sum of positive integers, ≤ n, can be added together to equal n, regardless of order. The function p(n) is used to denote the number of partitions for n. Below we show our 5 “coins” as addends to evaluate 7 partitions, that is p(5)=7.
5 = 5
= 4+1
= 3+2
= 3+1+1
= 2+2+1
= 2+1+1+1
= 1+1+1+1+1
We use a generating function to create the series until we find the required n.
The generating function requires at most 500 so-called generalized pentagonal numbers, given by n(3n – 1)/2 with 0, ± 1, ± 2, ± 3…, the first few of which are 0, 1, 2, 5, 7, 12, 15, 22, 26, 35, … (Sloane’s A001318).
We have the following generating function which uses our pentagonal numbers as exponents:
1 - q - q^2 + q^5 + q^7 - q^12 - q^15 + q^22 + q^26 + ...
my blog at blog.dreamshire.com has a perl program that solves this in under 10 sec.
Have you done problems 31 or 76 yet? They form a nice set that is an generalization of the same base problem each time. Doing the easier questions may give you insight into a solution for 78.
here some hints:
Divisibility by one million is not the same thing as just being larger than one million. 1 million = 1,000,000 = 10^6 = 2^6 * 5^6.
So the question is to find a lowest n so that the factors of p(n) contain six 2's and six 5's.