Solving equation. Counting (x,y)

Question

I'm having a trouble with my math:

Assume that we have a function: F(x,y) = P; And my question is: what would be the most efficient way of counting up suitable (x,y) plots for this function ? It means that I don't need the coordinates themself, but I need a number of them. P is in a range: [0 ; 10^14]. "x" and "y" are integers. Is it solved using bruteforce or there are some advanced tricks(math / programming language(C,C++)) to solve this fast enough ?

To be more concrete, the function is: x*y - ((x+y)/2) + 1.

Do you mean that you want to count the number of integer solutions to *F(x,y) = P* for some fixed *P*? — Oliver Charlesworth, Jun 03 '11 at 09:00
Do you want to solve F(x,y) = , or do you just want to plot it, automatically findint a suitable range? — csl, Jun 03 '11 at 09:02
C or C++ - I still dont know what to use. Depends if C++ has any special techniques, that can simplify the problem. If not - no reason for using it. — Dmitri, Jun 03 '11 at 09:05
Neither C nor C++ has any built-in help for solving Diophantine equations, so you'll need to apply some mathematical reasoning. A very simple observation: your equation factorizes to `(x - 1/2) * (y - 1/2) == P - 3/4`, so even if you do need to brute-force, you only have to do so in one of the variables. Obviously, though, you can only truly brute force over the limited range of an integer type in your chosen language, not over all integers. — Steve Jessop, Jun 03 '11 at 09:08

Steve Jessop · Accepted Answer · 2011-06-03T09:26:06.730

10

x*y - ((x+y)/2) + 1 == P is equivalent to (2x-1)(2y-1) == (4P-3).

So, you're basically looking for the number of factorizations of 4P-3. How to factor a number in C or C++ is probably a different question, but each factorization yields a solution to the original equation. [Edit: in fact two solutions, since if A*B == C then of course (-A)*(-B) == C also].

As far as the programming languages C and C++ are concerned, just make sure you use a type that's big enough to contain 4 * 10^14. int won't do, so try long long.

edited Jun 03 '11 at 09:26

answered Jun 03 '11 at 09:20

Steve Jessop

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IMO it is almost complete solution. +1 (@Dmitri: Why don't you mark this answer as the right one?) – Serge Dundich Jun 03 '11 at 12:15
These two answers both seem to be right, atleast for me :) I will take look at both of them. – Dmitri Jun 03 '11 at 12:50

score 2 · Answer 2 · answered Jun 03 '11 at 09:14

2

You have a two-parameter function and want to solve it for a given constant.

This is a pretty big field in mathematics, and there are probably dozens of algorithms of solving your equation. One key idea that many use is the fact that if you find a point where F<P and then a point F>P, then somewhere between these two points, F must equal P.

One of the most basic algorithms for finding a root (or zero, which you of course can convert to by taking F'=F-P) is Newton's method. I suggest you start with that and read your way up to more advanced algorithms. This is a farily large field of study, so happy reading!

Wikipedia has a list of root-finding algorithms that you can use as a starting place.

answered Jun 03 '11 at 09:14

csl

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Newton's method and most other root-finding algorithms aren't really focused on integer-valued functions, but of course you can still use many them (for instance, simply solve the floating-point equation, then round the solution to integer, try if it is still a solution for the integer equation, if not try the adjacent values and so on). – leftaroundabout Jun 03 '11 at 09:49
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The "key idea" assumes a functions that's continuous. A classic counterexample is `y=1/x`. There's no point where y=0, even though y=-1 at x=-1 and y=+1 at x=+1. – MSalters Jun 03 '11 at 10:20
IMO this answer has almost nothing to do with the original question. The fact that `x, y` are integers matters very much. – Serge Dundich Jun 03 '11 at 12:10
@MSalters: Take any _constant_ C, then even for y=1/x you can find a value for y=C, but singularities are special, yes. – csl Jun 03 '11 at 15:27
@Serge: He asks how to plot them, a very general question, not how to solve Diophantine equations. – csl Jun 03 '11 at 15:27

Solving equation. Counting (x,y)

2 Answers2