ࡱ> FHE ubjbjBrBr .x  &@     LiTcccc>>>$-!#4 >>>>>4  cc> c c>c 80q$pq$q$ T>>>>>>>446p>>>>>>>q$>>>>>>>>> : Womens Programming Team Mathematics and Geometry Lecture 10/23/12 I. Base Conversion To convert from base A to base 10: If the base A value is dndn-1dn-2... d0, we simply compute the polynomial dnAn + dn-1An-1 + ... + d1A + d0. To convert from base 10 to base A: assume we are storing the converted value in an array newVal and that this array is "big enough" to store the converted value. Then we convert the base ten value Val as follows: // Compute the successive digits in the converted value in reverse order. place = 0; while (Val > 0) { newVal[place] = Val%A; Val = Val/A; place++; } // Reverse the contents of newVal. for (i=0; i 4) val++; return val/pow(10, decimals); } VII. Counting: Inclusion-Exclusion Principle is the following: |A ( B ( C| = |A| + |B| + |C| |A ( B| |B ( C| |A ( C| + |A ( B ( C|. Hopefully, you can see how it generalizes to a larger number of sets. This is used when you are counting from multiple sets that may overlap. Permutations with repetition: n!/a!b!c! where n is the total number of objects and a, b and c are the number of times each object is repeated. There are  EMBED Equation.3 subsets of size k of a set of size n. And a set of size n has a total of 2n subsets. With permutations and combinations, always be aware of double counting! When computing binomial coefficients, recursion is too expensive, rather, create a table with pascal's triangle: 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 etc. You simply initialize the table with 1's and then compute everything else, row-by-row, adding the two elements above the value being computed to compute the current value. This adding amounts to the identity:  EMBED Equation.3 . Finally, here's an example of a problem that directly uses binomial coefficients: On a grid with m+1 NS streets and n+1 EW streets, how many different paths can you take travelling from the SW corner of the grid to the NE corner of the grid assuming you always move North or East? The answer is  EMBED Equation.3  or  EMBED Equation.3 . The way to look at it is this: You have to walk n+m blocks total. Each block you walk is either N or E. In particular, you walk n blocks N and m blocks E. So, any set of directions to travel between corners can be expressed as a string of n N's and m E's. Each different permutation of these corresponds to another way to walk. The number of permutations is the combination above. (Or more directly, you know that of the m+n "steps" you are taking, you can choose any m of them to move East.) X. Number Theory Determining if a number n is prime: Loop through all the integers from 2 to  EMBED Equation.3 , and see if any of them divide evenly into n. GCD - I think this has been covered before in another lecture. It may be useful to code up the extended euclidean algorithm for your hackpack in case you need to find the inverse of a value mod n. (This is probably something that's much more relevant for finals than regionals though.) (GCD = Greatest Common Divisor) LCM - The LCM(a,b)*GCD(a,b) = a*b, so that gives you an easy way to calculate the LCM. Watch out for overflow here, might want to calculate (a/gcd(a,b))*b to avoid it. Modular Arithmetic: mod works very nicely with other rules of algebra. (x+y) mod n = ((x mod n) + (y mod n)) mod n (x - y) mod n = ((x mod n) - (y mod n)) mod n xy mod n = ((x mod n)(y mod n)) mod n If you know that a value you get from a mod n operations is in between -n+1 and 0, to find the correct corresponding value in between 0 and n-1, simply add n: -41 mod 100 = -41+100 = 59 mod 100. (I apologize for my bad notation!) A completely random fact about primes and factorials: The number of times a prime p divides evenly into n! is  EMBED Equation.3  Couple notes: the brackets stand for the floor function. The sum doesn't really need to go to n. You'll notice that pk fairly quickly exceeds the value of n. When it does, all subsequent terms in the sum are 0. So you just have to sum all the terms until one is zero. A brief explanation as to why this works is imagine dividing out p from the written out expression n! You would cancel out one out of every p values. But this would leave some extra terms, since when you got to p2 or any mutliple thereof, you would have only cancelled out one of the two p's in that term. That's where the rest of the sum comes in. k=2 knocks out the extra factors in each term that is divisible by p2, k=3 knocks out the extra factors in each term that is divisible by p3, etc. XI. Solving a system of equations Kramer's rule is good to solve a system of equations, particularly 2 or 3. The key when implementing this is to check for division by 0. In Kramer's rule, you divide by a determinant. If that's equal to 0 and you don't check, you get an error. If this determinant is not 0, that means the system does not have an unique solution. It either has no solution or an infinite number of solutions. Be aware of this when looking at input restrictions. Here is Kramer's rule: Let the equations be: ax + by + cz = j dx + ey + fz = k gx + hy + iz = l Then x =  EMBED Equation.3  , y =  EMBED Equation.3 , z =  EMBED Equation.3 , same pattern works for 2 eqns. XII. Geometry Triangle Area: bcsinA, "(s(s-a)(s-b)(s-c)), where s = (a+b+c)/2 Use of atan2: Gives you one unambiguous answer in between  PI and +PI. Problem with the rest of the inverse trig functions is that their ranges are smaller, so when you get an answer (the principle answer of the inverse), it may not be the one you want. In particular, watch out for asin, which can give incorrect results in the ambiguous case of the Law of Sines. When solving a triangle, law of cosines yields an unambiguous answer always. If you can't use it, keep in mind that law of sines has an ambiguous case. In particular, if you solve an equation and get sin x = 0.5,where x is an angle in a triangle, x could be 30 degrees or 150 degree. Sometimes you can rule out 150 if one of the other angles in the triangle is greater than 30. Other times, you can't. Know a couple of the different formulas for the area of a triangle. Kramer's rule is good to solve a system of equations, particularly 2 or 3. The key when implementing this is to check for division by 0. In Kramer's rule, you divide by a determinant. If that's equal to 0 and you don't check, you get an error. If this determinant is not 0, that means the system does not have an unique solution. It either has no solution or an infinite number of solutions. Be aware of this when looking at input restrictions. Dot Product: The dot product between two vectors is useful for calculating the angle between them. By definition, for two vectors a and b, a b = |a||b|cos, where  is the angle between the two vectors. Cross Product: The cross product of two vectors in three space yields a vector that is perpendicular to the two original vectors that has a magnitude that is the area of the parallelogram defined by the two vectors. |a x b| = |a||b|sin, where  is the angle between the two vectors. Polygon area: If we have a polygon given in cyclic order( (x1, y1), (x2, y2), (x3, y3), , (xn, yn) ) , we can calculate the area as follows: | (x1y2 y1x2) + (x2y3 y2x3) + + (xny1 ynx1)| / 2 Picks Theorem: If we create a simple polygon with all vertices on integer coordinates on the x-y plane, then the area of the polygon A = i + b/2 1, where I are the number of interior lattice points and b is the number of boundary lattice points. Lattice points are those with integer coordinates. This is typically useful if you want to count the number of interior points, because we can obtain the area of such a polygon using alternative methods. Line intersection in three dimensions: Usually lines in three dimensions don't intersect. You can use vector equations of a line to check: Let p1 and p2 be position vectors in three space, and d1 and d2 be direction vectors in three space and t and s are parameters: r1 = (p1) + t(d1) r2 = (p2) + s(d2) For there to be an intersection r1=r2 must have a solution for t and s. Solving we have: (p1) + t(d1) = (p2)+s(d2). Here is a concrete example: r1 = (1, 2, 3) + t(4,-3,1) r2 = (6, -1, 5) + s(3, -5, -3) (1, 2, 3) + t(4,-3,1) = (6, -1, 5) + s(3, -3, -1) (1+4t, 2-3t, 3+t) = (6+3s, -1-3s, 5-s) So, we have 1+4t = 6+3s solving these first two we get t=2, s=1 2-3t = -1-3s 3+t = 5-s, but this solution doesn't satisfy this equation, so there is no solution. Had the third equation been consistent with the solution from the first two, there would be a point of intersection, which could be determined by plugging either the solution for t into the first equation or the solution to s into the second equation. Line, Plane Intersection: The equation of a plane ax+by+cz = D encapsulates important information about the plane. In particular, (a,b,c) is the normal vector to the plane, and D/sqrt(a2+b2+c2) is the distance of the plane to the origin. The vector equation of the same plane is r%(a,b,c) = D. Thus, to solve the intersection of a line and a plane, just plug in an arbitrary point on the plane in parametric form into the plane equation. Example: r%(3,1,2) = 7 r1 = (1, 2, 3) + t(4,-3,1) = (1+4t, 2-3t, 3+t) We want r = r1, so we have (1+4t, 2-3t, 3+t) %(3,1,2) = 7 3(1+4t) + 1(2-3t)+2(3+t) = 7 3 + 12t + 2  3t + 6 + 2t = 7 11t = -4 t = -4/11, so the point of intersection is (-5/11, 34/11, 29/11). Finding the equation of a plane given three non-collinear points: calculate two vectors on the plane, one from the first point to the second, the other from the first point to the third. Take the cross product of these two vectors. This gives you the normal to the plane. Once you have this, plug in one of the three points on the plane to solve for D. Example: Points (3, 1, 6), (2, 8, 5), (-1, -3, 0). Vectors are (-1, 7, -1) and (-4, -4, -6). The cross product of these is (-38,-2, 32). The plane eqn is 38x-2y+32z = D, plug in (3,1,6) to get 38x-2y+32z = 76. Plane, Plane intersection: Typically, this is a line. It's not is when the normal vectors for both planes are parallel to each other. Otherwise, plug in an arbitrary value of x into both planes. This leaves you two equations in y and z. Solve for both of these for the point. Now, do this for a different value of x to yield a second point of intersection. Now, you have your line. The only time this won't work is when the intersection line is contained with a plane of the form x=c. You'll recognize this since the system of equations won't have a unique solution, meaning that a determinant in Kramer's rule will be zero. If this is the case, simply plug in two arbitrary values for y and solve for x and z. Example: Planes x+3y+z = 12 and 4x-5y+z = -8. Plug in x=0 to yield 3y+z=12 and 5y+z =-8. The solution here is y=2.5, z=4.5, soothe point (0, 2.5, 4.5) is on the line. Now, plug in x=-2 to yield 2+3y+z = 12 and 8-5y+z = -8. These simplify to 3y+z=14 and 5y+z = 0. Solving, we get y=1.75,z=8.75, so the point (-2,1.75,8.75) is also on the line of intersection. The equation of this line of intersection is r = (0, 2.5, 4.5) + t(-2, -.75, 4.25). Note on calculating intersections of parametric equations: Sometimes the parameter in two parametric equations is the same, other times it is different. Make sure you know the difference. If we describe the motion of two different airplanes in terms of t, and are asked to calculate whether the airplanes crash, the parameter we plug into the two equations can't be different. (In real life, my car goes through the exact same spot in space as yours all the time, but hopefully never at the same exact time. Only the latter infers a wreck.) With the intersection of two lines, the two parameters are different. Point-Line Distance: While there is a formula for this, here are two strategies: 1) Calculate the perpendicular line to the given line that goes through our point. 2) Calculate the intersection of this perpendicular line. 3) Use the distance formula between the two points. 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Plane-Plane Distance: Given two parallel planes, they must be of the form Ax + By + Cx = D1 and Ax + By + Cx = D2. The distance between these planes is | D1 D2 |/"(A2 + B2 + C2). Point-Plane Distance: Given a point, calculate the plane that goes through it that is parallel to the given plane, then do Plane-Plane distance. Line-Plane Distance: These two don t intersect iff the normal vector to the plane is perpendicular to the direction vector of the line. In this case, to get a distance, just pick a point on the line and use the Point-Plane distance method. Calculating a circle given three points on the circle: Calculate two perpendicular bisectors between pairs of the given points and find the intersection. This is the center of the circle. The radius is just the distance from this center point to any of the other three points. tu$a$,1h/ =!"#$% |Dd @J  C A? 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