## Euler 208- Robot Walk

http://projecteuler.net/index.php?section=forum&id;=208 –

This is a fun one, which I approached in a way which can be classified as ‘entirely more painful then it had to be’. I should probably add a disclaimer to these posts “Not neccessarily a good way to solve the problem”; but alas, the solution is here irregaurdless of its suitability.

In this problem we have a robot; and we want to know, given that it will move 70 units, turning left or right each time (at 72 degrees, or one fifth a circular arc), how many unique paths will end back at the origin. I thought about it off and on for a few weeks. The immediate thing that jumps out are there are 2^70 distinct paths. Clearly immune to normal brute forcing.

First odd thoughts were that there might be something in the realm of combinatorics… I experimented with establishing if there was an obvious tie in between number of left/right turns and solutions; I couldn’t’ find one. I made the discovery that there should be equal number of end points facing all directions: (70 end points, so 70/5=14 points should face in each direction, 0 degrees, 72, 144, 216, 288, 360=0). But, I couldn’t figure out how to combine this knowledge with the fact that only two possible values can occur after a specific value (e.g. you can’t go from facing 0 to facing 216), while also maintaining that each end point needs to occur equally. Also, I’m sorry for the fictional people that read that last paragraph. I didn’t even understand it. All that said, there actually is an odd combination of combinatorics and dynamic programming which I didn’t find; I’ll chat about it after I give my own terrible solution.

So, we already discussed that we can’t brute force it. We can take a different approach though. While there are 2^70 distinct paths, there are considerably fewer unique ending points to paths. As an example, after 5 movements, there are two paths that end at the origin; after 10, there are at least 4 (and possibly more); similar occurs for several other end points.

So, my approach is to build a list of ending points (direction, and ending x and y coordinates, assuming start from origin), and the number of paths that end there. When we have that, we can combine any two paths by rotating the second path according to the first paths direction, and we know how many paths will end there: number of the first path * number of second path. The end result is we save alot of duplication of calculation. That said, this is by no means a fast solution.

We’ll start with the definition of a Location:

class Location { public double ActualX; public double ActualY; public int CloseX; public int CloseY; public int Rotation; public Location(double x, double y, int rotation) { ActualX = x; ActualY = y; CloseX = (int)(100 * x); CloseY = (int)(100 * y); Rotation = rotation % 360; if (Rotation < 0) Rotation += 360; } public static Location operator +(Location left, Location right) { double angle=left.Rotation*Math.PI/180; double co=Math.Cos(angle); double si=Math.Sin(angle); double rotatedX = right.ActualX * co - right.ActualY * si; double rotatedY = right.ActualX * si + right.ActualY * co; return new Location(left.ActualX + rotatedX, left.ActualY + rotatedY, left.Rotation + right.Rotation); } public override bool Equals(object obj) { Location loc = (Location)obj; return loc.CloseX == CloseX && loc.CloseY == CloseY && loc.Rotation == Rotation; } public override int GetHashCode() { return ((Rotation/72) << 28) + (CloseX << 14) + CloseY; } } |

We see a few things about this. First, we’ve set it up to work in a Dictionary (as HashMaps are the most useful ADT ever created) by writing the GetHashCode and Equals methods; and we performed a minor ‘optimization’ of these, by using an integer approximation of the actual floating point location. This also solves a common problem in that checking equality on doubles is often spotty at best: 1.0000000001 != 1.0000000002. Basically, because of how we chose the CloseX and CloseY values, we assume that no two paths will get within 1/100th of a unit distance from each other, and not actually be at the same point. Experimentally, this results in the same number of distinct paths as using a 10,000 multiplier; so it probably works.

This can also sort of be verified mathamatically: because there are only 5 basic movements, the number of areas that can be visited is much decreased: you can’t visit every point on the plane using this robot. Or, somewhat closer to math terms: the movements that can be made do not form a Basis for the plane. (I’m fairly sure that’s a terrible butchering of Linear Algebra terminology; but it works for me). This is mostly because it can’t move at arbitrary factors of the directions it can move in: ie it can’t reverse, it can’t move pi steps in a given direction, and so on.

With those boring parts explained, we also have the plus operator. Given two Locations, it returns the Location that would be formed by following the path that lead to the first Location; then following the path that lead to the second Location given the new starting orientation and position. Basically, what it’s doing is rotating the second point around the origin, then adding that to the first point, and recalculating the rotation. Fairly simple. The rotation is stolen from Linear Algebra, and more specifically, I beleive I stole it from the wikipedia article on Rotation. (Sad fact: College Linear Algebra spent about 5 minutes on the topic, and never came up again).

With our Location defined, let’s look at the algorithms/code:

class E208 { private static Dictionary<int, Dictionary<Location, long>> paths; public static void Main() { paths = new Dictionary<int, Dictionary<Location, long>>(); GenerateStartingPaths(); GeneratePaths(70); Console.WriteLine("Count of 0's: "+paths[70][new Location(0, 0, 0)]); GeneratePaths(25); Console.WriteLine("Count of 0's: " + paths[25][new Location(0, 0, 0)]); } private static void GenerateStartingPaths() { paths[1] = new Dictionary<Location, long>(); double x = Math.Cos(72 * Math.PI / 180); double y = Math.Sin(72 * Math.PI / 180); paths[1][new Location(x, y, 72)] = 1; paths[1][new Location(x, -y, 4*72)] = 1; Combine(1, 1); //generate length 2 paths } private static void GeneratePaths(int length) { while(!paths.ContainsKey(length)) { int maxKey = 1; foreach (int key in paths.Keys) if (key < length) maxKey = Math.Max(maxKey, key); int otherKey=Math.Min(length-maxKey,2); Combine(maxKey, otherKey); } } private static void Combine(int firstKey, int secondKey) { Dictionary<Location, long> first = paths[firstKey]; Dictionary<Location, long> second = paths[secondKey]; Console.WriteLine("Combining " + firstKey + " " + secondKey+" with "+first.Count+" and "+second.Count+" paths"); Dictionary<Location, long> result = new Dictionary<Location, long>(2*first.Count); foreach (KeyValuePair<Location, long> firstKvp in first) foreach (KeyValuePair<Location, long> secondKvp in second) { Location newPoint = firstKvp.Key + secondKvp.Key; long count = firstKvp.Value * secondKvp.Value; long currentValue; if (result.TryGetValue(newPoint, out currentValue)) result[newPoint] = currentValue + count; else result[newPoint] = count; } paths[firstKey + secondKey] = result; } } |

So, we have a few large things here. First note that we keep track of all paths of a given length that were created. So, in that paths Dictionary, we have a list of all paths of length 1, length 2, length 4 … length 70. With that out of the way, let’s get started.

First thing we’ll chat about is combining. Combining takes two sets of paths of given lengths, and then finds all combinations of those two sets as a result set. This gives you all paths of length firstLength + secondLength. It does this by going through each path in the first set, and adding to it each path in the second set. This gives us a new location, and we know that the number of ways to get to that new location are number of ways in first set * number of ways in second set.

Not the cleanest of explanations I understand. A better way to think about it might be to say ‘if I wanted to get to Location X + Y, I could take any path from this first set of paths that take me to X, then any path in the second set of paths that take me to Y. I can thus take number of first paths * number of second paths unique routes.’ Basically.

Moving on from my terrible explanation attempts, we’ll look at GeneratePaths. This has an interesting story to it. When I first wrote this, I would go through, and select the largest two paths that would sum to near the desired length. It made sense to me: take this route, you perform only log n combine operations, and everyone ends up happy. With this approach, calculating paths for length 70 took ~280 seconds. And I spent a while trying to improve this to no avail. As it turns out, this would happen because combining a 32 length path with a 32 length path requires 18,000*18,000 = ~350,000,000 sets of operations (a set of operations being dictionary lookups, cos/sin, memmory allocations, math). What I found was, that by doing only paths of length 2, we decrease the number of operations needed to get to path length 70 by about a factor of 100. More specifically, path length 68 had 324,000 unique paths; if we assume all paths (except length 2, having 4 elements) had 324,000 unique elements, and we needed perform 68/2 Combinations, we would need to perform only 324,000*68/2*4 = 44,000,000 operations. It’s actually a fair bit less, I’d say on the order of 20,000,000. This leads to a total calculation time of only about 8 seconds. Very nice.

As to what it’s actually doing, it goes through, finds the largest key smaller then the length we’re looking for, then tries to combine it with the paths of length 2, or paths of length 1 if it happens that the desired length and max key are only different by a length of 1.

GenerateStartingPaths has the first two paths hard coded, and then goes on to generate paths of length 2, to make the code for generating arbitrary length paths a bit nicer.

Main is fairly obvious: Generates the starting paths, and calls for paths of length 70, then looks up the value of Location 0,0; facing North. Nice and simple. Sort of. This post makes me question the economics of >100 line solutions in a blog format. But, it was fun.

Now, for that nice combinatoric/DP solution.

static Dictionary<long, long> values=new Dictionary<long,long>(); public static void Main() { Console.WriteLine(Calculate(5, 5, 5, 5, 5)); Console.WriteLine(Calculate(14, 14, 14, 14, 14)); } static long Calculate(long n1, long n2, long n3, long n4, long n5) { if (n1 < 0 || n2 < 0 || n3 < 0 || n4 < 0 || n5 < 0) return 0; if ((n1 + n2 + n3 + n4 + n5) == 0) return 1; long key = n1 + 15 * (n2 + 15 * (n3 + 15 * (n4 + 15 * n5))); if (values.ContainsKey(key)) return values[key]; long result = Calculate(n2, n3, n4, n5, n1-1) + Calculate(n5-1, n1, n2, n3, n4); values[key] = result; return result; } |

It executes in under a second (and is based off code by ‘hirak99’ in the Project Euler forums). I almost/barely understand what it is doing. First, consider each movement as a move in x,y; rather then having a rotation component, and other complexities. If you look at it that way, then you need equal number of moves in each of 5 directions to get back to the origin. But, you can only move in one of two directions when at a specific location; this algorithm accounts for that by keeping track of how many moves we’ve made, and rotating them arround to account for ‘I’m turning left’ or ‘I’m turning right’.

As a slightly better (or worse) example, envision you start at the origin, turn left, then turn right. You were at 0,0; you moves to a point at cos(-72),sin(72), then move again to in that same basic direction (actually you move at cos(0),sin(0) rotated 72 degrees). You end up at the point 2x,2y, as the algorithm dictates. Which was the hardest bit for me to understand about it. And I still couldn’t explain this to (non-tech/math relative)… but I found it cool none-the-less.

## Euler 077- First value that can be written as sum of more then x primes

From Euler 077–

The goal is to find the first number that can be written as a sum of primes in 5,000 different ways. As an example, 10 can be written as 7+3, 2+2+2+2+2, 5+5, 5+3+2, etc.

Going about this the slow way. The first and immediate thoughts is ‘gee, I can try brute force’. The maximum number that the answer can be is 10,000 (2+2+…+2=2*5000). It’s actually a fair amount less (2+2+2=3+3, but, my math skills are too terrible to figure out a better upper bound).

Now, I envision that determining all unique sums is quite slow for a single number. Can’t prove but envision. And slow being relative. To speed this up, we use an interesting property of these numbers.

Assume P(k) is the number of unique sums for k; P(k) <= P(k+2), as k+2 can be given simply by 2+(all solutions of k). Similar for P(k+3). I can’t prove it for P(k+1) though, sadly. This means that we could do one of two things: Either do a binary search on all even numbers, then all odd numbers to find the first number with P(k)==5000, (should be about 2*log2(10000)=26 checks, roughly); or, we can do a binary search down to a resolution of about 3, then check those final 3 numbers. It’d be about 16-19 comparisons. As the binary search across all evens then all odds is much simpler/less sticky, we’ll go that route.

Now, humorous fact: I went through all this thinking, and then did a test to see a. how slow finding unique sums is, and b. verify my ‘find unique sums’ algorithm works. The answer is less then 100, and can be found without any of the cool binary searching. I got to do precisely none of the cool things I was planning on. None of them. I feel like life is such a terrible waste, as I probably spent hours working out ways to solve the problem which would be equivalent to ‘find the first k such that P(k) is 10^(something huge)’… very disappointing. And also, P(k)<=P(k+1) for all observed points.

Edit: Slight explanation of how to find the number of unique sums.

What we do is attempt all summations using primes in decreasing value of the prime. This ensures that any summation found will be unique. We do this recursively, and it can actually be considered dynamic programming; but, alas, the answer is already found.

Code:

static List primes; public static void Main() { BuildPrimes(10000); for (int i = 2; i <= 100; i++) Console.WriteLine(i+" "+NumberOfUniqueSums(i, 0)); } static int NumberOfUniqueSums(int k, int primeStartIndex) { if(k<0) return 0; if (k == 0) return 1; int sum = 0; for (int primeIndex = primeStartIndex; primeIndex < primes.Count; primeIndex++) sum += NumberOfUniqueSums(k - primes[primeIndex], primeIndex); return sum; } static void BuildPrimes(int maxSquare) { primes = new List { 2}; int maxI=((int)Math.Sqrt(maxSquare))+1; for (int primeCandidate = 3; primeCandidate < maxI; primeCandidate += 2) { bool isPrime = true; for (int i = 0; i < primes.Count && isPrime; i++) if (primeCandidate % primes[i] == 0) isPrime = false; if (isPrime) primes.Add(primeCandidate); } //max prime is now at index 0 primes.Reverse(); } |

## Euler 206- Concealed Square

Find the unique positive integer whose square has the form 1_2_3_4_5_6_7_8_9_0,

where each “_” is a single digit.

First thought was brute force by replacing blanks with numbers, and checking for a square root. This comes to a mere 10^9 possibilities, with a lot of parsing longs from string. After a bit of analysis though, you realize the last digit of the number we are looking for has to be 0. (0 being the only a*a to result in a 0 digit at the end). This also means the first blank is a 0, giving a mere 10^8 possibilities. Next is to realize that 9 can only be formed by a 3, or a 7 in the number. (Try not to forget the 7, as I did the first time through). So, we now know the target square ends in either 70 or 30.

We can move away a lot of parsing longs from strings, and sqrt’s, by just iterating over the possible answers, rather then the possible answers^2. This basically means iterating from 10^9+30 to something less then 10^10 (sqrt of 1929394959697989900; though I’m far too boring to plug that into a calculator). As already stated, we’re looking for something ending in 30 or 70, so we could start with 10^9+30 and increment by 100, then switch to +70, or alternatively alter increments of 40 and 60.

Probably the most complex thing is checking whether or not a square is of the desired format. For that I get the string representation of the square, and then in a for loop check that every other character is incrementing; with a special check for the 0 case. I envision there are better ways, but this was good enough for this problem.

Code:

static void Main(string[] args) { long start = 1000000000; //add 100 to long on the off chance that there's a terrible loss // of precision in the constant double long end = (long)(Math.Sqrt(1929394959697989900))+100; Test(start + 30, end, 100); Test(start + 70, end, 100); } static void Test(long start, long end, long incr) { for (long current = start; current < end; current += incr) { string squareString = (current * current).ToString(); bool isValid = true; for (int i = 1; i <= 9 && isValid; i++) if (squareString[2 * (i - 1)] != '0' + i) isValid = false; if (squareString[2 * (10 - 1)] != '0') isValid = false; if (isValid) Console.WriteLine("Solution: " + current); } } |

## Euler 001- Starting out Easy

At least something to start the blog with.

Project Euler problem #1; sum all natural numbers less then 1000 that are multiples of 3 or 5. Natural numbers meaning positive integers. The simplest way to go about this is a for loop with a sum, something akin to:

int sum=0; for(int i=1; i<1000; i++) if(i%3==0 || i%5==0) sum+=i; Console.WriteLine(sum); |

Which gives us a nice and specific answer, which is correct. Fun optimization side of this:

You can consider the sum of all 3’s to be 3*sum(1..333), which can be computed as 3*(333*332/2 + 333), if math holds.

Similarly, sum of 5 is 5*sum(1..199) = 5*(199*198/2 + 199).

Next, you need to determine the sum of natural numbers divisible by 15: 15*sum(1..66)=15*67*66/2;

Finally, take the sum of the first two values, and subtract from that the last value. In this case: 166833 + 99500 – 33165 = the answer. This follows from a little too much math, and a bit of set theory: numbers divisible by 15 are counted in both the 3 and 5 cases, effectively double counted. We subtract one of these double counts to end up with a single count of all numbers; and, an O(1) solution.