I'm working on strengthening my F#-fu and decided to tackle the Facebook Hacker Cup Double Squares problem. I'm having some problems with the run-time and was wondering if anyone could help me figure out why it is so much slower than my C# equivalent.
There's a good description from another post;
Source: Facebook Hacker Cup Qualification Round 2011
A double-square number is an integer X which can be expressed as the sum of two perfect squares. For example, 10 is a double-square because 10 = 3^2 + 1^2. Given X, how can we determine the number of ways in which it can be written as the sum of two squares? For example, 10 can only be written as 3^2 + 1^2 (we don't count 1^2 + 3^2 as being different). On the other hand, 25 can be written as 5^2 + 0^2 or as 4^2 + 3^2.
You need to solve this problem for 0 ≤ X ≤ 2,147,483,647.
Examples:
10 => 1
25 => 2
3 => 0
0 => 1
1 => 1
Numbers From Competition
1740798996
1257431873
2147483643
602519112
858320077
1048039120
415485223
874566596
1022907856
65
421330820
1041493518
5
1328649093
1941554117
4225
2082925
0
1
3
My basic strategy (which I'm open to critique on) is to;
Here is the F# code (See code changes at bottom) I've written that I believe corresponds to this strategy (Runtime: ~8:10);
open System
open System.Collections.Generic
open System.IO
/// Get a sequence of values
let rec range min max =
seq { for num in [min .. max] do yield num }
/// Get a sequence starting from 0 and going to max
let rec zeroRange max = range 0 max
/// Find the maximum number in a list with a starting accumulator (acc)
let rec maxNum acc = function
| [] -> acc
| p::tail when p > acc -> maxNum p tail
| p::tail -> maxNum acc tail
/// A helper for finding max that sets the accumulator to 0
let rec findMax nums = maxNum 0 nums
/// Build a collection of combinations; ie [1,2,3] = (1,1), (1,2), (1,3), (2,2), (2,3), (3,3)
let rec combos range =
seq {
let count = ref 0
for inner in range do
for outer in Seq.skip !count range do
yield (inner, outer)
count := !count + 1
}
let rec squares nums =
let dict = new Dictionary<int, int>()
for s in nums do
dict.[s] <- (s * s)
dict
/// Counts the number of possible double squares for a given number and keeps track of other counts that are provided in the memo dict.
let rec countDoubleSquares (num: int) (memo: Dictionary<int, int>) =
// The highest relevent square is the square root because it squared plus 0 squared is the top most possibility
let maxSquare = System.Math.Sqrt((float)num)
// Our relevant squares are 0 to the highest possible square; note the cast to int which shouldn't hurt.
let relSquares = range 0 ((int)maxSquare)
// calculate the squares up front;
let calcSquares = squares relSquares
// Build up our square combinations; ie [1,2,3] = (1,1), (1,2), (1,3), (2,2), (2,3), (3,3)
for (sq1, sq2) in combos relSquares do
let v = calcSquares.[sq1] + calcSquares.[sq2]
// Memoize our relevant results
if memo.ContainsKey(v) then
memo.[v] <- memo.[v] + 1
// return our count for the num passed in
memo.[num]
// Read our numbers from file.
//let lines = File.ReadAllLines("test2.txt")
//let nums = [ for line in Seq.skip 1 lines -> Int32.Parse(line) ]
// Optionally, read them from straight array
let nums = [1740798996; 1257431873; 2147483643; 602519112; 858320077; 1048039120; 415485223; 874566596; 1022907856; 65; 421330820; 1041493518; 5; 1328649093; 1941554117; 4225; 2082925; 0; 1; 3]
// Initialize our memoize dictionary
let memo = new Dictionary<int, int>()
for num in nums do
memo.[num] <- 0
// Get the largest number in our set, all other numbers will be memoized along the way
let maxN = findMax nums
// Do the memoize
let maxCount = countDoubleSquares maxN memo
// Output our results.
for num in nums do
printfn "%i" memo.[num]
// Have a little pause for when we debug
let line = Console.Read()
And here is my version in C# (Runtime: ~1:40:
using System;
using System.Collections.Generic;
using System.Diagnostics;
using System.IO;
using System.Linq;
using System.Text;
namespace FBHack_DoubleSquares
{
public class TestInput
{
public int NumCases { get; set; }
public List<int> Nums { get; set; }
public TestInput()
{
Nums = new List<int>();
}
public int MaxNum()
{
return Nums.Max();
}
}
class Program
{
static void Main(string[] args)
{
// Read input from file.
//TestInput input = ReadTestInput("live.txt");
// As example, load straight.
TestInput input = new TestInput
{
NumCases = 20,
Nums = new List<int>
{
1740798996,
1257431873,
2147483643,
602519112,
858320077,
1048039120,
415485223,
874566596,
1022907856,
65,
421330820,
1041493518,
5,
1328649093,
1941554117,
4225,
2082925,
0,
1,
3,
}
};
var maxNum = input.MaxNum();
Dictionary<int, int> memo = new Dictionary<int, int>();
foreach (var num in input.Nums)
{
if (!memo.ContainsKey(num))
memo.Add(num, 0);
}
DoMemoize(maxNum, memo);
StringBuilder sb = new StringBuilder();
foreach (var num in input.Nums)
{
//Console.WriteLine(memo[num]);
sb.AppendLine(memo[num].ToString());
}
Console.Write(sb.ToString());
var blah = Console.Read();
//File.WriteAllText("out.txt", sb.ToString());
}
private static int DoMemoize(int num, Dictionary<int, int> memo)
{
var highSquare = (int)Math.Floor(Math.Sqrt(num));
var squares = CreateSquareLookup(highSquare);
var relSquares = squares.Keys.ToList();
Debug.WriteLine("Starting - " + num.ToString());
Debug.WriteLine("RelSquares.Count = {0}", relSquares.Count);
int sum = 0;
var index = 0;
foreach (var square in relSquares)
{
foreach (var inner in relSquares.Skip(index))
{
sum = squares[square] + squares[inner];
if (memo.ContainsKey(sum))
memo[sum]++;
}
index++;
}
if (memo.ContainsKey(num))
return memo[num];
return 0;
}
private static TestInput ReadTestInput(string fileName)
{
var lines = File.ReadAllLines(fileName);
var input = new TestInput();
input.NumCases = int.Parse(lines[0]);
foreach (var lin in lines.Skip(1))
{
input.Nums.Add(int.Parse(lin));
}
return input;
}
public static Dictionary<int, int> CreateSquareLookup(int maxNum)
{
var dict = new Dictionary<int, int>();
int square;
foreach (var num in Enumerable.Range(0, maxNum))
{
square = num * num;
dict[num] = square;
}
return dict;
}
}
}
Thanks for taking a look.
UPDATE
Changing the combos function slightly will result in a pretty big performance boost (from 8 min to 3:45):
/// Old and Busted...
let rec combosOld range =
seq {
let rangeCache = Seq.cache range
let count = ref 0
for inner in rangeCache do
for outer in Seq.skip !count rangeCache do
yield (inner, outer)
count := !count + 1
}
/// The New Hotness...
let rec combos maxNum =
seq {
for i in 0..maxNum do
for j in i..maxNum do
yield i,j }
Again, the number of integer solutions of x^2 + y^2 = k is either
Note that in the second alternative, you count a^2 + b^2 as a different solution to (-a)^2 + b^2 (and other signs), and to b^2 + a^2. So you may want to divide by four, and divide again by 2 (taking the floor, as @Wei Hu points out) if you want solutions as sets instead of ordered pairs.
Knowing this, writing a program which gives the number of solutions is easy: compute prime numbers up to 46341 once and for all.
Given k, compute the prime divisors of k by using the above list (test up to sqrt(k)). Count the ones which are equal to 1 mod 4, and sum. Multiply answer by 4 if needed.
All this is a one or two liner in any lazy functional language (I don't know enough f#, in Haskell it would be two lines long), once you have a primes
infinite sequence: counting the divisors = 1 mod 4 (filterby |> count
or something along these lines) is something quite natural.
I suspect it is faster than brute forcing the decompositions.
Your F# combos
function has awful perf. Something like
let rec combos range =
let a = range |> Seq.toArray
seq {
for i in 0..a.Length-1 do
for j in i..a.Length-1 do
yield i,j }
should be a big speedup.
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With