## #StackBounty: #algorithm #apache-spark #mapreduce #graph-theory #disjoint-sets Disjoint sets on apache spark

### Bounty: 50

I trying to find algorithm of searching disjoint sets (connected components/union-find) on large amount of data with apache spark.
Problem is amount of data. Even Raw representation of graph vertex doesn’t fit in to ram on single machine. Edges also doesn’t fit in to the ram.

Source data is text file of graph edges on hdfs: “id1 t id2”.

id present as string value, not int.

Naive solution that I found is:

1. take rdd of edges -> [id1:id2] [id3:id4] [id1:id3]
2. group edges by key. -> [id1:[id2;id3]][id3:[id4]]
3. for each record set minimum id to each group -> (flatMap) [id1:id1][id2:id1][id3:id1][id3:id3][id4:id3]
4. reverse rdd from stage 3 [id2:id1] -> [id1:id2]
5. leftOuterJoin of rdds from stage 3 and 4
6. repeat from stage 2 while size of rdd on step 3 wouldn’t change

But this results in the transfer of large amounts of data between nodes
(shuffling)

Get this bounty!!!

## #HackerEarth: #BattleOfBots 9: Taunt

Problem

Taunt is a two player board game which is played on a 10X4 grid of cells and is played on opposite sides of the game-board. Each player has an allocated color, Orange ( First Player ) or Green ( Second Player ) being conventional. Each player has nine piece in total. The players move their pieces towards to his / her opponent’s area by moving their pieces strategically. Each piece has a different moving feature and is one of the 3 types of pieces.

Piece 1: It can move to horizontally or vertically adjacent cell, if the cell doesn’t contain a piece of same color.

Piece 2: It can move to horizontally adjacent cell or can move two steps forward, if the cell doesn’t contain a piece of same color (except the piece itself).

This type of piece can move to its own position if its in the second last row of the grid and going downward or if its in the second row of the grid and going upward.

Piece 3: It can move two step diagonally in the forward direction, if the cell doesn’t contain a piece of same color (except the piece itself).

This type of piece can move to its own position if its in the second last row of the grid and going downward or if its in the second row of the grid and going upward.

Players take turns involving moves of pieces as mentioned above and can captures opponent’s piece by jumping on or over opponent’s pieces.

Note: Forward direction for first player is downward and for second player is upward.

If a piece (except piece 1) is moving downward and touches the last row, its direction will change i.e. now it will move upward. Similarly, once if a piece (except piece 1) is moving upward and touches the first row, its direction will change i.e. now it will move downward.

Rules:

• Player can only move according to the moves mentioned above.
• A player may not move an opponent’s piece.
• A player can captures opponent’s piece by jumping on or over opponent pieces.

The game will end after 100 moves ( 50 moves for each player ) or when any of the players don’t have any move left. At the end of the game the player with majority of pieces will win.

We will play it on an 10X4 grid. The top left of the grid is [0,0] and the bottom right is [9,3].

Input:
The input will be a 10X4 matrix consisting only of 0,1or2. Next line will contain an integer denoting the total number of moves till the current state of the board. Next line contains an integer – 1 or 2 which is your player id.

In the given matrix, top-left is [0,0] and bottom-right is [9,3]. The y-coordinate increases from left to right, and x-coordinate increases from top to bottom.

A cell is represented by 3 integers.

First integer denotes the player id (1 or 2).
Second integer denotes the type of piece (1, 2 or 3).
Third integer denotes the direction of the piece (0 (upward) or 1 (downward)). When the piece is of first type, direction doesn’t matter as the piece is free to move to horizontally or vertically adjacent cell, if the cell doesn’t contain a piece of same color.

Empty cell is represented by 000.

Output:
In the first line print the coordinates of the cell separated by space, the piece you want to move.
In second line print the coordinates of the cell in which the piece will jump.
You must take care that you don’t print invalid coordinates. For example, [1,1] might be a valid coordinate in the game play if the piece is able to jump to [1,1], but [9,10] will never be. Also if you play an invalid move or your code exceeds the time/memory limit while determining the move, you lose the game.

Starting state
The starting state of the game is the state of the board before the game starts.

131 131 131 121
121 121 111 111
111 000 000 000
000 000 000 000
000 000 000 000
000 000 000 000
000 000 000 000
000 000 000 210
210 210 220 220
220 230 230 230


First Input
This is the input give to the first player at the start of the game.

131 131 131 121
121 121 111 111
111 000 000 000
000 000 000 000
000 000 000 000
000 000 000 000
000 000 000 000
000 000 000 210
210 210 220 220
220 230 230 230
0
1


SAMPLE INPUT
000 000 000 000
000 000 000 111
000 000 111 130
000 000 000 000
000 000 000 000
000 220 000 000
131 000 000 000
121 000 210 000
000 210 131 000
000 210 000 000
58
1
SAMPLE OUTPUT
8 2
8 0


Explanation

This is player 1’s turn, and the player will move the piece at [8,2] and will take two steps diagonally in downward direction and will be at [8,0]
After his/her move the state of game becomes:

000 000 000 000
000 000 000 111
000 000 111 130
000 000 000 000
000 000 000 000
000 220 000 000
131 000 000 000
121 000 210 000
130 210 000 000
000 000 000 000
59
2


Note: Direction of the piece is also changed from 1 to 0 as the piece was moving downward and touches the last row. This state will be fed as input to program of player 2.

Here is the code of the default bot.

Time Limit:1.0 sec(s) for each input file.
Memory Limit:256 MB
Source Limit:1024 KB

Sample Game

## #StackBounty: #c++ #algorithm #object-oriented #c++11 #interview-questions Solve a set of "restricted" linear equations effic…

### Bounty: 100

I was recently asked to solve the following challenge (in C++) as part of the interview process. However, I haven’t heard from them at all afterwards, and based on past experiences of unsuccessful applicants that I’ve read online, my submission didn’t meet their standards. Since I did solve the challenge to the best of my abilities, I’m at a loss to understand in what ways I could have made a better solution. I’m posting the problem statement (in my own words) and my solution here. Please critique it as you would for a potential applicant to your team (as a means for gauging whether it’s worthwhile to have a subsequent phone-screen with such an applicant).

## Input Details

The utility would take as input an input file containing a list of
equations, one per line. Each equation has the following format:
<LHS> = <RHS>, where LHS is the left-hand side of the equation and is always a variable name.
RHS is the right hand side of the equation and can be composed of the following only:

• Variables
• Unsigned integers
• The + operator

### Assumptions

Input is well-formed i.e.

• Number of variables = Number of equations, with each variable occurring on the LHS of exactly one equation.
• The system of equations has an unique solution, and does not have circular dependencies.
• There are one or more white spaces between each token (numbers, + operator, variables).
• A variable name can only be composed of letters from the alphabet (e.g. for which isalpha(c) is true).
• All integers will fit in a C unsigned long.

## Output Format

The utility would print the value of each variable after evaluating the set of equations, in the format <variable name> = <unsigned integer value>. The variables would be sorted in ascending (lexicographic) order.

### Sample Input Output

Input file:

off = 4 + r + 1
l   = 1 + or + off
or  = 3 + 5
r   = 2


Expected output for the above input:

l   = 16
off = 7
or  = 8
r   = 2


## Implementation Notes

Due to the simplified nature of the input equations, a full-blown linear
equation solver is not required in my opinion (as such a solver would have at least quadratic complexity). A much simplified (and asymptotically faster) solution can be arrived at by modeling the set of input equations as a Directed Acyclic Graph (DAG), by observing the dependencies of the variables from the input equations. Once we can model the system as a DAG, the steps to derive the variable values are as follows:

• Construct the dependency DAG, where each node in the graph corresponds to a variable, and $(a, b)$ is a directed edge from $a$ to $b$ if and only if the variable $a$ needs to be fully evaluated before evaluating $b$.
• Order the vertices in the DAG thus constructed using topological sort.
• For each vertex in the sorted order, evaluate its corresponding variable fully before moving on to the next vertex.

The algorithm above has a linear complexity, which is the best we could achieve under the current assumptions. I’ve encapsulated the algorithm in the following class (I’ve used Google’s C++ Style Guide in my code – not sure it’s the best choice, but I preferred to follow a style guide that’s at least recognized by and arrived at by a non-trivial number of engineers.)

//
// Class that encapsulates a (constrained) linear equation solver. See README.md
// for assumptions on input restrictions.
//
#include <unordered_map>
#include <vector>
#include <list>

#ifndef _EVALUATOR
#define _EVALUATOR

class Evaluator
{
private:
// Stores the values of each variable throughout algorithm
std::vector<UL>                      variable_values_;

// Hash tables that store the correspondence between variable name and index
std::unordered_map<std::string, UL>  variable_index_map_;
std::unordered_map<UL, std::string>  index_variable_map_;

// Adjacency list for DAG that stores the dependency information amongst
// variables. If A[i, j] is an edge, it implies variable 'i' appears on the
// RHS of definition of variable 'j'.

// List of equations stored as indices. If the list corresponding to eq[i]
// contains 'j', then variable 'j' appears on the RHS of variable 'i'.
std::vector<std::list<UL> >          equation_list_;

// For efficiency, this list stores the number of dependencies for each
// variable, which is useful while executing a topological sort.
std::vector<UL>                      num_dependencies_;

// Resets all internal data structures
void  Clear();

// Prints values of internal data structures to aid in debugging
void  PrintState();

// Adds an entry corresponding to each new variable detected while parsing input

// Parse the input equations from filename given as argument, and build the
// internal data structures coressponsing to the input.
bool  ParseEquationsFromFile(const std::string&);

// If DAG in dependency_adj_list_ has a valid topological order, returns
// true along with the ordered vertices in the input vector
bool  GetTopologicalVarOrder(std::vector<UL>&);

public:
Evaluator() {};

/**
* @brief Evaluate the set of constrained linear equations and returns the
*        values of the variables as a list.
*
* @param[in]  string: Filename containing list of constrained linear equations.
* @param[in]  vector<string>: If solution exists, returns the values of
*             variables in lexicographic order (ascending).
*
* @return True if solution exists (always exists for valid input), false if
*              input is not well-formed (See README.md for more details about input
*              format).
*/
bool SolveEquationSet(const std::string&, std::vector<std::string>& );
};
#endif


The main class file:

#include "evaluator.h"
#include <sstream>
#include <unordered_set>
#include <set>
#include <queue>
#include <algorithm>
#include <cassert>

#ifdef _EVALUATOR

// Used for early returns if the expression is false
#define TRUE_OR_RETURN(EXPR, MSG)
do
{
bool status = (EXPR);
if (status != true)
{
cerr << __FUNCTION__
<< ": " << MSG << endl;
return false;
}
} while(0)
#endif

using namespace std;
//****  Helper functions local to the file ****

// Returns true if each character in the non-empty string is a digit
bool IsNumber(string s)
{
return !s.empty() && std::all_of(s.begin(), s.end(), ::isdigit);
}

// Given a string, returns a vector of tokens separated by whitespace
vector<string> ParseTokensFromString(const string& s)
{
istringstream   iss(s);
vector<string>  token_list;
string          token;
while (iss >> token)
token_list.push_back(token);
}

// Returns true if the string can be a valid variable name (i.e has
// only alphabetical characters in it).
bool IsValidVar(string& v)
{
for (auto& c: v)
TRUE_OR_RETURN(isalpha(c), "Non-alphabetical char in variable: " + v);
return true;
}

//********************************************

void Evaluator::Clear()
{
variable_values_.clear();
variable_index_map_.clear();
index_variable_map_.clear();
equation_list_.clear();
num_dependencies_.clear();
}

void Evaluator::PrintState()
{
for (auto i = 0U; i < dependency_adj_list_.size(); ++i)
cout << index_variable_map_[i] << "(" << i << ") =>"
<< "Value(" << variable_values_[i] << "), Deps("
<< num_dependencies_[i] << ")" << endl;
}

// Ensures all data structures correctly set aside an entry for the new variable
{
if (variable_index_map_.count(v) == 0)
{
equation_list_.push_back(list<UL>());
variable_values_.push_back(0);
num_dependencies_.push_back(0);

assert(num_dependencies_.size() == variable_values_.size() &&
variable_index_map_.size() == variable_values_.size() &&
}
return variable_index_map_[v];
}

// Parses equation from given input file line-by-line, checking
// for validity of input at each step and returning true only if
// all equations were successfully parsed.
bool Evaluator::ParseEquationsFromFile(const string& sEqnFile)
{
string    line;
ifstream  infile(sEqnFile);

// This LUT serves as a sanity check for duplicate definitions of vars
// As per spec, only ONE definition (appearance as LHS) per variable is handled
unordered_set<string>  defined_vars;
while (getline(infile, line))
{
vector<string> tokens = ParseTokensFromString(line);
string         lhs    = tokens[0];

// Check if equation is adhering to spec
TRUE_OR_RETURN(tokens.size() >= 3 && IsValidVar(lhs) && tokens[1] == "=",
"Invalid equation: " + line);

// Check if variable on LHS was previously defined - this would make the
// current approach untenable, and require general equation solver.
TRUE_OR_RETURN(defined_vars.count(lhs) == 0, "Multiple defn for: " + lhs);
defined_vars.insert(lhs);

// The operands appear in alternate positions in RHS, tracked by isOp
for (size_t i = 2, isOp = 0; i < tokens.size(); ++i, isOp ^= 1)
{
string token = tokens[i];
if (isOp)
TRUE_OR_RETURN(token == "+", "Unsupported operator: " + token);
else
{
if (IsNumber(token))
variable_values_[lhs_idx] += stol(token);
else
{
TRUE_OR_RETURN(IsValidVar(token), "Invalid variable name: " + token);

// Token variable must be evaluated before LHS.
// Hence adding token => LHS edge, and adding token to RHS of
// equation_list_[lhs]
assert(lhs_idx < equation_list_.size());
equation_list_[lhs_idx].push_back(token_idx);
num_dependencies_[lhs_idx]++;
}
}
}
}
}

// Execute the BFS version of topological sorting, using queue
bool Evaluator::GetTopologicalVarOrder(vector<UL>& ordered_vertices)
{
ordered_vertices.clear();
queue<UL> q;
for (auto i = 0U; i < dependency_adj_list_.size(); ++i)
if (num_dependencies_[i] == 0)
q.push(i);

while (!q.empty())
{
UL var_idx = q.front();
ordered_vertices.push_back(var_idx);
q.pop();
{
assert(num_dependencies_[nbr] >= 0);
num_dependencies_[nbr]--;
if (num_dependencies_[nbr] == 0)
q.push(nbr);
}
}
}

// Solves the constrained set of linear equations in 3 phases:
// 1) Parsing equations and construction of the dependency DAG
// 2) Topological sort on the dependency DAG to get the order of vertices
// 3) Substituting the values of variables according to the sorted order,
//    to get the final values for each variable.
bool Evaluator::SolveEquationSet(const string& eqn_file, vector<string>& solution_list)
{
Clear();
vector<UL> order;
TRUE_OR_RETURN(ParseEquationsFromFile(eqn_file), "Parsing Equations Failed");

// Populate variable values in topological order
for (auto& idx: order)
for (auto& nbr: equation_list_[idx])
variable_values_[idx] += variable_values_[nbr];

// Get keys from the LUT sorted in ascending order
set<pair<string, UL> > sorted_var_idx;
for (auto& vi_pair: variable_index_map_)
sorted_var_idx.insert(vi_pair);
for (auto& vi_pair: sorted_var_idx)
solution_list.push_back(vi_pair.first + " = " +
to_string(variable_values_[vi_pair.second]));

return true;
}
#endif


The usage of the class is as follows:

   string          eqn_file, log_file;
Evaluator       evaluate;
vector<string>  solution_list;

// Logic to get input filename from user - skipping it here
bool bStatus = evaluate.SolveEquationSet(eqn_file, solution_list);

for (auto& s: solution_list)
cout << s << endl;


Get this bounty!!!

## #StackBounty: #c++ #algorithm #boost #numerical-methods Double exponential quadrature

### Bounty: 50

I’m trying to lighten the code review load for the maintainers of boost.math, and I was hoping you guys could help me out. I have a pull request which implements tanh-sinh quadrature, which is provably optimal for integrands in Hardy spaces.

Here’s my code:

which is also reproduced below.

I have a few design worries.

1. It is a class and not a function. This is a bit confusing; I worry that people will not recognize that the constructor is doing some one-time calculations to make integrations faster.
2. It takes a long time to compile. I generated the abscissas and weights to 100 digits, and then they must be lexically cast to their correct type. I could keep fewer levels of abscissas and weights in the .hpp, but then the runtime would longer for complex integrands.
3. For integrands in Hardy spaces, the number of correct digits always doubles on each refinement. However, we want to do just the right amount of work to deliver the requested accuracy, which is almost always overshot.

Interface:

#ifndef BOOST_MATH_QUADRATURE_TANH_SINH_HPP

#include <cmath>
#include <limits>
#include <memory>

namespace boost{ namespace math{

template<class Real>
class tanh_sinh
{
public:
tanh_sinh(Real tol = sqrt(std::numeric_limits<Real>::epsilon()), size_t max_refinements = 20);

template<class F>
Real integrate(F f, Real a, Real b, Real* error = nullptr);

private:
std::shared_ptr<detail::tanh_sinh_detail<Real>> m_imp;
};

template<class Real>
tanh_sinh<Real>::tanh_sinh(Real tol, size_t max_refinements) : m_imp(std::make_shared<detail::tanh_sinh_detail<Real>>(tol, max_refinements))
{
return;
}

template<class Real>
template<class F>
Real tanh_sinh<Real>::integrate(F f, Real a, Real b, Real* error)
{
using std::isfinite;
using boost::math::constants::half;
using boost::math::detail::tanh_sinh_detail;

if (isfinite(a) && isfinite(b))
{
if (b <= a)
{
throw std::domain_error("Arguments to integrate are in wrong order; integration over [a,b] must have b > a.n");
}
Real avg = (a+b)*half<Real>();
Real diff = (b-a)*half<Real>();
auto u = [=](Real z) { return f(avg + diff*z); };
return diff*m_imp->integrate(u, error);
}

// Infinite limits:
if (a <= std::numeric_limits<Real>::lowest() && b >= std::numeric_limits<Real>::max())
{
auto u = [=](Real t) { auto t_sq = t*t; auto inv = 1/(1 - t_sq); return f(t*inv)*(1+t_sq)*inv*inv; };
return m_imp->integrate(u, error);
}

// Right limit is infinite:
if (isfinite(a) && b >= std::numeric_limits<Real>::max())
{
auto u = [=](Real t) { auto z = 1/(t+1); auto arg = 2*z + a - 1; return f(arg)*z*z; };
return 2*m_imp->integrate(u, error);
}

if (isfinite(b) && a <= std::numeric_limits<Real>::lowest())
{
auto u = [=](Real t) { return f(b-t);};
auto v = [=](Real t) { auto z = 1/(t+1); auto arg = 2*z - 1; return u(arg)*z*z; };
return 2*m_imp->integrate(v, error);
}

throw std::logic_error("The domain of integration is not sensible; please check the bounds.n");
}

}}
#endif


Implementation (with some layers of abscissas and weights removed, for brevity):

#ifndef BOOST_MATH_QUADRATURE_DETAIL_TANH_SINH_DETAIL_HPP

#include <cmath>
#include <vector>
#include <typeinfo>
#include <boost/math/constants/constants.hpp>
#include <boost/math/special_functions/next.hpp>

namespace boost{ namespace math{ namespace detail{

// Returns the tanh-sinh quadrature of a function f over the open interval (-1, 1)

template<class Real>
class tanh_sinh_detail
{
public:
tanh_sinh_detail(Real tol = sqrt(std::numeric_limits<Real>::epsilon()), size_t max_refinements = 15);

template<class F>
Real integrate(F f, Real* error = nullptr);

private:
Real m_tol;
Real m_t_max;
size_t m_max_refinements;

const std::vector<std::vector<Real>> m_abscissas{
{
lexical_cast<Real>("0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"), // g(0)
lexical_cast<Real>("0.951367964072746945727055362904639667492765811307212865380079106867050650113429723656597692697291568999499"), // g(1)
lexical_cast<Real>("0.999977477192461592864899636308688849285982470957489530009950811164291603926066803141755297920571692976244"), // g(2)
lexical_cast<Real>("0.999999999999957058389441217592223051901253805502353310119906858210171731776098247943305316472169355401371"), // g(3)
lexical_cast<Real>("0.999999999999999999999999999999999999883235110249013906725663510362671044720752808415603434458608954302982"), // g(4)
lexical_cast<Real>("0.999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999988520"), // g(5)
},
{
lexical_cast<Real>("0.674271492248435826080420090632052144267244477154740136576044169121421222924677035505101849981126759183585"), // g(0.5)
lexical_cast<Real>("0.997514856457224386832717419238820368149231215381809295391585321457677448277585202664213469426402672227688"), // g(1.5)
lexical_cast<Real>("0.999999988875664881984668015033322737014902900245095922058323073481945768209599289672119775131473502267885"), // g(2.5)
lexical_cast<Real>("0.999999999999999999999946215084086063112254432391666747077319911504923816981286361121293026490456057993796"), // g(3.5)
lexical_cast<Real>("0.999999999999999999999999999999999999999999999999999999999999920569786807778838966034206923747918174840316"), // g(4.5)
},
};

const std::vector<std::vector<Real>> m_weights{
{
lexical_cast<Real>("1.570796326794896619231321691639751442098584699687552910487472296153908203143104499314017412671058533991074"), // g'(0)
lexical_cast<Real>("0.230022394514788685000412470422321665303851255802659059205889049267344079034811721955914622224110769925389"), // g'(1)
lexical_cast<Real>("0.000266200513752716908657010159372233158103339260303474890448151422406465941700219597184051859505048039379"), // g'(2)
lexical_cast<Real>("0.000000000001358178427453909083422196787474500211182703205221379182701148467473091863958082265061102415190"), // g'(3)
lexical_cast<Real>("0.000000000000000000000000000000000010017416784066252963809895613167040078319571113599666377944404151233916"), // g'(4)
lexical_cast<Real>("0.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000002676308"), // g'(5)
},
{
lexical_cast<Real>("0.965976579412301148012086924538029475282925173953839319280640651228612016942374156051084481340637789686773"), // g'(0.5)
lexical_cast<Real>("0.018343166989927842087331266912053799182780844891859123704139702537001454134135727608312038892655885289502"), // g'(1.5)
lexical_cast<Real>("0.000000214312045569430393576972333072321177878392994404158296872167216420137367626015660887389125958297359"), // g'(2.5)
lexical_cast<Real>("0.000000000000000000002800315101977588958258001699217015336310581249269449114860803391121477177123095973212"), // g'(3.5)
lexical_cast<Real>("0.000000000000000000000000000000000000000000000000000000000011232705345486918789827474356787339538750684404"), // g'(4.5)
},
};

};

template<class Real>
tanh_sinh_detail<Real>::tanh_sinh_detail(Real tol, size_t max_refinements)
{
m_tol = tol;
m_max_refinements = max_refinements;
/*
* Our goal is to calculate t_max such that tanh(pi/2 sinh(t_max)) < 1 in the requested precision.
* What follows is a good estimate for t_max, but in fact we can get closer by starting with this estimate
* and then calculating tanh(pi/2 sinh(t_max + eps)) until it = 1 (the second to last is t_max).
* However, this is computationally expensive, so we can't do it.
* An alternative is to cache the value of t_max for various types (float, double, long double, float128, cpp_bin_float_50, cpp_bin_float_100)
* and then simply use them, but unfortunately the values will be platform dependent.
* As such we are then susceptible to the catastrophe where we evaluate the function at x = 1, when we have promised we wouldn't do that.
* Other authors solve this problem by computing the abscissas in double the requested precision, and then returning the result at the request precision.
* This seems to be overkill to me, but presumably it's an option if we find integrals on which this method struggles.
*/

using std::tanh;
using std::sinh;
using std::asinh;
using std::atanh;
using boost::math::constants::half_pi;
using boost::math::constants::two_div_pi;

auto g = [](Real t) { return tanh(half_pi<Real>()*sinh(t)); };
auto g_inv = [](Real x) { return asinh(two_div_pi<Real>()*atanh(x)); };

Real x = float_prior((Real) 1);
m_t_max = g_inv(x);
while(!isnormal(m_t_max))
{
// Although you might suspect that we could be in this loop essentially for ever, in point of fact it is only called once
// even for 100 digit accuracy, and isn't called at all up to float128.
x = float_prior(x);
m_t_max = g_inv(x);
}
// This occurs once on 100 digit arithmetic:
while(!(g(m_t_max) < (Real) 1))
{
x = float_prior(x);
m_t_max = g_inv(x);
}
}

template<class Real>
template<class F>
Real tanh_sinh_detail<Real>::integrate(F f, Real* error)
{
using std::abs;
using std::floor;
using std::tanh;
using std::sinh;
using std::sqrt;
using boost::math::constants::half;
using boost::math::constants::half_pi;
Real h = 1;
Real I0 = half_pi<Real>()*f(0);
Real IL0 = abs(I0);
for(size_t i = 1; i <= m_t_max; ++i)
{
Real x = m_abscissas[0][i];
Real w = m_weights[0][i];
Real yp = f(x);
Real ym = f(-x);
I0 += (yp + ym)*w;
IL0 += (abs(yp) + abs(ym))*w;
}

Real I1 = half<Real>()*I0;
Real IL1 = abs(I1);
h /= 2;
Real sum = 0;
Real absum = 0;
for(size_t i = 0; i < m_weights[1].size() && h + i <= m_t_max; ++i)
{
Real x = m_abscissas[1][i];
Real w = m_weights[1][i];

Real yp = f(x);
Real ym = f(-x);
sum += (yp + ym)*w;
absum += (abs(yp) + abs(ym))*w;
}
I1 += sum*h;
IL1 += sum*h;

size_t k = 2;
Real err = abs(I0 - I1);
while (k < 4 || (k < m_weights.size() && k < m_max_refinements) )
{
I0 = I1;
IL0 = IL1;

I1 = half<Real>()*I0;
IL1 = half<Real>()*IL0;
h = (Real) 1/ (Real) (1 << k);
Real sum = 0;
Real absum = 0;
auto const& abscissa_row = m_abscissas[k];
auto const& weight_row = m_weights[k];

// We have no guarantee that round-off error won't cause the function to be evaluated strictly at the endpoints.
// However, making sure x < 1 - eps is a reasonable compromise between accuracy and usability..
for(size_t j = 0; j < weight_row.size() && abscissa_row[j] < (Real) 1 - std::numeric_limits<Real>::epsilon(); ++j)
{
Real x = abscissa_row[j];
Real w = weight_row[j];

Real yp = f(x);
Real ym = f(-x);
Real term = (yp + ym)*w;
sum += term;

// A question arises as to how accurately we actually need to estimate the L1 integral.
// For simple integrands, computing the L1 norm makes the integration 20% slower,
// but for more complicated integrands, this calculation is not noticeable.
Real abterm = (abs(yp) + abs(ym))*w;
absum += abterm;
}

I1 += sum*h;
IL1 += absum*h;
++k;
err = abs(I0 - I1);
if (err <= m_tol*IL1)
{
if (error)
{
*error = err;
}
return I1;
}
}

if (!isnormal(I1))
{
throw std::domain_error("The tanh_sinh integrate evaluated your function at a singular point. Please narrow the bounds of integration or chech your function for singularities.n");
}

while (k < m_max_refinements && err > m_tol*IL1)
{
I0 = I1;
IL0 = IL1;

I1 = half<Real>()*I0;
IL1 = half<Real>()*IL0;
h *= half<Real>();
Real sum = 0;
Real absum = 0;
for(Real t = h; t < m_t_max - std::numeric_limits<Real>::epsilon(); t += 2*h)
{
Real s = sinh(t);
Real c = sqrt(1+s*s);
Real x = tanh(half_pi<Real>()*s);
Real w = half_pi<Real>()*c*(1-x*x);

Real yp = f(x);
Real ym = f(-x);
Real term = (yp + ym)*w;
sum += term;
Real abterm = (abs(yp) + abs(ym))*w;
absum += abterm;
// There are claims that this test improves performance,
// however my benchmarks show that it's slower!
// However, I leave this comment here because it totally stands to reason that this *should* help:
//if (abterm < std::numeric_limits<Real>::epsilon()) { break; }
}

I1 += sum*h;
IL1 += absum*h;
++k;
err = abs(I0 - I1);
if (!isnormal(I1))
{
throw std::domain_error("The tanh_sinh integrate evaluated your function at a singular point. Please narrow the bounds of integration or chech your function for singularities.n");
}
}

if (error)
{
*error = err;
}
return I1;
}

}}}
#endif


For those of you interested in performance, I have used google benchmark to measure the runtime, which is reproduced below:

#include <cmath>
#include <iostream>
#include <random>
#include <boost/math/constants/constants.hpp>
#include <boost/multiprecision/float128.hpp>
#include <boost/multiprecision/cpp_bin_float.hpp>
#include <boost/multiprecision/cpp_dec_float.hpp>
#include <benchmark/benchmark.h>

using boost::math::tanh_sinh;
using boost::math::constants::half_pi;
using boost::multiprecision::float128;
using boost::multiprecision::cpp_bin_float_50;
using boost::multiprecision::cpp_bin_float_100;
using boost::multiprecision::cpp_dec_float_50;
using boost::multiprecision::cpp_dec_float_100;

template<class Real>
static void BM_tanh_sinh_sqrt_log(benchmark::State& state)
{
using std::sqrt;
using std::log;

auto f = [](Real t) { return sqrt(t)*log(t); };
Real Q;
Real tol = 100*std::numeric_limits<Real>::epsilon();
tanh_sinh<Real> integrator(tol, 20);
while(state.KeepRunning())
{
benchmark::DoNotOptimize(Q = integrator.integrate(f, (Real) 0, (Real) 1));
}
}

template<class Real>
static void BM_tanh_sinh_linear(benchmark::State& state)
{
auto f = [](Real t) { return 5*t + 7; };
Real Q;
Real tol = 100*std::numeric_limits<Real>::epsilon();
tanh_sinh<Real> integrator(tol, 20);
while(state.KeepRunning())
{
benchmark::DoNotOptimize(Q = integrator.integrate(f, (Real) 0, (Real) 1));
}
}

template<class Real>
{
auto f = [](Real t) { return 5*t*t + 3*t + 7; };
Real Q;
Real tol = 100*std::numeric_limits<Real>::epsilon();
tanh_sinh<Real> integrator(tol, 20);
while(state.KeepRunning())
{
benchmark::DoNotOptimize(Q = integrator.integrate(f, (Real) -1, (Real) 1));
}
}

template<class Real>
static void BM_tanh_sinh_runge(benchmark::State& state)
{
auto f = [](Real t) { return 1/(1+25*t*t); };
Real Q;
Real tol = 100*std::numeric_limits<Real>::epsilon();
tanh_sinh<Real> integrator(tol, 20);
while(state.KeepRunning())
{
benchmark::DoNotOptimize(Q = integrator.integrate(f, (Real) -1, (Real) 1));
}
}

template<class Real>
static void BM_tanh_sinh_runge_move_pole(benchmark::State& state)
{
auto f = [](Real t) { Real tmp = t/5; return 1/(1+tmp*tmp); };
Real Q;
Real tol = 100*std::numeric_limits<Real>::epsilon();
tanh_sinh<Real> integrator(tol, 20);
while(state.KeepRunning())
{
benchmark::DoNotOptimize(Q = integrator.integrate(f, (Real) -1, (Real) 1));
}
}

template<class Real>
static void BM_tanh_sinh_exp_cos(benchmark::State& state)
{
auto f = [](Real t) { return exp(t)*cos(t); };
Real Q;
Real tol = 100*std::numeric_limits<Real>::epsilon();
tanh_sinh<Real> integrator(tol, 20);
while(state.KeepRunning())
{
benchmark::DoNotOptimize(Q = integrator.integrate(f, (Real) 0, half_pi<Real>()));
}
}

template<class Real>
static void BM_tanh_sinh_horrible(benchmark::State& state)
{
auto f = [](Real x) { return x*sin(2*exp(2*sin(2*exp(2*x) ) ) ); };
Real Q;
Real tol = 100*std::numeric_limits<Real>::epsilon();
tanh_sinh<Real> integrator(tol, 20);
while(state.KeepRunning())
{
benchmark::DoNotOptimize(Q = integrator.integrate(f, (Real) -1, (Real) 1));
}
}

BENCHMARK_TEMPLATE(BM_tanh_sinh_sqrt_log, float);
BENCHMARK_TEMPLATE(BM_tanh_sinh_sqrt_log, double);
BENCHMARK_TEMPLATE(BM_tanh_sinh_sqrt_log, long double);
BENCHMARK_TEMPLATE(BM_tanh_sinh_sqrt_log, float128);
BENCHMARK_TEMPLATE(BM_tanh_sinh_sqrt_log, cpp_bin_float_50);
BENCHMARK_TEMPLATE(BM_tanh_sinh_sqrt_log, cpp_bin_float_100);

BENCHMARK_TEMPLATE(BM_tanh_sinh_linear, float);
BENCHMARK_TEMPLATE(BM_tanh_sinh_linear, double);
BENCHMARK_TEMPLATE(BM_tanh_sinh_linear, long double);
BENCHMARK_TEMPLATE(BM_tanh_sinh_linear, float128);
BENCHMARK_TEMPLATE(BM_tanh_sinh_linear, cpp_bin_float_50);
BENCHMARK_TEMPLATE(BM_tanh_sinh_linear, cpp_bin_float_100);

BENCHMARK_TEMPLATE(BM_tanh_sinh_runge, float);
BENCHMARK_TEMPLATE(BM_tanh_sinh_runge, double);
BENCHMARK_TEMPLATE(BM_tanh_sinh_runge, long double);
BENCHMARK_TEMPLATE(BM_tanh_sinh_runge, float128);

BENCHMARK_TEMPLATE(BM_tanh_sinh_runge_move_pole, float);
BENCHMARK_TEMPLATE(BM_tanh_sinh_runge_move_pole, double);
BENCHMARK_TEMPLATE(BM_tanh_sinh_runge_move_pole, long double);
BENCHMARK_TEMPLATE(BM_tanh_sinh_runge_move_pole, float128);

BENCHMARK_TEMPLATE(BM_tanh_sinh_exp_cos, float);
BENCHMARK_TEMPLATE(BM_tanh_sinh_exp_cos, double);
BENCHMARK_TEMPLATE(BM_tanh_sinh_exp_cos, long double);
BENCHMARK_TEMPLATE(BM_tanh_sinh_exp_cos, float128);
BENCHMARK_TEMPLATE(BM_tanh_sinh_exp_cos, cpp_bin_float_50);
BENCHMARK_TEMPLATE(BM_tanh_sinh_exp_cos, cpp_bin_float_100);

BENCHMARK_TEMPLATE(BM_tanh_sinh_horrible, float);
BENCHMARK_TEMPLATE(BM_tanh_sinh_horrible, double);
BENCHMARK_TEMPLATE(BM_tanh_sinh_horrible, long double);
BENCHMARK_TEMPLATE(BM_tanh_sinh_horrible, float128);
BENCHMARK_TEMPLATE(BM_tanh_sinh_horrible, cpp_bin_float_50);
BENCHMARK_TEMPLATE(BM_tanh_sinh_horrible, cpp_bin_float_100);

BENCHMARK_MAIN();
/*
---------------------------------------------------------------------------------
Benchmark                                          Time           CPU Iterations
---------------------------------------------------------------------------------
BM_tanh_sinh_sqrt_log<float>                     550 ns        550 ns    1214044
BM_tanh_sinh_sqrt_log<double>                   9004 ns       9003 ns      77327
BM_tanh_sinh_sqrt_log<long double>              3635 ns       3635 ns     192432
BM_tanh_sinh_sqrt_log<float128>               342661 ns     342653 ns       2043
BM_tanh_sinh_sqrt_log<cpp_bin_float_50>      5940813 ns    5940664 ns        117
BM_tanh_sinh_sqrt_log<cpp_bin_float_100>    41784341 ns   41783310 ns         17
BM_tanh_sinh_linear<float>                        33 ns         33 ns   20972925
BM_tanh_sinh_linear<double>                      150 ns        150 ns    4655756
BM_tanh_sinh_linear<long double>                 347 ns        347 ns    2019473
BM_tanh_sinh_linear<float128>                  41586 ns      41585 ns      16807
BM_tanh_sinh_linear<cpp_bin_float_50>         147107 ns     147104 ns       4753
BM_tanh_sinh_linear<cpp_bin_float_100>        482590 ns     482581 ns       1452
BM_tanh_sinh_quadratic<float>                     79 ns         79 ns    8846856
BM_tanh_sinh_quadratic<double>                   183 ns        183 ns    3828752
BM_tanh_sinh_quadratic<long double>              424 ns        424 ns    1651417
BM_tanh_sinh_quadratic<float128>               58832 ns      58831 ns      11778
BM_tanh_sinh_runge<float>                        340 ns        340 ns    2061682
BM_tanh_sinh_runge<double>                     17312 ns      17311 ns      40403
BM_tanh_sinh_runge<long double>                36697 ns      36696 ns      19071
BM_tanh_sinh_runge<float128>                 2984174 ns    2984116 ns        236
BM_tanh_sinh_runge_move_pole<float>              158 ns        158 ns    4431412
BM_tanh_sinh_runge_move_pole<double>             777 ns        777 ns     896286
BM_tanh_sinh_runge_move_pole<long double>       1095 ns       1095 ns     636425
BM_tanh_sinh_runge_move_pole<float128>         80297 ns      80295 ns       8678
BM_tanh_sinh_exp_cos<float>                     5685 ns       5685 ns     121337
BM_tanh_sinh_exp_cos<double>                   55022 ns      55021 ns      12558
BM_tanh_sinh_exp_cos<long double>              71875 ns      71874 ns       9663
BM_tanh_sinh_exp_cos<float128>                379522 ns     379514 ns       1848
BM_tanh_sinh_exp_cos<cpp_bin_float_50>       4538073 ns    4537984 ns        156
BM_tanh_sinh_exp_cos<cpp_bin_float_100>     33965946 ns   33965260 ns         21
BM_tanh_sinh_horrible<float>                  427490 ns     427478 ns       1633
BM_tanh_sinh_horrible<double>                 572976 ns     572966 ns       1214
BM_tanh_sinh_horrible<long double>           1346058 ns    1346033 ns        516
BM_tanh_sinh_horrible<float128>             22030156 ns   22029403 ns         32
BM_tanh_sinh_horrible<cpp_bin_float_50>    635390418 ns  635373654 ns          1
BM_tanh_sinh_horrible<cpp_bin_float_100>  4867960245 ns 4867844329 ns          1
*/


These numbers were created using an Intel Xeon E3-1230, and the benchmarks can be compiled with

CXX = g++
INC=-I/path/to/boost

all: perf.x

perf.x: perf.o
$(CXX) -o$@ $< -lbenchmark -pthread -lquadmath perf.o: perf.cpp$(CXX) --std=c++14 -fext-numeric-literals -Wfatal-errors -g -O3 $(INC) -c$< -o \$@

clean:
rm -f *.x *.o


Get this bounty!!!

## #StackBounty: #python #algorithm #python-2.7 #sudoku Sudoku solver recursive solution

### Bounty: 50

Here is my code in Python 2.7 for a Sudoku resolver. Any advice on performance improvement, code bugs or general code style advice is appreciated.

My major idea is:

1. Using method generate some random data, then using check_invalid to see if it is a valid Sudoku
2. If from 1, it is a valid Sudoku, then I will call resolve_sudoku to fill valid values

I find my code sometimes run a long time without completion. Are there any code bugs you can find?

import random
from collections import defaultdict
found = False
def check_invalid(matrix):
# check for each row
for row in range(len(matrix)):
cur_row = set()
for col in range(len(matrix[0])):
if matrix[row][col] == 0:
continue
elif 1 <= matrix[row][col] <= 9:
if matrix[row][col] in cur_row:
return False
else:
else:
return False # invalid number
# check each col
for col in range(len(matrix[0])):
cur_col = set()
for row in range(len(matrix)):
if matrix[row][col] == 0:
continue
elif 1 <= matrix[row][col] <= 9:
if matrix[row][col] in cur_col:
return False
else:
else:
return False # invalid number
# check each 3*3 square
for start_row in [0,3,6]:
for start_col in [0,3,6]:
cur_square = set()
for row in range(start_row, start_row+3):
for col in range(start_col, start_col + 3):
if matrix[row][col] == 0:
continue
elif 1 <= matrix[row][col] <= 9:
if matrix[row][col] not in cur_square:
else:
return False
else:
return False # invalid value
return True

def resolve_sudoku(matrix, row_map, col_map, square_map, cur_row, cur_col):
global found
if found:
return
if cur_row == len(matrix):
found = True
for r in matrix:
print r
elif cur_col == len(matrix[0]):
resolve_sudoku(matrix, row_map, col_map, square_map, cur_row+1, 0)
elif matrix[cur_row][cur_col] != 0:
resolve_sudoku(matrix, row_map, col_map, square_map, cur_row, cur_col+1)
else:
for val in range(1,10):
square_x = cur_row / 3
square_y = cur_col / 3
if val in row_map[cur_row] or val in col_map[cur_col] or val in square_map[(square_x, square_y)]:
continue
else:
matrix[cur_row][cur_col] = val
resolve_sudoku(matrix, row_map, col_map, square_map, cur_row, cur_col+1)
row_map[cur_row].remove(val)
col_map[cur_col].remove(val)
square_map[(square_x, square_y)].remove(val)
matrix[cur_row][cur_col] = 0
if found:
return
if __name__ == "__main__":
matrix = []
for row in range(9):
cur_row = []
for col in range(9):
if random.random() < 0.1:
cur_row.append(random.randint(1,9))
else:
cur_row.append(0)
matrix.append(cur_row)
for r in matrix:
print r
re = check_invalid(matrix)
print re
if re:
# init for row map and col map
row_map = defaultdict(set)
col_map = defaultdict(set)
square_map = defaultdict(set)
for row in range(len(matrix)):
for col in range(len(matrix[0])):
square_x = row / 3
square_y = row / 3
if matrix[row][col] != 0:
resolve_sudoku(matrix, row_map, col_map, square_map, 0, 0)


Get this bounty!!!

## K random combinations of N elements in List in Java

Given a List of N Strings, generate and print all possible combinations of R elements in array and return X random combinations from the result. Following is the code for implementing it:

## HackerRank: Repeated String

### Problem

Lilah has a string, s, of lowercase English letters that she repeated infinitely many times.

Given an integer, n, find and print the number of letter a‘s in the first letters of Lilah’s infinite string.

### Input Format

The first line contains a single string, s.
The second line contains an integer, n.

### Constraints

• 1<=|s|<=100
• 1<=|n|<=10^12
• For 25% of the test cases, n <= 10^6

### Output Format

Print a single integer denoting the number of letter a’s in the first letters of the infinite string created by repeating infinitely many times.

aba
10

7

### Explanation 0

The first n = 10 letters of the infinite string are abaabaabaa. Because there are 7 a‘s, we print on a new line.

a
1000000000000

1000000000000

### Explanation 1

Because all of the first n=1000000000000 letters of the infinite string are a, we print 1000000000000 on a new line.

## Apache Commons DbUtils Mini Wrapper

This is a very small DB Connector code in Java as a wrapper class to Apache DBUtils.

The Commons DbUtils library is a small set of classes designed to make working with JDBC easier. JDBC resource cleanup code is mundane, error prone work so these classes abstract out all of the cleanup tasks from your code leaving you with what you really wanted to do with JDBC in the first place: query and update data.

Some of the advantages of using DbUtils are:

• No possibility for resource leaks. Correct JDBC coding isn’t difficult but it is time-consuming and tedious. This often leads to connection leaks that may be difficult to track down.
• Cleaner, clearer persistence code. The amount of code needed to persist data in a database is drastically reduced. The remaining code clearly expresses your intention without being cluttered with resource cleanup.
• Automatically populate Java Bean properties from Result Sets. You don’t need to manually copy column values into bean instances by calling setter methods. Each row of the Result Set can be represented by one fully populated bean instance.

DbUtils is designed to be:

• Small – you should be able to understand the whole package in a short amount of time.
• Transparent – DbUtils doesn’t do any magic behind the scenes. You give it a query, it executes it and cleans up for you.
• Fast – You don’t need to create a million temporary objects to work with DbUtils.

DbUtils is not:

• An Object/Relational bridge – there are plenty of good O/R tools already. DbUtils is for developers looking to use JDBC without all the mundane pieces.
• A Data Access Object (DAO) framework – DbUtils can be used to build a DAO framework though.
• An object oriented abstraction of general database objects like a Table, Column, or Primary Key.
• A heavyweight framework of any kind – the goal here is to be a straightforward and easy to use JDBC helper library.

## HackerRank: Circular Array Rotation

### Problem

John Watson performs an operation called a right circular rotation on an array of integers, [a(0),a(1).a(2)...a(n-2),a(n-1)]. After performing one right circular rotation operation, the array is transformed from

[a(0),a(1).a(2)...a(n-2),a(n-1)]

to

[a(n-1),a(0),a(1).a(2)...a(n-2)].

Watson performs this operation k times. To test Sherlock’s ability to identify the current element at a particular position in the rotated array, Watson asks q queries, where each query consists of a single integer, m, for which you must print the element at index in the rotated array (i.e., the value of a(m)).

#### Input Format

The first line contains space-separated integers, n, k, and q, respectively.
The second line contains space-separated integers, where each integer i describes array element a(i)(where 0 <= i <= n).
Each of the q subsequent lines contains a single integer denoting m.

#### Constraints

• 0 <= i <= 10^5
• 0 <= a(i) <= 10^5
• 0 <= k <= 10^5
• 0 <= q <= 500
• 0 <= m <= N-1

#### Output Format

For each query, print the value of the element at index m of the rotated array on a new line.

##### Sample Input
3 2 3
1 2 3
0
1
2

##### Sample Output
2
3
1


#### Explanation

After the first rotation, the array becomes [3,1,2].
After the second (and final) rotation, the array becomes [2,3,1].

Let’s refer to the array’s final state as array b. For each query, we just have to print the value of b(m) on a new line:

• m=0 , so we print 2 on a new line.
• m=1 , so we print 3 on a new line.
• m=2 , so we print 1 on a new line.

## Problem:

Sample Game

Draughts is a two player board game which is played on a 8X8 grid of cells and is played on opposite sides of the game-board. Each player has an allocated color, Red ( First Player ) or White ( Second Player ) being conventional. Players take turns involving diagonal moves of uniform game pieces in the forward direction only and mandatory captures by jumping over opponent pieces.

Rules:

• Player can only move diagonally to the adjacent cell and in forward direction, if the diagonally adjacent cell is vacant.
• A player may not move an opponent’s piece.
• If the diagonally adjacent cell contains an opponent’s piece, and the cell immediately beyond it is vacant, the opponent’s piece may be captured (and removed from the game) by jumping over it in the forward direction only.
• If a player made a jump, then its mandatory to keep on jumping as long as the jump is possible.
• Player cannot move to the diagonally adjacent cell once the player made a jump.

The game will end when any of the players don’t have any move left. At the end of the game the player with majority of pieces will win.

We will play it on an 8X8 grid. The top left of the grid is [0,0] and the bottom right is [7,7].

Input:
The input will be a 8X8 matrix consisting only of 0o2. Then another line will follow which will contain a number –  1 or 2 which is your player id. In the given matrix, top-left is [0,0] and bottom-right is [7,7]. The x-coordinate increases from left to right, and y-coordinate increases from top to bottom.

The cell marked 0 means it doesn’t contain any stones. The cell marked 1 means it contains first player’s stone which is Red in color. The cell marked 2 means it contains second player’s stone which is white in color.

Output:
In the first line print the coordinates of the cell separated by space, the piece he / she wants to move.
In second line print an integer N, number of steps or jumps the piece will make in one move.
In the next N lines print the coordinates of the cells in which the piece will make jump.
You must take care that you don’t print invalid coordinates. For example, [1,1] might be a valid coordinate in the game play if [1,1] in diagonal to the piece in which is going to jump, but [9,10] will never be. Also if you play an invalid move or your code exceeds the time/memory limit while determining the move, you lose the game.

Starting state
The starting state of the game is the state of the board before the game starts.

0 1 0 1 0 1 0 1
1 0 1 0 1 0 1 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 2 0 2 0 2 0 2
2 0 2 0 2 0 2 0


First Input
This is the input give to the first player at the start of the game.

0 1 0 1 0 1 0 1
1 0 1 0 1 0 1 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 2 0 2 0 2 0 2
2 0 2 0 2 0 2 0
1

SAMPLE INPUT
0 1 0 1 0 1 0 1
1 0 1 0 1 0 0 0
0 0 0 0 0 1 0 0
0 0 0 0 2 0 0 0
0 0 0 0 0 0 0 0
0 0 2 0 0 0 0 0
0 0 0 2 0 0 0 2
2 0 2 0 2 0 2 0
1
SAMPLE OUTPUT
2 5
2
4 3
6 1

Explanation

This is player 1’s turn, and the player will move the piece at [2,5] and will make two jumps. First jump will be at [4,3and second jump will be at [6,1]

After his/her move the state of game becomes:

0 1 0 1 0 1 0 1
1 0 1 0 1 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 1 0 2 0 2 0 2
2 0 2 0 2 0 2 0


This state will be fed as input to program of player 2.

Other valid move for the first player is

2 5
1
3 6


But the following are invalid moves.
Case 1:

2 5
1
4 3


Because after making a jump its possible to jump again and its mandatory to jump as long as its possible to jump.

Case 2:

2 5
2
4 3
5 4


Because after making a jump its invalid to move to diagonally adjacent cell.

Here is the code of the Random Bot.

Time Limit:1.0 sec(s) for each input file.
Memory Limit:256 MB
Source Limit:1024 KB

## Solution

This is the solution submitted by me