Mathematical Expression Showdown!

Question

You are given 6 numbers: 5 digits [0-9] and a target number. Your goal is to intersperse operators between the digits to get as close as you can to the target. You have to use each digit exactly once, and can use the following operators as many times as you want: + - * / () ^ sqrt sin cos tan. For example, if I'm given 8 2 4 7 2 65 I can output 82-(2*7)-4. This evaluates to 64, thus giving me a score of 1 since I was 1 away from the target. Note: You can not put a decimal point between digits.

I am using the code from this StackOverflow answer to evaluate the mathematical expressions. At the bottom of this question there are programs you can use to test it out.

Chaining Functions (Update!)

@mdahmoune has revealed a new level of complexity to this challenge. As such, I'm adding a new feature: chaining unary functions. This works on sin, cos, tan, and sqrt. Now instead of writing sin(sin(sin(sin(10)))), you can write sin_4(10). Try it out in the evaluator!

Input

200 line-separated test cases of 5 digits and a target number that are space-separated. You can use the program at the bottom of the question to make sample test cases, but I will have my own test cases for official scoring. The test cases are broken up into 5 sections of 40 tests with the following ranges for the target number:

• Section 1: [0,1] (to 5 decimal points)
• Section 2: [0,10] (to 4 decimal points)
• Section 3: [0,1000] (to 3 decimal points)
• Section 4: [0,10⁶] (to 1 decimal point)
• Section 5: [0,10⁹] (to 0 decimal points)

Output

200 line separated mathematical expressions. For example, if the test case is 5 6 7 8 9 25.807, a possible output could be 78-59+6

Scoring

The goal each round is to get closer to the target number than the other competing programs. I'm going to use Mario Kart 8 scoring, which is: 1st: 15 2nd: 12 3rd: 10 4th: 9 5th: 8 6th: 7 7th: 6 8th: 5 9th: 4 10th: 3 11th: 2 12th: 1 13th+: 0. If multiple answers get the same exact score, the points are split evenly, rounded to the nearest int. For example, if the programs in 5th-8th place are tied, they each get (8+7+6+5)/4 = 6.5 => 7 points that round. At the end of 200 rounds, the program that got the most points wins. If two programs have the same number of points at the end, the tie-breaker is the program that finished running faster.

Rules

1. You can only use one of the languages commonly pre-installed on Mac like C, C++, Java, PhP, Perl, Python (2 or 3), Ruby, and Swift. If you have a language you want to use with a compiler/interpreter that is a relatively small download I may add it. You can also use a language with an online interpreter, but that will not run as fast.
2. Specify in your answer if you want trig functions to be calculated in degrees or radians.
3. Your program must output its solutions to all 200 test cases (to a file or STDOUT) within 60 seconds on my Mac.
4. Randomness must be seeded.
5. Your total output for all test cases can't be more than 1 MB.
6. If you have improved your solution and would like to be re-scored, add Re-Score at the top of your answer in bold.

Programs

(change the "deg" argument to "rad" if you want radians)

1. Test out evaluator
2. Score your program's output for test cases
3. Generate test cases:

document.getElementById("but").onclick = gen;
var checks = document.getElementById("checks");
for(var i = 1;i<=6;i++) {
var val = i<6 ? i : "All";
var l = document.createElement("label");
l.for = "check" + val;
l.innerText = " "+val+" ";
checks.appendChild(l);
  var check = document.createElement("input");
  check.type = "checkBox";
  check.id = "check"+val;
  if(val == "All") {
  check.onchange = function() {
  if(this.checked == true)  {
  for(var i = 0;i<5;i++) {
    this.parentNode.elements[i].checked = true;
  }
  }
};  
  }
  else {
  check.onchange = function() {
    document.getElementById("checkAll").checked = false;
  }
  }
  checks.appendChild(check);
  
}



function gen() {
var tests = [];
var boxes = checks.elements;
if(boxes[0].checked)genTests(tests,1,5,40);
if(boxes[1].checked)genTests(tests,10,4,40);
if(boxes[2].checked)genTests(tests,1000,3,40);
if(boxes[3].checked)genTests(tests,1e6,1,40);
if(boxes[4].checked)genTests(tests,1e9,0,40);
document.getElementById("box").value =  tests.join("\n");
}

function genTests(testArray,tMax,tDec,n) {
for(var i = 0;i<n;i++) {
  testArray.push(genNums(tMax,tDec).join(" "));
}
}

function genNums(tMax,tDec) {
var nums = genDigits();
nums.push(genTarget(tMax,tDec));
return nums;
}

function genTarget(tMax,tDec) {
  return genRand(tMax,tDec);
}

function genRand(limit,decimals) {
  var r = Math.random()*limit;
  return r.toFixed(decimals);
}

function genDigits() {
  var digits = [];
   for(var i = 0;i<5;i++) {
    digits.push(Math.floor(Math.random()*10));
   }
   return digits;
}

textarea {
  font-size: 14pt;
  font-family: "Courier New", "Lucida Console", monospace;
}

div {
text-align: center;
}

<div>
<label for="checks">Sections: </label><form id="checks"></form>
<input type="button" id="but" value="Generate Test Cases" /><br/><textarea id="box" cols=20 rows=15></textarea>
</div>

Leaderboard

1. user202729 (C++): 2856, 152 wins
2. mdahmoune (Python 2) [v2]: 2544, 48 wins

Section scores (# of wins):

1. [0-1] user202729: 40, mdahmoune: 0
2. [0-10] user202729: 40, mdahmoune: 0
3. [0-1000] user202729: 39, mdahmoune: 1
4. [0-10⁶] user202729: 33, mdahmoune: 7
5. [0-10⁹] user202729:0, mdahmoune: 40

Related: Generate a valid equation using user-specified numbers

Answer 1

C++

// This program use radian mode

//#define DEBUG

#ifdef DEBUG
#define _GLIBCXX_DEBUG
#include <cassert>
#else
#define assert(x) void(0)
#endif

namespace std {
    /// Used for un-debug.
    struct not_cerr_t {
    } not_cerr;
}

template <typename T>
std::not_cerr_t& operator<<(std::not_cerr_t& not_cerr, T) {return not_cerr;}

#include <iostream>
#include <iomanip>
#include <cmath>
#include <limits>
#include <array>
#include <bitset>
#include <string>
#include <sstream>

#ifndef DEBUG
#define cerr not_cerr
#endif // DEBUG


// String conversion functions, because of some issues with MinGW
template <typename T>
T from_string(std::string st) {
    std::stringstream sst (st);
    T result;
    sst >> result;
    return result;
}

template <typename T>
std::string to_string(T x) {
    std::stringstream sst;
    sst << x;
    return sst.str();
}

template <typename T> int sgn(T val) {
    return (T(0) < val) - (val < T(0));
}


const int N_ITER = 1000, N_DIGIT = 5, NSOL = 4;
std::array<int, N_DIGIT> digits;
double target;

typedef std::bitset<N_ITER> stfunc; // sin-tan expression
// where sin = 0, tan = 1

double eval(const stfunc& fn, int length, double value) {
    while (length --> 0) {
        value = fn[length] ? std::tan(value) : std::sin(value);
    }
    return value;
}

struct stexpr { // just a stfunc with some information
    double x = 0, val = 0; // fn<length>(x) == val
    int length = 0;
    stfunc fn {};
//    bool operator[] (const int x) {return fn[x];}
    double eval() {return val = ::eval(fn, length, x);}
};

struct expr { // general form of stexpr
    // note that expr must be *always* atomic.
    double val = 0;
    std::string expr {};

    void clear() {
        val = 0;
        expr.clear();
    }

    // cos(cos(x)) is in approx 0.5 - 1,
    // so we can expect that sin(x) and tan(x) behaves reasonably nice
    private: void wrapcos2() {
        expr = "(cos_2 " + expr + ")"; // we assume that all expr is atomic
        val = std::cos(std::cos(val));
    }

    public: void wrap1() {
        if (val == 0) {
            expr = "(cos " + expr + ")"; // we assume that all expr is atomic
            val = std::cos(val);
        }
        if (val == 1) return;
        wrapcos2(); // range 0.54 - 1
        int cnt_sqrt = 0;
        for (int i = 0; i < 100; ++i) {
            ++cnt_sqrt;
            val = std::sqrt(val);
            if (val == 1) break;
        }
        expr = "(sqrt_" + to_string(cnt_sqrt) + " " + expr + ")"; // expr must be atomic
    }
};

stexpr nearest(double initial, double target) {
    stexpr result; // built on the fn of that
    result.x = initial;
    double value [N_ITER + 1];
    value[0] = initial;
    for (result.length = 1; result.length <= N_ITER; ++result.length) {
        double x = value[result.length-1];
        if (x < target) {
            result.fn[result.length-1] = 1;
        } else if (x > target) {
            result.fn[result.length-1] = 0;
        } else { // unlikely
            --result.length;
//            result.val = x;
            result.eval();
            assert(result.val == x);
            return result;
        }
        value[result.length] = result.eval(); // this line takes most of the time
        if (value[result.length] == value[result.length-1])
            break;
    }

//    for (int i = 0; i < N_ITER; ++i) {
//        std::cerr << i << '\t' << value[i] << '\t' << (value[i] - target) << '\n';
//    }

    double mindiff = std::numeric_limits<double>::max();
    int resultlength = -1;
    result.length = std::min(N_ITER, result.length);
    for (int l = 0; l <= result.length; ++l) {
        if (std::abs(value[l] - target) < mindiff) {
            mindiff = std::abs(value[l] - target);
            resultlength = l;
        }
    }

    result.length = resultlength;
    double val = value[resultlength];
    assert(std::abs(val - target) == mindiff);
    if (val != target) { // second-order optimization
        for (int i = 1; i < result.length; ++i) {
            // consider pair (i-1, i)
            if (result.fn[i-1] == result.fn[i]) continue; // look for (sin tan) or (tan sin)
            if (val < target && result.fn[i-1] == 0) { // we need to increase val : sin tan -> tan sin
                result.fn[i-1] = 1;
                result.fn[i] = 0;
                double newvalue = result.eval();
//                if (!(newvalue >= val)) std::cerr << "Floating point sin-tan error 1\n";
                if (std::abs(newvalue - target) < std::abs(val - target)) {
//                    std::cerr << "diff improved from " << std::abs(val - target) << " to " << std::abs(newvalue - target) << '\n';
                    val = newvalue;
                } else {
                    result.fn[i-1] = 0;
                    result.fn[i] = 1; // restore
                    #ifdef DEBUG
                    result.eval();
                    assert(val == result.val);
                    #endif // DEBUG
                }
            } else if (val > target && result.fn[i-1] == 1) {
                result.fn[i-1] = 0;
                result.fn[i] = 1;
                double newvalue = result.eval();
//                if (!(newvalue <= val)) std::cerr << "Floating point sin-tan error 2\n";
                if (std::abs(newvalue - target) < std::abs(val - target)) {
//                    std::cerr << "diff improved from " << std::abs(val - target) << " to " << std::abs(newvalue - target) << '\n';
                    val = newvalue;
                } else {
                    result.fn[i-1] = 1;
                    result.fn[i] = 0; // restore
                    #ifdef DEBUG
                    result.eval();
                    assert(val == result.val);
                    #endif // DEBUG
                }
            }
        }
    }
    double newdiff = std::abs(val - target);
    if (newdiff < mindiff) {
        mindiff = std::abs(val - target);
        std::cerr << "ok\n";
    } else if (newdiff > mindiff) {
        std::cerr << "Program error : error value = " << (newdiff - mindiff) << " (should be <= 0 if correct) \n";
        std::cerr << "mindiff = " << mindiff << ", newdiff = " << newdiff << '\n';
    }
    result.eval(); // set result.result
    assert(val == result.val);

    return result;
}

expr nearest(const expr& in, double target) {
    stexpr tmp = nearest(in.val, target);
    expr result;
    for (int i = 0; i < tmp.length; ++i)
        result.expr.append(tmp.fn[i] ? "tan " : "sin ");

    result.expr = "(" + result.expr + in.expr + ")";
    result.val = tmp.val;
    return result;
}

int main() {
    double totalscore = 0;

    assert (std::numeric_limits<double>::is_iec559);
    std::cerr << std::setprecision(23);

//    double initial = 0.61575952241185627;
//    target = 0.6157595200093855;
//    stexpr a = nearest(initial, target);
//    std::cerr << a.val << ' ' << a.length << '\n';
//    return 0;

    while (std::cin >> digits[0]) {
        for (unsigned i = 1; i < digits.size(); ++i) std::cin >> digits[i];
        std::cin >> target;

/*        std::string e;
//        int sum = 0;
//        for (int i : digits) {
//            sum += i;
//            e.append(to_string(i)).push_back('+');
//        }
//        e.pop_back(); // remove the last '+'
//        e = "cos cos (" + e + ")";
//        double val = std::cos(std::cos((double)sum));
//
//        stexpr result = nearest(val, target); // cos(cos(x)) is in approx 0.5 - 1,
//        // so we can expect that sin(x) and tan(x) behaves reasonably nice
//        std::string fns;
//        for (int i = 0; i < result.length; ++i) fns.append(result.fn[i] ? "tan" : "sin").push_back(' ');
//
//        std::cout << (fns + e) << '\n';
//        continue;*/

        std::array<expr, NSOL> sols;
        expr a, b, c, d; // temporary for solutions

        /* ----------------------------------------
           solution 1 : nearest cos cos sum(digits) */

        a.clear();
        for (int i : digits) {
            a.val += i; // no floating-point error here
            a.expr.append(to_string(i)).push_back('+');
        }
        a.expr.pop_back(); // remove the last '+'
        a.expr = "(" + a.expr + ")";
        a.wrap1();

        sols[0] = nearest(a, target);


        /* -----------------------------------------
              solution 2 : a * tan(b) + c (also important) */

        // find b first, then a, then finally c
        a.clear(); b.clear(); c.clear(); // e = a, b = e1, c = e2

        a.expr = to_string(digits[0]);
        a.val = digits[0];
        a.wrap1();

        b.expr = "(" + to_string(digits[1]) + "+" + to_string(digits[2]) + ")";
        b.val = digits[1] + digits[2];
        b.wrap1();

        c.expr = to_string(digits[3]);
        c.val = digits[3];
        c.wrap1();

        d.expr = to_string(digits[4]);
        d.val = digits[4];
        d.wrap1();

        b = nearest(b, std::atan(target));

        double targetA = target / std::tan(b.val);
        int cnt = 0;
        while (targetA < 1 && targetA > 0.9) {
            ++cnt;
            targetA = targetA * targetA;
        }
        a = nearest(a, targetA);
        while (cnt --> 0) {
            a.val = std::sqrt(a.val);
            a.expr = "sqrt " + a.expr;
        }
        a.expr = "(" + a.expr + ")"; // handle number of the form 0.9999999999

        /// partition of any number to easy-to-calculate sum of 2 numbers
        {{{{{{{{{{{{{{{{{{{{{{{{{{{{}}}}}}}}}}}}}}}}}}}}}}}}}}}}

        double targetC, targetD; // near 1, not in [0.9, 1), >= 0.1
        // that is, [0.1, 0.9), [1, inf)

        double target1 = target - (a.val * std::tan(b.val));

        double ac = std::abs(target1), sc = sgn(target1);
        if (ac < .1) targetC = 1 + ac, targetD = -1;
        else if (ac < 1) targetC = 1 + ac/2, targetD = ac/2 - 1;
        else if (ac < 1.8 || ac > 2) targetC = targetD = ac/2;
        else targetC = .8, targetD = ac - .8;

        targetC *= sc; targetD *= sc;

        c = nearest(c, std::abs(targetC)); if (targetC < 0) c.val = -c.val, c.expr = "(-" + c.expr + ")";
        d = nearest(d, std::abs(targetD)); if (targetD < 0) d.val = -d.val, d.expr = "(-" + d.expr + ")";

        sols[1].expr = a.expr + "*tan " + b.expr + "+" + c.expr + "+" + d.expr;
        sols[1].val = a.val * std::tan(b.val) + c.val + d.val;

        std::cerr
        << "\n---Method 2---"
        << "\na = " << a.val
        << "\ntarget a = " << targetA
        << "\nb = " << b.val
        << "\ntan b = " << std::tan(b.val)
        << "\nc = " << c.val
        << "\ntarget c = " << targetC
        << "\nd = " << d.val
        << "\ntarget d = " << targetD
        << "\n";

        /* -----------------------------------------
              solution 3 : (b + c) */

        target1 = target / 2;
        b.clear(); c.clear();

        for (int i = 0; i < N_DIGIT; ++i) {
            expr &ex = (i < 2 ? b : c);
            ex.val += digits[i];
            ex.expr.append(to_string(digits[i])).push_back('+');
        }
        b.expr.pop_back();
        b.expr = "(" + b.expr + ")";
        b.wrap1();

        c.expr.pop_back();
        c.expr = "(" + c.expr + ")";
        c.wrap1();

        b = nearest(b, target1);
        c = nearest(c, target - target1); // approx. target / 2

        sols[2].expr = "(" + b.expr + "+" + c.expr + ")";
        sols[2].val = b.val + c.val;

        /* -----------------------------------------
              solution 4 : a (*|/) (b - c)  (important) */

        a.clear(); b.clear(); c.clear(); // a = a, b = e1, c = e2

        a.expr = to_string(digits[0]);
        a.val = digits[0];
        a.wrap1();

        b.expr = "(" + to_string(digits[1]) + "+" + to_string(digits[2]) + ")";
        b.val = digits[1] + digits[2];
        b.wrap1();

        c.expr = "(" + to_string(digits[3]) + "+" + to_string(digits[4]) + ")";
        c.val = digits[3] + digits[4];
        c.wrap1();


        // (b-c) should be minimized
        bool multiply = target < a.val;
        double factor = multiply ? target / a.val : a.val / target;

        target1 = 1 + 2 * factor; // 1 + 2 * factor and 1 + factor

        std::cerr << "* Method 4 :\n";
        std::cerr << "b initial = " << b.val << ", target = " << target1 << ", ";
        b = nearest(b, target1);
        std::cerr << " get " << b.val << '\n';

        std::cerr << "c initial = " << c.val << ", target = " << b.val - factor << ", ";
        c = nearest(c, b.val - factor); // factor ~= e1.val - e2.val
        std::cerr << " get " << c.val << '\n';

        sols[3].expr = "(" + a.expr + (multiply ? "*(" : "/(") +
        ( b.expr + "-" + c.expr )
        + "))";
        factor = b.val - c.val;
        sols[3].val = multiply ? a.val * factor : a.val / factor;

        std::cerr << "a.val = " << a.val << '\n';

        /* ----------------------------------
                    Final result */

        int minindex = 0;
        assert(NSOL != 0);
        for (int i = 0; i < NSOL; ++i) {
            if (std::abs(target - sols[i].val) < std::abs(target - sols[minindex].val)) minindex = i;
            std::cerr << "Sol " << i << ", diff = " << std::abs(target - sols[i].val) << "\n";
        }
        std::cerr << "Choose " << minindex << "; target = " << target << '\n';
        totalscore += std::abs(target - sols[minindex].val);

        std::cout << sols[minindex].expr << '\n';
    }

    // #undef cerr // in case no-debug
    std::cerr << "total score = " << totalscore << '\n';
}

Input from standard input, output to standard output.

Answer 2

Python 2, radians, score 0.0032 on official test

This is the second draft solution gives an average score of 0.0032 points. As it uses a composition of a lot of sin I used the following compact notation for the output formula:

• sin_1 x=sin(x)
• sin_2 x=sin(sin(x))
• ...
• sin_7 x=sin(sin(sin(sin(sin(sin(sin(x)))))))
• ...

import math
import bisect
s1=[[float(t) for t in e.split()] for e in s0.split('\n')]
maxi=int(1e7)
A=[]
B=[]
C=[]
D=[]
a=1
for i in range(maxi):
	A.append(a)
	C.append(1/a)
	b=math.sin(a)
	c=a-b
	B.append(1/c)
	D.append(c)
	a=b
B.sort() 
C.sort() 
A.sort() 
D.sort() 
d15={0:'sqrt_100 tan_4 cos_2 sin 0',1:'sqrt_100 tan_4 cos_2 sin 1',2:'sqrt_100 tan_2 cos_2 sin 2',3:'sqrt_100 tan_4 cos_2 sin 3',4:'sqrt_100 tan_4 cos_2 sin 4',5:'sqrt_100 tan_4 cos_2 sin 5',6:'sqrt_100 tan_4 cos_2 sin 6',7:'sqrt_100 tan_2 cos_2 sin 7',8:'sqrt_100 tan_2 cos_2 sin 8',9:'sqrt_100 tan_4 cos_2 sin 9'}
def d16(d):return '('+d15[d]+')'

def S0(l):
	cpt=0
	d=l[:-1]
	r=l[-1]
	a1,a2,a3,a4,a5=[int(t) for t in d]
	i1=bisect.bisect(B, r)-1
	w1=abs(r-B[i1])
	i2=bisect.bisect(C, w1)-1
	w2=abs(w1-C[i2]) 
	s='('+d16(a1)+'/(sin_'+str(i1)+' '+d16(a2)+'-'+'sin_'+str(i1+1)+' '+d16(a3)+')'+'+'+d16(a4)+'/sin_'+str(i2)+' '+d16(a5)+')'
	return (w2,s)

def S1(l):
	cpt=0
	d=l[:-1]
	r=l[-1]
	a1,a2,a3,a4,a5=[int(t) for t in d]
	i1=bisect.bisect(C, r)-1
	w1=abs(r-C[i1])
	i2=bisect.bisect(A, w1)-1
	w2=abs(w1-A[i2]) 
	s='('+d16(a1)+'/sin_'+str(i1)+' '+d16(a2)+'+sin_'+str(maxi-i2-1)+' ('+d16(a3)+'*'+d16(a4)+'*'+d16(a5)+')'
	return (w2,s)

def S2(l):
	cpt=0
	d=l[:-1]
	r=l[-1]
	a1,a2,a3,a4,a5=[int(t) for t in d]
	i1=bisect.bisect(A, r)-1
	w1=abs(r-A[i1])
	i2=bisect.bisect(D, w1)-1
	w2=abs(w1-D[i2]) 
	s='('+'(sin_'+str(maxi-i2-1)+' '+d16(a1)+'-'+'sin_'+str(maxi-i2)+' '+d16(a2)+')'+'+sin_'+str(maxi-i1-1)+' ('+d16(a3)+'*'+d16(a4)+'*'+d16(a5)+'))'
	return (w2,s)

def S3(l):
	cpt=0
	d=l[:-1]
	r=l[-1]
	a1,a2,a3,a4,a5=[int(t) for t in d]
	i1=bisect.bisect(A, r)-1
	w2=abs(r-A[i1])
	s='('+'sin_'+str(maxi-i1-1)+' ('+d16(a1)+'*'+d16(a2)+'*'+d16(a3)+'*'+d16(a4)+'*'+d16(a5)+'))'
	return (w2,s)

def S4(l):
	cpt=0
	d=l[:-1]
	r=l[-1]
	a1,a2,a3,a4,a5=[int(t) for t in d]
	i1=bisect.bisect(B, r)-1
	w2=abs(r-B[i1])
	s='('+d16(a1)+'/(sin_'+str(i1)+' '+d16(a2)+'-'+'sin_'+str(i1+1)+' '+d16(a3)+'*'+d16(a4)+'*'+d16(a5)+')'+')'
	return (w2,s)

def S5(l):
	cpt=0
	d=l[:-1]
	r=l[-1]
	a1,a2,a3,a4,a5=[int(t) for t in d]
	i1=bisect.bisect(C, r)-1
	w2=abs(r-C[i1])
	s='('+d16(a1)+'/sin_'+str(i1)+' '+d16(a2)+'*'+d16(a3)+'*'+d16(a4)+'*'+d16(a5)+')'
	return (w2,s)

def S6(l):
	cpt=0
	d=l[:-1]
	r=l[-1]
	a1,a2,a3,a4,a5=[int(t) for t in d]
	i1=bisect.bisect(D, r)-1
	w2=abs(r-D[i1])
	s='(sin_'+str(maxi-i1-1)+' '+d16(a1)+'-'+'sin_'+str(maxi-i1)+' '+d16(a2)+'*'+d16(a3)+'*'+d16(a4)+'*'+d16(a5)+')'
	return (w2,s)

def all4(s1):
	s=0
	for l in s1:
		f=min(S0(l),S1(l),S2(l),S3(l),S4(l),S5(l),S6(l))
		print f[1]
		s+=f[0]
	s/=len(s1)
	print 'average unofficial score:',s
all4(s1)

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