Tags: #◇Computer Science #◇CSALGCM #◇Algorithms and Complexity #◇HackerRank #◇Priestess Arcana #◇Bisection Method #◇Binary Search

Let's Plant Rice

◇Problem Description

A game uses a complex evaporation function controlled by daily factors $a$ , $b$ , and $c$ . The water level $W$ in the rice paddy ranges from $0$ (empty) to $1$ (full). Given these factors and a target "sweet spot" evaporation rate $E$ , the task is to determine the lowest starting water level $W$ that hits $E$ within an acceptable error margin ( $10^{-6}$ ). If the sweet spot is unreachable, we must output a specific failure string.

Input Format:
The first line of the input contains an integer $T$ , indicating the number of days you would need to figure out the starting water level for.
$T$ lines follow, each containing four space-separated real numbers that correspond to the value of $a$ , $b$ , $c$ , and $E$ . $E$ is the sweet spot evaporation rate. $a$ , $b$ , and $c$ are described in the problem statement.

Output Format:
There will be $T$ lines of outputs, each line being the lowest starting water level that would get to within the acceptable margins for the sweet spot evaporation rate $E$ . Absolute or relative error must not exceed $10^{-6}$ . Print "Sweet spot cannot be reached! Those cheeky developers!" (without the quotation marks) if the sweet spot is impossible to be reached.

Constraints:

Sample:

Input	Output
`5` `0.7119 -0.1206 2.9930 4.0577` `0.9219 0.1963 2.7140 4.3201` `0.2692 -0.0695 2.9132 1.6116` `0.3601 -0.1771 0.6213 1.1870` `0.3492 -0.3865 3.8889 2.5649`	`0.697784503037` `0.887908531876` `0.921929270250` `0.645118392190` `0.599166042326`
`7` `0.2772 0.2253 4.2550 1.9385` `0.2807 -0.2682 1.5963 1.6372` `0.9055 0.0522 2.5857 2.2600` `0.3196 -0.2520 2.7628 0.9047` `0.6853 -0.2366 3.6572 2.5567` `0.7148 -0.1846 0.9737 4.1721` `0.9411 -0.1370 2.0860 2.6794`	`Sweet spot cannot be reached! Those cheeky developers!` `0.793247141838` `0.387402764833` `0.340837109300` `0.353158012405` `0.817511673620` `Sweet spot cannot be reached! Those cheeky developers!`

◇Approach & Intuition

** Abstract the problem**
A rice paddy evaporates water at a rate that depends on how full it is, following a sigmoid-shaped formula. Given the target evaporation rate s and formula parameters, find the exact starting water level x ∈ [0,1] that produces that rate.

** Dumb sol?**
Try every possible value of x from 0.000000 to 1.000000 in tiny steps (say 0.000001).
O(1,000,000 per query) — slow and imprecise; you'd need millions of steps for 10⁻⁶ precision.

** Smart sol?**
The formula is monotone (always decreasing in x), so use bisection — halve the search interval each iteration. After 200 halvings, precision is 2⁻²⁰⁰ ≈ 10⁻⁶⁰ — absurdly more precise than needed.
O(200) per query — instant.

◇The Actual Formula

$rate(x) = 1 - \left(\frac{1}{1+e^{-k(x-0.5)}}\right)^a + d$

Where x = starting water level ∈ [0.0, 1.0]. Parameters d, k, a are given per day.

◇Understanding the Formula Component by Component

Part 1: The Sigmoid (S-curve)

『』

TEXT

sigma(x) = 1 / (1 + exp(-k * (x - 0.5)))

The sigmoid maps x → a number strictly between 0 and 1:

『』

TIMELINE // GANTT

sigma(x) for k=3, centered at x=0.5:

1.0 |                                           *****
    |                                      *****
    |                                 *****
0.5 |                        **********
    |                   *****
    |              *****
    |         *****
0.0 |*****
    +---+---+---+---+---+---+---+---+---+---+---
      0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
                          ^
                          Always = 0.5 here, regardless of k

Key facts:

At x = 0.5: sigma = 0.5 always (center is fixed)
Larger k → steeper S-curve (more abrupt transition)
As x increases → sigma increases (monotone)

Part 2: sigma raised to power `a`

Since sigma ∈ (0,1), raising to a power reshapes the range:

『』

TEXT

sigma = 0.7:
  a=0.5  -> 0.7^0.5 = 0.836  (pulled toward 1)
  a=1.0  -> 0.7^1.0 = 0.700  (unchanged)
  a=2.0  -> 0.7^2.0 = 0.490  (pulled toward 0)
  a=4.0  -> 0.7^4.0 = 0.240  (much closer to 0)

Part 3: 1 − sigma^a + d

『』

TEXT

rate(x) = 1 - sigma(x)^a + d

As x increases → sigma↑ → sigma^a↑ → 1−sigma^a ↓
So rate(x) is monotone DECREASING in x (for k > 0)
The +d shifts the whole curve up (d>0) or down (d<0)

『』

TIMELINE // GANTT

rate(x) shape (d=-0.1206, k=2.993, a=4.0577):

HIGH  0.877 |* (x=0.0, empty paddy — high evaporation)
            | *
            |  *
            |   *
target 0.711|------*--- <- we want THIS x (answer ≈ 0.698)
            |       *
            |        *
            |         *
LOW   0.271 |          * (x=1.0, full paddy — lower evaporation)
            +----------->
              0         1

◇Why Bisection Works Here

The function is strictly monotone → at most ONE solution exists. We check if the target s is achievable by seeing if rate(0) and rate(1) bracket it:

The Bisection Step by Step (target s=0.7119)

『』

TEXT

Iter 1: lo=0.000, hi=1.000, mid=0.500
  g(lo)=+0.165 (+), g(mid)=+0.038 (+) → same sign → lo=0.500

Iter 2: lo=0.500, hi=1.000, mid=0.750
  g(lo)=+0.038 (+), g(mid)=-0.139 (-) → opposite! → hi=0.750

Iter 3: lo=0.500, hi=0.750, mid=0.625
  g(lo)=+0.038, g(mid)=-0.076 → opposite → hi=0.625

...200 iterations...
  lo = hi = 0.697784503037 ← converged

Each iteration cuts the interval in half → after 200 steps: width = 1/2²⁰⁰ ≈ 10⁻⁶⁰.

◇Code: Every Line Explained

『』

CPP

void solve(double s, double d, double k, double a) {
    // s = target evaporation rate (what we want rate(x) to equal)
    // d = vertical shift of the curve
    // k = sigmoid steepness
    // a = power applied to sigmoid

    // g(x) = rate(x) - s  (we want to find x where g(x) = 0)
    auto g = [&](double x) -> double {
        // Step 1: compute sigmoid centered at 0.5
        double sig = 1.0 / (1.0 + exp(-k * (x - 0.5)));

        // Step 2: full formula minus the target → "residual"
        return 1.0 - pow(sig, a) + d - s;
    };

    double g0 = g(0.0);   // residual at x=0 (empty paddy)
    double g1 = g(1.0);   // residual at x=1 (full paddy)

    // Same sign = target is above both OR below both endpoints
    // = impossible to reach with any x ∈ [0,1]
    if (g0 * g1 > 0) {
        printf("Sweet spot cannot be reached! Those cheeky developers!\n");
        return;
    }

    // Root is guaranteed between lo and hi → bisect
    double lo = 0.0, hi = 1.0;
    for (int iter = 0; iter < 200; iter++) {
        double mid = (lo + hi) / 2.0;

        // If g(lo) and g(mid) have opposite signs: root is in [lo, mid]
        // Otherwise: root is in [mid, hi]
        if (g(lo) * g(mid) <= 0) hi = mid;
        else lo = mid;
    }

    printf("%.12f\n", lo);   // 12 decimal places (required: 10^-6 precision)
}

◇Feasibility Table (Sample Inputs)

s (target)	d	k	a	rate(0)	rate(1)	s in [rate1,rate0]?	Answer
0.7119	-0.1206	2.993	4.058	0.877	0.271		0.697784503037
0.9219	0.1963	2.714	4.320	1.073	0.413		0.887908531876
0.2772	0.2253	4.255	1.939	1.212	0.421	s < rate(1)	Sweet spot!
0.9411	-0.1370	2.086	2.679	0.836	0.418	s > rate(0)	Sweet spot!

◇Complexity Analysis

	Complexity
Time	O(200 × n)
Space	O(1)

◇Original Stub (HackerRank Template)

『』

CPP

#include <bits/stdc++.h>
using namespace std;

double evaporate(double W, double a, double b, double c) {
    // Write your code here
}

int main() {
    int T;
    cin >> T;
    while (T--) {
        double a, b, c, E;
        cin >> a >> b >> c >> E;

        // Write your code here
    }
    return 0;
}

◇Full Solution

『』

CPP

#include <bits/stdc++.h>

#define EPS 1e-8

using namespace std;

void solve(double s, double d, double k, double a){
 	// compute and print answer here
    // Formula: rate(x) = 1 - (1 / (1 + exp(-k*(x-0.5))))^a + d
    // Find lowest x in [0,1] such that rate(x) = s

    auto g = [&](double x) -> double {
        double sig = 1.0 / (1.0 + exp(-k * (x - 0.5)));
        return 1.0 - pow(sig, a) + d - s;
    };

    double g0 = g(0.0), g1 = g(1.0);

    // If both endpoints are on the same side of zero, sweet spot is unreachable
    if (g0 * g1 > 0) {
        printf("Sweet spot cannot be reached! Those cheeky developers!\n");
        return;
    }

    // Binary search for root in [0, 1]
    double lo = 0.0, hi = 1.0;
    for (int iter = 0; iter < 200; iter++) {
        double mid = (lo + hi) / 2.0;
        if (g(lo) * g(mid) <= 0) hi = mid;
        else lo = mid;
    }
    printf("%.12f\n", lo);
}

int main() {
    int n;
    scanf("%d",&n);
    
    vector<vector<double>> vals(n,vector<double>(4));
    for(int i = 0; i < n; i++){
      	double s,d,k,a;
      	scanf("%lf%lf%lf%lf",&s,&d,&k,&a);
      	solve(s,d,k,a);
    }
}

Related Notes: ◇Divide and Conquer Introduction
Previous Problem: ◇02 - Laughter Is The Best Medicine
Next Problem: ◇04 - POWAH UNLIMITED POWAH
Parent Set: ◇Priestess Arcana Index

03 - LETS PLANT RICE

Let's Plant Rice

◇Problem Description

◇Approach & Intuition

◇The Actual Formula

◇Understanding the Formula Component by Component

Part 1: The Sigmoid (S-curve)

Part 2: sigma raised to power `a`

Part 3: 1 − sigma^a + d

◇Why Bisection Works Here

The Bisection Step by Step (target s=0.7119)

◇Code: Every Line Explained

◇Feasibility Table (Sample Inputs)

◇Complexity Analysis

◇Original Stub (HackerRank Template)

◇Full Solution

◇References & Navigation

03 - LETS PLANT RICE

Let's Plant Rice

◇Problem Description

◇Approach & Intuition

◇The Actual Formula

◇Understanding the Formula Component by Component

Part 1: The Sigmoid (S-curve)

Part 2: sigma raised to power a

Part 3: 1 − sigma^a + d

◇Why Bisection Works Here

The Bisection Step by Step (target s=0.7119)

◇Code: Every Line Explained

◇Feasibility Table (Sample Inputs)

◇Complexity Analysis

◇Original Stub (HackerRank Template)

◇Full Solution

◇References & Navigation

Part 2: sigma raised to power `a`