使用 Javascript 接近暴力算法

1. 下面是一些从简单到进阶的例子（旅行商问题和0/1背包问题）
2. 这些示例基于暴力算法

我的笔记：-

1. 这种暴力算法有几个缺点，但在直接进入动态编程和其他方法之前
2. 你应该对这种方法有想法，并且你必须找出为什么我们需要动态编程模式（递归+记忆）

如果你仔细观察暴力破解的模式

const wrapper = (value) => {
    const helper = (combinedArray, depth) => {
       if (depth == 3) {
          // operation
           return ;
       }

       for (let coin of coins) {
           if (value - coin >=0) {
               combinedArray.push(coin);
               helper(combinedArray, label+1);
               combinedArray.pop();
           }
       }
    }

    helper([], 0);
    return result;
};

const res = wrapper(value);
console.log(res);

Q1。从 2 个硬币组合开始

const wrapper = () => {
    const coinSide = ['head', 'tail']
    const result = [];
    const helper = (currentCombination, depth) => {
        if (depth == 2) {
            result.push([...currentCombination]);
            return ;
        }

        for (side of coinSide) {
            currentCombination.push(side);
            helper(currentCombination, depth +1);
            currentCombination.pop()
        }
    }

    helper([], 0);

    return result;
};

const res = wrapper();

console.log(res);

第二季度。从 3 个硬币组合开始

const wrapper = () => {
    const coinSide = ['head', 'tail']
    const result = [];
    const helper = (currentCombination, depth) => {
        if (depth == 3) {
            result.push([...currentCombination]);
            return ;
        }

        for (side of coinSide) {
            currentCombination.push(side);
            helper(currentCombination, depth +1);
            currentCombination.pop()
        }
    }

    helper([], 0);

    return result;
};

const res = wrapper();

console.log(res);

/*
[
  [ 'head', 'head', 'head' ],
  [ 'head', 'head', 'tail' ],
  [ 'head', 'tail', 'head' ],
  [ 'head', 'tail', 'tail' ],
  [ 'tail', 'head', 'head' ],
  [ 'tail', 'head', 'tail' ],
  [ 'tail', 'tail', 'head' ],
  [ 'tail', 'tail', 'tail' ]
]
*/

第三季度。座位安排

const wrapper = () => {
    const result = [];
    const group = ['b1', 'b2', 'g1']
    const helper = (combination, depth) => {
        if (depth == 3) {
            result.push([...combination]);
            return;
        }

        for (let item of group) {
            if (combination.indexOf(item) < 0) {
                combination.push(item);
            helper(combination, depth +1);
            combination.pop();
            }
        }
    }

    helper([], 0);

    return result;
};

/*
[
  [ 'b1', 'b2', 'g1' ],
  [ 'b1', 'g1', 'b2' ],
  [ 'b2', 'b1', 'g1' ],
  [ 'b2', 'g1', 'b1' ],
  [ 'g1', 'b1', 'b2' ],
  [ 'g1', 'b2', 'b1' ]
]
*/

第四季度。硬币/总和问题

// Minimum coin Problem
const wrapper = (value) => {
    let result = 99999;
    let resultArr = [];
    const coins = [10, 6, 1];
    const helper = (value, label, combinedArray) => {
       if (value == 0) {
           if (result > label) {
               result = label;
               resultArr = [...combinedArray]
           }
           return ;
       }

       for (let coin of coins) {
           if (value - coin >=0) {
               combinedArray.push(coin);
               helper(value-coin, label+1, combinedArray);
               combinedArray.pop();
           }
       }
    }

    helper(value, 0, []);
    console.log(resultArr)

    return result;
};

const res = wrapper(12);

console.log(res);
/*
[ 6, 6 ]
2
*/

Q5.集合生成

// Problem 1: Generating All Subsets of a Set
// Problem Statement:
// Given a set of unique elements, generate all possible subsets (the power set).
// This solution need more enhancement.
// Example:
// Input: [1, 2, 3]
// Output: [[], [1], [2], [3], [1, 2], [1, 3], [2, 3], [1, 2, 3]]


const wrapper = () => {
    const result = [[]];
    const input = [1,2,3];
    input.forEach(item => result.push([item]));

    const helper = (combination, depth) => {
        if (depth == 2) {
            if (result.indexOf(combination) < 0) {
                result.push([...combination]);
            }


            return;
        }

        for (let item of input) {
           if (combination.indexOf(item) < 0) {
                combination.push(item);
            helper(combination, depth+1);
            combination.pop()
           }
        }

    }

    helper([], 0);
    result.push([...input])
    return result;
}

const test = wrapper();

console.log(test);
/*
[
  [],          [ 1 ],
  [ 2 ],       [ 3 ],
  [ 1, 2 ],    [ 1, 3 ],
  [ 2, 1 ],    [ 2, 3 ],
  [ 3, 1 ],    [ 3, 2 ],
  [ 1, 2, 3 ]
]
*/

Q6.使用暴力算法的旅行推销员问题

// Travelling sales man problem using brut force algorithm

function calculateDistance(matrix, path) {
  let totalDistance = 0;
  for (let i = 0; i < path.length - 1; i++) {
    totalDistance += matrix[path[i]][path[i + 1]];
  }
  // Return to the starting city
  totalDistance += matrix[path[path.length - 1]][path[0]];
  return totalDistance;
}

function permute(arr) {
  const result = [];

  const helper = (combination, depth) => {
      if (depth == 4) {
          result.push([...combination]);

          return;
      }

      for (let item of arr) {
          if (combination.indexOf(item) < 0) {
              combination.push(item);
              helper(combination, depth +1);
              combination.pop()
          }
      }
  }
  helper([], 0);

  return result;
}

function tsp(matrix) {
  const cities = Array.from({length: matrix.length}, (_, index) => index)
  console.log(cities)
  const permutations = permute(cities);
  console.log(permutations)
  let minDistance = Infinity;
  let bestPath = [];

  for (let path of permutations) {
    const distance = calculateDistance(matrix, path);
    if (distance < minDistance) {
      minDistance = distance;
      bestPath = path;
    }
  }

  return { minDistance, bestPath };
}

// Example usage:
const distanceMatrix = [
  [0, 10, 15, 20],
  [10, 0, 35, 25],
  [15, 35, 0, 30],
  [20, 25, 30, 0]
];

const result = tsp(distanceMatrix);

console.log(`The shortest distance is: ${result.minDistance}`);
console.log(`The best path is: ${result.bestPath}`);
/*
Initialization:  Calculate Distance Explanation

totalDistance is initialized to 0.
First Iteration (i = 0):

From city 0 to city 1: matrix[0][1] is 10.
Add 10 to totalDistance, making it 10.
Second Iteration (i = 1):

From city 1 to city 3: matrix[1][3] is 25.
Add 25 to totalDistance, making it 35.
Third Iteration (i = 2):

From city 3 to city 2: matrix[3][2] is 30.
Add 30 to totalDistance, making it 65.
Return to Starting City:

From city 2 back to city 0: matrix[2][0] is 15.
Add 15 to totalDistance, making it 80.
Return Total Distance:

The function returns 80, which is the total distance of the path [0, 1, 3, 2, 0].

// Output
[ 0, 1, 2, 3 ]
[
  [ 0, 1, 2, 3 ], [ 0, 1, 3, 2 ],
  [ 0, 2, 1, 3 ], [ 0, 2, 3, 1 ],
  [ 0, 3, 1, 2 ], [ 0, 3, 2, 1 ],
  [ 1, 0, 2, 3 ], [ 1, 0, 3, 2 ],
  [ 1, 2, 0, 3 ], [ 1, 2, 3, 0 ],
  [ 1, 3, 0, 2 ], [ 1, 3, 2, 0 ],
  [ 2, 0, 1, 3 ], [ 2, 0, 3, 1 ],
  [ 2, 1, 0, 3 ], [ 2, 1, 3, 0 ],
  [ 2, 3, 0, 1 ], [ 2, 3, 1, 0 ],
  [ 3, 0, 1, 2 ], [ 3, 0, 2, 1 ],
  [ 3, 1, 0, 2 ], [ 3, 1, 2, 0 ],
  [ 3, 2, 0, 1 ], [ 3, 2, 1, 0 ]
]
The shortest distance is: 80
The best path is: 0,1,3,2

*/

Q7。 0/1背包蛮力问题

// 0/1 knapsack Brut force Problem
function knapsackBruteForce(weights, values, capacity) {
  let n = weights.length;
  let maxValue = 0;
  const subsetResult = [];
  const binaryVals = [0, 1];

  // Function to calculate the total weight and value of a subset
  function calculateSubset(subset) {
    let totalWeight = 0;
    let totalValue = 0;
    for (let i = 0; i < subset.length; i++) {
      if (subset[i]) {
        totalWeight += weights[i];
        totalValue += values[i];
      }
    }
    return { totalWeight, totalValue };
  }

  const helper = (combination, depth) => {
      if (depth == 4) {
          subsetResult.push([...combination]);
          return ;
      }

      for (let item of binaryVals) {
          combination.push(item);
          helper(combination, depth +1);
          combination.pop()
      }

  }

    helper([], 0);
    console.log(subsetResult)
  // Generate all subsets using binary representation
  for (let subset of subsetResult) {
    let { totalWeight, totalValue } = calculateSubset(subset);
    if (totalWeight <= capacity && totalValue > maxValue) {
      maxValue = totalValue;
    }
  }

  return maxValue;
}

// Example usage:
const weights = [2, 3, 4, 5];
const values = [3, 4, 5, 6];
const capacity = 5;
const maxVal = knapsackBruteForce(weights, values, capacity);

console.log(`The maximum value in the knapsack is: ${maxVal}`);
/*
[
  [ 0, 0, 0, 0 ], [ 0, 0, 0, 1 ],
  [ 0, 0, 1, 0 ], [ 0, 0, 1, 1 ],
  [ 0, 1, 0, 0 ], [ 0, 1, 0, 1 ],
  [ 0, 1, 1, 0 ], [ 0, 1, 1, 1 ],
  [ 1, 0, 0, 0 ], [ 1, 0, 0, 1 ],
  [ 1, 0, 1, 0 ], [ 1, 0, 1, 1 ],
  [ 1, 1, 0, 0 ], [ 1, 1, 0, 1 ],
  [ 1, 1, 1, 0 ], [ 1, 1, 1, 1 ]
]
The maximum value in the knapsack is: 7
*/

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