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我正在制作自己的 android 应用程序，可以帮助解决魔方问题。我买了小米的 Giiker Cube 并研究了 Herbert Kociemba 的两阶段算法。

所以问题从这里开始。对于 Giiker Cube，我查看了下面的 api。但是这个 api 给出了 20 字节数据的多维数据集的状态。 https://github.com/Vexu/SuperCube-API

我还从这里查看了相同的 React 库及其依赖库，以研究该状态到立方体面和移动的转换。但是那个库的解码逻辑对我不起作用。

我也没有使用本机反应。 https://www.npmjs.com/package/react-native-giiker

我需要帮助将 20 字节的立方体状态转换为 Herbert Kociemba 的算法和移动的立方体面

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1 回答 1

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我有新的 Giiker cube i3s，它具有加密状态。Giiker cube i3 并非如此。

可以在此处找到解码加密状态的逻辑。

https://github.com/kabelbinder/giiker/blob/50db5d58e0417749fe5815e72856b90a1afa43b1/index.js#L326

下面是解密和解析的逻辑。

const turns = {
  0: 1,
  1: 2,
  2: -1,
  8: -2,
};

const faces = ['B', 'D', 'L', 'U', 'R', 'F'];

_parseCubeValue (value) {
const state = {
  cornerPositions: [],
  cornerOrientations: [],
  edgePositions: [],
  edgeOrientations: []
};
const moves = [];
if (value.getUint8(18) == 0xa7) { // decrypt
    var key = [176, 81, 104, 224, 86, 137, 237, 119, 38, 26, 193, 161, 210, 126, 150, 81, 93, 13, 236, 249, 89, 235, 88, 24, 113, 81, 214, 131, 130, 199, 2, 169, 39, 165, 171, 41];
        var k = value.getUint8(19);
        var k1 = k >> 4 & 0xf;
        var k2 = k & 0xf;
    for (let i = 0; i < value.byteLength; i++) {
        const move = (value.getUint8(i) + key[i + k1] + key[i + k2]) & 0xff;
        const highNibble = move >> 4;
        const lowNibble = move & 0b1111;
        if (i < 4) {
            state.cornerPositions.push(highNibble, lowNibble);
        } else if (i < 8) {
            state.cornerOrientations.push(highNibble, lowNibble);
        } else if (i < 14) {
            state.edgePositions.push(highNibble, lowNibble);
        } else if (i < 16) {
            state.edgeOrientations.push(!!(move & 0b10000000));
            state.edgeOrientations.push(!!(move & 0b01000000));
            state.edgeOrientations.push(!!(move & 0b00100000));
            state.edgeOrientations.push(!!(move & 0b00010000));
            if (i === 14) {
                state.edgeOrientations.push(!!(move & 0b00001000));
                state.edgeOrientations.push(!!(move & 0b00000100));
                state.edgeOrientations.push(!!(move & 0b00000010));
                state.edgeOrientations.push(!!(move & 0b00000001));
            }
        } else {
            moves.push(this._parseMove(highNibble, lowNibble));
        }
    }
}
 else { // not encrypted
     for (let i = 0; i < value.byteLength; i++) {
         const move = value.getUint8(i)
         const highNibble = move >> 4;
         const lowNibble = move & 0b1111;
         if (i < 4) {
             state.cornerPositions.push(highNibble, lowNibble);
         } else if (i < 8) {
             state.cornerOrientations.push(highNibble, lowNibble);
         } else if (i < 14) {
             state.edgePositions.push(highNibble, lowNibble);
         } else if (i < 16) {
             state.edgeOrientations.push(!!(move & 0b10000000));
             state.edgeOrientations.push(!!(move & 0b01000000));
             state.edgeOrientations.push(!!(move & 0b00100000));
             state.edgeOrientations.push(!!(move & 0b00010000));
             if (i === 14) {
                 state.edgeOrientations.push(!!(move & 0b00001000));
                 state.edgeOrientations.push(!!(move & 0b00000100));
                 state.edgeOrientations.push(!!(move & 0b00000010));
                 state.edgeOrientations.push(!!(move & 0b00000001));
             }
         } else {
             moves.push(this._parseMove(highNibble, lowNibble));
         }
     } 
 }
return {state, moves};
}

_parseMove (faceIndex, turnIndex) {
    const face = faces[faceIndex - 1];
    const amount = turns[turnIndex - 1];
    let notation = face;

    switch (amount) {
      case 2: notation = `${face}2`; break;
      case -1: notation = `${face}'`; break;
      case -2: notation = `${face}2'`; break;
    }

    return {face, amount, notation};
  }
于 2020-06-11T07:15:50.550 回答