英文:
Algorithm for guaranteeing win (or draw) in row of coins problem
问题
以下是您提供的代码的翻译部分：
// 该方法选择两个硬币中较大的那一个，前提是较大硬币旁边的硬币能够在最坏情况下保证获胜 - 所有剩余的最大数字都被 Tamar 选走。
public static void win(int[] arr){
int beginning = 0, end=arr.length-1, sumAmir = 0, sumTamar = 0;
int[] arrClone = arr.clone();
java.util.Arrays.sort (arrClone, beginning, end+1);
while (beginning < end){
if ((arr[end] >= arr[beginning] && !loses(arr, arrClone, end-1, sumAmir, sumTamar, beginning, end))
|| (arr[end] < arr[beginning] && loses(arr, arrClone, beginning+1, sumAmir, sumTamar, beginning, end))) {
System.out.println("Amir took " + arr[end]);
sumAmir += arr[end];
end--;
}
else{
System.out.println("Amir took " + arr[beginning]);
sumAmir += arr[beginning];
beginning++;
}
if (arr[beginning] > arr[end]){
System.out.println("Tamar took " + arr[beginning]);
sumTamar += arr[beginning];
beginning++;
}
else{
System.out.println("Tamar took " + arr[end]);
sumTamar += arr[end];
end--;
}
}
System.out.println("Amir total " + sumAmir + "\nTamar total " + sumTamar);
}
private static boolean loses(int[] arr, int[] arrClone, int place, int sumAmir, int sumTamar, int beginning, int end) {
int currPlace = arr[place];
if (end - beginning == 1) {
return false;
}
if (place == beginning + 1) {
sumAmir += arr[beginning];
beginning += 2;
} else {
sumAmir += arr[end];
end -= 2;
}
// 对数组的最低和最高一半求和（排除已检查的二元组）
int lowestValsSum = 0;
int k = 0;
int highestValsSum = 0;
for (int i = beginning; i < (beginning+end)/2 + 1; i++) {
k++;
lowestValsSum += arrClone[i];
highestValsSum += arrClone[end-k];
}
return sumAmir + highestValsSum < sumTamar + currPlace + lowestValsSum;
}
请注意，这只是您提供的代码的中文翻译，不包含额外的解释或讨论。如果您有任何问题或需要进一步的帮助，请随时提问。
英文:
In a row of n coins (>=0, n=even) each player (Amir and Tamar) must pick a coin from one of the EDGES of the row. Player with largest amount wins. Need to guarantee at least a draw for Amir. Amir goes first. My algorithm for this was Make Amir take the largest coin unless the IMMEDIATELY FOLLOWING coin guarantees a loss (i.e. if Amir takes all the highest possible remaining coins - he will STILL lose). Do you think this is right?
If so I would appreciate if you took a look at my code (Java) - getting random wins Amir/Tamar.
//This method takes the largest of the two coins providing the coin adjacent to the larger coin leaves a chance of winning assuming
//worse case scenario - all highest (remaining numbers) are picked by - Tamar.
public static void win(int[] arr){
int beginning = 0, end=arr.length-1, sumAmir = 0, sumTamar = 0;
int[] arrClone = arr.clone();
java.util.Arrays.sort (arrClone, beginning, end+1);
while (beginning<end){
if ((arr[end]>=arr[beginning] && !loses(arr,arrClone, end-1,sumAmir,sumTamar,beginning,end))
|| (arr[end]<arr[beginning] && loses(arr, arrClone,beginning+1,sumAmir,sumTamar,beginning,end))) {
System.out.println ("Amir took " + arr[end]);
sumAmir+=arr[end];
end--;
}
else{
System.out.println ("Amir took "+ arr[beginning]);
sumAmir+=arr[beginning];
beginning++;
}
if (arr[beginning]> arr[end]){
System.out.println ("Tamar took " + arr[beginning]) ;
sumTamar+=arr[beginning];
beginning++;
}
else{
System.out.println ("Tamar took " + arr[end]) ;
sumTamar+=arr[end];
end--;
}
}
System.out.println ("Amir total " + sumAmir + "\nTamar total " + sumTamar);
}
private static boolean loses(int[] arr,int[] arrClone, int place, int sumAmir, int sumTamar, int beginning, int end) {
int currPlace =arr[place];
if (end - beginning == 1) {
return false;
}
if (place == beginning + 1) {
sumAmir += arr[beginning];
beginning += 2;
} else {
sumAmir += arr[end];
end -= 2;
}
//Sums lowest & highest halves of (sorted) array. (excluding checked duo)
int lowestValsSum = 0;
int k=0;
int highestValsSum = 0;
for (int i = beginning; i < (beginning+end)/2 + 1; i++) {
k++;
lowestValsSum+=arrClone[i];
highestValsSum+=arrClone[end-k];
}
return sumAmir + highestValsSum < sumTamar + currPlace + lowestValsSum;
}
TX
答案1
得分: 0
你可以尝试为你的游戏实现极小化极大算法，以确保你的玩家进行最优游戏（无论是获胜还是平局）。
英文:
You can try to implement the minimax algorithm for your game to guarantee optimal play for your player(either win or draw).
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