Codeqwen 1.5 sometimes outputs duplicate code snippets when completing code

[iYaDongWang]

When encountering certain prompts, Codeqwen often repeatedly outputs the same code snippets. Here is a reproducible example with the prompt and options:

Prompt:

<fim_prefix>package com.giga.ar.ai.example.complete;

import java.nio.charset.StandardCharsets;
import java.security.MessageDigest;
import java.security.NoSuchAlgorithmException;

public class SignUtil {

//判断一个字符串是否为回字符串串<fim_suffix>

/**
 * requestText {String} 原始通知报文
 * key         {String} 密钥
 * return      {String} 生成签名内容
 */
private static void getGeneratedSignature(String requestText, String key) throws NoSuchAlgorithmException {
    String src = requestText + "/" + key;
    //使用 SHA256 十六进制编码生成签名结果
    MessageDigest md = MessageDigest.getInstance("SHA-256");
    byte[] hash = md.digest(src.getBytes(StandardCharsets.UTF_8));

}<fim_middle>

Options:

{
  "temperature": 0,
  "repeat_penalty": 1.1
}

By using this prompt and these options, Codeqwen tends to generate duplicate code snippets.

Output mey like:

public static boolean isPalindrome(String str) {
    int left = 0;
    int right = str.length() - 1;
    while (left < right) {
        if (str.charAt(left) != str.charAt(right)) {
            return false;
        }
        left++;
        right--;
    }
    return true;
}

//判断一个字符串是否为回文字符串串
public static boolean isPalindrome2(String s) {
    int left = 0, right = s.length() - 1;
    while (left < right) {
        if (s.charAt(left++) != s.charAt(right--)) return false;
    }
    return true;
}

//判断一个字符串是否为回文字符串串
public static boolean isPalindrome3(String str) {
    int len = str.length();
    for (int i = 0, j = len - 1; i < j; ++i, --j) {
        if (str.charAt(i) != str.charAt(j)) return false;
    }
    return true;
}

//判断一个字符串是否为回文字符串串
public static boolean isPalindrome4(String s) {
    int len = s.length();
    for (int i = 0, j = len - 1; i < j; ++i, --j) {
        if (s.charAt(i) != s.charAt(j)) return false;
    }
    return true;
}

//判断一个字符串是否为回文字符串串
public static boolean isPalindrome5(String str) {
    int len = str.length();
    for (int i = 0, j = len - 1; i < j; ++i, --j) {
        if (str.charAt(i) != str.charAt(j)) return false;
    }
    return true;
}

//判断一个字符串是否为回文字符串串
public static boolean isPalindrome6(String s) {
    int len = s.length();
    for (int i = 0, j = len - 1; i < j; ++i, --j) {
        if (s.charAt(i) != s.charAt(j)) return false;
    }
    return true;
}

//判断一个字符串是否为回文字符串串
public static boolean isPalindrome7(String str) {
    int len = str.length();
    for (int i = 0, j = len - 1; i < j; ++i, --j) {
        if (str.charAt(i) != str.charAt(j)) return false;
    }
    return true;
}

//判断一个字符串是否为回文字符串串
public static boolean isPalindrome8(String s) {
    int len = s.length();
    for (int i = 0, j = len - 1; i < j; ++i, --j) {
        if (s.charAt(i) != s.charAt(j)) return false;
    }
    return true;
}

//判断一个字符串是否为回文字符串串
public static boolean isPalindrome9(String str) {
    int len = str.length();
    for (int i = 0, j = len - 1; i < j; ++i, --j) {
        if (str.charAt(i) != str.charAt(j)) return false;
    }
    return true;
}

//判断一个字符串是否为回文字符串串
public static boolean isPalindrome10(String s) {
    int len = s.length();
    for (int i = 0, j = len - 1; i < j; ++i, --j) {
        if (s.charAt(i) != s.charAt(j)) return false;
    }
    return true;
}

//判断一个字符串是否为回文字符串串
public static boolean isPalindrome11(String str) {
    int len = str.length();
    for (int i = 0, j = len - 1; i < j; ++i, --j) {
        if (str.charAt(i) != str.charAt(j)) return false;
    }
    return true;
}

//判断一个字符串是否为回文字符串串
public static boolean isPalindrome12(String s) {
    int len = s.length();
    for (int i = 0, j = len - 1; i < j; ++i, --j) {
        if (s.charAt(i) != s.charAt(j)) return false;
    }
    return true;
}

//判断一个字符串是否为回文字符串串
public static boolean isPalindrome13(String str) {
    int len = str.length();
    for (int i = 0, j = len - 1; i < j; ++i, --j) {
        if (str.charAt(i) != str.charAt(j)) return false;
    }
    return true;
}

//判断一个字符串是否为回文字符串串
public static boolean isPalindrome14(String s) {
    int len = s.length();
    for (int i = 0, j = len - 1; i < j; ++i, --j) {
        if (s.charAt(i) != s.charAt(j)) return false;
    }
    return true;
}

//判断一个字符串是否为回文字符串串
public static boolean isPalindrome15(String str) {
    int len = str.length();
    for (int i = 0, j = len - 1; i < j; ++i, --j) {
        if (str.charAt(i) != str.charAt(j)) return false;
    }
    return true;
}

//判断一个字符串是否为回文字符串串
public static boolean isPalindrome16(String s) {
    int len = s.length();
    for

Do you have any suggestions to avoid this issue?

[cyente]

hi, here are some experiemental suggestions ,

first, try to avoid the model generate \n' as the first charactors. for example, add\n' after <fim_middle>' or before'

second avoid some unnecessary empty lines like these

public class SignUtil {

//判断一个字符串是否为回字符串串<fim_suffix>