C 消息同步算法

我们有两个无限重复的消息，由字符a-z组成。每个字符的传输时间单位不同；a=1，b=2，c=4，d=8，e=16…，字符|告诉我们消息中的当前位置。我们的工作是确定同步这两条消息需要多少时间单位。同步意味着两条消息同时开始

例如：

我们有信息1：ea | babab

信息2:d|abaca

因此，我们知道信息1是bababea，信息2是abacad

消息将在42个时间单位内同步：

ea | bababea |=42

d | abacad | abacad |=42

例2： 消息1:| acabbaaa消息2:| dcbabaaaa

解决方案：0，因为它们已同步

我们想提出一个算法来计算第一次同步发生之前的时间

我是用C写的。我基本上做了所有的事情，除了算法本身

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我认为这可以用扩展的欧几里德算法来实现。

我回答了一个不同的问题，我认为这个问题的解决方案是完全相同的。您需要解方程m1Offset+（m1Len*intN1）=m2Offset+（m2Len*intN2）

我们需要找到满足上述方程的intN1和intN2当且仅当m1Len和m2Len的GCD除以（m2Offset-m1Offset）时，才会有解决方案。

在下面的代码中

• m1Len和m2Len：消息m1和m2的长度。例如：“ea | babab”的长度是“bababa”的长度=25，而在消息“d | abaca”中，“abacad”的长度=17
• m1Offset和m2Offset：初始偏移。例如：在消息“ea | babab”中，“ea”是偏移量，等于17。类似地，在“d | abaca”中，“d”是偏移量，等于8

• m1Time和m2Time应该相等，这是第一次同步时间

这是我的密码

#include <stdio.h>

void findVal(unsigned int m1Offset, unsigned int m1Len, unsigned int m2Offset, unsigned int m2Len) ;
unsigned int getGCD(unsigned int n1, unsigned int n2);

int main()
{
    findVal(17, 25, 8, 17);
    return 0;
}

void findVal(unsigned int m1Offset, unsigned int m1Len, unsigned int m2Offset, unsigned int m2Len) {

    unsigned int  n1       = 0;
    unsigned int  n2       = 0;
    unsigned char foundVal = 1;
    unsigned int  m1Time   = m1Offset;
    unsigned int  m2Time   = m2Offset;

    //No need to find n1 and n2 if msgs are starting from beginning.
    if(((m1Offset == m1Len) && (m2Offset == m2Len)) || ((0 == m1Offset) && (0 == m2Offset)))
    {
        m1Time = 0;
        m2Time = 0;
    }

    //No need to find n1 and n2 if offset times are same.
    else if(m1Offset != m2Offset)
    {
       //Offset times are not same.
       foundVal = 0;

       //Find GCD of m1Len and m2Len.
       unsigned int gcd = getGCD(m1Len, m2Len);

       //There is a solution only if the difference of offsets is divisible by gcd.
       if(0 == (m2Offset-m1Offset) % gcd)
       {
           for(n2=1; n2<(unsigned int)-1; n2++)
           {
               unsigned int temp1 = (m2Len*n2)+(m2Offset-m1Offset);
               if(0 == temp1 % m1Len)
               {
                    n1 = temp1/m1Len;
                    m1Time = m1Offset + n1*m1Len;
                    m2Time = m2Offset + n2*m2Len;

                    foundVal = 1;
                    break;
               }
           }
       }
    }

    if(1 == foundVal)
    {
        printf("Found n1[%u] n2[%u] m1Time[%u] m2Time[%u]\n", n1, n2, m1Time, m2Time);
    }
    else
    {
        printf("Could not find n1, n2, m1Time, m2Time\n");
    }
}

unsigned int getGCD(unsigned int n1, unsigned int n2)
{
    while(n1!=n2)
    {
        if(n1 > n2)
            n1 -= n2;
        else
            n2 -= n1;
    }
    printf("GCD = %u\n",n1);

    return n1;
}

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如果我正确理解了您的问题陈述，那么解决方案就是直接应用，非常感谢！虽然我发现下面的输入创建了无限循环：findVal（16,20,19,24）@Mykybo：它不是无限循环，它检查从0到UNSIGNED_INT_MAX，这需要时间。你可以参考我对那个问题的回答。如果消息长度不是成对共素数，即GCD（m1Len，m2Len）大于1，您可能找不到解决方案。实际上，findVal（52，56，60，72）：m1Len是56，m2Len是72，它有GCD 8（大于1），您可以找到解决方案：276。是的。如果GCD>1，则无法保证找到同步时间。是否没有确定的方法来说明解决方案是否存在？一直运行到UNSIGNED_INT_MAX需要很多时间。

#include <stdio.h>

void findVal(unsigned int m1Offset, unsigned int m1Len, unsigned int m2Offset, unsigned int m2Len) ;
unsigned int getGCD(unsigned int n1, unsigned int n2);

int main()
{
    findVal(17, 25, 8, 17);
    return 0;
}

void findVal(unsigned int m1Offset, unsigned int m1Len, unsigned int m2Offset, unsigned int m2Len) {

    unsigned int  n1       = 0;
    unsigned int  n2       = 0;
    unsigned char foundVal = 1;
    unsigned int  m1Time   = m1Offset;
    unsigned int  m2Time   = m2Offset;

    //No need to find n1 and n2 if msgs are starting from beginning.
    if(((m1Offset == m1Len) && (m2Offset == m2Len)) || ((0 == m1Offset) && (0 == m2Offset)))
    {
        m1Time = 0;
        m2Time = 0;
    }

    //No need to find n1 and n2 if offset times are same.
    else if(m1Offset != m2Offset)
    {
       //Offset times are not same.
       foundVal = 0;

       //Find GCD of m1Len and m2Len.
       unsigned int gcd = getGCD(m1Len, m2Len);

       //There is a solution only if the difference of offsets is divisible by gcd.
       if(0 == (m2Offset-m1Offset) % gcd)
       {
           for(n2=1; n2<(unsigned int)-1; n2++)
           {
               unsigned int temp1 = (m2Len*n2)+(m2Offset-m1Offset);
               if(0 == temp1 % m1Len)
               {
                    n1 = temp1/m1Len;
                    m1Time = m1Offset + n1*m1Len;
                    m2Time = m2Offset + n2*m2Len;

                    foundVal = 1;
                    break;
               }
           }
       }
    }

    if(1 == foundVal)
    {
        printf("Found n1[%u] n2[%u] m1Time[%u] m2Time[%u]\n", n1, n2, m1Time, m2Time);
    }
    else
    {
        printf("Could not find n1, n2, m1Time, m2Time\n");
    }
}

unsigned int getGCD(unsigned int n1, unsigned int n2)
{
    while(n1!=n2)
    {
        if(n1 > n2)
            n1 -= n2;
        else
            n2 -= n1;
    }
    printf("GCD = %u\n",n1);

    return n1;
}

Found n1[1] n2[2] m1Time[42] m2Time[42]