我正在使用递归公式研究Rabin Karp算法。以下是代码。 在代码中我正在检查正常方式和递归公式计算的哈希值。两个值都不匹配。我花了足够的时间调试近3个小时,不确定是什么问题。请求帮助找到错误。
#include <iostream>
#include <fstream>
#include <streambuf>
#include <cstdint>
#include <string>
#include <vector>
const std::uint64_t uiLargePrime = 1000000007;
const unsigned int uiXValue = 263;
const unsigned int uiHashTableSize = 79;
struct sCalcHash {
std::uint64_t operator() (const std::string& strText) {
// user horners method.
unsigned int uiStrLength = strText.length();
std::uint64_t uiResult = 0;
// calculate hash value
for(int uiIdx = (uiStrLength - 1); uiIdx >= 0; uiIdx--) {
uiResult = (((uiResult * uiXValue) % uiLargePrime) + strText[uiIdx]) % uiLargePrime ;
}
// return uiResult % uiHashTableSize;
return uiResult;
}
};
// calculate x ^ uiPatternLength % uiLargePrime.
unsigned int expValueOfX(unsigned int uiXVal, unsigned int uiPower) {
// get X value in range of prime;
uiXVal = uiXVal % uiLargePrime;
unsigned int uiResult = 1;
while (uiPower > 0 ) {
// check if power is odd
if (uiPower & 1) {
uiResult = ((uiResult % uiLargePrime) * (uiXVal % uiLargePrime) ) % uiLargePrime;
}
// now uiPower is even
uiPower = uiPower >> 1;
uiXVal = ((uiXVal % uiLargePrime) * (uiXVal % uiLargePrime)) % uiLargePrime;
}
return uiResult;
}
// Rabin Karp Algorithm
void RabinKarpAlgo(std::string& Text, std::string& pattern) {
std::vector<unsigned int> vecPostions;
//calculate hash value of pattern.
sCalcHash hash;
std::uint64_t hashValPattern = hash(pattern);
std::cout << "Hash Value of pattern: " << hashValPattern << std::endl;
unsigned int uiPatternLength = pattern.length();
// calculate x ^ uiPatternLength % uiLargePrime.
unsigned int uiXExpVal = expValueOfX(uiXValue, uiPatternLength);
//std::cout << "Exponential value " << uiXExpVal << std::endl;
// calculate hash value
unsigned int uiStrLength = Text.length();
// calculate hash value of last part of string of pattern length.
unsigned int uiLastIdx = uiStrLength - uiPatternLength;
std::uint64_t hashValLastIdx = hash(Text.substr(uiLastIdx));
std::cout << "Hash Value of last indx of text: " << hashValLastIdx << std::endl;
// if hash value is same then compare string
if (hashValLastIdx == hashValPattern) {
if(pattern == Text.substr(uiLastIdx)) {
std::cout << "Pushing index: " << uiLastIdx << std::endl;
vecPostions.push_back(uiLastIdx);
}
}
for(int uiIdx = uiLastIdx - 1; uiIdx >= 0; uiIdx--) {
// calculate hash value of string
std::int64_t iHashValRecur = ( (Text[uiIdx] % uiLargePrime) +
((hashValLastIdx % uiLargePrime) * (uiXValue % uiLargePrime)) % uiLargePrime -
((Text[uiIdx + uiPatternLength] % uiLargePrime) * (uiXExpVal % uiLargePrime) ) % uiLargePrime
) % uiLargePrime;
unsigned int iHashVal = hash(Text.substr(uiIdx, uiPatternLength));
std::cout << "Hash Value of with recurr " << uiIdx << " is " << iHashValRecur << " and with hash func: " << iHashVal << std::endl;
if(iHashValRecur == hashValPattern) {
// compare string
if(pattern == Text.substr(uiIdx, uiPatternLength) ) {
std::cout << "Pushing index: " << uiIdx << std::endl;
vecPostions.push_back(uiIdx);
}
}
hashValLastIdx = iHashValRecur;
}
// print vectors
for( int uiIdx = vecPostions.size() - 1; uiIdx >= 0; uiIdx--) {
std::cout << vecPostions[uiIdx] << " ";
}
std::cout << std::endl;
return ;
}
int main() {
std::ifstream inputFile("rabinkarp.in");
std::streambuf *pCinbuf = std::cin.rdbuf();
std::cin.set_rdbuf(inputFile.rdbuf());
std::string strText;
std::string strPattern;
std::cin >> strPattern;
std::cin >> strText;
std::cout << "Text: " << strText << std::endl;
std::cout << "Pattern: " << strPattern << std::endl;
RabinKarpAlgo(strText, strPattern);
return 0;
}
Text: baaaaaaa
Pattern: aaaaa
Hash Value of pattern: 853306522
Hash Value of last indx of text: 853306522
Pushing index: 3
Hash Value of with recurr 2 is 435650523 and with hash func: 853306522
Hash Value of with recurr 1 is 9779548 and with hash func: 853306522
Hash Value of with recurr 0 is 5713908 and with hash func: 853306523
3
Press any key to continue . . .
预期答案是:1 2 3
答案 0 :(得分:0)
问题是我用无符号std64整数计算mod,所以mod变为正值,这是依赖于实现的
参考:C ++ 03第5.6段第4条:
二元/运算符产生商,二元%运算符从第一个表达式除以第二个表达式得到余数。如果/或%的第二个操作数为零,则行为未定义;否则(a / b)* b + a%b等于a。如果两个操作数都是非负的,那么余数是非负的;如果没有,余数的符号是实现定义的。
所以为了避免这个问题,我尝试下面改变了大的素数数据类型到int64_t它工作并在下面完成
std::int64_t iHashValRecur = ( (Text[uiIdx] % uiLargePrime) +
((uiXValue % uiLargePrime * hashValLastIdx % uiLargePrime) % uiLargePrime) -
((Text[uiIdx + uiPatternLength] % uiLargePrime * (uiXExpVal % uiLargePrime)) % uiLargePrime)
);
std::int64_t iHashVal = hash(Text.substr(uiIdx, uiPatternLength));
iHashValRecur = iHashValRecur % uiLargePrime;
while (iHashValRecur < 0) {
iHashValRecur += uiLargePrime;
}
iHashValRecur = iHashValRecur % uiLargePrime;