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diff --git a/challenge-287/ulrich-rieke/cpp/ch-1.cpp b/challenge-287/ulrich-rieke/cpp/ch-1.cpp
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+++ b/challenge-287/ulrich-rieke/cpp/ch-1.cpp
@@ -0,0 +1,71 @@
+#include <iostream>
+#include <string>
+#include <regex>
+#include <vector>
+#include <numeric>
+
+int main( ) {
+   std::cout << "Enter a password!\n" ;
+   std::string word ;
+   std::cin >> word ;
+   int totalsteps = 0 ;
+   const std::regex rgx1("[a-z]") ;
+   const std::regex rgx2("[A-Z]") ;
+   const std::regex rgx3("\\d") ;
+   int len = word.length( ) ;
+   if ( len < 6 ) {
+      totalsteps += 6 - len ;
+   }
+   //beware : it took me quite some time to find out that 
+   //std::regex_match does not return the right result!!
+   //if you want to check whether a given regular expression
+   //such as \d matches I had to use std::regex_search!!!
+   if ( ! std::regex_search( word , rgx1 )) {
+      if ( len >= 6 ) {
+	 totalsteps++ ;
+      }
+   }
+   if ( ! std::regex_search( word , rgx2 )) {
+      if ( len >= 6 ) {
+	 totalsteps++ ;
+      }
+   }
+   if ( ! std::regex_search( word , rgx3 )) {
+      if ( len >= 6 ) {
+	 totalsteps++ ;
+      }
+   }
+   //std::regex appears not to allow backreferences...?
+   //so, to prevent too many identical neighbouring letters
+   //I group the letters , divide their lengths by 3 and sum the
+   //quotients up ( inspired by Haskell where this operation is so easy
+   //to perform!!!
+   std::vector<std::string> neighbouring_letters ;
+   std::string currentWord ;
+   //imitate Haskell's group function!
+   for ( auto it = word.begin( ) ; it != word.end( ) ; ++it ) {
+      if ( currentWord.empty( ) ) 
+	 currentWord.push_back( *it ) ;
+      else {
+	 char last_char = currentWord.back( ) ;
+	 if ( last_char == *it ) {
+	    currentWord.push_back( *it ) ;
+	 }
+	 else {
+	    neighbouring_letters.push_back( currentWord ) ;
+	    currentWord.clear( ) ;
+	    currentWord.push_back( *it ) ;
+	 }
+      }
+   }
+   neighbouring_letters.push_back( currentWord ) ;
+   std::vector<int> grouplengths ;
+   for ( auto v : neighbouring_letters ) {
+      grouplengths.push_back( v.length( ) / 3 ) ;
+   }
+   int replacements = std::accumulate( grouplengths.begin( ) , 
+	 grouplengths.end( ) , 0 ) ;
+   totalsteps += replacements ;
+   std::cout << totalsteps << '\n' ;
+   return 0 ;
+}