1A)
bool IsMode(const apvector<int> &data, int k) {
    return ((data[k-1] < data[k]) && (data[k] > data[k+1]));
}

1B)
int ModeIndex(const apvector<int> &data) {
    int a, result = -1;
    for (a = 0; a < data.length(); a++)
       if (IsMode(data, a) result = a;
    return result;
}

1C)

void PrintHistogram(const apvector<int> &data, int longestBar, char barChar) {
    int mode;
    double barspervalue;
    mode = data[ModeIndex(data)];
    barspervalue = longestBar / mode;
    for (int a = 0; a < data.length(); a++) {
       for (int b = 0; b < (data[a] * barspervalue); b++)
           cout << barChar;
       cout << endl;
       }
}

#2 is skipped since this deals with another case study

3A)

int Occurrences(const WordCollection &C, const apstring &word) {
    int count=0;
    for (int a = 0; a < C.Size(); a++)
       if (C.FindKth(a+1) == word) count++; // note, the counting for FindKth starts at 1
    return count;
    }

3B)

void RemoveDuplicates(WordCollection &C, const apstring &word) {
    int instances;
    apstring word;
    for (int a = C.Size(); a > 0; a--) {
       word = C.FindKth(a);
       instances = Occurrences(C,word);
       if (instances > 1) {
          for (int b = 1; b < instances; b++) {
             C.Remove(word);
             a--; // we have one less item in the collection now
             }
           }
       }
}

3C)
// This is not the most efficient imaginable answer, but it'll work perfectly fine
apstring MostCommon(const WordCollection &C) {
      int instances, mostinstances;
      apstring word, biggestword;
    biggestword = C.FindKth(C, 1);
    mostinstances = Occurrences(C, biggestword);
    for (int a = 0; a < C.Size(); a++) {
       word = C.FindKth(a);
       instances = Occurrences(C,word);
       if (instances > mostinstances) {
          mostinstances = instances;
          biggestword = word;
          }
       }
    return biggestword;
    }

4A)
Point Encryptor::GetCoordinates(char ch) const {
    Point result;
    for (int a = 0; a < 6; a++)
       for (int b = 0; b < 6; b++)
          if (myMat[a][b]==ch) {
             result.row = a;
             result.col = b;
             }
    return result;
    }

4B)
// note: this can be done in a single line!
apstring Encryptor::EncryptTwo(const apstring &pair) const {
    Point position1, position2;
    apstring result;
    position1 = GetCoordinates(pair[0]);
    position2 = GetCoordinates(pair[1]);
    result = myMat[position1.row][position2.col];
    result = result + myMat[position2.row][position1.col];
    return result;
}

4C)
apstring Encryptor::EncryptWord(const apstring &word) const {
    apstring result = "";
    apstring pair;
    for (int a = 0; a < word.length()-1; a+=2) {// -1 means we're not going to try to process the stray last character (in case it exists)
       pair = word[0];
       pair = pair + word[1];
       result = result + EncryptTwo(pair);
    }
    if (a != word.length()) result = result + word[word.length()-1]; // add in the stray character if you should
    return result;
    }