12.9. Bisection search¶

If the cards in the deck are not in order, there is no way to search that is faster than the linear search. We have to look at every card, since otherwise there is no way to be certain the card we want is not there.

But when you look for a word in a dictionary, you don’t search linearly through every word. The reason is that the words are in alphabetical order. As a result, you probably use an algorithm that is similar to a bisection search:

Start in the middle somewhere.
Choose a word on the page and compare it to the word you are looking for.
If you found the word you are looking for, stop.
If the word you are looking for comes after the word on the page, flip to somewhere later in the dictionary and go to step 2.
If the word you are looking for comes before the word on the page, flip to somewhere earlier in the dictionary and go to step 2.

If you ever get to the point where there are two adjacent words on the page and your word comes between them, you can conclude that your word is not in the dictionary. The only alternative is that your word has been misfiled somewhere, but that contradicts our assumption that the words are in alphabetical order.

In the case of a deck of cards, if we know that the cards are in order, we can write a version of find that is much faster. The best way to write a bisection search is with a recursive function. That’s because bisection is naturally recursive.

The trick is to write a function called findBisect that takes two indices as parameters, low and high, indicating the segment of the vector that should be searched (including both low and high).

To search the vector, choose an index between low and high, and call it mid. Compare the card at mid to the card you are looking for.
If you found it, stop.
If the card at mid is higher than your card, search in the range from low to mid-1.
If the card at mid is lower than your card, search in the range from mid+1 to high.

Steps 3 and 4 look suspiciously like recursive invocations. Here’s what this all looks like translated into C++:

int findBisect (const Card& card, const vector<Card>& deck,
                int low, int high) {
  int mid = (high + low) / 2;

  // if we found the card, return its index
  if (equals (deck[mid], card)) return mid;

  // otherwise, compare the card to the middle card
  if (deck[mid].isGreater (card)) {
    // search the first half of the deck
    return findBisect (card, deck, low, mid-1);
  } else {
    // search the second half of the deck
    return findBisect (card, deck, mid+1, high);
  }
}

Although this code contains the kernel of a bisection search, it is still missing a piece. As it is currently written, if the card is not in the deck, it will recurse forever. We need a way to detect this condition and deal with it properly (by returning -1).

The easiest way to tell that your card is not in the deck is if there are no cards in the deck, which is the case if high is less than low. Well, there are still cards in the deck, of course, but what I mean is that there are no cards in the segment of the deck indicated by low and high.

With that line added, the function works correctly:

int findBisect (const Card& card, const vector<Card>& deck,
                int low, int high) {

  cout << low << ", " << high << endl;

  if (high < low) return -1;

  int mid = (high + low) / 2;

  if (equals (deck[mid], card)) return mid;

  if (deck[mid].isGreater (card)) {
    return findBisect (card, deck, low, mid-1);
  } else {
    return findBisect (card, deck, mid+1, high);
  }
}

I added an output statement at the beginning so I could watch the sequence of recursive calls and convince myself that it would eventually reach the base case. I tried out the following code:

cout << findBisect (deck[23], deck, 0, 51);

And got the following output:

0, 51
0, 24
13, 24
19, 24
22, 24
I found the card at index = 23

Then I made up a card that is not in the deck (the 15 of Diamonds), and tried to find it. I got the following:

0, 51
0, 24
13, 24
13, 17
13, 14
13, 12
I found the card at index = -1

These tests don’t prove that this program is correct. In fact, no amount of testing can prove that a program is correct. On the other hand, by looking at a few cases and examining the code, you might be able to convince yourself.

The code below searches finds the same card from the same deck we used on the previous page. This time, it uses bisection search to locate the card.

#include <iostream>
#include <string>
#include <vector>
using namespace std;

struct Card {
    int suit, rank;

Card ();
    Card (int s, int r);
    void print () const;
    bool isGreater (const Card& c2) const;
};

vector<Card> buildDeck();

bool equals (const Card& c1, const Card& c2){
    return (c1.rank == c2.rank && c1.suit == c2.suit);
}

void printDeck(const vector<Card>& deck);
int find (const Card& card, const vector<Card>& deck);
int findBisect (const Card& card, const vector<Card>& deck, int low, int high);

int main() {
    vector<Card> deck = buildDeck();
    Card card (3, 6);
    // We need to sort from the first card (0) to the last card (size-1)
    cout << findBisect(card, deck, 0, deck.size() - 1);
}

====

Card::Card () {
   suit = 0;  rank = 0;
}

Card::Card (int s, int r) {
   suit = s;  rank = r;
}

void Card::print () const {
   vector<string> suits (4);
   suits[0] = "Clubs";
   suits[1] = "Diamonds";
   suits[2] = "Hearts";
   suits[3] = "Spades";

vector<string> ranks (14);
   ranks[1] = "Ace";
   ranks[2] = "2";
   ranks[3] = "3";
   ranks[4] = "4";
   ranks[5] = "5";
   ranks[6] = "6";
   ranks[7] = "7";
   ranks[8] = "8";
   ranks[9] = "9";
   ranks[10] = "10";
   ranks[11] = "Jack";
   ranks[12] = "Queen";
   ranks[13] = "King";

cout << ranks[rank] << " of " << suits[suit] << endl;
 }

vector<Card> buildDeck() {
   vector<Card> deck (52);
   int i = 0;
   for (int suit = 0; suit <= 3; suit++) {
      for (int rank = 1; rank <= 13; rank++) {
         deck[i].suit = suit;
         deck[i].rank = rank;
         i++;
      }
   }
   return deck;
}

void printDeck (const vector<Card>& deck) {
   for (size_t i = 0; i < deck.size(); i++) {
     deck[i].print ();
   }
 }

int find (const Card& card, const vector<Card>& deck) {
   for (size_t i = 0; i < deck.size(); i++) {
    if (equals (deck[i], card)) return i;
   }
   return -1;
}

int findBisect (const Card& card, const vector<Card>& deck, int low, int high) {

cout << low << ", " << high << endl;

if (high < low) return -1;

int mid = (high + low) / 2;

if (equals (deck[mid], card)) return mid;

if (deck[mid].isGreater (card)) {
      return findBisect (card, deck, low, mid-1);
   } else {
      return findBisect (card, deck, mid+1, high);
   }
}
bool Card::isGreater (const Card& c2) const {
  if (suit > c2.suit) return true;
  if (suit < c2.suit) return false;

if (rank > c2.rank) return true;
  if (rank < c2.rank) return false;

return false;
}

The number of recursive calls is fairly small, typically 6 or 7. That means we only had to call equals and isGreater 6 or 7 times, compared to up to 52 times if we did a linear search. In general, bisection is much faster than a linear search, especially for large vectors.

Two common errors in recursive programs are forgetting to include a base case and writing the recursive call so that the base case is never reached. Either error will cause an infinite recursion, in which case C++ will (eventually) generate a run-time error.

Q-2: You are given a list of spelling words where the words are not sorted in any way. What search method should you use?

linear search
Correct! No search is faster than linear search when elements are not sorted.
bisection search
Incorrect! Bisection sort does not work on unsorted elements.
both methods will work, but linear search is more efficient
Incorrect! Bisection sort does not work on unsorted elements.
both methods will work, but bisection search is more efficient
Incorrect! Bisection sort does not work on unsorted elements.

Q-3: You are given the same list of spelling words, but this time the words are sorted alphabetically. What search method should you use this time?

linear search
Incorrect! You could use linear search, but it is not the only option.
bisection search
Incorrect! You could use bisection search, but it is not the only option.
both methods will work, but linear search is more efficient
Incorrect! Both methods will work, but linear search is not the most efficient method.
both methods will work, but bisection search is more efficient
Correct! When elements are sorted, bisection search is much quicker.

Q-4: When writing a recursive function, which of the following will result in infinite recursion?

having more than one recursive call
Incorrect! You are allowed to make multiple recursive calls inside of a function! You might do this if there is more than one condition.
not including a base case
Correct! You always need a base case!
writing recursive calls such that the base case is never reached
Correct! If you never reach the base case, the program will never stop making recursive calls.
having more than one base case
Incorrect! You are allowed to have multiple base cases. This is often necessary!

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12.9. Bisection search¶