# Lecture 20 — Searching¶

## Overview¶

• Notion of an algorithm
• Problems:
• Finding the two smallest values in a list
• Finding the index of a particular value in a list
• Finding the index of a particular value in a sorted list
• Sorting a list (Lecture 21)
• Analyzing our solutions:
• Mathematically
• Experimental timing

Material for Lectures 20 and 21 is in Chapters 12 and 13 of the text.

## Algorithm¶

• Precise description of the steps necessary to solve a computing problem
• Description is intended for people to read and understand
• Starts with English sentences
• Gradually, the sentences are made more detailed and more like programming statements
• Allows us to lay out the basic steps of the program before getting to the details.
• A program is an implementation of one or more algorithms.

## Multiple Algorithms¶

• Often there are many different algorithms that can solve a problem.
• They differ in:
• Ease of understanding
• Ease of implementation
• Efficiency
• All three considerations are important and their relative weight depends on the context.

## Problem 1: Finding the Two Smallest Values in a List¶

• Given a list of integers, floats, or any other values that can be compared with a less than operation, find the two smallest values in the list AND their indices in the list
• We need to be careful with this problem formulation: are duplicates allowed? does it matter?

## Brainstorming Session¶

1. Outline two or more approaches to finding the indices of the two smallest values in a list.
3. Write a more detailed description of the solutions.
4. How might your approaches change if we just have to find the values and not the indices?

## Evaluating Our Solutions Analytically¶

We’ve already covered this briefly in Lecture 15.

• Count the number of steps as a function of the size of the list.
• Usually we use $$N$$ as a variable to indicate this size.
• Informally, if the number of operations is (roughly) proportional to $$N$$ we write $$O(N)$$ (read as “order of N”)
• If the number of operations is proportional to $$N log N$$ we write $$O(N log N)$$.
• Importantly, the best sorting algorithms, including the one implemented in Python for lists, are $$O(N log N)$$.
• We will informally apply this analysis to our solution approaches.

## Evaluating Our Solutions Experimentally¶

• Needs:
• generate example data, and
• time our algorithm implementations.
• Experimental data can be generated using the random module. We will make use of
• randrange
• shuffle
• Timing uses the time module and its time function, which returns the number of seconds (as a float) since an arbitrary start time called an “epoch”.
• We will compute the difference between a start time and an end time as our timing measurement.

## Completing the Solutions¶

• We will implement two of the algorithms we came up with to find the indices of the two smallest values in the list:

import random
import time

def index_two_v1( values ):
pass  # not implemented yet

def index_two_v2( values ):
pass # not implemented yet

if __name__ == "__main__":
n = int(input("Enter the number of values to test ==> "))
values = list(range(0,n))
random.shuffle( values )

s1 = time.time()
(i0,i1) = index_two_v1(values)
t1 = time.time() - s1
print("Ver 1:  indices ({},{}); time {:.3f} seconds".format(i0,i1,t1))

s2 = time.time()
(j0,j1) = index_two_v2(values)
t2 = time.time() - s2
print("Ver 2:  indices ({},{}); time {:.3f} seconds".format(j0,j1,t2))


We will experiment with these implementations.

## Searching for a Value¶

• Problem: given a list of values, L, and given a single value, x, find the (first) index of x in L or determine that x is not in L.
• Basic algorithm is straightforward, and requires $$O(N)$$ steps
• We can solve this in Python using a combination of in and index, or by writing our own loop.
• The text book discusses a number of variations on the algorithm.
• We will implement our own variation as an exercise.

## Algorithm and Implementation¶

• We need to keep track of two indices:

• low: all values in the list at locations 0..low-1 are less than x
• high: all values in the list at locations high ..N are greater than or equal to x. Write N as the length of the list.
• Initialize low = 0 and high = N.

• In each iteration of a while loop

• Set mid to be the average of low and high.
• Update the value of low or high based on comparing x to L[mid].
• Here is the actual code:

def binary_search( x, L):
low = 0
high = len(L)
while low != high:
mid = (low+high)//2
if x > L[mid]:
low = mid+1
else:
high = mid
return low


## Practice¶

1. Using

L = [ 1.3, 7.9, 11.2, 15.3, 18.5, 18.9, 19.7 ]


what are the values of low, high and mid each time through the while loop for the calls

binary_search( 11.2, L )

binary_search( 19.1, L )

binary_search( -1, L)

binary_search( 25, L)

2. How many times will the loop execute for $$N = 1,000$$ or $$N = 1,000,000$$? (You will not be able to come up with an exact number, but you should be able to come close.) How does this compare to the linear search?

3. Would the code still work if we changed the > to the >=? Why?

4. Modify the code to return a tuple that includes both the index where x is or should be inserted and a boolean that indicates whether or not x is in the list.

We will also perform experimental timing runs if we have time at the end of class.

## Summary¶

• Algorithm vs. implementation
• Criteria for choosing an algorithm: speed, clarity, ease of implementation
• Timing/speed evaluations can be either analytical or experimental.
• Searching for indices of two smallest values
• Linear search
• Binary search of a list that is ordered.