Stanford splash spring 2016 basic programming

1. Stanford Splash Spring 2016
Basic Programming
Yu-Sheng (Yosen) Chen
chen.yosen@gmail.com

2. Self Introduction
Yosen Chen
● 1st year Master’s student in Stanford ICME (Institute for Computational &
Math Engineering). Got an EE Master’s degree in Taiwan in 2010.
● Will be a Facebook software intern this summer
● 5 years of industry working experience in Taiwan (software/algorithm
engineer)
● 2 years of research experience in Taiwan (computer vision)
● Publications: 3 camera patents (pending), 1 computer vision patent, 1 best
paper in IEEE ITC.
● Programming experience
○ C/C++ (7+ years), Python, Matlab, Perl, Java, ...

3. Why Programming?
● High salary?
● Change the way people communicate/work/build relationships?
● Redefine how a device can play a role in people’s lives?
● The world is flat: Everyone can learn programming once they have access to
the internet!
Google
services
FacebookFrom glassdoor.com

4. What Can You Learn from This Lecture?
● Basic programming concept (control flows, functions, iterative, recursive, ...)
● Basic data structure concept: what tools can we use to deal with problems
● Basic complexity concept: how much work does it take? how many times of
basic operations, ex:
○ 10*100=1000, less than 10 operations
○ 10+10+10+...+10=1000, at least 100 operations
● Be able to deal with basic programming problems
● How to learn on your own (Programming learning never ends!!)
Note that in this lecture, I will only use C++, but the concepts are generally the
same for all languages.

5. Basic Programming Concept

6. Control Flows
● if/else:
○ if (condition) statement1 else statement2
○
● While:
○ while (condition) statement
○
○
Output: 10, 9, 8, 7, 6, 5, 4, 3, 2, 1,
What if we write like this: while(true) { /*do something*/ }
→ it will never end unless we add some “break;” inside the loop
body.
Reference: http://www.cplusplus.com/doc/tutorial/control/

7. Control Flows (cont’)
● for:
○ for (initialization①; condition②; increase③) {statement④;}
○ How it works
■ A. initialization is executed. This is executed a single time, at the beginning of the loop.
■ B. condition is checked. If it is true, the loop continues; otherwise, the loop ends.
■ C. statement(s) is executed.
■ D. Increase (or decrease) is executed, then the loop gets back to step B.
○ It goes like:
■ Start → ①②{④③②}{④③②}{④③②}...{④③②} → End
■ Start → ①② → End
○ for (;;) is identical to while(1)
Reference http://www.cplusplus.com/doc/tutorial/control/

8. Control Flows (cont’)
● switch:
○ to check for a value among a number of possible constant expressions.
Reference http://www.cplusplus.com/doc/tutorial/control/

9. Functions
● A function is a group of statements that together perform a task. Every C++
program has at least one function, which is main(), and all the most trivial
programs can define additional functions. You can divide up your code into
separate functions.
● A function can have its input(s) and output(s)
● Functions can call other functions, including itself (recursive calls)
● We can create variables inside the functions, but they will be released when
we exit from the function.
Return_type Function_name (input1, input2, …, inputN)
{
Statement1; Statement2; ...
return Output;
}
Reference http://www.tutorialspoint.com/cplusplus/cpp_functions.htm

10. Recursive vs. Iterative Method
● Recursive function: a function who calls itself.
○ It will recursively call itself until it reaches some stopping condition (usually its base case)
● Iterative method uses a loop structure (while, for), it terminates when loop
condition fails.
● Let’s see how to solve a problem in both way. Problem: 1*2*3*...*n = ?
Execution result:
3628800
3628800

11. Complexity ( the smaller, the better)
● Complexity: What is the order of the number of basic operations?
● Given an input with size N, how many basic operations will be executed in this
function/algorithm? A measure of algorithm efficiency.
● Let’s see some examples
○ Calculate 1+2+3+...+N = ?
■ Naive method: 1+2+3+...+N, Complexity: (N-1) additions.
■ Better method: (1+N)*N/2, Complexity: 1 add, 1 div, and 1 mult.
○ Is N a prime number?
■ Naive method: check if all the remainders of N/k, where k = 2, 3, 4, ... N-1, are nonzero,
then it’s a prime number. Complexity: N-2 divisions
■ Better method: check if all the remainders of N/k, where k=2, 3, 4, … sqrt(N), are nonzero,
then it’s a prime number. Complexity: sqrt(N) -1 divisions
Note: A composite number N is a positive integer which is not prime (i.e., which has
factors other than 1 and itself, at least one factor <= sqrt(N))

12. Basic Data Structures

13. Array vs. Linked-List
Array
Linked-List
Remove element
Insert element
Picture credit: java67.blogspot.com https://en.wikipedia.org/wiki/Linked_list
Array Linked-List
Adv. random access,
no extra memory
Distributed memory;
fast insert, remove, resize
Disadv
.
Slow insert, remove, resize iterating access;
need memory to store next
address

14. Queue vs. Stack
Queue: First in first out (operations: enqueue, dequeue, see back/front)
Stack: First in last out, like crowded caltrain or elevators (operations: push/pop, see top)
Picture credit: En.wikipedia.org www.creativebloq.com quincypublicschools.com

15. More!
● Trees
● Hash table
Picture credit: Collegelabs.co http://www.tutorialspoint.com/data_structures_algorithms/tree_data_structure.htm
class TreeNode
{
int data;
TreeNode* leftChild;
TreeNote* rightChild;
};
By Jorge Stolfi - Own work, CC BY-SA 3.0, https://commons.wikimedia.org/w/index.php?curid=6471238
A small phone book as a
hash table
The beauty of hash table is
search/insert/delete in constant time,
regardless of the size of the table

16. Let’s solve some coding
problems!

17. Let’s See What Problems We Can Solve Now!
● Surprisingly, we now are able to solve some real interview questions!!
● Problem#1: Number of the trailing zeros in a factorial
● Problem#2: valid parenthesis expression → symmetricity problem
● Problem#3: Fibonacci sequence
● All the above have been asked in real job interviews. (software engineer, data
scientist, trader, …)

18. Problem#1: Num of Trailing Zeros
● Given an integer n, return the number of trailing zeroes in n!
● Analysis:
○ Don’t try to calculate the n factorial (it could be very large, overflow!)
○ Idea A: number of trailing zeros = min{p, q} where p and q are the exponent numbers of
2p
and 5q
in n!’s prime factor decomposition.
○ Ex: 210
*56
= 24
x(2x5)6
= 24
x(10)6
→ 6 trailing zeros
○ So we need to count the 2’s and 5’s exponent numbers for all numbers between 1 and
n?
○ Actually, we just have to count 5’s exponents, since num of 5’s is less than 2’s
EX: 10!
1x2x3x4x5x6x7x8x9x10, we have 1+2+1+3+1=8 2’s, but only 2 5’s
min{8, 2} is 2.
2’s multiples are way more than 5’s multiples!

19. Problem#1: Num of Trailing Zeros (cont’)
● So, how to count the number of 5’s?
● Idea B: for i=1 to n, we accumulate the exponents of 5 in i’s prime factor
decomposition
Convert this into code:
EX: 28!
1x2x3x4x5x6x7x8x9x10x...x15x...x20x...x25x…x28
Count: 1 1 1 1 2
Sum: 1 2 3 4 6 → 6 trailing zeros

20. Problem#1: Num of Trailing Zeros (cont’)
In fact, we have a better way to do it (no need to iterate all number from 1 to n)
Idea C: EX: 28!
1x2x3x4x5x6x7x8x9x10x...x15x...x20x...x25x…x28
Count: 1 1 1 1 2 Sum = 6 → 6 trailing zeros
28/5 = 5 … 3, which means among 1 to 28, 5 of them are 5’s multiples (exponent of 5 >=1)
28/25 = 1 … 3, which means among this 5 multiples, 1 of them have exponent of 5 >=2
So the sum = 5 + 1 = 6.
One more example, EX: 152!
1x...x5x...x10x...x15x...x20x...x25x…x50x...x100x...x125x...x150x151x152
1 1 1 1 2 2 2 3 2
152/5 = 30 … 2, which means among 1 to 152, 30 of them are 5’s multiples (exponent of 5 >=1)
152/25 = 6 … 2, which means among this 30 multiples, 6 of them have exponent of 5 >=2
152/125 = 1...27, which means among this 6 multiples, 1 of them has exponent of 5 >=3
So the sum = 30+6+1 = 37.

21. Problem#1: Num of Trailing Zeros (cont’)
Idea C:
Let’s convert it into code (iterative) i=5 cnt = 0 + 30
i=25 cnt = 30 + 6
i=125 cnt = 36 + 1
i=625 Condition fails
Return cnt=37
One more example, EX: 152!
1x...x5x...x10x...x15x...x20x...x25x…x50x...x100x...x125x...x150x151x152
1 1 1 1 2 2 2 3 2
152/5 = 30 … 2, which means among 1 to 152, 30 of them are 5’s multiples (exponent of 5 >=1)
152/25 = 6 … 2, which means among this 30 multiples, 6 of them have exponent of 5 >=2
152/125 = 1...27, which means among this 6 multiples, 1 of them has exponent of 5 >=3
So the sum = 30+6+1 = 37.

22. Problem#1: Num of Trailing Zeros (cont’)
Interpret idea C into a recursive method (only 1 line!)
Complexity comparison between Idea A, idea B, Idea C
Ex: n = 152
trailingZeros(152) = 152/5(=30) + trailingZeros(30);
trailingZeros(30) = 30/5(=6) + trailingZeros(6);
trailingZeros(6) = 6/5(=1) + trailingZeros(1);
trailingZeros(1) = 1/5(=0) + trailingZeros(0);
trailingZeros(0) = 0; so finally we have 30+6+1+0+0=37
Idea A Idea B Idea C
Details Count both exp of 2 and
5, find min of them
Only count exp of 5 Count exp of 5 in a
clever way
Complexity Double of Idea B’s Iterate every number
from 1 to n
Iterate only 5, 25, 125,
…, n

23. Problem#2: Valid Parenthesis Expression
● A good example to make use of stack
● Given a string, tell if it’s a valid parenthesis expression (all are paired properly)
○ () → OK
○ (()) → OK
○ (()()(())) → OK
○ ()( → X
○ ))(( → X
○ (()()()()()()()()()()()()()()()()()()()()()()())) → X
● Solution: use a stack to deal with all paired parenthesis, if finally the stack is
not empty, then it’s not a valid expression

24. Problem#2: Valid Parenthesis Expression (cont’)
● Solution:
○ Iterate through the input string, from the first characters to the last one
■ If we see a ‘(’, then push it into the stack (waiting for being paired)
■ If we see a ‘)’, then check the top of the stack,
● if it’s a ‘(’, good, then pop it, which means the incoming ‘)’ is paired with a ‘(’
● Otherwise (i.e., stack is empty), the incoming ‘)’ cannot be paired, return false
○ In the end
■ If the stack is not empty, then it means some are not paired, return false
■ If the stack is empty, then all the parentheses are paired, return true
Let’s see an example: ( ( ) ( ( ) ) ) )
iter# 1: ( 2: ( 3: ) 4: ( 5: ( 6: ) 7: ) 8: ) 9: )
Push ( Push ( Pop ( Push ( Push ( Pop ( Pop ( Pop ( Empty, can’t pop
return false!( (( ( (( ((( (( (

25. Problem#2: Valid Parenthesis Expression (cont’)
Convert it into code
● Iterate through the input string, from the first
characters to the last one
○ If we see a ‘(’, then push it into the
stack (waiting for being paired)
○ If we see a ‘)’, then check the top of
the stack,
■ if it’s a ‘(’, good, then pop it,
which means the incoming ‘)’ is
paired with a ‘(’
■ Otherwise (i.e., stack is empty),
the incoming ‘)’ cannot be
paired, return false
● In the end
○ If the stack is not empty, then it means
some are not paired, return false
○ If the stack is empty, then all the
parentheses are paired, return true

26. Problem#3: Fibonacci Sequence
● Fibonacci sequence, F(1)=F(2)=1; F(n) = F(n-1) + F(n-2)
● 1,1,2,3,5,8,13,21,34,55,89,144,.., how to write a function to find F(n) = ?
● Idea A: Intuitively, we can convert it into recursive functions
In fact, it’s very inefficient!
Lots of redundancy!
EX:
Why should we calculate F(3) 3
times and F(4) 2 times, and many
F(1)’s and F(2)’s.
In fact, F(n) has about 1.6n
recursive calls!

27. Problem#3: Fibonacci Sequence (cont’)
● Another approach to Fibonacci Sequence
● Idea B: for each iteration, we sum prev and prevprev as curr, then before next
iteration, we copy prev’s value to prevprev, and curr to prev
● Convert into iterative code (iterate n-2 times)
1 1 2 3 5 8 13 ...
pv=1, pvpv=1
curr = 1+1 = 2
Copy pv to pvpv (1)
Copy curr to pv (2)
pv=2, pvpv=1
curr = 2+1 = 3
Copy pv to pvpv (2)
Copy curr to pv (3)
pv=3, pvpv=2
curr = 3+2 = 5
Copy pv to pvpv (3)
Copy curr to pv (5)

28. Problem#3: Fibonacci Sequence (cont’)
● There are some even more advanced ideas!!
● F(n) can be solved in nearly const time! (regardless of how large n is)
● Involves Matrix computation and diagonalization! (topics in Colleges)

29. Wrap Up

30. Learning Programming is Free!!
● Some useful online resource
○ https://www.codecademy.com/ (Learn to code interactively, recommended for beginners)
○ https://codefights.com/home (Test your skills agains some company robots)
○ https://leetcode.com/ (LeetCode Online judges, and most important, find the best rated
solutions in its “discuss” tab!)
○ http://www.cplusplus.com/ (A good website to consult with for any C++ standard
function/library)
○ https://checkio.org/ (A good website to learn Python, like a RPG game)
○ https://docs.python.org (A good website to consult with for any Python standard libs)
● It’s not hard to find resources to learn coding, the problem is how to find
answers when you have problems.
○ Google it! It always works, except for using wrong keywords.

31. Any Advice for New Coders?
● Be expert in one single language (strong, complete), then know basics in other 2-3
languages
○ EX: like me, C++, Python, Matlab, Perl, Java, …
○ At least be familiar with 1 compiled language and a script one.
● Be patient for debugging
○ printing out all the values stage by stage is the dumbest, but most useful way.
● Learning never ends
○ new standard, new libs, new problems, new fields, new languages…
● If you can tell a computer how to solve a problem, that means you really understand the
solution. (because computers know nothing but only digits!)
● Programming is just an approach to solving a problem.
○ It is the knowledge of the solutions that really make you irreplaceable! Not the
code itself!!!

32. How to find this slides
Google: slideshare YosenChen, the first result!