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Programming Assignment 2: Algorithmic Warm-up
Introduction
Welcome to your second programming assignment of the Algorithmic Toolbox at Coursera! It consists of eight programming challenges. Three of them require you just to implement carefully the algorithms covered in the lectures. The remaining challenges will require you to first design an algorithm and then to implement it. For all the challenges, we provide starter solutions in C++, Java, and Python3. These solutions implement straightforward algorithms that usually work only for small inputs. To verify this, you may want to submit these solutions to the grader. This will usually give you a “time limit exceeded” message for Python starter files and either “time limit exceeded” or “wrong answer” message for C++ and Java solutions (the reason for wrong answer being an integer overflow issue). Your goal is to replace a naive algorithm with an efficient one. In particular, you may want to use the naive implementation for stress testing your efficient implementation.
In this programming assignment, the grader will show you the input data if your solution fails on any of the tests. This is done to help you to get used to the algorithmic problems in general and get some experience debugging your programs while knowing exactly on which tests they fail. However, for all the following programming assignments, the grader will show the input data only in case your solution fails on one of the first few tests.
Learning Outcomes
Upon completing this programming assignment you will be able to:
See the huge difference between a slow algorithm and a fast one.
Play with examples where knowing something interesting about a problem helps to design an algorithm that is much faster than a naive one.
Implement solutions that work much more faster than straightforward solutions for the following pro-gramming challenges:
compute a small Fibonacci number;
compute the last digit of a large Fibonacci number;
compute a huge Fibonacci number modulo ;
compute the last digit of a sum of Fibonacci numbers;
compute the last digit of a partial sum of Fibonacci numbers;
compute the greatest common divisor of two integers;
compute the least common multiple of two integers.
Implement the algorithms covered in the lectures, design new algorithms.
Practice implementing, testing, and debugging your solution. In particular, you will find out how in practice, when you implement an algorithm, you bump into unexpected questions and problems not covered by the general description of the algorithm. You will also check your understanding of the algorithm itself and most probably see that there are some aspects you did not think of before you had to actually implement it. You will overcome all those complexities, implement the algorithms, test them, debug, and submit to the system.
Passing Criteria: 4 out of 8
Passing this programming assignment requires passing at least 4 out of 8 programming challenges from this assignment. In turn, passing a programming challenge requires implementing a solution that passes all the tests for this problem in the grader and does so under the time and memory limits specified in the problem statement.
Contents
1
Fibonacci Number
3
2
Last Digit of a Large Fibonacci Number
4
3
Greatest Common Divisor
6
4
Least Common Multiple
7
5
Fibonacci Number Again
8
6
Last Digit of the Sum of Fibonacci Numbers
9
7
Last Digit of the Sum of Fibonacci Numbers Again
10
8
Last Digit of the Sum of Squares of Fibonacci Numbers
11
9
Appendix
12
9.1
Compiler Flags . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12
9.2
Frequently Asked Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Fibonacci Number
Problem Introduction
Recall the definition of Fibonacci sequence: 0 = 0, 1 = 1, and = −1 + −2 for ≥ 2. Your goal in this problem is to implement an efficient algorithm for computing Fibonacci numbers. The starter files for this problem contain an implementation of the following naive recursive algorithm for computing Fibonacci numbers in C++, Java, and Python3:
Fibonacci( ):
if ≤ 1:
return
return Fibonacci( − 1) + Fibonacci( − 2)
Try compiling and running a starter solution on your machine. You will see that computing, say, 40 already takes noticeable time.
Another way to appreciate the dramatic difference between an exponential time algo-rithm and a polynomial time algorithm is to use the following visualization by David Galles: http://www.cs.usfca.edu/~galles/visualization/DPFib.html. Try com-puting 20 by a recursive algorithm by entering “20” and pressing the “Fibonacci Re-cursive” button. You will see an endless number of recursive calls. Now, press “Skip Forward” to stop the current algorithm and call the iterative algorithm by pressing “Fibonacci Table”. This will compute 20 very quickly. (Note that the visualization uses a slightly different definition of Fibonacci numbers: 0 = 1 instead of 0 = 0. This of course has almost no influence on the running time.)
Problem Description
Task. Given an integer , find the th Fibonacci number .
Input Format. The input consists of a single integer .
Constraints. 0 ≤ ≤ 45.
Output Format. Output .
Sample 1.
Input:
10
Output:
55
10 = 55.
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Last Digit of a Large Fibonacci Number
Problem Introduction
Your goal in this problem is to find the last digit of -th Fibonacci number. Recall that Fibonacci numbers grow exponentially fast. For example,
200 = 280 571 172 992 510 140 037 611 932 413 038 677 189 525 .
Therefore, a solution like
[0]←0
[1]←1
for from 2 to :
[]← [ −1]+ [ −2]
print( [ ] mod 10)
will turn out to be too slow, because as grows the th iteration of the loop computes the sum of longer and longer numbers. Also, for example, 1000 does not fit into the standard C++ int type. To overcome this difficulty, you may want to store in [ ] not the th Fibonacci number itself, but just its last digit (that is, mod 10). Computing the last digit of is easy: it is just the last digit of the sum of the last digits of −1 and −2:
[ ] ← ( [ − 1] + [ − 2]) mod 10
This way, all [ ]’s are just digits, so they fit perfectly into any standard integer type, and computing a sum of [ − 1] and [ − 2] is performed very quickly.
Problem Description
Task. Given an integer , find the last digit of the th Fibonacci number (that is, mod 10).
Input Format. The input consists of a single integer .
Constraints. 0 ≤ ≤ 107.
Output Format. Output the last digit of .
Sample 1.
Input:
3
Output:
2
3=2.
Sample 2.
Input:
331
Output:
9
331 = 668 996 615 388 005 031 531 000 081 241 745 415 306 766 517 246 774 551 964 595 292 186 469.
Sample 3.
Input:
327305
Output:
5
327305 does not fit into one line of this pdf, but its last digit is equal to 5.
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Least Common Multiple
Problem Introduction
The least common multiple of two positive integers and is the least positive integer that is divisible by both and .
Problem Description
Task. Given two integers and , find their least common multiple.
Input Format. The two integers and are given in the same line separated by space.
Constraints. 1 ≤ , ≤ 107.
Output Format. Output the least common multiple of and .
Sample 1.
Input:
6 8
Output:
24
Among all the positive integers that are divisible by both 6 and 8 (e.g., 48, 480, 24), 24 is the smallest one.
Sample 2.
Input:
761457 614573
Output:
467970912861
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Fibonacci Number Again
Problem Introduction
In this problem, your goal is to compute modulo , where may be really huge: up to 1014. For such values of , an algorithm looping for iterations will not fit into one second for sure. Therefore we need to avoid such a loop.
To get an idea how to solve this problem without going through all for from 0 to , let’s see what happens when is small — say, = 2 or = 3.
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
0
1
1
2
3
5
8
13
21
34
55
89
144
233
377
610
mod 2
0
1
1
0
1
1
0
1
1
0
1
1
0
1
1
0
mod 3
0
1
1
2
0
2
2
1
0
1
1
2
0
2
2
1
Take a detailed look at this table. Do you see? Both these sequences are periodic! For = 2, the period is 011 and has length 3, while for = 3 the period is 01120221 and has length 8. Therefore, to compute, say, 2015 mod 3 we just need to find the remainder of 2015 when divided by 8. Since 2015 = 251 · 8 + 7, we conclude that 2015 mod 3 = 7 mod 3 = 1.
This is true in general: for any integer ≥ 2, the sequence mod is periodic. The period always starts with 01 and is known as Pisano period.
Problem Description
Task. Given two integers and , output mod (that is, the remainder of when divided by ).
Input Format. The input consists of two integers and given on the same line (separated by a space).
Constraints. 1 ≤ ≤ 1014, 2 ≤ ≤ 103.
Output Format. Output mod .
Sample 1.
Input:
239 1000
Output:
161
239 mod 1 000 = 39 679 027 332 006 820 581 608 740 953 902 289 877 834 488 152 161 (mod 1 000) = 161.
Sample 2.
Input:
2816213588 239
Output:
151
2 816 213 588 does not fit into one page of this file, but 2 816 213 588 mod 239 = 151.
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Last Digit of the Sum of Fibonacci Numbers
Problem Introduction
The goal in this problem is to find the last digit of a sum of the first Fibonacci numbers.
Problem Description
Task. Given an integer , find the last digit of the sum 0 + 1 + · · · + .
Input Format. The input consists of a single integer .
Constraints. 0 ≤ ≤ 1014.
Output Format. Output the last digit of 0 + 1 + · · · + .
Sample 1.
Input:
3
Output:
4
0+ 1+ 2+ 3=0+1+1+2=4.
Sample 2.
Input:
100
Output:
5
The sum is equal to 927 372 692 193 078 999 175, the last digit is 5.
What To Do
Instead of computing this sum in a loop, try to come up with a formula for 0 + 1 + 2 + · · · + . For this, play with small values of . Then, use a solution for the previous problem.
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Solution
A detailed solution (with Python code) for this challenge is covered in the companion MOOCBook. We strongly encourage you to do your best to solve the challenge yourself before looking into the book! There are at least three good reasons for this.
By solving this challenge, you practice solving algorithmic problems similar to those given at technical interviews.
The satisfaction and self confidence that you get when passing the grader is priceless =)
Even if you fail to pass the grader yourself, the time will not be lost as you will better understand the solution from the book and better appreciate the beauty of the underlying ideas.
Last Digit of the Sum of Fibonacci Numbers Again
Problem Introduction
Now, we would like to find the last digit of a partial sum of Fibonacci numbers: + +1 + · · · + .
Problem Description
Task. Given two non-negative integers and , where ≤ , find the last digit of the sum + +1 +
···+ .
Input Format. The input consists of two non-negative integers and separated by a space.
Constraints. 0 ≤ ≤ ≤ 1014.
Output Format. Output the last digit of + +1 + · · · + .
Sample 1.
Input:
3 7
Output:
1
3+ 4+ 5+ 6+ 7=2+3+5+8+13=31.
Sample 2.
Input:
10 10
Output:
5
10 = 55.
Sample 3.
Input:
10 200
Output:
2
10 + 11 + · · · + 200 = 734 544 867 157 818 093 234 908 902 110 449 296 423 262
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Last Digit of the Sum of Squares of Fibonacci Numbers
Problem Description
Task. Compute the last digit of 02 + 12 + · · · + 2.
Input Format. Integer .
Constraints. 0 ≤ ≤ 1014.
Output Format. The last digit of 02 + 12 + · · · + 2.
Sample 1.
Input:
7
Output:
3
02+ 12+···+ 72 =0+1+1+4+9+25+64+169=273.
Sample 2.
Input:
73
Output:
1
02 + · · · + 732 = 1 052 478 208 141 359 608 061 842 155 201.
Sample 3.
Input:
1234567890
Output:
0
What To Do
Since the brute force search algorithm for this problem is too slow ( may be as large as 1018), we need to come
up with a simple formula for 02 + 12 + · · ·+ 2. The figure below represents the sum 12 + 22 + 32 + 42 + 52 as the area of a rectangle with vertical side 5 = 5 and horizontal side 5 + 4 = 3 + 5 = 6.
3
5
1
2
1
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Appendix
9.1 Compiler Flags
(gcc 7.4.0). File extensions: .c. Flags:
gcc - pipe - O2 - std = c11 < filename > - lm
C++ (g++ 7.4.0). File extensions: .cc, .cpp. Flags:
g ++ - pipe - O2 - std = c ++14 < filename > - lm
If your C/C++ compiler does not recognize -std=c++14 flag, try replacing it with -std=c++0x flag or compiling without this flag at all (all starter solutions can be compiled without it). On Linux and MacOS, you most probably have the required compiler. On Windows, you may use your favorite compiler or install, e.g., cygwin.
C# (mono 4.6.2). File extensions: .cs. Flags:
mcs
Go (golang 1.13.4). File extensions: .go. Flags
go
Haskell (ghc 8.0.2). File extensions: .hs. Flags:
ghc - O2
Java (OpenJDK 1.8.0_232). File extensions: .java. Flags:
javac - encoding UTF -8
java - Xmx1024m
JavaScript (NodeJS 12.14.0). File extensions: .js. No flags:
nodejs
Kotlin (Kotlin 1.3.50). File extensions: .kt. Flags:
kotlinc
java - Xmx1024m
Python (CPython 3.6.9). File extensions: .py. No flags:
python3
Ruby (Ruby 2.5.1p57). File extensions: .rb.
ruby
Rust (Rust 1.37.0). File extensions: .rs.
rustc
Scala (Scala 2.12.10). File extensions: .scala.
scalac
9.2 Frequently Asked Questions
Why My Submission Is Not Graded?
You need to create a submission and upload the source file(rather than the executable file) of your solution. Make sure that after uploading the file with your solution you press the blue “Submit” button at the bottom. After that, the grading starts, and the submission being graded is enclosed in an orange rectangle. After the testing is finished, the rectangle disappears, and the results of the testing of all problems are shown.
What Are the Possible Grading Outcomes?
There are only two outcomes: “pass” or “no pass.” To pass, your program must return a correct answer on all the test cases we prepared for you, and do so under the time and memory constraints specified in the problem statement. If your solution passes, you get the corresponding feedback "Good job!" and get a point for the problem. Your solution fails if it either crashes, returns an incorrect answer, works for too long, or uses too much memory for some test case. The feedback will contain the index of the first test case on which your solution failed and the total number of test cases in the system. The tests for the problem are numbered from 1 to the total number of test cases for the problem, and the program is always tested on all the tests in the order from the first test to the test with the largest number.
Here are the possible outcomes:
Good job! Hurrah! Your solution passed, and you get a point!
Wrong answer. Your solution outputs incorrect answer for some test case. Check that you consider all the cases correctly, avoid integer overflow, output the required white spaces, output the floating point numbers with the required precision, don’t output anything in addition to what you are asked to output in the output specification of the problem statement.
Time limit exceeded. Your solution worked longer than the allowed time limit for some test case. Check again the running time of your implementation. Test your program locally on the test of max-imum size specified in the problem statement and check how long it works. Check that your program doesn’t wait for some input from the user which makes it to wait forever.
Memory limit exceeded. Your solution used more than the allowed memory limit for some test case. Estimate the amount of memory that your program is going to use in the worst case and check that it does not exceed the memory limit. Check that your data structures fit into the memory limit. Check that you don’t create large arrays or lists or vectors consisting of empty arrays or empty strings, since those in some cases still eat up memory. Test your program locally on the tests of maximum size specified in the problem statement and look at its memory consumption in the system.
Cannot check answer. Perhaps the output format is wrong. This happens when you output something different than expected. For example, when you are required to output either “Yes” or “No”, but instead output 1 or 0. Or your program has empty output. Or your program outputs not only the correct answer, but also some additional information (please follow the exact output format specified in the problem statement). Maybe your program doesn’t output anything, because it crashes.
Unknown signal 6 (or 7, or 8, or 11, or some other). This happens when your program crashes. It can be because of a division by zero, accessing memory outside of the array bounds, using uninitialized variables, overly deep recursion that triggers a stack overflow, sorting with a contradictory comparator, removing elements from an empty data structure, trying to allocate too much memory, and many other reasons. Look at your code and think about all those possibilities. Make sure that you use the same compiler and the same compiler flags as we do.
∙ Internal error: exception... Most probably, you submitted a compiled program instead of a source code.
Grading failed. Something wrong happened with the system. Report this through Coursera or edX Help Center.
May I Post My Solution at the Forum?
Please do not post any solutions at the forum or anywhere on the web, even if a solution does not pass the tests (as in this case you are still revealing parts of a correct solution). Our students follow the Honor Code: “I will not make solutions to homework, quizzes, exams, projects, and other assignments available to anyone else (except to the extent an assignment explicitly permits sharing solutions).”
Do I Learn by Trying to Fix My Solution?
My implementation always fails in the grader, though I already tested and stress tested it a lot. Would not it be better if you gave me a solution to this problem or at least the test cases that you use? I will then be able to fix my code and will learn how to avoid making mistakes. Otherwise, I do not feel that I learn anything from solving this problem. I am just stuck.
First of all, learning from your mistakes is one of the best ways to learn.
The process of trying to invent new test cases that might fail your program is difficult but is often enlightening. Thinking about properties of your program makes you understand what happens inside your program and in the general algorithm you’re studying much more.
Also, it is important to be able to find a bug in your implementation without knowing a test case and without having a reference solution, just like in real life. Assume that you designed an application and an annoyed user reports that it crashed. Most probably, the user will not tell you the exact sequence of operations that led to a crash. Moreover, there will be no reference application. Hence, it is important to learn how to find a bug in your implementation yourself, without a magic oracle giving you either a test case that your program fails or a reference solution. We encourage you to use programming assignments in this class as a way of practicing this important skill.
If you have already tested your program on all corner cases you can imagine, constructed a set of manual test cases, applied stress testing, etc, but your program still fails, try to ask for help on the forum. We encourage you to do this by first explaining what kind of corner cases you have already considered (it may happen that by writing such a post you will realize that you missed some corner cases!), and only afterwards asking other learners to give you more ideas for tests cases.
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