Transcript Week 6 Tutorial

Week 6
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Lab 1 and 2 results
 Common mistakes in Style
 Lab 1 common mistakes in Design
 Lab 2 common mistakes in Design
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Tips on PE preparation
Tutorial Questions for Week 6
 Might not be able to cover all of them
 I’ll explain Question x, x, and x
 If you would like to explain another question, raise
hands and I’ll explain
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Discussion group? Name? Program
description? (-1)
Function description (-1)
 Did not write or write in weird places
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Bad identifiers Name (-2)
 If you want to use k, n, at least comment it
somewhere to tell what they are
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Superfluous comments (-1)
 //executable statement
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Indentation is wrong in all your programs
No deduction on marks this time
Try to utilize the auto-indentation feature in
your vim editor
Better don’t use tab for indentation; it causes
problem on Coursemarker.
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One exercise is 10 marks. In total 30 marks.
1 Attempt mark is given when you try all
exercises
The majority of you are in the range 29~30
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Did not write functions at all, when the
question explicitly asks to write some
functions. (-5 for each omitted function)
Declare a function which return integer; but
does not use the returned value in main
function at all (-2)
Complex or redundant if else statements (-2)
 If (x%2==0) {…} else if (x%2==1) {…} else {…}
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Output result in functions which are
supposed to do computation only
 Normally we do input and output in main
function; and complex computation is carried out
by functions. Unless the function is specifically
designed to output (print_cookies)
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Very Serious Mistakes
 Missed the Deadline
▪ 0 mark for everything…even the attempt mark.
 Used Recursion (explicitly forbidden in question)
▪ 0 mark for correctness and 0 mark for design
▪ But since this is first time, so I’m more lenient (30 marks for
correctness)
 Did not test at all before submission (failed all test
cases)
▪ 0 mark for correctness
▪ Even if some test cases happen to be true due to hard coding
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Take any natural number n. If n is even, divide
it by 2 to get n/2; if n is odd, triple it and add 1
to obtain 3n + 1. Repeat the process
indefinitely. No matter what number your
start with, you will always eventually reach 1.
While loop needed
n=1 is terminating condition
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//some function description here…
int count_iterations(int n)
{
int count = 0; // number of iterations
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while (n > 1)
{
if (n%2 == 1)
n = 3*n + 1;
else
n /= 2;
count++;
}
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return count;
}
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Alexandra has n candles. He burns them one
at a time and carefully collects all unburnt
residual wax. Out of the residual wax of
exactly k (where k > 1) candles, he can roll out
a new candle.
n/k gives new rolled out candles
n%k gives candles left
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// This function computes the total number of candles burned.
int count_candles (int n, int k)
{
int candles_burned = 0;
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while (n >= k) // n is the total number of candles, k is the residual wax.
{
candles_burned = candles_burned + ((n/k) * k);
// candles burned is the total number of residual wax sets multiply
//the number of candles in that set.
n = (n/k) + (n%k);
// n is the number of new candles plus the previous remaining
//candles.
}
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candles_burned = candles_burned + n;
return candles_burned;
}
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int count_candles(int candles, int residuals)
{
int actual_residuals = candles; // initial number of residuals
int new_candles;
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while (actual_residuals >= residuals)
{
new_candles = actual_residuals/residuals;
candles += new_candles;
actual_residuals = actual_residuals%residuals + new_candles;
}
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return candles;
}
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Inefficiency if you write this
int count_candles (int n, int k)
{
int i = n;
// Initial number of candles
int count = 0; // Number of new candles
while (n >= k) // Pre-condition, execution stops when n < k
{
n -= (k - 1); // After each set of k candles are burnt
count ++; // 1 new candle can be rolled out
}
return i + count;
// Total numbers of candles equals to initial number
// of candles plus new candles created.
}
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Write a program cookies.c to read in a
positive integer and add up its digits
repeatedly until the sum is a single digit. For
example, if the integer is 12345, then adding
its digits (1 + 2 + 3 + 4 + 5) yields 15, and
adding its digits again (1 + 5) yields 6. Hence
the answer is 6.
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Some of you uses only one while loop (did not
consider the case when yielded-sum is not a
single digit again)
int sum_digits (int number)
{
while (number >= 10)
{
number %= 10;
number += number;
}
return number;
}
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Some of you did
not consider the
case when the
input integer is
already a single
digit.
int sum_digits(int number)
{
int total;
int temp;
while(number >= 10)
{
total = 0;
while(number != 0)
{
temp = number % 10;
total += temp;
number /= 10;
}
number = total;
}
return total;
}
int sum_digits (int number)
{
int digit1, digit2, digit3, digit4, digit5, digit6,
digit7, digit8, digit9, digit10;
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Some of
you write
algorithm
which is
not
general
while (number >= 10)
{
digit1 = number / 1000000000;
digit2 = (number % 1000000000) / 100000000;
digit3 = (number % 100000000) / 10000000;
digit4 = (number % 10000000) / 1000000;
digit5 = (number % 1000000) / 100000;
digit6 = (number % 100000) / 10000;
digit7 = (number % 10000) / 1000;
digit8 = (number % 1000) / 100;
digit9 = (number % 100) /10;
digit10 = (number % 10);
number = digit1 + digit2 + digit3 + digit4 + digit5
+ digit6 + digit7 + digit8 +
digit9 +digit10;
}
return number;
}
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Read your Lab 1 and Lab 2 feedback
Understand all lecture notes covered
Do some hands-on practice
Good sleep before that day