I was reading book about competitive programming and was encountered to problem where we have to count all possible paths in the n*n matrix.
Now the conditions are :
`
1. All cells must be visited for once (cells must not be unvisited or visited more than once)
2. Path should start from (1,1) and end at (n,n)
3. Possible moves are right, left, up, down from current cell
4. You cannot go out of the grid
Now this my code for the problem :
typedef long long ll;
ll path_count(ll n,vector<vector<bool>>& done,ll r,ll c){
ll count=0;
done[r][c] = true;
if(r==(n-1) && c==(n-1)){
for(ll i=0;i<n;i++){
for(ll j=0;j<n;j++) if(!done[i][j]) {
done[r][c]=false;
return 0;
}
}
count++;
}
else {
if((r+1)<n && !done[r+1][c]) count+=path_count(n,done,r+1,c);
if((r-1)>=0 && !done[r-1][c]) count+=path_count(n,done,r-1,c);
if((c+1)<n && !done[r][c+1]) count+=path_count(n,done,r,c+1);
if((c-1)>=0 && !done[r][c-1]) count+=path_count(n,done,r,c-1);
}
done[r][c] = false;
return count;
}
Here if we define recurrence relation then it can be like: T(n) = 4T(n-1)+n2
Is this recurrence relation true? I don't think so because if we use masters theorem then it would give us result as O(4n*n2) and I don't think it can be of this order.
The reason, why I am telling, is this because when I use it for 7*7 matrix it takes around 110.09 seconds and I don't think for n=7 O(4n*n2) should take that much time.
If we calculate it for n=7 the approx instructions can be 47*77 = 802816 ~ 106. For such amount of instruction it should not take that much time. So here I conclude that my recurrene relation is false.
This code generates output as 111712 for 7 and it is same as the book's output. So code is right.
So what is the correct time complexity??
No, the complexity is not O(4^n * n^2).
Consider the 4^n in your notation. This means, going to a depth of at most n - or 7 in your case, and having 4 choices at each level. But this is not the case. In the 8th, level you still have multiple choices where to go next. In fact, you are branching until you find the path, which is of depth n^2.
So, a non tight bound will give us O(4^(n^2) * n^2). This bound however is far from being tight, as it assumes you have 4 valid choices from each of your recursive calls. This is not the case.
I am not sure how much tighter it can be, but a first attempt will drop it to O(3^(n^2) * n^2), since you cannot go from the node you came from. This bound is still far from optimal.
Related
Why is it that when I see example code for binary search there is never an if statement to check if the start of the array or end is the target?
import java.util.Arrays;
public class App {
public static int binary_search(int[] arr, int left, int right, int target) {
if (left > right) {
return -1;
}
int mid = (left + right) / 2;
if (target == arr[mid]) {
return mid;
}
if (target < arr[mid]) {
return binary_search(arr, left, mid - 1, target);
}
return binary_search(arr, mid + 1, right, target);
}
public static void main(String[] args) {
int[] arr = { 3, 2, 4, -1, 0, 1, 10, 20, 9, 7 };
Arrays.sort(arr);
for (int i = 0; i < arr.length; i++) {
System.out.println("Index: " + i + " value: " + arr[i]);
}
System.out.println(binary_search(arr, arr[0], arr.length - 1, -1));
}
}
in this example if the target was -1 or 20 the search would enter recursion. But it added an if statement to check if target is mid, so why not add two more statements also checking if its left or right?
EDIT:
As pointed out in the comments, I may have misinterpreted the initial question. The answer below assumes that OP meant having the start/end checks as part of each step of the recursion, as opposed to checking once before the recursion even starts.
Since I don't know for sure which interpretation was intended, I'm leaving this post here for now.
Original post:
You seem to be under the impression that "they added an extra check for mid, so surely they should also add an extra check for start and end".
The check "Is mid the target?" is in fact not a mere optimization they added. Recursively checking "mid" is the whole point of a binary search.
When you have a sorted array of elements, a binary search works like this:
Compare the middle element to the target
If the middle element is smaller, throw away the first half
If the middle element is larger, throw away the second half
Otherwise, we found it!
Repeat until we either find the target or there are no more elements.
The act of checking the middle is fundamental to determining which half of the array to continue searching through.
Now, let's say we also add a check for start and end. What does this gain us? Well, if at any point the target happens to be at the very start or end of a segment, we skip a few steps and end slightly sooner. Is this a likely event?
For small toy examples with a few elements, yeah, maybe.
For a massive real-world dataset with billions of entries? Hm, let's think about it. For the sake of simplicity, we assume that we know the target is in the array.
We start with the whole array. Is the first element the target? The odds of that is one in a billion. Pretty unlikely. Is the last element the target? The odds of that is also one in a billion. Pretty unlikely too. You've wasted two extra comparisons to speed up an extremely unlikely case.
We limit ourselves to, say, the first half. We do the same thing again. Is the first element the target? Probably not since the odds are one in half a billion.
...and so on.
The bigger the dataset, the more useless the start/end "optimization" becomes. In fact, in terms of (maximally optimized) comparisons, each step of the algorithm has three comparisons instead of the usual one. VERY roughly estimated, that suggests that the algorithm on average becomes three times slower.
Even for smaller datasets, it is of dubious use since it basically becomes a quasi-linear search instead of a binary search. Yes, the odds are higher, but on average, we can expect a larger amount of comparisons before we reach our target.
The whole point of a binary search is to reach the target with as few wasted comparisons as possible. Adding more unlikely-to-succeed comparisons is typically not the way to improve that.
Edit:
The implementation as posted by OP may also confuse the issue slightly. The implementation chooses to make two comparisons between target and mid. A more optimal implementation would instead make a single three-way comparison (i.e. determine ">", "=" or "<" as a single step instead of two separate ones). This is, for instance, how Java's compareTo or C++'s <=> normally works.
BambooleanLogic's answer is correct and comprehensive. I was curious about how much slower this 'optimization' made binary search, so I wrote a short script to test the change in how many comparisons are performed on average:
Given an array of integers 0, ... , N
do a binary search for every integer in the array,
and count the total number of array accesses made.
To be fair to the optimization, I made it so that after checking arr[left] against target, we increase left by 1, and similarly for right, so that every comparison is as useful as possible. You can try this yourself at Try it online
Results:
Binary search on size 10: Standard 29 Optimized 43 Ratio 1.4828
Binary search on size 100: Standard 580 Optimized 1180 Ratio 2.0345
Binary search on size 1000: Standard 8987 Optimized 21247 Ratio 2.3642
Binary search on size 10000: Standard 123631 Optimized 311205 Ratio 2.5172
Binary search on size 100000: Standard 1568946 Optimized 4108630 Ratio 2.6187
Binary search on size 1000000: Standard 18951445 Optimized 51068017 Ratio 2.6947
Binary search on size 10000000: Standard 223222809 Optimized 610154319 Ratio 2.7334
so the total comparisons does seem to tend to triple the standard number, implying the optimization becomes increasingly unhelpful for larger arrays. I'd be curious whether the limiting ratio is exactly 3.
To add some extra check for start and end along with the mid value is not impressive.
In any algorithm design the main concerned is moving around it's complexity either it is time complexity or space complexity. Most of the time the time complexity is taken as more important aspect.
To learn more about Binary Search Algorithm in different use case like -
If Array is not containing any repeated
If Array has repeated element in this case -
a) return leftmost index/value
b) return rightmost index/value
and many more point
I am working on developing an efficient iterative code to compute mn. After some thinking and googling I found this code;
public static int power(int n, int m)
// Efficiently calculates m to the power of n iteratively
{
int pow=m, acc=1, count=n;
while(count!=0)
{
if(count%2==1)
acc=acc*pow;
pow=pow*pow;
count=count/2;
}
return acc;
}
This logic makes sense to me except the fact that why are we squaring value of pow towards the end each time. I am familiar with similar recursive approach, but this squaring is not looking very intuitive to me. Can I kindly get some help her? An example with explanation will be really helpful.
The accumulator is being squared each iteration because count (which is the inverse cumulative power) is being halved each iteration.
If the count is odd, the accumulator is multiplied by the number. This algorithm relies on integer arithmetic, which discards the fractional part of a division, effectively further decrementing by 1 when the count is odd.
This is a very tricky solution to understand. I am solving this problem in leetcode and have found the iterative solution. I have spent a whole day to understand this beautiful iterative solution. The main problem is this iterative solution does not work as like its recursive solution.
Let's pick an example to demonstrate. But first I have to re-write your code by replacing some variable name as the name in your given code is very confusing.
// Find m^n
public static int power(int n, int m)
{
int pow=n, result=1, base=m;
while(pow > 0)
{
if(pow%2 == 1) result = result * base;
base = base * base;
pow = pow/2;
}
return result;
}
Let's understand the beauty step by step.
Let say, base = 2 and power = 10.
Calculation
Description
2^10= (2*2)^5 = 4^5
even
We have changed the base to 4 and power to 5. So it is now enough to find 4^5. [base multiplied with itself and power is half
4^5= 4*(4)^4 = 4^5
odd
We separate single 4 which is the base for current iteration. We store the base to result variable. We will now find the value of 4^4 and then multiply with the result variable.
4^4= (4*4)^2 = 16^2
even
We change the base to 16 and power to 2. It is now enough to find 16^2
16^2= (16*16)^1 = 256^1
even
We change the base to 256 and power to 1. It is now enough to find 256^1
256^1 = 256 * 256^0
odd
We separate single 256 which is the base for current iteration. This value comes from evaluation of 4^4.So, we have to multiply it with our previous result variable. And continue evaluating the rest value of 256^0.
256^0
zero
Power 0. So stop the iteration.
So, after translating the process in pseudo code it will be similar to this,
If power is even:
base = base * base
power /= 2
If power is odd:
result = result * base
power -= 1
Now, let have another observation. It is observed that floor(5 / 2) and (5-1) / 2 is same.
So, for odd power, we can directly set power / 2 instead of power -= 1. So, the pseudo code will be like the below,
if power is both odd or even:
base = base * base
power /= 2
If power is odd:
result = result * base
I hope you got the behind the scene.
Despite the last 30 minutes i spent on trying to understand time and space complexity better, i still can't confidently determine those for the algorithm below:
bool checkSubstr(std::string sub)
{
//6 OR(||) connected if statement(checks whether the parameter
//is among the items in the list)
}
void checkWords(int start,int end)
{
int wordList[2] ={0};
int j = 0;
if (start < 0)
{
start = 0;
}
if (end>cAmount)
{
end = cAmount -1;
}
if (end-start < 2)
{
return;
}
for (int i = start; i <= end-2; i++)
{
if (crystals[i] == 'I' || crystals[i] == 'A')
{
continue;
}
if (checkSubstr(crystals.substr(i,3)))
{
wordList[j] = i;
j++;
}
}
if (j==1)
{
crystals.erase(wordList[0],3);
cAmount -= 3;
checkWords(wordList[0]-2,wordList[0]+1);
}
else if (j==2)
{
crystals.erase(wordList[0],(wordList[1]-wordList[0]+3));
cAmount -= wordList[1]-wordList[0]+3;
checkWords(wordList[0]-2,wordList[0]+1);
}
}
The function basically checks a sub-string of the whole string for predetermined (3 letter, e.g. "SAN") combinations of letters. Sub-string length can be 4-6 no real way to determine, depends on the input(pretty sure it's not relevant, although not 100%).
My reasoning:
If there are n letters in the string, worst case scenario, we have to check each of them. Again depending on the input, this can be done 3 ways.
All 6 length sub-strings: If this is the case the function runs n/6 times, each running 8(or 10?) processes, which(i think) means that its time complexity is O(n).
All 4 length sub-strings: Pretty much the same reason above, O(n).
4 and 6 length sub-strings mixed: Can't see why this would be different than previous 2. O(n)
As for the space complexity, i am completely lost. However, i have an idea:
If the function recurs for maximum amount of time,it will require:
n/4 x The Amount Used In One Run
which made me think it should be O(n). Although, i'm not convinced this is correct. I thought maybe seeing someone else's thought process on this example would help me understand how to calculate time and space complexity better.
Thank you for your time.
EDIT: Let me provide clearer information. We read a combination of 6 different letters into a string, this can be (almost)any combination in any length. 'crystals' is the string, and we are looking for 6 different 3 letter combinations in that list of letters. Sort of like a jewel matching game. Now the starting list contains no matches(none of the 6 predetermined combinations exist in the first place). Therefore the only way matches can occur from then on is by swaps or matches disappearing. Once a swap is processed by top level code, the function is called to check for matches, and if a match is found the function recurs after deleting the "match" part of the string.
Now let's look at how the code is looking for a match. To demonstrate a swap of 2 letters:
ABA B-R ZIB(no spaces or '-' in the actual string, used for better demonstration),
B and R is being swapped. This swap only effects the 6 letters starting from 2nd letter and ending on 7th letter. In other words, the letters the first A and last B can form a match with are same, before and after the swap, thus no point checking for matches including those words. So a sub-string of 6 letters sent to the checking algorithm. Similarly, if a formed match disappears(gets deleted from the string) the range of effected letters is 4. So when i thought of a worst case scenario, i imagined either 1 swap creating a whole chain reaction and matching all the way till there are not enough letters to form a match, or each match happens with a swap. Again, i am not saying this is how we should think when calculating time and space complexity but this is how the code works. Hope this is clear enough if not let me know and i can provide more details. It's also important to note that swap amount and places are a part of the input we read.
EDIT: Here is how the function is called on top level for the first time:
checkWords(swaps[i]-2,swaps[i]+3);
Sub-string length can be 4-6 no real way to determine, depends on the
input (pretty sure it's not relevant, although not 100%).
That's not what the code shows; the line if (checkSubstr(crystals.substr(i,3))) conveys that substrings always have exactly 3 characters. If the substring length varies, it is relevant, since your naive substring match will degrade to O(N*M) in the general case, where N is start-end+1 (the size of the input string) and M is the size of the substring being searched. This happens because in the worst case you'll compare M characters for each of the N characters of the source string.
The rest of this answer assumes that substrings are of size 3, since that's what the code shows.
If substrings are always 3 characters long, it's different: you can essentially assume checkSubstr() is O(1) because you will always compare at most 3 characters. The bulk of the work happens inside the for loop, which is O(N), where N is end-1-start.
After the loop, in the worst case (when one of the ifs is entered), you erase a bunch of characters from crystal. Assuming this is a string backed by an array in memory, this is an O(cAmount) operation, because all elements after wordList[0] must be shifted. The recursive call always passes in a range of size 4; it does not grow nor shrink with the size of the input, so you can also say there are O(1) recursive calls.
Thus, time complexity is O(N+cAmount) (where N is end-1-start), and space complexity is O(1).
Introduction
I need to split an array filled with a certain type (let's take water buckets for example) with two values set (in this case weight and volume), while keeping the difference between the total of the weight to a minimum (preferred) and the difference between the total of the volumes less than 1000 (required). This doesn't need to be a full-fetched genetic algorithm or something similar, but it should be better than what I currently have...
Current Implementation
Due to not knowing how to do it better, I started by splitting the array in two same-length arrays (the array can be filled with an uneven number of items), replacing a possibly void spot with an item with both values being 0. The sides don't need to have the same amount of items, I just didn't knew how to handle it otherwise.
After having these distributed, I'm trying to optimize them like this:
func (main *Main) Optimize() {
for {
difference := main.Difference(WEIGHT)
for i := 0; i < len(main.left); i++ {
for j := 0; j < len(main.right); j++ {
if main.DifferenceAfter(i, j, WEIGHT) < main.Difference(WEIGHT) {
main.left[i], main.right[j] = main.right[j], main.left[i]
}
}
}
if difference == main.Difference(WEIGHT) {
break
}
}
for main.Difference(CAPACITY) > 1000 {
leftIndex := 0
rightIndex := 0
liters := 0
weight := 100
for i := 0; i < len(main.left); i++ {
for j := 0; j < len(main.right); j++ {
if main.DifferenceAfter(i, j, CAPACITY) < main.Difference(CAPACITY) {
newLiters := main.Difference(CAPACITY) - main.DifferenceAfter(i, j, CAPACITY)
newWeight := main.Difference(WEIGHT) - main.DifferenceAfter(i, j, WEIGHT)
if newLiters > liters && newWeight <= weight || newLiters == liters && newWeight < weight {
leftIndex = i
rightIndex = j
liters = newLiters
weight = newWeight
}
}
}
}
main.left[leftIndex], main.right[rightIndex] = main.right[rightIndex], main.left[leftIndex]
}
}
Functions:
main.Difference(const) calculates the absolute difference between the two sides, the constant taken as an argument decides the value to calculate the difference for
main.DifferenceAfter(i, j, const) simulates a swap between the two buckets, i being the left one and j being the right one, and calculates the resulting absolute difference then, the constant again determines the value to check
Explanation:
Basically this starts by optimizing the weight, which is what the first for-loop does. On every iteration, it tries every possible combination of buckets that can be switched and if the difference after that is less than the current difference (resulting in better distribution) it switches them. If the weight doesn't change anymore, it breaks out of the for-loop. While not perfect, this works quite well, and I consider this acceptable for what I'm trying to accomplish.
Then it's supposed to optimize the distribution based on the volume, so the total difference is less than 1000. Here I tried to be more careful and search for the best combination in a run before switching it. Thus it searches for the bucket switch resulting in the biggest capacity change and is also supposed to search for a tradeoff between this, though I see the flaw that the first bucket combination tried will set the liters and weight variables, resulting in the next possible combinations being reduced by a big a amount.
Conclusion
I think I need to include some more math here, but I'm honestly stuck here and don't know how to continue here, so I'd like to get some help from you, basically that can help me here is welcome.
As previously said, your problem is actually a constrained optimisation problem with a constraint on your difference of volumes.
Mathematically, this would be minimise the difference of volumes under constraint that the difference of volumes is less than 1000. The simplest way to express it as a linear optimisation problem would be:
min weights . x
subject to volumes . x < 1000.0
for all i, x[i] = +1 or -1
Where a . b is the vector dot product. Once this problem is solved, all indices where x = +1 correspond to your first array, all indices where x = -1 correspond to your second array.
Unfortunately, 0-1 integer programming is known to be NP-hard. The simplest way of solving it is to perform exhaustive brute force exploring of the space, but it requires testing all 2^n possible vectors x (where n is the length of your original weights and volumes vectors), which can quickly get out of hands. There is a lot of literature on this topic, with more efficient algorithms, but they are often highly specific to a particular set of problems and/or constraints. You can google "linear integer programming" to see what has been done on this topic.
I think the simplest might be to perform a heuristic-based brute force search, where you prune your search tree early when it would get you out of your volume constraint, and stay close to your constraint (as a general rule, the solution of linear optimisation problems are on the edge of the feasible space).
Here are a couple of articles you might want to read on this kind of optimisations:
UCLA Linear integer programming
MIT course on Integer programming
Carleton course on Binary programming
Articles on combinatorial optimisation & linear integer programming
If you are not familiar with optimisation articles or math in general, the wikipedia articles provides a good introduction, but most articles on this topic quickly show some (pseudo)code you can adapt right away.
If your n is large, I think at some point you will have to make a trade off between how optimal your solution is and how fast it can be computed. Your solution is probably suboptimal, but it is much faster than the exhaustive search. There might be a better trade off, depending on the exact configuration of your problem.
It seems that in your case, difference of weight is objective, while difference of volume is just a constraint, which means that you are seeking for solutions that optimize difference of weight attribute (as small as possible), and satisfy the condition on difference of volume attribute (total < 1000). In this case, it's a single objective constrained optimization problem.
Whereas, if you are interested in multi-objective optimization, maybe you wanna look at the concept of Pareto Frontier: http://en.wikipedia.org/wiki/Pareto_efficiency . It's good for keeping multiple good solutions with advantages in different objective, i.e., not losing diversity.
I remember many years back, when I was in school, one of my computer science teachers taught us that it was better to check for 'trueness' or 'equality' of a condition and not the negative stuff like 'inequality'.
Let me elaborate - If a piece of conditional code can be written by checking whether an expression is true or false, we should check the 'trueness'.
Example: Finding out whether a number is odd - it can be done in two ways:
if ( num % 2 != 0 )
{
// Number is odd
}
or
if ( num % 2 == 1 )
{
// Number is odd
}
(Please refer to the marked answer for a better example.)
When I was beginning to code, I knew that num % 2 == 0 implies the number is even, so I just put a ! there to check if it is odd. But he was like 'Don't check NOT conditions. Have the practice of checking the 'trueness' or 'equality' of conditions whenever possible.' And he recommended that I use the second piece of code.
I am not for or against either but I just wanted to know - what difference does it make? Please don't reply 'Technically the output will be the same' - we ALL know that. Is it a general programming practice or is it his own programming practice that he is preaching to others?
NOTE: I used C#/C++ style syntax for no reason. My question is equally applicable when using the IsNot, <> operators in VB etc. So readability of the '!' operator is just one of the issues. Not THE issue.
The problem occurs when, later in the project, more conditions are added - one of the projects I'm currently working on has steadily collected conditions over time (and then some of those conditions were moved into struts tags, then some to JSTL...) - one negative isn't hard to read, but 5+ is a nightmare, especially when someone decides to reorganize and negate the whole thing. Maybe on a new project, you'll write:
if (authorityLvl!=Admin){
doA();
}else{
doB();
}
Check back in a month, and it's become this:
if (!(authorityLvl!=Admin && authorityLvl!=Manager)){
doB();
}else{
doA();
}
Still pretty simple, but it takes another second.
Now give it another 5 to 10 years to rot.
(x%2!=0) certainly isn't a problem, but perhaps the best way to avoid the above scenario is to teach students not to use negative conditions as a general rule, in the hopes that they'll use some judgement before they do - because just saying that it could become a maintenance problem probably won't be enough motivation.
As an addendum, a better way to write the code would be:
userHasAuthority = (authorityLvl==Admin);
if (userHasAuthority){
doB();
else{
doA();
}
Now future coders are more likely to just add "|| authorityLvl==Manager", userHasAuthority is easier to move into a method, and even if the conditional is reorganized, it will only have one negative. Moreover, no one will add a security hole to the application by making a mistake while applying De Morgan's Law.
I will disagree with your old professor - checking for a NOT condition is fine as long as you are checking for a specific NOT condition. It actually meets his criteria: you would be checking that it is TRUE that a value is NOT something.
I grok what he means though - mostly the true condition(s) will be orders of magnitude smaller in quantity than the NOT conditions, therefore easier to test for as you are checking a smaller set of values.
I've had people tell me that it's to do with how "visible" the ping (!) character is when skim reading.
If someone habitually "skim reads" code - perhaps because they feel their regular reading speed is too slow - then the ! can be easily missed, giving them a critical mis-understanding of the code.
On the other hand, if a someone actually reads all of the code all of the time, then there is no issue.
Two very good developers I've worked with (and respect highily) will each write == false instead of using ! for similar reasons.
The key factor in my mind is less to do with what works for you (or me!), and more with what works for the guy maintaining the code. If the code is never going to be seen or maintained by anyone else, follow your personal whim; if the code needs to be maintained by others, better to steer more towards the middle of the road. A minor (trivial!) compromise on your part now, might save someone else a week of debugging later on.
Update: On further consideration, I would suggest factoring out the condition as a separate predicate function would give still greater maintainability:
if (isOdd(num))
{
// Number is odd
}
You still have to be careful about things like this:
if ( num % 2 == 1 )
{
// Number is odd
}
If num is negative and odd then depending on the language or implementation num % 2 could equal -1. On that note, there is nothing wrong with checking for the falseness if it simplifies at least the syntax of the check. Also, using != is more clear to me than just !-ing the whole thing as the ! may blend in with the parenthesis.
To only check the trueness you would have to do:
if ( num % 2 == 1 || num % 2 == -1 )
{
// Number is odd
}
That is just an example obviously. The point is that if using a negation allows for fewer checks or makes the syntax of the checks clear then that is clearly the way to go (as with the above example). Locking yourself into checking for trueness does not suddenly make your conditional more readable.
I remember hearing the same thing in my classes as well. I think it's more important to always use the more intuitive comparison, rather than always checking for the positive condition.
Really a very in-consequential issue. However, one negative to checking in this sense is that it only works for binary comparisons. If you were for example checking some property of a ternary numerical system you would be limited.
Replying to Bevan (it didn't fit in a comment):
You're right. !foo isn't always the same as foo == false. Let's see this example, in JavaScript:
var foo = true,
bar = false,
baz = null;
foo == false; // false
!foo; // false
bar == false; // true
!bar; // true
baz == false; // false (!)
!baz; // true
I also disagree with your teacher in this specific case. Maybe he was so attached to the generally good lesson to avoid negatives where a positive will do just fine, that he didn't see this tree for the forest.
Here's the problem. Today, you listen to him, and turn your code into:
// Print black stripe on odd numbers
int zebra(int num) {
if (num % 2 == 1) {
// Number is odd
printf("*****\n");
}
}
Next month, you look at it again and decide you don't like magic constants (maybe he teaches you this dislike too). So you change your code:
#define ZEBRA_PITCH 2
[snip pages and pages, these might even be in separate files - .h and .c]
// Print black stripe on non-multiples of ZEBRA_PITCH
int zebra(int num) {
if (num % ZEBRA_PITCH == 1) {
// Number is not a multiple of ZEBRA_PITCH
printf("*****\n");
}
}
and the world seems fine. Your output hasn't changed, and your regression testsuite passes.
But you're not done. You want to support mutant zebras, whose black stripes are thicker than their white stripes. You remember from months back that you originally coded it such that your code prints a black stripe wherever a white strip shouldn't be - on the not-even numbers. So all you have to do is to divide by, say, 3, instead of by 2, and you should be done. Right? Well:
#define DEFAULT_ZEBRA_PITCH 2
[snip pages and pages, these might even be in separate files - .h and .c]
// Print black stripe on non-multiples of pitch
int zebra(int num, int pitch) {
if (num % pitch == 1) {
// Number is odd
printf("*****\n");
}
}
Hey, what's this? You now have mostly-white zebras where you expected them to be mostly black!
The problem here is how think about numbers. Is a number "odd" because it isn't even, or because when dividing by 2, the remainder is 1? Sometimes your problem domain will suggest a preference for one, and in those cases I'd suggest you write your code to express that idiom, rather than fixating on simplistic rules such as "don't test for negations".