An interesting small programming puzzle.
Abubakar Sultan who was my senior at NetSol and the best Microsoft.NET programmer I have seen so far in Pakistani industry, has made this programming question and asked many to solve this.
For any given n(positive integer only), you have to print integers from 1 to n and then print from n-1 to 1 using only a loop without any explicit if-else or ternary operators and not using any other memory location except loop variable.
Example: if n = 7 then out put should be 1 2 3 4 5 6 7 6 5 4 3 2 1
He claimed no one have been able to solve this in first try. However I solved this problem and come up with the follwoing solution. I am pasting the C++ source to print the same.
int n = 7 ; // set any integer value here
for( int i = 1; i <= (2* n) ; i++)
cout << (( i <=n) * i) + ((i > n) * (i – ( i – n ) * 2));
This is really an interesting programming puzzle and I enjoyed working on it.
Haroon
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Firstly, as stated in description “print integers from 1 to n and then print from n-1 to 1” but your code is also printing 0 that is not required. So change for loop condition as
“i n)*(i-n)
I am replying again because post is not viewable correctly.
Firstly, as stated in description “print integers from 1 to n and then print from n-1 to 1” but your code is also printing 0 that is not required. So change for loop condition as
“i n)*(i-n)
I am replying 3rd time because post is not viewable correctly.
Firstly, as stated in description “print integers from 1 to n and then print from n-1 to 1” but your code is also printing 0 that is not required. So change for loop condition as
“i n)*(i-n)
Slaam,
Haroon bhai I was just reading your blog, suddenly I got this programming puzzle. I found it interesting and after almost half an hour, I had my own solution. I am sending you my code and I did not see your code until I had my own solution. One thing about my solutoin is that it contains only arithmetic operators and no comparison operators.
for(int i =1;i
I think there is some problem. I did paste my code , but that is not viewable 😦
Anyways, I also did it in my first attempt.
It’s going to sound nuts but I’m trying to locate a Win CE programmer named Zaheer Ahmad out of India. Would that be you? If so, please go to the web site for Giraffe Publications (add .com) about a project if you’re interested. Sorry for being off subject here everyone… Small world, big internet… -B.
main(){
int n=7,i;
for(i=1;in)*(2*n -i));
}
}
Is the abs() function allowed?
You can just do:
for i = 1 to 2n-1
print (n – abs(n-i));
Good question though!
int i, n = 7;
for( i = 1; i n)*(i-n)
Haroon bhai, you can delete this comment, i was just checking a thing.
int i, n = 7;
for( i = 1; i n)*(i-n)
Haroon bhai, you can delete this comment, i was just checking a thing.
int i, n = 7;for( i = 1; i n)*(i-n)
Just analysing the problem in blog viewer window
int i, n = 7;
for( i = 1; i “”n)*(i-n)”
This is what I came up with, I haven’t tested it…
int i, n=7;
for( i=-(n-1); i <= (n-1); ++i)
printf( “%d “, n-abs(i) );
Hi.
I do not think your answer is correct or maybe I’m being to strict about “using only a loop without any explicit if-else or ternary operators and not using any other memory location except loop variable.” …
because in c# you just could use
for (int i = 1; i < 2 * n ; i++) {
Console.WriteLine(“{0}”, (i <= 7)?i:(2*n-i));
}
i think it would be the same that when u compare, please let me know if u can.
btw, loved ur blog, several interesting things,
cheers.
bruno from Lima, Peru
Haroon is definitely wrong — he used a ternary operator that is explicitly disallowed.
Neil is also wrong. No fair using abs() which just hides the ternary operator inside a subroutine.
The solution you want is more like:
#include
int main() {
int i, n = 7;
for (i = 1; i < 2*n; i++) {
printf(“%d “, i + (((i%(n+1))-i) * 2 * (i-n) / (n+1)));
}
}
// pseudo code
for i = (-n+1) to (n-1)
print (n+i)*(i0) // prints n-1 to 1
//example:
for n = 5:
for i = -4 to +4
print (5+i)*(i0) // prints 4 to 1
But I think using (i0) is like using conditional statements. This does not work in Java for example.
Can be done is abs() but could be done without that too.
// pseudo code
for i = (-n+1) to (n-1)
print (n+i)*(i less than or equals 0) // prints 1 to n
+ (n-i)*(i greater than 0) // prints n-1 to 1
Hi!,
for(int i=1; i<n*2; i++)
System.out.println(i-2*((i-n)*(i/n)));
abs(i) === (i^2)^(1/2)
Here are a couple more programming puzzles to puzzle you…
Good Luck!!
hello
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