This programming problem...

[Nineshadow]

So I came across this problem on a computer on which some grade 5 kids had had programming the hour before I came into the class.

 

Determine all the numbers which have this property :

The square of the sum of its digits is equal to the number itself, which must have at least 2 digits and is natural.

 

A few minutes later, I realized what a dumb programming problem it was.

Seriously...the only number which fulfills that property is 81.

Yup.

Only 1 number.

My amazing solution in O(1) :

cout << 81;

I know that programming problems which can be solved with just 1 formula are kinda bad, but this one...lol.

 

 

Well, at least I discovered an interesting property of the number 81.

[79wjd]

Ummm, no? 

 

Let's assume it's 81. 

 

Then the square of the sum of it's digits (9) would be 3 =/= 81. 

[Guest]

11/10

[werto165]

Ummm, no? 

 

Let's assume it's 81. 

 

Then the square of the sum of it's digits (9) would be 3 =/= 81. 

square  not square root. 

[Nineshadow]

Ummm, no? 

 

Let's assume it's 81. 

 

Then the square of the sum of it's digits (9) would be 3 =/= 81.

It's squared, not sqrt.

[Enderman]

well cout<<81 is not actually a formula...thats just directly outputting the answer

 

you would need to actually write code to go through and check every number and then output the answer once it finds it

[Nineshadow]

well cout<<81 is not actually a formula...thats just directly outputting the answer

 

you would need to actually write code to go through and check every number and then output the answer once it finds it

Fine then.

int answer = 81;

[Jykos]

0 and 1

 

google it. 

[Enderman]

 

Fine then.

int answer = 81;

that's still not a formula...

it doesnt find anything

youre just giving it an answer

[Nineshadow]

0 and 1

 

google it. 

Numbers with at least 2 digits, forgot to mention.

[Jykos]

 

Fine then.

int answer = 81;

 

Less efficent

 

that's still not a formula...

it doesnt find anything

youre just giving it an answer

A good program is a lazy program. (least effort)

[Nineshadow]

that's still not a formula...

it doesnt find anything

youre just giving it an answer

But there is , mathematically , no other number with that property. What's the purpose of searching for others?

The formula is literally

f(x) = 81

That's the most efficient way to do it , of course.

You can still go ahead and check the proprieties for all numbers up to a certain limit, but that would be pointless.

[Enderman]

But there is , mathematically , no other number with that property. What's the purpose of searching for others?

The formula is literally

f(x) = 81

That's the most efficient way to do it , of course.

You can still go ahead and check the proprieties for all numbers up to a certain limit, but that would be pointless.

but its not the program solving that problem, its you

the only thing the program does is print it to the screen

[Jykos]

But there is , mathematically , no other number with that property. What's the purpose of searching for others?

The formula is literally

f(x) = 81

That's the most efficient way to do it , of course.

You can still go ahead and check the proprieties for all numbers up to a certain limit, but that would be pointless.

Programming competitions (yes they exist, NZ programming competition for example) are judged on how many problems you can solve (with varying points). The check to see if the program works simply throws a ton of expected and boundary inputs at it and sees if the out put is correct. The original program would do that, plus efficiency score for a tie breaker. 

[Nineshadow]

but its not the program solving that problem, its you

the only thing the program does is print it to the screen

So if I were given this problem :

Find all prime numbers which are also even.

I should go ahead and check through all numbers and see if it's prime and even at the same time  ? Of course not, the only solution is 2.

[Jykos]

but its not the program solving that problem, its you

the only thing the program does is print it to the screen

It,s just a bad problem.

 

Would make for a decent math one where you have to prove that 81 is the only answer but not a programming problem.

[Nineshadow]

Programming competitions (yes they exist, NZ programming competition for example) are judged on how many problems you can solve (with varying points). The check to see if the program works simply throws a ton of expected and boundary inputs at it and sees if the out put is correct. The original program would do that, plus efficiency score for a tie breaker. 

Exactly.

Here we focus on programming competitions and heavily on efficiency of programs.

[Jykos]

If you asked a computer to call a file named "file.file". Then it uses a directory to go strait there rather than search the entire computer.

[Jykos]

Exactly.

Here we focus on programming competitions and heavily on efficiency of programs.

And if it works, it works.

[79wjd]

So if I were given this problem :

Find all prime numbers which are also even.

I should go ahead and check through all numbers and see if it's prime and even at the same time  ? Of course not, the only solution is 2.

That wouldn't teach the same lesson (which was presumably bit shifting -- or being lazy and just separating numbers). (albeit there are definitely better ways to teach the same lesson...but whatever)

int main(int argc, char ** argv){    int x = 0, y=0;    for(int i=10 ; i<100; i++){ //yes, I could do something a bit more complex that will                                 //calculate the length of the number, but I'm lazy           x = i/10;           y = (i % x*10)/10;           if((x+y)*(x+y) == i)                 printf("ANSWER: %d\n", i);    }    return 0; }

[Philosobyte]

If a competition wants to ensure that the competitors actually used the concept, then one possibility would be to add the requirement of the code working for any non-decimal numeral system. 

[Enderman]

So if I were given this problem :

Find all prime numbers which are also even.

I should go ahead and check through all numbers and see if it's prime and even at the same time  ? Of course not, the only solution is 2.

yes you should

because a computer doesn't know that 2 is the only prime number

only you do

 

having a computer solve a problem by you putting the answer into it is not making the computer solve a problem

[Jykos]

If a competition wants to ensure that the competitors actually used the concept, then one possibility would be to add the requirement of the code working for any non-decimal numeral system.

which would make a quite good problem.

Input: base then number of digits

Output: possible numbers where their root=sum of digits.

Example:

10

2

81

[Nineshadow]

yes you should

because a computer doesn't know that 2 is the only prime number

only you do

 

having a computer solve a problem by you putting the answer into it is not making the computer solve a problem

But Maths is still Maths, doesn't matter if it's a computer doing it or a human.

[Enderman]

But Maths is still Maths, doesn't matter if it's a computer doing it or a human.

no...

if the point of the problem is to solve something using code, solving it yourself and inputting the answer so that the code prints it is not making the code solve the problem