Friday's Coding Challenge

Jason Hooper

Does the language have to already exist?

OriginalGriff

"GIMMEE CODEZ!!!" in Q&A should do it...

Ideological Purity is no substitute for being able to stick your thumb down a pipe to stop the water

BobJanova

Well, if not, it is obviously possible to invent a language where one symbol does this operation :P. Since "sort and take" seems to be how to do it, it would be easy to conceive of a language where putting two symbols for "take" and "sort" next to each other would create a composite function that did it, making two characters the sensible theoretical minimum.

PIEBALDconsult

Chris Maunder wrote:

How does SQL handle them?

You tell me how you want them handled and I'll tell SQL how. It's SQL, not LISP.

PIEBALDconsult

f(x)

Jason Hooper

Yeah exactly. Since I'm the first (i think) to propose it but I don't feel like spending 5 hours creating an interpreter, let's just pretend I'm the winner.

Jason

BobJanova

I could easily modify my Rowan interpreter (what I posted my solution in) to implement sort as a single symbol, but it isn't worth it just for something like this :-\ . I don't allow composition of functions by juxtaposition like that though because it's confusing (see J).

Rajesh R Subramanian

std::list li; //A list, so that it can be sorted in ascending order.
int i = 0, /*i is increment counter*/ n = 3; //n denotes how many small numbers to print.
while(i++<5) li.push_back(i*3); //fill the list with dummy data.
li.sort(); //Small numbers will go to the end in ascending order.
std::list::iterator it = li.begin();

**while(n-- && it!=li.end()) std::cout << \*(it++) << std::endl;** //This is the line you're looking for. :)

"Real men drive manual transmission" - Rajesh.

Roger Wright

X| If we're allowed to cheat by using third party software, I'd 1. Copy and paste to Minitab 2. Click on Graphs/Stem and Leaf The top line of numbers in the displayed graph are the smallest values in the set, ordered ascending. :-D

Will Rogers never met me.

Gary Wheeler

You win the "We Can't Judge Your Submission Because It's Incomprehensible" award.

Software Zen: delete this;

parths

void find_n_in_m(int *mData, int *nData, unsigned int m, unsigned int n)
{
unsigned int ctr, ctr2 = 0;
for(unsigned int i = 0; i < n; i++)
nData[i] = mData[i];
m--;
while (((ctr = m) && (m >= n))
&& ((mData[ctr] >= nData[ctr2]) ?
((++ctr2 < n) || (ctr2 = 0) || m--) :
(((mData[ctr] ^= nData[ctr2]) && (nData[ctr2] ^= mData[ctr]) && (mData[ctr] ^= nData[ctr2]) && (ctr2 = 0)) || 1)
));
}

- Doesn't give a sorted list - Modifies the input array (nothing mentioned against that in the specification but that can be avoided by adding an input argument for an temp / scratch buffer of the same size as the source buffer) - Tried it on VS2008 Express Ed and with a few basic data so I don't know if it's correct for all inputs. - What it does is fill the destination with the first n elements then goes about trying to see if it can place the elements from n to m into the new array. Thought I'd post it while I'm trying to improve ( :~ ) it.

"It was when I found out I could make mistakes that I knew I was on to something." -Ornette Coleman "Philosophy is a study that lets us be unhappy more intelligently." -Anon.

Single Step Debugger

void Test()
{
int arr[] = {1, 4, 6, 8, 9};
int nNumToGet = 3;

int \*arr2 = ReturnLowerN(arr, sizeof(arr)/sizeof(int), nNumToGet);

}

int* ReturnLowerN(int *arr, int arrSize, int nNum)
{
std::sort(arr, arr + 5, std::greater<int>());

return(arr + arrSize - nNum);

}

Of course if you want to use qsort instead, here is an article from someone you may know. :-D Using qsort on arrays of sequential data[^]

There is only one Vera Farmiga and Salma Hayek is her prophet! Advertise here – minimum three posts per day are guaranteed.

Daniel Grunwald

BobJanova wrote:

Well, if not, it is obviously possible to invent a language where one symbol does this operation

Why one symbol? If you're inventing a new language, you might as well invent one where the empty program solves this challenge.

Roger Wright

I wasn't aware that the Betty Ford Clinic has a hamster wing. How very nice. I hope they enjoy the rest. :-D

Will Rogers never met me.

Luc Pattyn

I was thinking about APL too. It has been over 30 years I touched it though. :)

Luc Pattyn [My Articles] Nil Volentibus Arduum

YvesDaoust

My attempt in plain C:

for (i= 0; i < m; i++)
{
int r= 0;
for (j= 0; j < m; j++)
r+= a[j] < a[i];
if (r < n)
printf("%d\n", a[i]);
}

It simply evaluates the rank of every element. Unfortunately, this method cannot meet the specs in case of equal elements. Actually, it reports all elements with rank less than n.

YvesDaoust

Slightly modified Straight Selection Sort will do the trick (moves n lowest elements first):

for (int i= 0; i < n; i++)
{
int k = i;
for (int j= i; j < m; j++)
{
if (a[k] > a[j])
{
k = j;
}
}

int swap= a\[i\]; a\[i\]= a\[k\]; a\[k\]= swap;

}

IMHO, allowing function calls makes the challenge nonsensical, as the solutions reduces to S(a, m, n), where S is the function that does just that.

YvesDaoust

If you don't care about time, sort the whole array and parameter n is virtually useless !

S(a, m); // Solution in a[0..n-1]

where S stands for some sorting algorithm on an array. Slightly shorter in Python, assuming a has length m:

S(a) # Solution in a[0..n-1]

George Danila

var nSmallestNumbers = numbers.OrderBy(x => x).Take(n);

BobJanova

Thanks but it's actually simple and the same as everyone else's:

f←{
sorted←⍵⌷⍨⍋⍵; // sorts the right argument
⍺↑sorted // first n items (n = left argument)
}