Comment On Nerds, Jocks, and Lockers

public static void main(String[] args) {
int num = 100;
int i=1;
int k=1;
while(k<num){
k=i*i;
i++;
System.out.println(k);
}

}

#!perl

sub lockers { return map $_**2, 1..sqrt shift }

i don't have a method of solving this problem, but i can clearly see the pattern when it's done - closed lockers in runs of 2,4,6,8,10,12,14,16,18 with a single open locker between them

how delightful!

Umm squares of 1 to 10?

Captcha "eros" .....I love you too!

Or to put it more simply- The open lockers are i^2 where i runs from 1-10.

The open lockers are perfect squares.

The reasoning is simple: non-perfect squares have an even number of factors. The jocks toggle a number for every factor it has. Therefore every door should be toggled an even number of times and should therefore end closed (the starting state).

However, perfect squares have an odd number of unique factors, because one of the factors is repeated. Because of this, perfect squares are toggled for their square root which has no pair to reverse the toggle, so the door ends up open (the opposite state it started in).

Make sense?

Paste this formula in Excel and use copy-down

=ROW(A1)^2

Welbog:

The open lockers are perfect squares.

The reasoning is simple: non-perfect squares have an even number of factors. The jocks toggle a number for every factor it has. Therefore every door should be toggled an even number of times and should therefore end closed (the starting state).

However, perfect squares have an odd number of unique factors, because one of the factors is repeated. Because of this, perfect squares are toggled for their square root which has no pair to reverse the toggle, so the door ends up open (the opposite state it started in).

Make sense?

Somehow I think that makes it less fun.

vbs file, work in nearly any windows

a = inputbox("locker?")

for i = 1 to a
l = l+1
b = b & i & " "
i = i + (2*l)
next

msgbox b

Addison:

Welbog:

The open lockers are perfect squares.

The reasoning is simple: non-perfect squares have an even number of factors. The jocks toggle a number for every factor it has. Therefore every door should be toggled an even number of times and should therefore end closed (the starting state).

However, perfect squares have an odd number of unique factors, because one of the factors is repeated. Because of this, perfect squares are toggled for their square root which has no pair to reverse the toggle, so the door ends up open (the opposite state it started in).

Make sense?

Somehow I think that makes it less fun.

That's good because I don't find solving solved problems to be fun in the first place.

Haskell:

lockers n = takeWhile (<= n) [x*x|x<-[1..]]

Looks like the most-toggled locker is the 96th one, with 12 toggles.

DOS, because I'm too lazy to SSH into my linux machine:

Edit.exe:

@ECHO OFF
SET /A LOCKERS = %1

IF %LOCKERS% LEQ 0 GOTO ERROR

SET /A COUNT = 1

:JOCKHACK

SET /A SQUARE = %COUNT%*%COUNT%

IF %SQUARE% GTR %LOCKERS% GOTO NERDSLAM

ECHO %SQUARE%

SET /A COUNT=%COUNT%+1
GOTO JOCKHACK

:NERDSLAM
ECHO Done
GOTO END

:ERROR
ECHO How can you have a negative amount of lockers? That's just silly.
GOTO END

:END
ECHO.

for me, I saw the pattern +3, +5 , +7, +9, +11, +13, +15, +17 and +19

not i^2

neat:

i don't have a method of solving this problem, but i can clearly see the pattern when it's done - closed lockers in runs of 2,4,6,8,10,12,14,16,18 with a single open locker between them

how delightful!

I also noticed this pattern and it is trivial to code:

void lockers(int n)

{

  int i, j, k;

  

  for (i = 1, k = 2; i <= n;)

  {

    printf ("O")

    i++;

    for (j = 0; (i <= n) && (j < k); i++, k++)

      printf ("X");

    k = k+k;

  }

}

From the explanation about the squares, we can deduce why this works:

(n+1)^2 = n^2 + 2n + 1

The header of the article shows up as white on white in chrome, maybe this category hasn't had a header color assigned to it yet?

Not exactly awesome:

import math; l=lambda x:map(lambda y: y**2,range(1,int(math.sqrt(x))+1))

Welbog:

That's good because I don't find solving solved problems to be fun in the first place.

So... Not having fun in your job as a programmer ?

neat:

i don't have a method of solving this problem, but i can clearly see the pattern when it's done - closed lockers in runs of 2,4,6,8,10,12,14,16,18 with a single open locker between them

how delightful!

Thanks for pointing out the rythm!

Classic ASP:

<%

function doLockers(numLocks)

	dim Lockers()

	redim Lockers(numLocks-1)	

	dim inc : inc = 1

	dim lastOpened : lastOpened = 0

        dim nextToOpen : nextToOpen = 0

	

	' closing all lockers

	for i = 0 to numLocks-1

		Lockers(i) = 0

	next

	

	' Jocking through the corridor...

	do while nextToOpen < numLocks		

		Lockers(nextToOpen) = 1		' open locker		

		lastOpened = nextToOpen

		inc = inc + 2					' skip ahead

		nextToOpen = lastOpened + inc

	loop

	

	' printing result

	dim ret : ret = "Open lockers: "

	for i = 0 to numLocks-1

		if Lockers(i) = 1 then

			ret = ret & "<br />" & cstr(i+1) 

		end if

	next

	

	doLockers = ret

end function



response.write "<HR />" 

response.write "<HR />" & doLockers(100)



%>

moltonel:

Welbog:

That's good because I don't find solving solved problems to be fun in the first place.

So... Not having fun in your job as a programmer ?

Not particularly.

Thinking just shortly (1-2 min) about it, I'd say that exactly the square numbers stay open.

input: totalLockerNum

max = floor(sqrt(totalLockerNum))

for i=1; i<=max; i++
{
result = i**2;
add result to array
}
return array

anonym:

for me, I saw the pattern +3, +5 , +7, +9, +11, +13, +15, +17 and +19

not i^2

Yes, that would be the pattern for perfect squares.

.NET Solution:

public static IEnumerable Lockers(int numLockers)

{

    for(int i = 1; i <= (int)Math.Sqrt(numLockers); i++) {

        yield return (i*i);

    }

}

i just watched the example and i feel like a jock

stupid nerds with their fancy "square" numbers, nobody needs em!

Welbog:

moltonel:

Welbog:

That's good because I don't find solving solved problems to be fun in the first place.

So... Not having fun in your job as a programmer ?

Not particularly.

How sad :(

what with the brilliant paula being brillant?

Answer:

perfect squares.

using formulae:

(let  ((i 0))

    (mapcar (lambda (x)

                (if (zerop (mod (sqrt (incf i)) 1))

                    "_"

                    x))

            (make-list 100 :initial-element "#")))

Brute force:

(defun visit-door (doors doornum value1 value2)

    "visits a door, swapping the value1 to value2 or vice-versa"

    (let ((d (copy-list doors))

          (n (- doornum 1)))

             (if (eq   (nth n d) value1)

                 (setf (nth n d) value2)

                 (setf (nth n d) value1))

             d))

 

(defun visit-every (doors num iter value1 value2)

    "visits every 'num' door in the list"

    (if (> (* iter num) (length doors))

        doors

        (visit-every (visit-door doors (* num iter) value1 value2)

                     num

                     (+ 1 iter)

                     value1

                     value2)))

 

(defun do-all-visits (doors cnt value1 value2)

    "Visits all doors changing the values accordingly"

    (if (< cnt 1)

        doors

        (do-all-visits (visit-every doors cnt 1 value1 value2)

                       (- cnt 1)

                       value1

                       value2)))

 

(defun print-doors (doors)

    "Pretty prints the doors list"

    (format T "~{~A ~A ~A ~A ~A ~A ~A ~A ~A ~A~%~}~%" doors))

 

(defun start (&optional (size 100))

    "Start the program"

    (let* ((open "_")

           (shut "#")

           (doors (make-list size :initial-element shut)))

               (print-doors (do-all-visits doors size open shut))))

Solved this ages ago for rosetta code: (http://rosettacode.org/wiki/100_doors#Common_Lisp)

Python:

import math; def getOpenLockers(i): return [x**2 for x in range(1, 1+int(math.sqrt(i)))]

Using this excuse to practice some Groovy:

public List getLockers(int numLockers) {
(1..numLockers).collect{ it*it }
}

Any groovy programmers that want to show me better ways to do this, please point them out. :D

Pretty simple :) after you watch the simulation...
getOpenLockers(n){open = 1;count = 3; while(open<=n) { print open; open += count;count += 2;} }
I'll leave the mathematical solution for somebody else.

Yup, only one small change....

groovy:000> public List getLockers( int numLockers ) {

groovy:001> (1..Math.sqrt(numLockers)).collect { it*it }

groovy:002> }

===> true

groovy:000> getLockers(1)

===> [1]

groovy:000> getLockers(100)

===> [1, 4, 9, 16, 25, 36, 49, 64, 81, 100]

groovy:000> getLockers(30) 

===> [1, 4, 9, 16, 25]

groovy:000> 

Might as well bandwagon on the languages thing:

You can download it here

I'm getting eaten by the spam detector, uh oh.

Ah, I think I see the WTF here...

Drew:

Using this excuse to practice some Groovy:

public List getLockers(int numLockers) {
(1..numLockers).collect{ it*it }
}

Any groovy programmers that want to show me better ways to do this, please point them out. :D

I AM NOT A GROOVY PROGRAMMER.

It looks like it will return the squares from 1 to numLockers**2, which might not be what you want.

SQL - though the use of numbers table may qualify as brute force...

create FUNCTION [dbo].[getOpenLockersList] ()

RETURNS @aReturnTable TABLE
( locker INT,
isOpen BIT,
myCount INT)
AS BEGIN

INSERT INTO @aReturnTable (locker, mycount)
SELECT locker, SUM(CASE WHEN locker%number = 0 THEN 1 ELSE 0 END)
FROM (SELECT number locker FROM numbers WHERE number <= 100) lcker
CROSS join
(SELECT number FROM numbers WHERE number <= 100) nbr
GROUP BY locker

return
END
GO
SELECT locker,
CASE WHEN mycount%2 = 1 THEN 'OPEN' ELSE 'closed' END lockerState ,
myCount
FROM dbo.getOpenLockersList()

SELECT SUM(mycount%2) FROM dbo.getOpenLockersList()
SELECT locker FROM dbo.getOpenLockersList()
WHERE mycount =
(SELECT MAX(mycount) FROM dbo.getOpenLockersList() )

Welbog:

The open lockers are perfect squares.

The reasoning is simple: non-perfect squares have an even number of factors. The jocks toggle a number for every factor it has. Therefore every door should be toggled an even number of times and should therefore end closed (the starting state).

However, perfect squares have an odd number of unique factors, because one of the factors is repeated. Because of this, perfect squares are toggled for their square root which has no pair to reverse the toggle, so the door ends up open (the opposite state it started in).

Make sense?

What about numbers like 1*2*3*4 = 24 or 2*3*4*7 = 168 then?

I don't see where it says n should be less than 100.

Zecc:

Welbog:

The open lockers are perfect squares.

The reasoning is simple: non-perfect squares have an even number of factors. The jocks toggle a number for every factor it has. Therefore every door should be toggled an even number of times and should therefore end closed (the starting state).

However, perfect squares have an odd number of unique factors, because one of the factors is repeated. Because of this, perfect squares are toggled for their square root which has no pair to reverse the toggle, so the door ends up open (the opposite state it started in).

Make sense?

What about numbers like 1*2*3*4 = 24 or 2*3*4*7 = 168 then?

I said factors. 24's factors are {1, 2, 3, 4, 6, 8, 12, 24}. There is an even number of them.

And the limit of 100 is irrelevant. That just gives us the maximum perfect square we care about.

Welbog:

Zecc:

Welbog:

The open lockers are perfect squares.

The reasoning is simple: non-perfect squares have an even number of factors. The jocks toggle a number for every factor it has. Therefore every door should be toggled an even number of times and should therefore end closed (the starting state).

However, perfect squares have an odd number of unique factors, because one of the factors is repeated. Because of this, perfect squares are toggled for their square root which has no pair to reverse the toggle, so the door ends up open (the opposite state it started in).

Make sense?

What about numbers like 1*2*3*4 = 24 or 2*3*4*7 = 168 then?

I said factors. 24's factors are {1, 2, 3, 4, 6, 8, 12, 24}. There is an even number of them.

Yeah, I was going to delete after posting, but it was to late.
I was thinking of a slight different problem, I guess; where you don't start toggling at the i-th locker for the i-th iteration.

Welbog:

Zecc:

Welbog:

The open lockers are perfect squares.

The reasoning is simple: non-perfect squares have an even number of factors. The jocks toggle a number for every factor it has. Therefore every door should be toggled an even number of times and should therefore end closed (the starting state).

However, perfect squares have an odd number of unique factors, because one of the factors is repeated. Because of this, perfect squares are toggled for their square root which has no pair to reverse the toggle, so the door ends up open (the opposite state it started in).

Make sense?

What about numbers like 1*2*3*4 = 24 or 2*3*4*7 = 168 then?

I said factors. 24's factors are {1, 2, 3, 4, 6, 8, 12, 24}. There is an even number of them.

Yeah, I was going to delete after posting, but it was too late.
I was thinking of a slight different problem, I guess; where you don't start toggling at the i-th locker for the i-th iteration.

Addendum (2009-08-05 10:12):
Addendum: never mind. I guess I'm just not thinking straight about this.

The locker toggled the most is the locker whose prime factors can create the greatest number of combinations. (locker ... combinations ... get it?).

At first, I think 64 might be a good choice, but (2,2,2,2,2,2) creates only 6 combinations. If I replace a 2 with a 3, I describe locker 96, and I can now create 11 combinations. Applying my knowledge of cribbage, I see that I can replace three 2s with two 3s to describe locker 72 (2,2,2,3,3), which also creates 11 combinations.

Wrote this quick in C#

Console.Write("How many lockers? ");
int lockers = Convert.ToInt32(Console.ReadLine());
int openlockers = 0;
int interval = 1;
Console.Write("Open Lockers: ");
while (openlockers + interval < lockers)
{
openlockers += interval;
Console.Write(openlockers + " ");
interval += 2;

}

Even worse is when the problem was already solved by Tom and Ray on Car Talk:

http://www.cartalk.com/content/puzzler/transcripts/200546/answer.html

I would bet that Mr. Zargas was the coach of the football team. This challenge is EXTREMELY biased towards the jocks. I respectfully submit that in order to figure out the function, you HAVE to iterate (brute force) at SOME point in time.

SHENANIGANS!

Zecc:

Welbog:

The open lockers are perfect squares.
(snip)

What about numbers like 1*2*3*4 = 24 or 2*3*4*7 = 168 then?

24 can be divided by 1, 2, 3, 4, 6, 8, 12 and 24. That's 8 factors, or 6 if we exclude the extrema -> even

25 can be divided by 1, 5 and 25. That's 3 factors -> odd.

And a slightly smaller version in JS

<SCRIPT>

	function doLockers(n)

	{

		var i = 1;

		while((i*i)<=n)

		{

			document.write("<BR />" + i*i);

			i++;

		}

	}

	doLockers(100);

</SCRiPT>

BTW, I'm getting an awful lot of errors when posting lately...

#!/usr/bin/python

from math import sqrt

def getOpenLockers(n):

	result = map(lambda x: x * x, range(1, (int)(sqrt(n)) + 1))

	return result

Re: "96 has the most"...

Other numbers not above 100 with 12 distinct factors:

60: 1 2 3 4 5 6 10 12 15 20 30 60
72: 1 2 3 4 6 8 9 12 18 24 36 72
84: 1 2 3 4 6 7 12 14 21 28 42 84

simply - squares of prime numbers (or perfect squares)

I'm amazed with amount of i^2 solutions. Think a bit longer before posting.