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here's a fun one for any computer/math nerd, how about a virtual enigma? no need for a plugboard, but 3 encoding disks plus the reflection disk.
FYI - for those who don't know the Enigma was a German encryption machine used in WWII. But it's relatively simple in construct.
It's been done ;)
http://homepages.tesco.net/~andycarl.../enigma_j.html
Hey now, is that yours or are you just showing someone else's final project? :D
A lot of these challenges have already been done, like the next perm one I did around the beginning of last year. But it gives people who want a challenge something to do....like myself.
Oops, sorry, I didn't mean to give the impression that that was mine. Never mind. I'll go away now. :D
;)
Random Quote program
#include <iostream>
#include <string>
#include <time>
#include <math>
#include <vector>
#include <fstream>
using namespace std;
void error(const char*p, const char *p2="")
{
cerr << p << " " << p2 << "\n";
exit(1);
}
int main(int argc, char *argv[])
{
vector<string> quotes(0);
int items=0;
string q;
if (argc < 2)
error("no filename given");
ifstream file(argv[1]);
if(!file)
error("can't open", argv[1]);
char ch;
while (file.get(ch)) {
if (ch=='\n'){
if(items % 100 == 0)
quotes.resize(quotes.size()+100); //increase the size of the vector
quotes[items++]=q;
q="";
}else
q+=ch;
}
int n;
srand(time(0));
n=rand()%items;
cout << quotes[n] <<endl;
return 0;
}
Quote:
Originally posted by Enlighted One
Random Quote program
Can it get any simpler than that?Code:#!/bin/bash
# pass the file with quotes as an argument
# range = the lenth of quote file
range=`cat $1 | wc -l`
# line = a random number between one and range
let "line=($RANDOM%$range)+1"
# output the random line
sed $line'!d' $1
I must admit that I'm pretty proud of myself. :D
Now somebody else will come along and make it simpler... heheh
:pCode:/usr/games/fortune
damn I love slackware.... making program challenges easier every day. :D
There's a theorem from graph theory that sums this um very nicely, no need to do brute search methods...
::goes to track down his graph theory book::
darn can't find it. Something to do with the degree of every vertex added together is 2 times the total number of edges. Therefore when doing person-person relationships, it's only possible for an even amount (it's impossible for there to be 7 people at a party and each one to know exactly 3 others).
Something like that :)
Quote:
Originally posted by duncanbojangles
Okay, I read this book "Archimede's Revenge" a couple weeks ago and it has a buttload of stuff like this, basically all this number and geometry thoery and stuff. Good reading. Here's one of the problems:
"Suppose there are 100 people in a group, and you want to find out if 50 of them know each other".
This can be done on paper by drawing 100 dots, connecting lines between people that know each other, and looking for all the people that know 49 others that all know the other 49.
Yeah, it's one of those problems that as the set of people gets larger, the amount of work that it would require to find the answer increases exponentially.
Find an algorithm that finds the answer in a manner such that the answer is found in an amount of work like n * number of people, rather than number of people ^ n.
I was working on a method where you make a table, with all the people in the columns and all the people in the rows. Mark an X in the appropriate spot where people know one another(including that person knowing themself). Then remove all people that have less than the needed number of associations. Then you are left with all the people knowing the amount of people needed to associate with. Then, if all the spaces are filled, the answer is 'yes', they all know each other. I belive this approach is not exponential, but I could be wrong.
Don't know if it has been said before, and I am really late on this thread...but anyway. Just learned how to do this in high school....Quote:
Originally posted by grady
You can waste alot of multiplications evaluating a polynomial such as a x^3+b x^2+ c x + d , but its possible to evaluate an n degree polynomial with only n multiplications. Can you figure it out? Don't forget that computing x^3 takes 3 multiplications in itself.
say z is what you are evaluating for
ax^3+bx^2+cx+d
a*z = m
m+b = n
n*z = o
o+c = p
p*z = q
q+d = r
and r is the answer.
example-
2x^3+x^3+4x+2
For 3
putting it in = 77
2 * 3 = 6
6 + 1 = 7
7 * 3 = 21
21 + 4 = 25
25 * 3 = 75
75 + 2 = 77
Anyway, it was probably already done, and I probably explained it way too much...but I'm bored anyway :D.
So in other words, you "factor" ax^3 + bx^2 + cx + d into:Quote:
Originally posted by gobeavers
say z is what you are evaluating for
ax^3+bx^2+cx+d
a*z = m
m+b = n
n*z = o
o+c = p
p*z = q
q+d = r
and r is the answer.
(((ax + b)x + c)x + d
It's not a true factorization, but it does reduce into n multiplications.
(FWIW: This question was asked so long ago that I don't remember if there was an answer to it posted already either! ;))
I don't konw...its something called "synthetic substitution" using some guys root theorem (or something like that)....we just learned it in pre-calc.Quote:
Originally posted by bwkaz
So in other words, you "factor" ax^3 + bx^2 + cx + d into:
(((ax + b)x + c)x + d
It's not a true factorization, but it does reduce into n multiplications.
(FWIW: This question was asked so long ago that I don't remember if there was an answer to it posted already either! ;))
Basically, you divide the polynomial by x-z, z being what you are evaluating for, and the remainder is the answer.
This piece of code is awesome! Compile it and run it!
Quote:
//The Beapanator!!!!!!!
#include <iostream>
using namespace std;
int main()
{
int w, n;
char y;
cout << "Would you like to annoy the people around you?";
cin >> y;
if(y=='y')
{
cout << "How many times?";
cin >> n;
for(n; n!=0; n--)
{
cout << "\a";
}
}
else
return 0;
}
bwkaz, don't you have anything to say about this? ;)Quote:
Enlighted One:
using namespace std;