1 September 2012
I have heard people say that there are endless possibilities (if you watch any TV you will know that this is in response to the curries advert about buying TVs "the endless possibilities") with information of various formats of transmission, despite it being implied that the object of data in question is of finite length (any other length is impractical for a human to read, given our average ageing state is mortal). However, as I can very easily demonstrate there are not endless possibilities with the inferred conditions.
Admittedly you have to go quite far down to arrive at 1 possibility (which is completely irrelevant as this is no words yet alone letters, oh the possibilities). However, you see my point, any non infinite number in addition to any other non infinite number is not infinity.
Possibilities : No. of letters 1. 0 26. 1 676. 2 17, 576. 3 456,976. 4 11,881,376. 5
When you get to one thousand words you get an massive number. Which is 1000000000000000008493621433689702976148869924598760615894999102702796905906176 = 26^1000
Although I have a suspicion that my computer within its elaborate calculations ran out of space for the first few digits.
The calculation is simple. If one letter can be any in the alphabet then there would be twenty six possibilities as there are twenty six letters. Then for two letters there are 26*26 possibilities because each set of letters before the last can have the last digit as 26 different possibilities. Hence the possibility of have the initial possibility can have twenty six others when an extra letter is appended to it. Hence P2 = P1 * 26. Or Pn = 26^n
If any of you are interested I made this objective c code specifically for this post; and my computer ran out of memory!
//-----------MAIN----------- ------START PROGRAM---- // // main.m // Power // // Created by James Thorneycroft on 01/09/2012. // Copyright 2012 James Thorneycroft. All rights reserved. //
#import <Foundation/Foundation.h>
int main (int argc, const char * argv[]) {
NSAutoreleasePool * pool = [[NSAutoreleasePool alloc] init];
// DEFINE long double result = 0.0f; long long int resultroundedtoint = 0; double resultminusrounded = 0.0f; double power = 0.0f; double base = 0.0f;
//Take in arguments NSLog(@"Please enter the base..."); scanf("%lf", &base); NSLog(@"Please enter below the power to raise the base to..."); scanf("%lf", &power);
//Do the calculation in question, y = b^x //Does this correspond to the argument order. result = pow(power, base);
//Find out wether the double result has an integer result i.e .0000000 resultroundedtoint = (long long int)round(result); resultminusrounded = result - resultroundedtoint;
//Place ruling on outcome dependant on the integer condition of the double result if ( resultminusrounded > 0 > resultminusrounded ) { NSLog(@"The result of %f being raised to %f is %Lf. y = b^x.", base, power, result); }
if (resultminusrounded == 0) { NSLog(@"The result of %i being raised to %i is %li. y = b^x.", (int) base, (int) power, (long long int) result); }
[pool drain]; return 0; }
//------------END------------
//NSLog(@"Sorry, I lied about the money.");
If You Read This Then I Will Give You Money • Opuss № I