Encryption | 23 Oct 2009

shabbir's Avatar, Join Date: Jul 2004
Go4Expert Founder
((run)/(these)) nails down to 187 / 5703 Then what would be for ((google)/(yahoo))
xpi0t0s's Avatar, Join Date: Aug 2004
Mentor
49/50.
xpi0t0s's Avatar, Join Date: Aug 2004
Mentor
No, 49/5. One dp out (d'oh!). Getting careless in my old age.

('r'-'a'+i)*('u'-'a'+i)*('n'-'a'+i)=187 when i=3
('t'-'a'+i)*('h'-'a'+i)*('e'-'a'+i)*('s'-'a'+i)*('e'-'a'+i)=5703 when i=3

So if i==3,
('g'-'a'+i)*('o'-'a'+i)*('o'-'a'+i)*('g'-'a'+i)*('l'-'a'+i)*('e'-'a'+i)=2294082
('y'-'a'+i)*('a'-'a'+i)*('h'-'a'+i)*('o'-'a'+i)*('o'-'a'+i)=234090
2294082/234090=9.8, or 98/10, or 49/5.
nimesh's Avatar, Join Date: Apr 2009
Invasive contributor
nice one xpt, but didn't get the logic

'r'-'a' ???
SaswatPadhi's Avatar, Join Date: May 2009
~ Б0ЯИ Τ0 С0δЭ ~
@xpi0t0s :

Yeah, even I didn't get the logic ... :?
The following code uses the "encryption" method you mentioned :
Code: CPP
#include <stdio.h>

int main()
{
    char Decoded[50];
    unsigned int Value;
    int delta;

    Decoded[0]='O';
    while(Decoded[0] != '0')
    {
        scanf("%s", Decoded);
        scanf("%d", &delta);
        Value = 1;
        for(int i = 0; Decoded[i] != '\0'; ++i)
            Value *= (Decoded[i]-'a'+delta);
        printf("%d\n\n", Value);
    }
    return 0;
}

When I run it ...
Code:
run 3
7360

these 3
226380

google 3
2294082

yahoo 3
234090
The values in red, don't match with 187 and 5703 respectively. Those is green match perfectly with the values you predict as per your encryption scheme.

And, 7360 / 226380 == 368 / 11319 in the simplest form. That might be close to 187/5703 but not equal to it.
shabbir's Avatar, Join Date: Jul 2004
Go4Expert Founder
Hint : Think in simple terms
xpi0t0s's Avatar, Join Date: Aug 2004
Mentor
yeah I'm way out. Anyway I'm sticking with my guess of 49/50.
How about posting your own guess instead of criticising mine.