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exam problem

Author

Topic

ace333
Starting Member

11 Posts

Posted - 2005-09-20 : 13:06:40

What initial values of a and c are required such that the final values of a and b are:

a = 32
b = 4

…

int a,b,c

a = ?
b = 0
c = ?

for( b=0; a<12; b++ )
{
a = (a+a) * c;
}

…
I know its not the area but i'm hoping someone has an idea

SamC
White Water Yakist

3467 Posts

Posted - 2005-09-20 : 13:20:58

If you want help with home exams, you should post it as a puzzler in the Yak Corral so it doesn't appear to be schoolwork.

ace333
Starting Member

11 Posts

Posted - 2005-09-20 : 13:23:04

no its not school work, its a sample of an exam question given in an interview, what is the yak corral (will google it)

SamC
White Water Yakist

3467 Posts

Posted - 2005-09-20 : 13:25:31

The Yak Corral is another forum here on SQL Team.

Here's a list of all the forums.

http://sqlteam.com/Forums/default.asp

SamC
White Water Yakist

3467 Posts

Posted - 2005-09-20 : 13:29:23

Since you *say* it isn't homework, here's my hack:

b=0; a = 2ca
b=1; a = 2c(2ca)
b=N; a = (2c)^(N+1)*a

Final condition: b=4, a=32
32 = (2c)^5 * a = 32 * c^5 * a

Looks like there are many initial conditions, so long as a = 1/(c^5), it'll work, such as c=a=1.

ace333
Starting Member

11 Posts

Posted - 2005-09-20 : 13:41:52

what language is that and what is the ^ mean. I dont understand ur answer......
i have to come up with numerical values for a and c......
and its not homework,,,,,, its annoying ...this is going to bug me all night

Seventhnight
Master Smack Fu Yak Hacker

2878 Posts

Posted - 2005-09-20 : 13:52:09

Sam - you need to reread the code in the question... it doesn't execute when b=4... it last executes with b=3.

ace - my guess is a=-4 and c=-1. Although, I think that Sam is correct in that any number of solutions could be true assuming non-integers could be used.

edit:
also a=.5 & c=2 works...

Corey

Co-worker on The Wizard of Oz "...those three midgets that came out and danced, the freaked me out when I was little. But they are ok now."

SamC
White Water Yakist

3467 Posts

Posted - 2005-09-20 : 14:51:02

quote:
Originally posted by ace333

what language is that and what is the ^ mean. I dont understand ur answer......

the language is math, and ^ means ** (raised to the power)

SamC
White Water Yakist

3467 Posts

Posted - 2005-09-20 : 14:56:37

quote:
Originally posted by Seventhnight

Sam - you need to reread the code in the question... it doesn't execute when b=4... it last executes with b=3.

OK, OK. The final value of b is 4, but the loop exit condition is a>=12, so that changes a few things...

We must satisfy the condition a*c**(4+1) = 1, and when b=3, a<12, and when b=4, a>=12

So... what are a and c initial values?

eyechart
Master Smack Fu Yak Hacker

3575 Posts

Posted - 2005-09-20 : 15:26:38

42

-ec

Seventhnight
Master Smack Fu Yak Hacker

2878 Posts

Posted - 2005-09-20 : 16:06:33

i still stand that integer wise:

a=-4 & b=-1

or with rationals:
a=.5 & b = 2
or
a=.0625 & b = 4
etc...

Corey

Co-worker on The Wizard of Oz "...those three midgets that came out and danced, the freaked me out when I was little. But they are ok now."

SamC
White Water Yakist

3467 Posts

Posted - 2005-09-20 : 16:08:05

quote:
Originally posted by eyechart

42

The answer to everything?

I don't think this one has a solution that meets the initial and final conditions.

Seventhnight
Master Smack Fu Yak Hacker

2878 Posts

Posted - 2005-09-20 : 16:16:08

Sam... are you ignoring me

I've posted 3 different solutions...

Corey

Co-worker on The Wizard of Oz "...those three midgets that came out and danced, the freaked me out when I was little. But they are ok now."

SamC
White Water Yakist

3467 Posts

Posted - 2005-09-20 : 16:21:32

quote:
Originally posted by Seventhnight

i still stand that integer wise:

a=-4 & b=-1

or with rationals:
a=.5 & b = 2
or
a=.0625 & b = 4

I read the problem as:

Initial conditions: b=0, a, c are to be determined
Final conditions: b=4, a=32, c can be calculated from the initial conditions

SamC
White Water Yakist

3467 Posts

Posted - 2005-09-20 : 16:25:50

quote:
Originally posted by Seventhnight

Sam... are you ignoring me I've posted 3 different solutions...

Sorry, I wasn't giving enought notice to your earlier posts.

I don't see a working solution. To get to 32, the loop would exit at b=3.

Seventhnight
Master Smack Fu Yak Hacker

2878 Posts

Posted - 2005-09-20 : 16:28:13

okay okay... so i mis-typed:

a=-1 & c=-1

or with rationals:
a=.5 & c=2
or
a=.0625 & c=4
...

i even think that the rational solutions could both be positive or negative...

Corey

Co-worker on The Wizard of Oz "...those three midgets that came out and danced, the freaked me out when I was little. But they are ok now."

SamC
White Water Yakist

3467 Posts

Posted - 2005-09-20 : 16:31:20

a=-1, c=-1 works...

SamC
White Water Yakist

3467 Posts

Posted - 2005-09-20 : 16:31:44

Hey! Fix your thumbnail image. I miss the entertainment.

SamC
White Water Yakist

3467 Posts

Posted - 2005-09-20 : 16:32:36

a=.5, c=2 gives a=512 when b=4

Seventhnight
Master Smack Fu Yak Hacker

2878 Posts

Posted - 2005-09-20 : 16:37:15

sorry.. i was under the mis-guided impression that the b would increment, but that the a would force it to not continue.

so i ran it as a javascript function...
and
a=2 & c=-1 works.


<html>
<script>
a=2;c=-1;
b=0;

for( b=0; a<12; b++ )
{a = (a+a) * c;} 

alert("a:" + a + " b:" + b)
</script>
<body>
</body>
</html>

Corey

Co-worker on The Wizard of Oz "...those three midgets that came out and danced, the freaked me out when I was little. But they are ok now."

Seventhnight
Master Smack Fu Yak Hacker

2878 Posts

Posted - 2005-09-20 : 16:40:33

also
a=.125 & c=2

Corey

Co-worker on The Wizard of Oz "...those three midgets that came out and danced, the freaked me out when I was little. But they are ok now."

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