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# Adding 1 without using =, -, *, /

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how would you do this function. I know what to do if its even thats easy. unsigned int AddOne(unsigned int x) { ... } that returns x + 1. The function must not contain the following symbols: +,-,*,/ also Imagine you are doing integer arithmatic and multiplication is computationally expensive. What''s a really quick way to multiply by the number 7?

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If this is homework, it doesn''t belong on this board.

Otherwise, premature optimization is the root of all evil.

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These sound very much like homework problems, and the rules here say answers to homework problems may not be given. Though it is allowed to give hints when the person asking shows they are trying, you did not do so.

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That doesn''t just look like homework, that IS homework. Either that or interview questions. Either way, we can''t help you.

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I actually got the question, "how do you multiply by 7 without using the multiply operator?" during an interview at EA. The answer to that one was easy, it was the follow up, "divide a number by 7 without using the divide operator" that I couldn''t figure out. No wonder I screwed up the interview.

--Wendy

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Not homework. I have tried a couple of things. But none have seemed to work I dont use alot of bit operators. I know how they work but I cant seem to solve this one.

For even you just flip the first bit. Hmm im going to keep trying if I get it ill repost

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Where I could research this would help. I dont need the answer. I just dont know where to begin looking for help.

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I could have sworn I''ve seen the answer to both of these right here on GameDev. Tried a search, though, and it was futile.

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yea there are other bitwise posts with these types of questions. But not on this one.

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quote:
Original post by wendy
I actually got the question, "how do you multiply by 7 without using the multiply operator?" during an interview at EA. The answer to that one was easy, it was the follow up, "divide a number by 7 without using the divide operator" that I couldn''t figure out. No wonder I screwed up the interview.

--Wendy

Division by seven is just as easy, you just have to basically go the other way with it (well not exactly, but if you play with it, you''ll get it).

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There may be more efficient ways of doing it but one way is to add the way you were probably taught in school. Start with the least significant digit, if it''s 0 make it 1 and you''re done. If it''s 1, make it 0 and carry one to the next position. Repeat until done.

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This (kind of) question is very common on a Programmer''s Test
during a job interview, before it for screening purpose.

I have seen this exact example in more than one of them.

Hint: It has to do with bitwise operators.

Kami no Itte ga ore ni zettai naru!

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Matt thanks for the help. Took me a while to get what you were saying but that worked.

unsigned int add(unsigned int a)
{
unsigned int i = 1;
while(a & i)
{
a = a & ~i;
i = 1 << i;
}
return a | i;
}

Sweet. I was trying all kinds of shifts then masks. sigh thanks again.

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I plugged 31 into your add function and got 65544 out...

I''m not entirely clear on how it''s supposed to work so I won''t critque it. Anyway, my first thought was to use a lookup table.

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oops thanks. change it to i = i << 1; should work

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im curious as to the answers to these questions, i know bit shifting to the right (i think it was) can effectively multiply by 2 if i remember right, but how the heck do the division by 7 and the multiplication by 7 work? oh and the add one... lol, seriously though im immensley curious
-Dan

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quote:
Original post by mmmmm
how would you do this function.
I know what to do if its even thats easy.
unsigned int AddOne(unsigned int x)
{
...
}
that returns x + 1. The function must not contain
the following symbols: +,-,*,/

also

Imagine you are doing integer arithmatic and
multiplication is computationally expensive.
What''s a really quick way to multiply by the
number 7?

simple

if(!x&1)
x&=1; else
{
test for some other bits, quite simple really but to much labor for me
}

multiply by 7? isnt that: (x<<3)-x

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I knew how to multiply with any value, but divide is another thing ? ( apart from subtracting the value till the result is < 0 )

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quote:
Original post by Anonymous Poster
I knew how to multiply with any value, but divide is another thing ? ( apart from subtracting the value till the result is &lt; 0 )

(n>>3)+1

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For 9 it gives

( 9 >> 3 ) + 1 == 2,

but 9 = 1 * 7 + 2

and 1 != 2

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Add one seventh, and divide by eight.

Of course, to find the one-seventh, you need to be able to divide by 7...

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quote:
Original post by Ademan555
im curious as to the answers to these questions, i know bit shifting to the right (i think it was) can effectively multiply by 2 if i remember right, but how the heck do the division by 7 and the multiplication by 7 work? oh and the add one... lol, seriously though im immensley curious
-Dan

x << 3 is like x*(23), which is like x*8, which is like x+x+x+x+x+x+x+x. We want one less x (x*7), so we subtract x, and get (x << 3) - x.

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// No addition! omg!int add_one(int num){    std::vector<int> myvec(num);    myvec.push_back(0);    return myvec.size();}// No multiplication! omg!int times_seven(int num){    std::valarray<int> myarr(num, 7);    return myarr.sum();}

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Beer Hunter, very creative, but unfortunately, your add_one()
doesn''t work with negative numbers.

Kami no Itte ga ore ni zettai naru!

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IA-32 multiplication returns a 64-bit result from 32-bit operands. Go from there.

MSN

angryzenmaster @ livejournal