# Thread: Hello, need help.. not sure where this field applies to

1. ## Hello, need help.. not sure where this field applies to

how to calculate the number of certain size picture/paper into one larger paper...

for example 2"x4" size paper/picture wanted to fit into to 8"x10"

how many size of 2"x4" can fit into a 8"x10"

is there a solution that will work for all different sizes?

Any help is highly appreciated... Thanks

2. Originally Posted by sptham30
how to calculate the number of certain size picture/paper into one larger paper...

for example 2"x4" size paper/picture wanted to fit into to 8"x10"

how many size of 2"x4" can fit into a 8"x10"

is there a solution that will work for all different sizes?

Any help is highly appreciated... Thanks
Consider that a piece of paper measured by its side lengths, (like 8 by 10), is actually telling you the area of the rectangle that is the paper, because area is $\displaystyle (length)(hiegth) = area$. So to find how many can fit in a certain other piece of paper, you have to see how many times the area divides evenly. for your problem the smaller paper's area is $\displaystyle AREA = (2)*(4)=8$ and the larger area is: $\displaystyle AREA = (8)*(10) = 80$. So we need to see how many times 8 goes into 80, which turns out to be: $\displaystyle \frac{80}{8} = 10$, so you can fit 10 of the smaller peices of paper onto the larger piece. This method works of papers of any size, just find the areas and divide.

3. If I'm understanding the OP's question, then this is not a trivial problem. See here for some more info.

Edit: here is some more info for the case when the identical rectangles must be placed orthogonally.

4. Correct me if I'm wrong, but even though the area of 10 pictures might fit into the area of the paper, that might mean having to cut up the pictures. In this particular case, since you can turn the pictures side-ways and print two columns of five pictures (since 2 divides 10 evenly by 5, and 4 divides 8 evenly by 2), then the answer comes out right: 10 pictures. But in other situations this might fail.

5. Originally Posted by ragnar
Correct me if I'm wrong, but even though the area of 10 pictures might fit into the area of the paper, that might mean having to cut up the pictures. In this particular case, since you can turn the pictures side-ways and print two columns of five pictures (since 2 divides 10 evenly by 5, and 4 divides 8 evenly by 2), then the answer comes out right: 10 pictures. But in other situations this might fail.
Agreed. I was addressing the general question and assumed that cutting up pictures is not allowed.

6. thanks guys... i guess the solution i'm looking for includes the not breaking the picture into sizes to fit the whole thing...

anyone have other solutions?

7. Originally Posted by sptham30
thanks guys... i guess the solution i'm looking for includes the not breaking the picture into sizes to fit the whole thing...

anyone have other solutions?
Well what kind of other solution are you looking for? For arbitrary 4-tuple $\displaystyle (L,W,l,w)$, literature suggests that there is no known formula for the maximum number of smaller rectangles that will fit into one large rectangle, even when the smaller rectangles are restricted to orthogonal placements (meaning the sides of the smaller rectangles are parallel to those of the large rectangle). The solutions are given in terms of an algorithm. For not-necessarily-orthogonal arrangements the problem is even harder and I've had a hard time finding literature on it.

So, the only simple formulas you can get are for non-optimal solutions. For example, a non-optimal formula is: $\displaystyle \displaystyle\left\lfloor \frac{L}{l} \right \rfloor\cdot \left\lfloor \frac{W}{w} \right \rfloor$. This could be used as a lower bound.

Edit: Perhaps non-orthogonal arrangements are only optimal when side lengths are non-integer.