Floor Tiles: Newsletter Challenge (10/30/07)

The diagonal line will cross 135 tiles on the 81 x 63 grid.

I drew a horizontal line in AutoCAD with length of 81 units, then added a vertical line at one end with a length of 63 units, and a diagonal. I used the Array command to make copies of the horizontal and vertical originals at 1 unit intervals.

I didn't bother counting all the blocks intersected by the diagonal. 81 and 63 are the squares of 9 and 7, respectively. So, I just zoomed in on one corner and counted the number of blocks crossed by the diagonal in a 9 x 7 section. That gave a result of 15. I multiplied this by 9 to find 135.

I haven't yet done the larger grid, but a similar method should work.

3D,

As AH states in post #4, we don't know the grout width. this will likely affect the array you drew by forcing tile outward and away the the central line (creating a different drawing than the one you reviewed).

Dwight

That was just a joke. However, I would assume a grout width is zero for the puzzle.

Dwight,

Some tiles don't need grout or adhesives, such as these. I believe they have interlocking edges.

BTW, thanks for identifying yourself. If you read the prevous thread, you will find out that I will refer to any anonymus guest as Edgar, and the reasons why.

These tiles still have spacing between them according to the manufacturer.

"No adhesives or grout are required. The small gaps between the tiles are not filled but are specifically designed to allow water to drain away quickly and freely under the tiles." This does not pertain to the question at hand, But I would not want this type of tile INSIDE my home.

grout wouldn't matter as long as it's uniform around each tile.

The tiles could be asphalt tiles which abut each other. In either case you have to assume any grout scaring is taken into account by the size.

Technically, the problem does't state that the tiles were cemented down... it just says the tiles were laid out. I know it sounds silly, but some people do that, especially if complex patterns are being created. A "professional" would automatically leave space for the grout, and they have tools ( grout spacers) just for that purpose.

When tiles are laid out they are laid out with wedges between them to hold a consistent spacing.

Hi Doug,

"81 and 63 are the squares of 9 and 7, respectively."

I'll give you 1 out of 2 right.

9²=81 but 7²=49

This may throw your answer off. I think that 81 and 63 were chosen because they only have 3 as a common factor and you still have a 27 x 21 grid.

I keep coming back to the idea that we're supposed to solve this without counting somehow - derive some formula to get the answer, then count to confirm.

Mike

"I keep coming back to the idea that we're supposed to solve this without counting somehow - derive some formula to get the answer, then count to confirm."

Mikey,

You really should read all the posts before yours, especially if on the same sub-thread, before you post. Read my post #9 where I had already called him on the "81 and 63 are the squares of 9 and 7..." thing.

And I did try to solve it without counting in that post. Thought I had found a nice procedure, but I missed a few and was a little off! <grin>

By the way Mikey, it looks like MPM nailed it in #19 as far as the descriptive method goes, and the tkot formulated the math from #19 in #38.

KUDOS guys!

And I give you 3 out of 9.

You said 81 and 63 only have 3 as a common factor, but they actually have 9 as a common factor, which leaves you with a base pattern of 9x7.

But it doesn't matter for solving the general problem, and that is to put the whole thing in a neat equation.

Has anyone considered that the diagnonal line crosses a corner 8 times. At these times you must consider all four tiles as being contacted by the line.

Would pythagoras help?

cr3

Pythagoras says if A and B are the sides, the diagonal C

C=√A²+B²

Thus C²=81²+63²

C²=6651+3969=10530

C=102.62

We are crossing 103 Tiles

Likewise for the larger floor we find 557 Tiles

It seems counterinturitive, but it verifies graphically

My math skills prevent me from making any guess other than my stated 0. But I feel that you are over looking one fact. When you use Pythagoras as a reference, you need to take into account that you are using a non dimensional term for measurement. The problem as I see it is that the diagonal measurement of these units is longer than the side measurement. The figure that you derive would be correct only if the diagonal and side were the same length. That is not so with the stated square tiles.

Perhaps my speech skills are due for a tune up also.

I think that jfcayron is making the same mistake that another poster made.

See my response in #84 to the post in #67 above.

Actually, they have 9 as their greatest common factor.

Any floor made of m by n many tiles will have a diagonal that crosses m + n - gcf(m,n) many tiles. For example a 2 by 3 floor will have a diagonal that crosses 2 + 3 -1 = 4 tiles. And for the record, I was thinking of wooden tiles that have no grout spacing. It really doesn't matter if the line is the actual diagonal. As long as the line (curve for the more mathematically orthodox) crosses the intervening vertexes (i.e. were the width and height are evenly divisible), the line connecting these vertexes can be any convex curve that remains inside their collective edges. So, it could just run along the bottom row of tiles to the last one and then head up the last column to the opposite corner. This number of tiles crossed then counts the width plus the height minus one for the corner tile. Scaling this back up to the whole floor requires multiplying by the greatest common factor, thereby reproducing the original floor dimensions in the sum and the GCF in the difference.

- Sully.

I think your "convex" constraint is that the curve does not pass through the same position along the width or along the length more than once. (A looser, but equally good constraint would be that the curve crosses each gridline only once). But you need one more constraint - that the the curve intersects precisely the same number total of corners as would a straight line (=HCF(m,n)+1).
Subject to those constraints, the curve can wiggle all it likes. But I think the second constraint is getting a bit artificial for anything other than a straight line.

Sully, I'll accept your premise that the line could be convex, if you can show me a straight convex line, as the Challenge Question specifies.

I'm surprized STL didn't jump on this because he is from the "Show Me" state.

81 and 63 are the squares of 9 and 7, respectively.

Am I reading it wrong? 81 is the square of 9 but 63 is the product of 9 and 7; certainly not the square of 7.

7 squared is 49 not 63!!! You got the 9 part right though. Shouldn't we be using the formula a^2 + b^2 = c^2??? Then you come up with 103 tiles...

NO, Pythagoras is not the solution here. See my post #84.

OK,

You are right.

The correct answer is 135 tiles.

Your solution is correct but your explanation is rotten !

7 is NOT the square root of 63. What you REALLY wanted to say ( but didn't !) was:

I have scaled the problem by the largest common factor of the two sides (9) sence my rectangle now becomes 9 by 7. I have counted the intersected tiles by counting them (boring !) and found to be 15. By scaling up to the original size (x9) the large sqare will have 9x15=135 tiles interesected ! That would have been perfect.

Good thinker but cannot express himeslf 8/10 !

Goa too skool. gho direktly two skull. doughnut parse Gow. doo nought corlect too hunderd parnds.

Translation:
I thought I was the expert at gratuitous condescension, but I I find I have much to learn.

Very Southpark!

Sence yew Haff kritisized me foist PoaSt, Ah asoom yue dadn't red inny uv ma Othah wuns.

An howe du yu Kno whut aye waNned too sae? Kin yew Reed mindz? Due yoo hab XtrasensaBull Perseptshun?

Sew uh messtated Sumthin. Dedn't eye admet ma Mistook? Wall, dinnit iye?

I was going to say:

wie doan cha shud da fork ub?

but I changed my mind and I won't say that.

It might be taken as rude.

In the larger grid (472 x 296) the diagonal will cross 784 tiles.

Both numbers given for this grid are even, so repeated divison by 2 reduced the "sample" grid to 59 x 37.

I created the sample grid and counted 98 tiles. Because I divided by 2 three times, I multiplied 98 by 8 (2³) to reach 784.

Now, let's see if someone has a different method, and what the answers will be.

Hi 3Doug,

Are you sure you counted 98? My method would yield 95 in this reduced grid size.

Can you recount for me please. I do not have a copy of Autocad.

Maths Physics Maniac,

Your answer & method are correct as I've also derived same. I've checked on autoCAD too.

Well. a 59 x 37 grid is hard to count through. I did a recount as MPM requested, and used the text command to number the tiles. This time I came up with 97. So, 97 x 8 = 776.

I believe we can ignore the grout. I did a Google on "floor tiles no grout", and I used the website at the top of the list. IIRC, I seen an episode of "This Old House" where they used floor tiles with interlocking edges and they had no spaces between tiles. I should have looked a little more through the sites to find one for the type of product I was thinking of.

It seems that either your line isn't completely straight and narrow, or the tiles are not regular (identical similarly oriented rectangles would do equally well as squares).

You can calculate the number based on the fact that the line must cross two internal edges of every tile except the corner ones (which each have only one internal edge), that each internal edge crossing crosses two tiles, and that every internal grid-line must be crossed exactly once. That gives l+w-1, or 95 tiles for this case.

(That only applies once the diagonal don't pass through any internal corners. You used this already, of course - and that the condition for this is that the number of tiles on a side have no common factors).

Another question that is a play on words, so my interpretation is as follows. There is a room who's size is 296 units by 472 units, and a line is drawn across that room from corner to corner, then a portion of the floor is layed with tiles, each tile 1 unit square, numbering 81 in the long direction by 63. The answer then is that the line will cross the layed tiles sometimes crossing 1 tile per column, and sometimes 2 tiles per column. Number of tiles crosses = (472 * 81)/296 = 129.2. rounded up = 130 tiles.

Regards JD.

How thick is the grout?

Actually, I was trying to figure a simpler way to do this, but have not found it, yet. The first poster got the right answer, but the method is a little tedious. However, there is a mathematical method to do this and this is what I have done.

If you arrange the problem so that it can be thought of as a grid of 1-unit squares (with the grout thickness = 0), then you could use the slope of the diagonal line and the equation y = mx + b to calculate the number of tiles intersected. For simplicity, arrange the grid so that the longer side is up (taller than wide). Place the X/Y origin at the lower left corner of the grid (0,0).

If the slope of the diagonal line = 1, then the number of crossed tiles equals the number of vertical or horizontal tiles (the number of vertical tiles = the number of horizontal tiles, i.e., a square grid). This is a special case.

If the slope is greater than 1, then you can test the intercept of the diagonal line with the first vertical line segment. The Y intercept point for that line is then divided by the unit length of one tile. To make it easy, let the unit length of a tile = 1.

There are two possible scenarios for the intercept of the first vertical line.

If the divided results is equal to exactly a whole number, then the total number of tiles crossed is equal to the Y intercept * the number of horizontal tiles. It is that simple.

If the divided results is not a whole number, then you always round that result up. For example, if it is 2.3, then there are 3 tiles in the vertical direction that the diagonal crosses plus one extra tile in the horizontal direction (the tile adjacent to the 3rd tile up). For 2.3, the total number of tiles crossed is therefore 4 for the first vertical line. This includes the one extra tile to the right of the third tile up. You must repeat the calculation for each subsequent vertical line and subtract the previous rounded up Y intercept point from the calculated result. Again, round up the result and add one for the additional horizontal tile to the right. The total number of tiles crossed is equal to the sum of each vertical line tested. The last vertical line will always have the diagonal ending at the upper right, so the result of the delta Y intercept is always a whole number, so you do not add one horizontal tile to the right. This also applies to any intermediate tests of a vertical line where the delta Y intercept is a whole number.

Anyone have a simpler mathematical method without graphing?

I don't know why people are making it so difficult. If you the tiles are square which most are then using the right angle triangle you would find that 102.6 (103) units or tiles would be intersected. My point is also proved by measuring out a 9" * 7" area and measusing the diagional (11.375 est) and mulitply that * 9 which ='s 102.375 rounded up to 103. If this is wrong someone please explain to me what I am missing.

The harder we look, the more it is staring us in the eyes. Owen 2007

The harder we look, the more it is staring us in the eyes. Owen 2007

I guess, like "you can't see the forest for all the trees", right?

Owen, wake up and smell the grout! Yes, the diagonal line across the large grid would be about 103 "units" long and, yes, the diagonal of a small grid is about 11.4 units, Pythagoras says so, however what we are seeking is not the length of the line but the number of tiles the line must cross. Pythagoras does not apply here. You are comparing apples to oranges. Just look at any of the graphs drawn in other posts and you will see. Count the tiles yourself.

What I am trying to say in #89 is that I see your train of thought, and I agree with it.

Thanks, Bob!

Use 3/8" and see what you come up with.

Since this is tile used on a floor it tends to have wider space for grout then for tile on counter tops which is about 1/8" to 3/16".

My answers are 144 for the first question and 928 for the second question. Below I will explain how I got there.

OK, from what I have seen, 3Doug started out OK, but somehow went astray. His logic of simplifying the grids into smaller sections was OK but when he said that, "81 and 63 are the squares of 9 and 7, respectively.", I knew that something was amiss. 81 is the square of 9, but 63 is NOT the square of 7.

So, following 3Doug's lead but avoiding the mathematical error, I decided to take a look logically at the problem and how to simplify it. I concluded it was not for no reason that the numbers chosen, if put into a fractional form, could be simplified. For example a grid that is 63H x 81L would have a diagonal slope of 63/81. Now if you find the simplest form of this fraction it would be 7/9. Since 63 is 7x9 and 81 is 9x9, dividing the fraction 63/81 by 9/9 yields 7/9. So, in other words the large grid can be divided by 9 in both height and length, giving a smaller grid of 7H x 9L.

Now since the ratios of the grids are the same the diagonal across the large grid also touches the opposite corners of the smaller grid. Normally, a line with a slope between 0.5 and 1.0 (7/9=0.77777...) will cross two squares in the same column, except at the points where the diagonal goes EXACTLY through the corner point. Therefore in every small grid the number of squares crossed is the number of columns (9) times 2 squares and THEN subtract off one square for each corner the diagonal passes through. This gives 16 squares (no counting necessary, 3Doug). Moving along the diagonal we go through 9 small grids (16 squares each) to get to the far corner of the large grid. So the diagonal passes through exactly 16x9=144 squares. QED

Now the second problems was just a repeat but with different fractions and different size grids, since the slope this time, 296/472=0.6271..., is still between 0.5 and 1.0.

296/472 may be simplified to 37/59 with 8 being the factor. So, there are now 8 x 8 smaller grids. The diagonal of the larger grid will pass through exactly 8 of these, since, again, the ratios of height to length are the same as the large grid, giving corner points where the diagonal passes and therefore cutting through only one square in each column at the start and end of the smaller diagonal. Again each smaller grid has a number of columns (59) of which each except both ends have the diagonal passing through two squares. Therefore, the number of squares crossed is (2x59)-2=116. Now since there are 8 small grids that must be crossed, the total number of squares is 116x8=928. QED2

Perhaps someone more mathematical than I can put this into a formula or equation form. I only know that this procedure is logical and works for the parameters given in challenge. If the slope is between 1.0 and 2.0 I would just turn it 90 degrees to invert the slope and then solve the same way. I would have to do more work to come up with a more general case where the slope is NOT between 0.5 and 2.0 but I think I have taxed my poor brain enough on this one.

Oh, by the way, if the ratio can NOT be simplified, the problem is even easier since every column of the large grid will have two squares crossed except the start and end. So if the slope is still between 0.5 and 1.0, the number of squares crossed will be twice the length, minus 2.

Pretty cool!

Why limit the slope to .5 to 1? What if the array was 3 by 10? How many squares are crossed? That was a rhetorical question. Better ask what is wrong with slopes less than .5, which are easily possible.

Like I said, I solved the problem at hand. The challenge did not ask for a general solution for all possible slopes.

I'll let you have a go at the corollary.

Normally, a line with a slope between 0.5 and 1.0 (7/9=0.77777...) will cross two squares in the same column, except at the points where the diagonal goes EXACTLY through the corner point.

The 3 columns with the white below them only have the diagonal going through 1 sq.

I think 3D Doug is right. But it is a brute force method and not elegant.

Hmmmmm..... I see what you mean.

Nevermind!

Busted

What do you think about this answer: 2 (for each grid) or 1 (depending on the two corners measured across)

I don't believe the statement is true - consider for example a grid 2 wide and 5 high.

The detail is hard to put into words, so I'll doubtless fail. The best I can come up with is that (except where the diagonal passes through the corners) the diagonal must cross all horizontal dividing lines and all vertical dividing lines exactly once. Except where it passes exactly through the corners, it will pass through two squares for each intersection. Similarly (and again except where the diagonal passes through the corners), each square** will have exactly two of its edges intersected. Corner squares have exactly one such internal crossing. If we now equate the crossings calculated from the internal lines to those calculated from the squares we have:

2x(h-1+w-1) = 2xNinternalsquares+1xcorners = 2xNinternalsquares+2, or
Ninternalsquares = h+w-3, or
Total squares crossed = h+w-1

The only thing we need in addition is that iff the height and width are co-prime, the only squares that the diagonal will cross at the corners are those at the two outermost corners of the grid. This corresponds to 3Doug and Maths_Physics_Maniac dividing by the highest common factor of the sides of the larger grids to find sections where the diagonal passes through the corners.

If it works, it should result in 4 for a 2X3, 8 for a 4X6, and 12 for a 6X9, simple 1, 2, and 3. Your hxw-1 works for any minimum module. Not for 2. h x w -m(no of modules,). each module of no common factors. I don't know of a formula to reduce other than thinking about it. Think about it.

Rich

Hi DDK,

I found some ineteresting observation. If we make the floor in the multiples of 9X7, the no of tiles cut by diagonal is coming n(9+7) i.e. if the floor is 9X7 the tiles cut are 15, if the floor is 18X14 the tiles cut are31 and so on. only care is that we have to see this in enlarged view whether the line is cutting only one tile or touching both tiles. Thus the answer for this queation is 81+63-1=143. Can anybody devote some time to find out the formula for this?

Also your observation is right that the line is cutting 2 blocks in every column except the first and every fourth after that. Fourth because 4 is the square root of 9+7.

Thank you, thank you, thank you, ddk!

I may have been off on my factoring, but I wasn't off on my method.

STL, you might have something there about the slope. What you have to count is the number of times the diagonal crosses a grid line. You will need some kind of routine that computes x- and y-intercepts at integer values and increments a count.

An interesting observation is that 7 + 9 - 1 = 15. But then, 37 + 59 -1 = 95.

The obvious answer is that the diagonal crosses 81 tiles in one direction plus 63 in the right. Total 144. and a diagonal will never cross a tile corner to corner, the diagona; os not 45°.

Rich

And 768 for the larger floor.

Rich

I agree with Maths_Physics_Maniac. Good answer.

Hi Maniac!

I think this is the best answer so far. I will just try to rewrite the answer in a more mathematical form, at least as mathematical as I can personally get...

First of all, lets name A and B as the dimensions of the rectangle - A being the number of rows and B the number of columns. Assume A<B (otherwise we switch rows and columns). The line along its route will eventually cross A rows and B columns, so A+B times in total. Is this the answer? Not yet, because we have to substract the number of times the line passes over a crosspoint, in which case it passes from row to row or column to column without crossing a tile.

But when the line passes over a crosspoint? It does so in the n-th column if n.A mod B = 0. At this point, lets define the function:

f(n) = .... 1 if n.A mod B=0

................ 0 if n.A mod B≠0

The answer will now be:

number of crossings = A + B - ∑ f(n) (n is running from 1 to B)

Now, if HCF(A,B)=k (k can also be 1 in the case A and B are primes to each other) then it must be something like A=k.a, therefore:

n.A mod B = n.k.a mod B = k (n.a mod B)

n.a mod B will be ≠0 (and consequently f(n)=0) for all values of n (because 'a' has no common factor with B), except for n=B, therefore ∑ f(n) = k.

Summing up, the final answer is:

number of crossings = A + B - HCF(A,B)

Applying this to the problems at hand, we get:

rectangle 81x63: 81+63-9 = 135 crossings

rectangle 472x296: 472+296-8 = 760 crossings

Hi All, I'm an engineer, and I have done a bit of building work, and some tiling around the home, one of the things you learn after a few years is that the devil is in the detail. And there may be more than one correct answer.

As earlier posters have noted, the basic tiling module is a block of 7 x 9 , and there are 9 such modules on the diagonal (and 16 off the diagonal).Doing it in my head, the rough answer is between 126 and 162, perhaps around144. While there might some elegant solutions, I just put it in a CAD program to determine the following:

The basic answer (assuming the line width is thinner than the grout width which is thinner than 8.8% of tile width) is 15 x 9 = 135

Where the line is thicker than grout width (but less than 8.8% of tile width), you pick up an extra 2 tiles where modules touch so 15 x 9 + 8 x 2 = 151

Where the grout is just thicker than 8.8% of tile width you lose two tiles per module, so 13 x 9 = 117
(I think the above is true for grout width thicker than line width by more than 8.8% tile width)

There are more different answers if the grout width or line thickness are ludicrously large, likewise if you consider "line is touching tile" rather than "entire width of line visible on each tile" .

Typical floor tile gaps would be 6mm to 10mm. Typical line widths might be 1mm (pencil), or 3mm (permanent marker or diamond saw)

PS Who would want to know how many tiles are on diagonal? Because if you want to drain the floor of a bathroom , you need to make a "valley" in it, this is done by cutting the tiles on one or two diagonals, so you need to know how many tiles to push through the diamond saw, in real life one cuts the diagonals at 45 degrees (so only 63 tiles to cut per diagonal for our example, that's a lot of effort saved!).

Cheers, BobT

Good answer in simple explanation, no formula or complex theory. Thanks.

A rectangular floor is laid out made up of square tiles of the same size. 81 tiles are laid out in the long direction and 63 in the other.

The question is: If a straight line is drawn diagonally across the floor from one corner to the other, how many tiles will it cross?

If the floor is 472 x 296 tiles, what is the answer?

My Answer: The diagonal line on 81 x 63 floor would cross 159 tiles and the diagonal line on 472 x 296 floor would cross 941 tiles

Just use trigometry; this turns out to be the Hypotomus = square root of long side squared by short side squared . Something like this

have a good day

#tiles = 7/9[sqrt(x^2+y^2)] rounded up to the nearest whole tile

*2 - 9

Scratch that:
#tiles = 14/9{sqrt(x^2+y^2)} - 7 rounded to the whole tile

One last try:
#tiles = 14/9[sqrt(x^2+y^2)] - x/9 -2

The line crosses each volumn once, and only once, meaning it crosses L tiles laterally. It also crosses each row once and only once, for H rows. The line cannot either skip a row nor cross into a row twice. Therefor, the number of tiles it can cross is L + H The only alternate would be a square, where it would completely cross all columns AND rows joinlty and thus cross only L tiles.

Rich

Delete 9 + 1 from the 144 and the 134 is correct. The line will cross through the corner joint at every 9 X 7 th tiles, which then crosses a colunn And a row simultaneously, thus crossing two tile for two rows and two columns simultaneously, losing 1 tile from the total. Each module this happens, and the starting tile in the entire floor doesn't count twice, so gross - 1 is net.

468 X 296 is 4 modules of 118 X 74, thus 4 corners crossed, 3 entered from previous grids, so 764 - 3 for modules, - starting tile is 760.

Ricch

472, noty 468, so 472 + 196 -3 -3 is 764.

Rich

Add the numbers together.

Now that everyone has busted this question, I'd like to know how many tiles came in a box, and what happened to the extras ? What percentage was wasted by 'nibbling' them to fit around difficult bits of the (undescribed) room, and how many 'intact' tyles did a skilled craftsman/woman have left over for 'safety' ? The wastage rate on cutting is proportional to cost. Sorry, but I've got a probability hang-over. Possibly.

If I have a room that is 81 units long and 61 units wide, and I draw a line diagonally across it, and than lay out the tiles from this line, I will not have crossed any tiles.

Same results if tiles on the other floor are set diagonally.

Does the question define how the tiles are laid out?

Bob, that makes no sense in regards to the question. Was it meant to be funny?

Yes it was meant as a joke. But if the floor tiles were laid in a diagonal pattern, the diagonal line would not cross any tiles. Haven't you ever seen a floor patterned that way?

Actually, it is a valid question. And would make a lot of SENSE if, in fact, it were true that the edges of the tiles aren't laid parallel/perpendicular to the rectangular floor.

The question does not specify any type of orientation - and would blow every last single post out of the water since it was assumed to be otherwise.

O great STL! (no cool sunglasses for you) <grin>

Agreed. The question says the rectangular floor is made up of square tiles. If you wanted to lay the tiles on the slant relative to the outline, you would have to use some tiles that were not square (using a part of a tile that was originally square does not cut it as far as I am concerned).

I've drawn up some grids to test my theory, and I believe the equation should be:

x + y - common divider

Don't forget 1; if there's no other common divider (17 x 37), there's still 1:

17 + 37 - 1 = 53

so: 81x63 (both divisible by 9)-- 81 + 63 - 9 = 135

for 472x296 (both divisible by 8)-- 472 + 296 - 8 = 760

(Edit: dooh! looks like someone else figured out the math while I was testing my theory ;P)

Agreed - but would it be clearer of we used the standard term "highest common factor" (HCF).