Minimum Flooring Boards and Wastage

 Lemon Quarter
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Minimum Flooring Boards and Wastage
Room is 4m long and 3m wide. Floor boards will be laid along the longer side. Board length is 1.285m ie 4m side takes 3.11 boards ie 3 full planks and a plank of 14cm of length. However, 30cm length is the minimum allowed for a plank and joints must be staggered and not closer than 30cm from each other.
The planks are 19.2cm wide so the width takes 16 rows of planks of which one is cut to roughly half lengthwise.
1. What is the number of planks used which produces minimum amount of wastage?
2. What are the lengths in each row 1 to 16 and in which order?
The planks are 19.2cm wide so the width takes 16 rows of planks of which one is cut to roughly half lengthwise.
1. What is the number of planks used which produces minimum amount of wastage?
2. What are the lengths in each row 1 to 16 and in which order?

 Lemon Slice
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Re: Minimum Flooring Boards and Wastage
This is a sort of optimisation problem. I never found that the easiest of subjects dealing as it does with some slightly obscure modelling and then applying calculus. Prefer pure maths myself but surely someone has the necessary for it?
Dod
Dod

 Lemon Quarter
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Re: Minimum Flooring Boards and Wastage
I thought there might be an elegant mathematical solution instead of doodling with a square ruled paper. Whole, 1/3, 2/3 starting the three first rows seems to work relatively well.

 Lemon Slice
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Re: Minimum Flooring Boards and Wastage
JMN2 wrote:Room is 4m long and 3m wide. Floor boards will be laid along the longer side. Board length is 1.285m ie 4m side takes 3.11 boards ie 3 full planks and a plank of 14cm of length. However, 30cm length is the minimum allowed for a plank and joints must be staggered and not closer than 30cm from each other.
The planks are 19.2cm wide so the width takes 16 rows of planks of which one is cut to roughly half lengthwise.
1. What is the number of planks used which produces minimum amount of wastage?
2. What are the lengths in each row 1 to 16 and in which order?
Spoiler:
First, note that the length of the room is a shade less than 25/8ths of a plank. One eighth of a plank is 16 cm long.
Based on this observation, we note that we can easily cut the planks into lengths of 5/8ths and 3/8ths of a plank, and into quarters i.e. 2/8ths and halves, 4/8ths. Full length planks (8/8ths) can be retained too.
The total number of planks you need is (15*25)/8 for the fullwidth planks plus 12.5/8 for the halfwidth one, i.e 49 planks should be sufficient. [In fact, it can be shown that 48 planks is too few, if you divide total area of floor by area of each plank.] The shortest lengths are now just over 30 cm, satisfying the minimum length requirement. If we also ensure joins are 2/8ths of a plank apart on adjacent rows, which is more than the required 30cm, we should be able to stagger the planks easily.
It now becomes a problem of how to arrange the planks. If two full length planks are used in each row, they take up 16/25ths of the rooom length. There are two ways to fill the remaining 9/25ths: a five and two twos, or a three and three twos.
Let's try to arrange rows using these lengths of wood, such that they meet the overlap requirement. These are the cut length combinations we'll use for two rows A and B making up the pattern.
Row A: 2,2,5,8,8 Total 25/8ths
Row B: 2,4,3,8,8 Total 25/8ths
As a check: from inspection of the above we see there will be eight joins across two rows. If each of the joins is required to be 2/8ths of a plank apart, then we need the room to be at least 16/8ths long, which it is.
So let's try to arrange the planks with the correct minimum overlap. I shall do this by inspection, writing the order of the planks, and below that the cumulative distance across the floor, showing where the joins will be.
Row A: 2, 8, 2, 8, 5  Plank order
Row B: 4, 2, 8, 8, 3  Plank order
Row A: 2, 10, 12, 20, 25  Cumulative distance
Row B: 4, 6, 14, 22, 25  Cumulative distance
This pattern repeated every two rows seems to fulfill the requrements for the full width section as there is never less than 2/8ths of a plank between joins. Now let's see how the cutting goes:
We need 8 row As and 8 row Bs, so we cut the planks up as follows:
32 full length planks = 256/8ths
4 planks each cut in half = 32/8ths
6 planks each cut into quarters = 48/8ths
8 planks each cut into a 3/8th and a 5/8th section = 64/8ths
TOTAL = 400/8ths or 50 planks
However, the last row is half width, so we can reuse the other half to reduce the total planks needed. To get rid of one plank, take one of the whole planks from the final row and cut in half lengthwise. This provides two of the lengths of 8 needed.
Solution:
So to revise and conclude the above inventory, our final count is:
31 full length planks = 248/8ths, one of which will be cut in half lengthwise
4 planks each cut in half = 32/8ths
6 planks each cut into quarters = 48/8ths
8 planks each cut into a 3/8th and a 5/8th section = 64/8ths
TOTAL = 392/8ths or 49 planks
This is the final pattern, with length of planks represented in eighths of a plank:
2288888888228888888855555 A
4444228888888888888888333 B
2288888888228888888855555 A
4444228888888888888888333 B
2288888888228888888855555 A
4444228888888888888888333 B
2288888888228888888855555 A
4444228888888888888888333 B
2288888888228888888855555 A
4444228888888888888888333 B
2288888888228888888855555 A
4444228888888888888888333 B
2288888888228888888855555 A
4444228888888888888888333 B
2288888888228888888855555 A
4444228888888888888888333 B Halfwidth row, using one full length plank cut in half lengthwise; discard halfwidth of other pieces in this row.
Length of each row = 25/8 * 1.285m = 4.01m.
There are other patterns but this is my quick answer. The number of cuts can be reduced by having planks of length 6 instead of 2+4 at the start of the B rows.
GS

 Lemon Quarter
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Re: Minimum Flooring Boards and Wastage
A substantial (though not complete) spoiler:
Let's try to approach the optimum solution from above and from below  from below first:
Clearly the total area of the floor boards needs to be at least the area of the room, which is 3m x 4m = 12m^2 (*). The area of a plank is 0.192m x 1.285m = 0.24672m^2, so one needs at least 12m^2/0.24672m^2 = ~48.64 floor boards to cover the area. And since they come in whole numbers, that means at least 49 floor boards.
From above, a very crude solution is to lay out 64 floor boards in a 4 lengthwise by 16 widthwise grid, producing a rectangle 4*1.285m = 5.14m by 16*0.192m = 3.072m. Then trim that down to size in three steps, in which I'll refer to the 4 lengthwise by 1 widthwise sets of floor boards as "rows", and the 1 lengthwise by 16 widthwise sets as "columns":
1) Cut 98.5cm off the short edge of one of the outside columns, to reduce each floor board in that column to 30cm long and the rectangle to 4.155m x 3.072m.
2) Cut 15.5cm off the short edge of the other outside column, to reduce each floor board in that column to 113cm long and the rectangle to 4m x 3.072m.
3) Cut 7.2cm off the long edge of one of the outside rows to reduce the rectangle to 4m x 3m.
That layout doesn't comply with the "joints must be staggered and not closer than 30cm from each other" restriction, but that's easily fixed: reverse the order of the board pieces in every odd row, to produce a layout that is 8 repetitions of the following unit (minus 7.2cm along one long edge):
So 64 floor boards are clearly enough. But that solution is clearly very wasteful, and we can easily improve it to 52 floor boards by simply observing that we used sixteen floor boards to produce sixteen 30cm pieces, when they can equally easily be produced by cutting 4 floor boards each into four 30cm pieces and one 8.5cm piece of scrap. So we now know that we need at least 49 floor boards and that we can do the job with 52, so the answer to question 1 is one of 49, 50, 51 or 52.
We're still wasting quite a lot of floor board with that 52board solution  most particularly, sixteen 15.5cm pieces of scrap from cutting the 113cm pieces, and the four 8.5cm pieces of scrap from cutting the sixteen 30cm lengths. That's enough to make up eight 30cm pieces, saving two boards out of the 52  if we can get the scrap in bigger pieces...
Which we can, at least to some extent, by observing that there is some exploitable flexability in the 52board solution. In particular, we can replace the 30cm and 113cm pieces by (30+X)cm and (113X)cm pieces, as long as we ensure that the joints implies by them are still at least 30cm apart  i.e. that (30+X) <= (113X)30, or equivalently that X <= 26.5. Doing that to all the 30cm and 113cm pieces doesn't help  the scrap pieces we get are still shorter than the pieces we want  but doing it to some of them will create longer scrap pieces from those that we do it to, which are long enough to be used in the ones we haven't done it to.
A simple example is that using X=14.5 on half the units will create 30cm pieces from cutting the (11314.5)cm = 98.5cm pieces, which can be used in the other half of the repetitions. The result is a solution which is four repetitions of the following 4row unit:
We do of course need to check that these units can be stacked on top of each other  the crucial point is that stacking 44.5cm and 98.5cm pieces on top of each other at the join results in joints 54cm apart  comfortably bigger than the 30cm minimum.
Per 4row unit, that contains eight whole boards, two 113cm pieces that can each be generated from a whole board by cutting off a 15.5cm piece of scrap, two 98.5cm and two 30cm pieces that can each be generated by cutting up a whole board without wastage, and two 44.5cm pieces. Unfortunately, those 44.5cm pieces don't come out of cutting up whole boards at all efficiently  we can get two of them per whole board, but that leaves a 39.5cm piece of scrap that we cannot use in the solution as outlined  we've already got all the 30cm pieces we need. The result is that each 4row unit requires 8+2+2+1 = 13 whole boards, producing a 4*13 = 52board solution  no improvement in the number of boards used.
But we have produced fewer, longer pieces of scrap: eight 15.5cm pieces and four 39.5cm pieces, compared with sixteen 15.5cm pieces and four 8.5cm pieces in the previous 52board solution. That's generally progress, not just in that it is reflected in having less cutting to do (24 cuts compared with 32), but also because it suggests that more of the same approach might do better. That's also suggested by the fact that we came a lot closer to having useful pieces of scrap  the longest (39.5cm) pieces were only 5cm short of the 44.5cm pieces we wanted, whereas originally the longest (15.5cm) pieces were 14.5cm short of the 30cm pieces we wanted.
Another way of looking at the newer 52board solution is that each of the top and bottom halves of a 4row unit is the result of the following way of cutting six boards placed in a row, plus the emboldened 44.5cm piece at the left end, 'folded up' around the point I've marked with an asterisk:
So a repetition consists of two sets of six whole boards cut up in that way, plus a 13th whole board cut up to produce the two 44.5cm pieces plus the 39.5cm piece of scrap. What if we do the same thing without the 44.5cm pieces, but with three more whole boards added to the left end in its place? I won't try to draw the resulting row of nine whole boards (the last diagram was quite wide enough!), but the result is that the rightmost of the three extra boards is cut into 44.5cm and 84cm pieces (with an extra 'folding up' point at the cut), and we need an extra piece of length 4m  2*128.5cm  84cm = 59cm at the left end  which is nice because while two 59cm pieces aren't anything like a perfect match to coming out of a 128.5cm board, they're a much better match than two 44.5cm pieces.
So after doing the 'folding up', we get a 6row unit:
There are however two problems with this unit. The minor one is that sixteen rows don't divide up cleanly into 6row units; the more major one is that the 6row units don't stack on top of each other, as that stacks 59cm and 84cm pieces on top of each other, resulting in joints 25 cm apart  too close.
But fortunately, the two problems have the same solution: the 6row and 4row units do stack on top of each other, as that stacks 98.5cm and 59cm pieces on each other, and 84cm and 44.5cm pieces on each other, in each case resulting in joints that are 39.5cm apart, and two 6row units with a 4row unit between them neatly fills sixteen rows. That produces the following 51board solution:
At this point, the total amount of scrap we have is 6*15.5cm + 2*10.5cm + 39.5cm = 153.5cm length of board. Continued use of the same sort of tweaking would have to get it down to 25 cm to reduce the number of boards used to 50, which looks pretty challenging... I haven't yet found a way to do it, and suspect there isn't one, but that's long way from being a proof that it's impossible!
Continued use of the same type of tweaking also cannot possibly get the amount of scrap down to the 103.5cm needed to reduce the number of boards used further to the minimum on area considerations of 49. However, I have been ignoring the scrap produced by trimming 7.2cm off one long edge, which might prove useful if we're allowed sufficiently general use of narrowerthan19.2cm boards. But totally general use of them seems to lead to somewhat silly conclusions, such as that we probably ought as our first step to cut each 19.2cmwide board into eight 2.4cmwide boards, which allow us to fill 3m with 125 of them (2.4cm having been calculated as the gcd of 19.2cm and 300cm). I.e. that way we can eliminate any possibility of producing scrap because of the problem of crossing the long edges of the room and reduce the issue to solely how little scrap we can produce along the length of the rows. And in fact, that produces a way of doing it with 50 boards: split them into 400 2.4cmwide boards, then construct 25 5row units each made up of 16 of those 2.4cmwide boards as follows:
These units don't stack on top of each other straightforwardly, as that produces joints separated by 59cm44.5cm = 98.5cm84cm = 14.5cm, but they do stack on top of each other if every alternate one is first reflected leftright, producing joints separated by 84cm44.5cm = 98.5cm59cm = 39.5cm. So reflect every other one of them leftright, then stack them all on top of each other to cover the entire 4m by 3m area.
Or a somewhat less 'silly extreme' 50board solution is to use 48 of the boards to construct 3 units of that type made up out of 19.2cmwide material, each covering an area of 4m by 5*19.2cm = 96cm. Then split each of the last two boards into 8 2.4cmwide boards and use the resulting 16 such boards to construct one more such unit, covering an area of 4m by 5*2.4cm = 12cm. Stack the four units we now have together, reflecting every alternate one leftright, and we're now covering an area of 4m by 3*96cm+12cm = 3m  but I can't help feeling that the result is still somewhat silly!
One might incidentally notice that two 5row units stacked on top of each other in this way produce the same pattern as the previous 6row unit stacked on top of the previous 4row unit. Indeed, that's how I discovered it  I asked myself how many boards alternating 6row and 4row units would require. That however produced a bit of a problem counting the pieces of various lengths, since the obvious repeat of a 6row unit and a 4row unit only produces counts easily for multiples of 10 rows. However, the observation that the first 5 and last 5 of those 10 rows were identical apart from a reflection made easy counts possible for multiples of 5 rows, and fortunately the number of rows I was interested in was 125, a multiple of 5...
Anyway, to sum the above up, if we're sufficiently unrestricted about cutting boards along their length to produce narrower boards, 50board solutions are certainly possible, and 49board solutions might be  though I'm doubtful, having tried quite hard to find one without success (nut equally with no hint that I might be on the track of proving it impossible). It's provable that <=48board solutions are not possible, on a simple area argument.
If instead we're sufficiently restricted about cutting boards along their length to produce narrower boards (we can't be totally restricted about it, otherwise it simply isn't possible to solve the problem at all regardless of the number of boards), all those numbers increase by 1: I've only found 51board solutions, am similarly doubtful about the existence of 50board solutions, and can prove that <=49board solutions are impossible by a simple (though not quite as simple) area argument. An example of such a restriction is if we're not allowed to produce boards narrower than 10cm.
(*) At least assuming that floor boards must not be cut in half  or indeed more bits!  to make multiple "floor boards" each with the same area, but probably too thin to produce a "floor" capable of performing a floor's primary function of safely supporting the people and other things that are on it... If that assumption is wrong, the answer is clearly that one plank will do the job, as long as one has the precision cutting equipment needed to slice it into 49+ nearpaperthin layers!
Gengulphus
Let's try to approach the optimum solution from above and from below  from below first:
Clearly the total area of the floor boards needs to be at least the area of the room, which is 3m x 4m = 12m^2 (*). The area of a plank is 0.192m x 1.285m = 0.24672m^2, so one needs at least 12m^2/0.24672m^2 = ~48.64 floor boards to cover the area. And since they come in whole numbers, that means at least 49 floor boards.
From above, a very crude solution is to lay out 64 floor boards in a 4 lengthwise by 16 widthwise grid, producing a rectangle 4*1.285m = 5.14m by 16*0.192m = 3.072m. Then trim that down to size in three steps, in which I'll refer to the 4 lengthwise by 1 widthwise sets of floor boards as "rows", and the 1 lengthwise by 16 widthwise sets as "columns":
1) Cut 98.5cm off the short edge of one of the outside columns, to reduce each floor board in that column to 30cm long and the rectangle to 4.155m x 3.072m.
2) Cut 15.5cm off the short edge of the other outside column, to reduce each floor board in that column to 113cm long and the rectangle to 4m x 3.072m.
3) Cut 7.2cm off the long edge of one of the outside rows to reduce the rectangle to 4m x 3m.
That layout doesn't comply with the "joints must be staggered and not closer than 30cm from each other" restriction, but that's easily fixed: reverse the order of the board pieces in every odd row, to produce a layout that is 8 repetitions of the following unit (minus 7.2cm along one long edge):
+++++
 113  128.5  128.5  30 
++++++++
 30  128.5  128.5  113 
+++++
So 64 floor boards are clearly enough. But that solution is clearly very wasteful, and we can easily improve it to 52 floor boards by simply observing that we used sixteen floor boards to produce sixteen 30cm pieces, when they can equally easily be produced by cutting 4 floor boards each into four 30cm pieces and one 8.5cm piece of scrap. So we now know that we need at least 49 floor boards and that we can do the job with 52, so the answer to question 1 is one of 49, 50, 51 or 52.
We're still wasting quite a lot of floor board with that 52board solution  most particularly, sixteen 15.5cm pieces of scrap from cutting the 113cm pieces, and the four 8.5cm pieces of scrap from cutting the sixteen 30cm lengths. That's enough to make up eight 30cm pieces, saving two boards out of the 52  if we can get the scrap in bigger pieces...
Which we can, at least to some extent, by observing that there is some exploitable flexability in the 52board solution. In particular, we can replace the 30cm and 113cm pieces by (30+X)cm and (113X)cm pieces, as long as we ensure that the joints implies by them are still at least 30cm apart  i.e. that (30+X) <= (113X)30, or equivalently that X <= 26.5. Doing that to all the 30cm and 113cm pieces doesn't help  the scrap pieces we get are still shorter than the pieces we want  but doing it to some of them will create longer scrap pieces from those that we do it to, which are long enough to be used in the ones we haven't done it to.
A simple example is that using X=14.5 on half the units will create 30cm pieces from cutting the (11314.5)cm = 98.5cm pieces, which can be used in the other half of the repetitions. The result is a solution which is four repetitions of the following 4row unit:
+++++
 98.5  128.5  128.5  44.5
++++++++
 30  128.5  128.5  113 
++++++++
 113  128.5  128.5  30 
++++++++
44.5  128.5  128.5  98.5 
+++++
We do of course need to check that these units can be stacked on top of each other  the crucial point is that stacking 44.5cm and 98.5cm pieces on top of each other at the join results in joints 54cm apart  comfortably bigger than the 30cm minimum.
Per 4row unit, that contains eight whole boards, two 113cm pieces that can each be generated from a whole board by cutting off a 15.5cm piece of scrap, two 98.5cm and two 30cm pieces that can each be generated by cutting up a whole board without wastage, and two 44.5cm pieces. Unfortunately, those 44.5cm pieces don't come out of cutting up whole boards at all efficiently  we can get two of them per whole board, but that leaves a 39.5cm piece of scrap that we cannot use in the solution as outlined  we've already got all the 30cm pieces we need. The result is that each 4row unit requires 8+2+2+1 = 13 whole boards, producing a 4*13 = 52board solution  no improvement in the number of boards used.
But we have produced fewer, longer pieces of scrap: eight 15.5cm pieces and four 39.5cm pieces, compared with sixteen 15.5cm pieces and four 8.5cm pieces in the previous 52board solution. That's generally progress, not just in that it is reflected in having less cutting to do (24 cuts compared with 32), but also because it suggests that more of the same approach might do better. That's also suggested by the fact that we came a lot closer to having useful pieces of scrap  the longest (39.5cm) pieces were only 5cm short of the 44.5cm pieces we wanted, whereas originally the longest (15.5cm) pieces were 14.5cm short of the 30cm pieces we wanted.
Another way of looking at the newer 52board solution is that each of the top and bottom halves of a 4row unit is the result of the following way of cutting six boards placed in a row, plus the emboldened 44.5cm piece at the left end, 'folded up' around the point I've marked with an asterisk:
++++++++++
 44.5  128.5  128.5  98.5 : 30  128.5  128.5  113 :15.5
++++++++++
* (scrap)
So a repetition consists of two sets of six whole boards cut up in that way, plus a 13th whole board cut up to produce the two 44.5cm pieces plus the 39.5cm piece of scrap. What if we do the same thing without the 44.5cm pieces, but with three more whole boards added to the left end in its place? I won't try to draw the resulting row of nine whole boards (the last diagram was quite wide enough!), but the result is that the rightmost of the three extra boards is cut into 44.5cm and 84cm pieces (with an extra 'folding up' point at the cut), and we need an extra piece of length 4m  2*128.5cm  84cm = 59cm at the left end  which is nice because while two 59cm pieces aren't anything like a perfect match to coming out of a 128.5cm board, they're a much better match than two 44.5cm pieces.
So after doing the 'folding up', we get a 6row unit:
+++++
 84  128.5  128.5  59 
++++++++
44.5  128.5  128.5  98.5 
++++++++
 113  128.5  128.5  30 
++++++++
 30  128.5  128.5  113 
++++++++
 98.5  128.5  128.5  44.5
++++++++
 59  128.5  128.5  84 
+++++
There are however two problems with this unit. The minor one is that sixteen rows don't divide up cleanly into 6row units; the more major one is that the 6row units don't stack on top of each other, as that stacks 59cm and 84cm pieces on top of each other, resulting in joints 25 cm apart  too close.
But fortunately, the two problems have the same solution: the 6row and 4row units do stack on top of each other, as that stacks 98.5cm and 59cm pieces on each other, and 84cm and 44.5cm pieces on each other, in each case resulting in joints that are 39.5cm apart, and two 6row units with a 4row unit between them neatly fills sixteen rows. That produces the following 51board solution:
+++++ Uses:
 84  128.5  128.5  59  32 uncut boards
++++++++ 6 boards each cut into:
44.5  128.5  128.5  98.5  one 113cm piece
++++++++ one 15.5cm piece (scrap)
 113  128.5  128.5  30  6 boards each cut into:
++++++++ one 98.5cm piece
 30  128.5  128.5  113  one 30cm piece
++++++++ 4 boards each cut into:
 98.5  128.5  128.5  44.5 one 84cm piece
++++++++ one 44.5cm piece
 59  128.5  128.5  84  2 boards each cut into:
++++++++ two 59cm pieces
 98.5  128.5  128.5  44.5 one 10.5cm piece (scrap)
++++++++ 1 board cut into:
 30  128.5  128.5  113  two 44.5cm pieces
++++++++ one 39.5cm piece (scrap)
 113  128.5  128.5  30 
++++++++
44.5  128.5  128.5  98.5 
++++++++
 84  128.5  128.5  59 
++++++++
44.5  128.5  128.5  98.5 
++++++++
 113  128.5  128.5  30 
++++++++
 30  128.5  128.5  113 
++++++++
 98.5  128.5  128.5  44.5
++++++++
 59  128.5  128.5  84 
+++++
At this point, the total amount of scrap we have is 6*15.5cm + 2*10.5cm + 39.5cm = 153.5cm length of board. Continued use of the same sort of tweaking would have to get it down to 25 cm to reduce the number of boards used to 50, which looks pretty challenging... I haven't yet found a way to do it, and suspect there isn't one, but that's long way from being a proof that it's impossible!
Continued use of the same type of tweaking also cannot possibly get the amount of scrap down to the 103.5cm needed to reduce the number of boards used further to the minimum on area considerations of 49. However, I have been ignoring the scrap produced by trimming 7.2cm off one long edge, which might prove useful if we're allowed sufficiently general use of narrowerthan19.2cm boards. But totally general use of them seems to lead to somewhat silly conclusions, such as that we probably ought as our first step to cut each 19.2cmwide board into eight 2.4cmwide boards, which allow us to fill 3m with 125 of them (2.4cm having been calculated as the gcd of 19.2cm and 300cm). I.e. that way we can eliminate any possibility of producing scrap because of the problem of crossing the long edges of the room and reduce the issue to solely how little scrap we can produce along the length of the rows. And in fact, that produces a way of doing it with 50 boards: split them into 400 2.4cmwide boards, then construct 25 5row units each made up of 16 of those 2.4cmwide boards as follows:
+++++ Uses:
 84  128.5  128.5  59  10 uncut boards
++++++++ 2 boards each cut into:
44.5  128.5  128.5  98.5  one 113cm piece
++++++++ one 15.5cm piece (scrap)
 113  128.5  128.5  30  2 boards each cut into:
++++++++ one 98.5cm piece
 30  128.5  128.5  113  one 30cm piece
++++++++ 1 board cut into:
 98.5  128.5  128.5  44.5 one 84cm piece
+++++ one 44.5cm piece
1 board cut into:
one 59cm piece
one 44.5cm piece
one 25cm piece (scrap)
These units don't stack on top of each other straightforwardly, as that produces joints separated by 59cm44.5cm = 98.5cm84cm = 14.5cm, but they do stack on top of each other if every alternate one is first reflected leftright, producing joints separated by 84cm44.5cm = 98.5cm59cm = 39.5cm. So reflect every other one of them leftright, then stack them all on top of each other to cover the entire 4m by 3m area.
Or a somewhat less 'silly extreme' 50board solution is to use 48 of the boards to construct 3 units of that type made up out of 19.2cmwide material, each covering an area of 4m by 5*19.2cm = 96cm. Then split each of the last two boards into 8 2.4cmwide boards and use the resulting 16 such boards to construct one more such unit, covering an area of 4m by 5*2.4cm = 12cm. Stack the four units we now have together, reflecting every alternate one leftright, and we're now covering an area of 4m by 3*96cm+12cm = 3m  but I can't help feeling that the result is still somewhat silly!
One might incidentally notice that two 5row units stacked on top of each other in this way produce the same pattern as the previous 6row unit stacked on top of the previous 4row unit. Indeed, that's how I discovered it  I asked myself how many boards alternating 6row and 4row units would require. That however produced a bit of a problem counting the pieces of various lengths, since the obvious repeat of a 6row unit and a 4row unit only produces counts easily for multiples of 10 rows. However, the observation that the first 5 and last 5 of those 10 rows were identical apart from a reflection made easy counts possible for multiples of 5 rows, and fortunately the number of rows I was interested in was 125, a multiple of 5...
Anyway, to sum the above up, if we're sufficiently unrestricted about cutting boards along their length to produce narrower boards, 50board solutions are certainly possible, and 49board solutions might be  though I'm doubtful, having tried quite hard to find one without success (nut equally with no hint that I might be on the track of proving it impossible). It's provable that <=48board solutions are not possible, on a simple area argument.
If instead we're sufficiently restricted about cutting boards along their length to produce narrower boards (we can't be totally restricted about it, otherwise it simply isn't possible to solve the problem at all regardless of the number of boards), all those numbers increase by 1: I've only found 51board solutions, am similarly doubtful about the existence of 50board solutions, and can prove that <=49board solutions are impossible by a simple (though not quite as simple) area argument. An example of such a restriction is if we're not allowed to produce boards narrower than 10cm.
(*) At least assuming that floor boards must not be cut in half  or indeed more bits!  to make multiple "floor boards" each with the same area, but probably too thin to produce a "floor" capable of performing a floor's primary function of safely supporting the people and other things that are on it... If that assumption is wrong, the answer is clearly that one plank will do the job, as long as one has the precision cutting equipment needed to slice it into 49+ nearpaperthin layers!
Gengulphus

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Re: Minimum Flooring Boards and Wastage
Thanks GS, mathematically correct answer I am sure but unfortunately not a practical one if I read it correctly, and the fault is mine, apologies.
I forgot to define in the problem something very practical, which doesn't make your answer any less elegant.
I forgot to define in the problem something very practical, which doesn't make your answer any less elegant.

 Lemon Slice
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Re: Minimum Flooring Boards and Wastage
Gengulphus wrote:If instead we're sufficiently restricted about cutting boards along their length to produce narrower boards (we can't be totally restricted about it, otherwise it simply isn't possible to solve the problem at all regardless of the number of boards)
Gengulphus
While not stated in the question, a practical restriction would be that the boards are probably tongued and grooved. This would allow the last row to be cut lengthways but would not allow, for example, three boards of width 6cm to be cut from one board and placed sidebyside (or one each side of the floor) to make up the odd 12cm. The OP does state the following, however, which applies the same restriction:
JMN2 wrote:The planks are 19.2cm wide so the width takes 16 rows of planks of which one is cut to roughly half lengthwise.
I am guessing that this flooring is decorative and is applied over an existing floor otherwise we would have to ensure that any joins were centred over joists.
Julian F. G. W.

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Re: Minimum Flooring Boards and Wastage
GoSeigen, close but no cigar. I'm afraid you've ignored the "roughly" in JMN2's statement that "The planks are 19.2cm wide so the width takes 16 rows of planks of which one is cut to roughly half lengthwise." In fact, 15 rows of 19.2cm wide planks total 288cm, so 12cm more of width is required to fill 3m  which is 2.4cm more than a halfwidth plank. Distributed over 15 small gaps between rows of planks, that's 1.6mm per gap, which I would view as not the best of flooring jobs even viewing it as a practical DIY problem! But viewed as a puzzle, any nonzero gap is unacceptable...
You're trying to cover that last partial row with one plank fewer than the 25/8 planks that you use to cover a fullwidth row, i.e. with 17/8 planks, or 17/25ths = 0.680 times the area of planking needed to precisely cover a full row. The area of the partial row is 12cm/19.2cm = 0.625 times that of a fullwidth row. So it's not obviously impossible, but you can't discard more than 0.055 times the area of a fullwidth row. And your comment "Halfwidth row, using one full length plank cut in half lengthwise; discard halfwidth of other pieces in this row." ends up discarding 9/25ths horizontally by 3/8ths vertically = 0.135 times the area of a fullwidth row... The net result is that your solution could potentially be rescued, but only by discarding quite a bit less than half the area of halfwidth pieces that it does. Whether that can actually be done, I don't know  but the fact that the partial row is more than half a fullwidth row was a problem I struggled with in my own attempts to find a 49board solution, without success...
Not even after noting that 25/8 planks are 401.625cm long, giving a further 1.625cm potentially recoverable from each of the other 15 rows if one can find a good way of bringing it all together in worthwhilesized pieces!
Gengulphus
You're trying to cover that last partial row with one plank fewer than the 25/8 planks that you use to cover a fullwidth row, i.e. with 17/8 planks, or 17/25ths = 0.680 times the area of planking needed to precisely cover a full row. The area of the partial row is 12cm/19.2cm = 0.625 times that of a fullwidth row. So it's not obviously impossible, but you can't discard more than 0.055 times the area of a fullwidth row. And your comment "Halfwidth row, using one full length plank cut in half lengthwise; discard halfwidth of other pieces in this row." ends up discarding 9/25ths horizontally by 3/8ths vertically = 0.135 times the area of a fullwidth row... The net result is that your solution could potentially be rescued, but only by discarding quite a bit less than half the area of halfwidth pieces that it does. Whether that can actually be done, I don't know  but the fact that the partial row is more than half a fullwidth row was a problem I struggled with in my own attempts to find a 49board solution, without success...
Not even after noting that 25/8 planks are 401.625cm long, giving a further 1.625cm potentially recoverable from each of the other 15 rows if one can find a good way of bringing it all together in worthwhilesized pieces!
Gengulphus
Last edited by Gengulphus on February 13th, 2018, 6:27 pm, edited 1 time in total.

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Re: Minimum Flooring Boards and Wastage
JMN2 wrote:Thanks GS, mathematically correct answer I am sure but unfortunately not a practical one if I read it correctly, and the fault is mine, apologies.
I forgot to define in the problem something very practical, which doesn't make your answer any less elegant.
I am guessing that you would like as few sawcuts as possible (especially if you are cutting them by hand)
Some flooring is tongued and grooved on the ends so another restriction may be that all cut ends would have to be at the ends of the floor. If this is the case, if you trim a board to fit the righthand end, the offcut can only be used on the lefthand end.
Julian F. G. W.

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Re: Minimum Flooring Boards and Wastage
Spolier...
Scroll down for answer
Buy more than you need.
Keep receipt.
Return unused packs.
Scroll down for answer
Buy more than you need.
Keep receipt.
Return unused packs.

 Lemon Quarter
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Re: Minimum Flooring Boards and Wastage
Yes, I've got loads and more ordered, and if there's a lot left over I'll use the same for the living room too and not swop styles.
What I was thinking was in a practical solution one would have all the sawn edges against the skirting board.
What I was thinking was in a practical solution one would have all the sawn edges against the skirting board.

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Re: Minimum Flooring Boards and Wastage
Gengulphus wrote:GoSeigen, close but no cigar. I'm afraid you've ignored the "roughly" in JMN2's statement that "The planks are 19.2cm wide so the width takes 16 rows of planks of which one is cut to roughly half lengthwise."
Gengulphus
You're quite right. I wavered, but finally ignored the word "roughly" and didn't bother to check the width, thinking the OP would not have mentioned the "half" aspect unless it were useful/significant.
Agree that as a puzzle solution my answer is therefore unacceptable. It did seem a bit too easy at the time!
GS

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Re: Minimum Flooring Boards and Wastage
jfgw wrote:While not stated in the question, a practical restriction would be that the boards are probably tongued and grooved. This would allow the last row to be cut lengthways but would not allow, for example, three boards of width 6cm to be cut from one board and placed sidebyside (or one each side of the floor) to make up the odd 12cm. ...
Agreed, but there are plenty of other practical issues that affect a realworld solution  e.g. the questions you raise about positioning with respect to joists and about whether the ends are also tongued and grooved (which doesn't just affect where cut ends must go, but also which wall they must be against and hence when a piece has been trimmed both lengthwise and widthwise, which corner it must go into), the fact that in real life, a saw cut has a nonzero width and so wastes a small amount of material, and the difficulty of sawing a very small amount off the end of a plank (obviously that doesn't affect the cuts that take off pieces you're going to use, but it could affect taking off small pieces of scrap).
So basically, one can either treat the question as a puzzle, or as a practical issue that requires quite a few extra details to be given before it can really be answered. Given the title of this board and the fact that the question supplied enough detail for a puzzle but not enough for a practical issue, I basically chose the former interpretation, only noting practical issues when ignoring them led to truly silly 'solutions' such as the sliceinto50paperthinlayers one.
By the way, that's not saying that I think the board's title forbids practical questions, just that if a question doesn't say clearly that it is a practical issue and it could be taken either as a puzzle or as a practical issue, the default is to treat it as a puzzle. I.e. if one wants a question to be treated as a practical issue, say that it is one! Though it might actually be better to post it to an appropriate other board, such as the Building and DIY one in this case, to get a better chance that the practical details will be addressed properly, with a crosspost here when an issue has strongly puzzlelike aspects. And having thought to look for such a board for the purpose of this post, I note that JMN2 has in fact recently posted about flooring there. It isn't totally clear to me whether this is about the same flooring job, but the 19.2cm width of the boards does look to be a common feature...
jfgw wrote:... The OP does state the following, however, which applies the same restriction:JMN2 wrote:The planks are 19.2cm wide so the width takes 16 rows of planks of which one is cut to roughly half lengthwise.
That's not a restriction, at least as I read it  it's just an observation about what fits the room's width. It would need to read as something like "The solution is to use 16 rows of 19.2cmwide planks of which one is to be cut to roughly half lengthwise" to be a restriction.
Gengulphus

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Re: Minimum Flooring Boards and Wastage
Gengulphus wrote: jfgw wrote:
... The OP does state the following, however, which applies the same restriction:
JMN2 wrote:
The planks are 19.2cm wide so the width takes 16 rows of planks of which one is cut to roughly half lengthwise.
That's not a restriction, at least as I read it  it's just an observation about what fits the room's width. It would need to read as something like "The solution is to use 16 rows of 19.2cmwide planks of which one is to be cut to roughly half lengthwise" to be a restriction.
Fair comment. The word "so" makes it a false statement rather than a stipulation.
Referring to the thread on Building and DIY, it appears that the OP had laminate flooring in mind. The board dimensions support this notion. A more practical puzzle would specify tongued and grooved boards (including the ends) but would not have restrictions due to joist positions. Another restriction may be that joins on nearby runs should not be aligned (the pattern should look fairly random, not like a brick wall). What is acceptable would need to be clarified in order to be incorporated into a puzzle like this.
For the given area and board length, and where laminate flooring is stipulated, the saw kerf may be allowed for by assuming that the floor is one kerf width longer, e.g., 4.003m for a 3mm kerf. This is possible because both ends of each run have to be cut (otherwise one end would have a piece less than 300mm long) so the amount lost is the same for each run.
Julian F. G. W.
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