Friday Quiz problem 18 Mar 22 - how many beans make 5?

H

ere's today's question:

I played 40 games of backgammon and scored 25 points.  A win counts as one point, a draw counts as half a point, and a loss counts as zero points.  How many more games did I win than lose?

At first glance it looks as if you can 'formularise' this (ie w + d/2 + 0l = 25)  but you can't solve that, it seems as if there are many values that could work.

My title refers to an old kids' joke - "how many beans make 5?" The answer I knew was "two beans, a bean and a half, half a bean and a bean" (said quickly!) But other kids had their own version of this and you can have many combinations of beans and half-beans. 

 

The question is curiously-worded. "how many more times did I win than lose?"  

My starting point was that if you could see all the possible values of w, d and l  that total 25 points, the pattern would be obvious.

So the problem becomes one of printing out these values.

The easiest brute-force way is to nest 3 loops 0 - 40, and ignore unless the 3 add up to 40. But that's inefficient and inelegant.

We're going to win anywhere between 0 and 40 games:

10 FOR W = 0 TO 40

(I'm using BASIC, it's neat when it comes to nested loops and printing variables).

For each of those cases, the remainder (40-W) will be split between draws and losses. If you know the wins and draws, then the loses will be 40-(D+W). That's only two loops, with the inner one becoming smaller each time.

Then we're only interested in those combinations where the points add up to 25 (see line 40) 

10 FOR W = 0 TO 40                                                              

20 FOR D = 0 TO (40-W)                                                            

30 LET L = 40 - (D+W)                                                     

40 IF (W*2)+D=50 THEN PRINT W; D; L; (W-L)                                                    

50 NEXT:NEXT

(I've doubled the values in 40 so that we're dealing with integers. Using a floating-point value when you're testing equality is asking for trouble.)

That's still a lot of looping. But it's done in a couple of seconds.

Answer

Seeing the output from this program makes it obvious that it doesn't matter that we don't know the actual number of games won/lost/drawn, because the answer is always 10 for all possible values.

I had to look at the results with my human eyes to see the answer. The question asked about the difference between wins and loses. So perhaps 40 could also print that difference, to make it more obvious that it's always 10:

40 IF (W*2)+D=50 THEN PRINT W; D; L; (W-L) 

Efficiency

To make this more efficient, we can think about upper and lower limits. It's obvious now that if we win any more than 25 games, that makes more than 25 points, so that's our upper limit for W in line 10

There's a lower limit too - we can't make 25 points entirely from draws (half-points) because there aren't enough games. We can see from the result of the program that if we draw 30 (15 points) that's the maximum number of draws. Any more than that and it's not possible to make enough wins to make up the points.  But I'm not sure whether it's possible to calculate that, it only becomes clear when you see the values. 

So 

10 FOR W = 0 TO 25 gives us the same result and is much quicker. 

The question that remains is: is it possible to arrive at the answer - that wins are always 10 more than losses - using reason and without actually looking at the numbers and observing that fact?