I am trying to make an app that has the four basic mathematical operations. Addition, subtraction, multiplication and division.
Each operation has a series of exercises and for each correct exercise, assign a point in a score counter.
I clarify that both, the exercises and the answers are selected at random. They are not questions and answers stored in a list.
Everything is ready and finished, but I have a problem with the division and it is as follows.
If, for example, the result of the division has exactly two decimals, the score counter takes as correct the answer that is selected. But if the result of the division has more than two decimals, the score counter does not take the answer as correct.
Example:
20/8 = 1.25 No more decimals, then the score counter takes it as the correct answer
9/7 = 1.28571428571 This answer has many decimals, then the score counter does not take it as the correct answer
The problem is not in rounding up the figures or in formatting the number of decimals. The problem is that for some reason, answers with more than two decimal numbers are not taken as correct.
Not matter if I round the result to a integer or if I set only a 2 decimals for each result, for some reason, the score counter do not show the result as correct.
For example, if I take the division 9/7 = 1.28571428571 and I set only 2 decimals for the result, leaving it as 1.28, the score counter do not take this result as a correct result.
Even if I round the result to 1, occur the same problem.
How can this be fixed?
Many thanks to anyone who can help me find a solution.
P.S.: I'm not a programmer I'm just an amateur and nothing else, that just starts, so I appreciate, please, answers for a layman like me. Thanks in advance.
Here are the blocks
Related
I've been reading a lot about multicollinearity but am still unsure whether to use the Durbin-Watson score, the eigenvalues or the variance inflation factor. I only have three independent variables and the eigenvalues are:
1.81768828 0.95241948 0.22989225
How I understood it, only values too close to zero indicate multicollinearity. I wasn't sure if the last one (0.22) counts as "close to zero" but when checking its eigenvector, the result is:
-0.53977799 -0.44013805 0.71757802
and each of them would indicate collinearity as they are NOT close to zero (this time it's the other way around). Am I correct until here?
The Durbin-Watson score is 1.93 (calculated through the summary() function from statsmodels with an added intercept). This does NOT show strong multicollinearity, right?
As nobody gives clear "cutoff" values, I am a bit confusing as to which values count as "close to zero" or not.
Should I calculate the VIF as well, just to be extra sure?
Any help is much appreciated!
I was going through Google Interview Questions. to implement the random number generation from 1 to 7.
I did write a simple code, I would like to understand if in the interview this question asked to me and if I write the below code is it Acceptable or not?
import time
def generate_rand():
ret = str(time.time()) # time in second like, 12345.1234
ret = int(ret[-1])
if ret == 0 or ret == 1:
return 1
elif ret > 7:
ret = ret - 7
return ret
return ret
while 1:
print(generate_rand())
time.sleep(1) # Just to see the output in the STDOUT
(Since the question seems to ask for analysis of issues in the code and not a solution, I am not providing one. )
The answer is unacceptable because:
You need to wait for a second for each random number. Many applications need a few hundred at a time. (If the sleep is just for convenience, note that even a microsecond granularity will not yield true random numbers as the last microsecond will be monotonically increasing until 10us are reached. You may get more than a few calls done in a span of 10us and there will be a set of monotonically increasing pseudo-random numbers).
Random numbers have uniform distribution. Each element should have the same probability in theory. In this case, you skew 1 more (twice the probability for 0, 1) and 7 more (thrice the probability for 7, 8, 9) compared to the others in the range 2-6.
Typically answers to this sort of a question will try to get a large range of numbers and distribute the ranges evenly from 1-7. For example, the above method would have worked fine if u had wanted randomness from 1-5 as 10 is evenly divisible by 5. Note that this will only solve (2) above.
For (1), there are other sources of randomness, such as /dev/random on a Linux OS.
You haven't really specified the constraints of the problem you're trying to solve, but if it's from a collection of interview questions it seems likely that it might be something like this.
In any case, the answer shown would not be acceptable for the following reasons:
The distribution of the results is not uniform, even if the samples you read from time.time() are uniform.
The results from time.time() will probably not be uniform. The result depends on the time at which you make the call, and if your calls are not uniformly distributed in time then the results will probably not be uniformly distributed either. In the worst case, if you're trying to randomise an array on a very fast processor then you might complete the entire operation before the time changes, so the whole array would be filled with the same value. Or at least large chunks of it would be.
The changes to the random value are highly predictable and can be inferred from the speed at which your program runs. In the very-fast-computer case you'll get a bunch of x followed by a bunch of x+1, but even if the computer is much slower or the clock is more precise, you're likely to get aliasing patterns which behave in a similarly predictable way.
Since you take the time value in decimal, it's likely that the least significant digit doesn't visit all possible values uniformly. It's most likely a conversion from binary to some arbitrary number of decimal digits, and the distribution of the least significant digit can be quite uneven when that happens.
The code should be much simpler. It's a complicated solution with many special cases, which reflects a piecemeal approach to the problem rather than an understanding of the relevant principles. An ideal solution would make the behaviour self-evident without having to consider each case individually.
The last one would probably end the interview, I'm afraid. Perhaps not if you could tell a good story about how you got there.
You need to understand the pigeonhole principle to begin to develop a solution. It looks like you're reducing the time to its least significant decimal digit for possible values 0 to 9. Legal results are 1 to 7. If you have seven pigeonholes and ten pigeons then you can start by putting your first seven pigeons into one hole each, but then you have three pigeons left. There's nowhere that you can put the remaining three pigeons (provided you only use whole pigeons) such that every hole has the same number of pigeons.
The problem is that if you pick a pigeon at random and ask what hole it's in, the answer is more likely to be a hole with two pigeons than a hole with one. This is what's called "non-uniform", and it causes all sorts of problems, depending on what you need your random numbers for.
You would either need to figure out how to ensure that all holes are filled equally, or you would have to come up with an explanation for why it doesn't matter.
Typically the "doesn't matter" answer is that each hole has either a million or a million and one pigeons in it, and for the scale of problem you're working with the bias would be undetectable.
Using the same general architecture you've created, I would do something like this:
import time
def generate_rand():
ret = str(time.time()) # time in second like, 12345.1234
ret = ret % 8 # will return pseudorandom numbers 0-7
if ret == 0:
return 1 # or you could also return the result of another call to generate_rand()
return ret
while 1:
print(generate_rand())
time.sleep(1)
I am making some software that need to work with integers.
Also I need to apply some formula to those integers, repeatedly over time (example, do x/=z several times in a row for a indefinite amount).
All tools, algorithms and formulas I could think or find, or don't work with integers at all, or work as approximations at best.
For example the x/=z several times in a row for example, you can theoretically calculate what x will be in the 10th time by doing x = x/(z^10), but that will be wrong if the result is fractional, you can use floor(x/(z^10)), but the result will STILL be wrong.
Plotting software that I found also don't have integers at all, or has "floor()/ceil()" functions support, at best, and still the result would fall in the problem of the previous paragraph.
So how I do it?
Here's something to get you going for the iteration of x/=z:
(that should have ended in "all three terms are 0 with regard to integer division")
Now if x or z are negative, you can try and see whether this still holds; I did not invest the time to make the necessary case distinctions, but they should be fairly analogous.
As Karoly Horvath mentions in a comment, without a clear specification of the kinds of functions for which you would like to find a shortcut to replace iterative evaluation, helping you out won't be possible since there are uncountably many functions over the integers, and the same approach won't work for all of them.
Why this code 7.30 - 7.20 in ruby returns 0.0999999999999996, not 0.10?
But if i'll write 7.30 - 7.16, for example, everything will be ok, i'll get 0.14.
What the problem, and how can i solve it?
What Every Computer Scientist Should Know About Floating-Point Arithmetic
The problem is that some numbers we can easily write in decimal don't have an exact representation in the particular floating point format implemented by current hardware. A casual way of stating this is that all the integers do, but not all of the fractions, because we normally store the fraction with a 2**e exponent. So, you have 3 choices:
Round off appropriately. The unrounded result is always really really close, so a rounded result is invariably "perfect". This is what Javascript does and lots of people don't even realize that JS does everything in floating point.
Use fixed point arithmetic. Ruby actually makes this really easy; it's one of the only languages that seamlessly shifts to Class Bignum from Fixnum as numbers get bigger.
Use a class that is designed to solve this problem, like BigDecimal
To look at the problem in more detail, we can try to represent your "7.3" in binary. The 7 part is easy, 111, but how do we do .3? 111.1 is 7.5, too big, 111.01 is 7.25, getting closer. Turns out, 111.010011 is the "next closest smaller number", 7.296875, and when we try to fill in the missing .003125 eventually we find out that it's just 111.010011001100110011... forever, not representable in our chosen encoding in a finite bit string.
The problem is that floating point is inaccurate. You can solve it by using Rational, BigDecimal or just plain integers (for example if you want to store money you can store the number of cents as an int instead of the number of dollars as a float).
BigDecimal can accurately store any number that has a finite number of digits in base 10 and rounds numbers that don't (so three thirds aren't one whole).
Rational can accurately store any rational number and can't store irrational numbers at all.
That is a common error from how float point numbers are represented in memory.
Use BigDecimal if you need exact results.
result=BigDecimal.new("7.3")-BigDecimal("7.2")
puts "%2.2f" % result
It is interesting to note that a number that has few decimals in one base may typically have a very large number of decimals in another. For instance, it takes an infinite number of decimals to express 1/3 (=0.3333...) in the base 10, but only one decimal in the base 3. Similarly, it takes many decimals to express the number 1/10 (=0.1) in the base 2.
Since you are doing floating point math then the number returned is what your computer uses for precision.
If you want a closer answer, to a set precision, just multiple the float by that (such as by 100), convert it to an int, do the math, then divide.
There are other solutions, but I find this to be the simplest since rounding always seems a bit iffy to me.
This has been asked before here, you may want to look for some of the answers given before, such as this one:
Dealing with accuracy problems in floating-point numbers
I have an array of numbers that potentially have up to 8 decimal places and I need to find the smallest common number I can multiply them by so that they are all whole numbers. I need this so all the original numbers can all be multiplied out to the same scale and be processed by a sealed system that will only deal with whole numbers, then I can retrieve the results and divide them by the common multiplier to get my relative results.
Currently we do a few checks on the numbers and multiply by 100 or 1,000,000, but the processing done by the *sealed system can get quite expensive when dealing with large numbers so multiplying everything by a million just for the sake of it isn’t really a great option. As an approximation lets say that the sealed algorithm gets 10 times more expensive every time you multiply by a factor of 10.
What is the most efficient algorithm, that will also give the best possible result, to accomplish what I need and is there a mathematical name and/or formula for what I’m need?
*The sealed system isn’t really sealed. I own/maintain the source code for it but its 100,000 odd lines of proprietary magic and it has been thoroughly bug and performance tested, altering it to deal with floats is not an option for many reasons. It is a system that creates a grid of X by Y cells, then rects that are X by Y are dropped into the grid, “proprietary magic” occurs and results are spat out – obviously this is an extremely simplified version of reality, but it’s a good enough approximation.
So far there are quiet a few good answers and I wondered how I should go about choosing the ‘correct’ one. To begin with I figured the only fair way was to create each solution and performance test it, but I later realised that pure speed wasn’t the only relevant factor – an more accurate solution is also very relevant. I wrote the performance tests anyway, but currently the I’m choosing the correct answer based on speed as well accuracy using a ‘gut feel’ formula.
My performance tests process 1000 different sets of 100 randomly generated numbers.
Each algorithm is tested using the same set of random numbers.
Algorithms are written in .Net 3.5 (although thus far would be 2.0 compatible)
I tried pretty hard to make the tests as fair as possible.
Greg – Multiply by large number
and then divide by GCD – 63
milliseconds
Andy – String Parsing
– 199 milliseconds
Eric – Decimal.GetBits – 160 milliseconds
Eric – Binary search – 32
milliseconds
Ima – sorry I couldn’t
figure out a how to implement your
solution easily in .Net (I didn’t
want to spend too long on it)
Bill – I figure your answer was pretty
close to Greg’s so didn’t implement
it. I’m sure it’d be a smidge faster
but potentially less accurate.
So Greg’s Multiply by large number and then divide by GCD” solution was the second fastest algorithm and it gave the most accurate results so for now I’m calling it correct.
I really wanted the Decimal.GetBits solution to be the fastest, but it was very slow, I’m unsure if this is due to the conversion of a Double to a Decimal or the Bit masking and shifting. There should be a
similar usable solution for a straight Double using the BitConverter.GetBytes and some knowledge contained here: http://blogs.msdn.com/bclteam/archive/2007/05/29/bcl-refresher-floating-point-types-the-good-the-bad-and-the-ugly-inbar-gazit-matthew-greig.aspx but my eyes just kept glazing over every time I read that article and I eventually ran out of time to try to implement a solution.
I’m always open to other solutions if anyone can think of something better.
I'd multiply by something sufficiently large (100,000,000 for 8 decimal places), then divide by the GCD of the resulting numbers. You'll end up with a pile of smallest integers that you can feed to the other algorithm. After getting the result, reverse the process to recover your original range.
Multiple all the numbers by 10
until you have integers.
Divide
by 2,3,5,7 while you still have all
integers.
I think that covers all cases.
2.1 * 10/7 -> 3
0.008 * 10^3/2^3 -> 1
That's assuming your multiplier can be a rational fraction.
If you want to find some integer N so that N*x is also an exact integer for a set of floats x in a given set are all integers, then you have a basically unsolvable problem. Suppose x = the smallest positive float your type can represent, say it's 10^-30. If you multiply all your numbers by 10^30, and then try to represent them in binary (otherwise, why are you even trying so hard to make them ints?), then you'll lose basically all the information of the other numbers due to overflow.
So here are two suggestions:
If you have control over all the related code, find another
approach. For example, if you have some function that takes only
int's, but you have floats, and you want to stuff your floats into
the function, just re-write or overload this function to accept
floats as well.
If you don't have control over the part of your system that requires
int's, then choose a precision to which you care about, accept that
you will simply have to lose some information sometimes (but it will
always be "small" in some sense), and then just multiply all your
float's by that constant, and round to the nearest integer.
By the way, if you're dealing with fractions, rather than float's, then it's a different game. If you have a bunch of fractions a/b, c/d, e/f; and you want a least common multiplier N such that N*(each fraction) = an integer, then N = abc / gcd(a,b,c); and gcd(a,b,c) = gcd(a, gcd(b, c)). You can use Euclid's algorithm to find the gcd of any two numbers.
Greg: Nice solution but won't calculating a GCD that's common in an array of 100+ numbers get a bit expensive? And how would you go about that? Its easy to do GCD for two numbers but for 100 it becomes more complex (I think).
Evil Andy: I'm programing in .Net and the solution you pose is pretty much a match for what we do now. I didn't want to include it in my original question cause I was hoping for some outside the box (or my box anyway) thinking and I didn't want to taint peoples answers with a potential solution. While I don't have any solid performance statistics (because I haven't had any other method to compare it against) I know the string parsing would be relatively expensive and I figured a purely mathematical solution could potentially be more efficient.
To be fair the current string parsing solution is in production and there have been no complaints about its performance yet (its even in production in a separate system in a VB6 format and no complaints there either). It's just that it doesn't feel right, I guess it offends my programing sensibilities - but it may well be the best solution.
That said I'm still open to any other solutions, purely mathematical or otherwise.
What language are you programming in? Something like
myNumber.ToString().Substring(myNumber.ToString().IndexOf(".")+1).Length
would give you the number of decimal places for a double in C#. You could run each number through that and find the largest number of decimal places(x), then multiply each number by 10 to the power of x.
Edit: Out of curiosity, what is this sealed system which you can pass only integers to?
In a loop get mantissa and exponent of each number as integers. You can use frexp for exponent, but I think bit mask will be required for mantissa. Find minimal exponent. Find most significant digits in mantissa (loop through bits looking for last "1") - or simply use predefined number of significant digits.
Your multiple is then something like 2^(numberOfDigits-minMantissa). "Something like" because I don't remember biases/offsets/ranges, but I think idea is clear enough.
So basically you want to determine the number of digits after the decimal point for each number.
This would be rather easier if you had the binary representation of the number. Are the numbers being converted from rationals or scientific notation earlier in your program? If so, you could skip the earlier conversion and have a much easier time. Otherwise you might want to pass each number to a function in an external DLL written in C, where you could work with the floating point representation directly. Or you could cast the numbers to decimal and do some work with Decimal.GetBits.
The fastest approach I can think of in-place and following your conditions would be to find the smallest necessary power-of-ten (or 2, or whatever) as suggested before. But instead of doing it in a loop, save some computation by doing binary search on the possible powers. Assuming a maximum of 8, something like:
int NumDecimals( double d )
{
// make d positive for clarity; it won't change the result
if( d<0 ) d=-d;
// now do binary search on the possible numbers of post-decimal digits to
// determine the actual number as quickly as possible:
if( NeedsMore( d, 10e4 ) )
{
// more than 4 decimals
if( NeedsMore( d, 10e6 ) )
{
// > 6 decimal places
if( NeedsMore( d, 10e7 ) ) return 10e8;
return 10e7;
}
else
{
// <= 6 decimal places
if( NeedsMore( d, 10e5 ) ) return 10e6;
return 10e5;
}
}
else
{
// <= 4 decimal places
// etc...
}
}
bool NeedsMore( double d, double e )
{
// check whether the representation of D has more decimal points than the
// power of 10 represented in e.
return (d*e - Math.Floor( d*e )) > 0;
}
PS: you wouldn't be passing security prices to an option pricing engine would you? It has exactly the flavor...