I am currently making a random math quiz which generates random addition, subtraction, and multiplication fine. Those 3 are simple. I would like to know just how to make a randomly generated division question that always equals a whole number, using 2 int variables, num1 and num2 but I just cannot figure it out Any help would be greatly appreciated!
It should be pretty simple. The logic may go like this:
Randomly generate a number and call it divisor
Multiply your divisor with a randomly generated number and call it dividend
Now you've got two integer numbers (dividend & divisor) that should solve into a whole number quotient.
I would construct the the two integers using a pool of random generated factors, making sure the two have at least one factor in common. Or to make it even simpler, pick a random integer n, then multiply it by another random integer m and show the division m*n/n.
Related
I am making an anagram solver in Visual Basic that gives you every possible combination when you enter a string. I need to work out how many combinations there are depending on the amount of characters in the string and how many different characters there are.
E.G.
Sample string:
abc
Total characters: 3, Different Characters: 3
Possible combinations: 6
abc, acb, bac, bca, cab, cba
I need an equation (using the number of characters and different characters) to link this to a string that contains a different amount of characters.
I've been using trial and error to try and figure is out, but I can't quite get my head around it. So far I have:
((letters - 1) ^ (different letters - 1)) + (letters - 1)
which works for a few different letter counts but now for all.
Help please???
I'll lead you to the answer, but I'll try to explain along the way. Let's say you had 10 different letters. You'd have 10 choices for the first, 9 for the second, 8 for the third, etc. Ultimately, there would be 10*9*8*7*6...*2*1 = 10! possibilities. However, sometimes you'll have multiple instances of the same letter. For example, using that for the string "aaabcd" would overcount possibilities, because it counts each of the a's as distinct letters, even though they're not. To correct for that, you would have to divide by the factorial of the number of repeated letters. A good way to calculate the total number of possibilities would be (total number of letters factorial)/ (product of the factorials of the number of repeated instances of each letter).
For example:
There are 6!/(3!) ways to arrange the letters in "aaabcd"
There are 6! ways to arrange the letters is "abcdef"
There are 6!/(3!*2!) ways to arrange the letters in "aaabbc"
There are 10!/(5!*3!*2!) ways to arrange the letters in "aaaaabbbcc"
I hope this helps.
For the possible counting number, it's exactly the same as computing Multinomial Coefficient
A simple explanation is that, for no repeating characters,
It's simply permutation = n!
(It is easy to understand if you draw a tree diagram, with first character has n choices, second character has n-1choices...etc.)
However as you may have repeating characters, you will double count many of them.
Let's see an simple example: for aaa, how many possible arrangements IF WE COUNT EVEN THE OUTCOME IS THE SAME?
Answer is 3!(aaa,aaa,aaa,aaa,aaa,aaa)
This gives us an idea that, when we have a character appearing for m times, we will count m! instead of 1
So the counting is just n!(all possible arrangements, including same outcome) / m! (a character appear for m times)
Same for more characters repeating: n!/a!b!c!.. (first character appear a times, another appear for b times...)
If you understand the concept behind, then you will find that, actually for those "non-repeating" characters, it's just dividing an 1!. For eg, character (multi)set = {a,a,a,b,b,c}, #a = 3, #b = 2, #c = 1, so the answer (without repeating count) is (3+2+1)!/3!2!1! and fraction of this format is named multinomial coefficient as stated above.
In programming point of view, you can just pre-compute all factorials (with a pretty small n though as n~30 is already too large for a variable to store) with simple for loop
declare frac = array(n);
frac[0] = 1;
FOR i=1; i<=n;i++
frac[i] = i*frac[i-1]
For a larger n, you may just calculate double/float division on the fly in the loop to avoid overflow..you may face precision problem though.
If you further need to output the different strings, you may use DFS to backtrack all the possible outcomes. Or if you could use another language like C++, you can use built-in function like next_permutation() after sort the character set.
Well, it is a low-level question
Suppose I store a number (of course computer store number in binary format)
How can I print it in decimal format. It is obvious in high-level program, just print it and the library does it for you.
But how about in a very low-level situation where I don't have this library.
I can just tell what 'character' to output. How to convert the number into decimal characters?
I hope you understand my question. Thank You.
There are two ways of printing decimals - on CPUs with division/remainder instructions (modern CPUs are like that) and on CPUs where division is relatively slow (8-bit CPUs of 20+ years ago).
The first method is simple: int-divide the number by ten, and store the sequence of remainders in an array. Once you divided the number all the way to zero, start printing remainders starting from the back, adding the ASCII code of zero ('0') to each remainder.
The second method relies on the lookup table of powers of ten. You define an array of numbers like this:
int pow10 = {10000,1000,100,10,1}
Then you start with the largest power, and see if you can subtract it from the number at hand. If you can, keep subtracting it, and keep the count. Once you cannot subtract it without going negative, print the count plus the ASCII code of zero, and move on to the next smaller power of ten.
If integer, divide by ten, get both the result and the remainder. Repeat the process on the result until zero. The remainders will give you decimal digits from right to left. Add 48 for ASCII representation.
Basically, you want to tranform a number (stored in some arbitrary internal representation) into its decimal representation. You can do this with a few simple mathematical operations. Let's assume that we have a positive number, say 1234.
number mod 10 gives you a value between 0 and 9 (4 in our example), which you can map to a character¹. This is the rightmost digit.
Divide by 10, discarding the remainder (an operation commonly called "integer division"): 1234 → 123.
number mod 10 now yields 3, the second-to-rightmost digit.
continue until number is zero.
Footnotes:
¹ This can be done with a simple switch statement with 10 cases. Of course, if your character set has the characters 0..9 in consecutive order (like ASCII), '0' + number suffices.
It doesnt matter what the number system is, decimal, binary, octal. Say I have the decimal value 123 on a decimal computer, I would still need to convert that value to three characters to display them. Lets assume ASCII format. By looking at an ASCII table we know the answer we are looking for, 0x31,0x32,0x33.
If you divide 123 by 10 using integer math you get 12. Multiply 12*10 you get 120, the difference is 3, your least significant digit. we go back to the 12 and divide that by 10, giving a 1. 1 times 10 is 10, 12-10 is 2 our next digit. we take the 1 that is left over divide by 10 and get zero we know we are now done. the digits we found in order are 3, 2, 1. reverse the order 1, 2, 3. Add or OR 0x30 to each to convert them from integers to ascii.
change that to use a variable instead of 123 and use any numbering system you like so long as it has enough digits to do this kind of work
You can go the other way too, divide by 100...000, whatever the largest decimal you can store or intend to find, and work your way down. In this case the first non zero comes with a divide by 100 giving a 1. save the 1. 1 times 100 = 100, 123-100 = 23. now divide by 10, this gives a 2, save the 2, 2 times 10 is 20. 23 - 20 = 3. when you get to divide by 1 you are done save that value as your ones digit.
here is another given a number of seconds to convert to say hours and minutes and seconds, you can divide by 60, save the result a, subtract the original number - (a*60) giving your remainder which is seconds, save that. now take a and divide by 60, save that as b, this is your number of hours. subtract a - (b*60) this is the remainder which is minutes save that. done hours, minutes seconds. you can then divide the hours by 24 to get days if you want and days and then that by 7 if you want weeks.
A comment about divide instructions was brought up. Divides are very expensive and most processors do not have one. Expensive in that the divide, in a single clock, costs you gates and power. If you do the divide in many clocks you might as well just do a software divide and save the gates. Same reason most processors dont have an fpu, gates and power. (gates mean larger chips, more expensive chips, lower yield, etc). It is not a case of modern or old or 64 bit vs 8 bit or anything like that it is an engineering and business trade off. the 8088/86 has a divide with a remainder for example (it also has a bcd add). The gates/size if used might be better served than for a single instruction. Multiply falls into that category, not as bad but can be. If operand sizes are not done right you can make either instruction (family) not as useful to a programmer. Which brings up another point, I cant find the link right now but a way to avoid divides but convert from a number to a string of decimal digits is that you can multiply by .1 using fixed point. I also cant find the quote about real programmers not needing floating point related to keeping track of the decimal point yourself. its the slide rule vs calculator thing. I believe the link to the article on dividing by 10 using a multiply is somewhere on stack overflow.
I am trying to write the syntax to add up ten user-inputted numbers in a Console application, and then divide the final product by 10 in order to get the average number. So far I am able to allow the user to input the numbers properly, and I have the program set up to allow the user time to read the result, however I am slightly stuck on the syntax to add up the numbers overall. I know this is very simple, but the operation for creating this code is escaping me. I have tried finding the answer online already, but so far my only results have been overly-complex or just downright wrong.
Any and all help would be greatly appreciated.
There are lots of ways to acheive this, but the easiest is to just keep a running total. You'll need to cast the string input as a double using the double.Parse() method.
runningTotal = runningTotal + double.Parse(Console.ReadLine())
After the last input, simply divide runningTotal by 10 to display the result.
Exactly how depends on what you're using to store the numbers. A List(of Double) would be good for this because it'll store an arbitrary amount of numbers. Then to add them you can use a simple loop. Assuming your numbers are stored in a List called "numbers":
Dim total as Double = 0;
Dim average as Double = 0;
For Each number as Double in numbers
total += number
Next
average = total / numbers.Count()
What this does is goes through numbers, and for each number in it adds it to the total. At the end it divides the total by the count of numbers (however many you were given) to get the average. Note that you divide by the number of numbers to get the average, not necessarily 10. This one will still work if they only give you 7 numbers. :)
I have read that it has something to do with time, also you get from including time.h, so I assumed that much, but how does it work exactly? Also, does it have any tendencies towards odd or even numbers or something like that? And finally is there something with better distribution in the C standard library or the Foundation framework?
Briefly:
You use time.h to get a seed, which is an initial random number. C then does a bunch of operations on this number to get the next random number, then operations on that one to get the next, then... you get the picture.
rand() is able to touch on every possible integer. It will not prefer even or odd numbers regardless of the input seed, happily. Still, it has limits - it repeats itself relatively quickly, and in almost every implementation only gives numbers up to 32767.
C does not have another built-in random number generator. If you need a real tough one, there are many packages available online, but the Mersenne Twister algorithm is probably the most popular pick.
Now, if you are interested on the reasons why the above is true, here are the gory details on how rand() works:
rand() is what's called a "linear congruential generator." This means that it employs an equation of the form:
xn+1 = (*a****xn + ***b*) mod m
where xn is the nth random number, and a and b are some predetermined integers. The arithmetic is performed modulo m, with m usually 232 depending on the machine, so that only the lowest 32 bits are kept in the calculation of xn+1.
In English, then, the idea is this: To get the next random number, multiply the last random number by something, add a number to it, and then take the last few digits.
A few limitations are quickly apparent:
First, you need a starting random number. This is the "seed" of your random number generator, and this is where you've heard of time.h being used. Since we want a really random number, it is common practice to ask the system what time it is (in integer form) and use this as the first "random number." Also, this explains why using the same seed twice will always give exactly the same sequence of random numbers. This sounds bad, but is actually useful, since debugging is a lot easier when you control the inputs to your program
Second, a and b have to be chosen very, very carefully or you'll get some disastrous results. Fortunately, the equation for a linear congruential generator is simple enough that the math has been worked out in some detail. It turns out that choosing an a which satisfies *a***mod8 = 5 together with ***b* = 1 will insure that all m integers are equally likely, independent of choice of seed. You also want a value of a that is really big, so that every time you multiply it by xn you trigger a the modulo and chop off a lot of digits, or else many numbers in a row will just be multiples of each other. As a result, two common values of a (for example) are 1566083941 and 1812433253 according to Knuth. The GNU C library happens to use a=1103515245 and b=12345. A list of values for lots of implementations is available at the wikipedia page for LCGs.
Third, the linear congruential generator will actually repeat itself because of that modulo. This gets to be some pretty heady math, but the result of it all is happily very simple: The sequence will repeat itself after m numbers of have been generated. In most cases, this means that your random number generator will repeat every 232 cycles. That sounds like a lot, but it really isn't for many applications. If you are doing serious numerical work with Monte Carlo simulations, this number is hopelessly inadequate.
A fourth much less obvious problem is that the numbers are actually not really random. They have a funny sort of correlation. If you take three consecutive integers, (x, y, z), from an LCG with some value of a and m, those three points will always fall on the lattice of points generated by all linear combinations of the three points (1, a, a2), (0, m, 0), (0, 0, m). This is known as Marsaglia's Theorem, and if you don't understand it, that's okay. All it means is this: Triplets of random numbers from an LCG will show correlations at some deep, deep level. Usually it's too deep for you or I to notice, but its there. It's possible to even reconstruct the first number in a "random" sequence of three numbers if you are given the second and third! This is not good for cryptography at all.
The good part is that LCGs like rand() are very, very low footprint. It typically requires only 32 bits to retain state, which is really nice. It's also very fast, requiring very few operations. These make it good for noncritical embedded systems, video games, casual applications, stuff like that.
PRNGs are a fascinating topic. Wikipedia is always a good place to go if you are hungry to learn more on the history or the various implementations that are around today.
rand returns numbers generated by a pseudo-random number generator (PRNG). The sequence of numbers it returns is deterministic, based on the value with which the PRNG was initialized (by calling srand).
The numbers should be distributed such that they appear somewhat random, so, for example, odd and even numbers should be returned at roughly the same frequency. The actual implementation of the random number generator is left unspecified, so the actual behavior is specific to the implementation.
The important thing to remember is that rand does not return random numbers; it returns pseudo-random numbers, and the values it returns are determined by the seed value and the number of times rand has been called. This behavior is fine for many use cases, but is not appropriate for others (for example, rand would not be appropriate for use in many cryptographic applications).
How does rand() work?
http://en.wikipedia.org/wiki/Pseudorandom_number_generator
I have read that it has something to
do with time, also you get from
including time.h
rand() has nothing at all to do with the time. However, it's very common to use time() to obtain the "seed" for the PRNG so that you get different "random" numbers each time your program is run.
Also, does it have any tendencies
towards odd or even numbers or
something like that?
Depends on the exact method used. There's one popular implementation of rand() that alternates between odd and even numbers. So avoid writing code like rand() % 2 that depends on the lowest bit being random.
I would like to use newId to generate random numbers. Usually you would use it just once, but I might be generating up to 10 random numbers per newId.
Is it random enough?
Usually you would use it just once, but I might be generating up to 10 random numbers per newId. Is it random enough?
It depends on how you extract the numbers from the newid. You cannot treat it as 128 independently random bits.
For example if you use the first 8 bits for generating one random number between 0 and 255, use the next 8 bits to generate another number, etc. then you will see that your numbers will not be uniformly random.
v
E058D654-35A8-47F2-AE40-1C4EEBBDC549
01461481-ED8D-4B85-90FA-C08621D98DAE
AE861E4E-3469-4BDB-A38B-0031DACC8DAE
AF8905D0-E41B-4300-94F2-33BB45698CD1
003308A6-AE0A-4E20-9F24-047A6955E748
76F9B7ED-79AB-4EB1-B361-8C0AF5177CE3
B8F1CAC0-591D-436B-BB21-FAAD9EECA983
7FBEAEFD-2163-4315-A783-8106909E47D8
85E2FC60-E7B3-400F-B20A-CEFBECAEE4F9
17ED0A03-ADAD-4521-97EE-04815A867B32
^
|
always 4
You should also try to avoid reusing the same bits to generate different random numbers as your numbers will become related. If in doubt, don't reuse the same number.
Note that there is also a RAND function which you can call. This returns numbers from a uniform distribution.
Yes, it's statistically random. It's simply a GUID.
How do you plan on generating 10 numbers from one seed though? CHECKSUM(NEWID()) is normally how you'd do it for one value, perhaps with modulo and ABS
NewID generates a GUID. It is random enough.
Random enough for what? When you say use it to generate, are you just going to use it to seed a PNRG? I'm not sure it's any better than a timestamp for that. Or are you going to extract bits from the GUID - that's a bad idea.
http://www.random.org/randomness/