Here is the link of problem
https://www.hackerrank.com/challenges/equal
I read its editorial and unable to understand it. And if you are not make any account on hackerrank then surely you will not see it's editorial so here is some lines of editorial.
This is equivalent to saying, christy can take away the chocolates of
one coworker by 1, 2 or 5 while keeping others' chocolate untouched.
Let's consider decreasing a coworker's chocolate as an operation. To minimize the number of operations, we should try to make the number of chocolates of every coworker equal to the minimum one in the group(min). We have to decrease the number of chocolates the ith person A[i] by (A[i] - min). Let this value be x.
This can be done in k operations.
k = x/5 +(x%5)/2 + (x%5)%2
and from here i unable to understand
Let f(min) be sum of operations performed over all coworkers to reduce
each of their chocolates to min. However, sometimes f(min) might not
always give the correct answer. It can also be a case when
f(min) > f(min-1)
f(min) < f(min-5)
as f(min-5) takes N operations more than f(min) where N is the number
of coworkers. Therefore, if
A = {min,min-1,min-2,min-3,min-4}
then f(A) <= f(min) < f(min-5)
can someone help me to understand why this is necessary to check f(min),f(min-1),...,f(min-4)
Consider the case A = [1,5,5]
As the editorial said, it is intuitive to think it is optimal to change A to [1,1,1] with 4 (2 minus 2) operations, but it is better to change it to [0,0,0] with 3 (1 minus 1, 2 minus 5) operations.
Hence if min = minimum element in array, then change all elements to min may not be optimal.
The part you do not understand is to cater this situation, we know min may not be optimal as min-x maybe better, but how large is x? Well it is 4. The editorial is saying if we know x is at most 4, we can just simply brute force min, min-1...min-4 to see which one is the minimum without thinking too much.
Reasoning (Not proof!) for x <= 4
If x >= 5, then you have to use at least extra N type 3 (minus 5) operations on all elements which is definitely not worth.
Basically it is not a matter of the type of operation, it is because you need to use same operation on ALL elements, after you do that, the problem is not reduced, the relative difference between elements is still the same while you aim to make the relative difference to 0, you cost N operations for nothing.
In other words, if x >= 5, then x-5 must be a more optimal choice of goal, indeed x%5 must be the best goal.
(Below is TL;DR part: Version 2) Jump to the Last Section If You are Not Interested in the proof
In the process of writing the original solution, I suspect x <= 2 indeed, and I have tried to submit a code on HackerRank which only check minimum for f(min-x) where x <= 2, and it got ACed.
More formally, I claim
If 5> (z-min)%5 >= 3 and (z-min')%5==0, then F(min')< F(min)
where min'=min-x for x<=2, F(k) = min # of operation for element z to become k
(Beware the notation, I use F(), it is different meaning from f() in the question)
Here is the proof:
If (z-min)%5 = 1 or 2, then it needs at least (z-min)/5 + 1 operations, while (z-min')%5 == 0 needs (z-min')/5 = (z-min)/5 + 1 operation, means F(min') = F(min)
If(z-min)%5 == 3 or 4, then it needs at least (z-min)/5 + 2 operations, while (z-min')%5 == 0 needs (z-min')/5 = (z-min)/5 + 1 operation, means F(min') < F(min) (or F(min') = F(min)+1)
So we proof
If 5> (z-min)%5 >= 3 and (z-min')%5==0, then F(min')< F(min)
where min'=min-x
Now let's proof the range of x
As we assume (z-min)%5 >= 3 and (z-min')%5 == 0,
so (z-min')%5 = (z-min+x)%5 = ((z-min)%5 + x%5)%5 == 0
Now, if x >= 3, then (z-min)%5 can never be >= 3 in order to make ((z-min)%5 + x%5)%5 == 0
If x = 2, then (z-min)%5 can be 3; if x = 1 then (z-min)%5 can be 4, to meet both conditions: 5> (z-min)%5 >= 3 and (z-min')%5==0
Thus together we show
If 5> (z-min)%5 >= 3 and (z-min')%5==0, then F(min')< F(min)
where min'=min-x for x<=2
Note one can always generate array P, such that f(min') < f(min), as you can always repeat integer which can be improved by such method until it out number those integers cannot. This is because for elements that cannot be improved, they will always need exactly 1 more operations
eg: Let P = [2,2,2,10] f(min) = 0+3 = 3, f(min-2) = 3+2 = 5
Here 10 is the element which can be improved, while 2 cannot, so we can just add more 10 in the array. Each 2 will use 1 more operation to get to min' = min-2, while each 10 will save 1 operation to get min'. So we only have to add more 10 until it out number (compensate) the "waste" of 2:
P = [2,2,2,10,10,10,10,10], then f(min) = 0+15 = 15, f(min-2) = 3+10=13
or simply just
P = [2,10,10], f(min) = 6, f(min-2) = 5
(End of TL;DR part!)
EDITED
OMG THE TEST CASE ON HACKERRANK IS WEAK!
Story is when I arrive my office this morning, I keep thinking this problem a bit, and think that there maybe a problem in my code (which got ACed!)
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
int T, n, a[10005], m = 1<<28;
int f(int m){
m = max(0, m);
int cnt = 0;
for(int i=0; i<n;i++){
cnt += (a[i]-m)/5 + (a[i]-m)%5/2 + (a[i]-m)%5%2;
}
return cnt;
}
int main() {
cin >> T;
while(T--){
m = 1<<28;
cin >> n;
for(int i=0; i<n;i++) cin >> a[i], m = min(m,a[i]);
cout << min(min(f(m), f(m-1)),f(m-2)) << endl;
}
return 0;
}
Can you see the problem?
The problem is m = max(0, m); !
It ensure that min-x must be at least 0, but wait, my proof above did not say anything about the range of min-x! It can be negative indeed!
Remember the original question is about "adding", so there is no maximum value of the goal; while we model the question to "subtracting", there is no minimum value of the goal as well (but I set it to 0!)
Try this test case with the code above:
1
3
0 3 3
It forces min-x = 0, so it gives 4 as output, but the answer should be 3
(If we use "adding" model, the goal should be 10, with +5 on a[0],a[2], +5 on a[0],a[1], +2 on a[1], a[2])
So everything finally got right (I think...) when I remove the line m = max(0, m);, it allows min-x to get negative and give 3 as a correct output, and of course the new code get ACed as well...
Related
I have seen this question asked a lot but never seen a true concrete answer to it. So I am going to post one here which will hopefully help people understand why exactly there is "modulo bias" when using a random number generator, like rand() in C++.
So rand() is a pseudo-random number generator which chooses a natural number between 0 and RAND_MAX, which is a constant defined in cstdlib (see this article for a general overview on rand()).
Now what happens if you want to generate a random number between say 0 and 2? For the sake of explanation, let's say RAND_MAX is 10 and I decide to generate a random number between 0 and 2 by calling rand()%3. However, rand()%3 does not produce the numbers between 0 and 2 with equal probability!
When rand() returns 0, 3, 6, or 9, rand()%3 == 0. Therefore, P(0) = 4/11
When rand() returns 1, 4, 7, or 10, rand()%3 == 1. Therefore, P(1) = 4/11
When rand() returns 2, 5, or 8, rand()%3 == 2. Therefore, P(2) = 3/11
This does not generate the numbers between 0 and 2 with equal probability. Of course for small ranges this might not be the biggest issue but for a larger range this could skew the distribution, biasing the smaller numbers.
So when does rand()%n return a range of numbers from 0 to n-1 with equal probability? When RAND_MAX%n == n - 1. In this case, along with our earlier assumption rand() does return a number between 0 and RAND_MAX with equal probability, the modulo classes of n would also be equally distributed.
So how do we solve this problem? A crude way is to keep generating random numbers until you get a number in your desired range:
int x;
do {
x = rand();
} while (x >= n);
but that's inefficient for low values of n, since you only have a n/RAND_MAX chance of getting a value in your range, and so you'll need to perform RAND_MAX/n calls to rand() on average.
A more efficient formula approach would be to take some large range with a length divisible by n, like RAND_MAX - RAND_MAX % n, keep generating random numbers until you get one that lies in the range, and then take the modulus:
int x;
do {
x = rand();
} while (x >= (RAND_MAX - RAND_MAX % n));
x %= n;
For small values of n, this will rarely require more than one call to rand().
Works cited and further reading:
CPlusPlus Reference
Eternally Confuzzled
Keep selecting a random is a good way to remove the bias.
Update
We could make the code fast if we search for an x in range divisible by n.
// Assumptions
// rand() in [0, RAND_MAX]
// n in (0, RAND_MAX]
int x;
// Keep searching for an x in a range divisible by n
do {
x = rand();
} while (x >= RAND_MAX - (RAND_MAX % n))
x %= n;
The above loop should be very fast, say 1 iteration on average.
#user1413793 is correct about the problem. I'm not going to discuss that further, except to make one point: yes, for small values of n and large values of RAND_MAX, the modulo bias can be very small. But using a bias-inducing pattern means that you must consider the bias every time you calculate a random number and choose different patterns for different cases. And if you make the wrong choice, the bugs it introduces are subtle and almost impossible to unit test. Compared to just using the proper tool (such as arc4random_uniform), that's extra work, not less work. Doing more work and getting a worse solution is terrible engineering, especially when doing it right every time is easy on most platforms.
Unfortunately, the implementations of the solution are all incorrect or less efficient than they should be. (Each solution has various comments explaining the problems, but none of the solutions have been fixed to address them.) This is likely to confuse the casual answer-seeker, so I'm providing a known-good implementation here.
Again, the best solution is just to use arc4random_uniform on platforms that provide it, or a similar ranged solution for your platform (such as Random.nextInt on Java). It will do the right thing at no code cost to you. This is almost always the correct call to make.
If you don't have arc4random_uniform, then you can use the power of opensource to see exactly how it is implemented on top of a wider-range RNG (ar4random in this case, but a similar approach could also work on top of other RNGs).
Here is the OpenBSD implementation:
/*
* Calculate a uniformly distributed random number less than upper_bound
* avoiding "modulo bias".
*
* Uniformity is achieved by generating new random numbers until the one
* returned is outside the range [0, 2**32 % upper_bound). This
* guarantees the selected random number will be inside
* [2**32 % upper_bound, 2**32) which maps back to [0, upper_bound)
* after reduction modulo upper_bound.
*/
u_int32_t
arc4random_uniform(u_int32_t upper_bound)
{
u_int32_t r, min;
if (upper_bound < 2)
return 0;
/* 2**32 % x == (2**32 - x) % x */
min = -upper_bound % upper_bound;
/*
* This could theoretically loop forever but each retry has
* p > 0.5 (worst case, usually far better) of selecting a
* number inside the range we need, so it should rarely need
* to re-roll.
*/
for (;;) {
r = arc4random();
if (r >= min)
break;
}
return r % upper_bound;
}
It is worth noting the latest commit comment on this code for those who need to implement similar things:
Change arc4random_uniform() to calculate 2**32 % upper_bound as
-upper_bound % upper_bound. Simplifies the code and makes it the
same on both ILP32 and LP64 architectures, and also slightly faster on
LP64 architectures by using a 32-bit remainder instead of a 64-bit
remainder.
Pointed out by Jorden Verwer on tech#
ok deraadt; no objections from djm or otto
The Java implementation is also easily findable (see previous link):
public int nextInt(int n) {
if (n <= 0)
throw new IllegalArgumentException("n must be positive");
if ((n & -n) == n) // i.e., n is a power of 2
return (int)((n * (long)next(31)) >> 31);
int bits, val;
do {
bits = next(31);
val = bits % n;
} while (bits - val + (n-1) < 0);
return val;
}
Definition
Modulo Bias is the inherent bias in using modulo arithmetic to reduce an output set to a subset of the input set. In general, a bias exists whenever the mapping between the input and output set is not equally distributed, as in the case of using modulo arithmetic when the size of the output set is not a divisor of the size of the input set.
This bias is particularly hard to avoid in computing, where numbers are represented as strings of bits: 0s and 1s. Finding truly random sources of randomness is also extremely difficult, but is beyond the scope of this discussion. For the remainder of this answer, assume that there exists an unlimited source of truly random bits.
Problem Example
Let's consider simulating a die roll (0 to 5) using these random bits. There are 6 possibilities, so we need enough bits to represent the number 6, which is 3 bits. Unfortunately, 3 random bits yields 8 possible outcomes:
000 = 0, 001 = 1, 010 = 2, 011 = 3
100 = 4, 101 = 5, 110 = 6, 111 = 7
We can reduce the size of the outcome set to exactly 6 by taking the value modulo 6, however this presents the modulo bias problem: 110 yields a 0, and 111 yields a 1. This die is loaded.
Potential Solutions
Approach 0:
Rather than rely on random bits, in theory one could hire a small army to roll dice all day and record the results in a database, and then use each result only once. This is about as practical as it sounds, and more than likely would not yield truly random results anyway (pun intended).
Approach 1:
Instead of using the modulus, a naive but mathematically correct solution is to discard results that yield 110 and 111 and simply try again with 3 new bits. Unfortunately, this means that there is a 25% chance on each roll that a re-roll will be required, including each of the re-rolls themselves. This is clearly impractical for all but the most trivial of uses.
Approach 2:
Use more bits: instead of 3 bits, use 4. This yield 16 possible outcomes. Of course, re-rolling anytime the result is greater than 5 makes things worse (10/16 = 62.5%) so that alone won't help.
Notice that 2 * 6 = 12 < 16, so we can safely take any outcome less than 12 and reduce that modulo 6 to evenly distribute the outcomes. The other 4 outcomes must be discarded, and then re-rolled as in the previous approach.
Sounds good at first, but let's check the math:
4 discarded results / 16 possibilities = 25%
In this case, 1 extra bit didn't help at all!
That result is unfortunate, but let's try again with 5 bits:
32 % 6 = 2 discarded results; and
2 discarded results / 32 possibilities = 6.25%
A definite improvement, but not good enough in many practical cases. The good news is, adding more bits will never increase the chances of needing to discard and re-roll. This holds not just for dice, but in all cases.
As demonstrated however, adding an 1 extra bit may not change anything. In fact if we increase our roll to 6 bits, the probability remains 6.25%.
This begs 2 additional questions:
If we add enough bits, is there a guarantee that the probability of a discard will diminish?
How many bits are enough in the general case?
General Solution
Thankfully the answer to the first question is yes. The problem with 6 is that 2^x mod 6 flips between 2 and 4 which coincidentally are a multiple of 2 from each other, so that for an even x > 1,
[2^x mod 6] / 2^x == [2^(x+1) mod 6] / 2^(x+1)
Thus 6 is an exception rather than the rule. It is possible to find larger moduli that yield consecutive powers of 2 in the same way, but eventually this must wrap around, and the probability of a discard will be reduced.
Without offering further proof, in general using double the number
of bits required will provide a smaller, usually insignificant,
chance of a discard.
Proof of Concept
Here is an example program that uses OpenSSL's libcrypo to supply random bytes. When compiling, be sure to link to the library with -lcrypto which most everyone should have available.
#include <iostream>
#include <assert.h>
#include <limits>
#include <openssl/rand.h>
volatile uint32_t dummy;
uint64_t discardCount;
uint32_t uniformRandomUint32(uint32_t upperBound)
{
assert(RAND_status() == 1);
uint64_t discard = (std::numeric_limits<uint64_t>::max() - upperBound) % upperBound;
RAND_bytes((uint8_t*)(&randomPool), sizeof(randomPool));
while(randomPool > (std::numeric_limits<uint64_t>::max() - discard)) {
RAND_bytes((uint8_t*)(&randomPool), sizeof(randomPool));
++discardCount;
}
return randomPool % upperBound;
}
int main() {
discardCount = 0;
const uint32_t MODULUS = (1ul << 31)-1;
const uint32_t ROLLS = 10000000;
for(uint32_t i = 0; i < ROLLS; ++i) {
dummy = uniformRandomUint32(MODULUS);
}
std::cout << "Discard count = " << discardCount << std::endl;
}
I encourage playing with the MODULUS and ROLLS values to see how many re-rolls actually happen under most conditions. A sceptical person may also wish to save the computed values to file and verify the distribution appears normal.
Mark's Solution (The accepted solution) is Nearly Perfect.
int x;
do {
x = rand();
} while (x >= (RAND_MAX - RAND_MAX % n));
x %= n;
edited Mar 25 '16 at 23:16
Mark Amery 39k21170211
However, it has a caveat which discards 1 valid set of outcomes in any scenario where RAND_MAX (RM) is 1 less than a multiple of N (Where N = the Number of possible valid outcomes).
ie, When the 'count of values discarded' (D) is equal to N, then they are actually a valid set (V), not an invalid set (I).
What causes this is at some point Mark loses sight of the difference between N and Rand_Max.
N is a set who's valid members are comprised only of Positive Integers, as it contains a count of responses that would be valid. (eg: Set N = {1, 2, 3, ... n } )
Rand_max However is a set which ( as defined for our purposes ) includes any number of non-negative integers.
In it's most generic form, what is defined here as Rand Max is the Set of all valid outcomes, which could theoretically include negative numbers or non-numeric values.
Therefore Rand_Max is better defined as the set of "Possible Responses".
However N operates against the count of the values within the set of valid responses, so even as defined in our specific case, Rand_Max will be a value one less than the total number it contains.
Using Mark's Solution, Values are Discarded when: X => RM - RM % N
EG:
Ran Max Value (RM) = 255
Valid Outcome (N) = 4
When X => 252, Discarded values for X are: 252, 253, 254, 255
So, if Random Value Selected (X) = {252, 253, 254, 255}
Number of discarded Values (I) = RM % N + 1 == N
IE:
I = RM % N + 1
I = 255 % 4 + 1
I = 3 + 1
I = 4
X => ( RM - RM % N )
255 => (255 - 255 % 4)
255 => (255 - 3)
255 => (252)
Discard Returns $True
As you can see in the example above, when the value of X (the random number we get from the initial function) is 252, 253, 254, or 255 we would discard it even though these four values comprise a valid set of returned values.
IE: When the count of the values Discarded (I) = N (The number of valid outcomes) then a Valid set of return values will be discarded by the original function.
If we describe the difference between the values N and RM as D, ie:
D = (RM - N)
Then as the value of D becomes smaller, the Percentage of unneeded re-rolls due to this method increases at each natural multiplicative. (When RAND_MAX is NOT equal to a Prime Number this is of valid concern)
EG:
RM=255 , N=2 Then: D = 253, Lost percentage = 0.78125%
RM=255 , N=4 Then: D = 251, Lost percentage = 1.5625%
RM=255 , N=8 Then: D = 247, Lost percentage = 3.125%
RM=255 , N=16 Then: D = 239, Lost percentage = 6.25%
RM=255 , N=32 Then: D = 223, Lost percentage = 12.5%
RM=255 , N=64 Then: D = 191, Lost percentage = 25%
RM=255 , N= 128 Then D = 127, Lost percentage = 50%
Since the percentage of Rerolls needed increases the closer N comes to RM, this can be of valid concern at many different values depending on the constraints of the system running he code and the values being looked for.
To negate this we can make a simple amendment As shown here:
int x;
do {
x = rand();
} while (x > (RAND_MAX - ( ( ( RAND_MAX % n ) + 1 ) % n) );
x %= n;
This provides a more general version of the formula which accounts for the additional peculiarities of using modulus to define your max values.
Examples of using a small value for RAND_MAX which is a multiplicative of N.
Mark'original Version:
RAND_MAX = 3, n = 2, Values in RAND_MAX = 0,1,2,3, Valid Sets = 0,1 and 2,3.
When X >= (RAND_MAX - ( RAND_MAX % n ) )
When X >= 2 the value will be discarded, even though the set is valid.
Generalized Version 1:
RAND_MAX = 3, n = 2, Values in RAND_MAX = 0,1,2,3, Valid Sets = 0,1 and 2,3.
When X > (RAND_MAX - ( ( RAND_MAX % n ) + 1 ) % n )
When X > 3 the value would be discarded, but this is not a vlue in the set RAND_MAX so there will be no discard.
Additionally, in the case where N should be the number of values in RAND_MAX; in this case, you could set N = RAND_MAX +1, unless RAND_MAX = INT_MAX.
Loop-wise you could just use N = 1, and any value of X will be accepted, however, and put an IF statement in for your final multiplier. But perhaps you have code that may have a valid reason to return a 1 when the function is called with n = 1...
So it may be better to use 0, which would normally provide a Div 0 Error, when you wish to have n = RAND_MAX+1
Generalized Version 2:
int x;
if n != 0 {
do {
x = rand();
} while (x > (RAND_MAX - ( ( ( RAND_MAX % n ) + 1 ) % n) );
x %= n;
} else {
x = rand();
}
Both of these solutions resolve the issue with needlessly discarded valid results which will occur when RM+1 is a product of n.
The second version also covers the edge case scenario when you need n to equal the total possible set of values contained in RAND_MAX.
The modified approach in both is the same and allows for a more general solution to the need of providing valid random numbers and minimizing discarded values.
To reiterate:
The Basic General Solution which extends mark's example:
// Assumes:
// RAND_MAX is a globally defined constant, returned from the environment.
// int n; // User input, or externally defined, number of valid choices.
int x;
do {
x = rand();
} while (x > (RAND_MAX - ( ( ( RAND_MAX % n ) + 1 ) % n) ) );
x %= n;
The Extended General Solution which Allows one additional scenario of RAND_MAX+1 = n:
// Assumes:
// RAND_MAX is a globally defined constant, returned from the environment.
// int n; // User input, or externally defined, number of valid choices.
int x;
if n != 0 {
do {
x = rand();
} while (x > (RAND_MAX - ( ( ( RAND_MAX % n ) + 1 ) % n) ) );
x %= n;
} else {
x = rand();
}
In some languages ( particularly interpreted languages ) doing the calculations of the compare-operation outside of the while condition may lead to faster results as this is a one-time calculation no matter how many re-tries are required. YMMV!
// Assumes:
// RAND_MAX is a globally defined constant, returned from the environment.
// int n; // User input, or externally defined, number of valid choices.
int x; // Resulting random number
int y; // One-time calculation of the compare value for x
y = RAND_MAX - ( ( ( RAND_MAX % n ) + 1 ) % n)
if n != 0 {
do {
x = rand();
} while (x > y);
x %= n;
} else {
x = rand();
}
There are two usual complaints with the use of modulo.
one is valid for all generators. It is easier to see in a limit case. If your generator has a RAND_MAX which is 2 (that isn't compliant with the C standard) and you want only 0 or 1 as value, using modulo will generate 0 twice as often (when the generator generates 0 and 2) as it will generate 1 (when the generator generates 1). Note that this is true as soon as you don't drop values, whatever the mapping you are using from the generator values to the wanted one, one will occurs twice as often as the other.
some kind of generator have their less significant bits less random than the other, at least for some of their parameters, but sadly those parameter have other interesting characteristic (such has being able to have RAND_MAX one less than a power of 2). The problem is well known and for a long time library implementation probably avoid the problem (for instance the sample rand() implementation in the C standard use this kind of generator, but drop the 16 less significant bits), but some like to complain about that and you may have bad luck
Using something like
int alea(int n){
assert (0 < n && n <= RAND_MAX);
int partSize =
n == RAND_MAX ? 1 : 1 + (RAND_MAX-n)/(n+1);
int maxUsefull = partSize * n + (partSize-1);
int draw;
do {
draw = rand();
} while (draw > maxUsefull);
return draw/partSize;
}
to generate a random number between 0 and n will avoid both problems (and it avoids overflow with RAND_MAX == INT_MAX)
BTW, C++11 introduced standard ways to the the reduction and other generator than rand().
With a RAND_MAX value of 3 (in reality it should be much higher than that but the bias would still exist) it makes sense from these calculations that there is a bias:
1 % 2 = 1
2 % 2 = 0
3 % 2 = 1
random_between(1, 3) % 2 = more likely a 1
In this case, the % 2 is what you shouldn't do when you want a random number between 0 and 1. You could get a random number between 0 and 2 by doing % 3 though, because in this case: RAND_MAX is a multiple of 3.
Another method
There is much simpler but to add to other answers, here is my solution to get a random number between 0 and n - 1, so n different possibilities, without bias.
the number of bits (not bytes) needed to encode the number of possibilities is the number of bits of random data you'll need
encode the number from random bits
if this number is >= n, restart (no modulo).
Really random data is not easy to obtain, so why use more bits than needed.
Below is an example in Smalltalk, using a cache of bits from a pseudo-random number generator. I'm no security expert so use at your own risk.
next: n
| bitSize r from to |
n < 0 ifTrue: [^0 - (self next: 0 - n)].
n = 0 ifTrue: [^nil].
n = 1 ifTrue: [^0].
cache isNil ifTrue: [cache := OrderedCollection new].
cache size < (self randmax highBit) ifTrue: [
Security.DSSRandom default next asByteArray do: [ :byte |
(1 to: 8) do: [ :i | cache add: (byte bitAt: i)]
]
].
r := 0.
bitSize := n highBit.
to := cache size.
from := to - bitSize + 1.
(from to: to) do: [ :i |
r := r bitAt: i - from + 1 put: (cache at: i)
].
cache removeFrom: from to: to.
r >= n ifTrue: [^self next: n].
^r
Modulo reduction is a commonly seen way to make a random integer generator avoid the worst case of running forever.
When the range of possible integers is unknown, however, there is no way in general to "fix" this worst case of running forever without introducing bias. It's not just modulo reduction (rand() % n, discussed in the accepted answer) that will introduce bias this way, but also the "multiply-and-shift" reduction of Daniel Lemire, or if you stop rejecting an outcome after a set number of iterations. (To be clear, this doesn't mean there is no way to fix the bias issues present in pseudorandom generators. For example, even though modulo and other reductions are biased in general, they will have no issues with bias if the range of possible integers is a power of 2 and if the random generator produces unbiased random bits or blocks of them.)
The following answer of mine discusses the relationship between running time and bias in random generators, assuming we have a "true" random generator that can produce unbiased and independent random bits. The answer doesn't even involve the rand() function in C because it has many issues. Perhaps the most serious here is the fact that the C standard does not explicitly specify a particular distribution for the numbers returned by rand(), not even a uniform distribution.
How to generate a random integer in the range [0,n] from a stream of random bits without wasting bits?
As the accepted answer indicates, "modulo bias" has its roots in the low value of RAND_MAX. He uses an extremely small value of RAND_MAX (10) to show that if RAND_MAX were 10, then you tried to generate a number between 0 and 2 using %, the following outcomes would result:
rand() % 3 // if RAND_MAX were only 10, gives
output of rand() | rand()%3
0 | 0
1 | 1
2 | 2
3 | 0
4 | 1
5 | 2
6 | 0
7 | 1
8 | 2
9 | 0
So there are 4 outputs of 0's (4/10 chance) and only 3 outputs of 1 and 2 (3/10 chances each).
So it's biased. The lower numbers have a better chance of coming out.
But that only shows up so obviously when RAND_MAX is small. Or more specifically, when the number your are modding by is large compared to RAND_MAX.
A much better solution than looping (which is insanely inefficient and shouldn't even be suggested) is to use a PRNG with a much larger output range. The Mersenne Twister algorithm has a maximum output of 4,294,967,295. As such doing MersenneTwister::genrand_int32() % 10 for all intents and purposes, will be equally distributed and the modulo bias effect will all but disappear.
I just wrote a code for Von Neumann's Unbiased Coin Flip Method, that should theoretically eliminate any bias in the random number generation process. More info can be found at (http://en.wikipedia.org/wiki/Fair_coin)
int unbiased_random_bit() {
int x1, x2, prev;
prev = 2;
x1 = rand() % 2;
x2 = rand() % 2;
for (;; x1 = rand() % 2, x2 = rand() % 2)
{
if (x1 ^ x2) // 01 -> 1, or 10 -> 0.
{
return x2;
}
else if (x1 & x2)
{
if (!prev) // 0011
return 1;
else
prev = 1; // 1111 -> continue, bias unresolved
}
else
{
if (prev == 1)// 1100
return 0;
else // 0000 -> continue, bias unresolved
prev = 0;
}
}
}
I encountered a problem recently I have a hard time finding the answer.
This is the question:
Consider a set of numbers.There are tree kinds of input:
1 x
2 x
3
The first command adds integer x to the set.
The second one means for every element y in list, put:
y = y xor x
and The last command prints the biggest number in the set. for instance:
10
3
1 7
3
2 4
2 8
2 3
1 10
1 3
3
2 1
results:
0
7
15
if n is the number of commands in input:
and:
also there is a 1 second execution time limit!
My solution so far:
lets call the set S and have an integer m which initially is 0.as you know:
number = number xor x xor x
meaning that if we apply xor twice on something then the its effect is reversed and the original number doesn't change. That being said if we every time we insert a number(command 1) we do the following:
y = y xor m
add y to S
and every time we want to get a number from the set:
find y
y = y xor m
return y
and if command two comes to the following:
m = m xor x
then the problem is almost solved, since initially save the XORed version of the numbers and when needed we do the revers!
But the problem here is to find the largest number in the set( pay attention that the numbers in the set are different from original numbers) so command 3 works right. I don't know how to do this in an efficient time.but I have an idea here:
if we save the binary representation of the numbers in the set in a trie data structure at first the maybe we can quickly find the biggest number. I don't really know how but this idea occurred to me.
so to sum up these are my issues:
problem 1:
how to find the biggest number in the revised list
problem 2:
is this trie idea good?
problem 3:
how can I implement it in code(the language is not very important here) so that it works time find?
also what is the time complexity needed to solve this problem in the first place?
Thanks for reading my question.
Yes your idea is correct, it can be solved in O(N log 10^9) using binary trie data structure.
The idea is to store numbers in binary notation yet putting biggest bits first, so while traversing the trie we can choose a branch that leads to greatest answer.
For determining which branch to choose we can determine this bit by bit, if from some trie node we have 2 branches with values 0 and 1 we choose the one which gives better result after xoring with m
Sample code (C++):
#include <bits/stdc++.h>
using namespace std;
int Trie[4000005][2];
int nxt = 2;
void Add(int x)
{
bitset<32>b(x);
int c = 1;
for(int j=31; j>=0; j--)
if(Trie[c][b[j]])c=Trie[c][b[j]];
else c = Trie[c][b[j]] = nxt++;
}
int Get(int x)
{
bitset<32>b(x),res(0);
int c = 1;
for(int j=31; j>=0; j--)
if(Trie[c][!b[j]])c=Trie[c][!b[j]],res[j]=!b[j];
else c = Trie[c][b[j]], res[j]=b[j];
return res.to_ullong()^x;
}
int main()
{
ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
int q,m=0;
cin>>q;
Add(0);
while(q--)
{
int type;
cin>>type;
if(type==1)
{
int x;
cin>>x;
Add(x^m);
}
else if(type==2)
{
int x;
cin>>x;
m^=x;
}
else cout<<Get(m)<<"\n";
}
}
This is very similar to this problem and should be solvable in O(n), because the number of bits for x is constant (for 10^9 you will have to look at the 30 lowest bits).
At start m = 0, each time you encounter the 2nd command you do m ^= x (m = m xor x).
Use a binary tree. Unlike for the linked question the amount of numbers in a bucket doesn't matter, you just need to be able to tell if there is a number that has a certain bit which is one or zero. E.g. for 3-bit numbers 1, 4 and 5 the tree could look like this (left means bit is 0, right means bit is 1):
*
/ \
1 1 there are numbers with highest bit 0 and 1
/ /
1 1 of the numbers with 1st bit 0, there is a number with 2nd bit 0 and ...
\ / \
1 1 1 of the numbers with 1st and 2nd bit 0, there is a number with 3rd bit 1,...
1 4 5 (the numbers just to clarify)
So adding a number just means adding some edges and nodes.
To get the highest number in the set you go down the tree and through the bits of m and calculate the max x as follows:
Initialize node n as the root of the tree, i = 29 the bit of m we are looking at and the solution x = 0.
mi = (m & (1 << i)) >> i (1 if the bit in m is 1, 0 otherwise).
If we look at n and there is only an edge denoting a 0 or if mi == 1 and we have a 0-edge: n becomes the node connected by that edge, x = 2 * x + mi (or more fancy: x = (x << 1) | mi).
Otherwise n becomes the node connected by the 1-edge and x = 2 * x + 1 - mi
If i > 0: decrease i by 1 and continue with step 2.
An example for 3-bit numbers m = 6 (110) and the numbers 1 (001), 4 (100) and 5 (101) in the set, the answer should be 7 (111), i.e. 1 xor 6: First step we go left and x = 1, then we can only go left and x = 3, then we can only go right and x = 7.
http://codeforces.com/contest/520/problem/B
Vasya has found a strange device. On the front panel of a device there are: a red button, a blue button and a display showing some positive integer. After clicking the red button, device multiplies the displayed number by two. After clicking the blue button, device subtracts one from the number on the display. If at some point the number stops being positive, the device breaks down. The display can show arbitrarily large numbers. Initially, the display shows number n.
Bob wants to get number m on the display. What minimum number of clicks he has to make in order to achieve this result?
Input
The first and the only line of the input contains two distinct integers n and m (1 ≤ n, m ≤ 10^4), separated by a space .
Output
Print a single number — the minimum number of times one needs to push the button required to get the number m out of number n.
I developed the following recursive solution. I know it will time out, but I will memoise it, and that will get my solution accepted. But as of now, I am getting a wrong answer in one of the inputs.
My code is:
int func (int n, int m);
int main (void)
{
int n,m;
cin>>n>>m;
int count = func(n,m);
cout<<count<<"\n";
return 0;
}
int func (int n, int m)
{
if (n == 0)
return INT_MAX; // this should be because we can never go to some
// other digit if we are at 0
if (n == m)
return 0;
else if (2*n == m || n == m+1)
return 1;
else if (n > m)
return func(n-1,m)+1;
else
return min(func(n-1,m),func(n*2,m))+1;
}
Now, when I enter the input as (1,3), my code shows Segmentation fault. I tried to debug it, and I found out that it sorts of go in an infinite loop because of which I get the Seg fault. However, I want to know, then how should I make the logic for this? What will be the recursive function for this? Thanks!
The SEG fault is due to calculating doing INT_MAX+1.
Actually I think this problem is better solved working this way.
For all cases n>m, the shortest count is n-m.
if (n<m)
return n-m;
For all cases n==m, the shortest count is 0.
else if (n==m)
return 0;
For all cases n < m, the shortest count can be calculated as:
let sequence Y= [(m/(2^1), m/(2^2), ... 1] // use the ceiling values
find X is the next number in the Y where n >= X.
return func(X*2) + n-X + 1;
For n = 57, m = 201, then Y = [101, 51, 26, 13, 7, 4, 2, 1], X would be 51.
So the answer can be calculated as
(57-51)+1 = 7 steps, result now 51*2 = 102
(102-101)+1 = 2 steps, result now 101*2 = 202
(202-201) = 1 steps
=====> Total steps 10
For n = 4, m = 6, then Y = [3, 2, 1], X would be 3.
So the answer can be calculated as
(4-3)+1 = 2 steps, result now 3*2=6
=====> Total steps 2
For n = 1, m = 3, then Y = [2, 1], X would be 1.
So the answer can be calculated as
already in Y= 1 steps, result now 1*2 = 2
already in Y= 1 steps, result now 2*2 = 4
(4-3) = 1 step
=====> Total steps 3
Notice you can precalcuate Y before entering your function and pass it in so you don't have to recompute each time.
Given: set A = {a0, a1, ..., aN-1} (1 ≤ N ≤ 100), with 2 ≤ ai ≤ 500.
Asked: Find the sum of all least common multiples (LCM) of all subsets of A of size at least 2.
The LCM of a setB = {b0, b1, ..., bk-1} is defined as the minimum integer Bmin such that bi | Bmin, for all 0 ≤ i < k.
Example:
Let N = 3 and A = {2, 6, 7}, then:
LCM({2, 6}) = 6
LCM({2, 7}) = 14
LCM({6, 7}) = 42
LCM({2, 6, 7}) = 42
----------------------- +
answer 104
The naive approach would be to simply calculate the LCM for all O(2N) subsets, which is not feasible for reasonably large N.
Solution sketch:
The problem is obtained from a competition*, which also provided a solution sketch. This is where my problem comes in: I do not understand the hinted approach.
The solution reads (modulo some small fixed grammar issues):
The solution is a bit tricky. If we observe carefully we see that the integers are between 2 and 500. So, if we prime factorize the numbers, we get the following maximum powers:
2 8
3 5
5 3
7 3
11 2
13 2
17 2
19 2
Other than this, all primes have power 1. So, we can easily calculate all possible states, using these integers, leaving 9 * 6 * 4 * 4 * 3 * 3 * 3 * 3 states, which is nearly 70000. For other integers we can make a dp like the following: dp[70000][i], where i can be 0 to 100. However, as dp[i] is dependent on dp[i-1], so dp[70000][2] is enough. This leaves the complexity to n * 70000 which is feasible.
I have the following concrete questions:
What is meant by these states?
Does dp stand for dynamic programming and if so, what recurrence relation is being solved?
How is dp[i] computed from dp[i-1]?
Why do the big primes not contribute to the number of states? Each of them occurs either 0 or 1 times. Should the number of states not be multiplied by 2 for each of these primes (leading to a non-feasible state space again)?
*The original problem description can be found from this source (problem F). This question is a simplified version of that description.
Discussion
After reading the actual contest description (page 10 or 11) and the solution sketch, I have to conclude the author of the solution sketch is quite imprecise in their writing.
The high level problem is to calculate an expected lifetime if components are chosen randomly by fair coin toss. This is what's leading to computing the LCM of all subsets -- all subsets effectively represent the sample space. You could end up with any possible set of components. The failure time for the device is based on the LCM of the set. The expected lifetime is therefore the average of the LCM of all sets.
Note that this ought to include the LCM of sets with only one item (in which case we'd assume the LCM to be the element itself). The solution sketch seems to sabotage, perhaps because they handled it in a less elegant manner.
What is meant by these states?
The sketch author only uses the word state twice, but apparently manages to switch meanings. In the first use of the word state it appears they're talking about a possible selection of components. In the second use they're likely talking about possible failure times. They could be muddling this terminology because their dynamic programming solution initializes values from one use of the word and the recurrence relation stems from the other.
Does dp stand for dynamic programming?
I would say either it does or it's a coincidence as the solution sketch seems to heavily imply dynamic programming.
If so, what recurrence relation is being solved? How is dp[i] computed from dp[i-1]?
All I can think is that in their solution, state i represents a time to failure , T(i), with the number of times this time to failure has been counted, dp[i]. The resulting sum would be to sum all dp[i] * T(i).
dp[i][0] would then be the failure times counted for only the first component. dp[i][1] would then be the failure times counted for the first and second component. dp[i][2] would be for the first, second, and third. Etc..
Initialize dp[i][0] with zeroes except for dp[T(c)][0] (where c is the first component considered) which should be 1 (since this component's failure time has been counted once so far).
To populate dp[i][n] from dp[i][n-1] for each component c:
For each i, copy dp[i][n-1] into dp[i][n].
Add 1 to dp[T(c)][n].
For each i, add dp[i][n-1] to dp[LCM(T(i), T(c))][n].
What is this doing? Suppose you knew that you had a time to failure of j, but you added a component with a time to failure of k. Regardless of what components you had before, your new time to fail is LCM(j, k). This follows from the fact that for two sets A and B, LCM(A union B} = LCM(LCM(A), LCM(B)).
Similarly, if we're considering a time to failure of T(i) and our new component's time to failure of T(c), the resultant time to failure is LCM(T(i), T(c)). Note that we recorded this time to failure for dp[i][n-1] configurations, so we should record that many new times to failure once the new component is introduced.
Why do the big primes not contribute to the number of states?
Each of them occurs either 0 or 1 times. Should the number of states not be multiplied by 2 for each of these primes (leading to a non-feasible state space again)?
You're right, of course. However, the solution sketch states that numbers with large primes are handled in another (unspecified) fashion.
What would happen if we did include them? The number of states we would need to represent would explode into an impractical number. Hence the author accounts for such numbers differently. Note that if a number less than or equal to 500 includes a prime larger than 19 the other factors multiply to 21 or less. This makes such numbers amenable for brute forcing, no tables necessary.
The first part of the editorial seems useful, but the second part is rather vague (and perhaps unhelpful; I'd rather finish this answer than figure it out).
Let's suppose for the moment that the input consists of pairwise distinct primes, e.g., 2, 3, 5, and 7. Then the answer (for summing all sets, where the LCM of 0 integers is 1) is
(1 + 2) (1 + 3) (1 + 5) (1 + 7),
because the LCM of a subset is exactly equal to the product here, so just multiply it out.
Let's relax the restriction that the primes be pairwise distinct. If we have an input like 2, 2, 3, 3, 3, and 5, then the multiplication looks like
(1 + (2^2 - 1) 2) (1 + (2^3 - 1) 3) (1 + (2^1 - 1) 5),
because 2 appears with multiplicity 2, and 3 appears with multiplicity 3, and 5 appears with multiplicity 1. With respect to, e.g., just the set of 3s, there are 2^3 - 1 ways to choose a subset that includes a 3, and 1 way to choose the empty set.
Call a prime small if it's 19 or less and large otherwise. Note that integers 500 or less are divisible by at most one large prime (with multiplicity). The small primes are more problematic. What we're going to do is to compute, for each possible small portion of the prime factorization of the LCM (i.e., one of the ~70,000 states), the sum of LCMs for the problem derived by discarding the integers that could not divide such an LCM and leaving only the large prime factor (or 1) for the other integers.
For example, if the input is 2, 30, 41, 46, and 51, and the state is 2, then we retain 2 as 1, discard 30 (= 2 * 3 * 5; 3 and 5 are small), retain 41 as 41 (41 is large), retain 46 as 23 (= 2 * 23; 23 is large), and discard 51 (= 3 * 17; 3 and 17 are small). Now, we compute the sum of LCMs using the previously described technique. Use inclusion-exclusion to get rid of the subsets whose LCM whose small portion properly divides the state instead of being exactly equal. Maybe I'll work a complete example later.
What is meant by these states?
I think here, states refer to if the number is in set B = {b0, b1, ..., bk-1} of LCMs of set A.
Does dp stand for dynamic programming and if so, what recurrence relation is being solved?
dp in the solution sketch stands for dynamic programming, I believe.
How is dp[i] computed from dp[i-1]?
It's feasible that we can figure out the state of next group of LCMs from previous states. So, we only need array of 2, and toggle back and forth.
Why do the big primes not contribute to the number of states? Each of them occurs either 0 or 1 times. Should the number of states not be multiplied by 2 for each of these primes (leading to a non-feasible state space again)?
We can use Prime Factorization and exponents only to present the number.
Here is one example.
6 = (2^1)(3^1)(5^0) -> state "1 1 0" to represent 6
18 = (2^1)(3^2)(5^0) -> state "1 2 0" to represent 18
Here is how we can get LMC of 6 and 18 using Prime Factorization
LCM (6,18) = (2^(max(1,1)) (3^ (max(1,2)) (5^max(0,0)) = (2^1)(3^2)(5^0) = 18
2^9 > 500, 3^6 > 500, 5^4 > 500, 7^4>500, 11^3 > 500, 13^3 > 500, 17^3 > 500, 19^3 > 500
we can use only count of exponents of prime number 2,3,5,7,11,13,17,19 to represent the LCMs in the set B = {b0, b1, ..., bk-1}
for the given set A = {a0, a1, ..., aN-1} (1 ≤ N ≤ 100), with 2 ≤ ai ≤ 500.
9 * 6 * 4 * 4 * 3 * 3 * 3 * 3 <= 70000, so we only need two of dp[9][6][4][4][3][3][3][3] to keep tracks of all LCMs' states. So, dp[70000][2] is enough.
I put together a small C++ program to illustrate how we can get sum of LCMs of the given set A = {a0, a1, ..., aN-1} (1 ≤ N ≤ 100), with 2 ≤ ai ≤ 500. In the solution sketch, we need to loop through 70000 max possible of LCMs.
int gcd(int a, int b) {
int remainder = 0;
do {
remainder = a % b;
a = b;
b = remainder;
} while (b != 0);
return a;
}
int lcm(int a, int b) {
if (a == 0 || b == 0) {
return 0;
}
return (a * b) / gcd(a, b);
}
int sum_of_lcm(int A[], int N) {
// get the max LCM from the array
int max = A[0];
for (int i = 1; i < N; i++) {
max = lcm(max, A[i]);
}
max++;
//
int dp[max][2];
memset(dp, 0, sizeof(dp));
int pri = 0;
int cur = 1;
// loop through n x 70000
for (int i = 0; i < N; i++) {
for (int v = 1; v < max; v++) {
int x = A[i];
if (dp[v][pri] > 0) {
x = lcm(A[i], v);
dp[v][cur] = (dp[v][cur] == 0) ? dp[v][pri] : dp[v][cur];
if ( x % A[i] != 0 ) {
dp[x][cur] += dp[v][pri] + dp[A[i]][pri];
} else {
dp[x][cur] += ( x==v ) ? ( dp[v][pri] + dp[v][pri] ) : ( dp[v][pri] ) ;
}
}
}
dp[A[i]][cur]++;
pri = cur;
cur = (pri + 1) % 2;
}
for (int i = 0; i < N; i++) {
dp[A[i]][pri] -= 1;
}
long total = 0;
for (int j = 0; j < max; j++) {
if (dp[j][pri] > 0) {
total += dp[j][pri] * j;
}
}
cout << "total:" << total << endl;
return total;
}
int test() {
int a[] = {2, 6, 7 };
int n = sizeof(a)/sizeof(a[0]);
int total = sum_of_lcm(a, n);
return 0;
}
Output
total:104
The states are one more than the powers of primes. You have numbers up to 2^8, so the power of 2 is in [0..8], which is 9 states. Similarly for the other states.
"dp" could well stand for dynamic programming, I'm not sure.
The recurrence relation is the heart of the problem, so you will learn more by solving it yourself. Start with some small, simple examples.
For the large primes, try solving a reduced problem without using them (or their equivalents) and then add them back in to see their effect on the final result.
Actually, this question can be generalized as below:
Find all possible combinations from a given set of elements, which meets
a certain criteria.
So, any good algorithms?
There are only 16 possibilities (and one of those is to add together "none of them", which ain't gonna give you 24), so the old-fashioned "brute force" algorithm looks pretty good to me:
for (unsigned int choice = 1; choice < 16; ++choice) {
int sum = 0;
if (choice & 1) sum += elements[0];
if (choice & 2) sum += elements[1];
if (choice & 4) sum += elements[2];
if (choice & 8) sum += elements[3];
if (sum == 24) {
// we have a winner
}
}
In the completely general form of your problem, the only way to tell whether a combination meets "certain criteria" is to evaluate those criteria for every single combination. Given more information about the criteria, maybe you could work out some ways to avoid testing every combination and build an algorithm accordingly, but not without those details. So again, brute force is king.
There are two interesting explanations about the sum problem, both in Wikipedia and MathWorld.
In the case of the first question you asked, the first answer is good for a limited number of elements. You should realize that the reason Mr. Jessop used 16 as the boundary for his loop is because this is 2^4, where 4 is the number of elements in your set. If you had 100 elements, the loop limit would become 2^100 and your algorithm would literally take forever to finish.
In the case of a bounded sum, you should consider a depth first search, because when the sum of elements exceeds the sum you are looking for, you can prune your branch and backtrack.
In the case of the generic question, finding the subset of elements that satisfy certain criteria, this is known as the Knapsack problem, which is known to be NP-Complete. Given that, there is no algorithm that will solve it in less than exponential time.
Nevertheless, there are several heuristics that bring good results to the table, including (but not limited to) genetic algorithms (one I personally like, for I wrote a book on them) and dynamic programming. A simple search in Google will show many scientific papers that describe different solutions for this problem.
Find all possible combinations from a given set of elements, which
meets a certain criteria
If i understood you right, this code will helpful for you:
>>> from itertools import combinations as combi
>>> combi.__doc__
'combinations(iterable, r) --> combinations object\n\nReturn successive r-length
combinations of elements in the iterable.\n\ncombinations(range(4), 3) --> (0,1
,2), (0,1,3), (0,2,3), (1,2,3)'
>>> set = range(4)
>>> set
[0, 1, 2, 3]
>>> criteria = range(3)
>>> criteria
[0, 1, 2]
>>> for tuple in list(combi(set, len(criteria))):
... if cmp(list(tuple), criteria) == 0:
... print 'criteria exists in tuple: ', tuple
...
criteria exists in tuple: (0, 1, 2)
>>> list(combi(set, len(criteria)))
[(0, 1, 2), (0, 1, 3), (0, 2, 3), (1, 2, 3)]
Generally for a problem as this you have to try all posebilities, the thing you should do have the code abort the building of combiantion if you know it will not satesfie the criteria (if you criteria is that you do not have more then two blue balls, then you have to abort calculation that has more then two). Backtracing
def perm(set,permutation):
if lenght(set) == lenght(permutation):
print permutation
else:
for element in set:
if permutation.add(element) == criteria:
perm(sett,permutation)
else:
permutation.pop() //remove the element added in the if
The set of input numbers matters, as you can tell as soon as you allow e.g. negative numbers, imaginary numbers, rational numbers etc in your start set. You could also restrict to e.g. all even numbers, all odd number inputs etc.
That means that it's hard to build something deductive. You need brute force, a.k.a. try every combination etc.
In this particular problem you could build an algoritm that recurses - e.g. find every combination of 3 Int ( 1,22) that add up to 23, then add 1, every combination that add to 22 and add 2 etc. Which can again be broken into every combination of 2 that add up to 21 etc. You need to decide if you can count same number twice.
Once you have that you have a recursive function to call -
combinations( 24 , 4 ) = combinations( 23, 3 ) + combinations( 22, 3 ) + ... combinations( 4, 3 );
combinations( 23 , 3 ) = combinations( 22, 2 ) + ... combinations( 3, 2 );
etc
This works well except you have to be careful around repeating numbers in the recursion.
private int[][] work()
{
const int target = 24;
List<int[]> combos = new List<int[]>();
for(int i = 0; i < 9; i++)
for(int x = 0; x < 9; x++)
for(int y = 0; y < 9; y++)
for (int z = 0; z < 9; z++)
{
int res = x + y + z + i;
if (res == target)
{
combos.Add(new int[] { x, y, z, i });
}
}
return combos.ToArray();
}
It works instantly, but there probably are better methods rather than 'guess and check'. All I am doing is looping through every possibility, adding them all together, and seeing if it comes out to the target value.
If i understand your question correctly, what you are asking for is called "Permutations" or the number (N) of possible ways to arrange (X) numbers taken from a set of (Y) numbers.
N = Y! / (Y - X)!
I don't know if this will help, but this is a solution I came up with for an assignment on permutations.
You have an input of : 123 (string) using the substr functions
1) put each number of the input into an array
array[N1,N2,N3,...]
2)Create a swap function
function swap(Number A, Number B)
{
temp = Number B
Number B = Number A
Number A = temp
}
3)This algorithm uses the swap function to move the numbers around until all permutations are done.
original_string= '123'
temp_string=''
While( temp_string != original_string)
{
swap(array element[i], array element[i+1])
if (i == 1)
i == 0
temp_string = array.toString
i++
}
Hopefully you can follow my pseudo code, but this works at least for 3 digit permutations
(n X n )
built up a square matrix of nxn
and print all together its corresponding crossed values
e.g.
1 2 3 4
1 11 12 13 14
2 .. .. .. ..
3 ..
4 .. ..