I am solving this problem where we need to reach from X=0 to X=N.We can only take a step of 2 or 3 at a time.
For each step of 2 we have a probability of 0.2 and for each step of 3 we have a probability of 0.8.How can we find the total probability to reach N.
e.g. for reaching 5,
2+3 with probability =0.2 * 0.8=0.16
3+2 with probability =0.8 * 0.2=0.16 total = 0.32.
My initial thoughts:
Number of ways can be found out by simple Fibonacci sequence.
f(n)=f(n-3)+f(n-2);
But how do we remember the numbers so that we can multiply them to find the probability?
This can be solved using Dynamic programming.
Lets call the function F(N) = probability to reach 0 using only 2 and 3 when the starting number is N
F(N) = 0.2*F(N-2) + 0.3*F(N-3)
Base case:
F(0) = 1 and F(k)= 0 where k< 0
So the DP code would be somthing like that:
F[0] = 1;
for(int i = 1;i<=N;i++){
if(i>=3)
F[i] = 0.2*F[i-2] + 0.8*F[i-3];
else if(i>=2)
F[i] = 0.2*F[i-2];
else
F[i] = 0;
}
return F[N];
This algorithm would run in O(N)
Some clarifications about this solution: I assume the only allowed operation for generating the number from 2s and 3s is addition (your definition would allow substraction aswell) and the input-numbers are always valid (2 <= input). Definition: a unique row of numbers means: no other row with the same number of 3s and 2s in another order is in scope.
We can reduce the problem into multiple smaller problems:
Problem A: finding all sequences of numbers that can sum up to the given number. (Unique rows of numbers only)
Start by finding the minimum-number of 3s required to build the given number, which is simply input % 2. The maximum-number of 3s that can be used to build the input can be calculated this way:
int max_3 = (int) (input / 3);
if(input - max_3 == 1)
--max_3;
Now all sequences of numbers that sum up to input must hold between input % 2 and max_3 3s. The 2s can be easily calculated from a given number of 3s.
Problem B: calculating the probability for a given list and it's permutations to be the result
For each unique row of numbers, we can easily derive all permutations. Since these consist of the same number, they have the same likeliness to appear and produce the same sum. The likeliness can be calculated easily from the row: 0.8 ^ number_of_3s * 0.2 ^ number_of_2s. Next step would be to calculate the number of different permuatations. Calculating all distinct sets with a specific number of 2s and 3s can be done this way: Calculate all possible distributions of 2s in the set: (number_of_2s + number_of_3s)! / (number_of_3s! * numer_of_2s!). Basically just the number of possible distinct permutations.
Now from theory to praxis
Since the math is given, the rest is pretty straight forward:
define prob:
input: int num
output: double
double result = 0.0
int min_3s = (num % 2)
int max_3s = (int) (num / 3)
if(num - max_3 == 1)
--max_3
for int c3s in [min_3s , max_3s]
int c2s = (num - (c3s * 3)) / 2
double p = 0.8 ^ c3s * 0.2 * c2s
p *= (c3s + c2s)! / (c3s! * c2s!)
result += p
return result
Instead of jumping into the programming, you can use math.
Let p(n) be the probability that you reach the location that is n steps away.
Base cases:
p(0)=1
p(1)=0
p(2)=0.2
Linear recurrence relation
p(n+3)=0.2 p(n+1) + 0.8 p(n)
You can solve this in closed form by finding the exponential solutions to the linear recurrent relation.
c^3 = 0.2 c + 0.8
c = 1, (-5 +- sqrt(55)i)/10
Although this was cubic, c=1 will always be a solution in this type of problem since there is a constant nonzero solution.
Because the roots are distinct, all solutions are of the form a1(1)^n + a2((-5+sqrt(55)i) / 10)^n + a3((-5-sqrt(55)i)/10)^n. You can solve for a1, a2, and a3 using the initial conditions:
a1=5/14
a2=(99-sqrt(55)i)/308
a3=(99+sqrt(55)i)/308
This gives you a nonrecursive formula for p(n):
p(n)=5/14+(99-sqrt(55)i)/308((-5+sqrt(55)i)/10)^n+(99+sqrt(55)i)/308((-5-sqrt(55)i)/10)^n
One nice property of the non-recursive formula is that you can read off the asymptotic value of 5/14, but that's also clear because the average value of a jump is 2(1/5)+ 3(4/5) = 14/5, and you almost surely hit a set with density 1/(14/5) of the integers. You can use the magnitudes of the other roots, 2/sqrt(5)~0.894, to see how rapidly the probabilities approach the asymptotics.
5/14 - (|a2|+|a3|) 0.894^n < p(n) < 5/14 + (|a2|+|a3|) 0.894^n
|5/14 - p(n)| < (|a2|+|a3|) 0.894^n
f(n, p) = f(n-3, p*.8) + f(n -2, p*.2)
Start p at 1.
If n=0 return p, if n <0 return 0.
Instead of using the (terribly inefficient) recursive algorithm, start from the start and calculate in how many ways you can reach subsequent steps, i.e. using 'dynamic programming'. This way, you can easily calculate the probabilities and also have a complexity of only O(n) to calculate everything up to step n.
For each step, memorize the possible ways of reaching that step, if any (no matter how), and the probability of reaching that step. For the zeroth step (the start) this is (1, 1.0).
steps = [(1, 1.0)]
Now, for each consecutive step n, get the previously computed possible ways poss and probability prob to reach steps n-2 and n-3 (or (0, 0.0) in case of n < 2 or n < 3 respectively), add those to the combined possibilities and probability to reach that new step, and add them to the list.
for n in range(1, 10):
poss2, prob2 = steps[n-2] if n >= 2 else (0, 0.0)
poss3, prob3 = steps[n-3] if n >= 3 else (0, 0.0)
steps.append( (poss2 + poss3, prob2 * 0.2 + prob3 * 0.8) )
Now you can just get the numbers from that list:
>>> for n, (poss, prob) in enumerate(steps):
... print "%s\t%s\t%s" % (n, poss, prob)
0 1 1.0
1 0 0.0
2 1 0.2
3 1 0.8
4 1 0.04
5 2 0.32 <-- 2 ways to get to 5 with combined prob. of 0.32
6 2 0.648
7 3 0.096
8 4 0.3856
9 5 0.5376
(Code is in Python)
Note that this will get you both the number of possible ways of reaching a certain step (e.g. "first 2, then 3" or "first 3, then 2" for 5), and the probability to reach that step in one go. Of course, if you need only the probability, you can just use single numbers instead of tuples.
Related
Is there a way to find programmatically the consecutive natural numbers?
On the Internet I found some examples using either factorization or polynomial solving.
Example 1
For n(n−1)(n−2)(n−3) = 840
n = 7, -4, (3+i√111)/2, (3-i√111)/2
Example 2
For n(n−1)(n−2)(n−3) = 1680
n = 8, −5, (3+i√159)/2, (3-i√159)/2
Both of those examples give 4 results (because both are 4th degree equations), but for my use case I'm only interested in the natural value. Also the solution should work for any sequences size of consecutive numbers, in other words, n(n−1)(n−2)(n−3)(n−4)...
The solution can be an algorithm or come from any open math library. The parameters passed to the algorithm will be the product and the degree (sequences size), like for those two examples the product is 840 or 1640 and the degree is 4 for both.
Thank you
If you're interested only in natural "n" solution then this reasoning may help:
Let's say n(n-1)(n-2)(n-3)...(n-k) = A
The solution n=sthen verifies:
remainder of A/s = 0
remainder of A/(s-1) = 0
remainder of A/(s-2) = 0
and so on
Now, we see that s is in the order of t= A^(1/k) : A is similar to s*s*s*s*s... k times. So we can start with v= (t-k) and finish at v= t+1. The solution will be between these two values.
So the algo may be, roughly:
s= 0
t= (int) (A^(1/k)) //this truncation by leave out t= v+1. Fix it in the loop
theLoop:
for (v= t-k to v= t+1, step= +1)
{ i=0
while ( i <= k )
{ if (A % (v - k + i) > 0 ) // % operator to find the reminder
continue at theLoop
i= i+1
}
// All are valid divisors, solution found
s = v
break
}
if (s==0)
not natural solution
Assuming that:
n is an integer, and
n > 0, and
k < n
Then approximately:
n = FLOOR( (product ** (1/(k+1)) + (k+1)/2 )
The only cases I have found where this isn't exactly right is when k is very close to n. You can of course check it by back-calculating the product and see if it matches. If not, it almost certainly is only 1 or 2 in higher than this estimate, so just keep incrementing n until the product matches. (I can write this up in pseudocode if you need it)
Given a number K which is a product of two different numbers (A,B), find the maximum number(<=A & <=B) who's square divides the K .
Eg : K = 54 (6*9) . Both the numbers are available i.e 6 and 9.
My approach is fairly very simple or trivial.
taking the smallest of the two ( 6 in this case).Lets say A
Square the number and divide K, if its a perfect division, that's the number.
Else A = A-1 ,till A =1.
For the given example, 3*3 = 9 divides K, and hence 3 is the answer.
Looking for a better algorithm, than the trivial solution.
Note : The test cases are in 1000's so the best possible approach is needed.
I am sure someone else will come up with a nice answer involving modulus arithmetic. Here is a naive approach...
Each of the factors can themselves be factored (though it might be an expensive operation).
Given the factors, you can then look for groups of repeated factors.
For instance, using your example:
Prime factors of 9: 3, 3
Prime factors of 6: 2, 3
All prime factors: 2, 3, 3, 3
There are two 3s, so you have your answer (the square of 3 divides 54).
Second example of 36 x 9 = 324
Prime factors of 36: 2, 2, 3, 3
Prime factors of 9: 3, 3
All prime factors: 2, 2, 3, 3, 3, 3
So you have two 2s and four 3s, which means 2x3x3 is repeated. 2x3x3 = 18, so the square of 18 divides 324.
Edit: python prototype
import math
def factors(num, dict):
""" This finds the factors of a number recursively.
It is not the most efficient algorithm, and I
have not tested it a lot. You should probably
use another one. dict is a dictionary which looks
like {factor: occurrences, factor: occurrences, ...}
It must contain at least {2: 0} but need not have
any other pre-populated elements. Factors will be added
to this dictionary as they are found.
"""
while (num % 2 == 0):
num /= 2
dict[2] += 1
i = 3
found = False
while (not found and (i <= int(math.sqrt(num)))):
if (num % i == 0):
found = True
factors(i, dict)
factors(num / i, dict)
else:
i += 2
if (not found):
if (num in dict.keys()):
dict[num] += 1
else:
dict[num] = 1
return 0
#MAIN ROUTINE IS HERE
n1 = 37 # first number (6 in your example)
n2 = 41 # second number (9 in your example)
dict = {2: 0} # initialise factors (start with "no factors of 2")
factors(n1, dict) # find the factors of f1 and add them to the list
factors(n2, dict) # find the factors of f2 and add them to the list
sqfac = 1
# now find all factors repeated twice and multiply them together
for k in dict.keys():
dict[k] /= 2
sqfac *= k ** dict[k]
# here is the result
print(sqfac)
Answer in C++
int func(int i, j)
{
int k = 54
float result = pow(i, 2)/k
if (static_cast<int>(result)) == result)
{
if(i < j)
{
func(j, i);
}
else
{
cout << "Number is correct: " << i << endl;
}
}
else
{
cout << "Number is wrong" << endl;
func(j, i)
}
}
Explanation:
First recursion then test if result is a positive integer if it is then check if the other multiple is less or greater if greater recursive function tries the other multiple and if not then it is correct. Then if result is not positive integer then print Number is wrong and do another recursive function to test j.
If I got the problem correctly, I see that you have a rectangle of length=A, width=B, and area=K
And you want convert it to a square and lose the minimum possible area
If this is the case. So the problem with your algorithm is not the cost of iterating through mutliple iterations till get the output.
Rather the problem is that your algorithm depends heavily on the length A and width B of the input rectangle.
While it should depend only on the area K
For example:
Assume A =1, B=25
Then K=25 (the rect area)
Your algorithm will take the minimum value, which is A and accept it as answer with a single
iteration which is so fast but leads to wrong asnwer as it will result in a square of area 1 and waste the remaining 24 (whatever cm
or m)
While the correct answer here should be 5. which will never be reached by your algorithm
So, in my solution I assume a single input K
My ideas is as follows
x = sqrt(K)
if(x is int) .. x is the answer
else loop from x-1 till 1, x--
if K/x^2 is int, x is the answer
This might take extra iterations but will guarantee accurate answer
Also, there might be some concerns on the cost of sqrt(K)
but it will be called just once to avoid misleading length and width input
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.
I need to write an algorithm for a given problem: You have infinite pennies, nickels, dimes, and quarters. Write a class method that will output all combinations of coins such that the total is 99 cents.
It seems like a permutation nPr problem. Any algoritham for it?
Regards,
Priyank
I think this problem is most easily answered using recursion w a table of denominations
{5000, 2000, ... 1} // $50's to one penny
You would start with:
WaysToMakeChange(10000, 0) // ie. $100...highest denomination index is 0 ($50)
WaysToMakeChange(amount, maxdenomindex) would calculate using 0 or more of the maxdenom
the recurance is something like
WaysToMakeChange(amount - usedbymaxdenom, maxdenomindex - 1)
I programmed this and it can be optimized in many ways:
1) multithreading
2) caching. This is very important. B/c of the way the algorithm works, WaysToMakeChange(m,n) will be called many times with the same initial values:
For example. Changing $100 can be done by:
1 $50 + 0 $20's + 0 $10's + ways to $50 dollars with highest currency $5 (ie. WaysToMakeChange(5000, index for $5)
0 $50 + 2 $20's + 1 $10's + ways to $50 dollars with highest currency $5 (ie. WaysToMakeChange(5000, index for $5)
Clearly WaysToMakeChange(5000, index for $5) can be cached so that the subsequent call does not need to be made
3) Shortcircuiting the lowest recursion.
Suppose static const int denom[] = {5000, 2000, 1000, 500, 200, 100, 50, 25, 10, 5, 1};
The first test for WaysToMakeChange(int total, int coinIndex) should be something like:
if( coins[_countof(coins)-1] == 1 && coinIndex == _countof(coins) - 2){
return total / coins[_countof(coins)-2] + 1;
}
What does this mean? Well if your lowest denom is 1 then you only have to go as far as the second lowest denom (say a nickel). Then there are 1+ total/second lowest denom left. For example:
49c -> 5 nickels + 4 pennies. 4 nickels + 9 pennies....49 pennies = 1+ total/second lowest denom left
The easiest way is probably to spend a few moments thinking about the problem. There is a relatively nice, recursive, algorithm that lends itself neatly to either memoization or reworking into a dynamic programming solution.
This problem is classic Dynamic Programming problem. You can read about it here
http://www.algorithmist.com/index.php/Coin_Change
the python code is:
def count( n, m ):
if n == 0:
return 1
if n < 0:
return 0
if m <= 0 and n >= 1:
return 0
return count( n, m - 1 ) + count( n - S[m], m )
Here S[m] gives the value of the denomination and S is a sorted array of denominations
This problem seems like it is a diophantine equation, i.e. for a*x + b*y + ... = n, find a solution, where all letters are integers. The simplest, but not the most elegant solution would be an iterative one (displayed in python, note that I skip variable l because it resembles the number 1):
dioph_combinations = list()
for i in range(0, 99, 25):
for j in range(0, 99-i, 10):
for k in range(0, 99-i-j, 5):
for m in range(0, 99-i-j-k, 1):
if i + j + k + m == 99:
dioph_combinations.append( (i/25, j/10, k/5, m) )
The resulting list dioph_combinations will contain the possible combinations.
This problem is taken from interviewstreet.com
Given array of integers Y=y1,...,yn, we have n line segments such that
endpoints of segment i are (i, 0) and (i, yi). Imagine that from the
top of each segment a horizontal ray is shot to the left, and this ray
stops when it touches another segment or it hits the y-axis. We
construct an array of n integers, v1, ..., vn, where vi is equal to
length of ray shot from the top of segment i. We define V(y1, ..., yn)
= v1 + ... + vn.
For example, if we have Y=[3,2,5,3,3,4,1,2], then v1, ..., v8 =
[1,1,3,1,1,3,1,2], as shown in the picture below:
For each permutation p of [1,...,n], we can calculate V(yp1, ...,
ypn). If we choose a uniformly random permutation p of [1,...,n], what
is the expected value of V(yp1, ..., ypn)?
Input Format
First line of input contains a single integer T (1 <= T <= 100). T
test cases follow.
First line of each test-case is a single integer N (1 <= N <= 50).
Next line contains positive integer numbers y1, ..., yN separated by a
single space (0 < yi <= 1000).
Output Format
For each test-case output expected value of V(yp1, ..., ypn), rounded
to two digits after the decimal point.
Sample Input
6
3
1 2 3
3
3 3 3
3
2 2 3
4
10 2 4 4
5
10 10 10 5 10
6
1 2 3 4 5 6
Sample Output
4.33
3.00
4.00
6.00
5.80
11.15
Explanation
Case 1: We have V(1,2,3) = 1+2+3 = 6, V(1,3,2) = 1+2+1 = 4, V(2,1,3) =
1+1+3 = 5, V(2,3,1) = 1+2+1 = 4, V(3,1,2) = 1+1+2 = 4, V(3,2,1) =
1+1+1 = 3. Average of these values is 4.33.
Case 2: No matter what the permutation is, V(yp1, yp2, yp3) = 1+1+1 =
3, so the answer is 3.00.
Case 3: V(y1 ,y2 ,y3)=V(y2 ,y1 ,y3) = 5, V(y1, y3, y2)=V(y2, y3, y1) =
4, V(y3, y1, y2)=V(y3, y2, y1) = 3, and average of these values is
4.00.
A naive solution to the problem will run forever for N=50. I believe that the problem can be solved by independently calculating a value for each stick. I still need to know if there is any other efficient approach for this problem. On what basis do we have to independently calculate value for each stick?
We can solve this problem, by figure out:
if the k th stick is put in i th position, what is the expected ray-length of this stick.
then the problem can be solve by adding up all the expected length for all sticks in all positions.
Let expected[k][i] be the expected ray-length of k th stick put in i th position, let num[k][i][length] be the number of permutations that k th stick put in i th position with ray-length equals to length, then
expected[k][i] = sum( num[k][i][length] * length ) / N!
How to compute num[k][i][length]? For example, for length=3, consider the following graph:
...GxxxI...
Where I is the position, 3 'x' means we need 3 sticks that are strictly lower then I, and G means we need a stick that are at least as high as I.
Let s_i be the number of sticks that are smaller then the k th the stick, and g_i be the number of sticks that are greater or equal to the k th stick, then we can choose any one of g_i to put in G position, we can choose any length of s_i to fill the x position, so we have:
num[k][i][length] = P(s_i, length) * g_i * P(n-length-1-1)
In case that all the positions before I are all smaller then I, we don't need a greater stick in G, i.e. xxxI...., we have:
num[k][i][length] = P(s_i, length) * P(n-length-1)
And here's a piece of Python code that can solve this problem:
def solve(n, ys):
ret = 0
for y_i in ys:
s_i = len(filter(lambda x: x < y_i, ys))
g_i = len(filter(lambda x: x >= y_i, ys)) - 1
for i in range(n):
for length in range(1, i+1):
if length == i:
t_ret = combination[s_i][length] * factorial[length] * factorial[ n - length - 1 ]
else:
t_ret = combination[s_i][length] * factorial[length] * g_i * factorial[ n - length - 1 - 1 ]
ret += t_ret * length
return ret * 1.0 / factorial[n] + n
This is the same question as https://cs.stackexchange.com/questions/1076/how-to-approach-vertical-sticks-challenge and my answer there (which is a little simpler than those given earlier here) was:
Imagine a different problem: if you had to place k sticks of equal heights in n slots then the expected distance between sticks (and the expected distance between the first stick and a notional slot 0, and the expected distance between the last stick and a notional slot n+1) is (n+1)/(k+1) since there are k+1 gaps to fit in a length n+1.
Returning to this problem, a particular stick is interested in how many sticks (including itself) as as high or higher. If this is k, then the expected gap before it is also (n+1)/(k+1).
So the algorithm is simply to find this value for each stick and add up the expectation. For example, starting with heights of 3,2,5,3,3,4,1,2, the number of sticks with a greater or equal height is 5,7,1,5,5,2,8,7 so the expectation is 9/6+9/8+9/2+9/6+9/6+9/3+9/9+9/8 = 15.25.
This is easy to program: for example a single line in R
V <- function(Y){(length(Y) + 1) * sum(1 / (rowSums(outer(Y, Y, "<=")) + 1) )}
gives the values in the sample output in the original problem
> V(c(1,2,3))
[1] 4.333333
> V(c(3,3,3))
[1] 3
> V(c(2,2,3))
[1] 4
> V(c(10,2,4,4))
[1] 6
> V(c(10,10,10,5,10))
[1] 5.8
> V(c(1,2,3,4,5,6))
[1] 11.15
As you correctly, noted we can solve problem independently for each stick.
Let F(i, len) is number of permutations, that ray from stick i is exactly len.
Then answer is
(Sum(by i, len) F(i,len)*len)/(n!)
All is left is to count F(i, len). Let a(i) be number of sticks j, that y_j<=y_i. b(i) - number of sticks, that b_j>b_i.
In order to get ray of length len, we need to have situation like this.
B, l...l, O
len-1 times
Where O - is stick #i. B - is stick with bigger length, or beginning. l - is stick with heigth, lesser then ith.
This gives us 2 cases:
1) B is the beginning, this can be achieved in P(a(i), len-1) * (b(i)+a(i)-(len-1))! ways.
2) B is bigger stick, this can be achieved in P(a(i), len-1)*b(i)*(b(i)+a(i)-len)!*(n-len) ways.
edit: corrected b(i) as 2nd term in (mul)in place of a(i) in case 2.