- #1
richardfrobey
- 1
- 0
I wasn't sure if this should go in the number theory section, but here goes:
Is there a formula for solving problems such as: If there are n coin denominations [itex]x_{1}[/itex],[itex]x_{2}[/itex]...[itex]x_{n}[/itex] that total p cents, how many combinations are possible? n and p are positive real numbers, of course.
On a side note, wikipedia says: "There is an explicit formula for the Frobenius number when there are only two different coin denominations. If the number of coin denominations is three or more, no explicit formula is known; but, for any fixed number of coin denominations, there is an algorithm computing the Frobenius number in polynomial time (in the logarithms of the coin denominations forming an input)."
Does anyone know what the algorithm that wikipedia mentions is? I know it cannot be written explicitly and mathematically (as per wikipedia), but can anyone write it in another format?
Feel free to repost this question on mathstackexchange, mathoverflow, yahoo! answers, etc. if it will help to get the answer.
Thanks in advance.
Is there a formula for solving problems such as: If there are n coin denominations [itex]x_{1}[/itex],[itex]x_{2}[/itex]...[itex]x_{n}[/itex] that total p cents, how many combinations are possible? n and p are positive real numbers, of course.
On a side note, wikipedia says: "There is an explicit formula for the Frobenius number when there are only two different coin denominations. If the number of coin denominations is three or more, no explicit formula is known; but, for any fixed number of coin denominations, there is an algorithm computing the Frobenius number in polynomial time (in the logarithms of the coin denominations forming an input)."
Does anyone know what the algorithm that wikipedia mentions is? I know it cannot be written explicitly and mathematically (as per wikipedia), but can anyone write it in another format?
Feel free to repost this question on mathstackexchange, mathoverflow, yahoo! answers, etc. if it will help to get the answer.
Thanks in advance.