Coin sums
In England the currency is made up of pound, £, and pence, p, and there are eight coins in general circulation:
1p, 2p, 5p, 10p, 20p, 50p, £1 (100p) and £2 (200p).It is possible to make £2 in the following way:
1×£1 + 1×50p + 2×20p + 1×5p + 1×2p + 3×1pHow many different ways can £2 be made using any number of coins?
Actually a lot of people are confused with the question. First of all let me explain to you the meaning of the question and then I will tell you the approach to solution.
As in case of every currency, also in England we have decimal places for a sum of money. For example 2.33, 5.05, 4.2 and so on. Now in 2.33, 2 is called the pound(£) and 33 is called as pence(p). To get 33 pence we can use 20 pence, 10 pence, 2 pence and 1 pence. We can also form the same 33 pence in many different ways. A similar question was asked to be solved.
Now coming back to our problem, firstly we will have to determine the maximum number of coins for each and every denomination.
| Denomination | Maximum Number of coins |
|---|---|
| 1p | 200 |
| 2p | 100 |
| 5p | 40 |
| 10p | 20 |
| 20p | 10 |
| 50p | 4 |
| 100p(1£) | 2 |
| 200p(2£) | 1 |
We can form 2£ in many ways from the above denominations. But for our calculation purposes I will assume some notations for each and every denomination.
| Denomination | Symbol |
|---|---|
| 1p | x1 |
| 2p | x2 |
| 5p | x3 |
| 10p | x4 |
| 20p | x5 |
| 50p | x6 |
| 100p(1£) | x7 |
| 200p(2£) | x8 |
So to write the question in mathematical equation format it can be written as:
x1+2x2+5x3+10x4+20x5+50x6+100x7+200x8 = 200
{x1,x2,x3,x4,x5,x6,x7,x8} ∈ Integers
As you can see the maximum value and the only value x8 can take is 1. So to reduce the complexity in our problem we can eliminate x8 in the equation. But to the final solution we will add the 1 possibility.
So for our solution we will take x8 = 0 and we can add 1 to the final count.So the equation will become
x1+2x2+5x3+10x4+20x5+50x6+100x7 = 200
Now we have one equation with 7 unknowns. But we know that with one equation we can only find the value of one variable.
Okay, Lets say I will first choose 100p denomination and give some a coins,
then the maximum number of 50p coins I can use will become
(200-100a)/50 = b(say)
Now the maximum number of 20p coins I can use will become
(200-100a-50b)/20 = c(say)
Maximum number of 10p coins I can use will become (200-100a-50b-20c)/10 = d(say)
maximum number of 5p coins I can use will become (200-100a-50b-20c-10d)/5 = e(say)
maximum number of 2p coins I can use will become (200-100a-50b-20c-10d-5e)/2 = f(say)
If you have observed out of 7 variables in the equation we have already fixed 6 variables and now the resulting will be the number of 1p coins which I can use directly.
So there is no need for any calculations at the end for the number of 1p coins.
The same has been used in the program. See the program and you will understand it easily.
Program
There are a series of for loops in this program.
I have added 1 with the counter at the end so that we can find the total value including the 200p case. If you remember we have left x8 from our equation.
As always you can download the source code from Github Gist pep31.py .
Output
Summary
This problem has been relatively easy for me to solve. But when I solved for the first time I wrote a program which would loop for around 128000000. If I had used that program then it would have taken hours to get executed. But after that I thought of starting from 100p instead of 1p then I ended up writing the above program which would loop only the required number of times(Answer). I am satisfied with the speed with which the program gets executed. As always I think there is a scope for improvement still. I was unable to eliminate the for loops. May be one can think in that point of view.
If you haven't understood anything or have any doubt then comment in the comment box below and I will be glad to help you.
Please do comment if you have found any typo or if you have any better program or different program or if you have any suggestion. I will be glad to view each and every one of them.
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Thank you. Have a nice day.