Pandigital products
We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once; for example, the 5-digit number, 15234, is 1 through 5 pandigital.
The product 7254 is unusual, as the identity, 39 × 186 = 7254, containing multiplicand, multiplier, and product is 1 through 9 pandigital.
Find the sum of all products whose multiplicand/multiplier/product identity can be written as a 1 through 9 pandigital.
The product 7254 is unusual, as the identity, 39 × 186 = 7254, containing multiplicand, multiplier, and product is 1 through 9 pandigital.
Find the sum of all products whose multiplicand/multiplier/product identity can be written as a 1 through 9 pandigital.
HINT: Some products can be obtained in more than one way so be sure to only include it once in your sum.
Even though I have written the correct code I didn't get the correct answer. The answer lies in the question. Read on to understand why?
One might consider that we will have to loop till 987654321 and then find all the pandigital numbers. Then find if any part of the number will form the product of the other two parts. While this might be a case, but looping 987654321 times is not really necessary. Even if you will loop these many times and find the numbers, then you would end up getting a list of 362880(9!) numbers for which you will have to do the same. We can scale down the number of times we will loop.
Instead of looking at the problem from the first sentence(in the question) point of view, lets look at the problem in the second sentence(in question) point of view. Now to form a nine digit there are many ways. But to get, the length of two numbers combined with their product to be 9 we only have few options. Mathematically we can write it as:
len(a)+len(b)+len(a*b) = 9To get the above we only have few options.
Also to not repeat the same combination we can still reduce the number of ways.
Lets consider that a will take 1 digit then b will take 4 digit number to get another 4 digit number, to sum up to 9.
Consider a taking 2 digit number, b will take 3 digit number to get 4 digit number again.
Consider a taking 3 digit number, then b should take 2 digit number to get 4 digit number again.
Consider a taking 4 digit number, then b should take 1 digit number to get 4 digit number.
If you have seen, the last two sentences(in the above sentences) can be eliminated because they are a similar repetition of the first two sentences.
Also if the length of two number and their product increases 9 then we can stop multiplying with other numbers because the value is just going to increase instead of decreasing.
For example consider that we will have to form a single digit number as the product. Even though we are not exactly going to consider pandigital numbers this example is for understanding the above problem
My loop is running through 3 now(say), and the limit for my second loop is 5(say), then the looping will go on like this
3 x 1 = 3
3 x 2 = 6
3 x 3 = 9
3 x 4 = 12
3 x 5 = 15
4 x 1 = 4
4 x 2 = 8
4 x 3 = 12
...
In the above looping process after I had seen that the number 12 is a two digit number, and if I have breaked the loop there itself, then we would have saved one iteration. Obviously as we are using python even though you will break from one loop another loop will be working. If you break the loop at 12(3 x 4) then we will again start at 4 x 1 = 4 and so on. Now in this we will break at 12(4 x 3) again.
Using the above technique we can reduce the number of iterations by almost 80% percent. This is what had been translated to python in the program. Hope you have understood the concept.
Program
And yes I have looped separately for a two digit multiplicand and for a single digit multiplicand to avoid unnecessary iterations.
Also as the question states not to add the same product twice to get the solution, I have used python sets to eliminate duplicates.
As always you can download the source code from Github Gist pep32.py.
Output
Summary
This problem was very simple, but had a very good twist in the question. It also tested us whether we would be using the first statement or the second statement to get the answer. Anyways I am satisfied with this optimized code which had a very good performance and ran in less than 0.1 second. I think still there is a scope for improvement.
If you have any doubt or didn't understand anything then comment in the comment box below and I will be glad to help you.
Please do comment in the comment box below if you have found any typo or have a different or better program or have any suggestions. I will be glad to view each of them.
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Thank you. Have a nice day😃.