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calculations and sums (logic or prolog)

Posted on 2001-07-05
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Last Modified: 2011-10-03
hi
i have a 32 long number like
32392382832738273827382715273987
and i need a code wich finds a calculation where the result is the number
i want it also as small at it can be
so dont need a calculation like

1+1+1+1+1+1+1+1+.....
i need it more like 15^20^52
please write it in prolog or in a text

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Question by:blizz50
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by:ygal02
ID: 6255819
The simplest way (if I understood you well) will be:

3*(10^31) + 2*(10^30) + 3*(10^29) + 9*(10^28) + ... + 8*10 + 7
Total of 32 parts.
You can add every two part if you want and you'll get something like:
32*(10^30) + 39*(10^28) + ... + 87
This will end up with only 16 parts, and so on. You can combine every two parts again so you'll have only 8 parts but the coefficients will become bigger and bigger, so you can decide which one is the best for you.

Good Luck !
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Author Comment

by:blizz50
ID: 6255961
nono
in your comment you get per diit 8 symbols
the 1st number 3 would be
3*(10^31) thats to much
i will show you a example
(its not 32 long)
387420489
after running the programm you get
9^9
thats 3 symbols
you understand now?
i need a programm that finds  such a calculation for any number

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by:edey
ID: 6256030
or, and I'm sorry but I just couldn't resist, in base 32392382832738273827382715273987 the answer would be simply a constant - 10

GL
Mike
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by:Neutron
ID: 6258791
Since it is Prolog you will use, you need the solution that has your 32digit number as a list.

If you want to make it simple, you can make 10 predicates which have a single digit as a list Head and rest of digits are the Tail, something like this:

[0|Rest]
[1|Rest]
[2|Rest]
...
[9|Rest]

So, head unifies with a digit. You would have an accumulating parameter which is in your initial call equal to 0 (zero).
If head unifies with 7, you just multiply accumulating parameter with 10 and add 7 and make a new call with tail and 10*oldAccumulatedParameter+7 as a new accumulated value. You end your recursion when the list is empty.

calculate( [], A ).
calculate( [0|Tail], A ):-calculate( Tail, A*10 ).
calculate( [1|Tail], A ):-calculate( Tail, A*10+1 ).
calculate( [2|Tail], A ):-calculate( Tail, A*10+2 ).
...
calculate( [9|Tail], A ):-calculate( Tail, A*10+9 ).
calculate( [H|Tail], A ):-!, fail.

I'm not using Prolog so in this code above there might be some syntax errors, but since you're programming in Prolog you should see my logic out of it anyway.

'Hope this helps,
    Ntr:)
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by:Neutron
ID: 6258803
If you want to control if the number is exactly 32digit, you can add another accumulating parameter, which you would increase in each new predicate call, so it would be a digit counter.
This parameter you would also set to zero in initial call.

Greetings,
    Ntr:)
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by:nico5038
ID: 7487613

No comment has been added lately, so it's time to clean up this TA.
I will leave a recommendation in the Cleanup topic area that this question is:
 - PAQ'd and pts removed
Please leave any comments here within the
next seven days.

PLEASE DO NOT ACCEPT THIS COMMENT AS AN ANSWER !

Nic;o)
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Netminder earned 0 total points
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Per recommendation, points NOT refunded and question closed.

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