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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

Procedural form?

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Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,376

Loops and such.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

There is no other way.At least not that I know of.But Maxima allows while loops.

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Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,376

While loops are procedural. Why are you not using math commands to do math?

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

What would you to sum those terms then?

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,376

Sum[x^k,k,1,6] yields

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

```
CombGF(list) := block([i, a, m, s],
i = 1,
m = 1,
while i <= length(list) do
[
a = get(list,i),
s=sum(x^k/k!,k,1,i),
m = m * s,
delete(a,list,1),
i = i + 1
],
print(m)
);
```

*Last edited by anonimnystefy (2012-04-30 02:11:39)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,376

You do not need the while loop and you are jumping ahead. Let's get the OGF's down first.

To have the probability gf for 2 die:

Sum[x^k/6,k,1,6]^2 yields

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

I want the function to do everything for me.I just have to input 6,6 instead of just 6 to get probability for 2 dice.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,376

I want the function to do everything for me.I just have to input 6,6 instead of just 6 to get probability for 2 dice.

That is overkill. You make functions that do the minimum and use other commands on them. You do not try to make functions to do everything.

You make a function that inputs the parameters you need and then you use the series inside the function.

Here is pseudocode:

gf(a,b,d)

Begin:

Sum[x^k/d,k,a,b]

End:

Now when you call gf(1,6,6) you get:

Congratulations you just threw a die!

To throw 2 you just go

gf(1,6,6)^2

Simple, no need to make a function throw n die when you just have to square or cube or raise to the 10th power...

I want you to work on creating a function gf that inputs the 3 parameters as I have indicated.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

This code both compiles and runs,but doesn't give a good output:

```
CombGF(list) := block([i, a, m, s],
i : 1,
m : 1,
while i <= length(list) do
[
a : first(list),
print(a),
s : sum(x^k, k, 0, a),
m : m * s,
list:rest(list),
i : i + 1
],
print(m)
);
```

and the output for:

list:[1,1,2,3];

CombGF(list);

is:

[1,1,2,3]

1

1

(x+1)^2

*Last edited by anonimnystefy (2012-04-30 03:51:46)*

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

Done it! Could you do a little test on the code:

Exp GF:

```
CombEGF(list) := block([i, a, m, s],
i : 1,
m : 1,
l : length(list),
while i <= l do
[
a : first(list),
s : sum(x^k/k!, k, 0, a),
m : m * s,
list : rest(list),
i : i + 1
],
m
);
©anonimnystefy 2012
```

Ordinary GF:

```
CombOGF(list) := block([i, a, m, s],
i : 1,
m : 1,
l : length(list),
while i <= l do
[
a : first(list),
s : sum(x^k, k, 0, a),
m : m * s,
list : rest(list),
i : i + 1
],
m
);
©anonimnystefy 2012
```

The coefficient of OGF:

```
OGFcoeff(list,n) := ratcoeff(CombOGF(list),x^n);
©anonimnystefy 2012
```

The coefficient of EGF:

```
EGFcoeff(list,n) := ratcoeff(CombEGF(list),x^n)*n!;
©anonimnystefy 2012
```

Notice that I multiply by n! in the EGFcoeff function to get the final result immediately.

*Last edited by anonimnystefy (2012-04-30 04:50:58)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,376

Hi;

Please look at my hidden text message in post 110.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

I did even more than that.The function CombOGF can create those generating functions.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,376

You missed alot of what I am trying to say by sticking to the procedural

approach. It is not uncommon for people being introduced into new

fields to try to keep what they already know.

While this is usually a good idea in the case of programming it can

be disastrous. I wanted you to tear down your Pascal or Basic famework

and learn another one.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

But this solution works.It is applicable in many situations.

*Last edited by anonimnystefy (2012-04-30 05:42:59)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,376

I understand that it is tough to do what I ask. Of the two people I taught neither could accomplish it. Perhaps you remember this:

Yibka: Once you start down the path of procedural programming forever will it

dominate your destiny, consume you it will as it did Bit One's apprentice.

Lukky: Is the procedural paradigm stronger?

Yibka: No, no no! Easier! More seductive...

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

I do not remember that.Where is that from?

But,I am willing to try if you have the patience.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,376

I do not remember that.Where is that from?

I think we have been lied to on every level by our masters. I was assured many times that that was standard knowledge.

But,I am willing to try if you have the patience.

I have the patience but do have I have the lifespan of several millenia?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

Well you have already live hundreds of millenia,so that won't be the problem.Let's get started if you want.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,376

Show me a full output of your best and most complicated example using your function.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

Hi bobbym

Inout:

```
list : [1,2,3,4,5,6,7,8,9,10,11,12,13];
CombOGF(list);
```

Output;

```
[1,2,3,4,5,6,7,8,9,10,11,12,13]
(%o74) (x+1)*(x^2+x+1)*(x^3+x^2+x+1)*(x^4+x^3+x^2+x+1)*(x^5+x^4+x^3+x^2+x+1)*(x^6+x^5+x^4+x^3+x^2+x+1)*
(x^7+x^6+x^5+x^4+x^3+x^2+x+1)*(x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1)*(x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1)*
(x^10+x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1)*(x^11+x^10+x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1)*
(x^12+x^11+x^10+x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1)*
(x^13+x^12+x^11+x^10+x^9+x^8+x^7+x^6+x^5+x^4+x^3+x^2+x+1)
```

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,376

Okay, how do you throw a die 4 times and get the probability of having the

sum of them equal 20?

Taking a break, be back in a while.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,522

Hi bobbym

35/1296?

Okay,see you later!

*Last edited by anonimnystefy (2012-04-30 06:51:12)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,376

That is correct! Okay, then use yours.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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