Display a linear combination: Difference between revisions

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;Task:
The purpose of this task is to display a finite [[wp:linear combination|linear combination]] in an infinite vector basis <math>(e_1, e_2,\ldots)</math>.
Display a finite &nbsp; [[wp:linear combination|linear combination]] &nbsp; in an infinite vector basis &nbsp; <big><big><math> (e_1, e_2,\ldots)</math></big></big>.


Write a function that, when given a finite list of scalars <math>(\alpha^1,\alpha^2,\ldots)</math>, creates a string representing the linear combination <math>\sum_i\alpha^i e_i</math> in an explicit format often used in mathematics, that is:
Write a function that, &nbsp; when given a finite list of scalars &nbsp; <big><big><math> (\alpha^1,\alpha^2,\ldots) </math>,</big></big>
<br>creates a string representing the linear combination &nbsp; <big><big><math> \sum_i\alpha^i e_i </math></big></big> &nbsp; in an explicit format often used in mathematics, that is:


:<math>\alpha^{i_1}e_{i_1}\pm|\alpha^{i_2}|e_{i_2}\pm|\alpha^{i_3}|e_{i_3}\pm\ldots</math>
::: <big><big><math> \alpha^{i_1}e_{i_1}\pm|\alpha^{i_2}|e_{i_2}\pm|\alpha^{i_3}|e_{i_3}\pm\ldots </math></big></big>
⚫
:::::::: where <big><big><math> \alpha^{i_k}\neq 0 </math></big></big>


<br>
⚫
where <math>\alpha^{i_k}\neq 0</math>
⚫
The output must comply to the following rules:

⚫
* &nbsp; don't show null terms, unless the whole combination is null.
⚫
In other words, and with examples and counter-examples, your output must comply to the following rules:
::::::: '''e(1)''' &nbsp; &nbsp; is fine, &nbsp; &nbsp; '''e(1) + 0*e(3)''' &nbsp; &nbsp; or &nbsp; &nbsp; '''e(1) + 0''' &nbsp; &nbsp; is wrong.

⚫
* &nbsp; don't show scalars when they are equal to one or minus one.
⚫
* you don't show null terms, unless the whole combination is null. For instance while the output "0" is fine, "e(1) + 0*e(3)" or "0 + e(1)" are wrong.
::::::: '''e(3)''' &nbsp; &nbsp; is fine, &nbsp; &nbsp; '''1*e(3)''' &nbsp; &nbsp; is wrong.
⚫
* you don't show scalars when they are equal to one or minus one. For instance the string "1*e(3)" is wrong.
* you don't prefix by '-' if it follows a preceding term. Instead you use subtraction. Thus "e(4) - e(5)" is correct while "e(4) + -e(5)" is wrong.
* &nbsp; don't prefix by a minus sign if it follows a preceding term. &nbsp; Instead you use subtraction.
::::::: '''e(4) - e(5)''' &nbsp; &nbsp; is fine, &nbsp; &nbsp; '''e(4) + -e(5)''' &nbsp; &nbsp; is wrong.


<br>
Show here output for the following lists of scalars:
Show here output for the following lists of scalars:
<pre>1) 1, 2, 3
<pre>
2) 0, 1, 2, 3
1) 1, 2, 3
3) 1, 0, 3, 4
2) 0, 1, 2, 3
4) 1, 2, 0
3) 1, 0, 3, 4
5) 0, 0, 0
4) 1, 2, 0
6) 0
5) 0, 0, 0
7) 1, 1, 1
6) 0
8) -1, -1, -1
7) 1, 1, 1
9) -1, -2, 0, -3
8) -1, -1, -1
10) -1</pre>
9) -1, -2, 0, -3
10) -1
</pre>
<br><br>


=={{header|EchoLisp}}==
=={{header|EchoLisp}}==
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