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Old 11-20-2012, 08:20 PM   #1
 
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Default ÇáÑÌÇÁ ÇáãÓÇÚÏå ÇáÓÑ*Úå ÌÏÇÇÇÇÇÇÇÇÇÇÇÇÇ :( :( Úä Çáprolog

<div>ÇáÓáÇã Úá*ßã
áæ ÓãÍ澂 ÚäÏ* ÓÄÇá*ä Úä ÇáÜ prolog æÇÈÏÇ ãæ ÚÇÑÝÉ Íáåã ÓÇÇÇÇÚÏæææææä* ÌÏÇ ãÖØÑå æÇáÊÓá*ã ÈßÑÇ

Exercise 2.7.
Write a predicate replace/4 to replace all occurrences of a given element
(second argument) by another given element (third argument) in a given list (first
argument). Example:

?- replace([1, 2, 3, 4, 3, 5, 6, 3], 3, x, List).
List = [1, 2, x, 4, x, 5, 6, x]
Yes

Exercise 2.8.
Prolog lists without duplicates can be interpreted as sets. Write a
program that given such a list computes the corresponding power set. Recall that the
power set of a set
S is the set of all subsets of S. This includes the empty set as well as
the set
S itself.
Define a predicate
power/2 such that, if the first argument is instantiated with a
list, the corresponding power set (i.e. a list of lists) is returned in the second position.
Example:

?- power([a, b, c], P).
P = [[a, b, c], [a, b], [a, c], [a], [b, c], [b], [c], []]
Yes
Note: The order of the sub-lists in your result doesn’t matter.
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