Combinations with repetitions: Difference between revisions
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Thundergnat (talk | contribs) Rename Perl 6 -> Raku, alphabetize, minor clean-up |
Thundergnat (talk | contribs) m Reverted edits by Thundergnat (talk) to last revision by E |
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Total Donuts: 6 |
Total Donuts: 6 |
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Total Tens: 220</pre> |
Total Tens: 220</pre> |
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=={{header|AppleScript}}== |
=={{header|AppleScript}}== |
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{{Trans|Haskell}} |
{{Trans|Haskell}} |
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Line 455: | Line 454: | ||
Were there ten donuts, we'd have had 220 choices of three</pre> |
Were there ten donuts, we'd have had 220 choices of three</pre> |
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=={{header|C++}}== |
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Non recursive version. |
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<lang cpp> |
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#include <cstdio> |
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#include <vector> |
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#include <string> |
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using namespace std; |
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void print_vector(const vector<int> &v, size_t n, const vector<string> &s){ |
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for (size_t i = 0; i < n; ++i) |
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printf("%s\t", s[v[i]].c_str()); |
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printf("\n"); |
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} |
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void combination_with_repetiton(int sabores, int bolas, const vector<string>& v_sabores){ |
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sabores--; |
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vector<int> v(bolas+1, 0); |
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while (true){ |
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for (int i = 0; i < bolas; ++i){ //vai um |
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if (v[i] > sabores){ |
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v[i + 1] += 1; |
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for (int k = i; k >= 0; --k){ |
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v[k] = v[i + 1]; |
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} |
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//v[i] = v[i + 1]; |
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} |
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} |
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if (v[bolas] > 0) |
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break; |
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print_vector(v, bolas, v_sabores); |
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v[0] += 1; |
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} |
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} |
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int main(){ |
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vector<string> options{ "iced", "jam", "plain" }; |
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combination_with_repetiton(3, 2, options); |
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return 0; |
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} |
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</lang> |
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{{out}} |
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<pre> |
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iced iced |
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jam iced |
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plain iced |
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jam jam |
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plain jam |
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plain plain |
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</pre> |
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=={{header|C sharp|C#}}== |
=={{header|C sharp|C#}}== |
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Line 556: | Line 609: | ||
</lang> |
</lang> |
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=={{header|C++}}== |
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Non recursive version. |
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<lang cpp> |
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#include <cstdio> |
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#include <vector> |
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#include <string> |
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using namespace std; |
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void print_vector(const vector<int> &v, size_t n, const vector<string> &s){ |
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for (size_t i = 0; i < n; ++i) |
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printf("%s\t", s[v[i]].c_str()); |
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printf("\n"); |
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} |
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void combination_with_repetiton(int sabores, int bolas, const vector<string>& v_sabores){ |
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sabores--; |
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vector<int> v(bolas+1, 0); |
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while (true){ |
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for (int i = 0; i < bolas; ++i){ //vai um |
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if (v[i] > sabores){ |
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v[i + 1] += 1; |
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for (int k = i; k >= 0; --k){ |
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v[k] = v[i + 1]; |
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} |
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//v[i] = v[i + 1]; |
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} |
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} |
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if (v[bolas] > 0) |
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break; |
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print_vector(v, bolas, v_sabores); |
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v[0] += 1; |
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} |
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} |
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int main(){ |
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vector<string> options{ "iced", "jam", "plain" }; |
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combination_with_repetiton(3, 2, options); |
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return 0; |
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} |
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</lang> |
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{{out}} |
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<pre> |
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iced iced |
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jam iced |
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plain iced |
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jam jam |
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plain jam |
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plain plain |
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</pre> |
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=={{header|Clojure}}== |
=={{header|Clojure}}== |
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Line 1,539: | Line 1,538: | ||
The ways to choose 3 items from 10 with replacement = 220 |
The ways to choose 3 items from 10 with replacement = 220 |
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</pre> |
</pre> |
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=={{header|JavaScript}}== |
=={{header|JavaScript}}== |
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===ES5=== |
===ES5=== |
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Line 1,867: | Line 1,865: | ||
((iced iced) (iced jam) (iced plain) (jam jam) (jam plain) (plain plain)) |
((iced iced) (iced jam) (iced plain) (jam jam) (jam plain) (plain plain)) |
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</lang> |
</lang> |
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=={{header|Lobster}}== |
=={{header|Lobster}}== |
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Line 2,160: | Line 2,159: | ||
my $count =()= combinations_with_repetition([1..10],7); |
my $count =()= combinations_with_repetition([1..10],7); |
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print "There are $count ways to pick 7 out of 10\n";</lang> |
print "There are $count ways to pick 7 out of 10\n";</lang> |
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=={{header|Perl 6}}== |
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One could simply generate all [[Permutations_with_repetitions#Perl_6|permutations]], and then remove "duplicates": |
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{{works with|Rakudo|2016.07}} |
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<lang perl6>my @S = <iced jam plain>; |
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my $k = 2; |
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.put for [X](@S xx $k).unique(as => *.sort.cache, with => &[eqv])</lang> |
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{{out}} |
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<pre> |
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iced iced |
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iced jam |
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iced plain |
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jam jam |
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jam plain |
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plain plain |
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</pre> |
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Alternatively, a recursive solution: |
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{{trans|Haskell}} |
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<lang perl6>proto combs_with_rep (UInt, @) {*} |
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multi combs_with_rep (0, @) { () } |
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multi combs_with_rep (1, @a) { map { $_, }, @a } |
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multi combs_with_rep ($, []) { () } |
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multi combs_with_rep ($n, [$head, *@tail]) { |
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|combs_with_rep($n - 1, ($head, |@tail)).map({ $head, |@_ }), |
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|combs_with_rep($n, @tail); |
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} |
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.say for combs_with_rep( 2, [< iced jam plain >] ); |
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# Extra credit: |
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sub postfix:<!> { [*] 1..$^n } |
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sub combs_with_rep_count ($k, $n) { ($n + $k - 1)! / $k! / ($n - 1)! } |
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say combs_with_rep_count( 3, 10 );</lang> |
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{{out}} |
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<pre>(iced iced) |
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(iced jam) |
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(iced plain) |
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(jam jam) |
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(jam plain) |
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(plain plain) |
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220</pre> |
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=={{header|Phix}}== |
=={{header|Phix}}== |
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Line 2,199: | Line 2,247: | ||
220 |
220 |
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</pre> |
</pre> |
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=={{header|PHP}}== |
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Non-recursive algorithm to generate all combinations with repetitons. Taken from here: [https://habrahabr.ru/post/311934/] |
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You must set k n variables and fill arrays b and c. |
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<lang PHP> |
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<?php |
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//Author Ivan Gavryushin @dcc0 |
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$k=3; |
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$n=5; |
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//set amount of elements as in $n var |
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$c=array(1,2,3,4,5); |
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//set amount of 1 as in $k var |
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$b=array(1,1,1); |
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$j=$k-1; |
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//Вывод |
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function printt($b,$k) { |
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$z=0; |
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while ($z < $k) print $b[$z++].' '; |
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print '<br>'; |
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} |
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printt ($b,$k); |
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while (1) { |
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//Увеличение на позиции K до N |
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if (array_search($b[$j]+1,$c)!==false ) { |
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$b[$j]=$b[$j]+1; |
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printt ($b,$k); |
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} |
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if ($b[$k-1]==$n) { |
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$i=$k-1; |
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//Просмотр массива справа налево |
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while ($i >= 0) { |
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//Условие выхода |
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if ( $i == 0 && $b[$i] == $n) break 2; |
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//Поиск элемента для увеличения |
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$m=array_search($b[$i]+1,$c); |
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if ($m!==false) { |
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$c[$m]=$c[$m]-1; |
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$b[$i]=$b[$i]+1; |
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$g=$i; |
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//Сортировка массива B |
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while ($g != $k-1) { |
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array_unshift ($c, $b[$g+1]); |
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$b[$g+1]=$b[$i]; |
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$g++; |
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} |
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//Удаление повторяющихся значений из C |
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$c=array_diff($c,$b); |
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printt ($b,$k); |
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array_unshift ($c, $n); |
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break; |
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} |
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$i--; |
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} |
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} |
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} |
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?> |
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</lang> |
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=={{header|PHP}}== |
=={{header|PHP}}== |
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Line 2,441: | Line 2,559: | ||
(combinations xs (- k 1))))])) |
(combinations xs (- k 1))))])) |
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</lang> |
</lang> |
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=={{header|Raku}}== |
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(formerly Perl 6) |
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One could simply generate all [[Permutations_with_repetitions#Perl_6|permutations]], and then remove "duplicates": |
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{{works with|Rakudo|2016.07}} |
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<lang perl6>my @S = <iced jam plain>; |
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my $k = 2; |
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.put for [X](@S xx $k).unique(as => *.sort.cache, with => &[eqv])</lang> |
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{{out}} |
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<pre> |
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iced iced |
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iced jam |
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iced plain |
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jam jam |
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jam plain |
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plain plain |
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</pre> |
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Alternatively, a recursive solution: |
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{{trans|Haskell}} |
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<lang perl6>proto combs_with_rep (UInt, @) {*} |
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multi combs_with_rep (0, @) { () } |
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multi combs_with_rep (1, @a) { map { $_, }, @a } |
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multi combs_with_rep ($, []) { () } |
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multi combs_with_rep ($n, [$head, *@tail]) { |
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|combs_with_rep($n - 1, ($head, |@tail)).map({ $head, |@_ }), |
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|combs_with_rep($n, @tail); |
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} |
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.say for combs_with_rep( 2, [< iced jam plain >] ); |
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# Extra credit: |
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sub postfix:<!> { [*] 1..$^n } |
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sub combs_with_rep_count ($k, $n) { ($n + $k - 1)! / $k! / ($n - 1)! } |
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say combs_with_rep_count( 3, 10 );</lang> |
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{{out}} |
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<pre>(iced iced) |
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(iced jam) |
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(iced plain) |
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(jam jam) |
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(jam plain) |
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(plain plain) |
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220</pre> |
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=={{header|REXX}}== |
=={{header|REXX}}== |
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Line 3,034: | Line 3,102: | ||
220 |
220 |
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</pre> |
</pre> |
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=={{header|TXR}}== |
=={{header|TXR}}== |
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<lang dos>txr -p "(rcomb '(iced jam plain) 2)"</lang> |
<lang dos>txr -p "(rcomb '(iced jam plain) 2)"</lang> |