Eban numbers: Difference between revisions
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=={{header|zkl}}== |
=={{header|zkl}}== |
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{{trans|Go}} |
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<lang zkl></lang> |
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<lang zkl>rgs:=T( T(2, 1_000, True), // (start,end,print) |
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<lang zkl></lang> |
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T(1_000, 4_000, True), |
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T(2, 1e4, False), |
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T(2, 1e5, False), |
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T(2, 1e6, False), |
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T(2, 1e7, False), |
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T(2, 1e8, False), // slow |
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T(2, 1e9, False), ); // very slow |
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foreach start,end,pr in (rgs){ |
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if(start==2) println("eban numbers up to and including %,d:".fmt(end)); |
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else println("eban numbers between %,d and %,d (inclusive):".fmt(start,end)); |
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count:=0; |
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foreach i in ([start..end,2]){ |
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b:=i/100_0000_000; |
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r:=i%1_000_000_000; |
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m:=r/1_000_000; |
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r:=i%1_000_000; |
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t:=r/1_000; |
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r=r%1_000; |
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if(30<=m<=66) m=m%10; |
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if(30<=t<=66) t=t%10; |
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if(30<=r<=66) r=r%10; |
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if(magic(b) and magic(m) and magic(t) and magic(r)){ |
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if(pr) print(i," "); |
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count+=1; |
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} |
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} |
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if(pr) println(); |
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println("count = %,d\n".fmt(count)); |
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} |
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fcn magic(z){ z.isEven and z<=6 }</lang> |
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{{out}} |
{{out}} |
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<pre style="height:35ex"> |
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<pre> |
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eban numbers up to and including 1,000: |
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2 4 6 30 32 34 36 40 42 44 46 50 52 54 56 60 62 64 66 |
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count = 19 |
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eban numbers between 1,000 and 4,000 (inclusive): |
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2000 2002 2004 2006 2030 2032 2034 2036 2040 2042 2044 2046 2050 2052 2054 2056 2060 2062 2064 2066 4000 |
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count = 21 |
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eban numbers up to and including 10,000: |
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count = 79 |
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eban numbers up to and including 100,000: |
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count = 399 |
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eban numbers up to and including 1,000,000: |
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count = 399 |
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eban numbers up to and including 10,000,000: |
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count = 1,599 |
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eban numbers up to and including 100,000,000: |
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count = 7,999 |
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eban numbers up to and including 1,000,000,000: |
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count = 7,999 |
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</pre> |
</pre> |
Revision as of 16:27, 1 July 2019
![Task](http://static.miraheze.org/rosettacodewiki/thumb/b/ba/Rcode-button-task-crushed.png/64px-Rcode-button-task-crushed.png)
You are encouraged to solve this task according to the task description, using any language you may know.
- Definition
An eban number is a number that has no letter e in it when the number is spelled in ENglish.
Or more literally, spelled numbers that contain the letter e are banned.
The American version of spelling numbers will be used here (as opposed to the British).
2,000,000,000 is two billion, not two milliard.
Only numbers less than one sextillion (1021) will be considered in/for this task.
This will allow optimizations to be used.
- Task
-
- show all eban numbers ≤ 1,000 (in a horizontal format), and a count
- show all eban numbers between 1,000 and 4,000 (inclusive), and a count
- show a count of all eban numbers up and including 10,000
- show a count of all eban numbers up and including 100,000
- show a count of all eban numbers up and including 1,000,000
- show a count of all eban numbers up and including 10,000,000
- show all output here.
- See also
-
- The MathWorld entry: eban numbers.
- The OEIS entry: A6933, eban numbers.
Go
<lang go>package main
import "fmt"
type Range struct {
start, end uint64 print bool
}
func main() {
rgs := []Range{ {2, 1000, true}, {1000, 4000, true}, {2, 1e4, false}, {2, 1e5, false}, {2, 1e6, false}, {2, 1e7, false}, {2, 1e8, false}, {2, 1e9, false}, } for _, rg := range rgs { if rg.start == 2 { fmt.Printf("eban numbers up to and including %d:\n", rg.end) } else { fmt.Printf("eban numbers between %d and %d (inclusive):\n", rg.start, rg.end) } count := 0 for i := rg.start; i <= rg.end; i += 2 { b := i / 1000000000 r := i % 1000000000 m := r / 1000000 r = i % 1000000 t := r / 1000 r %= 1000 if m >= 30 && m <= 66 { m %= 10 } if t >= 30 && t <= 66 { t %= 10 } if r >= 30 && r <= 66 { r %= 10 } if b == 0 || b == 2 || b == 4 || b == 6 { if m == 0 || m == 2 || m == 4 || m == 6 { if t == 0 || t == 2 || t == 4 || t == 6 { if r == 0 || r == 2 || r == 4 || r == 6 { if rg.print { fmt.Printf("%d ", i) } count++ } } } } } if rg.print { fmt.Println() } fmt.Println("count =", count, "\n") }
}</lang>
- Output:
eban numbers up to and including 1000: 2 4 6 30 32 34 36 40 42 44 46 50 52 54 56 60 62 64 66 count = 19 eban numbers between 1000 and 4000 (inclusive): 2000 2002 2004 2006 2030 2032 2034 2036 2040 2042 2044 2046 2050 2052 2054 2056 2060 2062 2064 2066 4000 count = 21 eban numbers up to and including 10000: count = 79 eban numbers up to and including 100000: count = 399 eban numbers up to and including 1000000: count = 399 eban numbers up to and including 10000000: count = 1599 eban numbers up to and including 100000000: count = 7999 eban numbers up to and including 1000000000: count = 7999
J
<lang J> Filter =: (#~`)(`:6)
itemAmend =: (29&< *. <&67)`(,: 10&|)} iseban =: [: *./ 0 2 4 6 e.~ [: itemAmend [: |: (4#1000)&#:
(;~ #) iseban Filter >: i. 1000
┌──┬─────────────────────────────────────────────────────┐ │19│2 4 6 30 32 34 36 40 42 44 46 50 52 54 56 60 62 64 66│ └──┴─────────────────────────────────────────────────────┘
NB. INPUT are the correct integers, head and tail shown ({. , {:) INPUT =: 1000 + i. 3001
1000 4000
(;~ #) iseban Filter INPUT
┌──┬────────────────────────────────────────────────────────────────────────────────────────────────────────┐ │21│2000 2002 2004 2006 2030 2032 2034 2036 2040 2042 2044 2046 2050 2052 2054 2056 2060 2062 2064 2066 4000│ └──┴────────────────────────────────────────────────────────────────────────────────────────────────────────┘
(, ([: +/ [: iseban [: >: i.))&> 10000 * 10 ^ i. +:2 10000 79
100000 399
1e6 399 1e7 1599
</lang>
Julia
I changed your program but not the output. You had used the common cut-n-paste error repeating the 10000 println("eban numbers up to and including 10000:
<lang Julia> function iseban(n::Integer)
b, r = divrem(n, oftype(n, 10 ^ 9)) m, r = divrem(r, oftype(n, 10 ^ 6)) t, r = divrem(r, oftype(n, 10 ^ 3)) m, t, r = (30 <= x <= 66 ? x % 10 : x for x in (m, t, r)) return all(in((0, 2, 4, 6)), (b, m, t, r))
end
println("eban numbers up to and including 1000:") println(join(filter(iseban, 1:100), ", "))
println("eban numbers between 1000 and 4000 (inclusive):") println(join(filter(iseban, 1000:4000), ", "))
println("eban numbers up to and including 10000: ", count(iseban, 1:10000)) println("eban numbers up to and including 100000: ", count(iseban, 1:100000)) println("eban numbers up to and including 1000000: ", count(iseban, 1:1000000)) println("eban numbers up to and including 10000000: ", count(iseban, 1:10000000)) println("eban numbers up to and including 100000000: ", count(iseban, 1:100000000)) println("eban numbers up to and including 1000000000: ", count(iseban, 1:1000000000)) </lang>
- Output:
eban numbers up to and including 1000: 2, 4, 6, 30, 32, 34, 36, 40, 42, 44, 46, 50, 52, 54, 56, 60, 62, 64, 66 eban numbers between 1000 and 4000 (inclusive): 2000, 2002, 2004, 2006, 2030, 2032, 2034, 2036, 2040, 2042, 2044, 2046, 2050, 2052, 2054, 2056, 2060, 2062, 2064, 2066, 4000 eban numbers up to and including 10000: 79 eban numbers up to and including 10000: 399 eban numbers up to and including 10000: 399 eban numbers up to and including 10000: 1599 eban numbers up to and including 10000: 7999 eban numbers up to and including 10000: 7999
Kotlin
<lang scala>// Version 1.3.21
typealias Range = Triple<Int, Int, Boolean>
fun main() {
val rgs = listOf<Range>( Range(2, 1000, true), Range(1000, 4000, true), Range(2, 10_000, false), Range(2, 100_000, false), Range(2, 1_000_000, false), Range(2, 10_000_000, false), Range(2, 100_000_000, false), Range(2, 1_000_000_000, false) ) for (rg in rgs) { val (start, end, prnt) = rg if (start == 2) { println("eban numbers up to and including $end:") } else { println("eban numbers between $start and $end (inclusive):") } var count = 0 for (i in start..end step 2) { val b = i / 1_000_000_000 var r = i % 1_000_000_000 var m = r / 1_000_000 r = i % 1_000_000 var t = r / 1_000 r %= 1_000 if (m >= 30 && m <= 66) m %= 10 if (t >= 30 && t <= 66) t %= 10 if (r >= 30 && r <= 66) r %= 10 if (b == 0 || b == 2 || b == 4 || b == 6) { if (m == 0 || m == 2 || m == 4 || m == 6) { if (t == 0 || t == 2 || t == 4 || t == 6) { if (r == 0 || r == 2 || r == 4 || r == 6) { if (prnt) print("$i ") count++ } } } } } if (prnt) println() println("count = $count\n") }
}</lang>
- Output:
Same as Go example.
Perl
Exhaustive search
A couple of 'e'-specific optimizations keep the running time reasonable. <lang perl>use strict; use warnings; use feature 'say'; use Lingua::EN::Numbers qw(num2en);
sub comma { reverse ((reverse shift) =~ s/(.{3})/$1,/gr) =~ s/^,//r }
sub e_ban {
my($power) = @_; my @n; for (1..10**$power) { next unless 0 == $_%2; next if $_ =~ /[789]/ or /[12].$/ or /[135]..$/ or /[135]...$/ or /[135].....$/; push @n, $_ unless num2en($_) =~ /e/; } @n;
}
my @OK = e_ban(my $max = 7);
my @a = grep { $_ <= 1000 } @OK; say "Number of eban numbers up to and including 1000: @{[1+$#a]}"; say join(', ',@a); say ;
my @b = grep { $_ >= 1000 && $_ <= 4000 } @OK; say "Number of eban numbers between 1000 and 4000 (inclusive): @{[1+$#b]}"; say join(', ',@b); say ;
for my $exp (4..$max) {
my $n = + grep { $_ <= 10**$exp } @OK; printf "Number of eban numbers and %10s: %d\n", comma(10**$exp), $n;
}</lang>
- Output:
eban numbers up to and including 1000: 2, 4, 6, 30, 32, 34, 36, 40, 42, 44, 46, 50, 52, 54, 56, 60, 62, 64, 66 eban numbers between 1000 and 4000 (inclusive): 2000, 2002, 2004, 2006, 2030, 2032, 2034, 2036, 2040, 2042, 2044, 2046, 2050, 2052, 2054, 2056, 2060, 2062, 2064, 2066, 4000 Number of eban numbers up to 10,000: 79 Number of eban numbers up to 100,000: 399 Number of eban numbers up to 1,000,000: 399 Number of eban numbers up to 10,000,000: 1599
Algorithmically generate / count
Alternately, a partial translation of Perl 6. Does not need to actually generate the e-ban numbers to count them. Display counts up to 10**21.
<lang perl>use strict; use warnings; use bigint; use feature 'say'; use Lingua::EN::Nums2Words 'num2word'; use List::AllUtils 'sum';
sub comma { reverse ((reverse shift) =~ s/(.{3})/$1,/gr) =~ s/^,//r }
sub nban {
my ($n, @numbers) = @_; grep { lc(num2word($_)) !~ /[$n]/i } @numbers;
}
sub enumerate {
my ($n, $upto) = @_; my @ban = nban($n, 1 .. 99); my @orders; for my $o (2 .. $upto) { push @orders, [nban($n, map { $_ * 10**$o } 1 .. 9)]; } for my $oom (@orders) { next unless +@$oom; my @these; for my $num (@$oom) { push @these, $num, map { $_ + $num } @ban; } push @ban, @these; } unshift @ban, 0 if nban($n, 0); @ban
}
sub count {
my ($n, $upto) = @_; my @orders; for my $o (2 .. $upto) { push @orders, [nban($n, map { $_ * 10**$o } 1 .. 9)]; } my @count = scalar nban($n, 1 .. 99); for my $o ( 0 .. $#orders - 1 ) { push @count, sum(@count) * (scalar @{$orders[$o]}) + (scalar @{$orders[$o]}); } ++$count[0] if nban($n, 0); for my $m ( 0 .. $#count - 1 ) { next unless scalar $orders[$m]; if (nban($n, 10**($m+2))) { $count[$m]++; $count[$m + 1]-- } } map { sum( @count[0..$_] ) } 0..$#count;
}
for my $t ('e') {
my @bans = enumerate($t, 4); my @count = count($t, my $max = 21);
my @j = grep { $_ <= 10 } @bans; unshift @count, @{[1+$#j]};
say "\n============= $t-ban: ============="; my @a = grep { $_ <= 1000 } @bans; say "$t-ban numbers up to 1000: @{[1+$#a]}"; say '[', join(' ',@a), ']'; say ;
my @b = grep { $_ >= 1000 && $_ <= 4000 } @bans; say "$t-ban numbers between 1,000 & 4,000 (inclusive): @{[1+$#b]}"; say '[', join(' ',@b), ']'; say ;
say "Counts of $t-ban numbers up to ", lc(num2word(10**$max));
for my $exp (1..$max) { my $nu = $count[$exp-1]; printf "Up to and including %23s: %s\n", lc(num2word(10**$exp)), comma($nu); }
}</lang>
============= e-ban: ============= e-ban numbers up to 1000: 19 [2 4 6 30 32 34 36 40 42 44 46 50 52 54 56 60 62 64 66] e-ban numbers between 1,000 & 4,000 (inclusive): 21 [2000 2002 2004 2006 2030 2032 2034 2036 2040 2042 2044 2046 2050 2052 2054 2056 2060 2062 2064 2066 4000] Counts of e-ban numbers up to one sextillion Up to and including ten: 3 Up to and including one hundred: 19 Up to and including one thousand: 19 Up to and including ten thousand: 79 Up to and including one hundred thousand: 399 Up to and including one million: 399 Up to and including ten million: 1,599 Up to and including one hundred million: 7,999 Up to and including one billion: 7,999 Up to and including ten billion: 31,999 Up to and including one hundred billion: 159,999 Up to and including one trillion: 159,999 Up to and including ten trillion: 639,999 Up to and including one hundred trillion: 3,199,999 Up to and including one quadrillion: 3,199,999 Up to and including ten quadrillion: 12,799,999 Up to and including one hundred quadrillion: 63,999,999 Up to and including one quintillion: 63,999,999 Up to and including ten quintillion: 255,999,999 Up to and including one hundred quintillion: 1,279,999,999 Up to and including one sextillion: 1,279,999,999
Perl 6
Modular approach, very little is hard coded. Change the $upto order-of-magnitude limit to adjust the search/display ranges. Change the letter(s) given to the enumerate / count subs to modify which letter(s) to disallow.
Will handle multi-character 'bans'. Demonstrate for e-ban, t-ban and subur-ban.
Directly find :
- OEIS:A006933 Numbers without e: Eban
- OEIS:A008521 Numbers without o: Oban
- OEIS:A008523 Numbers without t: Tban
- OEIS:A072954 Numbers without a, i, l, t: TALIban
- OEIS:A072955 Numbers without b, r, s, u: SUBURban
- OEIS:A072956 Numbers without r, t, u: TURban
- OEIS:A072957 Numbers without r, u: URban
- OEIS:A072958 Numbers without a, c, i, l: CALIban
- OEIS:A089589 Numbers without i: Iban
- OEIS:A089590 Numbers without u: Uban
- and so on...
Considering numbers up to 1021, as the task directions suggest.
<lang perl6>use Lingua::EN::Numbers; # Version 2.4.0 or higher
sub nban ($seq, $n = 'e') { ($seq).map: { next if .&cardinal.contains(any($n.lc.comb)); $_ } }
sub enumerate ($n, $upto) {
my @ban = [nban(1 .. 99, $n)],; my @orders; (2 .. $upto).map: -> $o { given $o % 3 { # Compensate for irregulars: 11 - 19 when 0 { @orders.push: [[flat (10**$o X* 1..9).map: *.&nban($n)], [flat (10**$o X* 11..19).map: *.&nban($n)]]} when 1 { @orders.push: [flat (10**$o X* 2..9).map: *.&nban($n)] } default { @orders.push: [flat (10**$o X* 1..9).map: *.&nban($n)] } } } my @those; ^@orders .map: -> $o { @ban.push: [] and next unless +@orders[$o]; my @these; my @lower = |@orders[$o]; my @upper = []; if @orders[$o][0].^name eq 'Array' { @lower = |@orders[$o][0]; @upper = |@orders[$o][1]; } @lower.map: -> $m { @these.push: $m; for ^@ban -> $b { next unless +@ban[$b]; @these.push: $_ for (flat @ban[$b]) »+» $m ; } } @ban.push: @those.clone if +@those; @those = (); @upper.map: -> $m { @those.push: $m; for ^(@ban) -> $b { next unless +@ban[$b]; @those.push: $_ for (flat @ban[$b]) »+» $m ; } } @ban.push: @these; } @ban.push: @those.clone if +@those; @ban.unshift(0) if nban(0, $n); flat @ban.map: *.flat;
}
sub count ($n, $upto) {
my @orders; (2 .. $upto).map: -> $o { given $o % 3 { # Compensate for irregulars: 11 - 19 when 0 { @orders.push: [[flat (10**$o X* 1 .. 9).map: *.&nban($n)], [flat (10**$o X* 11..19).map: *.&nban($n)]]} when 1 { @orders.push: [flat (10**$o X* 2 .. 9).map: *.&nban($n)] } default { @orders.push: [flat (10**$o X* 1 .. 9).map: *.&nban($n)] } } } my @count = +nban(1 .. 99, $n); my $those; ^@orders .map: -> $o { @count.push: 0 and next unless +@orders[$o]; my @lower = |@orders[$o]; my @upper = []; if @orders[$o][0].^name eq 'Array' { @lower = |@orders[$o][0]; @upper = |@orders[$o][1]; } my $prev = so (@lower.first( { $_ ~~ /^ '1' '0'+ $/ } ) // 0 ); my $sum = @count.sum; my $these = +@lower * $sum + @lower; @count[1 + $o] += $those if $those; $those = +@upper * $sum + @upper; $these-- if $prev; @count[1 + $o] += $these; ++@count[$o] if $prev; } @count[*-1] += $those if $those and $upto % 3 == 1; ++@count[0] if nban(0, $n); [\+] @count;
}
- for < e o t tali subur tur ur cali i u > -> $n { # All of them
for < e t subur > -> $n { # An assortment for demonstration
my $upto = 21; # 1e21 my @bans = enumerate($n, 4); my @counts = count($n, $upto);
# DISPLAY my @k = @bans.grep: * < 1000; my @j = @bans.grep: 1000 <= * <= 4000; put "\n============= {$n}-ban: =============\n" ~ "{$n}-ban numbers up to 1000: {+@k}\n[{@k».&comma}]\n\n" ~ "{$n}-ban numbers between 1,000 & 4,000: {+@j}\n[{@j».&comma}]\n" ~ "\nCounts of {$n}-ban numbers up to {cardinal 10**$upto}" ;
my $s = max (1..$upto).map: { (10**$_).&cardinal.chars }; @counts.unshift: @bans.first: * > 10, :k; for ^$upto -> $c { printf "Up to and including %{$s}s: %s\n", cardinal(10**($c+1)), comma(@counts[$c]); }
}</lang>
- Output:
============= e-ban: ============= e-ban numbers up to 1000: 19 [2 4 6 30 32 34 36 40 42 44 46 50 52 54 56 60 62 64 66] e-ban numbers between 1,000 & 4,000: 21 [2,000 2,002 2,004 2,006 2,030 2,032 2,034 2,036 2,040 2,042 2,044 2,046 2,050 2,052 2,054 2,056 2,060 2,062 2,064 2,066 4,000] Counts of e-ban numbers up to one sextillion Up to and including ten: 3 Up to and including one hundred: 19 Up to and including one thousand: 19 Up to and including ten thousand: 79 Up to and including one hundred thousand: 399 Up to and including one million: 399 Up to and including ten million: 1,599 Up to and including one hundred million: 7,999 Up to and including one billion: 7,999 Up to and including ten billion: 31,999 Up to and including one hundred billion: 159,999 Up to and including one trillion: 159,999 Up to and including ten trillion: 639,999 Up to and including one hundred trillion: 3,199,999 Up to and including one quadrillion: 3,199,999 Up to and including ten quadrillion: 12,799,999 Up to and including one hundred quadrillion: 63,999,999 Up to and including one quintillion: 63,999,999 Up to and including ten quintillion: 255,999,999 Up to and including one hundred quintillion: 1,279,999,999 Up to and including one sextillion: 1,279,999,999 ============= t-ban: ============= t-ban numbers up to 1000: 56 [0 1 4 5 6 7 9 11 100 101 104 105 106 107 109 111 400 401 404 405 406 407 409 411 500 501 504 505 506 507 509 511 600 601 604 605 606 607 609 611 700 701 704 705 706 707 709 711 900 901 904 905 906 907 909 911] t-ban numbers between 1,000 & 4,000: 0 [] Counts of t-ban numbers up to one sextillion Up to and including ten: 7 Up to and including one hundred: 9 Up to and including one thousand: 56 Up to and including ten thousand: 56 Up to and including one hundred thousand: 56 Up to and including one million: 57 Up to and including ten million: 392 Up to and including one hundred million: 393 Up to and including one billion: 2,801 Up to and including ten billion: 19,600 Up to and including one hundred billion: 19,601 Up to and including one trillion: 140,000 Up to and including ten trillion: 140,000 Up to and including one hundred trillion: 140,000 Up to and including one quadrillion: 140,001 Up to and including ten quadrillion: 980,000 Up to and including one hundred quadrillion: 980,001 Up to and including one quintillion: 7,000,000 Up to and including ten quintillion: 7,000,000 Up to and including one hundred quintillion: 7,000,000 Up to and including one sextillion: 7,000,000 ============= subur-ban: ============= subur-ban numbers up to 1000: 35 [1 2 5 8 9 10 11 12 15 18 19 20 21 22 25 28 29 50 51 52 55 58 59 80 81 82 85 88 89 90 91 92 95 98 99] subur-ban numbers between 1,000 & 4,000: 0 [] Counts of subur-ban numbers up to one sextillion Up to and including ten: 6 Up to and including one hundred: 35 Up to and including one thousand: 35 Up to and including ten thousand: 35 Up to and including one hundred thousand: 35 Up to and including one million: 36 Up to and including ten million: 215 Up to and including one hundred million: 1,259 Up to and including one billion: 1,259 Up to and including ten billion: 1,259 Up to and including one hundred billion: 1,259 Up to and including one trillion: 1,259 Up to and including ten trillion: 1,259 Up to and including one hundred trillion: 1,259 Up to and including one quadrillion: 1,259 Up to and including ten quadrillion: 1,259 Up to and including one hundred quadrillion: 1,259 Up to and including one quintillion: 1,259 Up to and including ten quintillion: 1,259 Up to and including one hundred quintillion: 1,259 Up to and including one sextillion: 1,259
Note that the limit to one sextillion is somewhat arbitrary and is just to match the task parameters.
This will quite happily count *-bans up to one hundred centillion. (10305) It takes longer, but still on the order of seconds, not minutes.
Counts of e-ban numbers up to one hundred centillion ... Up to and including one hundred centillion: 35,184,372,088,831,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999
Phix
Why count when you can calculate? <lang Phix>function count_eban(integer p10) -- returns the count of eban numbers 1..power(10,p10)
integer n = p10-floor(p10/3), p5 = floor(n/2), p4 = floor((n+1)/2) return power(5,p5)*power(4,p4)-1
end function
function eban(integer n) -- returns true if n is an eban number (only fully tested to 10e9)
if n=0 then return false end if while n do integer thou = remainder(n,1000) if floor(thou/100)!=0 then return false end if if not find(floor(thou/10),{0,3,4,5,6}) then return false end if if not find(remainder(thou,10),{0,2,4,6}) then return false end if n = floor(n/1000) end while return true
end function
sequence s = {} for i=0 to 1000 do
if eban(i) then s &= i end if
end for printf(1,"eban to 1000 : %v (%d items)\n",{s,length(s)}) s = {} for i=1000 to 4000 do
if eban(i) then s &= i end if
end for printf(1,"eban 1000..4000 : %v (%d items)\n\n",{s,length(s)})
atom t0 = time() for i=0 to 21 do
printf(1,"count_eban(10^%d) : %,d\n",{i,count_eban(i)})
end for ?elapsed(time()-t0)</lang>
- Output:
eban to 1000 : {2,4,6,30,32,34,36,40,42,44,46,50,52,54,56,60,62,64,66} (19 items) eban 1000..4000 : {2000,2002,2004,2006,2030,2032,2034,2036,2040,2042,2044,2046,2050,2052,2054,2056,2060,2062,2064,2066,4000} (21 items) count_eban(10^0) : 0 count_eban(10^1) : 3 count_eban(10^2) : 19 count_eban(10^3) : 19 count_eban(10^4) : 79 count_eban(10^5) : 399 count_eban(10^6) : 399 count_eban(10^7) : 1,599 count_eban(10^8) : 7,999 count_eban(10^9) : 7,999 count_eban(10^10) : 31,999 count_eban(10^11) : 159,999 count_eban(10^12) : 159,999 count_eban(10^13) : 639,999 count_eban(10^14) : 3,199,999 count_eban(10^15) : 3,199,999 count_eban(10^16) : 12,799,999 count_eban(10^17) : 63,999,999 count_eban(10^18) : 63,999,999 count_eban(10^19) : 255,999,999 count_eban(10^20) : 1,279,999,999 count_eban(10^21) : 1,279,999,999 "0.0s"
Python
<lang Python>
- Use inflect
"""
show all eban numbers <= 1,000 (in a horizontal format), and a count show all eban numbers between 1,000 and 4,000 (inclusive), and a count show a count of all eban numbers up and including 10,000 show a count of all eban numbers up and including 100,000 show a count of all eban numbers up and including 1,000,000 show a count of all eban numbers up and including 10,000,000
"""
import inflect import time
before = time.perf_counter()
p = inflect.engine()
- eban numbers <= 1000
print(' ') print('eban numbers up to and including 1000:') print(' ')
count = 0
for i in range(1,1001):
if not 'e' in p.number_to_words(i): print(str(i)+' ',end=) count += 1
print(' ') print(' ') print('count = '+str(count)) print(' ')
- eban numbers 1000 to 4000
print(' ') print('eban numbers between 1000 and 4000 (inclusive):') print(' ')
count = 0
for i in range(1000,4001):
if not 'e' in p.number_to_words(i): print(str(i)+' ',end=) count += 1
print(' ') print(' ') print('count = '+str(count)) print(' ')
- eban numbers up to 10000
print(' ') print('eban numbers up to and including 10000:') print(' ')
count = 0
for i in range(1,10001):
if not 'e' in p.number_to_words(i): count += 1
print(' ') print('count = '+str(count)) print(' ')
- eban numbers up to 100000
print(' ') print('eban numbers up to and including 100000:') print(' ')
count = 0
for i in range(1,100001):
if not 'e' in p.number_to_words(i): count += 1
print(' ') print('count = '+str(count)) print(' ')
- eban numbers up to 1000000
print(' ') print('eban numbers up to and including 1000000:') print(' ')
count = 0
for i in range(1,1000001):
if not 'e' in p.number_to_words(i): count += 1
print(' ') print('count = '+str(count)) print(' ')
- eban numbers up to 10000000
print(' ') print('eban numbers up to and including 10000000:') print(' ')
count = 0
for i in range(1,10000001):
if not 'e' in p.number_to_words(i): count += 1
print(' ') print('count = '+str(count)) print(' ')
after = time.perf_counter()
print(" ") print("Run time in seconds: "+str(after - before)) </lang>
Output:
eban numbers up to and including 1000: 2 4 6 30 32 34 36 40 42 44 46 50 52 54 56 60 62 64 66 count = 19 eban numbers between 1000 and 4000 (inclusive): 2000 2002 2004 2006 2030 2032 2034 2036 2040 2042 2044 2046 2050 2052 2054 2056 2060 2062 2064 2066 4000 count = 21 eban numbers up to and including 10000: count = 79 eban numbers up to and including 100000: count = 399 eban numbers up to and including 1000000: count = 399 eban numbers up to and including 10000000: count = 1599 Run time in seconds: 1134.289519125
REXX
Programming note: REXX has no shortcuts for if statements, so the multiple if statements weren't combined into one. <lang rexx>/*REXX program to display eban numbers (those that don't have an "e" their English name)*/ numeric digits 20 /*support some gihugic numbers for pgm.*/ parse arg $ /*obtain optional arguments from the cL*/ if $= then $= '1 1000 1000 4000 1 -10000 1 -100000 1 -1000000 1 -10000000'
do k=1 by 2 to words($) /*step through the list of numbers. */ call banE word($, k), word($, k+1) /*process the numbers, from low──►high.*/ end /*k*/
exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ banE: procedure; parse arg x,y,_; z= reverse(x) /*obtain the number to be examined. */
tell= y>=0 /*Is HI non-negative? Display eban #s.*/ #= 0 /*the count of eban numbers (so far).*/ do j=x to abs(y) /*probably process a range of numbers. */ if hasE(j) then iterate /*determine if the number has an "e". */ #= # + 1 /*bump the counter of eban numbers. */ if tell then _= _ j /*maybe add to a list of eban numbers. */ end /*j*/ if _\== then say strip(_) /*display the list (if there is one). */ say; say # ' eban numbers found for: ' x " " y; say copies('═', 105) return
/*──────────────────────────────────────────────────────────────────────────────────────*/ hasE: procedure; parse arg x; z= reverse(x) /*obtain the number to be examined. */
do k=1 by 3 /*while there're dec. digit to examine.*/ @= reverse( substr(z, k, 3) ) /*obtain 3 dec. digs (a period) from Z.*/ if @==' ' then return 0 /*we have reached the "end" of the num.*/ uni= right(@, 1) /*get units dec. digit of this period. */ if uni//2==1 then return 1 /*if an odd digit, then not an eban #. */ if uni==8 then return 1 /*if an eight, " " " " " */ tens=substr(@, 2, 1) /*get tens dec. digit of this period. */ if tens==1 then return 1 /*if teens, then not an eban #. */ if tens==2 then return 1 /*if twenties, " " " " " */ if tens>6 then return 1 /*if 70s, 80s, 90s, " " " " " */ hun= left(@, 1) /*get hundreds dec. dig of this period.*/ if hun==0 then iterate /*if zero, then there is more of number*/ if hun\==' ' then return 1 /*any hundrEd (not zero) has an "e". */ end /*k*/ /*A "period" is a group of 3 dec. digs */ return 0 /*in the number, grouped from the right*/</lang>
- output when using the default inputs:
2 4 6 30 32 34 36 40 42 44 46 50 52 54 56 60 62 64 66 19 eban numbers found for: 1 1000 ═════════════════════════════════════════════════════════════════════════════════════════════════════════ 2000 2002 2004 2006 2030 2032 2034 2036 2040 2042 2044 2046 2050 2052 2054 2056 2060 2062 2064 2066 4000 21 eban numbers found for: 1000 4000 ═════════════════════════════════════════════════════════════════════════════════════════════════════════ 79 eban numbers found for: 1 -10000 ═════════════════════════════════════════════════════════════════════════════════════════════════════════ 399 eban numbers found for: 1 -100000 ═════════════════════════════════════════════════════════════════════════════════════════════════════════ 399 eban numbers found for: 1 -1000000 ═════════════════════════════════════════════════════════════════════════════════════════════════════════ 1599 eban numbers found for: 1 -10000000 ═════════════════════════════════════════════════════════════════════════════════════════════════════════
Yabasic
<lang Yabasic>data 2, 100, true data 1000, 4000, true data 2, 1e4, false data 2, 1e5, false data 2, 1e6, false data 2, 1e7, false data 2, 1e8, false REM data 2, 1e9, false // it takes a lot of time data 0, 0, false
do
read start, ended, printable if not start break if start = 2 then Print "eban numbers up to and including ", ended else Print "eban numbers between ", start, " and ", ended, " (inclusive):" endif count = 0 for i = start to ended step 2 b = int(i / 1000000000) r = mod(i, 1000000000) m = int(r / 1000000) r = mod(i, 1000000) t = int(r / 1000) r = mod(r, 1000) if m >= 30 and m <= 66 m = mod(m, 10) if t >= 30 and t <= 66 t = mod(t, 10) if r >= 30 and r <= 66 r = mod(r, 10) if b = 0 or b = 2 or b = 4 or b = 6 then if m = 0 or m = 2 or m = 4 or m = 6 then if t = 0 or t = 2 or t = 4 or t = 6 then if r = 0 or r = 2 or r = 4 or r = 6 then if printable Print i; count = count + 1 endif endif endif endif next if printable Print Print "count = ", count, "\n"
loop</lang>
zkl
<lang zkl>rgs:=T( T(2, 1_000, True), // (start,end,print)
T(1_000, 4_000, True), T(2, 1e4, False), T(2, 1e5, False), T(2, 1e6, False), T(2, 1e7, False), T(2, 1e8, False), // slow T(2, 1e9, False), ); // very slow
foreach start,end,pr in (rgs){
if(start==2) println("eban numbers up to and including %,d:".fmt(end)); else println("eban numbers between %,d and %,d (inclusive):".fmt(start,end));
count:=0; foreach i in ([start..end,2]){ b:=i/100_0000_000; r:=i%1_000_000_000; m:=r/1_000_000; r:=i%1_000_000; t:=r/1_000; r=r%1_000; if(30<=m<=66) m=m%10; if(30<=t<=66) t=t%10; if(30<=r<=66) r=r%10;
if(magic(b) and magic(m) and magic(t) and magic(r)){ if(pr) print(i," ");
count+=1;
} } if(pr) println(); println("count = %,d\n".fmt(count));
} fcn magic(z){ z.isEven and z<=6 }</lang>
- Output:
eban numbers up to and including 1,000: 2 4 6 30 32 34 36 40 42 44 46 50 52 54 56 60 62 64 66 count = 19 eban numbers between 1,000 and 4,000 (inclusive): 2000 2002 2004 2006 2030 2032 2034 2036 2040 2042 2044 2046 2050 2052 2054 2056 2060 2062 2064 2066 4000 count = 21 eban numbers up to and including 10,000: count = 79 eban numbers up to and including 100,000: count = 399 eban numbers up to and including 1,000,000: count = 399 eban numbers up to and including 10,000,000: count = 1,599 eban numbers up to and including 100,000,000: count = 7,999 eban numbers up to and including 1,000,000,000: count = 7,999