Increasing gaps between consecutive Niven numbers: Difference between revisions
m (→add digits in chunks: added wording to the REXX section header.) |
m (→add digits in chunks: added some comments.) |
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call tell center('gap size', 12) center(@gsa "index", 29) center(@gsa, 29) |
call tell center('gap size', 12) center(@gsa "index", 29) center(@gsa, 29) |
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call tell copies('═' , 12) copies('═' , 29) copies('═' , 29) |
call tell copies('═' , 12) copies('═' , 29) copies('═' , 29) |
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@.= 0 /*set all values to zero for chunk sums*/ |
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@.= 0 |
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do j=1 for 99999 /*pre─compute sums for #a up to 5 digs.*/ |
do j=1 for 99999 /*pre─compute sums for #a up to 5 digs.*/ |
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parse var j 1 sum 2 q /*use the first decimal digit for SUM.*/ |
parse var j 1 sum 2 q /*use the first decimal digit for SUM.*/ |
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do while q\==''; parse var q x 2 q; sum= sum + x |
do while q\==''; parse var q x 2 q; sum= sum + x |
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end /*while*/ |
end /*while*/ /*do sum of digits the hard way for now*/ |
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@.j= sum /*assume a sum for a particular number.*/ |
@.j= sum /*assume a sum for a particular number.*/ |
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if j>9999 then iterate /*if J has five digits or more, skip.*/ |
if j>9999 then iterate /*if J has five digits or more, skip.*/ |
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do zz= length(j)+1 to 4 /*handle all J's with leading zeros. */ |
do zz= length(j)+1 to 4 /*handle all J's with leading zeros. */ |
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jz= right(j, zz, 0) /*also add leading zeros from some J's.*/ |
jz= right(j, zz, 0) /*also add leading zeros from some J's.*/ |
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if @.jz==0 then @.jz= sum |
if @.jz==0 then @.jz= sum /*assign a sun to 000xx for instance.*/ |
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end /*zz*/ |
end /*zz*/ |
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end /*j*/ |
end /*j*/ |
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#= 0 /*#: is the index of a Niven number. */ |
#= 0 /*#: is the index of a Niven number. */ |
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do n=1 /*◄───── let's go Niven number hunting.*/ |
do n=1 /*◄───── let's go Niven number hunting.*/ |
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parse var n q1 +5 q2 +5 q3 +5 q4 +5 q4 +5 q6 |
parse var n q1 +5 q2 +5 q3 +5 q4 +5 q4 +5 q6 /*break apart N into 5─digit chunks. */ |
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sum= @.q1 + @.q2 + @.q3 + @.q4 + @.q5 + @.q6 |
sum= @.q1 + @.q2 + @.q3 + @.q4 + @.q5 + @.q6 /*add the 5─digit chunks to compute sum*/ |
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if n//sum > 0 then iterate /*is N not divisible by its sum? Skip.*/ |
if n//sum > 0 then iterate /*is N not divisible by its sum? Skip.*/ |
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#= # + 1 /*bump the index of the Niven number.*/ |
#= # + 1 /*bump the index of the Niven number.*/ |
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Revision as of 06:53, 12 January 2020
Note: Niven numbers are also called Harshad numbers.
- They are also called multidigital numbers.
Niven numbers are positive integers which are evenly divisible by the sum of its
digits (expressed in base ten).
Evenly divisible means divisible with no remainder.
- Task
-
- find the gap (difference) of a Niven number from the previous Niven number
- if the gap is larger than the (highest) previous gap, then:
- show the index (occurrence) of the gap (the 1st gap is 1)
- show the index of the Niven number that starts the gap (1st Niven number is 1, 33rd Niven number is 100)
- show the Niven number that starts the gap
- show all numbers with comma separators where appropriate (optional)
- I.E.: the gap size of 60 starts at the 33,494th Niven number which is Niven number 297,864
- show all increasing gaps up to the ten millionth (10,000,000th) Niven number
- (optional) show all gaps up to whatever limit is feasible/practical/realistic/reasonable/sensible/viable on your computer
- show all output here, on this page
- Related task
- Also see
Go
This reuses code from the [Harshad or Niven series] task though converted to use 'uint64' rather than 'int' in case anyone is running Go on a 32-bit platform. <lang go>package main
import "fmt"
type is func() uint64
func newSum() is {
var ms is
ms = func() uint64 {
ms = newSum()
return ms()
}
var msd, d uint64
return func() uint64 {
if d < 9 {
d++
} else {
d = 0
msd = ms()
}
return msd + d
}
}
func newHarshard() is {
i := uint64(0)
sum := newSum()
return func() uint64 {
for i++; i%sum() != 0; i++ {
}
return i
}
}
func commatize(n uint64) string {
s := fmt.Sprintf("%d", n)
le := len(s)
for i := le - 3; i >= 1; i -= 3 {
s = s[0:i] + "," + s[i:]
}
return s
}
func main() {
fmt.Println("Gap Index of gap Starting Niven")
fmt.Println("=== ============= ==============")
h := newHarshard()
pg := uint64(0) // previous highest gap
pn := h() // previous Niven number
for i, n := uint64(1), h(); n <= 20e9; i, n = i+1, h() {
g := n - pn
if g > pg {
fmt.Printf("%3d %13s %14s\n", g, commatize(i), commatize(pn))
pg = g
}
pn = n
}
}</lang>
- Output:
Gap Index of gap Starting Niven === ============= ============== 1 1 1 2 10 10 6 11 12 7 26 63 8 28 72 10 32 90 12 83 288 14 102 378 18 143 558 23 561 2,889 32 716 3,784 36 1,118 6,480 44 2,948 19,872 45 4,194 28,971 54 5,439 38,772 60 33,494 297,864 66 51,544 478,764 72 61,588 589,860 88 94,748 989,867 90 265,336 2,879,865 99 800,054 9,898,956 108 3,750,017 49,989,744 126 6,292,149 88,996,914 135 44,194,186 689,988,915 144 55,065,654 879,987,906 150 61,074,615 989,888,823 153 179,838,772 2,998,895,823 192 399,977,785 6,998,899,824 201 497,993,710 8,889,999,624 234 502,602,764 8,988,988,866 258 547,594,831 9,879,997,824 276 1,039,028,518 18,879,988,824
REXX
add digits serially
<lang rexx>/*REXX program finds and displays the largest gap between Niven numbers (up to LIMIT).*/ parse arg lim . /*obtain optional arguments from the CL*/ if lim== | lim==',' then lim= 10000000 /*Not specified? Then use the default.*/ numeric digits 2 + max(8, length(lim) ) /*enable the use of any sized numbers. */ gap= 0; old= 0 /*initialize (largest) gap; old Niven #*/
@gsa= 'gap starts at Niven #'
call tell center('gap size', 12) center(@gsa "index", 29) center(@gsa, 29) call tell copies('═' , 12) copies('═' , 29) copies('═' , 29)
- = 0 /*#: is the index of a Niven number. */
do n=1 /*◄───── let's go Niven number hunting.*/
parse var n 1 sum 2 q /*use the first decimal digit for SUM.*/
do while q\==; parse var q x 2 q; sum= sum + x
end /*while*/ /* ↑ */
if n//sum >0 then iterate /* └──────◄ is destructively parsed.*/
#= # + 1 /*bump the index of the Niven number.*/
if n-old<=gap then do; old= n; iterate; end /*Is gap not bigger? Then keep looking*/
gap= n - old; old= n /*We found a bigger gap; define new gap*/
idx= max(1, #-1); san= max(1, n-gap) /*handle special case of the first gap.*/
call tell right(commas(gap), 7)left(, 5), /*center right─justified Niven gap size*/
right(commas(idx), 25)left(, 4), /* " " " Niven num idx.*/
right(commas(san), 25) /* " " " " number. */
if n >= lim then leave /*have we exceeded the (huge) LIMit ? */
end /*n*/
exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ commas: parse arg _; do c=length(_)-3 to 1 by -3; _=insert(',', _, c); end; return _ tell: say arg(1); return</lang>
- output when using the input of: 20000000000 (which is 20 billion)
gap size gap starts at Niven # index gap starts at Niven #
════════════ ═════════════════════════════ ═════════════════════════════
1 1 1
2 10 10
6 11 12
7 26 63
8 28 72
10 32 90
12 83 288
14 102 378
18 143 558
23 561 2,889
32 716 3,784
36 1,118 6,480
44 2,948 19,872
45 4,194 28,971
54 5,439 38,772
60 33,494 297,864
66 51,544 478,764
72 61,588 589,860
88 94,748 989,867
90 265,336 2,879,865
99 800,054 9,898,956
108 3,750,017 49,989,744
126 6,292,149 88,996,914
135 44,194,186 689,988,915
144 55,065,654 879,987,906
150 61,074,615 989,888,823
153 179,838,772 2,998,895,823
192 399,977,785 6,998,899,824
201 497,993,710 8,889,999,624
234 502,602,764 8,988,988,866
258 547,594,831 9,879,997,824
276 1,039,028,518 18,879,988,824
add digits in chunks
The method used is to break a number into 5-digit (decimal) chunks (or less), and those chunks have been pre-computed to find their digit sum. Also pre-computed were chunks that had various lengths of chucks with leading zeros (from none to four), such that 3 03 003 0003 and 00003 all have the same digital sum. Same thing with 478 0478 and 00478. This method could've been expanded to six-digit chunks, at the expense of real memory.
It is about four times faster than the 1st REXX version.
The "chunk" method is essentially a sum of several chunks, the sums are found by a table lookup (associative array in REXX). <lang rexx>/*REXX program finds and displays the largest gap between Niven numbers (up to LIMIT).*/ parse arg lim . /*obtain optional arguments from the CL*/ if lim== | lim==',' then lim= 1000000000000 /*Not specified? Then use the default.*/ numeric digits 2 + max(8, length(lim) ) /*enable the use of any sized numbers. */ gap= 0; old= 0 /*initialize (largest) gap; old Niven #*/
@gsa= 'gap starts at Niven #'
call tell center('gap size', 12) center(@gsa "index", 29) center(@gsa, 29) call tell copies('═' , 12) copies('═' , 29) copies('═' , 29) @.= 0 /*set all values to zero for chunk sums*/
do j=1 for 99999 /*pre─compute sums for #a up to 5 digs.*/
parse var j 1 sum 2 q /*use the first decimal digit for SUM.*/
do while q\==; parse var q x 2 q; sum= sum + x
end /*while*/ /*do sum of digits the hard way for now*/
@.j= sum /*assume a sum for a particular number.*/
if j>9999 then iterate /*if J has five digits or more, skip.*/
do zz= length(j)+1 to 4 /*handle all J's with leading zeros. */
jz= right(j, zz, 0) /*also add leading zeros from some J's.*/
if @.jz==0 then @.jz= sum /*assign a sun to 000xx for instance.*/
end /*zz*/
end /*j*/
- = 0 /*#: is the index of a Niven number. */
do n=1 /*◄───── let's go Niven number hunting.*/
parse var n q1 +5 q2 +5 q3 +5 q4 +5 q4 +5 q6 /*break apart N into 5─digit chunks. */
sum= @.q1 + @.q2 + @.q3 + @.q4 + @.q5 + @.q6 /*add the 5─digit chunks to compute sum*/
if n//sum > 0 then iterate /*is N not divisible by its sum? Skip.*/
#= # + 1 /*bump the index of the Niven number.*/
if n-old<=gap then do; old= n; iterate; end /*Is gap not bigger? Then keep looking*/
gap= n - old; old= n /*We found a bigger gap; define new gap*/
idx= max(1, #-1); san= max(1, n-gap) /*handle special case of the first gap.*/
call tell right(commas(gap), 7)left(, 5), /*center right─justified Niven gap size*/
right(commas(idx), 25)left(, 4), /* " " " Niven num idx.*/
right(commas(san), 25) /* " " " " number. */
if n >= lim then leave /*have we exceeded the (huge) LIMit ? */
end /*n*/
exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ commas: parse arg _; do c=length(_)-3 to 1 by -3; _=insert(',', _, c); end; return _ tell: say arg(1); return</lang>
- output is identical to the 1st REXX version.
zkl
<lang zkl>harshadW:=[1..].tweak(fcn(n){ if(n%(n.split().sum(0))) Void.Skip else n }); harshadW:=Walker.zero().tweak(fcn(go){ // faster than one liner, fewer calls
foreach h in ([go.value..]){ // spin
s,t := 0,h; while(t){ s+=t%10; t/=10 } // sum of digits
if(0 == h%s){ go.set(h+1); return(h) }
}
}.fp(Ref(1)));</lang> <lang zkl>println("gap size Niven index Niven #"); prev,gap := harshadW.next(),0; while(harshadW.n<=10_000_000){
if( (g:=(h:=harshadW.next()) - prev) > gap){
println("%5,d %14,d %15,d".fmt(g, harshadW.n - 1, prev));
gap=g;
}
prev=h;
}</lang>
- Output:
gap size Niven index Niven #
1 1 1
2 10 10
6 11 12
7 26 63
8 28 72
10 32 90
12 83 288
14 102 378
18 143 558
23 561 2,889
32 716 3,784
36 1,118 6,480
44 2,948 19,872
45 4,194 28,971
54 5,439 38,772
60 33,494 297,864
66 51,544 478,764
72 61,588 589,860
88 94,748 989,867
90 265,336 2,879,865
99 800,054 9,898,956
108 3,750,017 49,989,744
126 6,292,149 88,996,914