Project Euler Code Golf

Project Euler Problems
Rule I. If any Euler problem is unsolved and you solve it, you go on the board
Rule II. If you devise a solution shorter than what’s on the board, you may take that place on the board
Rule III. If you implement an agreed-upon change to the language, you get credit for all shortened solutions

1 +/&~&/(!1e3)!/:3 5 (Explanation: PE1 Explained) Kevin

Known bound:	`+/{x~x!2}32(\|+\)\1 1`
Unknown bound(b f/ x):	`+/{x@&~x!2}(4e6>+/-2#){x,+/-2#x}/1`

isawdrones

3 |/d@&&/'2_'f'd:&~(f:{x!'!1+_sqrt x})600851475143
Works for this case, but not in general. Returns 3 for largest factor of 15. isawdrones

4 |/b@&{x~|x}'$b:,/a*/:a:!1000
Faster:
|/b@&{x~|x}'$b:?,/a*/:a:100+!900 isawdrones

5 Euclid’s algorithm:
{x*y%{:[y;_f[y]x!y;x]}[x]y}/1+!20
Above solution works in Kona, but not in k2.8 or k3.2
{x*y%{:[_ y;_f[y]x!y;x]}[x]y}/1+!20
This solution works in Kona, k2.8 and k3.2 isawdrones

6 {_(_sqr+/a)-+/a*a}1+!100 isawdrones

7 p:{:[x<4;,2;r,1_&~|/x#'~!:'r: _f[_ _ceil _sqrt x]]}
b:2;while[1e4>#a:p b*:2];a[10000]
Alternative:

p:2;i:3;while[10001>#p;if[{&/x!'p} i;p,:i];i+:1];*|p

i+2 cuts computation time in half and still arrives at the correct answer.
removed 3 characters from the previous shortest.
p:2;i:3;while[10001>#p;:[&/i!'p;p,:i];i+:2];*|p isawdrones, nrapport, carkat

8 |/*/'{x@(!5)+/:!-4+#x}0$',/0:` isawdrones

9 Brute force:
*/_*c@&1000=+/'c:b,'(_sqrt+/)'b*b:,/a,/:\:a:!500 isawdrones

10 p:{2_&{:[x@y;x&@[1,-1_ z#(1_ y#1),0;y;:;1];x]}/[x#1;2_! __ceil _sqrt x;x]};+/p@_2e6 Kevin

11 p:.:'0:`
d:{x ./:/:m@-1_1_=+/'m:,/n,/:\:n:!#x}
i:|/,/{:[4>#x;x;*/'x@(!4)+/:!-3+#x]}'
|/(i p;i@+p;i d@p;i d@|p) isawdrones

12 Brute force:
f:{1__ x*(x+1)%2};c:1;{x<500}{#{d:&~x!'!1+_sqrt x;d,_ x%|d}f c+:1}\1;f@c hakank

13 \p 0 10#($+/0.0$’0:`)_dv"." isawdrones

clz:{:[x!2;1+3*x;_ x%2]}
a?|/a:{C:-1 1,(x-2)#0
  do[x+i:0
    if[~C@i
      b:&x>j:((i-1)<)clz\i
      C[j@b]:C[*|j]+(|!#j)@b]
    i+:1]
  C}@_1e6

Source

isawdrones

`(40{1+':x,0}/1)20`
`_/(-2+4x)%x:1+!20` – implementing formulae $C_{2n}^n$

isawdrones
danbst

16 +/1000{x*:2;(*&x)_ x:((x>9),0)+0,x!/:10}/1 /slower: !' Kevin

a:$`one`two`three`four`five`six`seven`eight`nine
b:$`ten`eleven`twelve,($`thir`four,v:`fif`six`seven`eigh`nine),\:"teen"  
c:($`twen`thir`for,v),\:"ty"
g:a,b,c,/c,/:\:a
h:a,\:"hundred"    
i:g,h,/h,/:\:"and",/:g
z:#,/$i,`onethousand

Kevin

18 |/{(y+0,x)|y+x,0}/.:'0:`t.txt /shorter 0:` Kevin

19 +/{(&/"01"=-2#$_dj x)&6=-x!7}'from+!-((from:_(_jd 19010101))-(_jd 20001231)) hakank


xx: 1+til 100;
uc:{[x;y] tO:last[x]+y;aO:tO mod 10;bO:`int$(tO-aO) % 10;(-1_x),aO,bO}/[enlist 0;];
pp:{[x;y] mO:max count each rO:rO{[x;y] (y#0),x}'til count rO: y */: x;
          uc sum {[x;y] x,(y-count x)#0}[;mO] each rO };
lst:{[x] {parse x,"i"} each x} each reverse each string each xx
sum {[x;y] pp[x;y]}/[lst]

kuentang

21 f:{d:&~x!'!1+_sqrt x;d,_1_ x%d};+/&{(~&/x=b)&x~*|b:2{+/f x}\x}'!1e4
Code above not working for some reason; below is another solution (longer):
d: {+/(!1+_ceil x%2)*(0=x!/:!1+_ceil x%2)};d 284;n : !10000;a:d' n b:d' a j:n*(b=n);j:j@&1<j;j:j@&0<_sqr j-d' j;+/j BonucciAndrea&isawdrones

22 +/s*1+!#s:+/'-64+_ic m@<(-|/#:'m)$'m:.:'1_'(0,1_&","=n)_ n:",",,/0:`names.txt isawdrones

23 l:28123;n:l#1 / (l)imit, (n)ot sum of abundant
{n[x@&l>x]:0}'a+/:a:1_&{x<+/{?1,d,x%d:c@&~x!'c:1+1_!_sqrt x} x}'!l / (a)bundant
+/&n inrick

24 ({:[1<x;,/(>:'(x,x)#1,x#0)[;0,'1+_f x-1];,!x]}10)[_1e6-1] hakank


c:{:[|/x>9;_f (x!10)+0,-1_ x%10;x]}
n:1000;i:2; {0=*|x[1]} {i+:1;{c[x+y]}\|x}/ (2,n)#0=!n; i

juster

26 1+*>#:'{[n] {r:(10**|x)!*x; w:x?r; :[w=#x;x,r;(*x),w _ x]}/(n,1)}'1+!999 adamschmideg

I have a solution in A+.  It's reasonably fast (900 ms), 
but visually clunky, and the APL characters won't display here.
A remarkably elegant solution in J (from Oleg Kobchenko):

   pr=: 1: = #@q:"0
   pc=: 4 : '(>:^:((pr@((x,y,1)&p.)) :: 0:))^:_] 1'
   ($ #: (i. >./)@,)(p:i.168)pc"0/i:999
163 938
   (p:163)*938{i:999 
_59231

tavmem

28 +/+\1,,/4#'2*1+!_1001%2 isawdrones

29 #?,/p^/:p:2+!99 silentbicycle

30 +/2_&{x=+/(0$'$x)^5}'!7*9^5 hakank

s:{+/{(y _ x),y#0}[x]'y*!_ceil(#x)%y+0.0}
e31:{*s/[(x#0),1;1 2 5 10 20 50 100 200]}

paul7

32 +/?,/,/'(!10000;!1000){r*"123456789"~{x@<x}@,/$x,y,r:x*y}/:\:'(1+!9;11+!89) nrapport


f:{x@&~|/’x=\:y}
g:{0$(f[$x;$y];f[$y;$x])}
check:{(1=&/#:’$a)&(b<1)&(b:x%y)=%/a:g[x;y]}
combos:{,/x,/:\:x}
v:a@&~0=(a:10 + ! 90)!10
g .’ a@&check .’ a:combos[v]
/do the remaining simplification by hand

nrapport

34 +/2_&{x=+/{*/1+!x}'0$'$x}'!50000 hakank

35 p:{&/ 2_ x!' !1+_ _sqrt x}
q:{*/ p' 0$(1!)\ $x}
#2_ v@&q' v:!1000000 nrapport

36 +/&{&/{y~|y:x _vs y}'[2 10;x]}'!1e6
Faster:
p:{&{x~|x}'x@y};+/n@p[2_vsx]n:p[$:;!1e6] Kevin

37 ? ?

38 0$,/$*|v@&{(,/$1+!9)~a@<a:,/$x}'v:(9e3+!_1e3)*\:1 2 nrapport

39 r is all permutations
f eliminates tuples with 0
t is all triangles
a is all triangles grouped by perimeter
take the first perimiter ordered by count of triangles
abc:{c:_sqrt(x^2)+y^2;x,y,:[c=(_ c);_ c;0]}
{**a[>#:'a:t[&~1=#:'t:{f:r[&3=+/'~0='r:{abc/',/x,/:\:x}x;];{x,f[&{x=+/y}[x]'f]}'x}x]]}(!1000) carkat

40 */ . 1_(,/" ",'((,/ $ !2e5)@_10^!7)) BonucciAndrea

41 ? ?

42 m:.:'1_'(0,1+&","=n)_",",n:*:0:`words.txt
+/{a=_ a:_sqrt 1+8*x}'+/'-64+_ic m@<(-|/#:'m)$'m isawdrones

43 ? ?

44 ? ?

45 *|{x@&x _lin y}/({x*((3*x)-1)*%2};{x*((2*x)-1)})@\:!_4e4 BonucciAndrea

46 p:2 3 5 7
c:{|/a=_ a:_sqrt %2%x-p}
q:{&/x!'p@&p<1+_ _sqrt x}
i:9;d:0;while[~d;if[q i;p,:i];if[~c i|i=*|p;d:1];i+:2];i-2 nrapport

47 ? ?

uc:{[x;y] tO:last[x]+y;aO:tO mod 10;bO:`int$(tO-aO) % 10;(-1_x),aO,bO}/[enlist 0;];
plus:{[x;y] mO:max count @/:(x;y) ;m:uc (x,(mO-count x)#0) 
+ (y,(mO-count y)#0);?[0=last m ;-1_m;m]};
pp:{[x;y] mO:max count each rO:rO{[x;y] (y#0),x}'til count rO: y */: x;
m:uc sum {[x;y] x,(y-count x)#0}[;mO] each rO ; ?[0=last m ;-1_m;m]};
lst:{[x;y] ll:{parse x,"i"} each reverse string y;
plus[x;{[x;y]tmp:pp[x;y]; ?[10>=count tmp;tmp;10#tmp] }/[y#enlist ll]]}/[enlist[0];1+til 1000];
reverse 10#last lst

kuentang

49 ? ?

50 ? ?

51 ? ?

52 ? ?

53 +/~1e6>,/{x{+':0,x,0}\1.}100 isawdrones

54 ? ?

55 ? ?

56 ? ?

57 ? ?

58 ? ?

59 ? ?

60 ? ?

61 ? ?

62 ? ?

63 ? ?

64 ? ?

65 ? ?

66 ? ?

67 |/{(y+0,x)|y+x,0}/.:'-1_'0:`triangle.txt Kevin

68 ? ?

69 ? ?

70 ? ?

71 {a:2;c:5;while[~x<c+7;a+:3;c+:7];a}1e6 isawdrones

72 ? ?

73 ? ?

74 ? ?

75 ? ?

76 ? ?

77 ? ?

78 ? ?

79 ? ?

80 ? ?

81 ? ?

82 ? ?

83 ? ?

84 ? ?

85 r:{{{[x;y] x*y*(x+1)*(y+1)*%4}\:[x;x]}[1+!x]} a:r 1000;b:,/ a;l:b@&b>1990000;l:l@&l<2010000;l:_abs l-2e6; i:(b?(2e6-(&/l)))&(b?(2e6+(&/l)));(1+i!(#a))*(1+_floor i%(#a)) BonucciAndrea

86 ? ?

87 ? ?

88 ? ?

A general solution with roman-arabic and arabic-roman number conversion

r:"IVXLCDM"
i:(!4)#\:0;i:(i,,0 1),(1,/:i),,0 2
ar:{d:10_vs x;,/d{r@(i+y)@x}'|(#d)#2*!4}
d:.+(`$/:r;*\1,6#5 2)
ra:{n:d[`$/:x];+/n*1-2*>':n,0}
answer:+/{r:-1_ x;r::["M"=*r;1_ r;r];(#r)-#ar ra r}/:0:"roman.txt"
/ "M" hack, because ar fails above 4000

adamschmideg

90 ? ?

91 ? ?

92 ? ?

93 ? ?

94 ? ?

95 ? ?

96 ? ?

97 ? ?

98 ? ?

99 ? ?

100 ? ?

119 \p 0 i : (? ,/ (!70) ^/: (!10)); f : {|/ x=(+/. 1_(,/" ",'$x))^!10}; f 512; i : i * f' i; i : i @& 0 < i; i : 9_{x@<x}i; i@29 BonucciAndrea

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