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Euler problems/161 to 170

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1 Problem 161

Triominoes

Solution: This does not seem Haskell code to me. If the argument: Learning Haskell were valid pure Haskell code would have been given.

#include <stdio.h>
#include <stdlib.h>
    int x[3][9]={{0,0,0,0,0,0,0,-1,0},
                 {0,0,0,1,0,0,1,0,0},
                 {0,0,0,2,1,1,1,0,0}};
    int y[3][9]={{0,0,0,0,0,0,0,1,0},
                {0,1,1,0,1,1,0,0,0},
                {0,0,2,0,1,0,1,1,0}};
 
    int num[9]={1,2,3,3,3,3,3,3,0};
    int M[16][4],A[16],rem3;
 
void printint64(long long int n)
{
  int i,len,digit[20];
 
  if(n<0)  printf("-"),n=-n;
 
  digit[0]=n%10,n/=10,len=1;
  while(n)  digit[len]=n%10,n/=10,len++;
  for(i=len-1;i>=0;i--)  printf("%d",digit[i]);
 
  return;
}
 
void re(int pos)  {
    int j,c=A[pos];
 
    for(j=0;j<num[c];j++)  M[pos+x[j][c]][y[j][c]]--;
    rem3-=num[c];
 
    return;
}
 
 
int main()  {
 
    int c,i,j,k,n,w1,w2,T,pos,pow3[9];
    int *u,*w,var=0;
 
    u=(int*) (malloc) (131072*sizeof(int));
    w=(int*) (malloc) (131072*sizeof(int));
 
    pow3[0]=1;
    for(i=1;i<9;i++)  pow3[i]=3*pow3[i-1];
 
    long long int p[19683],q[19683];
 
    p[0]=1;
    for(i=1;i<19683;i++)  p[i]=0;
    for(i=0;i<19683;i++)  q[i]=0;
 
        pos=0;
        A[0]=0;
        rem3=0;
        for(j=0;j<11;j++)
            for(k=0;k<3;k++)  M[j][k]=0;
 
        while(A[0]<7)  {
            if(M[pos][0])  A[pos]=8,pos++,A[pos]=0;
            else {
                  c=A[pos];
                  rem3+=num[c];
                  if((pos<8)||(rem3%3==0))  {
                     T=1;
                     for(j=0;j<num[c];j++)  {
                         M[pos+x[j][c]][y[j][c]]++;
                         if(M[pos+x[j][c]][y[j][c]]>1)  T=0;
                     }
                     if((T==0)||M[9][0]||M[9][1])  {
                         re(pos);
                         while(A[pos]>=7)  pos--,re(pos);
                         A[pos]++;
                     }
                     else pos++,A[pos]=0;
                  }
                  else {
                       rem3-=num[c];
                       while(A[pos]>=7)  pos--,re(pos);
                       A[pos]++;
                  }
           }
 
        if(pos==9)  {
              w1=0;
              for(j=0;j<9;j++)  {
                  if(A[j]==0)  w1+=pow3[j];
                  else if(A[j]==1)  w1+=2*pow3[j];
              }
              w2=0;
              for(j=0;j<9;j++)  {
                  c=0;
                  if(M[j][1])  c=1;
                  if(M[j][2])  c=2;
                  w2+=c*pow3[j];
              }
              u[var]=w2;
              w[var]=w1;
              var++;
              pos--,re(pos);
              while(A[pos]>=7)  pos--,re(pos);
              A[pos]++;
        }
      }
 
   for(n=1;n<=12;n++)  {
       if(n%2)  {
          for(i=0;i<var;i++)  q[u[i]]+=p[w[i]];
          for(i=0;i<19683;i++)  p[i]=0;
          printf("tiling[9][%d]=",n),printint64(q[0]),printf("\n");
       }
       else  {
          for(i=0;i<var;i++)  p[u[i]]+=q[w[i]];
          for(i=0;i<19683;i++)  q[i]=0;
          printf("tiling[9][%d]=",n),printint64(p[0]),printf("\n");
       }
   }
   return 1;
}
problem_161 = main

2 Problem 162

Hexadecimal numbers

Solution:

numdigit=['0'..'9']++['A'..'F']
digits n
    |n<16=[n]
    |otherwise= y:digits x 
    where
    (x,y)=divMod n 16
fun=(\k->15*16^(k-1)-15^(k)-2*14*15^(k-1)+13*14^(k-1)+2*14^k-13^k)
fsum::Integer
fsum=sum $map fun [3..16]
problem_162=map ((numdigit!!).fromInteger) $reverse $digits fsum

3 Problem 163

Cross-hatched triangles

Solution:

--http://www.math.uni-bielefeld.de/~sillke/SEQUENCES/grid-triangles
fun n= 
    sum[div  (2*n3 + 5*n2 + 2*n) 8  ,
    2*(div n3 2- div n 6)  , 
    6* sum[div ( n*(n+1)*(n+2)) 6 ,
        div (2*n3 + 5*n2 + 2*n) 8 ,
        div (2*n3 + 3*n2 - 3*n) 18 ,
        div (2*n3 + 3*n2 - 3*n) 10 ],
    3 * div(22*n3 + 45*n2 - 4*n) 48
    ]
    where
    n3=n*n*n
    n2=n*n
problem_163=fun 36

4 Problem 164

Numbers for which no three consecutive digits have a sum greater than a given value.

Solution:

addDigit x = [[sum [x !! b !! c | c <- [0..9-a-b]] | b <- [0..9-a]] | a<-[0..9]]
x3 = [[10-a-b | b <- [0..9-a]] | a <- [0..9]]
x20 = iterate addDigit x3 !! 17
problem_164 = sum [x20 !! a !! b | a <- [1..9], b <- [0..9-a]]

5 Problem 165

Intersections

Solution:

bbsGen x = (x * x) `mod` 50515093
 
bbsSeq = iterate bbsGen 290797
 
tValues = map (`mod` 500) (tail bbsSeq)
 
lineSeg n = take 4 (drop n tValues)
 
lineSegs' n = lineSeg n : lineSegs' (n + 4)
lineSegs = lineSegs' 0
 
implicitLine :: [Integer] -> (Integer, Integer, Integer)
implicitLine [x1,y1,x2,y2] = (a, b, d) where
	a = y2 - y1
	b = -(x2 - x1)
	d = x1*a + y1 * b
 
within :: (Ord a, Num a, Integral b) => a -> b -> b -> Bool
within a b c | b > c = within a c b
			 | otherwise = a >= fromIntegral b && a <= fromIntegral c
 
withinSeg :: (Ord a, Num a) => a -> a -> [Integer] -> Bool
withinSeg x y l@[x1,y1,x2,y2] = within x x1 x2 && within y y1 y2 && not (endpoint x y l)
 
endpoint :: (Ord a, Num a) => a -> a -> [Integer] -> Bool
endpoint x y [x1,y1,x2,y2] = ((x == fromIntegral x1) && (y == fromIntegral y1)) ||
                             ((x == fromIntegral x2) && (y == fromIntegral y2))
 
boundingBoxOverlap l1@[l1x1,l1y1,l1x2,l1y2] l2@[l2x1,l2y1,l2x2,l2y2]
				| min l1x1 l1x2 > max l2x1 l2x2 = False
				| max l1x1 l1x2 < min l2x1 l2x2 = False
				| min l1y1 l1y2 > max l2y1 l2y2 = False
				| max l1y1 l1y2 < min l2y1 l2y2 = False
				| otherwise = True
 
intersect :: (Fractional a, Ord a) => [Integer] -> [Integer] -> (Bool, a, a)
intersect l1 l2 | boundingBoxOverlap l1 l2 && 
                  d /= 0 && 
                  withinSeg x y l1 && withinSeg x y l2 = (True, x, y)
				| otherwise = (False, 0, 0)
	where
		(a1, b1, d1) = implicitLine l1
		(a2, b2, d2) = implicitLine l2
		d = fromIntegral (a1*b2 - a2*b1)
		x = fromIntegral (b2 * d1 - b1 * d2) / d
		y = fromIntegral (a1 * d2 - a2 * d1) / d
 
listIntersects l [] = []
listIntersects l (l1:ls) | b = [(x,y)] ++ (listIntersects l ls)
						  | otherwise = listIntersects l ls
						  where (b, x, y) = intersect l l1
 
allIntersectsList [] = []
allIntersectsList (l:ls) = listIntersects l ls ++ allIntersectsList ls
 
problem_165 = length . quickSort . allIntersectsList $ take 5000 lineSegs
 
quickSort :: Ord a => [(a,a)] -> [(a,a)]
quickSort [] = []
quickSort (l:ls) = quickSort (filter (< l) ls) ++ 
                   [l] ++ 
                   quickSort (filter (> l) ls)

6 Problem 166

Criss Cross

Solution:

problem_166 = 
    sum [ product (map count [[0, c, b-d, a-b-d], 
            [0, b-a, c+d-a, b+d-a], 
            [0, -b-c, a-b-c-d, -c-d],
            [0, a, d, c+d]])|
        a <- [-9..9], 
        b <- [-9+a..9+a],
        c <- [-9..9],
        d <- [-9+a-c..9+a-c]]
    where
    count xs 
        |u<l=0 
        |otherwise=u-l+1
        where
        l = -minimum xs 
        u = 9-maximum xs

7 Problem 167

Investigating Ulam sequences

8 Problem 168

Number Rotations

Solution:

fun e =
    sum[n*10+d|
    let t=[1..9],
    d<-t,
    p<-t,
    let (n,m)=divMod ((e-p)*d) (10*p-1) ,
    m==0,
    10*n>=e
    ]
problem_168=flip mod (10^5)$sum[fun e|i<-[1..99],let e=10^i]

9 Problem 169

Exploring the number of different ways a number can be expressed as a sum of powers of 2.

Solution:

fusc' 0=(1,0)
fusc' n
    |even n=(a+b, b)
    |odd n=(a,a+b)
    where
    (a,b)=fusc' $div n 2
fusc =fst.fusc'
problem_169=fusc (10^25)

10 Problem 170

Find the largest 0 to 9 pandigital that can be formed by concatenating products.

Solution:

{-
1) The first integer must be a multiple of 3 
(otherwise the digital root of the result is not 9).
2) The first integer contains at most 2 digits 
(otherwise the result contains more than 10 digits).
3) The first integer must be less than 49 
(otherwise the result contains more than 10 digits). 
4) maybe answer is 98xxxx
5) This number must be a multiple of the first factor (f).
In the numbers f and cp/f all digits 1..9 have to occour 
once and at least one zeros. 
 -}
import Data.List 
permutationsOf [] = [[]]
permutationsOf xs = [x:xs' | x <- xs, xs' <- permutationsOf (delete x xs)]
digits =reverse.digits' 
    where
    digits' n 
        |n<10=[n]
        |otherwise= y:digits' x 
        where
        (x,y)=divMod n 10
digitsToNum n=foldl dmm 0  n
    where
    dmm=(\x y->x*10+y)
fun k xs c=any id [n/=0 && n<100|a<-k,let n=c*xs!!(a+1)]
problem_170 =
    maximum[b|
    aa<-[7,6..4],
    a<-permutationsOf $delete aa [0..7],
    let b=digitsToNum $[9,8]++(aa:a),
    c<-[12,15..48],
    let (d,m)=divMod b c ,
    m==0,
    let xs=digits d,
    (digits c++xs) \\t==[0],
    let k=findIndices (==0) xs,
    last xs/=0,
    fun k xs c
    ]
    where
    t=[0..9]