DECLARE FUNCTION isPrime! (x!)
DATA 19,1,7,16,49
DATA 43,3,4,5,41
DATA 61,64,2,11,49
DATA 37,11,53,36,23
DATA 31,25,29,59,9
FOR r = 1 TO 5
FOR c = 1 TO 5
READ gr1(r, c)
NEXT
NEXT
CLS
DIM SHARED prime(50)
DATA 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , 31 , 37 , 41 , 43 , 47 , 53 , 59
DATA 61 , 67 , 71 , 73 , 79 , 83 , 89 , 97 , 101 , 103 , 107 , 109 , 113 , 127 , 131
DATA 137 , 139 , 149 , 151 , 157 , 163 , 167 , 173 , 179 , 181 , 191 , 193 , 197
DATA 199 , 211 , 223 , 227 , 229
FOR i = 1 TO 50: READ prime(i): NEXT
FOR r = 1 TO 5
FOR c = 1 TO 5
gr2(r, c) = gr1(r - 1, c) + gr1(r + 1, c) + gr1(r, c - 1) + gr1(r, c + 1)
LOCATE r + 3, c * 8
IF isPrime(gr2(r, c)) THEN PRINT gr2(r, c); : ELSE PRINT " . ";
NEXT
NEXT
FUNCTION isPrime (x)
good = 0
FOR i = 1 TO 50
IF x = prime(i) THEN good = 1: EXIT FOR
NEXT
isPrime = good
END FUNCTION
shows
. 29 . 61 .
83 . 17 . 103
. . . . .
103 179 . . .
. 71 137 . .
where only the prime totals are shown for the resulting matrix. There are nine of them.
Then:
10 dim Grnum(10),H(10),Used(10)
20 data 17,29,61,71,83,103,103,137,179
30 for I=1 to 9:read Grnum(I):Ovrtot=Ovrtot+Grnum(I):next
50 H(1)=1:gosub *Addon2(2)
999 end
2000 *Addon2(Ptr)
2010 local I,J,Strt
2015 if Ptr @ 3=1 then Strt=H(Ptr-3)+1:else Strt=H(Ptr-1)+1
2020 for I=Strt to 9
2021 if Used(I)>0 then goto *NotThis
2025 if Ptr @ 3=0 and prmdiv(Grnum(H(Ptr-2))+Grnum(H(Ptr-1))+Grnum(I))<Grnum(H(Ptr-2))+Grnum(H(Ptr-1))+Grnum(I) then goto *NotThis
2030 H(Ptr)=I:Used(I)=1
2040 if Ptr=9 then
2090 :for J=1 to 9:print Grnum(H(J));:next:print
2091 :for J1=0 to 6 step 3:Tt=0:for J2=1 to 3:Tt=Tt+Grnum(H(J1+J2)):next:print Tt;:next:print:print
2100 :else
2110 :gosub *Addon2(Ptr+1)
2120 Used(I)=0
2125 *NotThis
2130 next
2190 return
prints
17 29 61 71 83 103 103 137 179
107 257 419
17 29 61 71 83 103 103 137 179
107 257 419
17 29 103 61 83 137 71 103 179
149 281 353
17 29 103 61 83 137 71 103 179
149 281 353
17 61 179 29 103 137 71 83 103
257 269 257
17 61 179 29 103 137 71 83 103
257 269 257
17 71 103 29 61 83 103 137 179
191 173 419
17 71 103 29 103 179 61 83 137
191 311 281
17 71 103 29 61 83 103 137 179
191 173 419
17 71 103 29 103 179 61 83 137
191 311 281
17 103 137 29 61 83 71 103 179
257 173 353
17 103 137 29 61 179 71 83 103
257 269 257
17 103 137 29 61 83 71 103 179
257 173 353
17 103 137 29 61 179 71 83 103
257 269 257
In each solution, the first three are the first group; the middle three are the second group; and the last three are the third group.
Each solution is listed twice, as the two 103's are considered distinct by the program. So there are actually 7 solutions, not 14.
The grouping totals are shown in the line below the groupings.
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Posted by Charlie
on 2010-09-20 14:21:32 |
computer solution
