+-----------+---------+
| 1 4 6 7 3 | R R R B |
+-----------+---------+
| 0 2 6 3 7 | B B B |
+-----------+---------+
| 6 9 3 0 8 | R |
+-----------+---------+
| 5 6 0 7 3 | R R |
+-----------+---------+
| 5 9 4 1 8 | R B |
+-----------+---------+
Each of the rows of the table given above indicates an attempt to find out the secret number.
• Each key has, in the column to the right has the letters R and B.
• Each R indicates that this digit has one digit common with the secret number, but in a different position.
• Each B indicates that the number has one digit common with the secret number in the same position.
Determine the secret number, given that the 3-digit number constituted by concatenating the last three digits (reading left to right) is a prime number.
1 4 6 7 3 | R R R B
0 2 6 3 7 | B B B
6 9 3 0 8 | R
5 6 0 7 3 | R R
5 9 4 1 8 | R B
The program below reads the above file and tests all the combinations.
clc, clearvars
fid=fopen('c:\VB5 Projects\flooble\covert input vi.txt','r');
for i=1:5
l=fgetl(fid);
l=erase(l,' ');
ix=strfind(l,'|');
trial{i}=l(1:ix-1);
clues{i}=l(ix+1:end);
end
fclose(fid);
for n=0:99999
ctok=0; tt=false(1,5);
good=true;
s=char(string(n));
if length(s)<5
s=['00000' s];
s=s(end-4:end);
end
for i=1:5
trialC=trial{i};
did=false(1,5);
bCount=0; rCount=0;
for p=1:5
if s(p)==trialC(p)
bCount=bCount+1;
did(p)=true;
% trialC(p)=' ';
end
end
for p=1:5
if ~did(p)
f=strfind(trialC,s(p));
if f
rCount=rCount+1;
trialC(f(1))=' ';
end
end
end
result=[repmat('R',1,rCount) repmat('B',1,bCount)];
if ~isequal(result,clues{i})
good=false;
break
else
ctok=ctok+1;
tt(i)=true;
end
end
if ctok>4 && isprime(str2double(s(3:5)))
disp([n ctok])
disp(tt)
end
end
It finds the answer as 42617. Without the prime restriction, it would also find 12437.
The commented line in the program (led by %) has do do with handling duplicates, but was not really needed due to lack of duplicates in the hidden number.
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Posted by Charlie
on 2022-11-22 10:26:33 |
computer solution (spoiler)
