Pattern recognition using a, b, c, d, and e
Clash Royale CLAN TAG#URR8PPP
$begingroup$
I need to find the solution of this pattern:
a,b,c, c,d,e, ?, ?, ?, b,c,e, a,c,d, b,c,d, a,c,e, ?
The solution should be something like this:
a,b,c, c,d,e x,x,x x,x,x x,x,x,b,c,e,a,c,d, b,c,d, a,c,e, x,x,x
where instead of the x you should replace letters.
Source: http://zagaza.ru/za548.htm
logical-deduction pattern combinatorics
edited Jan 7 at 1:42
Peter Mortensen
1253
asked Jan 6 at 16:19
alnesialnesi
928
$endgroup$
add a comment |
$begingroup$
I need to find the solution of this pattern:
a,b,c, c,d,e, ?, ?, ?, b,c,e, a,c,d, b,c,d, a,c,e, ?
The solution should be something like this:
a,b,c, c,d,e x,x,x x,x,x x,x,x,b,c,e,a,c,d, b,c,d, a,c,e, x,x,x
where instead of the x you should replace letters.
Source: http://zagaza.ru/za548.htm
logical-deduction pattern combinatorics
edited Jan 7 at 1:42
Peter Mortensen
1253
asked Jan 6 at 16:19
alnesialnesi
928
$endgroup$
add a comment |
$begingroup$
I need to find the solution of this pattern:
a,b,c, c,d,e, ?, ?, ?, b,c,e, a,c,d, b,c,d, a,c,e, ?
The solution should be something like this:
a,b,c, c,d,e x,x,x x,x,x x,x,x,b,c,e,a,c,d, b,c,d, a,c,e, x,x,x
where instead of the x you should replace letters.
Source: http://zagaza.ru/za548.htm
logical-deduction pattern combinatorics
edited Jan 7 at 1:42
Peter Mortensen
1253
asked Jan 6 at 16:19
alnesialnesi
928
$endgroup$
I need to find the solution of this pattern:
a,b,c, c,d,e, ?, ?, ?, b,c,e, a,c,d, b,c,d, a,c,e, ?
The solution should be something like this:
a,b,c, c,d,e x,x,x x,x,x x,x,x,b,c,e,a,c,d, b,c,d, a,c,e, x,x,x
where instead of the x you should replace letters.
Source: http://zagaza.ru/za548.htm
logical-deduction pattern combinatorics
logical-deduction pattern combinatorics
edited Jan 7 at 1:42
Peter Mortensen
1253
asked Jan 6 at 16:19
alnesialnesi
928
edited Jan 7 at 1:42
Peter Mortensen
1253
asked Jan 6 at 16:19
alnesialnesi
928
edited Jan 7 at 1:42
Peter Mortensen
1253
edited Jan 7 at 1:42
Peter Mortensen
1253
edited Jan 7 at 1:42
Peter Mortensen
1253
1253
asked Jan 6 at 16:19
alnesialnesi
928
asked Jan 6 at 16:19
alnesialnesi
928
asked Jan 6 at 16:19
alnesialnesi
928
928
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
How about:
a,b,c, c,d,e, a,b,d, b,d,e, a,b,e, b,c,e, a,c,d, b,c,d, a,c,e, a,d,e.
Because:
We know there are $binom53=10$ variants, and there are $10$ slots, so a one-to-one correspondence seems likely.
The lexical order is a,b,c, a,b,d, a,b,e, a,c,d, a,c,e, a,d,e, b,c,d, b,c,e, b,d,e, c,d,e.
Alternate indices cover 1,2,3,4,5 and 10,9,8,7,6, i.e. 1,10,2,9,3,8,4,7,5,6.
answered Jan 6 at 18:33
JonMark PerryJonMark Perry
18k63786
$endgroup$
$begingroup$
This is logical. I was going for Permutations instead.
$endgroup$
– ABcDexter
Jan 6 at 18:34
$begingroup$
thanks for answering,could you please explain me how you understood the order of those variants,i mean inside the x.x.x at the beginning there could even be C as first letter,please explain that to an heretical guy xD
$endgroup$
– alnesi
Jan 6 at 18:55
$begingroup$
do u have any other contact,like discord,steam even a twitter account that you use to communicate thanks in advace
$endgroup$
– alnesi
Jan 6 at 19:28
$begingroup$
@alnesi I think the explanation is already very good and clear. maybe you don't see how the lexical order is determined? Just take three letters, and order them as if they were words: "abc", "abd",... Once that is established, the index alternation should pretty easy.
$endgroup$
– Silly Freak
Jan 6 at 19:31
$begingroup$
oh thank you,i just wanted to know that :D .Really thank you
$endgroup$
– alnesi
Jan 6 at 19:38
add a comment |
Your Answer
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
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active
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active
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$begingroup$
How about:
a,b,c, c,d,e, a,b,d, b,d,e, a,b,e, b,c,e, a,c,d, b,c,d, a,c,e, a,d,e.
Because:
We know there are $binom53=10$ variants, and there are $10$ slots, so a one-to-one correspondence seems likely.
The lexical order is a,b,c, a,b,d, a,b,e, a,c,d, a,c,e, a,d,e, b,c,d, b,c,e, b,d,e, c,d,e.
Alternate indices cover 1,2,3,4,5 and 10,9,8,7,6, i.e. 1,10,2,9,3,8,4,7,5,6.
answered Jan 6 at 18:33
JonMark PerryJonMark Perry
18k63786
$endgroup$
$begingroup$
This is logical. I was going for Permutations instead.
$endgroup$
– ABcDexter
Jan 6 at 18:34
$begingroup$
thanks for answering,could you please explain me how you understood the order of those variants,i mean inside the x.x.x at the beginning there could even be C as first letter,please explain that to an heretical guy xD
$endgroup$
– alnesi
Jan 6 at 18:55
$begingroup$
do u have any other contact,like discord,steam even a twitter account that you use to communicate thanks in advace
$endgroup$
– alnesi
Jan 6 at 19:28
$begingroup$
@alnesi I think the explanation is already very good and clear. maybe you don't see how the lexical order is determined? Just take three letters, and order them as if they were words: "abc", "abd",... Once that is established, the index alternation should pretty easy.
$endgroup$
– Silly Freak
Jan 6 at 19:31
$begingroup$
oh thank you,i just wanted to know that :D .Really thank you
$endgroup$
– alnesi
Jan 6 at 19:38
add a comment |
$begingroup$
How about:
a,b,c, c,d,e, a,b,d, b,d,e, a,b,e, b,c,e, a,c,d, b,c,d, a,c,e, a,d,e.
Because:
We know there are $binom53=10$ variants, and there are $10$ slots, so a one-to-one correspondence seems likely.
The lexical order is a,b,c, a,b,d, a,b,e, a,c,d, a,c,e, a,d,e, b,c,d, b,c,e, b,d,e, c,d,e.
Alternate indices cover 1,2,3,4,5 and 10,9,8,7,6, i.e. 1,10,2,9,3,8,4,7,5,6.
answered Jan 6 at 18:33
JonMark PerryJonMark Perry
18k63786
$endgroup$
$begingroup$
This is logical. I was going for Permutations instead.
$endgroup$
– ABcDexter
Jan 6 at 18:34
$begingroup$
thanks for answering,could you please explain me how you understood the order of those variants,i mean inside the x.x.x at the beginning there could even be C as first letter,please explain that to an heretical guy xD
$endgroup$
– alnesi
Jan 6 at 18:55
$begingroup$
do u have any other contact,like discord,steam even a twitter account that you use to communicate thanks in advace
$endgroup$
– alnesi
Jan 6 at 19:28
$begingroup$
@alnesi I think the explanation is already very good and clear. maybe you don't see how the lexical order is determined? Just take three letters, and order them as if they were words: "abc", "abd",... Once that is established, the index alternation should pretty easy.
$endgroup$
– Silly Freak
Jan 6 at 19:31
$begingroup$
oh thank you,i just wanted to know that :D .Really thank you
$endgroup$
– alnesi
Jan 6 at 19:38
add a comment |
$begingroup$
How about:
a,b,c, c,d,e, a,b,d, b,d,e, a,b,e, b,c,e, a,c,d, b,c,d, a,c,e, a,d,e.
Because:
We know there are $binom53=10$ variants, and there are $10$ slots, so a one-to-one correspondence seems likely.
The lexical order is a,b,c, a,b,d, a,b,e, a,c,d, a,c,e, a,d,e, b,c,d, b,c,e, b,d,e, c,d,e.
Alternate indices cover 1,2,3,4,5 and 10,9,8,7,6, i.e. 1,10,2,9,3,8,4,7,5,6.
answered Jan 6 at 18:33
JonMark PerryJonMark Perry
18k63786
$endgroup$
How about:
a,b,c, c,d,e, a,b,d, b,d,e, a,b,e, b,c,e, a,c,d, b,c,d, a,c,e, a,d,e.
Because:
We know there are $binom53=10$ variants, and there are $10$ slots, so a one-to-one correspondence seems likely.
The lexical order is a,b,c, a,b,d, a,b,e, a,c,d, a,c,e, a,d,e, b,c,d, b,c,e, b,d,e, c,d,e.
Alternate indices cover 1,2,3,4,5 and 10,9,8,7,6, i.e. 1,10,2,9,3,8,4,7,5,6.
answered Jan 6 at 18:33
JonMark PerryJonMark Perry
18k63786
answered Jan 6 at 18:33
JonMark PerryJonMark Perry
18k63786
answered Jan 6 at 18:33
JonMark PerryJonMark Perry
18k63786
answered Jan 6 at 18:33
JonMark PerryJonMark Perry
18k63786
18k63786
$begingroup$
This is logical. I was going for Permutations instead.
$endgroup$
– ABcDexter
Jan 6 at 18:34
$begingroup$
thanks for answering,could you please explain me how you understood the order of those variants,i mean inside the x.x.x at the beginning there could even be C as first letter,please explain that to an heretical guy xD
$endgroup$
– alnesi
Jan 6 at 18:55
$begingroup$
do u have any other contact,like discord,steam even a twitter account that you use to communicate thanks in advace
$endgroup$
– alnesi
Jan 6 at 19:28
$begingroup$
@alnesi I think the explanation is already very good and clear. maybe you don't see how the lexical order is determined? Just take three letters, and order them as if they were words: "abc", "abd",... Once that is established, the index alternation should pretty easy.
$endgroup$
– Silly Freak
Jan 6 at 19:31
$begingroup$
oh thank you,i just wanted to know that :D .Really thank you
$endgroup$
– alnesi
Jan 6 at 19:38
add a comment |
$begingroup$
This is logical. I was going for Permutations instead.
$endgroup$
– ABcDexter
Jan 6 at 18:34
$begingroup$
thanks for answering,could you please explain me how you understood the order of those variants,i mean inside the x.x.x at the beginning there could even be C as first letter,please explain that to an heretical guy xD
$endgroup$
– alnesi
Jan 6 at 18:55
$begingroup$
do u have any other contact,like discord,steam even a twitter account that you use to communicate thanks in advace
$endgroup$
– alnesi
Jan 6 at 19:28
$begingroup$
@alnesi I think the explanation is already very good and clear. maybe you don't see how the lexical order is determined? Just take three letters, and order them as if they were words: "abc", "abd",... Once that is established, the index alternation should pretty easy.
$endgroup$
– Silly Freak
Jan 6 at 19:31
$begingroup$
oh thank you,i just wanted to know that :D .Really thank you
$endgroup$
– alnesi
Jan 6 at 19:38
$begingroup$
This is logical. I was going for Permutations instead.
$endgroup$
– ABcDexter
Jan 6 at 18:34
$begingroup$
This is logical. I was going for Permutations instead.
$endgroup$
– ABcDexter
Jan 6 at 18:34
$begingroup$
thanks for answering,could you please explain me how you understood the order of those variants,i mean inside the x.x.x at the beginning there could even be C as first letter,please explain that to an heretical guy xD
$endgroup$
– alnesi
Jan 6 at 18:55
$begingroup$
thanks for answering,could you please explain me how you understood the order of those variants,i mean inside the x.x.x at the beginning there could even be C as first letter,please explain that to an heretical guy xD
$endgroup$
– alnesi
Jan 6 at 18:55
$begingroup$
do u have any other contact,like discord,steam even a twitter account that you use to communicate thanks in advace
$endgroup$
– alnesi
Jan 6 at 19:28
$begingroup$
do u have any other contact,like discord,steam even a twitter account that you use to communicate thanks in advace
$endgroup$
– alnesi
Jan 6 at 19:28
$begingroup$
@alnesi I think the explanation is already very good and clear. maybe you don't see how the lexical order is determined? Just take three letters, and order them as if they were words: "abc", "abd",... Once that is established, the index alternation should pretty easy.
$endgroup$
– Silly Freak
Jan 6 at 19:31
$begingroup$
@alnesi I think the explanation is already very good and clear. maybe you don't see how the lexical order is determined? Just take three letters, and order them as if they were words: "abc", "abd",... Once that is established, the index alternation should pretty easy.
$endgroup$
– Silly Freak
Jan 6 at 19:31
$begingroup$
oh thank you,i just wanted to know that :D .Really thank you
$endgroup$
– alnesi
Jan 6 at 19:38
$begingroup$
oh thank you,i just wanted to know that :D .Really thank you
$endgroup$
– alnesi
Jan 6 at 19:38
add a comment |
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