Difficult Puzzle: ABBA

[Ingafgrinn Macabre]

Not so difficult when seeing the logics.

Spoiler

Symmetrical:
\----------
--\-A-A----
--A-\-B-A--
--A-B\B-A--
--A-B-\-A--
----A-A-\--
----------\


     [A1] [A2]     

[A*]      [B1] [A3]

[A*] [B2] [B3] [A4]

[A*] [B4]      [A5]

     [A*] [A*]     


Step 1:
A1 --> B1
A2 --> B1
A3 --> B1, B3
A4 --> B1, B3
A5 --> B1, B3

Step 2:
B1 --> B2, B3
B3 --> B1, B2, B4

Step 3:
B1 --> 4A (A1~A4)
B2 --> 3A
B3 --> 3A (A3~A5)
B4 --> 4A


Count from step 2:
B1 -->  6 ( 5 at A3~A5)
B3 --> 11 (10 at A1~A4)

Count from step 1:
A1 --> 6
A2 --> 6
A3 --> 5 + 10 (6 + 11)
A4 --> 5 + 10 (6 + 11)
A5 --> 5 + 11 (6 + 11)

Total count not reusing the first A:
for first half: 6 + 6 + 15 + 15 + 16 = 58
Total: 116

Total count with reusing the first A:
for first half: 6 + 6 + 17 + 17 + 17 = 63
Total: 126

 

[Keriwin]

i swear its 270...

going around from top clockwise...


there are 10 A's... so for the first one it will be 9xAB... then because one has already been use it will be 8x AB and so forth -

a1+a2
a1+a3
a1+a4
a1+a5
a1+a6
a1+a7
a1+a8
a1+a9
a1+a10

then...

a2+a3
a3+a4 etc...

therefore...

9+8+7+6+5+4+3+2+1 = 45

now there are 4 B's... they can go into 6 different types of patterns...

b1-b2
b1-b3
b1-b4
b2-b3
b2-b4
b3-b4

so therefore tis 45x6 = 270

imo ofc

 

[Ingafgrinn Macabre]

i swear its 270...

going around from top clockwise...


there are 10 A's... so for the first one it will be 9xAB... then because one has already been use it will be 8x AB and so forth -

a1+a2
a1+a3
a1+a4
a1+a5
a1+a6
a1+a7
a1+a8
a1+a9
a1+a10

then...

a2+a3
a3+a4 etc...

therefore...

9+8+7+6+5+4+3+2+1 = 45

now there are 4 B's... they can go into 6 different types of patterns...

b1-b2
b1-b3
b1-b4
b2-b3
b2-b4
b3-b4

so therefore tis 45x6 = 270

imo ofc

You're counting connections that cannot be made there.

If you first see that it's symmetrical you'll notice that you'll only have to start at half the A's and that the other half will contain all the same routes mirrorred.

If you'd use the numbering I have selected the post above, you'll notice that A1 thru A5 can only connect to B1 and B3, so calculating the connections to B2 would result in pathways that cannot be taken.

 

[Keriwin]

You're counting connections that cannot be made there.

If you first see that it's symmetrical you'll notice that you'll only have to start at half the A's and that the other half will contain all the same routes mirrorred.

If you'd use the numbering I have selected the post above, you'll notice that A1 thru A5 can only connect to B1 and B3, so calculating the connections to B2 would result in pathways that cannot be taken.

ah you said how many abba's can be made... not without breaking the chain hehe