Topic: [ANN][iQ CONTEST] 100K iQCash mined! Celebration puzzle: 100 $Waves to winner. - page 2. (Read 541 times)
February 28, 2018, 09:35:44 AM
Wow, that's a really hard one. I don't think I will solve this without another hint. I already spent way too much time looking at that graph
February 27, 2018, 07:53:03 AM
There are 25 yellow tiles, I don't think you misscounted them as 24, but that's really the only possible answer I can think of at the moment.
Nope.
February 27, 2018, 06:54:16 AM
There are 25 yellow tiles, I don't think you misscounted them as 24, but that's really the only possible answer I can think of at the moment.
February 26, 2018, 09:18:53 PM
Bonus question 1:
How many steps are there from the beginning to the end if you also count in alpha and beta? -> 24 -> X
How many steps are there from the beginning to the end if you also count in alpha and beta? -> 24 -> X
Nope. That's not it.
Another hint: It's related to one of the round 1 puzzle types on iQCash.net
it's not easy to solve the puzzle. i break my head all the time, which could be x. hope the next hint give me a new idea.
February 26, 2018, 05:39:51 PM
Bonus question 1:
How many steps are there from the beginning to the end if you also count in alpha and beta? -> 24 -> X
How many steps are there from the beginning to the end if you also count in alpha and beta? -> 24 -> X
Nope. That's not it.
Another hint: It's related to one of the round 1 puzzle types on iQCash.net
February 26, 2018, 10:55:45 AM
Bonus question 1:
How many steps are there from the beginning to the end if you also count in alpha and beta? -> 24 -> X
How many steps are there from the beginning to the end if you also count in alpha and beta? -> 24 -> X
February 26, 2018, 12:54:10 AM
BTW I wrote all the T-matrices down so anyone can check that the algo is correct:
if you have matlab installed on your machine (you can also use octave-online.net) put this code into the command promt, W22 will be equal to 415:
T1 = [1; 1];
T2 = [1 0; 1 1; 0 1];
T3 = [1 0 0; 0 1 0; 0 0 1];
T4 = [1 0 0; 0 1 0; 0 0 1; 0 0 1];
T5 = [1 0 0 0; 1 0 0 0; 0 1 0 0; 0 1 1 0; 0 0 1 1; 0 0 0 1];
T6 = [1 1 0 0 0 0; 0 0 1 1 0 0; 0 0 0 1 1 0; 0 0 0 0 1 1];
T7 = [1 0 0 0; 0 1 0 0; 0 0 1 1];
T8 = [1 0 0; 1 0 0; 0 1 0; 0 0 1];
T9 = [1 0 0 0; 1 1 0 0; 0 1 1 0; 0 0 1 1; 0 0 0 1];
T10 = [1 0 0 0 0; 0 1 0 0 0; 0 1 1 0 0; 0 0 1 1 0; 0 0 0 1 0; 0 0 0 0 1];
T11 = [1 0 0 0 0 0; 0 1 0 0 0 0; 0 0 1 0 0 0; 0 0 0 1 0 0; 0 0 0 0 1 0; 0 0 0 0 0 1];
T12 = [1 1 0 0 0 0; 0 0 1 0 0 0; 0 0 0 1 0 0; 0 0 0 0 1 1];
T13 = [1 0 0 0; 1 0 0 0; 0 1 0 0; 0 0 1 0; 0 0 0 1; 0 0 0 1];
T14 = [1 1 0 0 0 0; 0 0 1 0 0 0; 0 0 0 1 0 0; 0 0 0 1 0 0; 0 0 0 0 1 1];
T15 = [1 0 0 0 0; 1 0 0 0 0; 0 1 0 0 0; 0 0 1 0 0; 0 0 0 1 1; 0 0 0 0 1];
T16 = [1 0 0 0 0 0; 0 1 0 0 0 0; 0 0 1 0 0 0; 0 0 0 1 0 0; 0 0 0 1 0 0; 0 0 0 0 1 1];
T17 = [1 0 0 0 0 0; 0 1 1 0 0 0; 0 0 1 1 0 0; 0 0 0 1 1 0; 0 0 0 0 1 0; 0 0 0 0 0 1];
T18 = [1 1 0 0 0 0; 0 1 1 0 0 0; 0 0 1 1 0 0; 0 0 0 1 1 0; 0 0 0 0 1 1];
T19 = [1 1 0 0 0; 0 0 1 0 0; 0 0 0 1 1];
T20 = [1 0 0; 0 1 0; 0 0 1];
T21 = [1 1 0; 0 1 1];
T22 = [1 1];
W22 = T22 * T21 * T20 * T19 * T18 * T17 * T16 * T15 * T14 * T13 * T12 * T11 * T10 * T9 * T8 * T7 * T6 * T5 * T4 * T3 * T2 * T1;
disp('TOTAL NUMBER OF PATHS LEADING TO BETA IS:');
disp(W22);
if you have matlab installed on your machine (you can also use octave-online.net) put this code into the command promt, W22 will be equal to 415:
T1 = [1; 1];
T2 = [1 0; 1 1; 0 1];
T3 = [1 0 0; 0 1 0; 0 0 1];
T4 = [1 0 0; 0 1 0; 0 0 1; 0 0 1];
T5 = [1 0 0 0; 1 0 0 0; 0 1 0 0; 0 1 1 0; 0 0 1 1; 0 0 0 1];
T6 = [1 1 0 0 0 0; 0 0 1 1 0 0; 0 0 0 1 1 0; 0 0 0 0 1 1];
T7 = [1 0 0 0; 0 1 0 0; 0 0 1 1];
T8 = [1 0 0; 1 0 0; 0 1 0; 0 0 1];
T9 = [1 0 0 0; 1 1 0 0; 0 1 1 0; 0 0 1 1; 0 0 0 1];
T10 = [1 0 0 0 0; 0 1 0 0 0; 0 1 1 0 0; 0 0 1 1 0; 0 0 0 1 0; 0 0 0 0 1];
T11 = [1 0 0 0 0 0; 0 1 0 0 0 0; 0 0 1 0 0 0; 0 0 0 1 0 0; 0 0 0 0 1 0; 0 0 0 0 0 1];
T12 = [1 1 0 0 0 0; 0 0 1 0 0 0; 0 0 0 1 0 0; 0 0 0 0 1 1];
T13 = [1 0 0 0; 1 0 0 0; 0 1 0 0; 0 0 1 0; 0 0 0 1; 0 0 0 1];
T14 = [1 1 0 0 0 0; 0 0 1 0 0 0; 0 0 0 1 0 0; 0 0 0 1 0 0; 0 0 0 0 1 1];
T15 = [1 0 0 0 0; 1 0 0 0 0; 0 1 0 0 0; 0 0 1 0 0; 0 0 0 1 1; 0 0 0 0 1];
T16 = [1 0 0 0 0 0; 0 1 0 0 0 0; 0 0 1 0 0 0; 0 0 0 1 0 0; 0 0 0 1 0 0; 0 0 0 0 1 1];
T17 = [1 0 0 0 0 0; 0 1 1 0 0 0; 0 0 1 1 0 0; 0 0 0 1 1 0; 0 0 0 0 1 0; 0 0 0 0 0 1];
T18 = [1 1 0 0 0 0; 0 1 1 0 0 0; 0 0 1 1 0 0; 0 0 0 1 1 0; 0 0 0 0 1 1];
T19 = [1 1 0 0 0; 0 0 1 0 0; 0 0 0 1 1];
T20 = [1 0 0; 0 1 0; 0 0 1];
T21 = [1 1 0; 0 1 1];
T22 = [1 1];
W22 = T22 * T21 * T20 * T19 * T18 * T17 * T16 * T15 * T14 * T13 * T12 * T11 * T10 * T9 * T8 * T7 * T6 * T5 * T4 * T3 * T2 * T1;
disp('TOTAL NUMBER OF PATHS LEADING TO BETA IS:');
disp(W22);
That's awesome. You'll be a star at some of the later rounds of iQCash mining!
But then again, the progress of the puzzle rounds is intentionally educational, so there may be a lot of miners who have caught up by that time.
February 26, 2018, 12:49:28 AM
Bonus question 1:
How many diamonds can you find in the first big Waves Logo (smaller and bigger ones)?: 10 -> 10 = X
How many diamonds can you find in the first big Waves Logo (smaller and bigger ones)?: 10 -> 10 = X
Unfortunately no. Hint: The last hidden puzzle doesn't use the Roman numeral for 10, so the number 10 is not your target!
February 25, 2018, 10:59:42 AM
Bonus question 1:
How many diamonds can you find in the first big Waves Logo (smaller and bigger ones)?: 10 -> 10 = X
How many diamonds can you find in the first big Waves Logo (smaller and bigger ones)?: 10 -> 10 = X
February 25, 2018, 06:29:22 AM
BTW I wrote all the T-matrices down so anyone can check that the algo is correct:
if you have matlab installed on your machine (you can also use octave-online.net) put this code into the command promt, W22 will be equal to 415:
T1 = [1; 1];
T2 = [1 0; 1 1; 0 1];
T3 = [1 0 0; 0 1 0; 0 0 1];
T4 = [1 0 0; 0 1 0; 0 0 1; 0 0 1];
T5 = [1 0 0 0; 1 0 0 0; 0 1 0 0; 0 1 1 0; 0 0 1 1; 0 0 0 1];
T6 = [1 1 0 0 0 0; 0 0 1 1 0 0; 0 0 0 1 1 0; 0 0 0 0 1 1];
T7 = [1 0 0 0; 0 1 0 0; 0 0 1 1];
T8 = [1 0 0; 1 0 0; 0 1 0; 0 0 1];
T9 = [1 0 0 0; 1 1 0 0; 0 1 1 0; 0 0 1 1; 0 0 0 1];
T10 = [1 0 0 0 0; 0 1 0 0 0; 0 1 1 0 0; 0 0 1 1 0; 0 0 0 1 0; 0 0 0 0 1];
T11 = [1 0 0 0 0 0; 0 1 0 0 0 0; 0 0 1 0 0 0; 0 0 0 1 0 0; 0 0 0 0 1 0; 0 0 0 0 0 1];
T12 = [1 1 0 0 0 0; 0 0 1 0 0 0; 0 0 0 1 0 0; 0 0 0 0 1 1];
T13 = [1 0 0 0; 1 0 0 0; 0 1 0 0; 0 0 1 0; 0 0 0 1; 0 0 0 1];
T14 = [1 1 0 0 0 0; 0 0 1 0 0 0; 0 0 0 1 0 0; 0 0 0 1 0 0; 0 0 0 0 1 1];
T15 = [1 0 0 0 0; 1 0 0 0 0; 0 1 0 0 0; 0 0 1 0 0; 0 0 0 1 1; 0 0 0 0 1];
T16 = [1 0 0 0 0 0; 0 1 0 0 0 0; 0 0 1 0 0 0; 0 0 0 1 0 0; 0 0 0 1 0 0; 0 0 0 0 1 1];
T17 = [1 0 0 0 0 0; 0 1 1 0 0 0; 0 0 1 1 0 0; 0 0 0 1 1 0; 0 0 0 0 1 0; 0 0 0 0 0 1];
T18 = [1 1 0 0 0 0; 0 1 1 0 0 0; 0 0 1 1 0 0; 0 0 0 1 1 0; 0 0 0 0 1 1];
T19 = [1 1 0 0 0; 0 0 1 0 0; 0 0 0 1 1];
T20 = [1 0 0; 0 1 0; 0 0 1];
T21 = [1 1 0; 0 1 1];
T22 = [1 1];
W22 = T22 * T21 * T20 * T19 * T18 * T17 * T16 * T15 * T14 * T13 * T12 * T11 * T10 * T9 * T8 * T7 * T6 * T5 * T4 * T3 * T2 * T1;
disp('TOTAL NUMBER OF PATHS LEADING TO BETA IS:');
disp(W22);
if you have matlab installed on your machine (you can also use octave-online.net) put this code into the command promt, W22 will be equal to 415:
T1 = [1; 1];
T2 = [1 0; 1 1; 0 1];
T3 = [1 0 0; 0 1 0; 0 0 1];
T4 = [1 0 0; 0 1 0; 0 0 1; 0 0 1];
T5 = [1 0 0 0; 1 0 0 0; 0 1 0 0; 0 1 1 0; 0 0 1 1; 0 0 0 1];
T6 = [1 1 0 0 0 0; 0 0 1 1 0 0; 0 0 0 1 1 0; 0 0 0 0 1 1];
T7 = [1 0 0 0; 0 1 0 0; 0 0 1 1];
T8 = [1 0 0; 1 0 0; 0 1 0; 0 0 1];
T9 = [1 0 0 0; 1 1 0 0; 0 1 1 0; 0 0 1 1; 0 0 0 1];
T10 = [1 0 0 0 0; 0 1 0 0 0; 0 1 1 0 0; 0 0 1 1 0; 0 0 0 1 0; 0 0 0 0 1];
T11 = [1 0 0 0 0 0; 0 1 0 0 0 0; 0 0 1 0 0 0; 0 0 0 1 0 0; 0 0 0 0 1 0; 0 0 0 0 0 1];
T12 = [1 1 0 0 0 0; 0 0 1 0 0 0; 0 0 0 1 0 0; 0 0 0 0 1 1];
T13 = [1 0 0 0; 1 0 0 0; 0 1 0 0; 0 0 1 0; 0 0 0 1; 0 0 0 1];
T14 = [1 1 0 0 0 0; 0 0 1 0 0 0; 0 0 0 1 0 0; 0 0 0 1 0 0; 0 0 0 0 1 1];
T15 = [1 0 0 0 0; 1 0 0 0 0; 0 1 0 0 0; 0 0 1 0 0; 0 0 0 1 1; 0 0 0 0 1];
T16 = [1 0 0 0 0 0; 0 1 0 0 0 0; 0 0 1 0 0 0; 0 0 0 1 0 0; 0 0 0 1 0 0; 0 0 0 0 1 1];
T17 = [1 0 0 0 0 0; 0 1 1 0 0 0; 0 0 1 1 0 0; 0 0 0 1 1 0; 0 0 0 0 1 0; 0 0 0 0 0 1];
T18 = [1 1 0 0 0 0; 0 1 1 0 0 0; 0 0 1 1 0 0; 0 0 0 1 1 0; 0 0 0 0 1 1];
T19 = [1 1 0 0 0; 0 0 1 0 0; 0 0 0 1 1];
T20 = [1 0 0; 0 1 0; 0 0 1];
T21 = [1 1 0; 0 1 1];
T22 = [1 1];
W22 = T22 * T21 * T20 * T19 * T18 * T17 * T16 * T15 * T14 * T13 * T12 * T11 * T10 * T9 * T8 * T7 * T6 * T5 * T4 * T3 * T2 * T1;
disp('TOTAL NUMBER OF PATHS LEADING TO BETA IS:');
disp(W22);
February 24, 2018, 06:59:55 PM
re Bonus Question 2:
the algo for the solution is kinda similar to how a NN (neural network) is built...
1. every column of nodes (not arrows) in the puzzle is a separate layer;
2. each layer consists of a number of nodes (ni), so i-th layer has ni nodes;
3. every node in i-th layer is connected to some nodes in layer i-1;
4. let Ti be the transformation matrix that describes how nodes from i-th layer are connected to nodes in layer i-1, Ti is size of (ni*ni-1);
a simple example:
↗ ↘ ↗ ↘
↘ ↗ ↗
↘ ↗
there are 4 layers - (layer 0 - the initial layer, containing only one node), layer 1 - contains 2 nodes, layer 2 - contains 2 nodes, layer 3 - contains 2 nodes and layer 4 - contains 1 node (final node);
if a j-th node in i-th layer is connected to a k-th node in layer i-1 you write 1 in place (j,k - j-th row, k-th column) of the Ti matrix, otherwise you write 0:
T1 = [ 1; 1 ], of size (2x1) or (n1*n0);
T2 = [ 1 1; 0 1 ], of size (2x2) or (n2*n1);
T3 = [ 1 0; 0 1 ], of size (2x2) or (n3*n2);
T4 = [ 1 1 ], of size (1x2) or (n4*n3);
5. let Wi be the vector that describes how many paths lead to every node in i-th layer, so Wi is size of (ni*1);
now the formula:
Wi = Ti * Wi-1; (1)
W0 = [1];
considering there are 22 layers in the puzzle and 22 transformation matrices, the solution can be calculated by repeating (1) for each layer (column) in the puzzle, here is a pseudo-code:
W0 = [1];
for (i = 1:22)
Wi = Ti * Wi-1;
Answer = W22;
or:
Answer == W22 = T22 * T21 * T20 * ... * T1 * W0;
and for my simple example:
W0 = [1];
W1 = T1 * W0 = [ 1; 1 ] * [1] = [ 1; 1];
W2 = T2 * W1 = [ 1 1; 0 1 ] * [ 1; 1 ] = [ 2; 1];
W3 = T3 * W2 = [ 1 0; 0 1 ] * [ 2; 1 ] = [ 2; 1];
W4 = T4 * W3 = [ 1 1 ] * [ 2; 1 ] = [ 3 ];
answer - there are 3 paths
the algo for the solution is kinda similar to how a NN (neural network) is built...
1. every column of nodes (not arrows) in the puzzle is a separate layer;
2. each layer consists of a number of nodes (ni), so i-th layer has ni nodes;
3. every node in i-th layer is connected to some nodes in layer i-1;
4. let Ti be the transformation matrix that describes how nodes from i-th layer are connected to nodes in layer i-1, Ti is size of (ni*ni-1);
a simple example:
↗ ↘ ↗ ↘
↘ ↗ ↗
↘ ↗
there are 4 layers - (layer 0 - the initial layer, containing only one node), layer 1 - contains 2 nodes, layer 2 - contains 2 nodes, layer 3 - contains 2 nodes and layer 4 - contains 1 node (final node);
if a j-th node in i-th layer is connected to a k-th node in layer i-1 you write 1 in place (j,k - j-th row, k-th column) of the Ti matrix, otherwise you write 0:
T1 = [ 1; 1 ], of size (2x1) or (n1*n0);
T2 = [ 1 1; 0 1 ], of size (2x2) or (n2*n1);
T3 = [ 1 0; 0 1 ], of size (2x2) or (n3*n2);
T4 = [ 1 1 ], of size (1x2) or (n4*n3);
5. let Wi be the vector that describes how many paths lead to every node in i-th layer, so Wi is size of (ni*1);
now the formula:
Wi = Ti * Wi-1; (1)
W0 = [1];
considering there are 22 layers in the puzzle and 22 transformation matrices, the solution can be calculated by repeating (1) for each layer (column) in the puzzle, here is a pseudo-code:
W0 = [1];
for (i = 1:22)
Wi = Ti * Wi-1;
Answer = W22;
or:
Answer == W22 = T22 * T21 * T20 * ... * T1 * W0;
and for my simple example:
W0 = [1];
W1 = T1 * W0 = [ 1; 1 ] * [1] = [ 1; 1];
W2 = T2 * W1 = [ 1 1; 0 1 ] * [ 1; 1 ] = [ 2; 1];
W3 = T3 * W2 = [ 1 0; 0 1 ] * [ 2; 1 ] = [ 2; 1];
W4 = T4 * W3 = [ 1 1 ] * [ 2; 1 ] = [ 3 ];
answer - there are 3 paths
Impressive. You're definitely in the lead for the 20 Waves for this question!
February 24, 2018, 06:56:51 PM
Maybe the answer for Question 1 is:
How many hearts are there? I can find 10 and like I said before 10 = X
How many hearts are there? I can find 10 and like I said before 10 = X
Another nice try, but you'd have to include sideways hearts which look more like arrows, so this one doesn't quite make it.
(You'd also have to accept hearts with different proportions to total up to 10 too.)
February 24, 2018, 06:42:00 PM
Bonus Question 1:
How many arrows have to change colour that all paths have the same speed. -> 10 blue in the middle. 10 = X
How many arrows have to change colour that all paths have the same speed. -> 10 blue in the middle. 10 = X
That's not the intended puzzle, but it does work as a puzzle, so I'll award a bonus prize for that one, either 5 Waves or 10 if no-one finds the other puzzle.
Well spotted. PM your Waves address!
February 24, 2018, 06:21:33 PM
re Bonus Question 2:
the algo for the solution is kinda similar to how a NN (neural network) is built...
1. every column of nodes (not arrows) in the puzzle is a separate layer;
2. each layer consists of a number of nodes (ni), so i-th layer has ni nodes;
3. every node in i-th layer is connected to some nodes in layer i-1;
4. let Ti be the transformation matrix that describes how nodes from i-th layer are connected to nodes in layer i-1, Ti is size of (ni*ni-1);
a simple example:
↗ ↘ ↗ ↘
↘ ↗ ↗
↘ ↗
there are 4 layers - (layer 0 - the initial layer, containing only one node), layer 1 - contains 2 nodes, layer 2 - contains 2 nodes, layer 3 - contains 2 nodes and layer 4 - contains 1 node (final node);
if a j-th node in i-th layer is connected to a k-th node in layer i-1 you write 1 in place (j,k - j-th row, k-th column) of the Ti matrix, otherwise you write 0:
T1 = [ 1; 1 ], of size (2x1) or (n1*n0);
T2 = [ 1 1; 0 1 ], of size (2x2) or (n2*n1);
T3 = [ 1 0; 0 1 ], of size (2x2) or (n3*n2);
T4 = [ 1 1 ], of size (1x2) or (n4*n3);
5. let Wi be the vector that describes how many paths lead to every node in i-th layer, so Wi is size of (ni*1);
now the formula:
Wi = Ti * Wi-1; (1)
W0 = [1];
considering there are 22 layers in the puzzle and 22 transformation matrices, the solution can be calculated by repeating (1) for each layer (column) in the puzzle, here is a pseudo-code:
W0 = [1];
for (i = 1:22)
Wi = Ti * Wi-1;
Answer = W22;
or:
Answer == W22 = T22 * T21 * T20 * ... * T1 * W0;
and for my simple example:
W0 = [1];
W1 = T1 * W0 = [ 1; 1 ] * [1] = [ 1; 1];
W2 = T2 * W1 = [ 1 1; 0 1 ] * [ 1; 1 ] = [ 2; 1];
W3 = T3 * W2 = [ 1 0; 0 1 ] * [ 2; 1 ] = [ 2; 1];
W4 = T4 * W3 = [ 1 1 ] * [ 2; 1 ] = [ 3 ];
answer - there are 3 paths
the algo for the solution is kinda similar to how a NN (neural network) is built...
1. every column of nodes (not arrows) in the puzzle is a separate layer;
2. each layer consists of a number of nodes (ni), so i-th layer has ni nodes;
3. every node in i-th layer is connected to some nodes in layer i-1;
4. let Ti be the transformation matrix that describes how nodes from i-th layer are connected to nodes in layer i-1, Ti is size of (ni*ni-1);
a simple example:
↗ ↘ ↗ ↘
↘ ↗ ↗
↘ ↗
there are 4 layers - (layer 0 - the initial layer, containing only one node), layer 1 - contains 2 nodes, layer 2 - contains 2 nodes, layer 3 - contains 2 nodes and layer 4 - contains 1 node (final node);
if a j-th node in i-th layer is connected to a k-th node in layer i-1 you write 1 in place (j,k - j-th row, k-th column) of the Ti matrix, otherwise you write 0:
T1 = [ 1; 1 ], of size (2x1) or (n1*n0);
T2 = [ 1 1; 0 1 ], of size (2x2) or (n2*n1);
T3 = [ 1 0; 0 1 ], of size (2x2) or (n3*n2);
T4 = [ 1 1 ], of size (1x2) or (n4*n3);
5. let Wi be the vector that describes how many paths lead to every node in i-th layer, so Wi is size of (ni*1);
now the formula:
Wi = Ti * Wi-1; (1)
W0 = [1];
considering there are 22 layers in the puzzle and 22 transformation matrices, the solution can be calculated by repeating (1) for each layer (column) in the puzzle, here is a pseudo-code:
W0 = [1];
for (i = 1:22)
Wi = Ti * Wi-1;
Answer = W22;
or:
Answer == W22 = T22 * T21 * T20 * ... * T1 * W0;
and for my simple example:
W0 = [1];
W1 = T1 * W0 = [ 1; 1 ] * [1] = [ 1; 1];
W2 = T2 * W1 = [ 1 1; 0 1 ] * [ 1; 1 ] = [ 2; 1];
W3 = T3 * W2 = [ 1 0; 0 1 ] * [ 2; 1 ] = [ 2; 1];
W4 = T4 * W3 = [ 1 1 ] * [ 2; 1 ] = [ 3 ];
answer - there are 3 paths
February 24, 2018, 01:18:58 PM
Maybe the answer for Question 1 is:
How many hearts are there? I can find 10 and like I said before 10 = X
How many hearts are there? I can find 10 and like I said before 10 = X
February 24, 2018, 09:32:44 AM
Bonus Question 1:
How many arrows have to change colour that all paths have the same speed. -> 10 blue in the middle. 10 = X
How many arrows have to change colour that all paths have the same speed. -> 10 blue in the middle. 10 = X
February 24, 2018, 04:24:58 AM
Bonus Question 2:
(I hope you can read it)
https://www.dropbox.com/s/iqijidhiwllapsb/solution.jpg?dl=0
(I hope you can read it)
https://www.dropbox.com/s/iqijidhiwllapsb/solution.jpg?dl=0
Sorry, no. What we are looking for for bonus ques. 2 is a mathematical formula to describe as best as possible the calculation that you have performed.
The formula should be written with standard mathematical notation. e.g. f(x) = f(xn-1) - xn+1
That's not the answer, and probably not even close, but just an example of the type of layout that is required.
February 24, 2018, 04:17:54 AM
Bonus question 2:
How many of those 24 squares are made by color mixer (Blue + Yellow)? 10, which means X in Roman
The other 14 are just from single colors
Is that right?
How many of those 24 squares are made by color mixer (Blue + Yellow)? 10, which means X in Roman
The other 14 are just from single colors
Is that right?
Ha! That's clever but it's not the correct puzzle.
If it was a color mixer puzzle there would be distinct boundaries for each square.
But nice try!
February 24, 2018, 04:11:50 AM
Bonus Question 2:
(I hope you can read it)
https://www.dropbox.com/s/iqijidhiwllapsb/solution.jpg?dl=0
(I hope you can read it)
https://www.dropbox.com/s/iqijidhiwllapsb/solution.jpg?dl=0
February 23, 2018, 08:45:40 PM
Bonus question 2:
How many of those 24 squares are made by color mixer (Blue + Yellow)? 10, which means X in Roman
The other 14 are just from single colors
Is that right?
In fact, it's the 2nd question from Bonus question 1 How many of those 24 squares are made by color mixer (Blue + Yellow)? 10, which means X in Roman
The other 14 are just from single colors
Is that right?
Jump to: