Which technique would you choose for the memory game? (Classic, playing cards, or with other figures).
Considering that they would have very little time to memorize and it would be a memory game with the cards arranged in 7x7
Maybe memorize the letters in sequence? Or associate each pair with a position and put it in the memory palace?
Considering about 30-45 seconds, not enough time to do much.
I would use a memory palace with 49 locations. Each location should be associated with a position on the 7x7 grid. Alternatively you can create a special memory palace with locations in 7x7 grid.
That way you can memorise cards regardless of order in with they are being opened. And you can remember position of any card by remembering where is it in your memory palace.
If you need to play some number of games, you can create more special memory palaces or use the same one. I guess you can link new images to old images from same location to avoid ghost images.
I had created what I believed to be an interesting response to your question for which I used some of my knowledge of concrete mathematics. It appears however that nobody including you has any interest in what I wrote.
I am still somewhat curious about the 7 x 7, like is one card not part of a pair?
If indeed you’re talking about the classic, “find pairs, remove from the game” type of memory game, I would probably use pegs from a 2-digit image (or PAO) list or a memory palace of suitable size. This should allow grids and decks of different size. All you need to do is calculate the card positions sufficiently fast to succeed in the game. Still, if playing with friends/family I’d just not “cheat” and use my work memory in order to keep the game fair.
What card positions are you talking about? I mean are you first searching for matching pairs and then memorizing that for example 1 is a pair with 29, 2 with 31 and so on? Or are you memorising that 1 is a picture of a teddy bear, 2 is a tennis ball and place those images in order in a memory palace or connect them to your 2 digit peglist.
I am noticing a trend on this forum that people that create a post in which they ask a question quite often show no sign of being interested in the answers they receive.
Card position == the place where individual card resides on the playing area.
Pretty much this. But because players can choose to flip any two cards at the time, the time spent in figuring out the card position reliably is probably the bottle neck in this method. And you have to keep a certain kind of an eye for the pairs, e.g. the tennis ball in the birthday cake is now tied to a rope. So on the next turn, you’ll reach for the locations 77 and 49 to match the tennis ball pair.
It looks like you and dmitriyRubikscube are not really addressing the problem that Natividade appears to be asking. The …
Considering about 30-45 seconds
… refers in my estimation to the time you have looking at all the images before they get turned over. So this variation of the game doesn’t start with a person turning over 2 random cards but with someone trying to turn over as many pairs as possible after 30-45 seconds of memorisation.
Buddy, forgive me for real, I saw some of your answers talking about my “lack of interest” in the answers.
I was in a bit of a tight week, I asked the question but I couldn’t get into the forum these days. So I’m really sorry.
And also, as I am Brazilian, I have to translate to interact in this forum, and some of the times the translations are not good enough, generating a misunderstanding.
But I’m super interested in your answer. Forgive mediocrity
When I said “7x7” I meant only the orders of the cards.
There would be 49 cards, 7 columns and 7 rows.
The pairs would be in random order in the game.
The game starts with all the cards turned up, after 30-40 seconds, the cards are turned face down, without being able to see them.
And then, you would need to know where the pairs are located in that pile of cards.
A person turns over a card, and tries to guess where the same card is. That would be a “Pair”
exactly.
But if you still want to share your answer, I guarantee I’m just as interested.
Well, this is not so relevant to memorize the rest of the pairs in the 7x7 grid, but even if there was a card that has no pair, it would not be a problem, the focus is on the cards that have pairs.
Well, maybe I faltered on this one, but the discussion is here for everyone to see whenever they want.
No problem.
And also, as I am Brazilian, I have to translate to interact in this forum, and some of the times the translations are not good enough, generating a misunderstanding.
I sort of have the idea that we are talking past one another regarding the 7x7 grid that you proposed. The memory game always has an even number (multitude of 2) of cards because every card is one half of a pair. I believe you thought of a 7x7 grid because it is symmetrical and thus easy to visualize for the reader. But I also imagined that you overlooked the fact that 7x7 = 49 and thus this grid cannot possibly be used in a memory game.
You reacted to my question about this grid, but your answer kind of leaves me guessing if you understood the problem.
But I’m super interested in your answer.
I have a tendency to delete my contribution to a thread if the person asking for a contribution does not react to it within a reasonable amount of time.
I no longer have my first reaction, but I did save my “concrete mathematics analysis” of the memory game problem. So here it is:
A concrete mathematics perspective on the memory game
Perhaps of some relevance to this problem is that I used to study discrete mathematics as a hobby a long time ago. A lot of problems in the books I had were similar to this memory game problem. The proper method to solve these problems is always to first find out how you can describe it in the most simple way. It seems obvious that the only information you need is the locations of (x-1) pairs (x being the total number of pairs) because the pair that you are not memorising automatically follows from the information you did memorise. So in a 2 x 2 grid you would only need to memorise the locations of 1 pair and in a 10 x 10 grid you need to memorise the locations of 49 pairs.
You don’t have to memorise the order of the pairs meaning that for example you don’t have to memorise that the first pair can be found on location 3 and 19 and the second pair on 7 and 45. This means that you only have to code where the 2nd cards are for every first card position.
So let’s look at the below 4 x 4 grid.
So the first object we encounter if we use the left-right and up-down protocol is a basketball. The 2nd basketball can be found in location 14 . The 2nd He-man is in location 4. We thus end up with the following number sequence of just 7 numbers: 14-4-11-5-7-12-15. The position of the 1st and 2nd helicopter automatically follows from these 7 numbers.
Every number in the above sequence is an element of the range 1 
15. So if we create an image collection of 15 x 15 (=225) images, we can code every variation as a sequence of only 4 images; we can choose to include the last pair (the helicopter), because memorising just 15 as opposed to 15 and 13 (the 2nd helicopter) makes no difference for us in this case.
We can also choose to code the positions of the 2nd image relative to the first image. This would not make a difference for the basketball, but 2nd He-man would now be represented by number 3 (3 places away from 1st He-man). We can also ignore images once they are coded, so that means for example that the 2nd tennis ball is the next image if we ignore 2nd He-man. So the location of the 2nd tennis ball can now be represented by number 1. If we code in this way we get: 14-3-8-1-1-3-3. Notice that the 2nd number can never be higher than 13 and the 3th number can never be higher than 11 and so on. This means that there are 15 x 13 x 11 x … = 2.027.025 possible variations to memorise. There is a possible way to benefit from the fact that the number of variations drops the more pairs you have memorised. There are only 7 x 5 x 3 = 105 possible variations for the last 3 pairs (the actual last pair doesn’t have to be coded as explained above). So that means that every possible variation of these 3 pairs can be associated with just one of the 225 images you already created for the above described more simple version of representation. So we would need 14-3 = 1 image, 8-1 = 1 image and 1-3-3 = 1 image and that means that with a collection of 225 images we can represent every variation of this 4 * 4 grid with just 3 images.
I became more interested in this part, it is easier for me to create these associations.
Your answers were very interesting, but in some I was in doubt if it would be possible to do it quickly.
Also, I didn’t quite understand how to relate the encodings of the 2nd cards to the 1st in the first example you gave, can you explain it to me better?
I used 7x7 only as an example of size, if you want you can replace it with a grid that has an even amount of cards, like 8x8 or 6x6 or 7x6.
Using relative positions and also ignoring already memorised images/pairs would allow for the most efficient mathematical representation of the locations of all the pairs. I do however, as you seem to imply, admit that doing so will not make the memorisation process any easier.
I am inclined to think that finding the matching card for the 1st image, the 2nd image and so on as represented by the 14-4-11-5-7-12-15 number sequence is the best strategy. If the memory game were to become a financially interesting sport and it was played for example using a 10x10 grid (50 pairs) in all likelihood some players will use a 2.500 (50x50) image collection to represent 2 pairs with only one image (similar to 1 image for 2 playing card using the Ben system in card memorisation competitions).
The grid size determines in part what possible ways there are to memorise the information needed to succeed in this memory game. If you give an example that is not even possible in the actual game you are creating confusion for the people wanting to articulate their best strategies for no reason. You would not, in my honest estimation, use a grid size with an uneven number of cards if you were aware that such a grid would not be possible in the game you talked about.
Thanks for the recommendation, as I’m new to the forum I end up making mistakes.
And your answer was satisfactory to me, about memorization strategies.
