Beyond the Ages

Published on Sunday, September 30, 2012 in , , ,

LaMenta3's binary pillow photoIn April, in the first of my age-guessing series of posts, I discussed the magic trick known as the “Age Cards.”

In that post, I focused solely on their use in finding someone's age, but today I'm going to look at the trick from several other perspectives.

As a quick note before we proceed, you should be familiar with the basics of the Age Cards from that original post to follow most of this post.

Binary is important in this trick, because all the numbers are being uniquely identified by whether or not each of the powers of 2 are involved in adding up to that number.

The numbers are distributed on the Age Cards in accordance with their binary equivalent. For example, 53 in binary becomes 110101. The 1s in 110101 appear only in the 1s, 4s, 16s, and 32s places, so 53 will only appear on the cards with 1, 4, 16, and 32 in the upper left corner. This is why you can add the numbers in the upper left corner, 1+4+16+32 in this case, to get the person's number.

What if the range of numbers were smaller? If I limit your choice to the numbers 1 through 7, I would only need 3 cards, since 7 is 111 in binary. However, the trick would seem far less impressive, as it wouldn't take that long to just look through the cards and find a number common to the cards containing the number, and making sure those numbers aren't on the remaining cards.

This seems to suggest that the trick becomes less impressive with smaller sets. However, this is actually only true when you're using cards filled solely with numbers. What if the numbers were represented with something else?

Let's try using months. January will replace 1, February will replace 2, and so on up to December replacing 12. How many cards will we need for the month version? The largest number we'll deal with is 12, which is 1100 in binary, a 4-digit number, so we'll only need 4 cards.

How do we distribute the months? We can test each bit using the AND operation with the first 4 powers of 2 to work out where they go. With Wolfram|Alpha, this is achieved using the BitAnd command.

Testing each of the numbers from 1 to 12 by performing the AND operation against each of the first 4 powers of two, we get the rather odd-looking result of {{1, 0, 0, 0}, {0, 2, 0, 0}, {1, 2, 0, 0}, {0, 0, 4, 0}, {1, 0, 4, 0}, {0, 2, 4, 0}, {1, 2, 4, 0}, {0, 0, 0, 8}, {1, 0, 0, 8}, {0, 2, 0, 8}, {1, 2, 0, 8}, {0, 0, 4, 8}}. What does this mean?

This is the distribution of months we need! Formatted more clearly, the answer would look like this:

January (1) = {1, 0, 0, 0}
February (2) = {0, 2, 0, 0}
March (3) = {1, 2, 0, 0}
April (4) = {0, 0, 4, 0}
May (5) = {1, 0, 4, 0}
June (6) = {0, 2, 4, 0}
July (7) = {1, 2, 4, 0}
August (8) = {0, 0, 0, 8}
September (9) = {1, 0, 0, 8}
October (10) = {0, 2, 0, 8}
November (11) = {1, 2, 0, 8}
December (12) = {0, 0, 4, 8}

In other words, if we consider the various cards as 1, 2, 4, and 8, as we would were we using just numbers, November would appear on cards 1, 2, and 8, while May would only appear on cards 1 and 4. The final lists, then, would look like this (each column represents a singe card):

January        February       April          August
March          March          May            September
May            June           June           October
July           July           July           November
September      October        December       December
November       November
As long as you can think of the January card as having a value of 1, February as 2, April as 4, and August as 8 (the first month on each of the respective cards), you can determine someone's secret month (perhaps their birthday month?) as easily as their age. If a month is only on the cards beginning with February (2) and April (4), then their month must be 2 + 4, which equals 6. The 6th month is June, so June must be the month of which they're thinking!

Compared to using cards with the numbers 1 through 12, the month version is more deceptive, as the underlying logic and mathematics are better disguised.

As long as you can develop a list you can somehow associate with numbers, you can work out your own version of the Age Cards in this manner. Do you know the position of each letter in the alphabet? Then you can use the alphabet! 26 in binary is 11010, a 5-digit number, so you'd need 5 cards, and have the letters distributed appropriately among those 5 cards.

Similarly, as long as you have specific numbers attached to each item, you could use 52 playing cards, the 50 states (you'd have to have memorized an order such admission to the US), and much, much more! As you can start to see, if you have a big list memorized, you can create a unique and impressive routine.

UPDATE (Oct. 2, 2012): Numberphile just posted a video showing variations of the age cards that don't use binary! Watch and learn how this can be done with Fibonacci numbers and prime numbers, as well!

If you come up with your own variation of the Age Cards, I'd love to hear about it in the comments!


Secrets of Nim (Abacan 2)

Published on Thursday, September 27, 2012 in , , , ,

NIM is WIN upside down!In my previous post, I introduced a new version of Nim called Abacan, and introduced one strategy for winning it.

In this post, I'll teach the rest of the strategy you'll need to win Abacan every time. Before we do that, though, I'll go through a quick review.

In Abacan, you have 5 rows of beads. There are rows of 1, 3, 5, 7, and 9 beads. Players alternate taking turns, and can only move 1, 2, or 3 beads, all in the same row, on their turn. The last player to move a bead is the loser.

The strategy taught last time involves first reducing the number of beads in each row to a number from 0 to 3. In situations where the rows reduced to all 1s and 0s, or all 1s and 0s with a single 2 or 3, you simply move so that the reduced rows become an odd number of 1s. For example, rows of 1, 1, 4, 6, and 8 beads can be reduced to rows of 1, 1, 0, 2, and 0 beads. From there, it's a simple matter to see that the “2” row (reduced from 6) should have 1 bead removed from it, leaving the other player with 1, 1, 4, 5, and 8 beads (which reduces to 1, 1, 0, 1, and 0 beads - the equivalent of 3 piles of 1).

You should always reduce the numbers of beads first, then see if you can apply the strategy above first. This “odd 1s” approach is effectively the endgame strategy of multi-pile Misère Nim, as taught in Secrets of Nim (Part 3). In Abacan, however, it can be used well before the end.

What happens, though, if you reduce the rows of beads and find that there's more than one row containing the equivalent of 2 or 3 beads? In that case, you'll need a second strategy.

Strategy 2: In the Secrets of Nim (Part 2) and Visualizing Multi-Pile Nim posts, I discuss the importance of pairing up similar values using the binary. Thankfully, there won't be any need to delve into binary conversion in the Abacan version.

If you have a set of reduced rows containing multiple rows of 2 and/or 3 beads, you'll use the second strategy. In this strategy, you'll want to leave your opponent with rows that, when reduced, have any 1s, 2s, and 3s paired up with each other. You don't need to worry about pairing up 0s.

This can sound confusing, so let's use the starting arrangement of 1, 3, 5, 7, and 9 beads as an example. The first step is to reduce the arrangement from 1, 3, 5, 7, and 9 beads to its equivalent, which is rows of 1, 3, 1, 3, and 1 beads.

Since we can't use strategy 1 from the previous post, we move on to strategy 2, pairing up all the 1s, 2s, and 3s. With the reduced rows of 1, 3, 1, 3, and 1, it's not difficult to see how we can leave our opponent with a pair of 1s and a pair of 3s.

Taking 1 bead from the first row would make leave your opponent with 3, 5, 7, and 9 beads, which reduces to 3, 1, 3, and 1. You could also remove 1 from the 3rd row, leaving your opponent with rows of 1, 3, 4, 7, and 9, which reduces to rows 1, 3, 0, 3, 1. Another alternative is leaving your opponent with rows of 1, 3, 5, 7, and 8 beads, which reduces to rows of 1, 3, 1, 3, and 0 beads.

Now you know that it's to your advantage to move first, and that your first move should be sliding over 1 bead from the row of either 1, 5, or 9. Since 1, 5, and 9 are all equivalent after performing modulus 4, all 3 of these moves have the same effect.

Full approach: This is almost everything you need to know to win Abacan:

1) Mentally reduce the number of beads in each row by using mod 4.
2) If the reduced rows are all 1s and 0s, or all 1s and 0s with a single 2 or 3, remove enough beads to make the reduced rows consist of an odd number of 1s (strategy 1).
3) If using strategy 1 doesn't apply, use strategy 2. Remove beads in such a way that all any 1s, 2s, and 3s remaining are paired up.
Practice, of course, is the best way to do this. As mentioned in the previous post, Wolfram|Alpha can generate random arrangements for you. You can also double-check your mental reductions of each row on Wolfram|Alpha.

As you do this, however, you'll still find certain arrangements you won't be able to solve. This is why I said you know almost everything you need to know.

Recognizing unsolvable arrangements: By playing first and practicing and using the strategies in these Abacan posts, you'll always be able to leave your opponent with an arrangement that will guarantee they will lose.

However, if the other person goes first, and makes the right move, it is possible that they'll leave you with a position from which you don't have a good move. Assuming they're not wise to the strategy and just got lucky, you'll likely be able to recover on a later move. When you can't make an effective move, the best thing to do is simply slide 1 bead in the row with the most beads remaining.

You'll still need to recognize arrangements you can't win. Fortunately, they all fall into 1 of 4 easily recognizable patterns when reduced:
A) An odd number of 1s: 11559 = 11111, 11458 = 11010
B) All numbers paired: 13578 = 13130
C) 1/2/3: 13468 = 13020
D) 1/2/3 plus paired numbers: 13569 = 13121
Once you've practiced to the point where you can mentally reduce the numbers, apply the strategies to give your opponent losing patterns, and recognize when you're given a losing pattern, you're ready and equipped to win Abacan!

For those who would like to delve deeper into the mathematics behind this strategy, you might enjoy this discussion of the game in the BoardGameGeek forums.

Just like any version of Nim, you can play Abacan with any number of objects, such as playing cards or coins. If, however, you enjoyed these tutorials on how to win it, please consider helping this blog out by purchasing Abacan through the Grey Matters Store.


Secrets of Nim (Abacan 1)

Published on Sunday, September 23, 2012 in , , , , ,

NIM is WIN upside down!I just added a Nim game called Abacan to the Grey Matters Store. The main piece is a triangular shape with 25 beads in it, distributed as rows of 1, 3, 5, 7, and 9.

Abacan has rules that are slightly different from the previous versions of Nim discussed here on Grey Matters. To make this simpler, I'll review some previous versions of Nim before moving on to Abacan's peculiarities.

Let's start with single-pile Misère Nim (last player to move loses), using 25 objects, and each player is only allowed to remove 1, 2, or 3 objects on their turn. With your previous knowledge and a little practice, or the use of the Single-Pile Nim Strategy Calculator, you should have no problem determining that the other person should go first, and you need to hit the key numbers of 21, 17, 13, 9, 5, and 1.

For multi-pile Misère Nim, with rows of 1, 3, 5, 7, and 9, you'll need the binary understanding of multi-pile Nim taught here (and made easy to visualize here) and the endgame play knowledge of multi-pile Misère Nim taught here. The Multi-Pile Nim Strategy Calculator can help you learn the ideal strategy for this version.

If Abacan went by either of these rules, you'd be all set. However, as you'll see in the video review below, there's a subtle difference. You're only allowed to move 1, 2, or 3 beads, all from the same row, on each turn. The last player to move a bead loses.

Reducing the problem: The first step in Abacan strategy is, not surprisingly, treating each row as if it only had 0, 1, 2, or 3 beads in it, regardless of the actual number of beads in each row. This works out as follows:

Actual number of beads:   1  2  3  4  5  6  7  8  9
Reduced number of beads:  1  2  3  0  1  2  3  0  1
So, a row of 4 beads would be treated as if it had 0 beads, 5 beads would be treated as 1 bead, and so on. Effectively, you're simply subtracting the nearest multiple of 4 equal to or less than the number of beads in a given row. 9 becomes 1 because 9 - 8 (the nearest multiple of 4 less than or equal to 9) = 1.

The mathematical term for this process would be taking modulo 4, or “mod 4.” You can learn more about modular arithmetic via this fun and clear post at BetterExplained.

Abacan strategy 1: Toward the end of multi-pile Misère Nim, the ideal strategy is to leave your opponent an odd number of row, such as 1, 3, or 5, that each contain a single object. That will also work in Abacan, except that row you're treating as a single object may actually contain 1, 5, or even 9 beads! Let's examine various types of Abacan arrangements to see how we can do this.

All 4s and 8s: The simplest situation is when all the rows in play are multiples of 4. Let's say it's your turn, and there are rows of 4, 4, and 8 beads. You start by reducing the problem as discussed above, so the rows translate into rows of 0, 0, and 0 beads. In a normal game of Nim, every row consisting of 0 beads would be the end of the game, but not in Abacan!

If we slide 1 bead from the row of 8, leaving 7 beads, we see that 7 reduces to 3, so by moving 1, we effectively added 3 to one of the piles. This is also true of the moving 1 from the 4 pile, which also creates a row of 3.

Similarly, moving 2 beads from any row of 4 or 8 effectively adds 2 beads to that row, and moving 3 beads in a row of 4 or 8, effectively adds 1 to that row.

Remember strategy 1 above. If we can create an odd number of rows containing a single object (or a number can be reduced to a single object), we're a step closer to winning!

So, with our example rows of 4, 4, and 8 (again, this reduces to 0, 0, and 0), sliding 3 from the first row would give us 1, 4, and 8, which reduced to 1, 0, and 0 - an odd number of 1s!

Yes, you could slide 3 from the middle row instead, resulting in 4, 1, and 8 (reduces to 0, 1, and 0), with the same result. What happens if you slide 3 from the row of 8? You'd wind up with 4, 4, and 5, which reduces to 0, 0, and 1 - still resulting in the odd number of 1s needed to satisfy the strategy!

Mixed 1s, 4s, 5s, 8s, and 9s: If all the rows in play reduce to 1s (1, 5, and 9 beads) and 0s (4 and 8 beads), this strategy will still work. As an example, let's try rows of 1, 4, 5, and 8, which reduces to rows of 1, 0, 1, and 0 respectively.

The obvious move here would be to remove 1 bead from either the row of 1 or 5, resulting in either rows of 4, 5, and 8 (reduces to 0, 1, and 0) or 1, 4, 4, and 8 (reduces to 1, 0, 0, and 0), giving us an effective single row of 1.

There's another less obvious option here, as well. Remember that you can take rows reduced to 0, and effectively add 1 to them by removing 3 beads. If we take 3 from the row containing 4 beads, we're left with 1, 1, 5, and 8 beads, which reduces to 1, 1, 1, and 0 beads. There are now 3 rows which reduce to a single bead, which still satisfies strategy 1.

Removing 3 beads from the row of 8 would also have the same effect. The resulting piles would be 1, 4, 5, and 5, which reduce to 1, 0, 1, and 1, giving us another 3 piles of single objects. In other words, there's often more than one way to get to a safe position.

Mixed 1s, 4s, 5s, 8s, and 9s with a single 2, 3, 6, or 7: 2 and 6, as explained under Reducing the problem above, both reduce to 2, and 3 and 7 both reduce to 3. If there's just a single row consisting of 2, 3, 6, or 7 beads, and all the rest reduce to 1s and 0s, the same strategy will still work.

Let's assume that you turn leaves you with rows of 1, 4, 6, and 8 beads. As always, the first step is to reduce the problem, which gives us 1, 0, 2, and 0 beads. We're looking for a way to get to an odd number of 1s, and so the simplest solution is remove 2, leaving us with a single 1. So, we slide 2 from the row of 6, leaving rows of 1, 4, 4, and 8, which reduces to 1, 0, 0, and 0 beads!

You can practice your skill at reducing the problem by having Wolfram|Alpha generate random bead arrangements for you. If a randomly-generated arrangement happens to satisfy the situations I've already explained, you can also practice finding the best move. Keep in mind, Wolfram|Alpha may also generate an arrangement from which there is no winning move for you. This is valuable, as you'll start learning how to discern arrangements you can win from ones you can't.

If there are several rows, each with 2, 3, 6, and/or 7 beads, however, you'll need to learn and use a different strategy. We'll cover that strategy in the next post.


New Goodies in the Store!

Published on Thursday, September 20, 2012 in , , , ,

New Grey Matters Store itemsWith everyone settling into school and the holidays approaching, this seemed like a good time to start adding some new items to the Grey Matters store!

Looking around, I've found several new items that belong in the store. Some have already been added, and others will be added in the coming days.

Looking to improve your memory? Thomas Harrington's The Magic of Memory was the first of the new additions. It features an expanded version of the classic Major/Peg System of memorizing numbers that is very easy to use.

Timothy Hyde's Magazine Memory, which I reviewed back in 2009, is an addition to the store that is long overdue. This is probably the best resource for learning to perform the magazine memory feat which I've ever come across.

Last year, I reviewed Persi Diaconis' and Ron Graham's Magical Mathematics. I'm thrilled to have added this impressive work both as a hardcover and as an ebook.

I've long featured Danica McKellar's previous books in the Grey Matters store, so her newest book, Girls Get Curves: Geometry Takes Shape was a definite and early must-add.

Not all the new additions are books and ebooks, as you'll see. Fans of my Nim strategy posts will definitely enjoy Abacan. The rules are a mix of single-pile and multi-pile Nim, as you'll see in the review video. You may want to read up on those classic versions of Nim before trying to work out a strategy specifically for Abacan.

As I mentioned, this is only the start of many new items to come. If you enjoyed reading Grey Matters, and would like to help support future posts, consider buying items through the Grey Matters store.


Quick Snippets

Published on Sunday, September 16, 2012 in , , , , , , , ,

Luc Viatour's plasma lamp pictureIt's time for September's snippets!

In this edition of snippets, we're going to catch up with the latest news related to math and memory feats taught here on Grey Matters.

Google announced a new feature this week: The “Bacon number” search. You can now find out how close any actor is to having starred with Kevin Bacon in a movie. For example, to find out how close Will Smith is to Kevin Bacon by searching for the term Will Smith Bacon number, and you'll learn that Will Smith starred with Josh Brolin in Men In Black 3, and Josh Brolin appeared with Kevin Bacon in Hollow Man, so Will Smith's Bacon number would be 2.

Back in 2006, I created a memory feat based on the classic 6 Degrees of Kevin Bacon game. In the memory feat, you have people pick 2 names from names written on slips of paper, and then you recall the shortest connection between them. Since you need connections between actors who aren't Kevin Bacon, The Oracle of Bacon is actually a better tool for this feat.

• Also in the news this week, people in the San Francisco Bay Area got treated to Pi in the Sky on Sept. 12. Five skywriting planes filled the skies with the first 1000 digits of Pi. If you've practiced memorizing 400 digits of Pi, you should be able to recognize some of the locations in Pi in videos such as the one in the Pi in the Sky link, as well as this video (below) and this video.

• If you like my text-memorization app Verbatim (tutorial video here), and are looking for some classic poems to memorize, YouTube user 0adam0ant0fan0 has posted a series titled Famous Poems that are well worth your consideration. At this writing, they include The Raven (below), The Road Not Taken, Jabberwocky, From Paradise Lost, and If.

Mental math fans will enjoy TheClickMyVideos' YouTube channel. TheClickMyVideos has reposted several mental math videos made by others, including many of the long lost totalbreeze videos I mentioned back in Feb. 2011. They could be removed, so view them while you have the chance!


Amazing Mental Feat Performances

Published on Thursday, September 13, 2012 in , , , , , , ,

Anton Zellman performing his Day For Any Date FeatWhile I often teach how to perform mental feats, especially over in the Mental Gym, it's also important to get an idea of how professionals present them.

In the past, you've seen how performers including Maths Busking and Dr. Arthur Benjamin. Today's post features some other professional performers of mental feats.

First, we have Anton Zellman, who has made a good long living as a trade show performers, using his mind to entertain and educate prospective clients.

Because it's one of my favorite mental challenges, check out his Day For Any Date video. Watch it once just to see the performance, then watch it again to pick up on the finer details. For example, note that he only gives dates in two recent years (1990 and 1991 in the video). I have no doubt he could handle many more years, but this limitation makes the feat current, quicker to do, and easier to verify using only the 2 calendars hanging behind him.

Also, notice that he teaches how to do the effect. He's not performing a magic trick, and he's not trying give the impression that he's superior to you. He's simply showing that the boundaries of the human mind are much bigger than we may think. The multiplication table analogy is a wonderful tool to get the idea across of how something that seems hard can quickly become much easier with practice.

Among Anton Zellman's other videos are ones on remembering names and memorizing lists. There are also other videos on Zellman's own website. Watch these, keeping in mind that he's working in an environment where you often have only seconds to attract and keep the attention of attendees. If you don't engage it, that's a potential lost sale.

Scott Flansburg, also known as the Human Calculator, performs amazing mathematical feats for business meeting, schools, and, fortunately for us, the occasional TV appearance. Here's his appearance on a Discovery Channel program called More Than Human:

A few of the feats you see on there can be learned right here on Grey Matters, including the long division feat and the cube root feat.

Again, the attitude here is important. Just like with Anton Zellman, he's sharing, not showing off. Indeed, the most recent tweet (well, retweet) from Scott Flansburg at this writing says:

Now, in the video clip above, there is a large show-off component, because that's the nature of the show. Even in that clip, however, there's footage of him showing kids how they can do impressive feats on their own, such as adding a large column of 2-digit numbers in their head.

You can get a better idea of Scott Flansburg's performances in the video section of his site, as well as his YouTube channel.

Even if you don't do any of the specific feats shown in any of these videos, take the time to look through and enjoy them. Also, step back and examine just how the audience is brought into the performance and engaged. Often, the tips you discover in this way can help boost the performance of a wide variety of feats.


Card Colm

Published on Sunday, September 09, 2012 in , , , , ,

Photo of Colm MulcahyTo the math students of Spelman College, he's simply known as Colm Mulcahy. On the internet, however, he's better known as Card Colm.

It's a well-earned nickname, as he's been examining and creating principles of mathematical magic in his MAA column since October 2004!

If you're not familiar with Card Colm's column, you can find an index to all of the columns here. Up through June 2011, the columns are hosted on the Mathematical Association of America's (MAA) own site, and the columns after that are hosted on Colm Mulcahy's blogspot site.

Regular Grey Matters readers will recognize Colm from his G4G10 lecture I posted back in April. For those who need an introduction to his work, however, this video will give you a quick idea of his style:

At this writing, his newest column is titled Gilbreeath Shuuffling (pronounced gil-BREEATH SHOE-fling), and focuses on original routines for the classic Gilbreath Shuffle principle involving guessing the orders of suits.

The column before that, Something Old, Something True, Something Borrowed, Something New focuses on the poker deal taught in the first video in my 2010 Cheating At Poker post.

Besides his own original takes on each routine, one of the best things about Colm's work is that he is very detailed in his resources, so you know where to learn more. For example, there's plenty of Martin Gardner-inspired works among them.

As a matter of fact, between the search engine here on Grey Matters, Colm Mulcahy's columns, and their references, you should be able to grasp the basics of any principle, follow Colm's original variations, and then start creating your own variations! You can follow his latest routines on Twitter as @CardColm.

With the links above and a deck of cards in your hand, you're equipped to amaze yourself and other. Start exploring and have fun!


Scam School Match Puzzles

Published on Thursday, September 06, 2012 in , , ,

Scam School logoOur old friend Scam School is back for a visit!

This week's episode features some ingenious match puzzles in which Grey Readers readers may have an advantage.

Let's get right to the episode. Besides finding it below, it's available on the Scam School site, as well as YouTube. Stop the video, and try and figure out the first one:

If you've read through Grey Matters enough, the first puzzle is effectively one of the strategies you learn for playing Notakto (Part 2 here). See if you can use just that knowledge to help in solving it.

The 2nd puzzle uses just the matches (straws, actually, in the video). Try and give it time. As you'll see in the video, some people come agonizingly close to the solution without realizing it. If you like this type of puzzle, check out these other match puzzles at Puzzles.com, including the Three Squares puzzle.

As seen in the video, it is possible for people to figure out these puzzles, so be ready to pay up if you've bet on them not finding the solution. Take a closer look at the game of Notakto (linked above), if you're looking for something a little more surefire. You can also widen your arsenal of match puzzles with the Puzzles.com link, so have fun exploring!


FREE iOS Nim Games

Published on Sunday, September 02, 2012 in , , , ,

NIM is WIN upside down!I've written quite a few posts about the game of Nim, but practice is the real key.

In this post, we'll take a look at iOS Nim apps that you can download and use to practice for free on the go! Yes, there are free Nim apps for Android devices, but the variety is a little wider for iOS devices.

Most of the Nim apps you find are based on either single-pile Nim or multi-pile Nim.

For single-pile nim, try out Take Apples and NeonNim: The Subtraction Game. Multi-pile Nim can be practiced with the Stick Showdown and myQuickGame Free apps. If you want to switch between the two, there's the simply-named Nim Game and the aptly-named Appointed Win Game.

Back in June 2010, Scam School featured an episode of Nim without props. You'd wouldn't think you could turn a propless Nim game into an app, but it has been done! The calendar Nim variation is available as The Date Game. The other version is available as Race To 100.

There's also the ingenious tic-tac-toe variation of Nim known as Notakto. Notakto's winning approach isn't as easily picked up as some of the other variations of Nim. However, you can learn via my tutorials over in the Mental Gym, How to Play and Win Notakto and How To Play and Win Notakto: 3+ Boards.

There are a few variations of Nim out there, however, for which there are few (if any) ready-made tutorials. If you're up for the challenge, try a version with unfamiliar rules and strategies, and see how well you can do at working them out yourself. There's TacTix, in which you remove items from any contiguous columns OR rows, or the similar Triangular Nim, which throws diagonals into the mix, as well!

Probably the most challenging version of Nim is Hakenbush, which involves erasing lines, and eliminating any lines which are no longer connected to the base. Since there are endless possible shapes and color combinations to use, working out a general strategy is tougher than it may first appear!

I hope you find these versions of Nim fun and thought-provoking. If you have a favorite free Nim app, for any device, let me know about it in the comments!