# II.4: Hackenstrings and 0.999…vs.1

A significant portion of this page was inspired by Dr. A. N. Walker’s Hackenstrings, And The 0.999… ?= 1 Faq, but the page seems to be gone. It’s linked as a reference at this much shorter page, but Dr. Walker’s treatment was very well done and I do not have a copy of it.

Basic definitions
Now that we have an order on some positions (at least the integer ones), I think we can fill in the “number line” a little bit. To make this less abstract, we can look at an actual game instead of merely Conway representations of positions. “Hackenstrings” is a game vaguely similar to Nim in which players hack away at strings. For now, an individual string will be comprised of “L”s and “R”s. On their turn, Left gets to remove an L (and everything to the right of it) and Right gets to remove an R (and everything to the right of it). The integers we’ve already discussed pop up as positions in Hackenstrings: “”=0, “LLL”=3, “RR”=-2, etc. However, there are lots of new positions like “LR”, “RLL”, and even non-short games like “LLLL…”.

What is “LR”?
Let’s consider the game “LR”. Right can remove the R, leaving “L”=1, and Left can remove the L (and the R to the right of it) leaving “”=0. Therefore, “LR”={0|1}. Our order allows us to say more about this position, though. Since Left can win by taking the “L”, “LR”>0. How does {0|1} compare to 1? In {0|1}-1=”LR”+”R”, Right can move to “L”+”R”=1-1=0 (so Left loses), and Left can move to “R”=-1 (which Left still loses). Thus, {0|1}-1=”LR”+”R” is an R-position and 0<{0|1}<1. But is it almost 0 or almost 1 or neither?

We can get an even better understanding of the value of “LR” by comparing “LR”+”LR” with “L”=1. We must play “LR”+”LR”-“L”=”LR”+”LR”+”R”. If Left goes first, they move to “LR”+”R”<0 and Left loses. If Right goes first and takes the lone “R” then “LR”+”LR”>0 remains and Right loses. If Right goes first another R, then they leave “LR”+”L”+”R”=”LR” and Right loses. In any case, the person who goes first loses (if the other player plays perfectly), so this is a P-position! “LR”+”LR”-“L”=0, so “LR”+”LR”=”L”=1. In summary, “LR” is a game between 0 and 1 such that “LR”+”LR”=1. These are pretty good reasons to call “LR” “1/2”, and that’s exactly what people do. As an aside, “LR”+* is also between 0 and 1 and (“LR”+*)+(“LR”+*)=1, but there are other reasons to give “LR” the title as opposed to “LR”+*. (Note: At this point, “1/2” is just a name; a priori there’s no concept of multiplication/division of games.)

Other fractions?
“LR”=1/2 and similarly, “RL”=-1/2. but what happens with more complicated strings? Consider “LRR”; it’s clearly positive since Left wins even if Right goes first, and it’s less than 1 since “LRR”+”R” is a win for Right. If you play “LRR”+”LRR”-“LR”=”LRR”+”LRR”+”RL”, then if Right goes first and they leave “LRR”+”LR”+”RL”=”LRR”>0, they lose. If Right goes first and they leave “LRR”+”LRR”>0, they lose. If Left goes first and they leave “LRR”+”RL” then Right can respond with “LR”+”RL”=0 and they lose. If Left goes first and they leave “LRR”+”LRR”+”R” then Right can move to “LRR”+”LR”+”R” and Left’s only moves are to “LR”+”R” (a win for Right) or “LRR”+”R” (also a win for Right). In all cases, the second player has a winning strategy, so “LRR”+”LRR”=”LR”. Thus, we can/should call “LRR” “1/4”. This pattern continues: “LRRR”=1/8, “LRRRR”=1/16, etc.

Let’s look at another family of positions, starting with “LLR”={0,1|2}. Left can take the middle L which the R is sitting on, but the bottom L intuitively seems independent of that top “LR” action. For this reason, I’d guess that “LLR”=”L”+”LR”. To test this, play “LLR”+”R”+”RL”. If Right goes first, they might take the first R, leaving “LL”+”R”+”RL”=”L”+”RL”>0, but they’d lose. If they take the third R, they’d leave “LLR”+”R” and Left could respond with “L”+”R”=0 and Right would still lose. “LLR”+”RL”>”LLR”+”R”, so Left would win if Right started by taking the other R, too. If Left goes first, they can move to “LLR”+”R”+”R”<“LL”+”R”+”R”=0, but they’d lose. They could also move to “L”+”R”+”RL”=”RL”<0, but they’d still lose. And “R”+”RL” is even worse! “LLR”+”R”+”RL” is a P-position, so “LLR”=”L”+”LR”=1+1/2. Similarly, “LLLR”=2+1/2, etc.

Finally, let’s look at one more family before I make a general claim. Consider “LRL”; this is potentially better for Left than “LR”, but worse than “L”, so it’s somewhere in between 1/2 and 1. It actually turns out to be 3/4. We can prove that “LRL”+”LRR”=”L” by playing “LRL”+”LRR”+”R”. It’s easy to see (from previously known facts) that all of Left’s starting moves result in negative games, so this is ≤0. If Right takes the first R, they leave “L”+”LRR”+”R”=”LRR”>0. If Right takes the third R (the second is worse) then Left moves to 0 by leaving “LR”+”LR”+”R”. If Right takes the last R, Left leaves “LRL”>0. There will be a way to avoid these case analyses with a general theorem later. “LRLL” is even better for left, and equals 7/8, etc.

General Hackenstring Values
Since the negative of a basic Hackenstring position switches all of the Ls and Rs, and we know “”=0, it suffices to describe the values of strings starting with L. We will focus only on finite strings for now. A string consisting only of n Ls just has the integer value n. Otherwise, suppose there are n Ls before the first R. Then the value is n-(1/2)+a1(1/4)+a2(1/8)+a3(1/16)… where ai=1 if the ith character after the first R is L, -1 if it’s R, and 0 if the string has no character i spaces after the first R. In this way, it’s relatively easy to convert finite binary representations of numbers (the so-called dyadic fractions) to Hackenstrings: 10.1100101(2)=”LLLRLLRRLR”, etc. The book Lessons in Play brought to my attention an equivalent method due to Berlekamp: the value is (n-1)+the binary number you get by reading “L”s as 1s and “R”s as 0s after the first “LR”, with a 1 appended. This can be seen in: 10.1100101(2)=”LLLRLLRRLR“. I make these claims without proof because they’d be really annoying to prove without some better general theorems we will cover later.

To help you visualize this pattern, here’s a tree diagram (except here they’re using ↿s instead of Ls and ⇃s instead of Rs):

0.999…=1?
We can use this idea to set up a funny comparison to the classic “.999…=1” argument. Note that “LR”=.1(2), “LRL”=.11(2), “LRLL”=.111(2), “LRLLL”=.1111(2), etc. Now consider the game “LRLLLL…” where Right’s only move is to “L” and Left can move to any finite “LRLLL…L” (or to 0). However, this game (which seems like it might have been .1111…(2)=1) is not equal to “L”. To see this, we have to play “LRLLL…”+”R”. If Left moves first, they either move to “R” and lose, or to something of the form “LRLLL…L”+”R” and Right moves to “L”+”R”=0 for the win. But if Right moves first, they just move to “L”+”R”=0 to win immediately. This is an R-position, so “LRLLL…”<“L”!

People often say things like “If .999… were less than 1, then their difference would be nonzero, but that’s ridiculous.” In this case, we know that “L”-“LRLLL…”>0, but there’s actually more that can be said. I claim that “L”-“LRLLL…”=”LRRRR…” We have to check that “LRRRR…”+”LRLLL…”+”R” is a P-position. Since “LR”=1/2 and “LRR”=1/4, etc. and “LRL”=3/4 and “LRLL”=7/8, etc. If Left moves to, say, “LRRRR…”+”LRLL”+”R” then Right moves to “LRRR”+”LRLL”+”R”=1/8+7/8-1=0 and Left loses. If Left takes a whole string then Right moves to “L”+”R”=0 and Left loses. If Right goes first, then they can’t leave “L”+”R”+”L…”=”L…”>0, but if they move to something like “LRR”+”LRLLL…”+”R” then Left can move to “LRR”+”LRL”+”R”=1/4+3/4-1=0 and Right loses. Therefore, the difference between “LRLLL…” and “L” is the infinitesimal “LRRR…”.

Exercises

• How do things like * and ±1 compare to fractional values like 1/2 or 1+1/2?
• The thing that should be 0.111…(2)=1, wasn’t. What about the thing that should be 1/3=0.010101…(2), “LRRLRLRLRLR…”? Does “LRRLRLR…”+”LRRLRLR…”+”LRRLRLR…”=”L”?