Whattup I played on this website about 9 years ago under some other account name and I remember it being way harder to get long words. I remember all the names that I see now like mega word and lalatan always holding records for long words on 5x5s and I feel like I remember being the first or second person to snag a 16 letter word from the big boards. Was just wondering whether boards were always that easy and we were just not as good at finding words or if they added new ones over time. anyways its cool to be back on this website bc this was a part of my childhood I forgot about. Getting back into old hobbies like scrabble and boggle has been good for the mind these past few months. hope yall have merry christmas
Announcement
Collapse
No announcement yet.
Were boards always this prone to long words
Collapse
X
-
The boards are definitely more manufactured now and probably crafted so that a certain percentage are made to contain 20+ letter words. In reality, the probability of a board having a 25 letter word from 25 randomy placed letters is so minute as to be statistically improbable. Add to that the way letters have to connect in the game to form a word, the odds would be even smaller i.e. so low as to be impossible.
It's also not unusual for a board to have over 1,500 words contained within the grid nowadays. Personally, I don't find those games fun, but other players do. My approach has always been to play every game and not to cherry pick boards. When there are thousands of words to play in a single board, it's kind of like shooting fish in a barrel. By that I mean there are so many words that you could blindly type letters, hit enter and make a valid word.
The dynamics of the game have definitely changed. Franken-games and franken-words are not as enjoyable to me, but others seem to enjoy them.
- Likes 1
-
A new round of long words (in particular the huge-number words) did come in fairly recently. However, looking over past records, the longest word found in 5x5 went from 16 letters in November 2015 to 23 letters in December. I joined after that, so I always had longer words to try for.
By the way, one of the last changes was to add to the dictionary and seeding boards with a boatload of words. Some boards have 1100+ ultra-rare words.
- Likes 1
Comment
-
Words are called "ultra-rare" for a reason. It's not "shooting fish in a barrel" to get a bunch of high-scoring words. Often enough, even if you know a word you simply won't see it in front of you. I can't type nearly as fast as the top players, so even if I did see the words fast enough I couldn't enter them; so I have to pick high-scoring words and type them instead. And to be perfectly honest, I don't see the words as fast as the best players, either. The game is always a challenge, whatever your skill level. You play it for entertainment. If it's no longer entertaining, then don't play it. Really, it's a rule for life, isn't it? If something is no longer worth doing, find something else to do.
- Likes 1
Comment
-
Here's an example of what I mean, for a game I just played:
Game Results
YOUR STATS:Total points: 687
Total words: 61 (28 common,11 wide,5 rare,17 ultra rare)
Best word: QUANTIZERS (25 pts.) New record!
Longest word: QUANTIZERS (10 letters)
PUZZLE STATS:
Played: 4 times
Average Score: 443 points
Average Words: 57 words
High Score: 700 points by undisclosed
Most Words: 126 words by undisclosed
Best Word: SQUINTERS (20 pts) by secret
Longest Word: SANITIZERS (10 letters) by someone
- Likes 1
Comment
-
Originally posted by bwt1213 View PostWords are called "ultra-rare" for a reason. It's not "shooting fish in a barrel" to get a bunch of high-scoring words.
Comment
-
Very roughly, the number of possible 4 letter combinations in 5x5 are ~25x7^3 ~ 8,000 . This will be high for ignoring edge effects and double counting, but I think those would be a minority of cases for only 4 letters out of 25. The number of combinations should probably greatly increase further for the next few larger words (5-, 6-, 7- letter words). Sticking to the weaker case of 4-letter words for illustration, there are 500 four letter words and you have a possible say 5,000 combinations you could type in the puzzle, I think you would lose a significant amount of time getting poor results and not score very high (10% success rate, maybe?) Even if you add in some smarts of ensuring a vowel in every combination, I don't think it is at all clear that the probabilities would be with you (i.e., >50%) to get a word with some pseudorandom typing of connected letters that appear in the grid and hitting enter. At this point, it really isn't "blind" anymore, anyway. If, then, you want to apply more and more discretion in your choices of letters, we would all probably agree you will likely see more words than you can type.
By the way, I edited this about a dozen times and it still posts!Last edited by BoggleOtaku; 12-28-2023, 12:54 AM.
- Likes 1
Comment
-
Originally posted by Exquizzicalities View PostWhattup I played on this website about 9 years ago under some other account name and I remember it being way harder to get long words. I remember all the names that I see now like mega word and lalatan always holding records for long words on 5x5s and I feel like I remember being the first or second person to snag a 16 letter word from the big boards. Was just wondering whether boards were always that easy and we were just not as good at finding words or if they added new ones over time. anyways its cool to be back on this website bc this was a part of my childhood I forgot about. Getting back into old hobbies like scrabble and boggle has been good for the mind these past few months. hope yall have merry christmas
Amazing bwt1213 - QUANTIZERS is certainly a great find.
Comment
-
Originally posted by 2cute View Post
Amazing bwt1213 - QUANTIZERS is certainly a great find.
- Likes 1
Comment
-
I thought more about the question of just how many possible configurations of letter positions there are for a given word size on a 5x5 board. It is a really interesting problem! I gave a shot at simulating it using Octave. Here is my result achieved so far on a 5x5 board after a few hours run time (after which I gave up waiting for further results).
A second letter can be connected to by 8 different directions, giving 8 configurations, while a third letter can be connected to in 7 different directions to avoid using a given letter position more than once, giving 8x7=56 configurations. A single letter can be arrayed 5x5=25 times, while two letters when vertically or horizontally aligned can be arrayed 5x4=20 times and diagonally, 4x4=16 times.Word Length #Configs # after Arraying Example # words Probability 1 1 25 N/A N/A 2 8 144 N/A N/A 3 56 768 N/A N/A 4 368 3768 350 9% 5 2336 17280 416 2% 6 13608 74072 425 0.6% 7 71752 296360 297 0.1%
The simulation models all possible configurations, validates which are possible (i.e., do not reuse letter positions). Then, it finds the area covered by each configuration and, from that, how many times that shape can be arrayed across the playing field. For instance, 3 letters aligned vertically downward in a column fit in a 1x3 array and can be arrayed 5x3 times across the playing field, giving 15 possible word count for that configuration.
The example # words are from an actual game having a high (but not highest possible) word count. By comparison to the simulation, probabilities of randomly tracing a given connected pattern and getting non-duplicate words are low (10%) for 4-letter words and decreasing rapidly as word length increases (by ~5X smaller per added letter). (Duplicate words raise the probability, not incorporated here how many actual duplicated words there are in a given puzzle.)
I don't know if anyone is interested, but I had fun trying this simulation. Maybe it can help to get a feel for the complexities of the game. A comparison to 4x4 would be interesting as well.Last edited by BoggleOtaku; 01-04-2024, 02:45 AM.
- Likes 2
Comment
-
I played with this idea before, and you're right. It is fun. I think getting an exact count of n-letter word candidates is probably pretty tricky, what with accounting for board edges, dead ends (no next letter that hasn't been used), etc. At one time I thought about how I'd program trying to count all words in a board, given a dictionary of maybe a few hundred thousand words. I started with the (naive) idea of forming all these letter combinations & trying to look them up. As you noted, that number gets huge with n, and the number of dictionary lookups gets impractically huge for larger n. It's better to go the other way, going through the dictionary and trying to find the word in the board. (Checking on some board statistics beforehand can save lots of work.)
- Likes 1
Comment
-
Thanks for the Reply, Spike1007.
I know a lot of people have written solvers, but I haven't looked into how they have been implemented. Just thinking about it now, I think I would go in parallel. For instance, find all "A"s in the puzzle as 1st letter and then all neighbors as potential 2nd letter. Then, take all these possiblilities and match against starting 1st two letters of all words in the dictionary, then downselect both the promising paths in the puzzle and the word pool in the dictionary and move on to the third letter. After all possible "A" words found, move on to B. It seems like substantial narrowing down of possibilities each pass from both sides.
I sped up the simulation by noticing that all configurations for 2+ letter words are evenly divisible by 8. A factor 4x is from C4 symmetry: for example, for every configuration that starts by proceeding downward, there is a symmetric by rotation configuration that starts toward the right, the left, or upwards, and same for the diagonals. So I only have to calculate configurations that start moving downward on in one diagonal direction and then multiply by 4. I'm not sure why there remains another factor of 2X. Maybe there is still some other symmetry that could be taken advantage of. For three letters, there are reflection symmetries that could be taken advantage of, but only in cases where the three letters are not along a straight line, where the reflection would be itself and double counting would result.
Here's the configurations I've simulated. The smaller boards are good checks for the simulation and interesting because they indicate edge effects we would see if I could do simulation out to longest possible words for 4x4 and 5x5. For an NxN board, as soon as there are N(N-1) +1 letters, any valid configuration can only be arrayed 1x1 times, so #Configs = #total when arrayed. The 3x3 board is interesting because it is a bit like a screen lock on a computer where you swipe in some continous pattern on a keypad to gain access.
1x1:
1 letter word only, 1 configuration, total of 1 when arrayed. (The trivial case)
2x2:3x3:Word Length #Configs Total # after Arraying 1 1 4 2 8 16 3 24 24 4 24 24 4x4 (up to 11 letter words)Word Length #Configs Total # after Arraying 1 1 9 2 8 49 3 56 160 4 264 496 5 888 1208 6 2048 2240 7 2984 2984 8 2384 2384 9 784 784 5x5 (up to 8 letter words)Word Length # Configs Total # when arrayed 1 1 16 2 8 100 3 56 408 4 368 1764 5 2016 6712 6 9296 22672 7 36560 68272 8 122648 183472 9 346920 436984 10 811152 905776 11 1532952 1594648 Word Length #Configs Total # after arraying 1 1 25 2 8 169 3 56 768 4 368 3768 5 2336 17280 6 13608 74072 7 71752 296360 8 344088 1110000
- Likes 2
Comment
-
Here's a stab at estimating the probability of getting big word (16 letters on 4x4 board) from purely random letters in each position.
Assumptions:
1. There are ~5000 16-letter words (from WordUnscrambler site).
2. The typical 16 letter word has 3 letters that are repeated once and two letters repeated three times (wild guess). For example, a letter repeated 3 times would have 3 available for first position needed, 2 for the 2nd and 1 for the 3rd, becoming 3! in the probability equation for arranging on the grid. Also affected would be probability of choosing the right letters, as to begin with there would only be 11 letters we are looking for out of 26 because of the duplications (this is the part I am most unsure of how to incorporate).
3. For 16 letters that uniquely spell one word, there are 343184 different configurations out of 16! = 20,922,789,888,000 possible arrangements of the letters on a 4x4 grid (16 choices for first position, leaving 15 for the 2nd, 14 for the 3rd, etc.) that can actually be traced out to spell the word (based on my table below).
4. Since purely random selection of letters, we are drawing from a pool of 26 letters 16 times. The first time we have 16 shots of getting one of the letters we want, then 15, then 14, etc.
So, I'm thinking this all combines to give a probability of
p = (# possible 16-letter words) x (probability to choose the right letters) x (probability to arrange OK on grid)
= (5000) x (11/26)^3 (10/26)^2 (9/26)^2 (8/26)^2 (7!/26^7) x ( 343184 x 3!2!2!2!/16! )
= 5000 x 1/500,000,000 x 1/1,000,000
probability = 1 out of 100 billion
Probabilities of choosing the right letters and arranging on the board trade off with each other depending on how many duplicated letters there are in a given 16-letter word.
Since actual chances of getting a 16-letter word board are now more like 1 out of 1000, the probabilities have been improved by a factor of 100 million! Thank you, admin!
------------
I sped up my simulation by not calculating reflected configurations for a factor of ~2X and reducing coordinate arrays from 64-bit double to 8-bit integer for a factor of 8x improvement in memory allocation. Here's the final result for 4x4 (still can't complete 5x5!)
4x4 board----Word Length # Configs Total # after arraying 1 1 16 2 8 100 3 56 408 4 368 1764 5 2016 6712 6 9296 22672 7 36560 68272 8 122648 183472 9 346920 436984 10 811152 905776 11 1532952 1594648 12 2291552 2310264 13 2644520 2644520 14 2250192 2250192 15 1260672 1260672 16 343184 343184
Finally, one further thought on a solver: Since the dictionary is conveniently arranged alphabetically, it could be indexed to some further level, say first 2 and first 3 letters of each word. This would be huge speed improvement over just matching first 2 letters to narrow down the dictionary based on first letter pairs in the puzzle.Last edited by BoggleOtaku; 01-06-2024, 02:56 AM.
- Likes 2
Comment
Comment