Word Puzzle: Find The Maximum Words (+Easter Egg)
$begingroup$
Similar to the heart of this, but with different rules, I have developed a new word puzzle.
Partial answers which answer part of it correctly will be upvoted!
Your objective is to find the maximum number of possible words which are 2 or more letters long.
The puzzle is designed using all of the consonants once each (except for z) of the English alphabet, excluding the vowels;
and then every other letter, the five primary vowels A, E, I, O, U are inserted into each line.
Your objective is also to find the most efficient equation for discovering these maximum number of words (rather than guessing),
And additionally, these words are to be found including every direction: Up, Down, Left, Right, and Each Diagonal (but not in circles).
Rules:
The words must contain a vowel. For example, BGN is not counted as an acceptable "word".
The words do not have to be real English words. For example, a
possible "word" is BAC.Zig zag is included, but not back toward itself. For example, BAAC works and BAAJ works but BAAG does not work. Alternatively, CAAG works and CAAB works but CAAJ does not.
SUMMARY OF OBJECTIVES:
- Maximum # of words 2+ letters long
- Most efficient equation for finding them
- Words in every direction (↑,↓,←,→,↗,↙,↘,↖)
RULES SUMMARY:
- Each word must contain a vowel
- Words do not have to be real words
- Zig zag included, away from itself
MOREOVER,
Hidden within the puzzle or results contains a mystery which may or may not be related to a hint to which could help you solve this or another mystery contained within my user page. This paragraph is also written in riddle to obfuscate the result. Even if you cannot uncover the answer, you may uncover the mystery.
riddle logical-deduction word pattern calculation-puzzle
asked 4 hours ago
RiddlerRiddler
68216
$endgroup$
|
show 1 more comment
$begingroup$
Similar to the heart of this, but with different rules, I have developed a new word puzzle.
Partial answers which answer part of it correctly will be upvoted!
Your objective is to find the maximum number of possible words which are 2 or more letters long.
The puzzle is designed using all of the consonants once each (except for z) of the English alphabet, excluding the vowels;
and then every other letter, the five primary vowels A, E, I, O, U are inserted into each line.
Your objective is also to find the most efficient equation for discovering these maximum number of words (rather than guessing),
And additionally, these words are to be found including every direction: Up, Down, Left, Right, and Each Diagonal (but not in circles).
Rules:
The words must contain a vowel. For example, BGN is not counted as an acceptable "word".
The words do not have to be real English words. For example, a
possible "word" is BAC.Zig zag is included, but not back toward itself. For example, BAAC works and BAAJ works but BAAG does not work. Alternatively, CAAG works and CAAB works but CAAJ does not.
SUMMARY OF OBJECTIVES:
- Maximum # of words 2+ letters long
- Most efficient equation for finding them
- Words in every direction (↑,↓,←,→,↗,↙,↘,↖)
RULES SUMMARY:
- Each word must contain a vowel
- Words do not have to be real words
- Zig zag included, away from itself
MOREOVER,
Hidden within the puzzle or results contains a mystery which may or may not be related to a hint to which could help you solve this or another mystery contained within my user page. This paragraph is also written in riddle to obfuscate the result. Even if you cannot uncover the answer, you may uncover the mystery.
riddle logical-deduction word pattern calculation-puzzle
asked 4 hours ago
RiddlerRiddler
68216
$endgroup$
$begingroup$
The "i" on the first column of the last row, is that supposed to be a "t"?
$endgroup$
– North
3 hours ago
$begingroup$
Also are words containing only vowels acceptable?
$endgroup$
– North
3 hours ago
$begingroup$
@North Yes, words with only vowels are acceptable :)
$endgroup$
– Riddler
3 hours ago
$begingroup$
HINT: There are no mistakes in the puzzle, which is also a riddle.
$endgroup$
– Riddler
3 hours ago
$begingroup$
aaaahhhh I just noticed that the i and the t are swapped goddamn it i have to recalculate everything screw it
$endgroup$
– North
1 hour ago
|
show 1 more comment
$begingroup$
Similar to the heart of this, but with different rules, I have developed a new word puzzle.
Partial answers which answer part of it correctly will be upvoted!
Your objective is to find the maximum number of possible words which are 2 or more letters long.
The puzzle is designed using all of the consonants once each (except for z) of the English alphabet, excluding the vowels;
and then every other letter, the five primary vowels A, E, I, O, U are inserted into each line.
Your objective is also to find the most efficient equation for discovering these maximum number of words (rather than guessing),
And additionally, these words are to be found including every direction: Up, Down, Left, Right, and Each Diagonal (but not in circles).
Rules:
The words must contain a vowel. For example, BGN is not counted as an acceptable "word".
The words do not have to be real English words. For example, a
possible "word" is BAC.Zig zag is included, but not back toward itself. For example, BAAC works and BAAJ works but BAAG does not work. Alternatively, CAAG works and CAAB works but CAAJ does not.
SUMMARY OF OBJECTIVES:
- Maximum # of words 2+ letters long
- Most efficient equation for finding them
- Words in every direction (↑,↓,←,→,↗,↙,↘,↖)
RULES SUMMARY:
- Each word must contain a vowel
- Words do not have to be real words
- Zig zag included, away from itself
MOREOVER,
Hidden within the puzzle or results contains a mystery which may or may not be related to a hint to which could help you solve this or another mystery contained within my user page. This paragraph is also written in riddle to obfuscate the result. Even if you cannot uncover the answer, you may uncover the mystery.
riddle logical-deduction word pattern calculation-puzzle
asked 4 hours ago
RiddlerRiddler
68216
$endgroup$
Similar to the heart of this, but with different rules, I have developed a new word puzzle.
Partial answers which answer part of it correctly will be upvoted!
Your objective is to find the maximum number of possible words which are 2 or more letters long.
The puzzle is designed using all of the consonants once each (except for z) of the English alphabet, excluding the vowels;
and then every other letter, the five primary vowels A, E, I, O, U are inserted into each line.
Your objective is also to find the most efficient equation for discovering these maximum number of words (rather than guessing),
And additionally, these words are to be found including every direction: Up, Down, Left, Right, and Each Diagonal (but not in circles).
Rules:
The words must contain a vowel. For example, BGN is not counted as an acceptable "word".
The words do not have to be real English words. For example, a
possible "word" is BAC.Zig zag is included, but not back toward itself. For example, BAAC works and BAAJ works but BAAG does not work. Alternatively, CAAG works and CAAB works but CAAJ does not.
SUMMARY OF OBJECTIVES:
- Maximum # of words 2+ letters long
- Most efficient equation for finding them
- Words in every direction (↑,↓,←,→,↗,↙,↘,↖)
RULES SUMMARY:
- Each word must contain a vowel
- Words do not have to be real words
- Zig zag included, away from itself
MOREOVER,
Hidden within the puzzle or results contains a mystery which may or may not be related to a hint to which could help you solve this or another mystery contained within my user page. This paragraph is also written in riddle to obfuscate the result. Even if you cannot uncover the answer, you may uncover the mystery.
riddle logical-deduction word pattern calculation-puzzle
riddle logical-deduction word pattern calculation-puzzle
asked 4 hours ago
RiddlerRiddler
68216
asked 4 hours ago
RiddlerRiddler
68216
edited 2 hours ago
Riddler
asked 4 hours ago
RiddlerRiddler
68216
asked 4 hours ago
RiddlerRiddler
68216
asked 4 hours ago
RiddlerRiddler
68216
68216
$begingroup$
The "i" on the first column of the last row, is that supposed to be a "t"?
$endgroup$
– North
3 hours ago
$begingroup$
Also are words containing only vowels acceptable?
$endgroup$
– North
3 hours ago
$begingroup$
@North Yes, words with only vowels are acceptable :)
$endgroup$
– Riddler
3 hours ago
$begingroup$
HINT: There are no mistakes in the puzzle, which is also a riddle.
$endgroup$
– Riddler
3 hours ago
$begingroup$
aaaahhhh I just noticed that the i and the t are swapped goddamn it i have to recalculate everything screw it
$endgroup$
– North
1 hour ago
|
show 1 more comment
$begingroup$
The "i" on the first column of the last row, is that supposed to be a "t"?
$endgroup$
– North
3 hours ago
$begingroup$
Also are words containing only vowels acceptable?
$endgroup$
– North
3 hours ago
$begingroup$
@North Yes, words with only vowels are acceptable :)
$endgroup$
– Riddler
3 hours ago
$begingroup$
HINT: There are no mistakes in the puzzle, which is also a riddle.
$endgroup$
– Riddler
3 hours ago
$begingroup$
aaaahhhh I just noticed that the i and the t are swapped goddamn it i have to recalculate everything screw it
$endgroup$
– North
1 hour ago
$begingroup$
The "i" on the first column of the last row, is that supposed to be a "t"?
$endgroup$
– North
3 hours ago
$begingroup$
The "i" on the first column of the last row, is that supposed to be a "t"?
$endgroup$
– North
3 hours ago
$begingroup$
Also are words containing only vowels acceptable?
$endgroup$
– North
3 hours ago
$begingroup$
Also are words containing only vowels acceptable?
$endgroup$
– North
3 hours ago
$begingroup$
@North Yes, words with only vowels are acceptable :)
$endgroup$
– Riddler
3 hours ago
$begingroup$
@North Yes, words with only vowels are acceptable :)
$endgroup$
– Riddler
3 hours ago
$begingroup$
HINT: There are no mistakes in the puzzle, which is also a riddle.
$endgroup$
– Riddler
3 hours ago
$begingroup$
HINT: There are no mistakes in the puzzle, which is also a riddle.
$endgroup$
– Riddler
3 hours ago
$begingroup$
aaaahhhh I just noticed that the i and the t are swapped goddamn it i have to recalculate everything screw it
$endgroup$
– North
1 hour ago
$begingroup$
aaaahhhh I just noticed that the i and the t are swapped goddamn it i have to recalculate everything screw it
$endgroup$
– North
1 hour ago
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Well accounting for just rows
There are a total of 90 in just one row. This is because you can fit 9 two letter words going in one direction, and 9 two letter words in the other direction. This continues in a mathematical manner where the number of words become shorter. So next, there are eight three letter words in one direction, seven four letter, etc. You get an equation of something like this: ([9*2]+[8*2]+[7*2]+[6*2]+[5*2]+[4*2]+[3*2]+[2*2]+[1*2]= 90). Now, I don't know the mathematical equation to express that is, since my math isn't very great. There are four rows, so multiply 90 to get 360 words in just columns alone. The shortcut to this is basically subtracting the length of the entire row by one and then multiplying the results by two. Then subtract the total length by an extra 1 (i.e., now two) take that number, and multiply it by two. You multiply it by two because the letters go both ways. 360 for rows.
An then now for columns:
The all vowel columns are easy. it's only three potential words (e.g. aa, aaa,and aaaa). Multiply 3 by the number of columns with that exact circumstance to get 15. Now, tackling that first column, we can figure this out by doing the same principal we applied in the first row, so (4-1=3 multiplied by 2 which is 6 words. BGNI, GNI, NI, IN, ING, INBG. Total number of words in column is 21
Finally for the diagonals.
It's the same rule-set for the rows. Total of 8 "words" in two length diagonals, 24 "words" in three-length diagonals, and 168 "word" in the four-length diagonals.
For total sum of all possible numbers it is:
200+21+90 which is equal to 311 words.
Wait... except it's not
Shout out to @DanielMathias for the comment!
answered 3 hours ago
NorthNorth
1,9631733
$endgroup$
2
$begingroup$
The sum of numbers $1$ to $n$ is $frac{n(n+1)}{2}$, going forward and backward doubles the sum for a single row: $n(n+1)=9*10=90$
$endgroup$
– Daniel Mathias
3 hours ago
$begingroup$
Great job, it's a good start.
$endgroup$
– Riddler
2 hours ago
add a comment |
Your Answer
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1 Answer
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active
oldest
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1 Answer
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active
oldest
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votes
active
oldest
votes
$begingroup$
Well accounting for just rows
There are a total of 90 in just one row. This is because you can fit 9 two letter words going in one direction, and 9 two letter words in the other direction. This continues in a mathematical manner where the number of words become shorter. So next, there are eight three letter words in one direction, seven four letter, etc. You get an equation of something like this: ([9*2]+[8*2]+[7*2]+[6*2]+[5*2]+[4*2]+[3*2]+[2*2]+[1*2]= 90). Now, I don't know the mathematical equation to express that is, since my math isn't very great. There are four rows, so multiply 90 to get 360 words in just columns alone. The shortcut to this is basically subtracting the length of the entire row by one and then multiplying the results by two. Then subtract the total length by an extra 1 (i.e., now two) take that number, and multiply it by two. You multiply it by two because the letters go both ways. 360 for rows.
An then now for columns:
The all vowel columns are easy. it's only three potential words (e.g. aa, aaa,and aaaa). Multiply 3 by the number of columns with that exact circumstance to get 15. Now, tackling that first column, we can figure this out by doing the same principal we applied in the first row, so (4-1=3 multiplied by 2 which is 6 words. BGNI, GNI, NI, IN, ING, INBG. Total number of words in column is 21
Finally for the diagonals.
It's the same rule-set for the rows. Total of 8 "words" in two length diagonals, 24 "words" in three-length diagonals, and 168 "word" in the four-length diagonals.
For total sum of all possible numbers it is:
200+21+90 which is equal to 311 words.
Wait... except it's not
Shout out to @DanielMathias for the comment!
answered 3 hours ago
NorthNorth
1,9631733
$endgroup$
2
$begingroup$
The sum of numbers $1$ to $n$ is $frac{n(n+1)}{2}$, going forward and backward doubles the sum for a single row: $n(n+1)=9*10=90$
$endgroup$
– Daniel Mathias
3 hours ago
$begingroup$
Great job, it's a good start.
$endgroup$
– Riddler
2 hours ago
add a comment |
$begingroup$
Well accounting for just rows
There are a total of 90 in just one row. This is because you can fit 9 two letter words going in one direction, and 9 two letter words in the other direction. This continues in a mathematical manner where the number of words become shorter. So next, there are eight three letter words in one direction, seven four letter, etc. You get an equation of something like this: ([9*2]+[8*2]+[7*2]+[6*2]+[5*2]+[4*2]+[3*2]+[2*2]+[1*2]= 90). Now, I don't know the mathematical equation to express that is, since my math isn't very great. There are four rows, so multiply 90 to get 360 words in just columns alone. The shortcut to this is basically subtracting the length of the entire row by one and then multiplying the results by two. Then subtract the total length by an extra 1 (i.e., now two) take that number, and multiply it by two. You multiply it by two because the letters go both ways. 360 for rows.
An then now for columns:
The all vowel columns are easy. it's only three potential words (e.g. aa, aaa,and aaaa). Multiply 3 by the number of columns with that exact circumstance to get 15. Now, tackling that first column, we can figure this out by doing the same principal we applied in the first row, so (4-1=3 multiplied by 2 which is 6 words. BGNI, GNI, NI, IN, ING, INBG. Total number of words in column is 21
Finally for the diagonals.
It's the same rule-set for the rows. Total of 8 "words" in two length diagonals, 24 "words" in three-length diagonals, and 168 "word" in the four-length diagonals.
For total sum of all possible numbers it is:
200+21+90 which is equal to 311 words.
Wait... except it's not
Shout out to @DanielMathias for the comment!
answered 3 hours ago
NorthNorth
1,9631733
$endgroup$
2
$begingroup$
The sum of numbers $1$ to $n$ is $frac{n(n+1)}{2}$, going forward and backward doubles the sum for a single row: $n(n+1)=9*10=90$
$endgroup$
– Daniel Mathias
3 hours ago
$begingroup$
Great job, it's a good start.
$endgroup$
– Riddler
2 hours ago
add a comment |
$begingroup$
Well accounting for just rows
There are a total of 90 in just one row. This is because you can fit 9 two letter words going in one direction, and 9 two letter words in the other direction. This continues in a mathematical manner where the number of words become shorter. So next, there are eight three letter words in one direction, seven four letter, etc. You get an equation of something like this: ([9*2]+[8*2]+[7*2]+[6*2]+[5*2]+[4*2]+[3*2]+[2*2]+[1*2]= 90). Now, I don't know the mathematical equation to express that is, since my math isn't very great. There are four rows, so multiply 90 to get 360 words in just columns alone. The shortcut to this is basically subtracting the length of the entire row by one and then multiplying the results by two. Then subtract the total length by an extra 1 (i.e., now two) take that number, and multiply it by two. You multiply it by two because the letters go both ways. 360 for rows.
An then now for columns:
The all vowel columns are easy. it's only three potential words (e.g. aa, aaa,and aaaa). Multiply 3 by the number of columns with that exact circumstance to get 15. Now, tackling that first column, we can figure this out by doing the same principal we applied in the first row, so (4-1=3 multiplied by 2 which is 6 words. BGNI, GNI, NI, IN, ING, INBG. Total number of words in column is 21
Finally for the diagonals.
It's the same rule-set for the rows. Total of 8 "words" in two length diagonals, 24 "words" in three-length diagonals, and 168 "word" in the four-length diagonals.
For total sum of all possible numbers it is:
200+21+90 which is equal to 311 words.
Wait... except it's not
Shout out to @DanielMathias for the comment!
answered 3 hours ago
NorthNorth
1,9631733
$endgroup$
Well accounting for just rows
There are a total of 90 in just one row. This is because you can fit 9 two letter words going in one direction, and 9 two letter words in the other direction. This continues in a mathematical manner where the number of words become shorter. So next, there are eight three letter words in one direction, seven four letter, etc. You get an equation of something like this: ([9*2]+[8*2]+[7*2]+[6*2]+[5*2]+[4*2]+[3*2]+[2*2]+[1*2]= 90). Now, I don't know the mathematical equation to express that is, since my math isn't very great. There are four rows, so multiply 90 to get 360 words in just columns alone. The shortcut to this is basically subtracting the length of the entire row by one and then multiplying the results by two. Then subtract the total length by an extra 1 (i.e., now two) take that number, and multiply it by two. You multiply it by two because the letters go both ways. 360 for rows.
An then now for columns:
The all vowel columns are easy. it's only three potential words (e.g. aa, aaa,and aaaa). Multiply 3 by the number of columns with that exact circumstance to get 15. Now, tackling that first column, we can figure this out by doing the same principal we applied in the first row, so (4-1=3 multiplied by 2 which is 6 words. BGNI, GNI, NI, IN, ING, INBG. Total number of words in column is 21
Finally for the diagonals.
It's the same rule-set for the rows. Total of 8 "words" in two length diagonals, 24 "words" in three-length diagonals, and 168 "word" in the four-length diagonals.
For total sum of all possible numbers it is:
200+21+90 which is equal to 311 words.
Wait... except it's not
Shout out to @DanielMathias for the comment!
answered 3 hours ago
NorthNorth
1,9631733
edited 1 hour ago
answered 3 hours ago
NorthNorth
1,9631733
answered 3 hours ago
NorthNorth
1,9631733
answered 3 hours ago
NorthNorth
1,9631733
1,9631733
2
$begingroup$
The sum of numbers $1$ to $n$ is $frac{n(n+1)}{2}$, going forward and backward doubles the sum for a single row: $n(n+1)=9*10=90$
$endgroup$
– Daniel Mathias
3 hours ago
$begingroup$
Great job, it's a good start.
$endgroup$
– Riddler
2 hours ago
add a comment |
2
$begingroup$
The sum of numbers $1$ to $n$ is $frac{n(n+1)}{2}$, going forward and backward doubles the sum for a single row: $n(n+1)=9*10=90$
$endgroup$
– Daniel Mathias
3 hours ago
$begingroup$
Great job, it's a good start.
$endgroup$
– Riddler
2 hours ago
2
2
$begingroup$
The sum of numbers $1$ to $n$ is $frac{n(n+1)}{2}$, going forward and backward doubles the sum for a single row: $n(n+1)=9*10=90$
$endgroup$
– Daniel Mathias
3 hours ago
$begingroup$
The sum of numbers $1$ to $n$ is $frac{n(n+1)}{2}$, going forward and backward doubles the sum for a single row: $n(n+1)=9*10=90$
$endgroup$
– Daniel Mathias
3 hours ago
$begingroup$
Great job, it's a good start.
$endgroup$
– Riddler
2 hours ago
$begingroup$
Great job, it's a good start.
$endgroup$
– Riddler
2 hours ago
add a comment |
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$begingroup$
The "i" on the first column of the last row, is that supposed to be a "t"?
$endgroup$
– North
3 hours ago
$begingroup$
Also are words containing only vowels acceptable?
$endgroup$
– North
3 hours ago
$begingroup$
@North Yes, words with only vowels are acceptable :)
$endgroup$
– Riddler
3 hours ago
$begingroup$
HINT: There are no mistakes in the puzzle, which is also a riddle.
$endgroup$
– Riddler
3 hours ago
$begingroup$
aaaahhhh I just noticed that the i and the t are swapped goddamn it i have to recalculate everything screw it
$endgroup$
– North
1 hour ago