How to implement MapThread with basic list mapping? - algorithm

Mathematica has a function MapThread that behaves like this:
MapThread[ f , { {a,b,c} , {d,e,f} } ]
-> { f[a,d] , f[b,e] , f[c,f] }
I'd like to implement this in TeX, which has very primitive programming facilities. I've basic facilities for iterating over lists, but no logical indexing into them. Given this restriction, is there an algorithm for looping synchronously though multiple lists?
I could write something like the following: (pseudocode)
ii = 0; jj = 0;
for elem1 in list1
ii = ii+1
for elem2 in list2
jj = jj+1
if ii == jj
return ( elem1 , elem2 )
fi
end
end
but it seems terribly inefficient. Note that the big restriction is that I can't access elements of lists numerically, so something like the following is too "high level":
for ii = 1:length(list1)
func ( list1(ii) , list2(ii) )
end
The reason this restriction is in place is that in order to implement list1(ii) I'd need to write something like the following in the first place:
jj = 0
for elem1 in list1
jj = jj+1
if ii=jj
return elem1
fi
end
Or is the inefficient case probably the best I'll be able to do with such a primitive language?

In pseudocode, as long as you can test whether a list is empty, you can do it like so:
while (list1 is not empty) and (list2 is not empty)
x1 = first(list1);
x2 = first(list2);
list1 = rest(list1);
list2 = rest(list2);
func(x1, x2);
end while
It's still not going to win any beauty contests, but it'll get the job done without being crushingly inefficient.

Related

A simple Increasing Mathematical Algorithm

I actually tried to search this, I'm sure this basic algorithm is everywhere on internet, CS textbooks etc, but I cannot find the right words to search it.
What I want from this algorithm to do is write "A" and "B" with the limit always increasing by 2. Like I want it to write A 3 times, then B 5 times, then A 7 times, then B 9 times and so on. And I plan to have 100 elements in total.
Like: AAABBBBBAAAAAAABBBBBBBBB...
I only want to use a single "for loop" for the entire 100 elements starting from 1 to 100. And just direct/sort "A" and "B" through "if/else if/ else".
I'm just asking for the basic mathematical algorithm behind it, showing it through any programming language would be better or redirecting me to such topic would also be fine.
You can do something like this:
There might be shorter answers, but I find this one easy to understand.
Basically, you keep a bool variable that will tell you if it's A's turn or Bs. Then we keep a variable switch that will tell us when we should switch between them. times is being updated with the repeated times we need to print the next character.
A_B = true
times = 3 // 3,5,7,9,...
switch = 3 // 3,8,15,24,...
for (i from 1 to 100)
if (A_B)
print 'A'
else
print 'B'
if (i == switch)
times += 2
switch += times
A_B = !A_B
Python:
for n in range(1, 101):
print "BA"[(int(sqrt(n)) % 2)],
The parity of the square roots of the integers follows that pattern. (Think that (n+1)²-n² = 2n+1.)
If you prefer to avoid the square root, it suffices to use an extra variable that represents the integer square root and keep it updated
r= 1
for n in range(1, 101):
if r * r <= n:
r+= 1
print "AB"[r % 2],
Here is the snippet you can test on this page. It is an example for about 500 letters totally, sure you can modify it for 100 letters. It is quite flexible that you can change the constants to produce lot of different strings in the same manner.
var toRepeat = ['A', 'B'];
var result='', j, i=3;
var sum=i;
var counter = 0;
while (sum < 500) {
j = counter % 2;
result = result + toRepeat[j].repeat(i);
sum = sum + i;
i = i + 2;
counter++;
}
document.getElementById('hLetters').innerHTML=result;
console.log(result);
<div id="hLetters"></div>
If you want it to be exactly 500 / 100 letters, just use a substring function to trim off the extra letters from the end.
To get 100 groups of A and B with increasing length of 3, 5, 7 and so on, you can run this Python code:
''.join(('B' if i % 2 else 'A') * (2 * i + 3) for i in range(100))
The output is a string of 10200 characters.
If you want the output to have only 100 characters, you can use:
import math
''.join(('B' if math.ceil(math.sqrt(i)) % 2 else 'A') for i in range(2, 102))
In js you can start with somethink like this :
$res ="";
count2 = 0;
for (i=2;i<100; i = i+2) {
count = 0;
alert(i);
while (count < i ) {
$res = $res.concat(String.fromCharCode(65+count2));
count++;
}
count2++;
}
alert ($res);

Fast way of checking if an element is ranked higher than another

I am writing in MATLAB a program that checks whether two elements A and B were exchanged in ranking positions.
Example
Assume the first ranking is:
list1 = [1 2 3 4]
while the second one is:
list2 = [1 2 4 3]
I want to check whether A = 3 and B = 4 have exchanged relative positions in the rankings, which in this case is true, since in the first ranking 3 comes before 4 and in the second ranking 3 comes after 4.
Procedure
In order to do this, I have written the following MATLAB code:
positionA1 = find(list1 == A);
positionB1 = find(list1 == B);
positionA2 = find(list2 == A);
positionB2 = find(list2 == B);
if (positionA1 <= positionB1 && positionA2 >= positionB2) || ...
(positionA1 >= positionB1 && positionA2 <= positionB2)
... do something
end
Unfortunately, I need to run this code a lot of times, and the find function is really slow (but needed to get the element position in the list).
I was wondering if there is a way of speeding up the procedure. I have also tried to write a MEX file that performs in C the find operation, but it did not help.
If the lists don't change within your loop, then you can determine the positions of the items ahead of time.
Assuming that your items are always integers from 1 to N:
[~, positions_1] = sort( list1 );
[~, positions_2] = sort( list2 );
This way you won't need to call find within the loop, you can just do:
positionA1 = positions_1(A);
positionB1 = positions_1(B);
positionA2 = positions_2(A);
positionB2 = positions_2(B);
If your loop is going over all possible combinations of A and B, then you can also vectorize that
Find the elements that exchanged relative ranking:
rank_diff_1 = bsxfun(#minus, positions_1, positions_1');
rank_diff_2 = bsxfun(#minus, positions_2, positions_2');
rel_rank_changed = sign(rank_diff_1) ~= sign(rank_diff_2);
[A_changed, B_changed] = find(rel_rank_changed);
Optional: Throw out half of the results, because if (3,4) is in the list, then (4,3) also will be, and maybe you don't want that:
mask = (A_changed < B_changed);
A_changed = A_changed(mask);
B_changed = B_changed(mask);
Now loop over only those elements that have exchanged relative ranking
for ii = 1:length(A_changed)
A = A_changed(ii);
B = B_changed(ii);
% Do something...
end
Instead of find try to compute something like this
Check if there is any exchanged values.
if logical(sum(abs(list1-list2)))
do something
end;
For specific values A and B:
if (list1(logical((list1-list2)-abs((list1-list2))))==A)&&(list1(logical((list1-list2)+abs((list1-list2))))==B)
do something
end;

algorithm for checking close by numbers for similarities in a list

I have a list and I need to find and extract all numbers in close proximity to a new list.
for example I have a list:
1,5,10,8,11,14,15,11,14,1,4,7,5,9
so if I want to extract all numbers that are close by 3(only 3, the gap must be 3, so 11,14 is correct, 11,13 is not.) near each other how can I design this without hard-coding the whole thing?
the result should look like:
8,11,14,11,14,1,4,7
This doesn't look too hard ,but I'm kind stuck, all I can come up with is a loop that checks n+1 member of the loop if it's more than n by 3 and include the n+1 member in a new list, however I don't know how to include the n member without making it appear on the new list twice if there is a string of needed numbers.
any ideas?
Just loop through the list, checking the next and previous element, and save the current one if it differs by 3 from either one. In Python, that's
>>> l = [1,5,10,8,11,14,15,11,14,1,4,7,5,9]
>>> # pad with infinities to ease the final loop
>>> l = [float('-inf')] + l + [float('inf')]
>>> [x for i, x in enumerate(l[1:-1], 1)
... if 3 in (abs(x - l[i-1]), abs(x - l[i+1]))]
[8, 11, 14, 11, 14, 1, 4, 7]
In Matlab
list = [1,5,10,8,11,14,15,11,14,1,4,7,5,9]
then
list(or([diff([0 diff(list)==3]) 0],[0 diff(list)==3]))
returns
8 11 14 11 14 1 4 7
For those who don't understand Matlab diff(list) returns the first (forward) differences of the elements in list. The expression [0 diff(list)] pads the first differences with a leading 0 to make the result the same length as the original list. The rest should be obvious.
In a nutshell: take forward differences and backward differences, select the elements where either difference is 3.
A simple C++ code below:
assuming ar is the array of the initial integers and mark is a boolean array
for(int i=1;i<N;i++){
if(ar[i]-ar[i-1]==3){
mark[i]=1;
mark[i-1]=1;
}
}
Now to print the interesting numbers,
for(int i=0;i<N;i++){
if(mark[i]==1)cout<<ar[i]<<" ";
}
The idea behind the implementation is, we mark a number as interesting if the difference from it to its previous one is 3 or if the difference between it and its next number is 3.
that's a single loop:
public List<int> CloseByN(int n, List<int> oldL)
{
bool first = true;
int last = 0;
bool isLstAdded = false;
List<int> newL = new List<int>();
foreach(int curr in oldL)
{
if(first)
{
first = false;
last = curr;
continue;
}
if(curr - last == n)
{
if(isLstAdded == false)
{
newL.Add(last);
isLstAdded = true;
}
newL.Add(curr);
}
else
{
isLstAdded = false;
}
last = curr;
}
return newL;
}
tested on your input and got your output
And a Haskell version:
f g xs = dropWhile (null . drop 1) $ foldr comb [[last xs]] (init xs) where
comb a bbs#(b:bs)
| abs (a - head b) == g = (a:head bbs) : bs
| otherwise =
if null (drop 1 b) then [a] : bs else [a] : bbs
Output:
*Main> f 3 [5,10,8,11,14,15,11,14,1,4,7,5,9]
[[8,11,14],[11,14],[1,4,7]]
*Main> f 5 [5,10,8,11,14,15,11,14,1,4,7,5,9]
[[5,10]]

Algorithm for series

A, B, C,…. Z, AA, AB, ….AZ, BA,BB,…. , ZZ,AAA, …., write a function that takes a integer n and returns the string presentation. Can somebody tell me the algorithm to find the nth value in the series?
Treat those strings as numbers in base 26 with A=0. It's not quite an exact translation because in real base 26 A=AA=AAA=0, so you have to make some adjustments as necessary.
Here's a Java implementation:
static String convert(int n) {
int digits = 1;
for (int j = 26; j <= n; j *= 26) {
digits++;
n -= j;
}
String s = "";
for (; digits --> 0 ;) {
s = (char) ('A' + (n % 26)) + s;
n /= 26;
}
return s;
}
This converts 0=A, 26=AA, 702=AAA as required.
Without giving away too much (since this question seems to be a homework problem), what you're doing is close to the same as translating that integer n into base 26. Good luck!
If, as others suspect, this is homework, then this answer probably won't be much help. If this is for a real-world project though, it might make sense to do make a generator instead, which is an easy and idiomatic thing to do in some languages, such as Python. Something like this:
def letterPattern():
pattern = [0]
while True:
yield pattern
pattern[0] += 1
# iterate through all numbers in the list *except* the last one
for i in range(0,len(pattern)-1):
if pattern[i] == 26:
pattern[i] = 0
pattern[i+1] += 1
# now if the last number is 26, set it to zero, and append another zero to the end
if pattern[-1] == 26:
pattern[-1] = 0
pattern.append(0)
Except instead of yielding pattern itself you would reverse it, and map 0 to A, 1 to B, etc. then yield the string. I've run the code above and it seems to work, but I haven't tested it extensively at all.
I hope you'll find this readable enough to implement, even if you don't know Python. (For the Pythonistas out there, yes the "for i in range(...)" loop is ugly and unpythonic, but off the top of my head, I don't know any other way to do what I'm doing here)

Algorithm Issue: letter combinations

I'm trying to write a piece of code that will do the following:
Take the numbers 0 to 9 and assign one or more letters to this number. For example:
0 = N,
1 = L,
2 = T,
3 = D,
4 = R,
5 = V or F,
6 = B or P,
7 = Z,
8 = H or CH or J,
9 = G
When I have a code like 0123, it's an easy job to encode it. It will obviously make up the code NLTD. When a number like 5,6 or 8 is introduced, things get different. A number like 051 would result in more than one possibility:
NVL and NFL
It should be obvious that this gets even "worse" with longer numbers that include several digits like 5,6 or 8.
Being pretty bad at mathematics, I have not yet been able to come up with a decent solution that will allow me to feed the program a bunch of numbers and have it spit out all the possible letter combinations. So I'd love some help with it, 'cause I can't seem to figure it out. Dug up some information about permutations and combinations, but no luck.
Thanks for any suggestions/clues. The language I need to write the code in is PHP, but any general hints would be highly appreciated.
Update:
Some more background: (and thanks a lot for the quick responses!)
The idea behind my question is to build a script that will help people to easily convert numbers they want to remember to words that are far more easily remembered. This is sometimes referred to as "pseudo-numerology".
I want the script to give me all the possible combinations that are then held against a database of stripped words. These stripped words just come from a dictionary and have all the letters I mentioned in my question stripped out of them. That way, the number to be encoded can usually easily be related to a one or more database records. And when that happens, you end up with a list of words that you can use to remember the number you wanted to remember.
It can be done easily recursively.
The idea is that to handle the whole code of size n, you must handle first the n - 1 digits.
Once you have all answers for n-1 digits, the answers for the whole are deduced by appending to them the correct(s) char(s) for the last one.
There's actually a much better solution than enumerating all the possible translations of a number and looking them up: Simply do the reverse computation on every word in your dictionary, and store the string of digits in another field. So if your mapping is:
0 = N,
1 = L,
2 = T,
3 = D,
4 = R,
5 = V or F,
6 = B or P,
7 = Z,
8 = H or CH or J,
9 = G
your reverse mapping is:
N = 0,
L = 1,
T = 2,
D = 3,
R = 4,
V = 5,
F = 5,
B = 6,
P = 6,
Z = 7,
H = 8,
J = 8,
G = 9
Note there's no mapping for 'ch', because the 'c' will be dropped, and the 'h' will be converted to 8 anyway.
Then, all you have to do is iterate through each letter in the dictionary word, output the appropriate digit if there's a match, and do nothing if there isn't.
Store all the generated digit strings as another field in the database. When you want to look something up, just perform a simple query for the number entered, instead of having to do tens (or hundreds, or thousands) of lookups of potential words.
The general structure you want to hold your number -> letter assignments is an array or arrays, similar to:
// 0 = N, 1 = L, 2 = T, 3 = D, 4 = R, 5 = V or F, 6 = B or P, 7 = Z,
// 8 = H or CH or J, 9 = G
$numberMap = new Array (
0 => new Array("N"),
1 => new Array("L"),
2 => new Array("T"),
3 => new Array("D"),
4 => new Array("R"),
5 => new Array("V", "F"),
6 => new Array("B", "P"),
7 => new Array("Z"),
8 => new Array("H", "CH", "J"),
9 => new Array("G"),
);
Then, a bit of recursive logic gives us a function similar to:
function GetEncoding($number) {
$ret = new Array();
for ($i = 0; $i < strlen($number); $i++) {
// We're just translating here, nothing special.
// $var + 0 is a cheap way of forcing a variable to be numeric
$ret[] = $numberMap[$number[$i]+0];
}
}
function PrintEncoding($enc, $string = "") {
// If we're at the end of the line, then print!
if (count($enc) === 0) {
print $string."\n";
return;
}
// Otherwise, soldier on through the possible values.
// Grab the next 'letter' and cycle through the possibilities for it.
foreach ($enc[0] as $letter) {
// And call this function again with it!
PrintEncoding(array_slice($enc, 1), $string.$letter);
}
}
Three cheers for recursion! This would be used via:
PrintEncoding(GetEncoding("052384"));
And if you really want it as an array, play with output buffering and explode using "\n" as your split string.
This kind of problem are usually resolved with recursion. In ruby, one (quick and dirty) solution would be
#values = Hash.new([])
#values["0"] = ["N"]
#values["1"] = ["L"]
#values["2"] = ["T"]
#values["3"] = ["D"]
#values["4"] = ["R"]
#values["5"] = ["V","F"]
#values["6"] = ["B","P"]
#values["7"] = ["Z"]
#values["8"] = ["H","CH","J"]
#values["9"] = ["G"]
def find_valid_combinations(buffer,number)
first_char = number.shift
#values[first_char].each do |key|
if(number.length == 0) then
puts buffer + key
else
find_valid_combinations(buffer + key,number.dup)
end
end
end
find_valid_combinations("",ARGV[0].split(""))
And if you run this from the command line you will get:
$ ruby r.rb 051
NVL
NFL
This is related to brute-force search and backtracking
Here is a recursive solution in Python.
#!/usr/bin/env/python
import sys
ENCODING = {'0':['N'],
'1':['L'],
'2':['T'],
'3':['D'],
'4':['R'],
'5':['V', 'F'],
'6':['B', 'P'],
'7':['Z'],
'8':['H', 'CH', 'J'],
'9':['G']
}
def decode(str):
if len(str) == 0:
return ''
elif len(str) == 1:
return ENCODING[str]
else:
result = []
for prefix in ENCODING[str[0]]:
result.extend([prefix + suffix for suffix in decode(str[1:])])
return result
if __name__ == '__main__':
print decode(sys.argv[1])
Example output:
$ ./demo 1
['L']
$ ./demo 051
['NVL', 'NFL']
$ ./demo 0518
['NVLH', 'NVLCH', 'NVLJ', 'NFLH', 'NFLCH', 'NFLJ']
Could you do the following:
Create a results array.
Create an item in the array with value ""
Loop through the numbers, say 051 analyzing each one individually.
Each time a 1 to 1 match between a number is found add the correct value to all items in the results array.
So "" becomes N.
Each time a 1 to many match is found, add new rows to the results array with one option, and update the existing results with the other option.
So N becomes NV and a new item is created NF
Then the last number is a 1 to 1 match so the items in the results array become
NVL and NFL
To produce the results loop through the results array, printing them, or whatever.
Let pn be a list of all possible letter combinations of a given number string s up to the nth digit.
Then, the following algorithm will generate pn+1:
digit = s[n+1];
foreach(letter l that digit maps to)
{
foreach(entry e in p(n))
{
newEntry = append l to e;
add newEntry to p(n+1);
}
}
The first iteration is somewhat of a special case, since p-1 is undefined. You can simply initialize p0 as the list of all possible characters for the first character.
So, your 051 example:
Iteration 0:
p(0) = {N}
Iteration 1:
digit = 5
foreach({V, F})
{
foreach(p(0) = {N})
{
newEntry = N + V or N + F
p(1) = {NV, NF}
}
}
Iteration 2:
digit = 1
foreach({L})
{
foreach(p(1) = {NV, NF})
{
newEntry = NV + L or NF + L
p(2) = {NVL, NFL}
}
}
The form you want is probably something like:
function combinations( $str ){
$l = len( $str );
$results = array( );
if ($l == 0) { return $results; }
if ($l == 1)
{
foreach( $codes[ $str[0] ] as $code )
{
$results[] = $code;
}
return $results;
}
$cur = $str[0];
$combs = combinations( substr( $str, 1, $l ) );
foreach ($codes[ $cur ] as $code)
{
foreach ($combs as $comb)
{
$results[] = $code.$comb;
}
}
return $results;}
This is ugly, pidgin-php so please verify it first. The basic idea is to generate every combination of the string from [1..n] and then prepend to the front of all those combinations each possible code for str[0]. Bear in mind that in the worst case this will have performance exponential in the length of your string, because that much ambiguity is actually present in your coding scheme.
The trick is not only to generate all possible letter combinations that match a given number, but to select the letter sequence that is most easy to remember. A suggestion would be to run the soundex algorithm on each of the sequence and try to match against an English language dictionary such as Wordnet to find the most 'real-word-sounding' sequences.

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