Find the alphabetic word product

If the input string s is a word like 'hello', then the output word product p is a number based on the correspondence a=1, b=2, ... z=26. Assume the input will be a single word, although it may mixed case. Note that A=a=1 and B=b=2.

[Rik]

This should not be too hard to write. What did you try already?

BTW, this sounds like homework. You can find guidelines for posting homework on this forum here.

[Rik]

Sending an e-mail with your question text repeated doesn't do anything. I have read your question and I have responded. If you don't bother replying to the question I posted, why would anyone bother to help you?

I will repeat my question: what have you tried so far? How have you tried tackling this? In what steps did you try to break up this problem? What issues did you have with that code?

[Muchoa]

SorrySorry @Rıkwisseling

I CaN Try It1II

[Walter Roberson]

Hint:

lookup(1:127) = nan
lookup('~!@#$%') = [0 1 2 3 4 5];
lookup(InputString)

[Walter Roberson]

It is not valid to have an expression on the left side of an "=" in MATLAB, except for the case where the expression is indexing a variable and so designating a destination to store a result into.

Also your variables a and b are not defined, and you are not storing the result of the input() anywhere. You should also read the options available to input()

[Muchoa]

Thanks But Can you give an hints for homework because my matlab lesson is not enough to solve

[Walter Roberson]

Learn

[Rik]

I have also given you some hints in the answer below. Did you see it?

[Muchoa]

ı looked it and I done 1. question thanks also how can solve word to get a word product One million ??

[Walter Roberson]

factor(1e6) . You will get 2s and 5s. So create words out of B and E and J.

[Rik]

for n=1:26
    if all(ismember(factor(n),[2 5]))
        fprintf('%s,',n-1+'a')
    end
end
fprintf('\b\n')

So a combination of these letters may return 1e6 as a product: b,d,e,h,j,p,t,y

[Walter Roberson]

good point Rik.

[Muchoa]

I used it but ıt is not appropriate my question I find word to get a word product One million

ı supposed that need to use @(n)char(n-1+'a') for number2letter

BuySjsjlalaö

[Rik]

The code I showed in my comment finds the letters you are allowed to use to form a word. From there on out it is either brute-forcing a solution by checking all 12 letter permutations of these letters, or thinking of ways to cleverly reduce the number of options you need to check, or start solving the puzzle manually.

[Walter Roberson]

Is the word of value 1 million supposed to be exactly 12 letters? If so then the set of characters to be used is unique if you do not allow 'a' to be used.

Rik, we both missed 'a' as a possibility.

[Muchoa]

HowHow can i solve the find word to get a word product One million by using your chosen or my way pls some hints same others question

[Muchoa]

ı can't understand

[Rik]

It is very difficult to find a way to solve this with code alone. You would probably need to use some kind of dictionary to try to find a word that works.

You should just play around with it. You have a function that can calculate the word product, and because of the prime factors of 1e6 you know what letters you can use: a,b,d,e,h,j,p,t,y (thank Walter for pointing out the a).

As an example: the word 'bay' contains only the specified letters and returns a product of 50. So 'bay' doesn't work, but it points out that we're probably looking for a longer word. Words like abbey (500), paddy (6400) and jaded (800) don't work either.

Do you see what I'm doing here? There is a more structured approach in which you would generate all combinations of these letters that result in 1e6, which is probably what your teacher is after, but I don't see how you can easily filter out an existing word, without looking at thousands of jumbled 'words'.

[Walter Roberson]

I ran over the standard unix dictionary; there was no natural word with that product. 'curing' was close.

Looking around online I do see some place names and a small number of rather uncommon English words

[Muchoa]

ı got some research and tetty is one million but how can ı find writing code to find tetty

[Walter Roberson]

Finding tetty as a possibility is not the main problem: the main problem is recognizing it as a valid word. Like why not accept ebdedebey instead?

[Muchoa]

ıtt can be our problem is finding 1e6 by using 1e6s factors for example 10*10*10*10*10*10 =jjjjjj the code is using factors and find 1000000 but how? RıKK doYou have Any idea

[Walter Roberson]

Recursion.

Create a function factorize that accepts a number N and is responsible for outputing a cell array of all of the unique ways of factoring the number with factors between 2 and 26. It can do this by selecting looping through all of those possible factors L, and at each step testing whether the factor L divides the number N, and if so then recursively call factorize to factor the N/L and then to take the cell array that is returned and put L at the beginning of each entry; continue on for all of the possible factors L, then return the augmented entries.

At any one point, the function is only responsible for finding one more factor, and calls itself on the smaller value N/L . By induction, as each step involves smaller values, then you will eventually get down to things simple exact factors. Note, though, that that induction does not work if you permit 1 as a value (corresponding to 'a'): you can put any number of 'a' in without changing the product: tttt has the same product as atataaaaaataaaaaaaaaaaaa .

You will find plenty of combinations that multiply out to 1E6. That isn't the hard part. The hard part is identifying whether each of the combinations is a valid English word. Like why isn't jetbedy a valid solution?

Answer 1

[Rik]

See the code below for some more guided hints in an almost finished form. You don't need to call the input function btw. You can call the function by using e.g.

word_product('hello')

If you see a function you don't know, read its documentation. For most common tasks there exists a Matlab function. As an example: the sum function will calculate the sum of an array. Generally a combination of Google, the documentation, and guesses will get you there.

function p=word_product(s)
%convert a word to a ceasar cypher product
%(i.e. convert letters to numbers (a=A=1, b=B=2, etc), and calculate the
%product of these numbers)

%convert all letters to lowercase so we don't need to worry about uppercase
s=SomeFunction(s);

%Now convert the letters to numbers
%This can be done with a tedious lookup for all 26 cases, but this is
%Matlab, there is probably an easier way; and luckily there is.
s=double(s);
%This is not yet correct (nor technically even needed for the next step),
%so you will have to do an extra edit.
%Check out what double() does to char arrays (run in command window):
%double('a')
%double('ab')

%Now we can calculate the product:
p=SomeOtherFunction(s);
end

Answer 2

[kushwantkumar chimmiti]

function product = word_product(string)
    
    small_alphabet = char(97:122);
    large_alphabet = char(65:90);
    
    product = 1;


    alphabet_value = strings(length(small_alphabet),2);
    
    for i = 1:length(small_alphabet)
        alphabet_value{i,1} = small_alphabet(i);
        alphabet_value{i,2} = large_alphabet(i);
    end


    for k = 1:length(string)
        [order,~] = find(strcmp(alphabet_value, string(k)));
        product = product*order;
    end
        
end