In this question, would dig be 2 inputs, such as 91 and 99? And lim can be any number the user calls?

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champions2015 2017-9-6

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编辑： Jan 2018-2-25

A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 × 99. Write a function that is called this way:

>> n = palin_product(dig,lim);

The function returns the largest palindrome smaller than lim that is the product of two dig digit numbers. If no such number exists, the function returns 0. (Inspired by Project Euler.)

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Image Analyst 2017-9-6

http://www.mathworks.com/matlabcentral/answers/8626-how-do-i-get-help-on-homework-questions-on-matlab-answers

http://www.mathworks.com/matlabcentral/answers/13205#answer_18099

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Answer 1

Stephen23 2017-9-7

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https://ww2.mathworks.cn/matlabcentral/answers/355491-in-this-question-would-dig-be-2-inputs-such-as-91-and-99-and-lim-can-be-any-number-the-user-calls#answer_280612

编辑：Stephen23 2017-9-7

在 MATLAB Online 中打开

I know that this is homework, but someone needed to demonstrate how this can be done quite simply, without using slow and unnecessary third-party functions: I believe that champions2015 and future readers deserve to be shown that MATLAB code can be neat, efficient, and straightforward. Note that this code actually does exactly what the question asks for!

function n = palin_product(dig,lim)
n = 0;
V = 10.^dig-1:-1:10.^(dig-1);
for k1 = V
    for k2 = V
        p = k1*k2;
        if p>n && p<lim
            s = sprintf('%d',p);
            if all(s==s(end:-1:1))
                n = p;
            end
        end
    end
end
end

And tested:

>> palin_product(2,10000)
ans =  9009
>> palin_product(2,9009)
ans =  8448
>> palin_product(2,8448)
ans =  8118
>> palin_product(2,8118)
ans =  8008
>> palin_product(2,8008)
ans =  7227

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Stephen23 2017-9-7

编辑：Stephen23 2017-9-7

"Is 's' a string/cell of numbers"

No, s is a 1xN character array, as the sprintf documentation clearly describes. The char array s contains a representation of the product p.

"which keeps having continuous values of 'p' added onto it?"

No, nothing gets "added onto it": s gets totally reassigned each time that line is run. Whatever data s contained on the last iteration is totally irrelevant to the current iteration.

"Could you also create a cell named 's' instead of using 'sprintf' function?"

I have no idea what you mean, or why you would want to do this.

"Why do you have to check if p is greater than n?"

To identify "the largest palindrome" as your question requests. Only if the current product p is larger than the last identified palindromic number n does the code then check if the current product is also palindromic. There is no point in checking if p is smaller than n, because we already have a larger value stored in n. Thus as well as giving you the right answer it also makes the code much more efficient.

"What are you checking with 'all(s==s(end:-1:1))'?"

If the 1xN character array s is a palindrome.

"Is it to check if s/p is symetrical?"

If you mean checking if s is a palindrome, then yes.

John BG 2017-9-7

编辑：John Kelly 2017-9-14

在 MATLAB Online 中打开

hi Champion and Cobeldick

this is John BG <mailto:jgb2012@sky.com jgb2012@sky.com>

The Champions League 2015 was won by Barcelona FC, wasn't it?

to Champions2015, may I call you culé?

I caught an error in Coberdick's loops, look:

lim=121 with dig=2 should give the solution 11 11, right?

however,

clear all;clc
dig=2;lim=121
n = 0;
V = 10.^(dig-1):10.^dig-1;
for k1 = V
    for k2 = V
        p = k1*k2;
        if p>n && p<lim
            s = sprintf('%d',p);
            if all(s==s(end:-1:1))
                n = p;
            end
        end
    end
end
n
=
0

error, ain't null yet Stephen's loops return n=0.

.

my function

clear all;clc
lim2=121
dig=2
% function [lim r1 r2]=palin_product(dig,lim)
r1=0;r2=0;
   ispal=0;
   decom=1;
   lim=lim2  % keep start value in lim2, modify lim
    while decom
       ispal=0;
       while ~ispal   % find next symmetric integer below input lim2
          L1=num2str(lim);
          n1=1;n2=length(L1);
          while L1(n1)==L1(n2) && n1<=floor(length(L1)/2)
          n1=n1+1;n2=n2-1; 
          end
          if L1(n1)==L1(n2) ispal=1; end
          if ~ispal  lim=lim-1; end
       end
       s1=10^(dig-1);s2=str2num(repmat('9',1,dig));   % find whether found lim can be decomposed
       S=[s1:1:s2];
       L=combinator(numel(S),dig,'p','r');
       p=1;
       s1=S(L(p,1));s2=S(L(p,2));
       while ~(s1*s2==lim) && p<length(L)
            s1=S(L(p,1));s2=S(L(p,2));
            p=p+1;
       end
      if p<=length(L) && s1*s2==lim
          r1=s1;r2=s2
      end
     if r1==0 && r2==0 && lim>1
    lim=lim-1; 
      end
    if lim==1
         return
    end
    if r1*r2>0
        decom=0;
    end
end
% end % function
r1
r2
r1 =
    11
r2 =
    11

.

and you get the right answer plus the 2 figures of dig digits decomposition, if exists.

Jan 2017-9-14

在 MATLAB Online 中打开

+1. This is a clean, compact and efficient solution.

Alternative to if all(s==s(end:-1:1)):

if isequal(s, s(end:-1:1))

Stephen23 2017-9-14

Thank you Jan Simon, that is a very neat idea.

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Answer 2

the cyclist 2017-9-6

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I would interpret it as follows.

You are trying to find p1 * p2 = n.

lim is any number the user chooses (just as you guessed). The output, n, must be smaller than lim.
dig is the number of digits that p1 and p2 each have. For example, if p1=91, and p2=99, then dig=2, because they are 2-digit numbers.

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Answer 3

Image Analyst 2017-9-6

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在 MATLAB Online 中打开

Hint:

n = 9009
s = sprintf('%d', n)
flipped_s = fliplr(s)

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Answer 4

John BG 2017-9-7

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编辑：Jan 2018-2-25

在 MATLAB Online 中打开

palin_product.m

hi champions2015

this is John BG <mailto:jgb2012@sky.com jgb2012@sky.com>

the following is a slight variation of what's been asked, this function

returns the nearest palindrome to the input
also calculates the 2 numbers of digit length dig that multiplied are lim
if the input is already a palindrome, it decomposes as requested lim, not seeking any smaller palindrome.

clear all;clc
lim2=8448
dig=2
function [lim r1 r2]=palin_product(dig,lim2)
     r1=0;r2=0;
     ispal=0;
     decom=1;
     lim=lim2  % keep start value in lim2, modify lim
    while decom
       ispal=0;
       while ~ispal   % find next symmetric integer below input lim2
          L1=num2str(lim);
          n1=1;n2=length(L1);
          while L1(n1)==L1(n2) && n1<=floor(length(L1)/2)
          n1=n1+1;n2=n2-1; 
          end
          if L1(n1)==L1(n2) ispal=1; end
          if ~ispal  lim=lim-1; end
       end
       s1=10^(dig-1);s2=str2num(repmat('9',1,dig));   % find whether found lim can be decomposed
       S=[s1:1:s2];
       L=combinator(numel(S),dig,'p','r');
       p=1;
       s1=S(L(p,1));s2=S(L(p,2));
       while ~(s1*s2==lim) && p<length(L)
            s1=S(L(p,1));s2=S(L(p,2));
            p=p+1;
       end
      if p<=length(L) && s1*s2==lim
          r1=s1;r2=s2
      end
     if r1==0 && r2==0
    lim=lim-1; 
      end
    if r1*r2>0
        decom=0;
    end
end
end % function

.

the most recent version of the function combinator and its support functions are all packed and freely available from

https://uk.mathworks.com/matlabcentral/fileexchange/24325-combinator-combinations-and-permutations

combinator_update.zip and palin_product.m are both attached to this answer.

.

If you find this answer useful would you please be so kind to consider marking my answer as Accepted Answer?

To any other reader, if you find this answer useful please consider clicking on the thumbs-up vote link

thanks in advance

John BG

<mailto:jgb2012@sky.com jgb2012@sky.com>

[EDITED: Copyrighted code removed]

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champions2015 2017-9-7

Hi John! Thanks for you answer! I'm not entirely sure if I understand how your code works, and it doesn't seem to work either when I run the function... Why are there 2 outputs? Wouldn't the output be simply somthing like 9009?

John BG 2017-9-7

编辑：John BG 2017-9-7

Hi Champions2015

this is John BG <mailto:jgb2012@sky.com jgb2012@sky.com>

When the input is already is such palindrome that can be decomposed as requested, why would you want to seek a smaller palindrome that can be decomposed into 2 figures of digit length dig?

My function also calculates the requested digit dig length decomposition.

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In this question, would dig be 2 inputs, such as 91 and 99? And lim can be any number the user calls?

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