Resizing and scaling image












0












$begingroup$


I'm trying to resize image bitmap in software development, in that my requirement is as following,




  • Get the original height of image

  • Get the original width of image

  • Set a max threshold, output height and width should be below that threshold only


My requirement is,




  • output height and width should be below set threshold

  • output values should be maximum possible

  • The ratio of final height and width should be same as original.


I am new to this community, so pardon me if anything is incorrect, any help will be appreciate.






image-processing programming







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Hi Kiran, welcome to the MSE community! In my opinion, your question might be answered if posted in the Stackoverflow community if it's about software implementation rather than mathematical concepts?
    $endgroup$
    – Macrophage
    Jan 18 at 11:28










  • $begingroup$
    Hi @Macrophage, thank you for the quick reply, I do not want software implantation, I need mathematical approach and formula for that. So I guess, this question will be answered here in better way.
    $endgroup$
    – Kiran Malvi
    Jan 18 at 11:32










  • $begingroup$
    I'm voting to close this question as off-topic because it suits Stackoverflow better as it is about implementing something using software
    $endgroup$
    – lioness99a
    Jan 18 at 11:32
















0












$begingroup$


I'm trying to resize image bitmap in software development, in that my requirement is as following,




  • Get the original height of image

  • Get the original width of image

  • Set a max threshold, output height and width should be below that threshold only


My requirement is,




  • output height and width should be below set threshold

  • output values should be maximum possible

  • The ratio of final height and width should be same as original.


I am new to this community, so pardon me if anything is incorrect, any help will be appreciate.






image-processing programming







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Hi Kiran, welcome to the MSE community! In my opinion, your question might be answered if posted in the Stackoverflow community if it's about software implementation rather than mathematical concepts?
    $endgroup$
    – Macrophage
    Jan 18 at 11:28










  • $begingroup$
    Hi @Macrophage, thank you for the quick reply, I do not want software implantation, I need mathematical approach and formula for that. So I guess, this question will be answered here in better way.
    $endgroup$
    – Kiran Malvi
    Jan 18 at 11:32










  • $begingroup$
    I'm voting to close this question as off-topic because it suits Stackoverflow better as it is about implementing something using software
    $endgroup$
    – lioness99a
    Jan 18 at 11:32














0












0








0





$begingroup$


I'm trying to resize image bitmap in software development, in that my requirement is as following,




  • Get the original height of image

  • Get the original width of image

  • Set a max threshold, output height and width should be below that threshold only


My requirement is,




  • output height and width should be below set threshold

  • output values should be maximum possible

  • The ratio of final height and width should be same as original.


I am new to this community, so pardon me if anything is incorrect, any help will be appreciate.






image-processing programming







share|cite|improve this question









$endgroup$




I'm trying to resize image bitmap in software development, in that my requirement is as following,




  • Get the original height of image

  • Get the original width of image

  • Set a max threshold, output height and width should be below that threshold only


My requirement is,




  • output height and width should be below set threshold

  • output values should be maximum possible

  • The ratio of final height and width should be same as original.


I am new to this community, so pardon me if anything is incorrect, any help will be appreciate.






image-processing programming




image-processing programming






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 18 at 11:24









Kiran MalviKiran Malvi

1033




1033








  • 1




    $begingroup$
    Hi Kiran, welcome to the MSE community! In my opinion, your question might be answered if posted in the Stackoverflow community if it's about software implementation rather than mathematical concepts?
    $endgroup$
    – Macrophage
    Jan 18 at 11:28










  • $begingroup$
    Hi @Macrophage, thank you for the quick reply, I do not want software implantation, I need mathematical approach and formula for that. So I guess, this question will be answered here in better way.
    $endgroup$
    – Kiran Malvi
    Jan 18 at 11:32










  • $begingroup$
    I'm voting to close this question as off-topic because it suits Stackoverflow better as it is about implementing something using software
    $endgroup$
    – lioness99a
    Jan 18 at 11:32














  • 1




    $begingroup$
    Hi Kiran, welcome to the MSE community! In my opinion, your question might be answered if posted in the Stackoverflow community if it's about software implementation rather than mathematical concepts?
    $endgroup$
    – Macrophage
    Jan 18 at 11:28










  • $begingroup$
    Hi @Macrophage, thank you for the quick reply, I do not want software implantation, I need mathematical approach and formula for that. So I guess, this question will be answered here in better way.
    $endgroup$
    – Kiran Malvi
    Jan 18 at 11:32










  • $begingroup$
    I'm voting to close this question as off-topic because it suits Stackoverflow better as it is about implementing something using software
    $endgroup$
    – lioness99a
    Jan 18 at 11:32








1




1




$begingroup$
Hi Kiran, welcome to the MSE community! In my opinion, your question might be answered if posted in the Stackoverflow community if it's about software implementation rather than mathematical concepts?
$endgroup$
– Macrophage
Jan 18 at 11:28




$begingroup$
Hi Kiran, welcome to the MSE community! In my opinion, your question might be answered if posted in the Stackoverflow community if it's about software implementation rather than mathematical concepts?
$endgroup$
– Macrophage
Jan 18 at 11:28












$begingroup$
Hi @Macrophage, thank you for the quick reply, I do not want software implantation, I need mathematical approach and formula for that. So I guess, this question will be answered here in better way.
$endgroup$
– Kiran Malvi
Jan 18 at 11:32




$begingroup$
Hi @Macrophage, thank you for the quick reply, I do not want software implantation, I need mathematical approach and formula for that. So I guess, this question will be answered here in better way.
$endgroup$
– Kiran Malvi
Jan 18 at 11:32












$begingroup$
I'm voting to close this question as off-topic because it suits Stackoverflow better as it is about implementing something using software
$endgroup$
– lioness99a
Jan 18 at 11:32




$begingroup$
I'm voting to close this question as off-topic because it suits Stackoverflow better as it is about implementing something using software
$endgroup$
– lioness99a
Jan 18 at 11:32










3 Answers
3






active

oldest

votes


















1












$begingroup$

The maths is quite simple




  1. Calculate the ratio between the width and height

  2. Set the width to be equal to the limit

  3. Calculate the height based on the ratio previously calculated

  4. If the height exceeds the limit, then repeat steps (2) and (3) using the height instead



For example, say you have an image with $text{width}=6$ and $text{height}=9$ and your threshold is $15$



The ratio between width and height is $2:3$, in other words, $$text{width}=frac 23times text{height}$$



If we increase the width to be equal to the limit, $15$, then our height becomes $$15times frac32 = 22.5$$



As this is above the limit of $15$, we instead set the height to be $15$ which gives us a width of $$15times frac 23 = 10$$



This is inside our limit so we have a new image with $text{width}=10$ and $text{height}=15$, which still obeys our $2:3$ ratio and is within the threshold.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for the simplest explanation. Solution worked like a charm.
    $endgroup$
    – Kiran Malvi
    Jan 18 at 12:01










  • $begingroup$
    No worries. It has also just occurred to me that you could just take the larger of width and height in step 2 and skip the whole checking step as you know the other will be smaller than your threshold (unless you have different width and height thresholds, in which case you will need to check both)
    $endgroup$
    – lioness99a
    Jan 18 at 12:04



















0












$begingroup$

So, essentially you have numbers $(w, h)$ and you want to scale them so that $max{(w_{text{new}}, h_{text{new}})}=M$, some maximum value. You simply have to calculate
$$
(w_{text{new}}, h_{text{new}}) = frac{M}{max{(w,h)}} (w,h)
$$
You can see that doing so replaces the bigger value by $M$ and scales the other by the same ratio. Done.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Let your image have size $w times h$, $t$ be the threshold, and $r=frac{w}{h}$ be the ratio. If $t>max(w,h)$ do nothing. Let $wgeq h$. You want to find a linear mapping $f : [0,w] rightarrow [0,t]$, that is $f(x) = ax + b, f(0) = 0, f(w) = t$, from the first condition $b= 0$, from the second $a=frac{t}{w}$. The new width is $w'= t$, since we want to preserve the ratio: $r=frac{w'}{h'}$, then $h' = frac{w'}{r}$. The mapping for $y$ can similarly be computed as: $g(y) = frac{h'y}{h}$. Your new image has dimensions $w' times h'$. Then for every point in the image $(x,y)$, the new transformed coordinates are $(x',y') = (frac{w'x}{w}, frac{h'y}{h})$. In case $w<h$, $h'=t, w' = rh'$.






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      The maths is quite simple




      1. Calculate the ratio between the width and height

      2. Set the width to be equal to the limit

      3. Calculate the height based on the ratio previously calculated

      4. If the height exceeds the limit, then repeat steps (2) and (3) using the height instead



      For example, say you have an image with $text{width}=6$ and $text{height}=9$ and your threshold is $15$



      The ratio between width and height is $2:3$, in other words, $$text{width}=frac 23times text{height}$$



      If we increase the width to be equal to the limit, $15$, then our height becomes $$15times frac32 = 22.5$$



      As this is above the limit of $15$, we instead set the height to be $15$ which gives us a width of $$15times frac 23 = 10$$



      This is inside our limit so we have a new image with $text{width}=10$ and $text{height}=15$, which still obeys our $2:3$ ratio and is within the threshold.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Thank you for the simplest explanation. Solution worked like a charm.
        $endgroup$
        – Kiran Malvi
        Jan 18 at 12:01










      • $begingroup$
        No worries. It has also just occurred to me that you could just take the larger of width and height in step 2 and skip the whole checking step as you know the other will be smaller than your threshold (unless you have different width and height thresholds, in which case you will need to check both)
        $endgroup$
        – lioness99a
        Jan 18 at 12:04
















      1












      $begingroup$

      The maths is quite simple




      1. Calculate the ratio between the width and height

      2. Set the width to be equal to the limit

      3. Calculate the height based on the ratio previously calculated

      4. If the height exceeds the limit, then repeat steps (2) and (3) using the height instead



      For example, say you have an image with $text{width}=6$ and $text{height}=9$ and your threshold is $15$



      The ratio between width and height is $2:3$, in other words, $$text{width}=frac 23times text{height}$$



      If we increase the width to be equal to the limit, $15$, then our height becomes $$15times frac32 = 22.5$$



      As this is above the limit of $15$, we instead set the height to be $15$ which gives us a width of $$15times frac 23 = 10$$



      This is inside our limit so we have a new image with $text{width}=10$ and $text{height}=15$, which still obeys our $2:3$ ratio and is within the threshold.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        Thank you for the simplest explanation. Solution worked like a charm.
        $endgroup$
        – Kiran Malvi
        Jan 18 at 12:01










      • $begingroup$
        No worries. It has also just occurred to me that you could just take the larger of width and height in step 2 and skip the whole checking step as you know the other will be smaller than your threshold (unless you have different width and height thresholds, in which case you will need to check both)
        $endgroup$
        – lioness99a
        Jan 18 at 12:04














      1












      1








      1





      $begingroup$

      The maths is quite simple




      1. Calculate the ratio between the width and height

      2. Set the width to be equal to the limit

      3. Calculate the height based on the ratio previously calculated

      4. If the height exceeds the limit, then repeat steps (2) and (3) using the height instead



      For example, say you have an image with $text{width}=6$ and $text{height}=9$ and your threshold is $15$



      The ratio between width and height is $2:3$, in other words, $$text{width}=frac 23times text{height}$$



      If we increase the width to be equal to the limit, $15$, then our height becomes $$15times frac32 = 22.5$$



      As this is above the limit of $15$, we instead set the height to be $15$ which gives us a width of $$15times frac 23 = 10$$



      This is inside our limit so we have a new image with $text{width}=10$ and $text{height}=15$, which still obeys our $2:3$ ratio and is within the threshold.






      share|cite|improve this answer











      $endgroup$



      The maths is quite simple




      1. Calculate the ratio between the width and height

      2. Set the width to be equal to the limit

      3. Calculate the height based on the ratio previously calculated

      4. If the height exceeds the limit, then repeat steps (2) and (3) using the height instead



      For example, say you have an image with $text{width}=6$ and $text{height}=9$ and your threshold is $15$



      The ratio between width and height is $2:3$, in other words, $$text{width}=frac 23times text{height}$$



      If we increase the width to be equal to the limit, $15$, then our height becomes $$15times frac32 = 22.5$$



      As this is above the limit of $15$, we instead set the height to be $15$ which gives us a width of $$15times frac 23 = 10$$



      This is inside our limit so we have a new image with $text{width}=10$ and $text{height}=15$, which still obeys our $2:3$ ratio and is within the threshold.







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Jan 18 at 11:41

























      answered Jan 18 at 11:34









      lioness99alioness99a

      3,7492727




      3,7492727












      • $begingroup$
        Thank you for the simplest explanation. Solution worked like a charm.
        $endgroup$
        – Kiran Malvi
        Jan 18 at 12:01










      • $begingroup$
        No worries. It has also just occurred to me that you could just take the larger of width and height in step 2 and skip the whole checking step as you know the other will be smaller than your threshold (unless you have different width and height thresholds, in which case you will need to check both)
        $endgroup$
        – lioness99a
        Jan 18 at 12:04


















      • $begingroup$
        Thank you for the simplest explanation. Solution worked like a charm.
        $endgroup$
        – Kiran Malvi
        Jan 18 at 12:01










      • $begingroup$
        No worries. It has also just occurred to me that you could just take the larger of width and height in step 2 and skip the whole checking step as you know the other will be smaller than your threshold (unless you have different width and height thresholds, in which case you will need to check both)
        $endgroup$
        – lioness99a
        Jan 18 at 12:04
















      $begingroup$
      Thank you for the simplest explanation. Solution worked like a charm.
      $endgroup$
      – Kiran Malvi
      Jan 18 at 12:01




      $begingroup$
      Thank you for the simplest explanation. Solution worked like a charm.
      $endgroup$
      – Kiran Malvi
      Jan 18 at 12:01












      $begingroup$
      No worries. It has also just occurred to me that you could just take the larger of width and height in step 2 and skip the whole checking step as you know the other will be smaller than your threshold (unless you have different width and height thresholds, in which case you will need to check both)
      $endgroup$
      – lioness99a
      Jan 18 at 12:04




      $begingroup$
      No worries. It has also just occurred to me that you could just take the larger of width and height in step 2 and skip the whole checking step as you know the other will be smaller than your threshold (unless you have different width and height thresholds, in which case you will need to check both)
      $endgroup$
      – lioness99a
      Jan 18 at 12:04











      0












      $begingroup$

      So, essentially you have numbers $(w, h)$ and you want to scale them so that $max{(w_{text{new}}, h_{text{new}})}=M$, some maximum value. You simply have to calculate
      $$
      (w_{text{new}}, h_{text{new}}) = frac{M}{max{(w,h)}} (w,h)
      $$
      You can see that doing so replaces the bigger value by $M$ and scales the other by the same ratio. Done.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        So, essentially you have numbers $(w, h)$ and you want to scale them so that $max{(w_{text{new}}, h_{text{new}})}=M$, some maximum value. You simply have to calculate
        $$
        (w_{text{new}}, h_{text{new}}) = frac{M}{max{(w,h)}} (w,h)
        $$
        You can see that doing so replaces the bigger value by $M$ and scales the other by the same ratio. Done.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          So, essentially you have numbers $(w, h)$ and you want to scale them so that $max{(w_{text{new}}, h_{text{new}})}=M$, some maximum value. You simply have to calculate
          $$
          (w_{text{new}}, h_{text{new}}) = frac{M}{max{(w,h)}} (w,h)
          $$
          You can see that doing so replaces the bigger value by $M$ and scales the other by the same ratio. Done.






          share|cite|improve this answer









          $endgroup$



          So, essentially you have numbers $(w, h)$ and you want to scale them so that $max{(w_{text{new}}, h_{text{new}})}=M$, some maximum value. You simply have to calculate
          $$
          (w_{text{new}}, h_{text{new}}) = frac{M}{max{(w,h)}} (w,h)
          $$
          You can see that doing so replaces the bigger value by $M$ and scales the other by the same ratio. Done.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 18 at 11:33









          Matti P.Matti P.

          2,020414




          2,020414























              0












              $begingroup$

              Let your image have size $w times h$, $t$ be the threshold, and $r=frac{w}{h}$ be the ratio. If $t>max(w,h)$ do nothing. Let $wgeq h$. You want to find a linear mapping $f : [0,w] rightarrow [0,t]$, that is $f(x) = ax + b, f(0) = 0, f(w) = t$, from the first condition $b= 0$, from the second $a=frac{t}{w}$. The new width is $w'= t$, since we want to preserve the ratio: $r=frac{w'}{h'}$, then $h' = frac{w'}{r}$. The mapping for $y$ can similarly be computed as: $g(y) = frac{h'y}{h}$. Your new image has dimensions $w' times h'$. Then for every point in the image $(x,y)$, the new transformed coordinates are $(x',y') = (frac{w'x}{w}, frac{h'y}{h})$. In case $w<h$, $h'=t, w' = rh'$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Let your image have size $w times h$, $t$ be the threshold, and $r=frac{w}{h}$ be the ratio. If $t>max(w,h)$ do nothing. Let $wgeq h$. You want to find a linear mapping $f : [0,w] rightarrow [0,t]$, that is $f(x) = ax + b, f(0) = 0, f(w) = t$, from the first condition $b= 0$, from the second $a=frac{t}{w}$. The new width is $w'= t$, since we want to preserve the ratio: $r=frac{w'}{h'}$, then $h' = frac{w'}{r}$. The mapping for $y$ can similarly be computed as: $g(y) = frac{h'y}{h}$. Your new image has dimensions $w' times h'$. Then for every point in the image $(x,y)$, the new transformed coordinates are $(x',y') = (frac{w'x}{w}, frac{h'y}{h})$. In case $w<h$, $h'=t, w' = rh'$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Let your image have size $w times h$, $t$ be the threshold, and $r=frac{w}{h}$ be the ratio. If $t>max(w,h)$ do nothing. Let $wgeq h$. You want to find a linear mapping $f : [0,w] rightarrow [0,t]$, that is $f(x) = ax + b, f(0) = 0, f(w) = t$, from the first condition $b= 0$, from the second $a=frac{t}{w}$. The new width is $w'= t$, since we want to preserve the ratio: $r=frac{w'}{h'}$, then $h' = frac{w'}{r}$. The mapping for $y$ can similarly be computed as: $g(y) = frac{h'y}{h}$. Your new image has dimensions $w' times h'$. Then for every point in the image $(x,y)$, the new transformed coordinates are $(x',y') = (frac{w'x}{w}, frac{h'y}{h})$. In case $w<h$, $h'=t, w' = rh'$.






                  share|cite|improve this answer









                  $endgroup$



                  Let your image have size $w times h$, $t$ be the threshold, and $r=frac{w}{h}$ be the ratio. If $t>max(w,h)$ do nothing. Let $wgeq h$. You want to find a linear mapping $f : [0,w] rightarrow [0,t]$, that is $f(x) = ax + b, f(0) = 0, f(w) = t$, from the first condition $b= 0$, from the second $a=frac{t}{w}$. The new width is $w'= t$, since we want to preserve the ratio: $r=frac{w'}{h'}$, then $h' = frac{w'}{r}$. The mapping for $y$ can similarly be computed as: $g(y) = frac{h'y}{h}$. Your new image has dimensions $w' times h'$. Then for every point in the image $(x,y)$, the new transformed coordinates are $(x',y') = (frac{w'x}{w}, frac{h'y}{h})$. In case $w<h$, $h'=t, w' = rh'$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 18 at 11:40









                  lightxbulblightxbulb

                  900211




                  900211






























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                      Mario Kart Wii

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                      Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro...
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                      Understanding the size os this class of aleatory events

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                      0 1 $begingroup$ Hi I am having problems to understand this: I have: $Omega = [0,1]$ , $Lambda =$ all subsets with defined length Let $A_{0}$ = {A $subset$ [0,1]: A is finite union of intervals} 1º: My doubt is that $A_{0}$ is a algebra but its not a sigma algebra. Why? Also, A = $(0,1/2) cup (1/2,3/4) cup (3/4,7/8) cup ... cup(1-1/2^{n},1-1/2^{n+1})...$ 2º: So the class of aleatory events would be bigger than $A_{0}$ . Why? How the set A is greater than $A_{0}$ ? measure-theory share | cite | improve this question asked Jan 8 at 19:17 Diogo Bastos Diogo Bastos 116 6 ...
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                      Partial Derivative Guidance.

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                      0 $begingroup$ I was given this tricky problem in which I'm confuse on how to solve it and I would like some help: Let $v(x,y)$ be continuously differentiable function. Let the function $u(s,t)$ be given by $u(s,t) = sin(s)*v(s-2t,t^2)$ . Compute $du/ds$ and $du/dt$ in terms of $v$ and it's partial derivatives. My attempt (which is wrong): Using chain rule: $du/ds=(du/dx)(dx/ds) + (du/dy)(dy/ds)$ $du/dt=(du/dx)(dx/dt) + (du/dx)(dy/dt)$ So, $du/ds = sin(s)*v(x,y)*s + sin(s)*v(x,y)*0$ $du/dt = 2t*sin(s)*v(x,y) - 2*sin(s)*v(x,y)$ Thanks, calculus multivariable-calculus share | cite | improve this question edited Jan 19 at 19:38...
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