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https://github.com/UpsilonNumworks/Upsilon.git
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8c6ffa6d42
Scenario: just zoom in in a graph Fix: - Do not truncate labels. - Label size is computed depending on available size and number of labels. - If labels cannot be displayed properly, just display the minimal and maximal labels!
128 lines
5.6 KiB
C++
128 lines
5.6 KiB
C++
#include "curve_view_range.h"
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#include "curve_view.h"
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#include <cmath>
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#include <ion.h>
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#include <assert.h>
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#include <stddef.h>
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#include <float.h>
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namespace Shared {
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uint32_t CurveViewRange::rangeChecksum() {
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float data[4] = {xMin(), xMax(), yMin(), yMax()};
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size_t dataLengthInBytes = 4*sizeof(float);
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assert((dataLengthInBytes & 0x3) == 0); // Assert that dataLengthInBytes is a multiple of 4
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return Ion::crc32((uint32_t *)data, dataLengthInBytes/sizeof(uint32_t));
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}
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float CurveViewRange::yGridUnit() {
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return 0.0f;
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}
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float CurveViewRange::computeGridUnit(Axis axis, float range1) {
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float range = axis == Axis::Y ? range1 : ((1.0f-2.0f*CurveView::k_labelsHorizontalMarginRatio)*range1);
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// TODO share the axis enum of curve_view and then call a static method CurveView::marginRatio(Axis)
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int a = 0;
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int b = 0;
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float maxNumberOfUnits = (axis == Axis::X) ? k_maxNumberOfXGridUnits : k_maxNumberOfYGridUnits;
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float minNumberOfUnits = (axis == Axis::X) ? k_minNumberOfXGridUnits : k_minNumberOfYGridUnits;
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constexpr int unitsCount = 3;
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float units[unitsCount] = {k_smallGridUnitMantissa, k_mediumGridUnitMantissa, k_largeGridUnitMantissa};
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for (int k = 0; k < unitsCount; k++) {
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float currentA = units[k];
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int b1 = std::floor(std::log10(range/(currentA*maxNumberOfUnits)));
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int b2 = std::floor(std::log10(range/(currentA*minNumberOfUnits)));
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if (b1 != b2) {
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b = b2;
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a = currentA;
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}
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}
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return a*std::pow(10.0f,b);
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/* Proof of the algorithm:
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*
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* We want to find gridUnit = a*10^b, with a in {1; 2; 5} and b an integer
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* We want: minNumberOfUnits <= range/gridUnit < maxNumberOfUnits
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*
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* A solution thus needs to verify:
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*
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* minNumberOfUnits/range <= 1/(a*10^b) < maxNumberOfUnits/range
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* => range/minNumberOfUnits >= a*10^b > range/maxNumberOfUnits
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* => range/(a*minNumberOfUnits) >= 10^b > range/(a*maxNumberOfUnits)
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* => log10(range/(a*minNumberOfUnits)) >= b > log10(range/(a*maxNumberOfUnits))
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* => (1) log10(range/(a*maxNumberOfUnits)) < b <= log10(range/(a*minNumberOfUnits))
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* And, because b must be an integer,
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* => floor(log10(range/(a*maxNumberOfUnits))) != floor(log10(range/(a*minNumberOfUnits)))
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* The solution is then b = floor(log10(range/(a*minNumberOfUnits)))
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*
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* Is there always at least one solution ?
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*
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* (1) also gives:
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* E1 = log10(range)-log10(a)-log10(maxNumberOfUnits) < b <= log10(range)-log10(a)-log10(maxNumberOfUnits) = E2
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*
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* Let's compute E2-E1:
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* E2-E1 = log10(maxNumberOfUnits) - log10(minNumberOfUnits)
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* For minNumberOfUnits=7 and maxNumberOfUnits=18, E2-E1 = 0.41...
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* For minNumberOfUnits=5 and maxNumberOfUnits=13, E2-E1 = 0.41...
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*
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* Let's compute the union of the [E1;E2] for a in {1; 2; 5}:
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* [E1;E2 for a=1] U [E1;E2 for a=2] U [E1;E2 for a=5]
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* = [e1;e2] U [e1-log10(2); e2-log10(2)] U [e1-log10(5); e2-log10(5)]
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* = [e1;e2] U [e1-0.3; e2-0.3] U [e1-0.7; e2-0.7]
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* = [e1-0.7; e2-0.7] U [e1-0.3; e2-0.3] U [e1;e2]
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* = [e1-0.7; e2] because e2-0.7 > e1-0.3 as e2-e1 > 0.7-0.3 and e2-e1 = E2-E1 = 0.41...
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* and e2-0.3 > e1 as e2-e1 > 0.3
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*
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* The union of the [E1;E2] for a in {1; 2; 5} is an interval of size
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* e2-e1+0.7 = 1.1 > 1.
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* We will thus always have at least one a in {1; 2; 5} for which E1 and E2
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* are on each side of an integer.
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*
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* Let's make a drawing.
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*
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* n n+1 n+2 n+3 n+4 n+5
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* |.........|.........|.........|.........|.........|...
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* E1^---^E2
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* 0.41
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* *
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* To have a solution, we need E1 and E2 to be on each side of an integer.
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*
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* --------------------------------------------------------------------------------------------
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* --------------------------------------------------------------------------------------------
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*
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* If e1 - floor(e1) > 1-(e2-e1) = 0.58..., a=1 is a solution
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*
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* n n+0.2 n+0.4 n+0.6 n+0.8 n+1
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* .........||....|....|....|....|....|....|....|....|....|....||... a=1
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* e1^--------------------^e2
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*
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* --------------------------------------------------------------------------------------------
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* --------------------------------------------------------------------------------------------
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*
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* If log10(5)-(e2-e1) = 0.29... < e1 - floor(e1) < log10(5) = 0.69..., a=5 is a solution
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*
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* n n+0.2 n+0.4 n+0.6 n+0.8 n+1
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* .........||....|....|....|....|....|....|....|....|....|....||...
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* e1^--------------------^e2
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*
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* n n+0.2 n+0.4 n+0.6 n+0.8 n+1
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* .........||....|....|....|....|....|....|....|....|....|....||... a=5
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* E1^--------------------^E2 <- shift by log10(5) = 0.7
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*
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* --------------------------------------------------------------------------------------------
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* --------------------------------------------------------------------------------------------
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*
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* If e1 - floor(e1) < log10(2) = 0.3..., a=2 is a solution
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* n n+0.2 n+0.4 n+0.6 n+0.8 n+1
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* .........||....|....|....|....|....|....|....|....|....|....||...
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* e1^--------------------^e2
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*
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* n n+0.2 n+0.4 n+0.6 n+0.8 n+1
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* .........||....|....|....|....|....|....|....|....|....|....||... a=2
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* E1^--------------------^E2 <- shift by log10(2) = 0.3
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*
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* */
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}
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}
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